Connes & Marcolli paper on renormalization [Archive]

pindare_poet

Sep27-04, 03:31 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nHello,\n\nin math.NT/0409306, Connes and Marcolli investigate the occurence of\n\'the action of a certain universal "Motivic Galois Group" on the set\nof physical theories\' and wrote:\n\n"...these facts altogether indicate that the divergences of Quantum\nField Theory, far from just being an unwanted nuisance, are a clear\nsign of totally unexpected symmetries of geometrical origin."\n\nCould anybody possibly explain in very low-brow terms what they are\ndoing ? (i.e. for someone who doesn\'t know neither Hopf algebras nor\nmixed Tate motives...).\n\nOr at least: is it really as far-reaching as the quoted paragraph\nseems to imply ?\n\nRegards,\n---\nPP\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hello,

in math.NT/0409306, Connes and Marcolli investigate the occurence of
'the action of a certain universal "Motivic Galois Group" on the set
of physical theories' and wrote:

"...these facts altogether indicate that the divergences of Quantum
Field Theory, far from just being an unwanted nuisance, are a clear
sign of totally unexpected symmetries of geometrical origin."

Could anybody possibly explain in very low-brow terms what they are
doing ? (i.e. for someone who doesn't know neither Hopf algebras nor
mixed Tate motives...).

Or at least: is it really as far-reaching as the quoted paragraph
seems to imply ?

Regards,
---
PP

Arnold Neumaier

Sep28-04, 02:27 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>pindare_poet wrote:\n>\n> in math.NT/0409306, Connes and Marcolli investigate the occurence of\n> \'the action of a certain universal "Motivic Galois Group" on the set\n> of physical theories\' and wrote:\n>\n> "...these facts altogether indicate that the divergences of Quantum\n> Field Theory, far from just being an unwanted nuisance, are a clear\n> sign of totally unexpected symmetries of geometrical origin."\n>\n> Could anybody possibly explain in very low-brow terms what they are\n> doing ? (i.e. for someone who doesn\'t know neither Hopf algebras nor\n> mixed Tate motives...).\n>\n> Or at least: is it really as far-reaching as the quoted paragraph\n> seems to imply ?\n\nThis is one of a series of papers by Connes and coauthors which recast\nthe renormalization procedure of QFT in algebraic terms, making\nhigh loop calculations and renormalizability proofs (beyond just\nsuperficial counting) more accessible to those who understand Hopf\nalgebras, and giving some insight into the structure of renormalization.\nOn this level, the renormalization group is embedded into a much larger\n(in fact huge) group. The authors equate group = symmetries, but these\nsymmetries are far, far away from those which - say - give rise to\nNoether currents.\n\nThe above paper deepens the relations between physics and abstract\nmathematics, but it does not seem to have any consequences for working\nwith the standard model, say.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>pindare_poet wrote:
>
> in math.NT/0409306, Connes and Marcolli investigate the occurence of
> 'the action of a certain universal "Motivic Galois Group" on the set
> of physical theories' and wrote:
>
> "...these facts altogether indicate that the divergences of Quantum
> Field Theory, far from just being an unwanted nuisance, are a clear
> sign of totally unexpected symmetries of geometrical origin."
>
> Could anybody possibly explain in very low-brow terms what they are
> doing ? (i.e. for someone who doesn't know neither Hopf algebras nor
> mixed Tate motives...).
>
> Or at least: is it really as far-reaching as the quoted paragraph
> seems to imply ?

This is one of a series of papers by Connes and coauthors which recast
the renormalization procedure of QFT in algebraic terms, making
high loop calculations and renormalizability proofs (beyond just
superficial counting) more accessible to those who understand Hopf
algebras, and giving some insight into the structure of renormalization.
On this level, the renormalization group is embedded into a much larger
(in fact huge) group. The authors equate group = symmetries, but these
symmetries are far, far away from those which - say - give rise to
Noether currents.

The above paper deepens the relations between physics and abstract
mathematics, but it does not seem to have any consequences for working
with the standard model, say.

Arnold Neumaier

Chris Oakley

Sep29-04, 02:25 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nMy personal "insight into the structure of renormalization" does not require\nany especially clever mathematics. Quantum field theory as normally\npractised simply does not work. Most of the loop diagrams are divergent\nintegrals. The connection between the naive theory and the renormalized\ntheory involves differencing infinite quantities, which is mathematically\nmeaningless.\n\nThere is therefore *no* connection between naive QFT and renormalized QFT.\nThe latter may be inspired by the former, but it is not derived from it. The\nlatter merely takes some constructs from the former and applies "reasonable"\nrules to it, such as Lorentz and gauge invariance. The fallaciousness of the\nprocedure is evident in that one *should not need* to have to apply Lorentz\ninvariance or any other principle twice.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>My personal "insight into the structure of renormalization" does not require
any especially clever mathematics. Quantum field theory as normally
practised simply does not work. Most of the loop diagrams are divergent
integrals. The connection between the naive theory and the renormalized
theory involves differencing infinite quantities, which is mathematically
meaningless.

There is therefore *no* connection between naive QFT and renormalized QFT.
The latter may be inspired by the former, but it is not derived from it. The
latter merely takes some constructs from the former and applies "reasonable"
rules to it, such as Lorentz and gauge invariance. The fallaciousness of the
procedure is evident in that one *should not need* to have to apply Lorentz
invariance or any other principle twice.

Matthew Nobes

Sep29-04, 12:18 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n"Chris Oakley" <coakley@cgoakley.demon.co.uk> writes:\n\n> My personal "insight into the structure of renormalization" does not require\n> any especially clever mathematics. Quantum field theory as normally\n> practised simply does not work. Most of the loop diagrams are divergent\n> integrals. The connection between the naive theory and the renormalized\n> theory involves differencing infinite quantities, which is mathematically\n> meaningless.\n\nWhat about doing QCD on a lattice? There one encounters not one single\ninfinite quantity. This is obviously the case since it\'s being done on\na computer.\n\n> There is therefore *no* connection between naive QFT and renormalized\n> QFT.\n\nYou should look into the renormalization group.\n\n> The latter may be inspired by the former, but it is not derived from\n> it.\n\nAgain, by doing QCD on the lattice, the connection between the bare\nparameters that you put into the computer (the bare charm quark mass,\nfor example) and the renormalized ones that you get out is direct.\n\n> The\n> latter merely takes some constructs from the former and applies "reasonable"\n> rules to it, such as Lorentz and gauge invariance. The fallaciousness of the\n> procedure is evident in that one *should not need* to have to apply Lorentz\n> invariance or any other principle twice.\n\nYour problem seems to be with *perturbative* QFT. But you had no right\nto expect that to work in a simple way when you wrote down an integral\nover momentum that went from -infinity to infinity. Furthermore, you\nknow there are problems with the perturbation series. Even if all the\nterms were convergent integrals, the whole series diverges. So it\ncannot be the whole story.\n\nFortunatley (at least for asymtotically free theories at least) you can\ndefine the thing in a way which is manifestly finite.\n\n--\nMatthew Nobes\nNewman Lab, Cornell Univesity, Ithaca, NY 14853\nhttp://www.lepp.cornell.edu/~nobes\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Chris Oakley" <coakley@cgoakley.demon.co.uk> writes:

> My personal "insight into the structure of renormalization" does not require
> any especially clever mathematics. Quantum field theory as normally
> practised simply does not work. Most of the loop diagrams are divergent
> integrals. The connection between the naive theory and the renormalized
> theory involves differencing infinite quantities, which is mathematically
> meaningless.

What about doing QCD on a lattice? There one encounters not one single
infinite quantity. This is obviously the case since it's being done on
a computer.

> There is therefore *no* connection between naive QFT and renormalized
> QFT.

You should look into the renormalization group.

> The latter may be inspired by the former, but it is not derived from
> it.

Again, by doing QCD on the lattice, the connection between the bare
parameters that you put into the computer (the bare charm quark mass,
for example) and the renormalized ones that you get out is direct.

> The
> latter merely takes some constructs from the former and applies "reasonable"
> rules to it, such as Lorentz and gauge invariance. The fallaciousness of the
> procedure is evident in that one *should not need* to have to apply Lorentz
> invariance or any other principle twice.

Your problem seems to be with *perturbative* QFT. But you had no right
to expect that to work in a simple way when you wrote down an integral
over momentum that went from -infinity to infinity. Furthermore, you
know there are problems with the perturbation series. Even if all the
terms were convergent integrals, the whole series diverges. So it
cannot be the whole story.

Fortunatley (at least for asymtotically free theories at least) you can
define the thing in a way which is manifestly finite.

--
Matthew Nobes
Newman Lab, Cornell Univesity, Ithaca, NY 14853
http://www.lepp.cornell.edu/~nobes

Frank Hellmann

Sep30-04, 05:15 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nArnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<415856B6.8050907@univie.ac.at>...\n> pindare_poet wrote:\n> >\n> > in math.NT/0409306, Connes and Marcolli investigate the occurence of\n> > \'the action of a certain universal "Motivic Galois Group" on the set\n> > of physical theories\' and wrote:\n> >\n> > "...these facts altogether indicate that the divergences of Quantum\n> > Field Theory, far from just being an unwanted nuisance, are a clear\n> > sign of totally unexpected symmetries of geometrical origin."\n> >\n> > Could anybody possibly explain in very low-brow terms what they are\n> > doing ? (i.e. for someone who doesn\'t know neither Hopf algebras nor\n> > mixed Tate motives...).\n> >\n> > Or at least: is it really as far-reaching as the quoted paragraph\n> > seems to imply ?\n>\n> This is one of a series of papers by Connes and coauthors which recast\n> the renormalization procedure of QFT in algebraic terms, making\n> high loop calculations and renormalizability proofs (beyond just\n> superficial counting) more accessible to those who understand Hopf\n> algebras, and giving some insight into the structure of renormalization.\n> On this level, the renormalization group is embedded into a much larger\n> (in fact huge) group. The authors equate group = symmetries, but these\n> symmetries are far, far away from those which - say - give rise to\n> Noether currents.\n>\n> The above paper deepens the relations between physics and abstract\n> mathematics, but it does not seem to have any consequences for working\n> with the standard model, say.\n>\n>\n> Arnold Neumaier\n\nAre they in fact thinking about a symmetry on the "space of physical\ntheories" here? What do they define as a physical theory then? And how\ndoes that space acquire geometry?\n\n---\nfrank\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<415856B6.8050907@univie.ac.at>...
> pindare_poet wrote:
> >
> > in math.NT/0409306, Connes and Marcolli investigate the occurence of
> > 'the action of a certain universal "Motivic Galois Group" on the set
> > of physical theories' and wrote:
> >
> > "...these facts altogether indicate that the divergences of Quantum
> > Field Theory, far from just being an unwanted nuisance, are a clear
> > sign of totally unexpected symmetries of geometrical origin."
> >
> > Could anybody possibly explain in very low-brow terms what they are
> > doing ? (i.e. for someone who doesn't know neither Hopf algebras nor
> > mixed Tate motives...).
> >
> > Or at least: is it really as far-reaching as the quoted paragraph
> > seems to imply ?
>
> This is one of a series of papers by Connes and coauthors which recast
> the renormalization procedure of QFT in algebraic terms, making
> high loop calculations and renormalizability proofs (beyond just
> superficial counting) more accessible to those who understand Hopf
> algebras, and giving some insight into the structure of renormalization.
> On this level, the renormalization group is embedded into a much larger
> (in fact huge) group. The authors equate group = symmetries, but these
> symmetries are far, far away from those which - say - give rise to
> Noether currents.
>
> The above paper deepens the relations between physics and abstract
> mathematics, but it does not seem to have any consequences for working
> with the standard model, say.
>
>
> Arnold Neumaier

Are they in fact thinking about a symmetry on the "space of physical
theories" here? What do they define as a physical theory then? And how
does that space acquire geometry?

---
frank

Chris Oakley

Oct1-04, 04:22 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nHi Matthew,\n\n> Your problem seems to be with *perturbative* QFT. But you had no right\n> to expect that to work in a simple way when you wrote down an integral\n> over momentum that went from -infinity to infinity. Furthermore, you\n> know there are problems with the perturbation series. Even if all the\n> terms were convergent integrals, the whole series diverges. So it\n> cannot be the whole story.\n\nJust a minute ... if I do a calculation in ordinary QM using e.g. Fermi\'s\nGolden Rule & the result is a divergent integral do I then say that I had\n"no right" to expect that it should be finite? I think not ... I am far more\nlikely to just assume that I had made a mistake.\n\n> Fortunately (at least for asymptotically free theories at least) you can\n> define the thing in a way which is manifestly finite.\n\nAll the methods I have seen for doing this: Broyles, Epstein-Glaser, or the\ngood old traditional renormalization using infinite subtractions or spurious\nlimits involve unaxiomatic tampering the the theory after the event. And\nthey are all mathematically ugly. Much better just to admit that the theory\ndoes not work and move on.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hi Matthew,

> Your problem seems to be with *perturbative* QFT. But you had no right
> to expect that to work in a simple way when you wrote down an integral
> over momentum that went from -infinity to infinity. Furthermore, you
> know there are problems with the perturbation series. Even if all the
> terms were convergent integrals, the whole series diverges. So it
> cannot be the whole story.

Just a minute ... if I do a calculation in ordinary QM using e.g. Fermi's
Golden Rule & the result is a divergent integral do I then say that I had
"no right" to expect that it should be finite? I think not ... I am far more
likely to just assume that I had made a mistake.

> Fortunately (at least for asymptotically free theories at least) you can
> define the thing in a way which is manifestly finite.

All the methods I have seen for doing this: Broyles, Epstein-Glaser, or the
good old traditional renormalization using infinite subtractions or spurious
limits involve unaxiomatic tampering the the theory after the event. And
they are all mathematically ugly. Much better just to admit that the theory
does not work and move on.

Matthew Nobes

Oct1-04, 09:43 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n"Chris Oakley" <coakley@cgoakley.demon.co.uk> writes:\n\n> Hi Matthew,\n>\n> > Your problem seems to be with *perturbative* QFT. But you had no right\n> > to expect that to work in a simple way when you wrote down an integral\n> > over momentum that went from -infinity to infinity. Furthermore, you\n> > know there are problems with the perturbation series. Even if all the\n> > terms were convergent integrals, the whole series diverges. So it\n> > cannot be the whole story.\n>\n> Just a minute ... if I do a calculation in ordinary QM using e.g. Fermi\'s\n> Golden Rule & the result is a divergent integral do I then say that I had\n> "no right" to expect that it should be finite?\n\nWell, you can always construct something that wouldn\'t be finite. But\nin nonrelativistic QM my "no right to expect it to work" would probably\nmean that you\'d get an answer incompatible with relativity. So, in that\nsense, you have no right to expect that a OM calculation will work to\ninfinitly high energy.\n\n> I think not ... I am far more\n> likely to just assume that I had made a mistake.\n\nIt\'s a good thing people didn\'t do that when they tried to compute the\nself-energy of a "classical electron". You do realize that the Maxwell\nelectrodynamics prediction for that is infinite, correct?\n\n> > Fortunately (at least for asymptotically free theories at least) you can\n> > define the thing in a way which is manifestly finite.\n>\n> All the methods I have seen for doing this: Broyles, Epstein-Glaser, or the\n> good old traditional renormalization using infinite subtractions or spurious\n> limits involve unaxiomatic tampering the the theory after the event.\n\nWell, you snipped out the manifestly finite method I described. Do the\nthing on a computer. You can do QFT on a computer, and get results that\nare finite at every stage.\n\n> And they are all mathematically ugly.\n\nBeauty is in the eye of the beholder. Wilson\'s lattice formulation is\nfar from ugly in my mind.\n\n> Much better just to admit that the theory\n> does not work and move on.\n\nFunny then that it agrees with all those pesky experiments. There is\nnothing wrong with the theory. The trouble is with a naive application\nof perturbation theory. There is more to a theory than its perturbative\nexpansion.\n\n--\nMatthew Nobes\nNewman Lab, Cornell Univesity, Ithaca, NY 14853\nhttp://www.lepp.cornell.edu/~nobes\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Chris Oakley" <coakley@cgoakley.demon.co.uk> writes:

> Hi Matthew,
>
> > Your problem seems to be with *perturbative* QFT. But you had no right
> > to expect that to work in a simple way when you wrote down an integral
> > over momentum that went from -infinity to infinity. Furthermore, you
> > know there are problems with the perturbation series. Even if all the
> > terms were convergent integrals, the whole series diverges. So it
> > cannot be the whole story.
>
> Just a minute ... if I do a calculation in ordinary QM using e.g. Fermi's
> Golden Rule & the result is a divergent integral do I then say that I had
> "no right" to expect that it should be finite?

Well, you can always construct something that wouldn't be finite. But
in nonrelativistic QM my "no right to expect it to work" would probably
mean that you'd get an answer incompatible with relativity. So, in that
sense, you have no right to expect that a OM calculation will work to
infinitly high energy.

> I think not ... I am far more
> likely to just assume that I had made a mistake.

It's a good thing people didn't do that when they tried to compute the
self-energy of a "classical electron". You do realize that the Maxwell
electrodynamics prediction for that is infinite, correct?

> > Fortunately (at least for asymptotically free theories at least) you can
> > define the thing in a way which is manifestly finite.
>
> All the methods I have seen for doing this: Broyles, Epstein-Glaser, or the
> good old traditional renormalization using infinite subtractions or spurious
> limits involve unaxiomatic tampering the the theory after the event.

Well, you snipped out the manifestly finite method I described. Do the
thing on a computer. You can do QFT on a computer, and get results that
are finite at every stage.

> And they are all mathematically ugly.

Beauty is in the eye of the beholder. Wilson's lattice formulation is
far from ugly in my mind.

> Much better just to admit that the theory
> does not work and move on.

Funny then that it agrees with all those pesky experiments. There is
nothing wrong with the theory. The trouble is with a naive application
of perturbation theory. There is more to a theory than its perturbative
expansion.

--
Matthew Nobes
Newman Lab, Cornell Univesity, Ithaca, NY 14853
http://www.lepp.cornell.edu/~nobes

Chris Oakley

Oct3-04, 03:49 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n> Well, you can always construct something that wouldn\'t be finite. But\n> in nonrelativistic QM my "no right to expect it to work" would probably\n> mean that you\'d get an answer incompatible with relativity. So, in that\n> sense, you have no right to expect that a QM calculation will work to\n> infinitely high energy.\n\nAre you really saying that in a relativistic theory every cross section or\ndecay rate has to be infinite?\nOne should be careful to make a distinction between rates that become\ninfinite in some limit of the parameters and something that is always\ninfinite because it contains a divergent integral.\n\n> It\'s a good thing people didn\'t do that when they tried to compute the\n> self-energy of a "classical electron". You do realize that the Maxwell\n> electrodynamics prediction for that is infinite, correct?\n\nThese problems, along with the clear evidence of electromagnetic waves\nhaving particle-like behaviour demonstrate that classical electrodynamics\nhas limitations. Quantum electrodynamics, on the other hand, was devised to\nsolve the various problems and we should therefore set higher standards.\n\n> Well, you snipped out the manifestly finite method I described. Do the\n> thing on a computer. You can do QFT on a computer, and get results that\n> are finite at every stage.\n\nI know very little about Lattice Gauge Theory, and am not in a position to\nargue with this. However, even if you people manage to calculate hadron\nmasses that accurately agree with experiment I will still be looking for an\nunderstanding based on algebra and analytics. And I am sure that I do not\nneed to tell you that infinities *can* appear in numerical simulations on a\ncomputer: they just manifest themselves as instabilities and/or nonsensical\nanswers.\n\n> Beauty is in the eye of the beholder.\n\nI think that most beholders would agree that if an accountant falsifies\nfigures in order to make a money-losing company look profitable, then that\nis not "beautiful". Not only that, but the accountant might end up going to\njail. Why should we apply different standards to theoretical physics?\n\n> > Much better just to admit that the theory\n> > does not work and move on.\n>\n> Funny then that it agrees with all those pesky experiments.\n\nHold on a minute ... is the anomalous magnetic moment of the electron\ninfinite? Is the Lamb shift infinite? The thing that is agreeing with\nexperiment to eleven places of decimals is not quantum field theory, but\nquantum field theory that has been messed with.\n\n> The trouble is with a naive application of perturbation theory. There is\nmore to a theory than its perturbative\n> expansion.\n\nYes, but it is this perturbation expansion that leads to the much-vaunted\nsuccesses of QED.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Well, you can always construct something that wouldn't be finite. But
> in nonrelativistic QM my "no right to expect it to work" would probably
> mean that you'd get an answer incompatible with relativity. So, in that
> sense, you have no right to expect that a QM calculation will work to
> infinitely high energy.

Are you really saying that in a relativistic theory every cross section or
decay rate has to be infinite?
One should be careful to make a distinction between rates that become
infinite in some limit of the parameters and something that is always
infinite because it contains a divergent integral.

> It's a good thing people didn't do that when they tried to compute the
> self-energy of a "classical electron". You do realize that the Maxwell
> electrodynamics prediction for that is infinite, correct?

These problems, along with the clear evidence of electromagnetic waves
having particle-like behaviour demonstrate that classical electrodynamics
has limitations. Quantum electrodynamics, on the other hand, was devised to
solve the various problems and we should therefore set higher standards.

> Well, you snipped out the manifestly finite method I described. Do the
> thing on a computer. You can do QFT on a computer, and get results that
> are finite at every stage.

I know very little about Lattice Gauge Theory, and am not in a position to
argue with this. However, even if you people manage to calculate hadron
masses that accurately agree with experiment I will still be looking for an
understanding based on algebra and analytics. And I am sure that I do not
need to tell you that infinities *can* appear in numerical simulations on a
computer: they just manifest themselves as instabilities and/or nonsensical
answers.

> Beauty is in the eye of the beholder.

I think that most beholders would agree that if an accountant falsifies
figures in order to make a money-losing company look profitable, then that
is not "beautiful". Not only that, but the accountant might end up going to
jail. Why should we apply different standards to theoretical physics?

> > Much better just to admit that the theory
> > does not work and move on.
>
> Funny then that it agrees with all those pesky experiments.

Hold on a minute ... is the anomalous magnetic moment of the electron
infinite? Is the Lamb shift infinite? The thing that is agreeing with
experiment to eleven places of decimals is not quantum field theory, but
quantum field theory that has been messed with.

> The trouble is with a naive application of perturbation theory. There is
more to a theory than its perturbative
> expansion.

Yes, but it is this perturbation expansion that leads to the much-vaunted
successes of QED.

Arnold Neumaier

Oct3-04, 03:49 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nChris Oakley wrote:\n\n> Just a minute ... if I do a calculation in ordinary QM using e.g. Fermi\'s\n> Golden Rule & the result is a divergent integral do I then say that I had\n> "no right" to expect that it should be finite? I think not ... I am far more\n> likely to just assume that I had made a mistake.\n\nIf you do perturbation theory with a Hamiltonian p^2/2m +g delta(x)\n(which, I think, corresponds to a 2-body problem for a nonrelativistic\nphi^4 field theory), you get the same infinities, although this\nproblem is so simple that it can be renormalized nonperturbatively.\nSo there is nothing wrong with renormalization.\n\n\nArnold Neumaier\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:

> Just a minute ... if I do a calculation in ordinary QM using e.g. Fermi's
> Golden Rule & the result is a divergent integral do I then say that I had
> "no right" to expect that it should be finite? I think not ... I am far more
> likely to just assume that I had made a mistake.

If you do perturbation theory with a Hamiltonian p^2/2m +g \delta(x)
(which, I think, corresponds to a 2-body problem for a nonrelativistic
\phi^4 field theory), you get the same infinities, although this
problem is so simple that it can be renormalized nonperturbatively.
So there is nothing wrong with renormalization.

Arnold Neumaier

Arnold Neumaier

Oct4-04, 12:21 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Chris Oakley wrote:\n\n> Hold on a minute ... is the anomalous magnetic moment of the electron\n> infinite? Is the Lamb shift infinite? The thing that is agreeing with\n> experiment to eleven places of decimals is not quantum field theory, but\n> quantum field theory that has been messed with.\n\nWhat you call \'quantum field theory that has been messed with\'\ncall more serious people \'quantum field theory\'. It is the calculus\nthat gives finite values that can be computed to a certain accuracy\nand are in (sometimes excellent) agreement with experiment.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:

> Hold on a minute ... is the anomalous magnetic moment of the electron
> infinite? Is the Lamb shift infinite? The thing that is agreeing with
> experiment to eleven places of decimals is not quantum field theory, but
> quantum field theory that has been messed with.

What you call 'quantum field theory that has been messed with'
call more serious people 'quantum field theory'. It is the calculus
that gives finite values that can be computed to a certain accuracy
and are in (sometimes excellent) agreement with experiment.

Arnold Neumaier

Chris Oakley

Oct4-04, 12:21 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>> If you do perturbation theory with a Hamiltonian p^2/2m +g delta(x)\n> (which, I think, corresponds to a 2-body problem for a nonrelativistic\n> phi^4 field theory), you get the same infinities, although this\n> problem is so simple that it can be renormalized nonperturbatively.\n> So there is nothing wrong with renormalization.\n\nReference? Is this renormalization infinite? This is what I have the problem\nwith.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>> If you do perturbation theory with a Hamiltonian p^2/2m +g \delta(x)
> (which, I think, corresponds to a 2-body problem for a nonrelativistic
> \phi^4 field theory), you get the same infinities, although this
> problem is so simple that it can be renormalized nonperturbatively.
> So there is nothing wrong with renormalization.

Reference? Is this renormalization infinite? This is what I have the problem
with.

Matthew Nobes

Oct4-04, 12:22 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n"Chris Oakley" <coakley@cgoakley.demon.co.uk> writes:\n\n> > Well, you can always construct something that wouldn\'t be finite. But\n> > in nonrelativistic QM my "no right to expect it to work" would probably\n> > mean that you\'d get an answer incompatible with relativity. So, in that\n> > sense, you have no right to expect that a QM calculation will work to\n> > infinitely high energy.\n>\n> Are you really saying that in a relativistic theory every cross section or\n> decay rate has to be infinite?\n\nNo, I\'m saying that you have no right to expect that the theory works to\ninfinite energy when you\'ve only tested it to ~1 TeV.\n\n> One should be careful to make a distinction between rates that become\n> infinite in some limit of the parameters and something that is always\n> infinite because it contains a divergent integral.\n\nWell, the point is you cannot assume something like the propagator\ndoesn\'t change. At high energy you might have something like\nsupersymmetry, where a fermion contribution cancels the boson one.\nThat\'s just an example that somethinge *can* happen, not an arguement\nfor supersymmetry.\n\n> > It\'s a good thing people didn\'t do that when they tried to compute the\n> > self-energy of a "classical electron". You do realize that the Maxwell\n> > electrodynamics prediction for that is infinite, correct?\n>\n> These problems, along with the clear evidence of electromagnetic waves\n> having particle-like behaviour demonstrate that classical electrodynamics\n> has limitations. Quantum electrodynamics, on the other hand, was devised to\n> solve the various problems and we should therefore set higher standards.\n\nAnd it meets the only higher standard I care about: it explains the\nresults of more experiments than classical E&M. And the Electroweak\ntheory explains even more.\n\n> > Well, you snipped out the manifestly finite method I described. Do the\n> > thing on a computer. You can do QFT on a computer, and get results that\n> > are finite at every stage.\n>\n> I know very little about Lattice Gauge Theory, and am not in a position to\n> argue with this. However, even if you people manage to calculate hadron\n> masses that accurately agree with experiment\n\nThis has been done.\n\n> I will still be looking for an understanding based on algebra and\n> analytics.\n\nAgain, you should learn about the renormalization group. It clairifies\nwhy all this worrying about divergent integrals is not really helpful.\n\n> And I am sure that I do not\n> need to tell you that infinities *can* appear in numerical simulations on a\n> computer: they just manifest themselves as instabilities and/or nonsensical\n> answers.\n\nOf course. You\'ll have to take my word for it that lattice QCD doesn\'t\nencounter these sorts of problems.\n\n[snip]\n> > > Much better just to admit that the theory\n> > > does not work and move on.\n> >\n> > Funny then that it agrees with all those pesky experiments.\n>\n> Hold on a minute ... is the anomalous magnetic moment of the electron\n> infinite?\n\nNo.\n\n> Is the Lamb shift infinite?\n\nNo.\n\n> The thing that is agreeing with\n> experiment to eleven places of decimals is not quantum field theory, but\n> quantum field theory that has been messed with.\n\nIt hasn\'t been "messed with". It has been regulated and renormalized.\nThe renormalization group shows us how to understand this. My point\nabout lattice QCD is relevent here. You can set the theory up in a\nmanifestly finite way, which has no divergences (since it\'s on a\nlattice). Then you renormalize, and finally take the continuum limit.\nNever an infinity seen.\n\n[snip]\n\nMatthew\n--\nMatthew Nobes\nNewman Lab, Cornell Univesity, Ithaca, NY 14853\nhttp://www.lepp.cornell.edu/~nobes\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Chris Oakley" <coakley@cgoakley.demon.co.uk> writes:

> > Well, you can always construct something that wouldn't be finite. But
> > in nonrelativistic QM my "no right to expect it to work" would probably
> > mean that you'd get an answer incompatible with relativity. So, in that
> > sense, you have no right to expect that a QM calculation will work to
> > infinitely high energy.
>
> Are you really saying that in a relativistic theory every cross section or
> decay rate has to be infinite?

No, I'm saying that you have no right to expect that the theory works to
infinite energy when you've only tested it to ~1 TeV.

> One should be careful to make a distinction between rates that become
> infinite in some limit of the parameters and something that is always
> infinite because it contains a divergent integral.

Well, the point is you cannot assume something like the propagator
doesn't change. At high energy you might have something like
supersymmetry, where a fermion contribution cancels the boson one.
That's just an example that somethinge *can* happen, not an arguement
for supersymmetry.

> > It's a good thing people didn't do that when they tried to compute the
> > self-energy of a "classical electron". You do realize that the Maxwell
> > electrodynamics prediction for that is infinite, correct?
>
> These problems, along with the clear evidence of electromagnetic waves
> having particle-like behaviour demonstrate that classical electrodynamics
> has limitations. Quantum electrodynamics, on the other hand, was devised to
> solve the various problems and we should therefore set higher standards.

And it meets the only higher standard I care about: it explains the
results of more experiments than classical E&M. And the Electroweak
theory explains even more.

> > Well, you snipped out the manifestly finite method I described. Do the
> > thing on a computer. You can do QFT on a computer, and get results that
> > are finite at every stage.
>
> I know very little about Lattice Gauge Theory, and am not in a position to
> argue with this. However, even if you people manage to calculate hadron
> masses that accurately agree with experiment

This has been done.

> I will still be looking for an understanding based on algebra and
> analytics.

Again, you should learn about the renormalization group. It clairifies
why all this worrying about divergent integrals is not really helpful.

> And I am sure that I do not
> need to tell you that infinities *can* appear in numerical simulations on a
> computer: they just manifest themselves as instabilities and/or nonsensical
> answers.

Of course. You'll have to take my word for it that lattice QCD doesn't
encounter these sorts of problems.

[snip]
> > > Much better just to admit that the theory
> > > does not work and move on.
> >
> > Funny then that it agrees with all those pesky experiments.
>
> Hold on a minute ... is the anomalous magnetic moment of the electron
> infinite?

No.

> Is the Lamb shift infinite?

No.

> The thing that is agreeing with
> experiment to eleven places of decimals is not quantum field theory, but
> quantum field theory that has been messed with.

It hasn't been "messed with". It has been regulated and renormalized.
The renormalization group shows us how to understand this. My point
about lattice QCD is relevent here. You can set the theory up in a
manifestly finite way, which has no divergences (since it's on a
lattice). Then you renormalize, and finally take the continuum limit.
Never an infinity seen.

[snip]

Matthew
--
Matthew Nobes
Newman Lab, Cornell Univesity, Ithaca, NY 14853
http://www.lepp.cornell.edu/~nobes

Chris Oakley

Oct5-04, 06:42 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>> > > Well, you can always construct something that wouldn\'t be finite. But\n> > > in nonrelativistic QM my "no right to expect it to work" would\nprobably\n> > > mean that you\'d get an answer incompatible with relativity. So, in\nthat\n> > > sense, you have no right to expect that a QM calculation will work to\n> > > infinitely high energy.\n> >\n> > Are you really saying that in a relativistic theory every cross section\nor\n> > decay rate has to be infinite?\n>\n> No, I\'m saying that you have no right to expect that the theory works to\n> infinite energy when you\'ve only tested it to ~1 TeV.\n\nI am not sure that I am following you here. Before renormalization QFT as\nnormally practised leads to infinite cross sections and decay rates. I have\nseen it claimed that the infiniteness is because there is something going on\nat extremely high energies that we do not fully understand that rescues the\nsituation at low-energy - the "ultraviolet completion". I thought that this\nis what you were talking about. But you seem rather to be saying that you\n*can* construct a relativistic theory that gives sensible finite numbers at\nlow energy, it is just that you cannot guarantee it at high energies.\nClarification - ?\n\n> Again, you should learn about the renormalization group.\n\nIf people were to say of the renormalization group: "this is the point of\ndeparture. This is where we start in the process of building our\nmathematical theory of fundamental particles", I would not have so many\nproblems with it. Although not completely free of ambiguities, it is a\nprogram that enables us to classify and calculate some important measurable\nquantities. My problem is that I do not see the connection between this and\nthe quantisation of the classical field, except in the most qualitative\nterms. I see the Feynman-Dyson approach to QFT as failing to give sensible\nor even finite answers to physical quantities, but I do not see this\nphoenix-like rebirth after all the infinities have been conveniently wiped\nout as anything more than just wishful thinking.\n\nIf someone can get to the renormalization group from first principles - and\nI stress *from first principles* - without mathematical sleight of hand,\nthen I will gladly retract my criticisms.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>> > > Well, you can always construct something that wouldn't be finite. But
> > > in nonrelativistic QM my "no right to expect it to work" would
probably
> > > mean that you'd get an answer incompatible with relativity. So, in
that
> > > sense, you have no right to expect that a QM calculation will work to
> > > infinitely high energy.
> >
> > Are you really saying that in a relativistic theory every cross section
or
> > decay rate has to be infinite?
>
> No, I'm saying that you have no right to expect that the theory works to
> infinite energy when you've only tested it to ~1 TeV.

I am not sure that I am following you here. Before renormalization QFT as
normally practised leads to infinite cross sections and decay rates. I have
seen it claimed that the infiniteness is because there is something going on
at extremely high energies that we do not fully understand that rescues the
situation at low-energy - the "ultraviolet completion". I thought that this
is what you were talking about. But you seem rather to be saying that you
*can* construct a relativistic theory that gives sensible finite numbers at
low energy, it is just that you cannot guarantee it at high energies.
Clarification - ?

> Again, you should learn about the renormalization group.

If people were to say of the renormalization group: "this is the point of
departure. This is where we start in the process of building our
mathematical theory of fundamental particles", I would not have so many
problems with it. Although not completely free of ambiguities, it is a
program that enables us to classify and calculate some important measurable
quantities. My problem is that I do not see the connection between this and
the quantisation of the classical field, except in the most qualitative
terms. I see the Feynman-Dyson approach to QFT as failing to give sensible
or even finite answers to physical quantities, but I do not see this
phoenix-like rebirth after all the infinities have been conveniently wiped
out as anything more than just wishful thinking.

If someone can get to the renormalization group from first principles - and
I stress *from first principles* - without mathematical sleight of hand,
then I will gladly retract my criticisms.

Arnold Neumaier

Oct5-04, 06:44 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Chris Oakley wrote:\n>>If you do perturbation theory with a Hamiltonian p^2/2m +g delta(x)\n>>(which, I think, corresponds to a 2-body problem for a nonrelativistic\n>>phi^4 field theory), you get the same infinities, although this\n>>problem is so simple that it can be renormalized nonperturbatively.\n>>So there is nothing wrong with renormalization.\n>\n>\n> Reference?\n\nThe renormalization of this particular example is treated in\nhep-th/9305052.\n\n\n> Is this renormalization infinite? This is what I have the problem\n> with.\n\nRenormalized quantities are never infinite.\n\nEssentially what happens is that one has a family of Hamiltonians\nH(Lambda,g) that depend on a scale parameter Lambda and and a coupling\nconstant g (or several). H(Lambda,g) has a good limit H(g) as Lambda\nto inf, with g fixed, but the corresponding limit of the resolvent\nG(Lambda,g) does not exist; hence if one tries to do calculations with\nH(g) directly (the 1930 way of doing things, which was a dead end),\none gets infinities all over the place. On the other hand, if one\nchooses a good parameterization g(Lambda,mu) then, although\nH(Lambda,g(Lambda,mu)) has no longer a good limit as Lambda to inf,\nits resolvent G(Lambda,g(Lambda,mu)) has a well-defined limit G(mu).\nSince all scattering information is in the resolvent, the latter\ndefines a good physical model for a scattering process.\n\nInfinities only appear if one takes the limit in a way it\nshould not be taken.\n\nOf course, the relativistic case is more involved, but there is\nno difference in principle.\n\n\nArnold Neumaier\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
>>If you do perturbation theory with a Hamiltonian p^2/2m +g \delta(x)
>>(which, I think, corresponds to a 2-body problem for a nonrelativistic
>>\phi^4 field theory), you get the same infinities, although this
>>problem is so simple that it can be renormalized nonperturbatively.
>>So there is nothing wrong with renormalization.
>
>
> Reference?

The renormalization of this particular example is treated in
http://www.arxiv.org/abs/hep-th/9305052.

> Is this renormalization infinite? This is what I have the problem
> with.

Renormalized quantities are never infinite.

Essentially what happens is that one has a family of Hamiltonians
H(\Lambda,g) that depend on a scale parameter \Lambda and and a coupling
constant g (or several). H(\Lambda,g) has a good limit H(g) as \Lambda
to inf, with g fixed, but the corresponding limit of the resolvent
G(\Lambda,g) does not exist; hence if one tries to do calculations with
H(g) directly (the 1930 way of doing things, which was a dead end),
one gets infinities all over the place. On the other hand, if one
chooses a good parameterization g(\Lambda,\mu) then, although
H(\Lambda,g(\Lambda,\mu)) has no longer a good limit as \Lambda to inf,
its resolvent G(\Lambda,g(\Lambda,\mu)) has a well-defined limit G(\mu).
Since all scattering information is in the resolvent, the latter
defines a good physical model for a scattering process.

Infinities only appear if one takes the limit in a way it
should not be taken.

Of course, the relativistic case is more involved, but there is
no difference in principle.

Arnold Neumaier

Matthew Nobes

Oct6-04, 08:03 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Chris Oakley <coakley@cgoakley.demon.co.uk> writes:\n\n> > > Are you really saying that in a relativistic theory every cross section\n> > > or decay rate has to be infinite?\n> >\n> > No, I\'m saying that you have no right to expect that the theory works to\n> > infinite energy when you\'ve only tested it to ~1 TeV.\n>\n> I am not sure that I am following you here. Before renormalization QFT as\n> normally practised leads to infinite cross sections and decay rates. I have\n> seen it claimed that the infiniteness is because there is something going on\n> at extremely high energies that we do not fully understand that rescues the\n> situation at low-energy - the "ultraviolet completion". I thought that this\n> is what you were talking about. But you seem rather to be saying that you\n> *can* construct a relativistic theory that gives sensible finite numbers at\n> low energy, it is just that you cannot guarantee it at high energies.\n> Clarification - ?\n\nSorry for the confusion. I\'m saying that you start with a bare theory,\nand you compute some perturbative correction. And it\'s infinite. Now\nyou look at this theory and you say "well, when I computed the\ncorrection I had to integrate up to infinite energy". Now you *know*\nthat\'s not a good thing to be doing. You have no idea what\'s going on\nup there.\n\nSo what do you do? Well, you can throw out the theory, but that seems a\nbit drastic. Or you can cut the theory off at some energy above what\nyou\'ve measured. If you do that then there are no infinities. Then you\ncompute some peruturbative corrections, and everything is finite. You\ndiscover that if you arrange things right you can write every prediction\nof the the theory in a way which does not depend on the cutoff.\n\nAt this point you\'re free to take the cutoff to infinity, if you wish.\nBut you don\'t have to, the only assumption you have to make about it is\nthat it\'s much larger than the energy scales you\'re worried about.\n\nNow a cause for worry here would be if you cut the theory off in\ndifferent ways, and you got different answers. *That* would be a\nproblem, no question. But that doesn\'t happen for QCD/QED/Electroweak.\nAnd the lattice formulation provides a cutoff that works\nnon-perturbativly.\n\nI really don\'t see what the problem with this program is. I\'ll grant\nthat rigorously proving limits exist (such as lattice spacing -> 0) has\nnot been done. But all evidence is that it works.\n\n--\nMatthew Nobes\nNewman Lab, Cornell Univesity, Ithaca, NY 14853\nhttp://www.lepp.cornell.edu/~nobes\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley <coakley@cgoakley.demon.co.uk> writes:

> > > Are you really saying that in a relativistic theory every cross section
> > > or decay rate has to be infinite?
> >
> > No, I'm saying that you have no right to expect that the theory works to
> > infinite energy when you've only tested it to ~1 TeV.
>
> I am not sure that I am following you here. Before renormalization QFT as
> normally practised leads to infinite cross sections and decay rates. I have
> seen it claimed that the infiniteness is because there is something going on
> at extremely high energies that we do not fully understand that rescues the
> situation at low-energy - the "ultraviolet completion". I thought that this
> is what you were talking about. But you seem rather to be saying that you
> *can* construct a relativistic theory that gives sensible finite numbers at
> low energy, it is just that you cannot guarantee it at high energies.
> Clarification - ?

Sorry for the confusion. I'm saying that you start with a bare theory,
and you compute some perturbative correction. And it's infinite. Now
you look at this theory and you say "well, when I computed the
correction I had to integrate up to infinite energy". Now you *know*
that's not a good thing to be doing. You have no idea what's going on
up there.

So what do you do? Well, you can throw out the theory, but that seems a
bit drastic. Or you can cut the theory off at some energy above what
you've measured. If you do that then there are no infinities. Then you
compute some peruturbative corrections, and everything is finite. You
discover that if you arrange things right you can write every prediction
of the the theory in a way which does not depend on the cutoff.

At this point you're free to take the cutoff to infinity, if you wish.
But you don't have to, the only assumption you have to make about it is
that it's much larger than the energy scales you're worried about.

Now a cause for worry here would be if you cut the theory off in
different ways, and you got different answers. *That* would be a
problem, no question. But that doesn't happen for QCD/QED/Electroweak.
And the lattice formulation provides a cutoff that works
non-perturbativly.

I really don't see what the problem with this program is. I'll grant
that rigorously proving limits exist (such as lattice spacing -> 0) has
not been done. But all evidence is that it works.

--
Matthew Nobes
Newman Lab, Cornell Univesity, Ithaca, NY 14853
http://www.lepp.cornell.edu/~nobes

Aaron Bergman

Oct6-04, 08:03 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <cju1aj\\$tsk\\$1@lfa222122.richmond.edu>,\nChri s Oakley <coakley@cgoakley.demon.co.uk> wrote:\n\n> But you seem rather to be saying that you\n> *can* construct a relativistic theory that gives sensible finite numbers at\n> low energy, it is just that you cannot guarantee it at high energies.\n> Clarification - ?\n\nCheck out the Nobel prize today.\n\n> > Again, you should learn about the renormalization group.\n>\n> If people were to say of the renormalization group: "this is the point of\n> departure. This is where we start in the process of building our\n> mathematical theory of fundamental particles", I would not have so many\n> problems with it. Although not completely free of ambiguities, it is a\n> program that enables us to classify and calculate some important measurable\n> quantities. My problem is that I do not see the connection between this and\n> the quantisation of the classical field, except in the most qualitative\n> terms. I see the Feynman-Dyson approach to QFT as failing to give sensible\n> or even finite answers to physical quantities, but I do not see this\n> phoenix-like rebirth after all the infinities have been conveniently wiped\n> out as anything more than just wishful thinking.\n>\n> If someone can get to the renormalization group from first principles - and\n> I stress *from first principles* - without mathematical sleight of hand,\n> then I will gladly retract my criticisms.\n\nThe renormalization group is there even in the classical theory.\n\nSee hep-th/0309149, maybe.\n\nAaron\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <cju1aj$tsk$1@lfa222122.richmond.edu>,
Chris Oakley <coakley@cgoakley.demon.co.uk> wrote:

> But you seem rather to be saying that you
> *can* construct a relativistic theory that gives sensible finite numbers at
> low energy, it is just that you cannot guarantee it at high energies.
> Clarification - ?

Check out the Nobel prize today.

> > Again, you should learn about the renormalization group.
>
> If people were to say of the renormalization group: "this is the point of
> departure. This is where we start in the process of building our
> mathematical theory of fundamental particles", I would not have so many
> problems with it. Although not completely free of ambiguities, it is a
> program that enables us to classify and calculate some important measurable
> quantities. My problem is that I do not see the connection between this and
> the quantisation of the classical field, except in the most qualitative
> terms. I see the Feynman-Dyson approach to QFT as failing to give sensible
> or even finite answers to physical quantities, but I do not see this
> phoenix-like rebirth after all the infinities have been conveniently wiped
> out as anything more than just wishful thinking.
>
> If someone can get to the renormalization group from first principles - and
> I stress *from first principles* - without mathematical sleight of hand,
> then I will gladly retract my criticisms.

The renormalization group is there even in the classical theory.

See http://www.arxiv.org/abs/hep-th/0309149, maybe.

Aaron

Arnold Neumaier

Oct6-04, 08:03 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Chris Oakley wrote:\n\n>>No, I\'m saying that you have no right to expect that the theory works to\n>>infinite energy when you\'ve only tested it to ~1 TeV.\n>\n> I am not sure that I am following you here. Before renormalization QFT as\n> normally practised leads to infinite cross sections and decay rates. I have\n> seen it claimed that the infiniteness is because there is something going on\n> at extremely high energies that we do not fully understand that rescues the\n> situation at low-energy - the "ultraviolet completion".\n\nThis is nonsense. Infinities arise because integrals taken over\nunbounded momenta don\'t exist; so doing it leads to nonsense.\nInstead, proper QFT takes regularized integrals, for example by\nadding an explicit cutoff Lambda. This simply means that they\ncalculate with an action that depends on Lambda as an additional\nparameter. Once this is done, everything is finite, but\nLambda-dependent.\n\nThe only problem with that is that the cutoff destroys Lorentz\ncovariance - apart from that it would be a completely respectable\nfield theory in itself. Now Lorentz invariance is violated only\nat momentum > O(Lambda); hence to have the theory conform to\nphysics that can be checked it suffices to take Lambda large.\nBut for aesthetic reasons or since we believe that symmetries are\nfundamental, we want to have fully invariant theories. This requires\nthat we let Lambda go to infinity.\n\nBut in order that the results have a finite limit we must at the\nsame time make the coupling constants g dependent on Lambda.\nIf this is done in a correct way (and the textbooks on QFT teach\none or more of the known correct ways), one encounters no infinities\nat all in the whole process.\n\n\n> I see the Feynman-Dyson approach to QFT as failing to give sensible\n> or even finite answers to physical quantities, but I do not see this\n> phoenix-like rebirth after all the infinities have been conveniently wiped\n> out as anything more than just wishful thinking.\n\nThe approach that gives infinities is the pre Feynman-Dyson approach,\nwhile what you call \'wishful thinking\' is the advance that earned\nFeynaman, Schwinger and Tomonaga the Nobel prize. They showed how to\n_avoid_ the infinities completely, on the perturbative level, and the\nrenormalization group explains why the result is still meaningful as\na \'perturbative\' approach, although at first sight, very large corrections\nseems to be involved.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:

>>No, I'm saying that you have no right to expect that the theory works to
>>infinite energy when you've only tested it to ~1 TeV.
>
> I am not sure that I am following you here. Before renormalization QFT as
> normally practised leads to infinite cross sections and decay rates. I have
> seen it claimed that the infiniteness is because there is something going on
> at extremely high energies that we do not fully understand that rescues the
> situation at low-energy - the "ultraviolet completion".

This is nonsense. Infinities arise because integrals taken over
unbounded momenta don't exist; so doing it leads to nonsense.
Instead, proper QFT takes regularized integrals, for example by
adding an explicit cutoff \Lambda. This simply means that they
calculate with an action that depends on \Lambda as an additional
parameter. Once this is done, everything is finite, but
\Lambda-dependent.

The only problem with that is that the cutoff destroys Lorentz
covariance - apart from that it would be a completely respectable
field theory in itself. Now Lorentz invariance is violated only
at momentum > O(\Lambda); hence to have the theory conform to
physics that can be checked it suffices to take \Lambda large.
But for aesthetic reasons or since we believe that symmetries are
fundamental, we want to have fully invariant theories. This requires
that we let \Lambda go to infinity.

But in order that the results have a finite limit we must at the
same time make the coupling constants g dependent on \Lambda.
If this is done in a correct way (and the textbooks on QFT teach
one or more of the known correct ways), one encounters no infinities
at all in the whole process.

> I see the Feynman-Dyson approach to QFT as failing to give sensible
> or even finite answers to physical quantities, but I do not see this
> phoenix-like rebirth after all the infinities have been conveniently wiped
> out as anything more than just wishful thinking.

The approach that gives infinities is the pre Feynman-Dyson approach,
while what you call 'wishful thinking' is the advance that earned
Feynaman, Schwinger and Tomonaga the Nobel prize. They showed how to
_avoid_ the infinities completely, on the perturbative level, and the
renormalization group explains why the result is still meaningful as
a 'perturbative' approach, although at first sight, very large corrections
seems to be involved.

Arnold Neumaier

Chris Oakley

Oct7-04, 07:04 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>> Well, you can throw out the theory, but that seems a bit drastic.\n\nI believe that one has no choice. One can in any case find holes in the\ncutoff procedure. Supposing, for example, we were to choose Lambda as a\nfunction rather than a constant. Furthermore, let us make it a function of\nvariables that never even entered the original equations. There is no reason\nin principle why we should not be allowed to do this. We then very rapidly\nget nonsense. You will then say that it was my own fault for not making it a\nconstant, but all that means is that you can find a path through the\nminefield. It does not mean that the minefield has ceased to exist.\n\nThere is no such thing as "reasonable" in theoretical physics, there is only\nmathematical logic. The pillars of our current understanding of physics,\nnamely quantum mechanics and special relativity, both defy common sense. One\nbecomes aware of this very quickly when one tries to explain them to an\nintelligent non-physicist. You and Dr. Neumaier say that it is\n"unreasonable" to have four-momentum integrals extending to infinity,\nleading you to want to cut them off. I say that if the mathematical logic\nleads one there, then one has no choice other that to accept it, and if this\nleads to nonsense then one has to accept that that something is\nfundamentally wrong with the approach.\n\nLet me be clear on this - I could not do any better. All my attempts to find\na rigorous way of doing Feynman-Dyson perturbation theory failed. This puts\nme in good company with Dirac, Pauli, Feynman, the axiomatic field theorists\nand many others. But in the end, I just decided that it was not worth\nrescuing, tore everything up and started again. And - intermittently, at\nleast - I am still working on that.\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>> Well, you can throw out the theory, but that seems a bit drastic.

I believe that one has no choice. One can in any case find holes in the
cutoff procedure. Supposing, for example, we were to choose \Lambda as a
function rather than a constant. Furthermore, let us make it a function of
variables that never even entered the original equations. There is no reason
in principle why we should not be allowed to do this. We then very rapidly
get nonsense. You will then say that it was my own fault for not making it a
constant, but all that means is that you can find a path through the
minefield. It does not mean that the minefield has ceased to exist.

There is no such thing as "reasonable" in theoretical physics, there is only
mathematical logic. The pillars of our current understanding of physics,
namely quantum mechanics and special relativity, both defy common sense. One
becomes aware of this very quickly when one tries to explain them to an
intelligent non-physicist. You and Dr. Neumaier say that it is
"unreasonable" to have four-momentum integrals extending to infinity,
leading you to want to cut them off. I say that if the mathematical logic
leads one there, then one has no choice other that to accept it, and if this
leads to nonsense then one has to accept that that something is
fundamentally wrong with the approach.

Let me be clear on this - I could not do any better. All my attempts to find
a rigorous way of doing Feynman-Dyson perturbation theory failed. This puts
me in good company with Dirac, Pauli, Feynman, the axiomatic field theorists
and many others. But in the end, I just decided that it was not worth
rescuing, tore everything up and started again. And - intermittently, at
least - I am still working on that.

Arnold Neumaier

Oct8-04, 06:19 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Chris Oakley wrote:\n>>Well, you can throw out the theory, but that seems a bit drastic.\n>\n>\n> I believe that one has no choice. One can in any case find holes in the\n> cutoff procedure. Supposing, for example, we were to choose Lambda as a\n> function rather than a constant. Furthermore, let us make it a function of\n> variables that never even entered the original equations. There is no reason\n> in principle why we should not be allowed to do this. We then very rapidly\n> get nonsense.\n\nNo. One gets results that depend on these variables only through Lambda.\nIf after adding the right counterterms depending on Lambda one performs\na limit of these variables that force Lambda to inf, one gets the\nstandard results. Renormalized QFT is very robust with respect to the\ndetailed renormalization procedure.\n\n\n> There is no such thing as "reasonable" in theoretical physics, there is only\n> mathematical logic. The pillars of our current understanding of physics,\n> namely quantum mechanics and special relativity, both defy common sense.\n\nThe do not defy mathematical logic, and neither does QFT.\nThe common sense of theoretical physics is different from the common sense\nof the intelligent non-physicist; it is presented in the textbooks\n(in more or less liberal form).\n\nThe renormalized perturbative expansion is mathematically completely\nwell-defined. Only the limit of the resulting asymptotic series is not.\nBut this is completely independent of the formal infinities arising in\nnaive QFT; it happens already in nonrelativist QM of systems having a\nbound state. Except that we there have alternative nonperturbative ways\nof defining the S-matrix...\n\n\nArnold Neumaier\n\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
>>Well, you can throw out the theory, but that seems a bit drastic.
>
>
> I believe that one has no choice. One can in any case find holes in the
> cutoff procedure. Supposing, for example, we were to choose \Lambda as a
> function rather than a constant. Furthermore, let us make it a function of
> variables that never even entered the original equations. There is no reason
> in principle why we should not be allowed to do this. We then very rapidly
> get nonsense.

No. One gets results that depend on these variables only through \Lambda.
If after adding the right counterterms depending on \Lambda one performs
a limit of these variables that force \Lambda to inf, one gets the
standard results. Renormalized QFT is very robust with respect to the
detailed renormalization procedure.

> There is no such thing as "reasonable" in theoretical physics, there is only
> mathematical logic. The pillars of our current understanding of physics,
> namely quantum mechanics and special relativity, both defy common sense.

The do not defy mathematical logic, and neither does QFT.
The common sense of theoretical physics is different from the common sense
of the intelligent non-physicist; it is presented in the textbooks
(in more or less liberal form).

The renormalized perturbative expansion is mathematically completely
well-defined. Only the limit of the resulting asymptotic series is not.
But this is completely independent of the formal infinities arising in
naive QFT; it happens already in nonrelativist QM of systems having a
bound state. Except that we there have alternative nonperturbative ways
of defining the S-matrix...

Arnold Neumaier

Matthew Nobes

Oct11-04, 03:52 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n"Chris Oakley" <coakley@cgoakley.demon.co.uk> wrote in message news:<ck2v5d\\$mg2\\$1\\$8300dec7@news.demon.co.uk >...\n> > Well, you can throw out the theory, but that seems a bit drastic.\n>\n> I believe that one has no choice.\n\nOf course one has a choice. You regulate the theory, and investigate\nthe consquences of doing that.\n\n> One can in any case find holes in the\n> cutoff procedure. Supposing, for example, we were to choose Lambda as a\n> function rather than a constant. Furthermore, let us make it a function of\n> variables that never even entered the original equations. There is no reason\n> in principle why we should not be allowed to do this. We then very rapidly\n> get nonsense. You will then say that it was my own fault for not making it a\n> constant, but all that means is that you can find a path through the\n> minefield. It does not mean that the minefield has ceased to exist.\n\nThe cutoff in *by*definition* an energy much much greater than the\nenergies you\'re interested in. It can be a function of any variables\nyou like, but it must always be much greater than the scale you\'re\ninterested in. I have no idea what you think this example proves.\n\n[snip]\n> You and Dr. Neumaier say that it is\n> "unreasonable" to have four-momentum integrals extending to infinity,\n\nYes, it\'s unreasonable to expect that QED is the be all and end all of\nphysical theories. I have no idea why you find that a controversial\npoint.\n\n> I say that if the mathematical logic\n> leads one there, then one has no choice other that to accept it, and if this\n> leads to nonsense then one has to accept that that something is\n> fundamentally wrong with the approach.\n\nWhat "logic" leads you to integrate to infinite momentum? Physics is\nnot pure mathematics, you\'re allowed to make arguements like "this\ntheory is not expected to work at 10 TeV, therefore I won\'t integrate\nout past that".\n\nMatt\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Chris Oakley" <coakley@cgoakley.demon.co.uk> wrote in message news:<ck2v5d$mg2$1$8300dec7@news.demon.co.uk>...
> > Well, you can throw out the theory, but that seems a bit drastic.
>
> I believe that one has no choice.

Of course one has a choice. You regulate the theory, and investigate
the consquences of doing that.

> One can in any case find holes in the
> cutoff procedure. Supposing, for example, we were to choose \Lambda as a
> function rather than a constant. Furthermore, let us make it a function of
> variables that never even entered the original equations. There is no reason
> in principle why we should not be allowed to do this. We then very rapidly
> get nonsense. You will then say that it was my own fault for not making it a
> constant, but all that means is that you can find a path through the
> minefield. It does not mean that the minefield has ceased to exist.

The cutoff in *by*definition* an energy much much greater than the
energies you're interested in. It can be a function of any variables
you like, but it must always be much greater than the scale you're
interested in. I have no idea what you think this example proves.

[snip]
> You and Dr. Neumaier say that it is
> "unreasonable" to have four-momentum integrals extending to infinity,

Yes, it's unreasonable to expect that QED is the be all and end all of
physical theories. I have no idea why you find that a controversial
point.

> I say that if the mathematical logic
> leads one there, then one has no choice other that to accept it, and if this
> leads to nonsense then one has to accept that that something is
> fundamentally wrong with the approach.

What "logic" leads you to integrate to infinite momentum? Physics is
not pure mathematics, you're allowed to make arguements like "this
theory is not expected to work at 10 TeV, therefore I won't integrate
out past that".

Matt

Chris Oakley

Oct11-04, 03:52 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n> If after adding the right counterterms depending on Lambda one performs\n> a limit of these variables that force Lambda to inf, one gets the\n> standard results. Renormalized QFT is very robust with respect to the\n> detailed renormalization procedure.\n\nExactly my point ... the "right counterterms". If you were to choose the\n"wrong" counterterms you would get a different answer. It is therefore only\n"robust" up to a point.\n\n> The renormalized perturbative expansion is mathematically completely\n> well-defined.\n\nI am not going to argue this point ... it could well be. In other words, it\nmay well be possible to devise a procedure that is well defined.\n\nMy point all along is, and has been, that however well-defined the procedure\nis, it does not derive from basic field quantization.\n\nI am using "derive" here in the strict mathematical sense: the one thing\ndoes not follow from the other. With your renormalized theory you are\neffectively starting again, from scratch.\n\nAnd yes, I know that this is not what is in most text books on the subject,\nbut the text books are *wrong*.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>If after adding the right counterterms depending on \Lambda one performs
> a limit of these variables that force \Lambda to inf, one gets the
> standard results. Renormalized QFT is very robust with respect to the
> detailed renormalization procedure.

Exactly my point ... the "right counterterms". If you were to choose the
"wrong" counterterms you would get a different answer. It is therefore only
"robust" up to a point.

> The renormalized perturbative expansion is mathematically completely
> well-defined.

I am not going to argue this point ... it could well be. In other words, it
may well be possible to devise a procedure that is well defined.

My point all along is, and has been, that however well-defined the procedure
is, it does not derive from basic field quantization.

I am using "derive" here in the strict mathematical sense: the one thing
does not follow from the other. With your renormalized theory you are
effectively starting again, from scratch.

And yes, I know that this is not what is in most text books on the subject,
but the text books are *wrong*.

Arnold Neumaier

Oct11-04, 01:06 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nChris Oakley wrote:\n>>If after adding the right counterterms depending on Lambda one performs\n>>a limit of these variables that force Lambda to inf, one gets the\n>>standard results. Renormalized QFT is very robust with respect to the\n>>detailed renormalization procedure.\n>\n>\n> Exactly my point ... the "right counterterms". If you were to choose the\n> "wrong" counterterms you would get a different answer. It is therefore only\n> "robust" up to a point.\n\nThe form of the counterterms is arbitrary as long as the limit exists,\nand the limit is then independent of the remaining freedom.\nThus it is as robust as is meaningful.\n\nThat you impose inappropriate standars is not the fault of physics.\n\n\n>>The renormalized perturbative expansion is mathematically completely\n>>well-defined.\n>\n> I am not going to argue this point ... it could well be. In other words, it\n> may well be possible to devise a procedure that is well defined.\n\nNot only well possible but well known.\n\n\n> My point all along is, and has been, that however well-defined the procedure\n> is, it does not derive from basic field quantization.\n\nIn the eyes of theoretical physicists, it defines what\n\'field quantizaion\' means.\n\n\n> And yes, I know that this is not what is in most text books on the subject,\n> but the text books are *wrong*.\n\nNo, only your way to look at it is...\n\nOne would not get Nobel prizes for wrong results.\n\n\nArnold Neumaier\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
>>If after adding the right counterterms depending on \Lambda one performs
>>a limit of these variables that force \Lambda to inf, one gets the
>>standard results. Renormalized QFT is very robust with respect to the
>>detailed renormalization procedure.
>
>
> Exactly my point ... the "right counterterms". If you were to choose the
> "wrong" counterterms you would get a different answer. It is therefore only
> "robust" up to a point.

The form of the counterterms is arbitrary as long as the limit exists,
and the limit is then independent of the remaining freedom.
Thus it is as robust as is meaningful.

That you impose inappropriate standars is not the fault of physics.

>>The renormalized perturbative expansion is mathematically completely
>>well-defined.
>
> I am not going to argue this point ... it could well be. In other words, it
> may well be possible to devise a procedure that is well defined.

Not only well possible but well known.

> My point all along is, and has been, that however well-defined the procedure
> is, it does not derive from basic field quantization.

In the eyes of theoretical physicists, it defines what
'field quantizaion' means.

> And yes, I know that this is not what is in most text books on the subject,
> but the text books are *wrong*.

No, only your way to look at it is...

One would not get Nobel prizes for wrong results.

Arnold Neumaier

Chris Oakley

Oct12-04, 10:49 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n> > My point all along is, and has been, that however well-defined the\nprocedure\n> > is, it does not derive from basic field quantization.\n>\n> In the eyes of theoretical physicists, it defines what\n> \'field quantizaion\' means.\n\nIf this is the case, then I cannot possibly argue. Indeed, many QFT text\nbooks now more or less *define* the theory as being the Feynman graphs\ncombined with a set of rules for taming the infinities.\n\nIt is not good enough for me, though. The divergent integrals are not there\nbecause I chose to be deliberately awkward. They follow from the reduction\nof the S-matrix in terms of normal-ordered products of free fields and are\ntherefore a necessary consequence of the theory. You cannot put artificial\nlimits on the integrals just because you do not like the result.\n\nAnd let us remind ourselves - there is a big difference between\napproximation and inconsistency.\n\nNewtonian mechanics is an *approximation*. When speeds are close to the\nspeed of light, it will break down. But it works well otherwise. It does not\nsay, for example, that all masses, energies and momenta are infinite.\n\nQuantum Electrodynamics, on the other hand, unless you tamper with it, gives\ninfinite results for most measurable quantities. There may be a more\ngrandiose scheme that rescues it - who knows, this might even involve\nSuperstrings - but there is no domain in which the theory is a good\napproximation. Infinity is not a close approximation of most measurable\nquantities.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>My point all along is, and has been, that however well-defined the
procedure
> > is, it does not derive from basic field quantization.
>
> In the eyes of theoretical physicists, it defines what
> 'field quantizaion' means.

If this is the case, then I cannot possibly argue. Indeed, many QFT text
books now more or less *define* the theory as being the Feynman graphs
combined with a set of rules for taming the infinities.

It is not good enough for me, though. The divergent integrals are not there
because I chose to be deliberately awkward. They follow from the reduction
of the S-matrix in terms of normal-ordered products of free fields and are
therefore a necessary consequence of the theory. You cannot put artificial
limits on the integrals just because you do not like the result.

And let us remind ourselves - there is a big difference between
approximation and inconsistency.

Newtonian mechanics is an *approximation*. When speeds are close to the
speed of light, it will break down. But it works well otherwise. It does not
say, for example, that all masses, energies and momenta are infinite.

Quantum Electrodynamics, on the other hand, unless you tamper with it, gives
infinite results for most measurable quantities. There may be a more
grandiose scheme that rescues it - who knows, this might even involve
Superstrings - but there is no domain in which the theory is a good
approximation. Infinity is not a close approximation of most measurable
quantities.

Matthew Nobes

Oct13-04, 03:45 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n"Chris Oakley" <coakley@cgoakley.demon.co.uk> wrote in message news:<ckev8c\\$10v\\$1\\$830fa79d@news.demon.co.uk >...\n> Let\'s just remember how we got to where we are:\n>\n[snip derivation of perturbation theory]\n>\n[snip]\n> If you could make your adjustments *before* the final step, so that the\n> integrals were *naturally* bounded, I might be more sympathetic.\n\nAnd right back at the start, I suggested *start* with a lattice.\nDefine the theory on a finite grid, via the path integral. You\n*never* encounter an infinity. You can, if you wish, define\nperturbation theory, and all integrals will be cutoff at pi/a. There\nare even ways known now to define Fermions on the lattice (overlap\nfermions).\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Chris Oakley" <coakley@cgoakley.demon.co.uk> wrote in message news:<ckev8c$10v$1$830fa79d@news.demon.co.uk>...
> Let's just remember how we got to where we are:
>
[snip derivation of perturbation theory]
>
[snip]
> If you could make your adjustments *before* the final step, so that the
> integrals were *naturally* bounded, I might be more sympathetic.

And right back at the start, I suggested *start* with a lattice.
Define the theory on a finite grid, via the path integral. You
*never* encounter an infinity. You can, if you wish, define
perturbation theory, and all integrals will be cutoff at \pi/a. There
are even ways known now to define Fermions on the lattice (overlap
fermions).

Aaron Bergman

Oct13-04, 03:45 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIn article <ckev8c\\$10v\\$1\\$830fa79d@news.demon.co.uk>,\n" Chris Oakley" <coakley@cgoakley.demon.co.uk> wrote:\n\n> Let\'s just remember how we got to where we are:\n>\n> (i) We solved free field theory, including the expansion in terms of\n> annihilation and creation operators.\n> (ii) We introduced an interaction term in the Lagrangian density.\n> (iii) We introduced a unitary transformation that turns free fields into\n> interacting fields using the Interaction Picture.\n> (iv) We wrote out the Interaction Picture time evolution operator as\n> time-ordered products of the interaction.\n> (v) We reduced the time-ordered products to normal-ordered products.\n> (vi) We obtained thereby the S-matrix for scattering processes involving a\n> finite number of incoming particles with definite four-momentum scattering\n> to a finite number of outgoing particles with definite four-momenta as a sum\n> of Feynman diagrams.\n>\n> Many of these Feynman diagrams are pathologically divergent integrals.\n>\n> My response: This is because the theory is fundamentally flawed. Go back to\n> (ii) and try again.\n>\n> Your response (and everyone else\'s, it seems): Steps (i)-(vi) cannot have\n> really meant that we have momentum integrals that go to infinity, so rather\n> than fixing anything before (vi) let\'s just cut off the integrals.\n>\n> It\'s not pretty, it\'s not nice, and I have never condoned it. I don\'t care\n> who has been awarded prizes, Nobel or otherwise for doing it.\n>\n> If you could make your adjustments *before* the final step, so that the\n> integrals were *naturally* bounded, I might be more sympathetic.\n\nThe problem is step ii. The interaction term leads to a theory that is\nnot well-defined, so it\'s hardly a surprise that everything you get\nlater on is nonsense. There are a few ways you can fix this. The first\nis to treat the interacting theory as an effective field theory with an\nexplicit cutoff. Renormalization tells us that the answers we get can be\nindependent of the high energy stuff, so this explains pretty much every\nexperimental result we have.\n\nBut, you might not like this. So, what we can do is try to take the\ncutoff to infinity. In some cases we can do this while keeping all\nobservable quantities finite. When we do this, we generally see that the\ncoefficients of the interaction terms in the Lagrangian go to zero, ie,\nthe theory is asymptotically free. Thus, the reason that we failed in\nour original attempt to quantize the interacting theory is that the\ninteraction terms cannot have a finite coefficient. They have to be\ninfinitesimal perturbations around the free theory.\n\nIn general, theories don\'t have to be asymptotically free to be well\ndefined. Instead, they can have nontrivial UV fixed points. So, in\ngeneral, I would define an interacting quantum field theory to be an\ninfinitesimal pertubation of a conformal field theory.\n\nAaron\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <ckev8c$10v$1$830fa79d@news.demon.co.uk>,
"Chris Oakley" <coakley@cgoakley.demon.co.uk> wrote:

> Let's just remember how we got to where we are:
>
> (i) We solved free field theory, including the expansion in terms of
> annihilation and creation operators.
> (ii) We introduced an interaction term in the Lagrangian density.
> (iii) We introduced a unitary transformation that turns free fields into
> interacting fields using the Interaction Picture.
> (iv) We wrote out the Interaction Picture time evolution operator as
> time-ordered products of the interaction.
> (v) We reduced the time-ordered products to normal-ordered products.
> (vi) We obtained thereby the S-matrix for scattering processes involving a
> finite number of incoming particles with definite four-momentum scattering
> to a finite number of outgoing particles with definite four-momenta as a sum
> of Feynman diagrams.
>
> Many of these Feynman diagrams are pathologically divergent integrals.
>
> My response: This is because the theory is fundamentally flawed. Go back to
> (ii) and try again.
>
> Your response (and everyone else's, it seems): Steps (i)-(vi) cannot have
> really meant that we have momentum integrals that go to infinity, so rather
> than fixing anything before (vi) let's just cut off the integrals.
>
> It's not pretty, it's not nice, and I have never condoned it. I don't care
> who has been awarded prizes, Nobel or otherwise for doing it.
>
> If you could make your adjustments *before* the final step, so that the
> integrals were *naturally* bounded, I might be more sympathetic.

The problem is step ii. The interaction term leads to a theory that is
not well-defined, so it's hardly a surprise that everything you get
later on is nonsense. There are a few ways you can fix this. The first
is to treat the interacting theory as an effective field theory with an
explicit cutoff. Renormalization tells us that the answers we get can be
independent of the high energy stuff, so this explains pretty much every
experimental result we have.

But, you might not like this. So, what we can do is try to take the
cutoff to infinity. In some cases we can do this while keeping all
observable quantities finite. When we do this, we generally see that the
coefficients of the interaction terms in the Lagrangian go to zero, ie,
the theory is asymptotically free. Thus, the reason that we failed in
our original attempt to quantize the interacting theory is that the
interaction terms cannot have a finite coefficient. They have to be
infinitesimal perturbations around the free theory.

In general, theories don't have to be asymptotically free to be well
defined. Instead, they can have nontrivial UV fixed points. So, in
general, I would define an interacting quantum field theory to be an
infinitesimal pertubation of a conformal field theory.

Aaron

Arnold Neumaier

Oct13-04, 08:44 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Chris Oakley wrote:\n\n> Newtonian mechanics is an *approximation*. When speeds are close to the\n> speed of light, it will break down. But it works well otherwise. It does not\n> say, for example, that all masses, energies and momenta are infinite.\n>\n> Quantum Electrodynamics, on the other hand, unless you tamper with it, gives\n> infinite results for most measurable quantities.\n\nWhat you call QED is not a theory at all, since it gives infinities all\nover the place.\n\nWhat physicists call QED is the theory that gives an asymptotic series\nwith finite, computable coefficients for all S-matrix elements, and which\nis explained in every textbook on QFT.\n\nIt does _not_ say that all masses, energies and momenta are infinite.\nQED works with finite, meaningful values of all all masses, energies\nand momenta. That the limit of some intermediate quantities are infinite\nonly means that it is pointless to take these limits, and not that the\nphysical limits which are actually taken are also infinite.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:

> Newtonian mechanics is an *approximation*. When speeds are close to the
> speed of light, it will break down. But it works well otherwise. It does not
> say, for example, that all masses, energies and momenta are infinite.
>
> Quantum Electrodynamics, on the other hand, unless you tamper with it, gives
> infinite results for most measurable quantities.

What you call QED is not a theory at all, since it gives infinities all
over the place.

What physicists call QED is the theory that gives an asymptotic series
with finite, computable coefficients for all S-matrix elements, and which
is explained in every textbook on QFT.

It does _not_ say that all masses, energies and momenta are infinite.
QED works with finite, meaningful values of all all masses, energies
and momenta. That the limit of some intermediate quantities are infinite
only means that it is pointless to take these limits, and not that the
physical limits which are actually taken are also infinite.

Arnold Neumaier

Arnold Neumaier

Oct13-04, 02:13 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nChris Oakley wrote:\n> Let\'s just remember how we got to where we are:\n>\n> (i) We solved free field theory, including the expansion in terms of\n> annihilation and creation operators.\n> (ii) We introduced an interaction term in the Lagrangian density.\n> (iii) We introduced a unitary transformation that turns free fields into\n> interacting fields using the Interaction Picture.\n> (iv) We wrote out the Interaction Picture time evolution operator as\n> time-ordered products of the interaction.\n> (v) We reduced the time-ordered products to normal-ordered products.\n> (vi) We obtained thereby the S-matrix for scattering processes involving a\n> finite number of incoming particles with definite four-momentum scattering\n> to a finite number of outgoing particles with definite four-momenta as a sum\n> of Feynman diagrams.\n>\n> Many of these Feynman diagrams are pathologically divergent integrals.\n>\n> My response: This is because the theory is fundamentally flawed. Go back to\n> (ii) and try again.\n>\n> Your response (and everyone else\'s, it seems): Steps (i)-(vi) cannot have\n> really meant that we have momentum integrals that go to infinity, so rather\n> than fixing anything before (vi) let\'s just cut off the integrals.\n>\n> It\'s not pretty, it\'s not nice, and I have never condoned it. I don\'t care\n> who has been awarded prizes, Nobel or otherwise for doing it.\n\nNobody cares whether or not you care or condone. People care whether\nthey can successfully model reality. That affects the prizes awarded,\nwhile your don\'t care attitude is stale.\n\n\n> If you could make your adjustments *before* the final step, so that the\n> integrals were *naturally* bounded, I might be more sympathetic.\n\nThis is indeed what is done, though physicists are usually careless\nin the way they discuss it. In step (ii), the local interaction is\nreplaced by a nonlocal interaction depending on the UV cutoff Lambda.\nThus one has V(g,Lambda) in place of V(g), where g are the coupling\nconstants. Then one can do everything without encountering any infinity\nat all. The result is an asymptotic series S(g,Lambda) for the S-matrix\nwith finite, computable coefficients. This S-matrix is unitary and has\nall properties one would like to have, except that it is only\napproximately Lorentz invariant.\n\nTo restore Lorentz invariance, one uses a running coupling constant\ng=g(Lambda,mu) which, for fixed renormalization point mu, is uniquely\ndetermined as the solution of a renormalization group equation whose\ncoefficients are also defined as a (presumably even convergent)\nasymptotic expansion. Having this, one can take the limit\nS(mu) = lim_{Lambda to inf} S(g(Lambda,mu),Lambda)\nwhich is an asymptotic series in hbar with finite, computable coefficients\nwhen the theory is renormalizable, and is Lorentz invariant.\n\nThus one gets the desired Lorentz invariant theory as a perturbatively\nwell-defined limit of perturbatively well-defined but not Lorentz\ninvariant theories. The only infinity encountered is not worse than\nthe infinity encountered in defining Riemann integrals over the\nreal line, where one also gets a finite limit by letting a finite\ncutoff go to infinity.\n\nThe real mathematical difficulties in QFT are not in the renormalization\nprocedure but in giving a nonperturbative definition of the S-matrix\nS(mu).\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
> Let's just remember how we got to where we are:
>
> (i) We solved free field theory, including the expansion in terms of
> annihilation and creation operators.
> (ii) We introduced an interaction term in the Lagrangian density.
> (iii) We introduced a unitary transformation that turns free fields into
> interacting fields using the Interaction Picture.
> (iv) We wrote out the Interaction Picture time evolution operator as
> time-ordered products of the interaction.
> (v) We reduced the time-ordered products to normal-ordered products.
> (vi) We obtained thereby the S-matrix for scattering processes involving a
> finite number of incoming particles with definite four-momentum scattering
> to a finite number of outgoing particles with definite four-momenta as a sum
> of Feynman diagrams.
>
> Many of these Feynman diagrams are pathologically divergent integrals.
>
> My response: This is because the theory is fundamentally flawed. Go back to
> (ii) and try again.
>
> Your response (and everyone else's, it seems): Steps (i)-(vi) cannot have
> really meant that we have momentum integrals that go to infinity, so rather
> than fixing anything before (vi) let's just cut off the integrals.
>
> It's not pretty, it's not nice, and I have never condoned it. I don't care
> who has been awarded prizes, Nobel or otherwise for doing it.

Nobody cares whether or not you care or condone. People care whether
they can successfully model reality. That affects the prizes awarded,
while your don't care attitude is stale.

> If you could make your adjustments *before* the final step, so that the
> integrals were *naturally* bounded, I might be more sympathetic.

This is indeed what is done, though physicists are usually careless
in the way they discuss it. In step (ii), the local interaction is
replaced by a nonlocal interaction depending on the UV cutoff \Lambda.
Thus one has V(g,\Lambda) in place of V(g), where g are the coupling
constants. Then one can do everything without encountering any infinity
at all. The result is an asymptotic series S(g,\Lambda) for the S-matrix
with finite, computable coefficients. This S-matrix is unitary and has
all properties one would like to have, except that it is only
approximately Lorentz invariant.

To restore Lorentz invariance, one uses a running coupling constant
g=g(\Lambda,\mu) which, for fixed renormalization point \mu, is uniquely
determined as the solution of a renormalization group equation whose
coefficients are also defined as a (presumably even convergent)
asymptotic expansion. Having this, one can take the limit
S(\mu) = lim_{\Lambda to inf} S(g(\Lambda,\mu),\Lambda)
which is an asymptotic series in \hbar with finite, computable coefficients
when the theory is renormalizable, and is Lorentz invariant.

Thus one gets the desired Lorentz invariant theory as a perturbatively
well-defined limit of perturbatively well-defined but not Lorentz
invariant theories. The only infinity encountered is not worse than
the infinity encountered in defining Riemann integrals over the
real line, where one also gets a finite limit by letting a finite
cutoff go to infinity.

The real mathematical difficulties in QFT are not in the renormalization
procedure but in giving a nonperturbative definition of the S-matrix
S(\mu).

Arnold Neumaier

Chris Oakley

Oct14-04, 11:07 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n> This is indeed what is done, though physicists are usually careless\n> in the way they discuss it. In step (ii), the local interaction is\n> replaced by a nonlocal interaction depending on the UV cutoff Lambda.\n> Thus one has V(g,Lambda) in place of V(g), where g are the coupling\n> constants. Then one can do everything without encountering any infinity\n> at all. The result is an asymptotic series S(g,Lambda) for the S-matrix\n> with finite, computable coefficients. This S-matrix is unitary and has\n> all properties one would like to have, except that it is only\n> approximately Lorentz invariant.\n>\n> To restore Lorentz invariance, one uses a running coupling constant\n> g=g(Lambda,mu) which, for fixed renormalization point mu, is uniquely\n> determined as the solution of a renormalization group equation whose\n> coefficients are also defined as a (presumably even convergent)\n> asymptotic expansion. Having this, one can take the limit\n> S(mu) = lim_{Lambda to inf} S(g(Lambda,mu),Lambda)\n> which is an asymptotic series in hbar with finite, computable coefficients\n> when the theory is renormalizable, and is Lorentz invariant.\n>\n> Thus one gets the desired Lorentz invariant theory as a perturbatively\n> well-defined limit of perturbatively well-defined but not Lorentz\n> invariant theories. The only infinity encountered is not worse than\n> the infinity encountered in defining Riemann integrals over the\n> real line, where one also gets a finite limit by letting a finite\n> cutoff go to infinity.\n\nI am not aware of any scheme like this, so please give a reference or two.\nWhat I am aware of is the rewriting of mass, coupling and wave function as\nfunctions of the (as yet unspecified) UV cutoff (with otherwise no\nmodification to the orginal theory - i.e. no "non-local" coupling). The\nintegrals in the S-matrix calculation are cut off using Lambda as a\nparameter and the functional form of the "bare" parameters is chosen to\neliminate the Lambda dependency. The whole procedure is meaningless as\nnothing in the theory we developed ever gave us the right to cut off the\nintegrals. If we can indeed do better than this I would love to know.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>This is indeed what is done, though physicists are usually careless
> in the way they discuss it. In step (ii), the local interaction is
> replaced by a nonlocal interaction depending on the UV cutoff \Lambda.
> Thus one has V(g,\Lambda) in place of V(g), where g are the coupling
> constants. Then one can do everything without encountering any infinity
> at all. The result is an asymptotic series S(g,\Lambda) for the S-matrix
> with finite, computable coefficients. This S-matrix is unitary and has
> all properties one would like to have, except that it is only
> approximately Lorentz invariant.
>
> To restore Lorentz invariance, one uses a running coupling constant
> g=g(\Lambda,\mu) which, for fixed renormalization point \mu, is uniquely
> determined as the solution of a renormalization group equation whose
> coefficients are also defined as a (presumably even convergent)
> asymptotic expansion. Having this, one can take the limit
> S(\mu) = lim_{\Lambda to inf} S(g(\Lambda,\mu),\Lambda)
> which is an asymptotic series in \hbar with finite, computable coefficients
> when the theory is renormalizable, and is Lorentz invariant.
>
> Thus one gets the desired Lorentz invariant theory as a perturbatively
> well-defined limit of perturbatively well-defined but not Lorentz
> invariant theories. The only infinity encountered is not worse than
> the infinity encountered in defining Riemann integrals over the
> real line, where one also gets a finite limit by letting a finite
> cutoff go to infinity.

I am not aware of any scheme like this, so please give a reference or two.
What I am aware of is the rewriting of mass, coupling and wave function as
functions of the (as yet unspecified) UV cutoff (with otherwise no
modification to the orginal theory - i.e. no "non-local" coupling). The
integrals in the S-matrix calculation are cut off using \Lambda as a
parameter and the functional form of the "bare" parameters is chosen to
eliminate the \Lambda dependency. The whole procedure is meaningless as
nothing in the theory we developed ever gave us the right to cut off the
integrals. If we can indeed do better than this I would love to know.

Chris Oakley

Oct14-04, 11:07 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nMatthew,\n\nI am well aware of the fact that you are a strong advocate of Lattice Gauge\nTheory.\n\nSome of us, however, are put off by\n\n(i) the need for a computer to do any meaningful calculation, and\n(ii) the loss of all the symmetries, except as discrete approximations.\n\nOf course the lattice spacing provides a natural cutoff. I cannot argue with\nthis.\nBut that is the point ... it is a natural cutoff *for the technique you are\nusing*.\n\nWhat I am objecting to is unnatural cutoffs introduced in the continuum\ntheory solely to dig oneself out of a hole.\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Matthew,

I am well aware of the fact that you are a strong advocate of Lattice Gauge
Theory.

Some of us, however, are put off by

(i) the need for a computer to do any meaningful calculation, and
(ii) the loss of all the symmetries, except as discrete approximations.

Of course the lattice spacing provides a natural cutoff. I cannot argue with
this.
But that is the point ... it is a natural cutoff *for the technique you are
using*.

What I am objecting to is unnatural cutoffs introduced in the continuum
theory solely to dig oneself out of a hole.

Matthew Nobes

Oct15-04, 01:55 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n"Chris Oakley" <coakley@cgoakley.demon.co.uk> wrote in message news:<ckle28\\$lah\\$1\\$8300dec7@news.demon.co.uk >...\n> Matthew,\n>\n> I am well aware of the fact that you are a strong advocate of Lattice Gauge\n> Theory.\n\nI\'m an advocate of quantum field theory. The lattice happens to be a\nvery nice way of thinking about it, since it provides a regulator\nvalid non-perturbativly. So you can make statements about the theory\nthat you cannot make otherwise.\n\nI\'m only talking about lattice QFT because it offers a convieniant\n(and obvious) counterexample to your arguements about the\nrenormalization program.\n\n> Some of us, however, are put off by\n>\n> (i) the need for a computer to do any meaningful calculation, and\n\nThis is not relevent for my point. I am discussing the lattice as a\nregulator. You certainly do not need a computer to do low order (one\nloop) perturbative calculations, and you can, using this regulator,\nshow that you never once encounter an infinite quantity.\n\nFurthermore, even if perturbation theory were defined in such a way to\nsatisfy you, you\'d *still* need a computer to calculate things in QCD,\nsince the theory at low energies is strongly coupled.\n\n> (ii) the loss of all the symmetries, except as discrete approximations.\n\nThis is simply not true, one symmetry is lost, Poincare invariance.\n\nWilson\'s formulation retains local gauge symmetry (indeed that\'s the\nwhole point of Wilson\'s construction). Not suprisingly it is gauge\nsymmetry that is *crucial* to the renormalization program with a\ndifferent regulator. Modern fermion actions also retain exact chiral\nsymmetry, which was previously a failing of many lattice methods.\n\nIt\'s even possible to construct lattice models with exact\nsupersymmetries, if that sort of thing floats your boat.\n\n> Of course the lattice spacing provides a natural cutoff. I cannot argue with\n> this.\n\nWell then your point about infinite integrals, and the nonsensical\nnature of QFT is moot. There *is* a way to do things which contains\nno infinities. Enjoy.\n\n> But that is the point ... it is a natural cutoff *for the technique you are\n> using*.\n\nWhat technique am I using?\n\n> What I am objecting to is unnatural cutoffs introduced in the continuum\n> theory solely to dig oneself out of a hole.\n\nWhat I thought you were objecting to is the use of infinite integrals\nin QFT. Well, you don\'t need to use them, there is a manifestly\nfinite method. What\'s more you can show (Riesz theorem) that you\'ll\nget the same results using lattice perturbation theory as you would\nget using perturbation theory with a different regulator.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Chris Oakley" <coakley@cgoakley.demon.co.uk> wrote in message news:<ckle28$lah$1$8300dec7@news.demon.co.uk>...
> Matthew,
>
> I am well aware of the fact that you are a strong advocate of Lattice Gauge
> Theory.

I'm an advocate of quantum field theory. The lattice happens to be a
very nice way of thinking about it, since it provides a regulator
valid non-perturbativly. So you can make statements about the theory
that you cannot make otherwise.

I'm only talking about lattice QFT because it offers a convieniant
(and obvious) counterexample to your arguements about the
renormalization program.

> Some of us, however, are put off by
>
> (i) the need for a computer to do any meaningful calculation, and

This is not relevent for my point. I am discussing the lattice as a
regulator. You certainly do not need a computer to do low order (one
loop) perturbative calculations, and you can, using this regulator,
show that you never once encounter an infinite quantity.

Furthermore, even if perturbation theory were defined in such a way to
satisfy you, you'd *still* need a computer to calculate things in QCD,
since the theory at low energies is strongly coupled.

> (ii) the loss of all the symmetries, except as discrete approximations.

This is simply not true, one symmetry is lost, Poincare invariance.

Wilson's formulation retains local gauge symmetry (indeed that's the
whole point of Wilson's construction). Not suprisingly it is gauge
symmetry that is *crucial* to the renormalization program with a
different regulator. Modern fermion actions also retain exact chiral
symmetry, which was previously a failing of many lattice methods.

It's even possible to construct lattice models with exact
supersymmetries, if that sort of thing floats your boat.

> Of course the lattice spacing provides a natural cutoff. I cannot argue with
> this.

Well then your point about infinite integrals, and the nonsensical
nature of QFT is moot. There *is* a way to do things which contains
no infinities. Enjoy.

> But that is the point ... it is a natural cutoff *for the technique you are
> using*.

What technique am I using?

> What I am objecting to is unnatural cutoffs introduced in the continuum
> theory solely to dig oneself out of a hole.

What I thought you were objecting to is the use of infinite integrals
in QFT. Well, you don't need to use them, there is a manifestly
finite method. What's more you can show (Riesz theorem) that you'll
get the same results using lattice perturbation theory as you would
get using perturbation theory with a different regulator.

Arnold Neumaier

Oct15-04, 01:55 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>NNTP-Posting-Host: lfa222122.richmond.edu\nX-Trace: rumor.richmond.edu 1097867146 20476 141.166.222.122 (15 Oct 2004 19:05:46 GMT)\nX-Complaints-To: usenet@rumor.richmond.edu\nNNTP-Posting-Date: Fri, 15 Oct 2004 19:05:46 +0000 (UTC)\nReceived-SPF: Received-SPF: none (mailbox5.ucsd.edu: domain of news@novembaer.univie.ac.at does not designate permitted sender hosts)\nXref: core-easynews sci.physics.research:59940\n\n\n\nChris Oakley wrote:\n>>This is indeed what is done, though physicists are usually careless\n>>in the way they discuss it. In step (ii), the local interaction is\n>>replaced by a nonlocal interaction depending on the UV cutoff Lambda.\n>>Thus one has V(g,Lambda) in place of V(g), where g are the coupling\n>>constants. Then one can do everything without encountering any infinity\n>>at all. The result is an asymptotic series S(g,Lambda) for the S-matrix\n>>with finite, computable coefficients. This S-matrix is unitary and has\n>>all properties one would like to have, except that it is only\n>>approximately Lorentz invariant.\n>>\n>>To restore Lorentz invariance, one uses a running coupling constant\n>>g=g(Lambda,mu) which, for fixed renormalization point mu, is uniquely\n>>determined as the solution of a renormalization group equation whose\n>>coefficients are also defined as a (presumably even convergent)\n>>asymptotic expansion. Having this, one can take the limit\n>> S(mu) = lim_{Lambda to inf} S(g(Lambda,mu),Lambda)\n>>which is an asymptotic series in hbar with finite, computable coefficients\n>>when the theory is renormalizable, and is Lorentz invariant.\n>>\n>>Thus one gets the desired Lorentz invariant theory as a perturbatively\n>>well-defined limit of perturbatively well-defined but not Lorentz\n>>invariant theories. The only infinity encountered is not worse than\n>>the infinity encountered in defining Riemann integrals over the\n>>real line, where one also gets a finite limit by letting a finite\n>>cutoff go to infinity.\n>\n\n> I am not aware of any scheme like this, so please give a reference or two.\n\nThis is nowhere spelled out explicitly, as far as I can tell; you have\nto read between the lines. But if you follow a standard textbook that\nregularizes with an explicit cutoff, it will not be difficult to\nrewrite everything they do along the lines indicated above.\n\n> What I am aware of is the rewriting of mass, coupling and wave function as\n> functions of the (as yet unspecified) UV cutoff (with otherwise no\n> modification to the orginal theory - i.e. no "non-local" coupling).\n\nWrite the (Euclidean = Wick rotated) field as\nPhi(x) = integral dp exp(-i p dot x) Phihat(p)\nand substitute it into the action. This gives an action in the\nmomentum representation. Then regularize the interaction term by\nthrowing away the momenta above some cutoff.\n\nIntroducing the cutoff makes the interaction nonlocal, as you can see by\ngoing from the momentum representation of the interaction term\n(with high momenta cut off) back to the position representation\nby substituting\nPhihat(p) = const * integral dx exp(i p dot x) Phi(x).\nBut actually you don\'t need to care about locality or not, since the\nregularized interaction in the momentum representation is mathematically\nok. From here, you can proceed with meaningful formulas throughout the\nwhole renormalization procedure. All contributions to the S-matrix\nelements of this regularized theory are finite, and give the S-matrix\nof the regularized interaction. At the very end you can pass to the limit,\nbut not earlier.\n\n\n> The\n> integrals in the S-matrix calculation are cut off using Lambda as a\n> parameter and the functional form of the "bare" parameters is chosen to\n> eliminate the Lambda dependency.\n\nNo, the Lambda dependence is not eliminated. The bare parameters are\nchosen as a function of Lambda such that one can afterwards pass to the\nlimit in the complete S-matrix elements.\n\n\n> The whole procedure is meaningless as\n> nothing in the theory we developed ever gave us the right to cut off the\n> integrals.\n\nNo. Since we are free to choose any interaction we like, the theory\nexplicitly gives us the right to cut off the integrals. Nothing on the\nmathematical side forbids to start with a nonlocal interaction.\n\n\nArnold Neumaier\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>NNTP-Posting-Host: lfa222122.richmond.edu
X-Trace: rumor.richmond.edu 1097867146 20476 141.166.222.122 (15 Oct 2004 19:05:46 GMT)
X-Complaints-To: usenet@rumor.richmond.edu
NNTP-Posting-Date: Fri, 15 Oct 2004 19:05:46 +0000 (UTC)
Received-SPF: Received-SPF: none (mailbox5.ucsd.edu: domain of news@novembaer.univie.ac.at does not designate permitted sender hosts)
Xref: core-easynews sci.physics.research:59940

Chris Oakley wrote:
>>This is indeed what is done, though physicists are usually careless
>>in the way they discuss it. In step (ii), the local interaction is
>>replaced by a nonlocal interaction depending on the UV cutoff \Lambda.
>>Thus one has V(g,\Lambda) in place of V(g), where g are the coupling
>>constants. Then one can do everything without encountering any infinity
>>at all. The result is an asymptotic series S(g,\Lambda) for the S-matrix
>>with finite, computable coefficients. This S-matrix is unitary and has
>>all properties one would like to have, except that it is only
>>approximately Lorentz invariant.
>>
>>To restore Lorentz invariance, one uses a running coupling constant
>>g=g(\Lambda,\mu) which, for fixed renormalization point \mu, is uniquely
>>determined as the solution of a renormalization group equation whose
>>coefficients are also defined as a (presumably even convergent)
>>asymptotic expansion. Having this, one can take the limit
>> S(\mu) = lim_{\Lambda to inf} S(g(\Lambda,\mu),\Lambda)
>>which is an asymptotic series in \hbar with finite, computable coefficients
>>when the theory is renormalizable, and is Lorentz invariant.
>>
>>Thus one gets the desired Lorentz invariant theory as a perturbatively
>>well-defined limit of perturbatively well-defined but not Lorentz
>>invariant theories. The only infinity encountered is not worse than
>>the infinity encountered in defining Riemann integrals over the
>>real line, where one also gets a finite limit by letting a finite
>>cutoff go to infinity.
>

> I am not aware of any scheme like this, so please give a reference or two.

This is nowhere spelled out explicitly, as far as I can tell; you have
to read between the lines. But if you follow a standard textbook that
regularizes with an explicit cutoff, it will not be difficult to
rewrite everything they do along the lines indicated above.

> What I am aware of is the rewriting of mass, coupling and wave function as
> functions of the (as yet unspecified) UV cutoff (with otherwise no
> modification to the orginal theory - i.e. no "non-local" coupling).

Write the (Euclidean = Wick rotated) field as
\Phi(x) = integral dp \exp(-i p dot x) Phihat(p)
and substitute it into the action. This gives an action in the
momentum representation. Then regularize the interaction term by
throwing away the momenta above some cutoff.

Introducing the cutoff makes the interaction nonlocal, as you can see by
going from the momentum representation of the interaction term
(with high momenta cut off) back to the position representation
by substituting
Phihat(p) = const * integral dx \exp(i p dot x) \Phi(x).
But actually you don't need to care about locality or not, since the
regularized interaction in the momentum representation is mathematically
ok. From here, you can proceed with meaningful formulas throughout the
whole renormalization procedure. All contributions to the S-matrix
elements of this regularized theory are finite, and give the S-matrix
of the regularized interaction. At the very end you can pass to the limit,
but not earlier.

> The
> integrals in the S-matrix calculation are cut off using \Lambda as a
> parameter and the functional form of the "bare" parameters is chosen to
> eliminate the \Lambda dependency.

No, the \Lambda dependence is not eliminated. The bare parameters are
chosen as a function of \Lambda such that one can afterwards pass to the
limit in the complete S-matrix elements.

> The whole procedure is meaningless as
> nothing in the theory we developed ever gave us the right to cut off the
> integrals.

No. Since we are free to choose any interaction we like, the theory
explicitly gives us the right to cut off the integrals. Nothing on the
mathematical side forbids to start with a nonlocal interaction.

Arnold Neumaier

News Admin

Oct19-04, 03:29 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nI do not think that this is going to work.\n\nIgnoring Haag\'s theorem, the unitary transformation defined by\n\n\\phi_{in}({\\bf x}, t) = U(t)\\phi({\\bf x},t)U^\\dagger(t)\n\nonly enables one to transform interactions to their equivalents in terms of\n"in" fields when they apply to the same time. Thus we can transform a local\ninteraction \\phi^3(x) to \\phi_{in}^3(x), but we cannot do this if the field\noperators are a smearing that is non-local in time, as would result from\nyour proposed range limitation in momentum space.\n\nIn all the references I have the arguments rely on the interaction being a\nlocal product of fields. I do not know what one can do to make the theory\nwork when this is not the case.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>I do not think that this is going to work.

Ignoring Haag's theorem, the unitary transformation defined by

\phi_{in}({\bf x}, t) = U(t)\phi({\bf x},t)U^\dagger(t)

only enables one to transform interactions to their equivalents in terms of
"in" fields when they apply to the same time. Thus we can transform a local
interaction \phi^3(x) to \phi_{in}^3(x), but we cannot do this if the field
operators are a smearing that is non-local in time, as would result from
your proposed range limitation in momentum space.

In all the references I have the arguments rely on the interaction being a
local product of fields. I do not know what one can do to make the theory
work when this is not the case.

Arnold Neumaier

Oct20-04, 03:18 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nNews Admin wrote:\n\n> I do not think that this is going to work.\n\nPlease quote next time a small piece of what you are referring to,\nso that one can follow without guessing.\n\n\n[Moderator\'s note: As it turns out, it was aparently Chris Oakley who accidentally\nwrote under the name "News Admin". Arnold Neumaier is right that it is good\nUSENET practice and highly encouraged to begin a followup message with a *brief*\nquote from the text one is replying to, so that readers will know what is going\non. More details on how to post are summarized here:\nhttp://www-stud.uni-essen.de/~sb0264/HowToPost.html . -usc]\n\n\nI assume \'this\' is my statement:\n\n\'\'Write the (Euclidean = Wick rotated) field as\nPhi(x) = integral dp exp(-i p dot x) Phihat(p)\nand substitute it into the action. This gives an action in the\nmomentum representation. Then regularize the interaction term by\nthrowing away the momenta above some cutoff. [...]\nFrom here, you can proceed with meaningful formulas throughout the\nwhole renormalization procedure. All contributions to the S-matrix\nelements of this regularized theory are finite, and give the S-matrix\nof the regularized interaction. At the very end you can pass to the limit,\nbut not earlier.\'\'\n\n\n> Ignoring Haag\'s theorem, the unitary transformation defined by\n\nHaag\'s theorem does not apply since the theory with cutoff is not\nLorentz invariant.\n\n\n> \\phi_{in}({\\bf x}, t) = U(t)\\phi({\\bf x},t)U^\\dagger(t)\n>\n> only enables one to transform interactions to their equivalents in terms of\n> "in" fields when they apply to the same time. Thus we can transform a local\n> interaction \\phi^3(x) to \\phi_{in}^3(x), but we cannot do this if the field\n> operators are a smearing that is non-local in time, as would result from\n> your proposed range limitation in momentum space.\n\nSmearing an interaction in time just amounts to using a different\ntime-dependent potential, constructed at time t from the original\ntime-dependent potential (in the interaction representation) at all other times.\nThis is best seen by using the representation of the S-matrix\nas time-ordered exponential. This holds for arbitrary time-dependent\nnonlocal potentials, and one can go back from any nonlocal 4D interaction\nto a time-dependent potential formulation. Thus smearing in time is not a\nreal problem: One starts with the smeared interaction defined by the\ncutoff, uses the time-ordered exponential to work out the corresponding\nHamiltonian interaction, and proceeds from there.\n\n\n> In all the references I have the arguments rely on the interaction being a\n> local product of fields.\n\nThis is not even the case in nonrelativistic field theories:\nCoulomb interactions are not describable by local products of fields.\n\n\nArnold Neumaier\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>News Admin wrote:

> I do not think that this is going to work.

Please quote next time a small piece of what you are referring to,
so that one can follow without guessing.

[Moderator's note: As it turns out, it was aparently Chris Oakley who accidentally
wrote under the name "News Admin". Arnold Neumaier is right that it is good
USENET practice and highly encouraged to begin a followup message with a *brief*
quote from the text one is replying to, so that readers will know what is going
on. More details on how to post are summarized here:
http://www-stud.uni-essen.de/~sb0264/HowToPost.html . -usc]

I assume 'this' is my statement:

''Write the (Euclidean = Wick rotated) field as
\Phi(x) = integral dp \exp(-i p dot x) Phihat(p)
and substitute it into the action. This gives an action in the
momentum representation. Then regularize the interaction term by
throwing away the momenta above some cutoff. [...]
From here, you can proceed with meaningful formulas throughout the
whole renormalization procedure. All contributions to the S-matrix
elements of this regularized theory are finite, and give the S-matrix
of the regularized interaction. At the very end you can pass to the limit,
but not earlier.''

> Ignoring Haag's theorem, the unitary transformation defined by

Haag's theorem does not apply since the theory with cutoff is not
Lorentz invariant.

> \phi_{in}({\bf x}, t) = U(t)\phi({\bf x},t)U^\dagger(t)
>
> only enables one to transform interactions to their equivalents in terms of
> "in" fields when they apply to the same time. Thus we can transform a local
> interaction \phi^3(x) to \phi_{in}^3(x), but we cannot do this if the field
> operators are a smearing that is non-local in time, as would result from
> your proposed range limitation in momentum space.

Smearing an interaction in time just amounts to using a different
time-dependent potential, constructed at time t from the original
time-dependent potential (in the interaction representation) at all other times.
This is best seen by using the representation of the S-matrix
as time-ordered exponential. This holds for arbitrary time-dependent
nonlocal potentials, and one can go back from any nonlocal 4D interaction
to a time-dependent potential formulation. Thus smearing in time is not a
real problem: One starts with the smeared interaction defined by the
cutoff, uses the time-ordered exponential to work out the corresponding
Hamiltonian interaction, and proceeds from there.

> In all the references I have the arguments rely on the interaction being a
> local product of fields.

This is not even the case in nonrelativistic field theories:
Coulomb interactions are not describable by local products of fields.

Arnold Neumaier

Chris Oakley

Oct20-04, 03:18 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nNot sure what happened there, but this message was from me.\n\n\n"News Admin" <news@news.demon.net> wrote in message\nnews:ckuh5i\\$hp5\\$1\\$830fa7a5@news.dem on.co.uk...\n>\n> I do not think that this is going to work.\n>\n> Ignoring Haag\'s theorem, the unitary transformation defined by\n>\n> \\phi_{in}({\\bf x}, t) = U(t)\\phi({\\bf x},t)U^\\dagger(t)\n>\n> only enables one to transform interactions to their equivalents in terms\nof\n> "in" fields when they apply to the same time. Thus we can transform a\nlocal\n> interaction \\phi^3(x) to \\phi_{in}^3(x), but we cannot do this if the\nfield\n> operators are a smearing that is non-local in time, as would result from\n> your proposed range limitation in momentum space.\n>\n> In all the references I have the arguments rely on the interaction being a\n> local product of fields. I do not know what one can do to make the theory\n> work when this is not the case.\n>\n>\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Not sure what happened there, but this message was from me.

"News Admin" <news@news.demon.net> wrote in message
news:ckuh5i$hp5$1$830fa7a5@news.demon.co.uk...
>
> I do not think that this is going to work.
>
> Ignoring Haag's theorem, the unitary transformation defined by
>
> \phi_{in}({\bf x}, t) = U(t)\phi({\bf x},t)U^\dagger(t)
>
> only enables one to transform interactions to their equivalents in terms
of
> "in" fields when they apply to the same time. Thus we can transform a
local
> interaction \phi^3(x) to \phi_{in}^3(x), but we cannot do this if the
field
> operators are a smearing that is non-local in time, as would result from
> your proposed range limitation in momentum space.
>
> In all the references I have the arguments rely on the interaction being a
> local product of fields. I do not know what one can do to make the theory
> work when this is not the case.
>
>

Eugene Stefanovich

Oct21-04, 02:47 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nArnold Neumaier wrote:\n> Chris Oakley wrote:\n>\n>\n>>>No, I\'m saying that you have no right to expect that the theory works to\n>>>infinite energy when you\'ve only tested it to ~1 TeV.\n>>\n>>I am not sure that I am following you here. Before renormalization QFT as\n>>normally practised leads to infinite cross sections and decay rates. I have\n>>seen it claimed that the infiniteness is because there is something going on\n>>at extremely high energies that we do not fully understand that rescues the\n>>situation at low-energy - the "ultraviolet completion".\n>\n>\n> This is nonsense. Infinities arise because integrals taken over\n> unbounded momenta don\'t exist; so doing it leads to nonsense.\n> Instead, proper QFT takes regularized integrals, for example by\n> adding an explicit cutoff Lambda. This simply means that they\n> calculate with an action that depends on Lambda as an additional\n> parameter. Once this is done, everything is finite, but\n> Lambda-dependent.\n>\n> The only problem with that is that the cutoff destroys Lorentz\n> covariance - apart from that it would be a completely respectable\n> field theory in itself. Now Lorentz invariance is violated only\n> at momentum > O(Lambda); hence to have the theory conform to\n> physics that can be checked it suffices to take Lambda large.\n> But for aesthetic reasons or since we believe that symmetries are\n> fundamental, we want to have fully invariant theories. This requires\n> that we let Lambda go to infinity.\n>\n> But in order that the results have a finite limit we must at the\n> same time make the coupling constants g dependent on Lambda.\n> If this is done in a correct way (and the textbooks on QFT teach\n> one or more of the known correct ways), one encounters no infinities\n> at all in the whole process.\n>\n>\n>\n>>I see the Feynman-Dyson approach to QFT as failing to give sensible\n>>or even finite answers to physical quantities, but I do not see this\n>>phoenix-like rebirth after all the infinities have been conveniently wiped\n>>out as anything more than just wishful thinking.\n>\n>\n> The approach that gives infinities is the pre Feynman-Dyson approach,\n> while what you call \'wishful thinking\' is the advance that earned\n> Feynaman, Schwinger and Tomonaga the Nobel prize. They showed how to\n> _avoid_ the infinities completely, on the perturbative level, and the\n> renormalization group explains why the result is still meaningful as\n> a \'perturbative\' approach, although at first sight, very large corrections\n> seems to be involved.\n>\n>\n> Arnold Neumaier\n\nYour setting Lambda to infinity is fine. That\'s what Feynman et al. did,\nand how they obtained a finite and accurate S-matrix. The problem with\nthis approach is that masses and coupling constants are infinite. In\nother words, the Hamiltonian of the theory is infinite. In other words,\nyou cannot do time evolution. You may say that nobody needs time\nevolution in high energy physics. And you will be partly correct: almost\nall existing experiments are well described by the S-matrix, i.e., the\nrelationship between remote past and remote future.\n\nHowever, I presume that every physicist believes that there is\nsome non-trivial time evolution of interacting physical systems\nbetween the two time extremes. Such time evolution is evident\nin macroscopic low-energy systems, for example.\nQFT ignores that. That\'s the main\nproblem, why renormalization approach of Feynman, Tomonaga, and\nSchwinger is not the final word in QFT.\n\nEugene\nwww.geocities.com/meopemuk\n\n\n\n\n>\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Chris Oakley wrote:
>
>
>>>No, I'm saying that you have no right to expect that the theory works to
>>>infinite energy when you've only tested it to ~1 TeV.
>>
>>I am not sure that I am following you here. Before renormalization QFT as
>>normally practised leads to infinite cross sections and decay rates. I have
>>seen it claimed that the infiniteness is because there is something going on
>>at extremely high energies that we do not fully understand that rescues the
>>situation at low-energy - the "ultraviolet completion".
>
>
> This is nonsense. Infinities arise because integrals taken over
> unbounded momenta don't exist; so doing it leads to nonsense.
> Instead, proper QFT takes regularized integrals, for example by
> adding an explicit cutoff \Lambda. This simply means that they
> calculate with an action that depends on \Lambda as an additional
> parameter. Once this is done, everything is finite, but
> \Lambda-dependent.
>
> The only problem with that is that the cutoff destroys Lorentz
> covariance - apart from that it would be a completely respectable
> field theory in itself. Now Lorentz invariance is violated only
> at momentum > O(\Lambda); hence to have the theory conform to
> physics that can be checked it suffices to take \Lambda large.
> But for aesthetic reasons or since we believe that symmetries are
> fundamental, we want to have fully invariant theories. This requires
> that we let \Lambda go to infinity.
>
> But in order that the results have a finite limit we must at the
> same time make the coupling constants g dependent on \Lambda.
> If this is done in a correct way (and the textbooks on QFT teach
> one or more of the known correct ways), one encounters no infinities
> at all in the whole process.
>
>
>
>>I see the Feynman-Dyson approach to QFT as failing to give sensible
>>or even finite answers to physical quantities, but I do not see this
>>phoenix-like rebirth after all the infinities have been conveniently wiped
>>out as anything more than just wishful thinking.
>
>
> The approach that gives infinities is the pre Feynman-Dyson approach,
> while what you call 'wishful thinking' is the advance that earned
> Feynaman, Schwinger and Tomonaga the Nobel prize. They showed how to
> _avoid_ the infinities completely, on the perturbative level, and the
> renormalization group explains why the result is still meaningful as
> a 'perturbative' approach, although at first sight, very large corrections
> seems to be involved.
>
>
> Arnold Neumaier

Your setting \Lambda to infinity is fine. That's what Feynman et al. did,
and how they obtained a finite and accurate S-matrix. The problem with
this approach is that masses and coupling constants are infinite. In
other words, the Hamiltonian of the theory is infinite. In other words,
you cannot do time evolution. You may say that nobody needs time
evolution in high energy physics. And you will be partly correct: almost
all existing experiments are well described by the S-matrix, i.e., the
relationship between remote past and remote future.

However, I presume that every physicist believes that there is
some non-trivial time evolution of interacting physical systems
between the two time extremes. Such time evolution is evident
in macroscopic low-energy systems, for example.
QFT ignores that. That's the main
problem, why renormalization approach of Feynman, Tomonaga, and
Schwinger is not the final word in QFT.

Eugene
www.geocities.com/meopemuk

>

Eugene Stefanovich

Oct21-04, 02:47 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nChris Oakley wrote:\n> Let\'s just remember how we got to where we are:\n>\n> (i) We solved free field theory, including the expansion in terms of\n> annihilation and creation operators.\n> (ii) We introduced an interaction term in the Lagrangian density.\n> (iii) We introduced a unitary transformation that turns free fields into\n> interacting fields using the Interaction Picture.\n> (iv) We wrote out the Interaction Picture time evolution operator as\n> time-ordered products of the interaction.\n> (v) We reduced the time-ordered products to normal-ordered products.\n> (vi) We obtained thereby the S-matrix for scattering processes involving a\n> finite number of incoming particles with definite four-momentum scattering\n> to a finite number of outgoing particles with definite four-momenta as a sum\n> of Feynman diagrams.\n>\n> Many of these Feynman diagrams are pathologically divergent integrals.\n>\n> My response: This is because the theory is fundamentally flawed. Go back to\n> (ii) and try again.\n>\n\nI would like to suggest a different path (similar to what Weinberg is\nwriting in his book):\n\n(i) We constructed the Fock space as direct sums of tensor products of\n1-particle spaces\n(ii) We defined a non-interacting representation of the Poincare group\nthere, including non-interacting Hamiltonian H_0 and\nboost operator K_0\n(iii) We defined interaction terms in the Hamiltonian and boost\noperator H= H_0+V, K = K_0+W (all operators are expressed through\nparticle creation and annihilation operators, no fields)\n(iv) We found that S-matrix is ultraviolet divergent\n(v) We added renormalization (infinite) counterterms to the\nHamiltonian H\' = H+R , so as to\nmake S-matrix finite and in agreement with experiment.\n\nThis is where QED stands now: finite and perfectly accurate S-matrix\nobtained from infinite (wrong) Hamiltonian H\'. My suggestion is\n\n(vi) Apply unitary dressing transformation U to the Hamiltonian H\',\nso that\n\n1) the new Hamiltonian\n\nH\'\' = U H\' U^{-1}\n\nis finite\n\n2) The S-matrices obtained with H\' and H\'\' are exactly the same.\n\nNow you can forget the nightmares of QFT, canonical quantization,\nHaags theorem, etc. You can use H\'\' in all usual quantum mechanical\nformulas (obtain time evolution operator, S-matrix, bound states via\ndiagonalization, etc) without any need of regularization,\nrenormalization, and other tricks.\n\nEugene\nwww.geocities.com/meopemuk\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
> Let's just remember how we got to where we are:
>
> (i) We solved free field theory, including the expansion in terms of
> annihilation and creation operators.
> (ii) We introduced an interaction term in the Lagrangian density.
> (iii) We introduced a unitary transformation that turns free fields into
> interacting fields using the Interaction Picture.
> (iv) We wrote out the Interaction Picture time evolution operator as
> time-ordered products of the interaction.
> (v) We reduced the time-ordered products to normal-ordered products.
> (vi) We obtained thereby the S-matrix for scattering processes involving a
> finite number of incoming particles with definite four-momentum scattering
> to a finite number of outgoing particles with definite four-momenta as a sum
> of Feynman diagrams.
>
> Many of these Feynman diagrams are pathologically divergent integrals.
>
> My response: This is because the theory is fundamentally flawed. Go back to
> (ii) and try again.
>

I would like to suggest a different path (similar to what Weinberg is
writing in his book):

(i) We constructed the Fock space as direct sums of tensor products of
1-particle spaces
(ii) We defined a non-interacting representation of the Poincare group
there, including non-interacting Hamiltonian H_0 and
boost operator K_0
(iii) We defined interaction terms in the Hamiltonian and boost
operator H= H_0+V, K = K_0+W (all operators are expressed through
particle creation and annihilation operators, no fields)
(iv) We found that S-matrix is ultraviolet divergent
(v) We added renormalization (infinite) counterterms to the
Hamiltonian H' = H+R , so as to
make S-matrix finite and in agreement with experiment.

This is where QED stands now: finite and perfectly accurate S-matrix
obtained from infinite (wrong) Hamiltonian H'. My suggestion is

(vi) Apply unitary dressing transformation U to the Hamiltonian H',
so that

1) the new Hamiltonian

H'' = U H' U^{-1}

is finite

2) The S-matrices obtained with H' and H'' are exactly the same.

Now you can forget the nightmares of QFT, canonical quantization,
Haags theorem, etc. You can use H'' in all usual quantum mechanical
formulas (obtain time evolution operator, S-matrix, bound states via
diagonalization, etc) without any need of regularization,
renormalization, and other tricks.

Eugene
www.geocities.com/meopemuk

Chris Oakley

Oct21-04, 02:47 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n> Haag\'s theorem does not apply since the theory with cutoff is not\n> Lorentz invariant.\n\nOne cannot be so glib. In the only case of physical interest (Lambda =\ninfinity) the theory *must be* Lorentz invariant.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Haag's theorem does not apply since the theory with cutoff is not
> Lorentz invariant.

One cannot be so glib. In the only case of physical interest (\Lambda =
infinity) the theory *must be* Lorentz invariant.

Eugene Stefanovich

Oct21-04, 02:47 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nChris Oakley wrote:\n>>Well, you can throw out the theory, but that seems a bit drastic.\n>\n>\n> I believe that one has no choice. One can in any case find holes in the\n> cutoff procedure. Supposing, for example, we were to choose Lambda as a\n> function rather than a constant. Furthermore, let us make it a function of\n> variables that never even entered the original equations. There is no reason\n> in principle why we should not be allowed to do this. We then very rapidly\n> get nonsense. You will then say that it was my own fault for not making it a\n> constant, but all that means is that you can find a path through the\n> minefield. It does not mean that the minefield has ceased to exist.\n>\n> There is no such thing as "reasonable" in theoretical physics, there is only\n> mathematical logic. The pillars of our current understanding of physics,\n> namely quantum mechanics and special relativity, both defy common sense. One\n> becomes aware of this very quickly when one tries to explain them to an\n> intelligent non-physicist. You and Dr. Neumaier say that it is\n> "unreasonable" to have four-momentum integrals extending to infinity,\n> leading you to want to cut them off. I say that if the mathematical logic\n> leads one there, then one has no choice other that to accept it, and if this\n> leads to nonsense then one has to accept that that something is\n> fundamentally wrong with the approach.\n>\n> Let me be clear on this - I could not do any better. All my attempts to find\n> a rigorous way of doing Feynman-Dyson perturbation theory failed. This puts\n> me in good company with Dirac, Pauli, Feynman, the axiomatic field theorists\n> and many others. But in the end, I just decided that it was not worth\n> rescuing, tore everything up and started again. And - intermittently, at\n> least - I am still working on that.\n>\n>\n\nHi Chris,\n\nI share your frustration with renormalized QFT. I think I know the way\nto fix the problem of infinities. The idea is very simple: QED simply\nuses a wrong Hamiltonian. There is a rigorous way how to find a new\nHamiltonian of QED (while preserving the relativistic invariance and\ncluster separability) which is\n\n1. finite\n2. yields the same accurate S-matrix as renormalized QED.\n(so it has the same predictions as far as experiment is concerned)\n\nThe details are given in\n\nE.V.Stefanovich "Quantum field theory without infinities", Ann. Phys.\n292 (2001), 139\n\nand in online book available at www.meopemuk.com\n\nI would appreciate any comments and criticism of this approach to QED.\nIf you don\'t think it completely solves the problem of ultraviolet\ninfinities, I would like to know why.\n\nMore papers can be found at www.geocities.com/meopemuk\n\nRegards.\nEugene.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
>>Well, you can throw out the theory, but that seems a bit drastic.
>
>
> I believe that one has no choice. One can in any case find holes in the
> cutoff procedure. Supposing, for example, we were to choose \Lambda as a
> function rather than a constant. Furthermore, let us make it a function of
> variables that never even entered the original equations. There is no reason
> in principle why we should not be allowed to do this. We then very rapidly
> get nonsense. You will then say that it was my own fault for not making it a
> constant, but all that means is that you can find a path through the
> minefield. It does not mean that the minefield has ceased to exist.
>
> There is no such thing as "reasonable" in theoretical physics, there is only
> mathematical logic. The pillars of our current understanding of physics,
> namely quantum mechanics and special relativity, both defy common sense. One
> becomes aware of this very quickly when one tries to explain them to an
> intelligent non-physicist. You and Dr. Neumaier say that it is
> "unreasonable" to have four-momentum integrals extending to infinity,
> leading you to want to cut them off. I say that if the mathematical logic
> leads one there, then one has no choice other that to accept it, and if this
> leads to nonsense then one has to accept that that something is
> fundamentally wrong with the approach.
>
> Let me be clear on this - I could not do any better. All my attempts to find
> a rigorous way of doing Feynman-Dyson perturbation theory failed. This puts
> me in good company with Dirac, Pauli, Feynman, the axiomatic field theorists
> and many others. But in the end, I just decided that it was not worth
> rescuing, tore everything up and started again. And - intermittently, at
> least - I am still working on that.
>
>

Hi Chris,

I share your frustration with renormalized QFT. I think I know the way
to fix the problem of infinities. The idea is very simple: QED simply
uses a wrong Hamiltonian. There is a rigorous way how to find a new
Hamiltonian of QED (while preserving the relativistic invariance and
cluster separability) which is

1. finite
2. yields the same accurate S-matrix as renormalized QED.
(so it has the same predictions as far as experiment is concerned)

The details are given in

E.V.Stefanovich "Quantum field theory without infinities", Ann. Phys.
292 (2001), 139

and in online book available at www.meopemuk.com

I would appreciate any comments and criticism of this approach to QED.
If you don't think it completely solves the problem of ultraviolet
infinities, I would like to know why.

More papers can be found at www.geocities.com/meopemuk

Regards.
Eugene.

Chris Oakley

Oct21-04, 02:47 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n> > In all the references I have the arguments rely on the interaction being\na\n> > local product of fields. I do not know what one can do to make the\ntheory\n> > work when this is not the case.\n>\n> One simply gets more complicated Feynman rules. It is not very difficult\n> to generalize them to arbitrary nonlocal interactions. In momentum space,\n> the formulas become the standard formulas, but with explicit cutoff\n> included.\n\nI have never, *ever* seen this. Can you give me a reference? I have only\nseen Feynman rules for *local* field equations being used and then the\nintegrals being cut off *after* we get the expressions for the loops and\nhave not liked what we saw.\n\n> Thus to do the suggested exercise, you should always work in the momentum\n> representation, and just apply the arguments of the textbooks in a\nsensible\n> way to the regularized situation.\n\nI am familiar with the momentum representation (which I use more or less\nexclusively in my own work), but cannot see how to adapt the arguments of\nFeynman-Dyson perturbation theory using this to achieve the required result.\nPlease enlighten me.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In all the references I have the arguments rely on the interaction being
a
> > local product of fields. I do not know what one can do to make the
theory
> > work when this is not the case.
>
> One simply gets more complicated Feynman rules. It is not very difficult
> to generalize them to arbitrary nonlocal interactions. In momentum space,
> the formulas become the standard formulas, but with explicit cutoff
> included.

I have never, *ever* seen this. Can you give me a reference? I have only
seen Feynman rules for *local* field equations being used and then the
integrals being cut off *after* we get the expressions for the loops and
have not liked what we saw.

> Thus to do the suggested exercise, you should always work in the momentum
> representation, and just apply the arguments of the textbooks in a
sensible
> way to the regularized situation.

I am familiar with the momentum representation (which I use more or less
exclusively in my own work), but cannot see how to adapt the arguments of
Feynman-Dyson perturbation theory using this to achieve the required result.
Please enlighten me.

Chris Oakley

Oct21-04, 02:47 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n> Smearing an interaction in time just amounts to using a different\n> time-dependent potential, constructed at time t from the original\n> time-dependent potential (in the interaction representation) at all other\ntimes.\n> This is best seen by using the representation of the S-matrix\n> as time-ordered exponential. This holds for arbitrary time-dependent\n> nonlocal potentials, and one can go back from any nonlocal 4D interaction\n> to a time-dependent potential formulation. Thus smearing in time is not a\n> real problem: One starts with the smeared interaction defined by the\n> cutoff, uses the time-ordered exponential to work out the corresponding\n> Hamiltonian interaction, and proceeds from there.\n\nThis may be obvious to you, but it is not obvious to me, I am afraid, so\nplease spell it out.\n\n> > In all the references I have the arguments rely on the interaction being\na\n> > local product of fields.\n>\n> This is not even the case in nonrelativistic field theories:\n> Coulomb interactions are not describable by local products of fields.\n\nI think we are talking cross purposes here. E.g. Weinberg, QFT Vol. 1,\nchapter 6. The arguments assume that (e.g.) H(x_1), etc. in equation 6.1.1\nis a local product of fields defined at the particular spacetime point. He\nis already considering this interaction Hamiltonian density H in the\nInteraction Picture. My point is that one can only transform a product of\nfields to the I.P. in the naive way if they are all for the same time, so\none cannot get to this equation if H is a smearing that is non-local in\ntime.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Smearing an interaction in time just amounts to using a different
> time-dependent potential, constructed at time t from the original
> time-dependent potential (in the interaction representation) at all other
times.
> This is best seen by using the representation of the S-matrix
> as time-ordered exponential. This holds for arbitrary time-dependent
> nonlocal potentials, and one can go back from any nonlocal 4D interaction
> to a time-dependent potential formulation. Thus smearing in time is not a
> real problem: One starts with the smeared interaction defined by the
> cutoff, uses the time-ordered exponential to work out the corresponding
> Hamiltonian interaction, and proceeds from there.

This may be obvious to you, but it is not obvious to me, I am afraid, so
please spell it out.

> > In all the references I have the arguments rely on the interaction being
a
> > local product of fields.
>
> This is not even the case in nonrelativistic field theories:
> Coulomb interactions are not describable by local products of fields.

I think we are talking cross purposes here. E.g. Weinberg, QFT Vol. 1,
chapter 6. The arguments assume that (e.g.) H(x_1), etc. in equation 6.1.1
is a local product of fields defined at the particular spacetime point. He
is already considering this interaction Hamiltonian density H in the
Interaction Picture. My point is that one can only transform a product of
fields to the I.P. in the naive way if they are all for the same time, so
one cannot get to this equation if H is a smearing that is non-local in
time.

Arnold Neumaier

Oct22-04, 12:01 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nEugene Stefanovich wrote:\n> Chris Oakley wrote:\n>\n>>Let\'s just remember how we got to where we are:\n>>\n>>(i) We solved free field theory, including the expansion in terms of\n>>annihilation and creation operators.\n>>(ii) We introduced an interaction term in the Lagrangian density.\n>>(iii) We introduced a unitary transformation that turns free fields into\n>>interacting fields using the Interaction Picture.\n>>(iv) We wrote out the Interaction Picture time evolution operator as\n>>time-ordered products of the interaction.\n>>(v) We reduced the time-ordered products to normal-ordered products.\n>>(vi) We obtained thereby the S-matrix for scattering processes involving a\n>>finite number of incoming particles with definite four-momentum scattering\n>>to a finite number of outgoing particles with definite four-momenta as a sum\n>>of Feynman diagrams.\n>>\n>>Many of these Feynman diagrams are pathologically divergent integrals.\n>>\n>>My response: This is because the theory is fundamentally flawed. Go back to\n>>(ii) and try again.\n>>\n>\n>\n> I would like to suggest a different path (similar to what Weinberg is\n> writing in his book):\n>\n> (i) We constructed the Fock space as direct sums of tensor products of\n> 1-particle spaces\n> (ii) We defined a non-interacting representation of the Poincare group\n> there, including non-interacting Hamiltonian H_0 and\n> boost operator K_0\n> (iii) We defined interaction terms in the Hamiltonian and boost\n> operator H= H_0+V, K = K_0+W (all operators are expressed through\n> particle creation and annihilation operators, no fields)\n> (iv) We found that S-matrix is ultraviolet divergent\n> (v) We added renormalization (infinite) counterterms to the\n> Hamiltonian H\' = H+R , so as to\n> make S-matrix finite and in agreement with experiment.\n>\n> This is where QED stands now: finite and perfectly accurate S-matrix\n> obtained from infinite (wrong) Hamiltonian H\'. My suggestion is\n>\n> (vi) Apply unitary dressing transformation U to the Hamiltonian H\',\n> so that\n>\n> 1) the new Hamiltonian\n>\n> H\'\' = U H\' U^{-1}\n>\n> is finite\n>\n> 2) The S-matrices obtained with H\' and H\'\' are exactly the same.\n>\n> Now you can forget the nightmares of QFT, canonical quantization,\n> Haags theorem, etc. You can use H\'\' in all usual quantum mechanical\n> formulas (obtain time evolution operator, S-matrix, bound states via\n> diagonalization, etc) without any need of regularization,\n> renormalization, and other tricks.\n\nWell, this is a research program - the question is whether you can\nactually carry it out. How do you construct U? It is there that all\nthe renormalization issues will recur. Since H\' is an ill-defined object\none must regularize it, then construct U for the regularized theory,\nand then go to the limit and show that it really exists. This poses\nthe same problems as the original renormalization program.\n\n\nArnold Neumaier\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Chris Oakley wrote:
>
>>Let's just remember how we got to where we are:
>>
>>(i) We solved free field theory, including the expansion in terms of
>>annihilation and creation operators.
>>(ii) We introduced an interaction term in the Lagrangian density.
>>(iii) We introduced a unitary transformation that turns free fields into
>>interacting fields using the Interaction Picture.
>>(iv) We wrote out the Interaction Picture time evolution operator as
>>time-ordered products of the interaction.
>>(v) We reduced the time-ordered products to normal-ordered products.
>>(vi) We obtained thereby the S-matrix for scattering processes involving a
>>finite number of incoming particles with definite four-momentum scattering
>>to a finite number of outgoing particles with definite four-momenta as a sum
>>of Feynman diagrams.
>>
>>Many of these Feynman diagrams are pathologically divergent integrals.
>>
>>My response: This is because the theory is fundamentally flawed. Go back to
>>(ii) and try again.
>>
>
>
> I would like to suggest a different path (similar to what Weinberg is
> writing in his book):
>
> (i) We constructed the Fock space as direct sums of tensor products of
> 1-particle spaces
> (ii) We defined a non-interacting representation of the Poincare group
> there, including non-interacting Hamiltonian H_0 and
> boost operator K_0
> (iii) We defined interaction terms in the Hamiltonian and boost
> operator H= H_0+V, K = K_0+W (all operators are expressed through
> particle creation and annihilation operators, no fields)
> (iv) We found that S-matrix is ultraviolet divergent
> (v) We added renormalization (infinite) counterterms to the
> Hamiltonian H' = H+R , so as to
> make S-matrix finite and in agreement with experiment.
>
> This is where QED stands now: finite and perfectly accurate S-matrix
> obtained from infinite (wrong) Hamiltonian H'. My suggestion is
>
> (vi) Apply unitary dressing transformation U to the Hamiltonian H',
> so that
>
> 1) the new Hamiltonian
>
> H'' = U H' U^{-1}
>
> is finite
>
> 2) The S-matrices obtained with H' and H'' are exactly the same.
>
> Now you can forget the nightmares of QFT, canonical quantization,
> Haags theorem, etc. You can use H'' in all usual quantum mechanical
> formulas (obtain time evolution operator, S-matrix, bound states via
> diagonalization, etc) without any need of regularization,
> renormalization, and other tricks.

Well, this is a research program - the question is whether you can
actually carry it out. How do you construct U? It is there that all
the renormalization issues will recur. Since H' is an ill-defined object
one must regularize it, then construct U for the regularized theory,
and then go to the limit and show that it really exists. This poses
the same problems as the original renormalization program.

Arnold Neumaier

Chris Oakley

Oct22-04, 12:01 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n> I share your frustration with renormalized QFT. I think I know the way\n> to fix the problem of infinities. The idea is very simple: QED simply\n> uses a wrong Hamiltonian. There is a rigorous way how to find a new\n> Hamiltonian of QED (while preserving the relativistic invariance and\n> cluster separability) which is\n>\n> 1. finite\n> 2. yields the same accurate S-matrix as renormalized QED.\n> (so it has the same predictions as far as experiment is concerned)\n>\n> The details are given in\n>\n> E.V.Stefanovich "Quantum field theory without infinities", Ann. Phys.\n> 292 (2001), 139\n>\n> and in online book available at www.meopemuk.com\n>\n> I would appreciate any comments and criticism of this approach to QED.\n> If you don\'t think it completely solves the problem of ultraviolet\n> infinities, I would like to know why.\n>\n> More papers can be found at www.geocities.com/meopemuk\n\nEugene,\n\nAs you may remember we met and talked in California three years ago. It is\nnice (and rare) to see someone who has bothered to figure everything out for\nthemselves.\n\nI don\'t claim to understand every detail of your work, but I do take Haag\'s\ntheorem seriously and therefore don\'t believe that the U, W and F operators\nin section 2 of your 2001 Ann. Phys. paper exist. Not in a fully\nrelativistic theory, anyway. Also I notice that the regularization scheme is\nstill exactly that ... i.e. you get the infinities and then devise a scheme\n(in your case a "clothing" operator) for dealing with them. This is still\nsubtracting infinity from infinity, and I really don\'t think that people\nought to do that.\n\nChris\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>I share your frustration with renormalized QFT. I think I know the way
> to fix the problem of infinities. The idea is very simple: QED simply
> uses a wrong Hamiltonian. There is a rigorous way how to find a new
> Hamiltonian of QED (while preserving the relativistic invariance and
> cluster separability) which is
>
> 1. finite
> 2. yields the same accurate S-matrix as renormalized QED.
> (so it has the same predictions as far as experiment is concerned)
>
> The details are given in
>
> E.V.Stefanovich "Quantum field theory without infinities", Ann. Phys.
> 292 (2001), 139
>
> and in online book available at www.meopemuk.com
>
> I would appreciate any comments and criticism of this approach to QED.
> If you don't think it completely solves the problem of ultraviolet
> infinities, I would like to know why.
>
> More papers can be found at www.geocities.com/meopemuk

Eugene,

As you may remember we met and talked in California three years ago. It is
nice (and rare) to see someone who has bothered to figure everything out for
themselves.

I don't claim to understand every detail of your work, but I do take Haag's
theorem seriously and therefore don't believe that the U, W and F operators
in section 2 of your 2001 Ann. Phys. paper exist. Not in a fully
relativistic theory, anyway. Also I notice that the regularization scheme is
still exactly that ... i.e. you get the infinities and then devise a scheme
(in your case a "clothing" operator) for dealing with them. This is still
subtracting infinity from infinity, and I really don't think that people
ought to do that.

Chris

Arnold Neumaier

Oct22-04, 12:01 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nChris Oakley wrote:\n>>Haag\'s theorem does not apply since the theory with cutoff is not\n>>Lorentz invariant.\n>\n>\n> One cannot be so glib. In the only case of physical interest (Lambda =\n> infinity) the theory *must be* Lorentz invariant.\n\nWe only know that the universe is governed by Lorentz invariance to\na certain accuracy, which means that the theories with finite but large\ncutoff (significantly larger that the largest energy we can create or\nmeasure) already are in full agreement with nature.\n\nThus Lambda=inf is not needed for results of physical interest.\n\n\nArnold Neumaier\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
>>Haag's theorem does not apply since the theory with cutoff is not
>>Lorentz invariant.
>
>
> One cannot be so glib. In the only case of physical interest (\Lambda =
> infinity) the theory *must be* Lorentz invariant.

We only know that the universe is governed by Lorentz invariance to
a certain accuracy, which means that the theories with finite but large
cutoff (significantly larger that the largest energy we can create or
measure) already are in full agreement with nature.

Thus \Lambda=inf is not needed for results of physical interest.

Arnold Neumaier

Arnold Neumaier

Oct22-04, 12:20 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nChris Oakley wrote:\n\n> I think we are talking cross purposes here. E.g. Weinberg, QFT Vol. 1,\n> chapter 6. The arguments assume that (e.g.) H(x_1), etc. in equation 6.1.1\n> is a local product of fields defined at the particular spacetime point. He\n> is already considering this interaction Hamiltonian density H in the\n> Interaction Picture. My point is that one can only transform a product of\n> fields to the I.P. in the naive way if they are all for the same time, so\n> one cannot get to this equation if H is a smearing that is non-local in\n> time.\n\nRewrite in each of Weinberg\'s formulas the fields in x as integral\nover fields in p. This gives you completely equivalent formulations,\nnow only involving momenta. Once you have done that, it is very easy\nto take care of a cutoff.\n\n\nArnold Neumaier\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:

> I think we are talking cross purposes here. E.g. Weinberg, QFT Vol. 1,
> chapter 6. The arguments assume that (e.g.) H(x_1), etc. in equation 6.1.1
> is a local product of fields defined at the particular spacetime point. He
> is already considering this interaction Hamiltonian density H in the
> Interaction Picture. My point is that one can only transform a product of
> fields to the I.P. in the naive way if they are all for the same time, so
> one cannot get to this equation if H is a smearing that is non-local in
> time.

Rewrite in each of Weinberg's formulas the fields in x as integral
over fields in p. This gives you completely equivalent formulations,
now only involving momenta. Once you have done that, it is very easy
to take care of a cutoff.

Arnold Neumaier

Arnold Neumaier

Oct22-04, 12:20 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nChris Oakley wrote:\n>>>In all the references I have the arguments rely on the interaction being a\n>>>local product of fields. I do not know what one can do to make the theory\n>>>work when this is not the case.\n>>\n>>One simply gets more complicated Feynman rules. It is not very difficult\n>>to generalize them to arbitrary nonlocal interactions. In momentum space,\n>>the formulas become the standard formulas, but with explicit cutoff\n>>included.\n>\n> I have never, *ever* seen this. Can you give me a reference?\n\nNo; I don\'t know of any. I had to work it out for myself; it was a hard\nroad to gain this understanding.\n\nI gave you instead an exercise. Physicists are too lazy to do things as\ncarefully as you wish, so you have to do it yourself. I gave you the\noutline, so that you can fill in the details.\n\n\n> I am familiar with the momentum representation (which I use more or less\n> exclusively in my own work), but cannot see how to adapt the arguments of\n> Feynman-Dyson perturbation theory using this to achieve the required result.\n> Please enlighten me.\n\nPlease start doing it and you\'ll see that everything can be done.\nBegin by writing the interaction in the momentum representation as a\nmultiple integral, add the cutoff, and then start copying in spirit\neach line of Peskin & Schroeder or your favorite textbook, adapting it\nto the new situation at hand. You can hardly go wrong, but of course\nit takes some effort to get familiar with the nonlocal case.\n\nIt is very hard writing out pages of details in ascii, and I have no\nintention to do so. Should you ever want to visit Vienna, I\'d show you\non the blackboard what to do.\n\n\nArnold Neumaier\n\n\n\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
>>>In all the references I have the arguments rely on the interaction being a
>>>local product of fields. I do not know what one can do to make the theory
>>>work when this is not the case.
>>
>>One simply gets more complicated Feynman rules. It is not very difficult
>>to generalize them to arbitrary nonlocal interactions. In momentum space,
>>the formulas become the standard formulas, but with explicit cutoff
>>included.
>
> I have never, *ever* seen this. Can you give me a reference?

No; I don't know of any. I had to work it out for myself; it was a hard
road to gain this understanding.

I gave you instead an exercise. Physicists are too lazy to do things as
carefully as you wish, so you have to do it yourself. I gave you the
outline, so that you can fill in the details.

> I am familiar with the momentum representation (which I use more or less
> exclusively in my own work), but cannot see how to adapt the arguments of
> Feynman-Dyson perturbation theory using this to achieve the required result.
> Please enlighten me.

Please start doing it and you'll see that everything can be done.
Begin by writing the interaction in the momentum representation as a
multiple integral, add the cutoff, and then start copying in spirit
each line of Peskin & Schroeder or your favorite textbook, adapting it
to the new situation at hand. You can hardly go wrong, but of course
it takes some effort to get familiar with the nonlocal case.

It is very hard writing out pages of details in ascii, and I have no
intention to do so. Should you ever want to visit Vienna, I'd show you
on the blackboard what to do.

Arnold Neumaier

Arnold Neumaier

Oct22-04, 12:31 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nEugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n\n>>But in order that the results have a finite limit we must at the\n>>same time make the coupling constants g dependent on Lambda.\n>>If this is done in a correct way (and the textbooks on QFT teach\n>>one or more of the known correct ways), one encounters no infinities\n>>at all in the whole process.\n>>\n> Your setting Lambda to infinity is fine. That\'s what Feynman et al. did,\n> and how they obtained a finite and accurate S-matrix.\n\nThey don\'t set it to infinity; they keep it finite and take the limit at\nthe end.\n\n> The problem with\n> this approach is that masses and coupling constants are infinite.\n\nThe bare masses and bare coupling constants are finite but depend on\nLambda. In the limit they would diverge. But one never needs to take\nlimits on bare objects. The renormalized masses and coupling constants\nare finite.\n\n> However, I presume that every physicist believes that there is\n> some non-trivial time evolution of interacting physical systems\n> between the two time extremes. Such time evolution is evident\n> in macroscopic low-energy systems, for example.\n> QFT ignores that. That\'s the main\n> problem, why renormalization approach of Feynman, Tomonaga, and\n> Schwinger is not the final word in QFT.\n\nOne, not necessarily the main one. The fact that the S-matrix exists\nonly as a formal series instead of as a unitary operator on some\nHilbert space is another important reason.\n\n\nArnold\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:

>>But in order that the results have a finite limit we must at the
>>same time make the coupling constants g dependent on \Lambda.
>>If this is done in a correct way (and the textbooks on QFT teach
>>one or more of the known correct ways), one encounters no infinities
>>at all in the whole process.
>>
> Your setting \Lambda to infinity is fine. That's what Feynman et al. did,
> and how they obtained a finite and accurate S-matrix.

They don't set it to infinity; they keep it finite and take the limit at
the end.

> The problem with
> this approach is that masses and coupling constants are infinite.

The bare masses and bare coupling constants are finite but depend on
\Lambda. In the limit they would diverge. But one never needs to take
limits on bare objects. The renormalized masses and coupling constants
are finite.

> However, I presume that every physicist believes that there is
> some non-trivial time evolution of interacting physical systems
> between the two time extremes. Such time evolution is evident
> in macroscopic low-energy systems, for example.
> QFT ignores that. That's the main
> problem, why renormalization approach of Feynman, Tomonaga, and
> Schwinger is not the final word in QFT.

One, not necessarily the main one. The fact that the S-matrix exists
only as a formal series instead of as a unitary operator on some
Hilbert space is another important reason.

Arnold

Igor Khavkine

Oct22-04, 12:40 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nOn Thu, 21 Oct 2004 07:47:32 +0000, Chris Oakley wrote:\n\n>\n>> > In all the references I have the arguments rely on the interaction\n>> > being\n> a\n>> > local product of fields. I do not know what one can do to make the\n> theory\n>> > work when this is not the case.\n>>\n>> One simply gets more complicated Feynman rules. It is not very difficult\n>> to generalize them to arbitrary nonlocal interactions. In momentum\n>> space, the formulas become the standard formulas, but with explicit\n>> cutoff included.\n>\n> I have never, *ever* seen this. Can you give me a reference? I have only\n> seen Feynman rules for *local* field equations being used and then the\n> integrals being cut off *after* we get the expressions for the loops and\n> have not liked what we saw.\n\nOne does not have to go far to see this. As Arnold mentioned in a previous\npost, the coulomb interaction in non-relativistic many body systems is a\nstandard example. When drawing diagrams in momentum space instead of a\npoint interaction vertex we an internal line which represents the\ninteraction. However, the internal line does not represent a photon\nas we would expect in full QED, but simply the Coulomb potential. The\nelectron vertices are allowed to have distinct positions and the Coulomb\npotential depends on their difference. Since the Coulomb interaction is\ntranslation invariant, there is still a nice picture of the interaction in\nmomentum space: the internal interaction line represents a momentum\ntransfer. All the necessary Feynman rules are worked out for both position\nand momentum space in _Methods of QFT in Statistical Physics_ by Abrikosov\net al. Even the cases when the interaction is not translation invariant\nare treated.\n\nIgor.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Thu, 21 Oct 2004 07:47:32 +0000, Chris Oakley wrote:

>
>> > In all the references I have the arguments rely on the interaction
>> > being
> a
>> > local product of fields. I do not know what one can do to make the
> theory
>> > work when this is not the case.
>>
>> One simply gets more complicated Feynman rules. It is not very difficult
>> to generalize them to arbitrary nonlocal interactions. In momentum
>> space, the formulas become the standard formulas, but with explicit
>> cutoff included.
>
> I have never, *ever* seen this. Can you give me a reference? I have only
> seen Feynman rules for *local* field equations being used and then the
> integrals being cut off *after* we get the expressions for the loops and
> have not liked what we saw.

One does not have to go far to see this. As Arnold mentioned in a previous
post, the coulomb interaction in non-relativistic many body systems is a
standard example. When drawing diagrams in momentum space instead of a
point interaction vertex we an internal line which represents the
interaction. However, the internal line does not represent a photon
as we would expect in full QED, but simply the Coulomb potential. The
electron vertices are allowed to have distinct positions and the Coulomb
potential depends on their difference. Since the Coulomb interaction is
translation invariant, there is still a nice picture of the interaction in
momentum space: the internal interaction line represents a momentum
transfer. All the necessary Feynman rules are worked out for both position
and momentum space in _Methods of QFT in Statistical Physics_ by Abrikosov
et al. Even the cases when the interaction is not translation invariant
are treated.

Igor.

Chris Oakley

Oct22-04, 12:40 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n> However, I presume that every physicist believes that there is\n> some non-trivial time evolution of interacting physical systems\n> between the two time extremes. Such time evolution is evident\n> in macroscopic low-energy systems, for example.\n> QFT ignores that. That\'s the main\n> problem, why renormalization approach of Feynman, Tomonaga, and\n> Schwinger is not the final word in QFT.\n\nIt is interesting that undergraduate quantum mechanics does better than QFT\nfor processes that involve a (finite) time variable. QFT ought to be an\nextension of this, not an alternative.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>However, I presume that every physicist believes that there is
> some non-trivial time evolution of interacting physical systems
> between the two time extremes. Such time evolution is evident
> in macroscopic low-energy systems, for example.
> QFT ignores that. That's the main
> problem, why renormalization approach of Feynman, Tomonaga, and
> Schwinger is not the final word in QFT.

It is interesting that undergraduate quantum mechanics does better than QFT
for processes that involve a (finite) time variable. QFT ought to be an
extension of this, not an alternative.

Chris Oakley

Oct24-04, 09:04 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n> >>>In all the references I have the arguments rely on the interaction\nbeing a\n> >>>local product of fields. I do not know what one can do to make the\ntheory\n> >>>work when this is not the case.\n> >>\n> >>One simply gets more complicated Feynman rules. It is not very difficult\n> >>to generalize them to arbitrary nonlocal interactions. In momentum\nspace,\n> >>the formulas become the standard formulas, but with explicit cutoff\n> >>included.\n> >\n> > I have never, *ever* seen this. Can you give me a reference?\n>\n> No; I don\'t know of any. I had to work it out for myself; it was a hard\n> road to gain this understanding.\n>\n> I gave you instead an exercise. Physicists are too lazy to do things as\n> carefully as you wish, so you have to do it yourself. I gave you the\n> outline, so that you can fill in the details.\n\nI have not done it because I do not believe that it can be done. Products of\nfield operators in the interaction Hamiltonian density must refer to the\nsame time if we are to be able to transform to the Interaction Picture.\n\nI might just add that I am not the first person to be bothered by the sloppy\nmathematics in QFT. People have been working on this, on and off, since\n1928. If there was a way of making the procedure rigorous, it would have\nfound its way into the text books by now. However the best QFT text book *I*\nknow of - Weinberg - makes no mention of the possibility of defeating\ninfinite subtractions via non-local field equations. Neither does the\nsupposedly more rigorous Scharf.\n\n> Should you ever want to visit Vienna, I\'d show you\n> on the blackboard what to do.\n\nA kind offer, but be careful - I might take you up on it.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>>>>In all the references I have the arguments rely on the interaction
being a
> >>>local product of fields. I do not know what one can do to make the
theory
> >>>work when this is not the case.
> >>
> >>One simply gets more complicated Feynman rules. It is not very difficult
> >>to generalize them to arbitrary nonlocal interactions. In momentum
space,
> >>the formulas become the standard formulas, but with explicit cutoff
> >>included.
> >
> > I have never, *ever* seen this. Can you give me a reference?
>
> No; I don't know of any. I had to work it out for myself; it was a hard
> road to gain this understanding.
>
> I gave you instead an exercise. Physicists are too lazy to do things as
> carefully as you wish, so you have to do it yourself. I gave you the
> outline, so that you can fill in the details.

I have not done it because I do not believe that it can be done. Products of
field operators in the interaction Hamiltonian density must refer to the
same time if we are to be able to transform to the Interaction Picture.

I might just add that I am not the first person to be bothered by the sloppy
mathematics in QFT. People have been working on this, on and off, since
1928. If there was a way of making the procedure rigorous, it would have
found its way into the text books by now. However the best QFT text book *I*
know of - Weinberg - makes no mention of the possibility of defeating
infinite subtractions via non-local field equations. Neither does the
supposedly more rigorous Scharf.

> Should you ever want to visit Vienna, I'd show you
> on the blackboard what to do.

A kind offer, but be careful - I might take you up on it.

Chris Oakley

Oct24-04, 09:04 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n> > I think we are talking cross purposes here. E.g. Weinberg, QFT Vol. 1,\n> > chapter 6. The arguments assume that (e.g.) H(x_1), etc. in equation\n6.1.1\n> > is a local product of fields defined at the particular spacetime point.\nHe\n> > is already considering this interaction Hamiltonian density H in the\n> > Interaction Picture. My point is that one can only transform a product\nof\n> > fields to the I.P. in the naive way if they are all for the same time,\nso\n> > one cannot get to this equation if H is a smearing that is non-local in\n> > time.\n>\n> Rewrite in each of Weinberg\'s formulas the fields in x as integral\n> over fields in p. This gives you completely equivalent formulations,\n> now only involving momenta. Once you have done that, it is very easy\n> to take care of a cutoff.\n\nYou are missing my point. Limiting the integral in p^0 space results in a\nsmearing in the time integral. This is just basic Fourier analysis. It is\nthe latter which prevents a transformation between the Heisenberg and\nInteraction pictures for the field operators.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>I think we are talking cross purposes here. E.g. Weinberg, QFT Vol. 1,
> > chapter 6. The arguments assume that (e.g.) H(x_1), etc. in equation
6.1.1
> > is a local product of fields defined at the particular spacetime point.
He
> > is already considering this interaction Hamiltonian density H in the
> > Interaction Picture. My point is that one can only transform a product
of
> > fields to the I.P. in the naive way if they are all for the same time,
so
> > one cannot get to this equation if H is a smearing that is non-local in
> > time.
>
> Rewrite in each of Weinberg's formulas the fields in x as integral
> over fields in p. This gives you completely equivalent formulations,
> now only involving momenta. Once you have done that, it is very easy
> to take care of a cutoff.

You are missing my point. Limiting the integral in p^0 space results in a
smearing in the time integral. This is just basic Fourier analysis. It is
the latter which prevents a transformation between the Heisenberg and
Interaction pictures for the field operators.

Eugene Stefanovich

Oct24-04, 09:04 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\n\n\nChris Oakley wrote:\n>>However, I presume that every physicist believes that there is\n>>some non-trivial time evolution of interacting physical systems\n>>between the two time extremes. Such time evolution is evident\n>>in macroscopic low-energy systems, for example.\n>>QFT ignores that. That\'s the main\n>>problem, why renormalization approach of Feynman, Tomonaga, and\n>>Schwinger is not the final word in QFT.\n>\n>\n> It is interesting that undergraduate quantum mechanics does better than QFT\n> for processes that involve a (finite) time variable. QFT ought to be an\n> extension of this, not an alternative.\n>\n\n\nThat\'s exactly right, and my approach (with finite well-defined Hamiltonian)\nmakes time evolution in QFT\nnot more difficult that in non-relativistic undergraduate QM. The only\ndifference is that 1) relativity is respected; 2) creation and\nannihilation of particles is allowed.\n\nEugene\nwww.geocities.com/meopemuk\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
>>However, I presume that every physicist believes that there is
>>some non-trivial time evolution of interacting physical systems
>>between the two time extremes. Such time evolution is evident
>>in macroscopic low-energy systems, for example.
>>QFT ignores that. That's the main
>>problem, why renormalization approach of Feynman, Tomonaga, and
>>Schwinger is not the final word in QFT.
>
>
> It is interesting that undergraduate quantum mechanics does better than QFT
> for processes that involve a (finite) time variable. QFT ought to be an
> extension of this, not an alternative.
>

That's exactly right, and my approach (with finite well-defined Hamiltonian)
makes time evolution in QFT
not more difficult that in non-relativistic undergraduate QM. The only
difference is that 1) relativity is respected; 2) creation and
annihilation of particles is allowed.

Eugene
www.geocities.com/meopemuk

Eugene Stefanovich

Oct24-04, 09:04 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>\n>\n>>>But in order that the results have a finite limit we must at the\n>>>same time make the coupling constants g dependent on Lambda.\n>>>If this is done in a correct way (and the textbooks on QFT teach\n>>>one or more of the known correct ways), one encounters no infinities\n>>>at all in the whole process.\n>>>\n>>\n>>Your setting Lambda to infinity is fine. That\'s what Feynman et al. did,\n>>and how they obtained a finite and accurate S-matrix.\n>\n>\n> They don\'t set it to infinity; they keep it finite and take the limit at\n> the end.\n\nThat\'s what I meant.\n\n>\n>\n>>The problem with\n>>this approach is that masses and coupling constants are infinite.\n>\n>\n> The bare masses and bare coupling constants are finite but depend on\n> Lambda. In the limit they would diverge. But one never needs to take\n> limits on bare objects. The renormalized masses and coupling constants\n> are finite.\n\nBut the Hamiltonian after adding counterterms contains infinite masses\nand charges, and this is the Hamiltonian which must be used for\nS-matrix calculations, if we want to keep the usual quantum mechanical\nformula connecting the S-matrix with the Hamiltonian.\n\n>\n>\n>>However, I presume that every physicist believes that there is\n>>some non-trivial time evolution of interacting physical systems\n>>between the two time extremes. Such time evolution is evident\n>>in macroscopic low-energy systems, for example.\n>>QFT ignores that. That\'s the main\n>>problem, why renormalization approach of Feynman, Tomonaga, and\n>>Schwinger is not the final word in QFT.\n>\n>\n> One, not necessarily the main one. The fact that the S-matrix exists\n> only as a formal series instead of as a unitary operator on some\n> Hilbert space is another important reason.\n\nThere is a version of perturbation theory which allows you\nto write the S-operator in manifestly unitary form in each perturbation\norder. This version was suggested by Magnus.\nIt is not well known. You can find references in my Ann. Phys. paper.\n\nEugene.\nwww.geocities.com/meopemuk\n\n>\n>\n> Arnold\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>
>
>>>But in order that the results have a finite limit we must at the
>>>same time make the coupling constants g dependent on \Lambda.
>>>If this is done in a correct way (and the textbooks on QFT teach
>>>one or more of the known correct ways), one encounters no infinities
>>>at all in the whole process.
>>>
>>
>>Your setting \Lambda to infinity is fine. That's what Feynman et al. did,
>>and how they obtained a finite and accurate S-matrix.
>
>
> They don't set it to infinity; they keep it finite and take the limit at
> the end.

That's what I meant.

>
>
>>The problem with
>>this approach is that masses and coupling constants are infinite.
>
>
> The bare masses and bare coupling constants are finite but depend on
> \Lambda. In the limit they would diverge. But one never needs to take
> limits on bare objects. The renormalized masses and coupling constants
> are finite.

But the Hamiltonian after adding counterterms contains infinite masses
and charges, and this is the Hamiltonian which must be used for
S-matrix calculations, if we want to keep the usual quantum mechanical
formula connecting the S-matrix with the Hamiltonian.

>
>
>>However, I presume that every physicist believes that there is
>>some non-trivial time evolution of interacting physical systems
>>between the two time extremes. Such time evolution is evident
>>in macroscopic low-energy systems, for example.
>>QFT ignores that. That's the main
>>problem, why renormalization approach of Feynman, Tomonaga, and
>>Schwinger is not the final word in QFT.
>
>
> One, not necessarily the main one. The fact that the S-matrix exists
> only as a formal series instead of as a unitary operator on some
> Hilbert space is another important reason.

There is a version of perturbation theory which allows you
to write the S-operator in manifestly unitary form in each perturbation
order. This version was suggested by Magnus.
It is not well known. You can find references in my Ann. Phys. paper.

Eugene.
www.geocities.com/meopemuk

>
>
> Arnold

Eugene Stefanovich

Oct24-04, 09:04 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nChris Oakley wrote:\n>>I share your frustration with renormalized QFT. I think I know the way\n>>to fix the problem of infinities. The idea is very simple: QED simply\n>>uses a wrong Hamiltonian. There is a rigorous way how to find a new\n>>Hamiltonian of QED (while preserving the relativistic invariance and\n>>cluster separability) which is\n>>\n>>1. finite\n>>2. yields the same accurate S-matrix as renormalized QED.\n>> (so it has the same predictions as far as experiment is concerned)\n>>\n>>The details are given in\n>>\n>>E.V.Stefanovich "Quantum field theory without infinities", Ann. Phys.\n>>292 (2001), 139\n>>\n>>and in online book available at www.meopemuk.com\n>>\n>>I would appreciate any comments and criticism of this approach to QED.\n>>If you don\'t think it completely solves the problem of ultraviolet\n>>infinities, I would like to know why.\n>>\n>>More papers can be found at www.geocities.com/meopemuk\n>\n\n\n>\n> I don\'t claim to understand every detail of your work, but I do take Haag\'s\n> theorem seriously and therefore don\'t believe that the U, W and F operators\n> in section 2 of your 2001 Ann. Phys. paper exist. Not in a fully\n> relativistic theory, anyway. Also I notice that the regularization scheme is\n> still exactly that ... i.e. you get the infinities and then devise a scheme\n> (in your case a "clothing" operator) for dealing with them. This is still\n> subtracting infinity from infinity, and I really don\'t think that people\n> ought to do that.\n>\n> Chris\n>\n\nHi Chris,\n\nNice to "talk" to you again!\n\nYou are right that QED has (at least) two problems. One of them is less\nserious.\nAnother is more serious.\n\nThe less serious problem is revealed by Haag\'s theorem. In a nutshell,\nit says that interacting fields cannot have Lorentz transformation\nproperties. In my approach I do not have interacting fields at all.\nThe interaction in my Hamiltonian is written in terms of particle\ncreation and annihilation operators. From this Hamiltonian I can\n(in principle) calculate S-matrix, bound states, time evolution,\nwhatever I like, without ever mentioning "interacting quantum fields".\nSo, the Haag\'s theorem does not apply.\n\nThe more serious problem is related to renormalization. You are right\nthat if we take usual QED Hamiltonian (e.g., (8.4.22) - (8.4.25) in\nWeinberg\'s book) and calculate elements of the S-matrix, we\'ll get\ninfinite results. My idea is that this Hamiltonian H is simply wrong.\nIt is wrong, but not completely wrong. It contain\'s a grain of truth.\nIn order to find the correct Hamiltonian H\'\', one should perform\ntwo steps:\n\n1. The first step is renormalization. We add infinite (mass and\ncharge renormalization) counterterms to H, so that in the process of\ncalculation of the S-matrix all infinities nicely cancel each other\nand we get agreement with experiment to 10^{-11}%. I agree that\nresulting Hamiltonian H\' is infinite and ugly, but this is the only\nway to calculate the S-matrix we have so far. We swept infinities\n"under the rug", this rug is Hamiltonian. That\'s were QED stands today.\n\n2. The second step is "dressing transformation". That\'s what I suggest\nin my book. We can apply a unitary transformation U to the Hamiltonian\nH\'\n\nH\'\' = U H\' U^{-1}\n\nThis transformation can be chosen so that\n1) the good S-matrix obtained in p. 1. is preserved\n2) H\'\' is finite.\n\nAll infinities from H\' "transfer" to U. We swept infinities under\nanother "rug". This time the rug is operator U. Since (in contrast\nto Hamiltonian) U has no physical meaning whatsoever, we can now\nthrow this rug away and forget the problems with ultraviolet infinities:\nWe have finite Hamiltonian H\'\' with interactions written in terms\nof particle creation and annihilation operators. We can choose H\'\',\nso that there are no more self-interaction effects in vacuum and\none-particle subspaces. So, one can work with H\'\' just as we work\nwith non-relativistic Hamiltonians (though, everything is relativistic,\nof course). The only difference is that H\'\' does not conserve the\nnumber of particles.\n\nThe Hamiltonian H\'\' is rather complex. It has interactions in\nall perturbation orders starting from 2nd. It has Coulomb,\nmagnetic, spin-orbit, etc. interactions between charged particles.\nIt has interaction responsible for the Lamb\'s shifts in 4th and\nhigher orders. It has interactions responsible for the Compton\nscattering, bremsstrahlung, pair creation, etc.\n\nIt would be great if we could write operator H\'\' somehow from\n"general principles". Then we could avoid the two bad-smelling\nsteps I wrote above. However, I simply don\'t know how it is possible.\n\nI think the original problem of QED dates way back to the end of 1920\'s\nwhen Dirac, Born, Heisenberg, and others guessed the form of QED\nHamiltonian (from canonical quantization of Maxwell\'s theory).\nTheir guess was fortunate and unfortunate at the same\ntime. It was unfortunate that their interaction had self-interaction\neffects, and we pay now with all these tricks (regularization,\nrenormalization, dressing, etc.) to get rid of these effects.\nIt was fortunate that their guess contained a grain of truth,\nwhich gives us accurate S-matrix and a hope to finally find\nthe correct Hamiltonian H\'\'.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
>>I share your frustration with renormalized QFT. I think I know the way
>>to fix the problem of infinities. The idea is very simple: QED simply
>>uses a wrong Hamiltonian. There is a rigorous way how to find a new
>>Hamiltonian of QED (while preserving the relativistic invariance and
>>cluster separability) which is
>>
>>1. finite
>>2. yields the same accurate S-matrix as renormalized QED.
>> (so it has the same predictions as far as experiment is concerned)
>>
>>The details are given in
>>
>>E.V.Stefanovich "Quantum field theory without infinities", Ann. Phys.
>>292 (2001), 139
>>
>>and in online book available at www.meopemuk.com
>>
>>I would appreciate any comments and criticism of this approach to QED.
>>If you don't think it completely solves the problem of ultraviolet
>>infinities, I would like to know why.
>>
>>More papers can be found at www.geocities.com/meopemuk
>

>
> I don't claim to understand every detail of your work, but I do take Haag's
> theorem seriously and therefore don't believe that the U, W and F operators
> in section 2 of your 2001 Ann. Phys. paper exist. Not in a fully
> relativistic theory, anyway. Also I notice that the regularization scheme is
> still exactly that ... i.e. you get the infinities and then devise a scheme
> (in your case a "clothing" operator) for dealing with them. This is still
> subtracting infinity from infinity, and I really don't think that people
> ought to do that.
>
> Chris
>

Hi Chris,

Nice to "talk" to you again!

You are right that QED has (at least) two problems. One of them is less
serious.
Another is more serious.

The less serious problem is revealed by Haag's theorem. In a nutshell,
it says that interacting fields cannot have Lorentz transformation
properties. In my approach I do not have interacting fields at all.
The interaction in my Hamiltonian is written in terms of particle
creation and annihilation operators. From this Hamiltonian I can
(in principle) calculate S-matrix, bound states, time evolution,
whatever I like, without ever mentioning "interacting quantum fields".
So, the Haag's theorem does not apply.

The more serious problem is related to renormalization. You are right
that if we take usual QED Hamiltonian (e.g., (8.4.22) - (8.4.25) in
Weinberg's book) and calculate elements of the S-matrix, we'll get
infinite results. My idea is that this Hamiltonian H is simply wrong.
It is wrong, but not completely wrong. It contain's a grain of truth.
In order to find the correct Hamiltonian H'', one should perform
two steps:

1. The first step is renormalization. We add infinite (mass and
charge renormalization) counterterms to H, so that in the process of
calculation of the S-matrix all infinities nicely cancel each other
and we get agreement with experiment to 10^{-11}%. I agree that
resulting Hamiltonian H' is infinite and ugly, but this is the only
way to calculate the S-matrix we have so far. We swept infinities
"under the rug", this rug is Hamiltonian. That's were QED stands today.

2. The second step is "dressing transformation". That's what I suggest
in my book. We can apply a unitary transformation U to the Hamiltonian
H'

H'' = U H' U^{-1}

This transformation can be chosen so that
1) the good S-matrix obtained in p. 1. is preserved
2) H'' is finite.

All infinities from H' "transfer" to U. We swept infinities under
another "rug". This time the rug is operator U. Since (in contrast
to Hamiltonian) U has no physical meaning whatsoever, we can now
throw this rug away and forget the problems with ultraviolet infinities:
We have finite Hamiltonian H'' with interactions written in terms
of particle creation and annihilation operators. We can choose H'',
so that there are no more self-interaction effects in vacuum and
one-particle subspaces. So, one can work with H'' just as we work
with non-relativistic Hamiltonians (though, everything is relativistic,
of course). The only difference is that H'' does not conserve the
number of particles.

The Hamiltonian H'' is rather complex. It has interactions in
all perturbation orders starting from 2nd. It has Coulomb,
magnetic, spin-orbit, etc. interactions between charged particles.
It has interaction responsible for the Lamb's shifts in 4th and
higher orders. It has interactions responsible for the Compton
scattering, bremsstrahlung, pair creation, etc.

It would be great if we could write operator H'' somehow from
"general principles". Then we could avoid the two bad-smelling
steps I wrote above. However, I simply don't know how it is possible.

I think the original problem of QED dates way back to the end of 1920's
when Dirac, Born, Heisenberg, and others guessed the form of QED
Hamiltonian (from canonical quantization of Maxwell's theory).
Their guess was fortunate and unfortunate at the same
time. It was unfortunate that their interaction had self-interaction
effects, and we pay now with all these tricks (regularization,
renormalization, dressing, etc.) to get rid of these effects.
It was fortunate that their guess contained a grain of truth,
which gives us accurate S-matrix and a hope to finally find
the correct Hamiltonian H''.

Eugene.

Eugene Stefanovich

Oct24-04, 09:04 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Chris Oakley wrote:\n>>\n>>\n>>>Let\'s just remember how we got to where we are:\n>>>\n>>>(i) We solved free field theory, including the expansion in terms of\n>>>annihilation and creation operators.\n>>>(ii) We introduced an interaction term in the Lagrangian density.\n>>>(iii) We introduced a unitary transformation that turns free fields into\n>>>interacting fields using the Interaction Picture.\n>>>(iv) We wrote out the Interaction Picture time evolution operator as\n>>>time-ordered products of the interaction.\n>>>(v) We reduced the time-ordered products to normal-ordered products.\n>>>(vi) We obtained thereby the S-matrix for scattering processes involving a\n>>>finite number of incoming particles with definite four-momentum scattering\n>>>to a finite number of outgoing particles with definite four-momenta as a sum\n>>>of Feynman diagrams.\n>>>\n>>>Many of these Feynman diagrams are pathologically divergent integrals.\n>>>\n>>>My response: This is because the theory is fundamentally flawed. Go back to\n>>>(ii) and try again.\n>>>\n>>\n>>\n>>I would like to suggest a different path (similar to what Weinberg is\n>>writing in his book):\n>>\n>>(i) We constructed the Fock space as direct sums of tensor products of\n>> 1-particle spaces\n>>(ii) We defined a non-interacting representation of the Poincare group\n>> there, including non-interacting Hamiltonian H_0 and\n>> boost operator K_0\n>>(iii) We defined interaction terms in the Hamiltonian and boost\n>> operator H= H_0+V, K = K_0+W (all operators are expressed through\n>> particle creation and annihilation operators, no fields)\n>>(iv) We found that S-matrix is ultraviolet divergent\n>>(v) We added renormalization (infinite) counterterms to the\n>> Hamiltonian H\' = H+R , so as to\n>> make S-matrix finite and in agreement with experiment.\n>>\n>>This is where QED stands now: finite and perfectly accurate S-matrix\n>>obtained from infinite (wrong) Hamiltonian H\'. My suggestion is\n>>\n>>(vi) Apply unitary dressing transformation U to the Hamiltonian H\',\n>> so that\n>>\n>>1) the new Hamiltonian\n>>\n>> H\'\' = U H\' U^{-1}\n>>\n>>is finite\n>>\n>>2) The S-matrices obtained with H\' and H\'\' are exactly the same.\n>>\n>>Now you can forget the nightmares of QFT, canonical quantization,\n>>Haags theorem, etc. You can use H\'\' in all usual quantum mechanical\n>>formulas (obtain time evolution operator, S-matrix, bound states via\n>>diagonalization, etc) without any need of regularization,\n>>renormalization, and other tricks.\n>\n>\n> Well, this is a research program - the question is whether you can\n> actually carry it out. How do you construct U? It is there that all\n> the renormalization issues will recur. Since H\' is an ill-defined object\n> one must regularize it, then construct U for the regularized theory,\n> and then go to the limit and show that it really exists. This poses\n> the same problems as the original renormalization program.\n>\n>\n> Arnold Neumaier\n\nYou are absolutely right. That\'s one way. This is a long way, but it\nreaches the goal. There is, however, a shortcut: we can simply fit\na finite interaction Hamiltonian H\'\' which yields the same (finite and\naccurate S-matrix of QED). The key condition on H\'\' is that interaction\nmust be "physical" - that\'s my terminology for saying that operator\nyields zero when acting on vacuum and one-particle states. This\nguarantees that the theory with H\'\' does not have self-interaction\neffects.\n\nYou are probably right that QED is a sick theory: there is Landau pole,\nthe perturbation series does not converge, etc. My approach does not\ncure all deseases. It achieves one important goal, though. It makes\nthe Hamiltonian of QED finite and well-behaving. So, with this\nHamiltonian you can do perturbative expansion of the S-matrix following\nstandard rules of quantum mechanics, and all terms in this expansion\nare finite. Few low order terms converge to extremely accurate QED\nresults for electron\'s magnetic moment, Lamb\'s shifts, etc.\nThis is guaranteed, because my approach does not change the S-matrix by\nconstruction.\n\nAnother advantage of having finite well-behaving Hamiltonian is that\nnow you can find bound states (both energies and wave functions)\nby simple diagonalization, and you can study time-dependent dynamics.\n\nYou can find all details on my web-site www.geocities.com/meopemuk,\nespecially in the book "Relativistic quantum dynamics" posted there.\n\nEugene Stefanovich.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Chris Oakley wrote:
>>
>>
>>>Let's just remember how we got to where we are:
>>>
>>>(i) We solved free field theory, including the expansion in terms of
>>>annihilation and creation operators.
>>>(ii) We introduced an interaction term in the Lagrangian density.
>>>(iii) We introduced a unitary transformation that turns free fields into
>>>interacting fields using the Interaction Picture.
>>>(iv) We wrote out the Interaction Picture time evolution operator as
>>>time-ordered products of the interaction.
>>>(v) We reduced the time-ordered products to normal-ordered products.
>>>(vi) We obtained thereby the S-matrix for scattering processes involving a
>>>finite number of incoming particles with definite four-momentum scattering
>>>to a finite number of outgoing particles with definite four-momenta as a sum
>>>of Feynman diagrams.
>>>
>>>Many of these Feynman diagrams are pathologically divergent integrals.
>>>
>>>My response: This is because the theory is fundamentally flawed. Go back to
>>>(ii) and try again.
>>>
>>
>>
>>I would like to suggest a different path (similar to what Weinberg is
>>writing in his book):
>>
>>(i) We constructed the Fock space as direct sums of tensor products of
>> 1-particle spaces
>>(ii) We defined a non-interacting representation of the Poincare group
>> there, including non-interacting Hamiltonian H_0 and
>> boost operator K_0
>>(iii) We defined interaction terms in the Hamiltonian and boost
>> operator H= H_0+V, K = K_0+W (all operators are expressed through
>> particle creation and annihilation operators, no fields)
>>(iv) We found that S-matrix is ultraviolet divergent
>>(v) We added renormalization (infinite) counterterms to the
>> Hamiltonian H' = H+R , so as to
>> make S-matrix finite and in agreement with experiment.
>>
>>This is where QED stands now: finite and perfectly accurate S-matrix
>>obtained from infinite (wrong) Hamiltonian H'. My suggestion is
>>
>>(vi) Apply unitary dressing transformation U to the Hamiltonian H',
>> so that
>>
>>1) the new Hamiltonian
>>
>> H'' = U H' U^{-1}
>>
>>is finite
>>
>>2) The S-matrices obtained with H' and H'' are exactly the same.
>>
>>Now you can forget the nightmares of QFT, canonical quantization,
>>Haags theorem, etc. You can use H'' in all usual quantum mechanical
>>formulas (obtain time evolution operator, S-matrix, bound states via
>>diagonalization, etc) without any need of regularization,
>>renormalization, and other tricks.
>
>
> Well, this is a research program - the question is whether you can
> actually carry it out. How do you construct U? It is there that all
> the renormalization issues will recur. Since H' is an ill-defined object
> one must regularize it, then construct U for the regularized theory,
> and then go to the limit and show that it really exists. This poses
> the same problems as the original renormalization program.
>
>
> Arnold Neumaier

You are absolutely right. That's one way. This is a long way, but it
reaches the goal. There is, however, a shortcut: we can simply fit
a finite interaction Hamiltonian H'' which yields the same (finite and
accurate S-matrix of QED). The key condition on H'' is that interaction
must be "physical" - that's my terminology for saying that operator
yields zero when acting on vacuum and one-particle states. This
guarantees that the theory with H'' does not have self-interaction
effects.

You are probably right that QED is a sick theory: there is Landau pole,
the perturbation series does not converge, etc. My approach does not
cure all deseases. It achieves one important goal, though. It makes
the Hamiltonian of QED finite and well-behaving. So, with this
Hamiltonian you can do perturbative expansion of the S-matrix following
standard rules of quantum mechanics, and all terms in this expansion
are finite. Few low order terms converge to extremely accurate QED
results for electron's magnetic moment, Lamb's shifts, etc.
This is guaranteed, because my approach does not change the S-matrix by
construction.

Another advantage of having finite well-behaving Hamiltonian is that
now you can find bound states (both energies and wave functions)
by simple diagonalization, and you can study time-dependent dynamics.

You can find all details on my web-site www.geocities.com/meopemuk,
especially in the book "Relativistic quantum dynamics" posted there.

Eugene Stefanovich.

Arnold Neumaier

Oct25-04, 08:09 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nEugene Stefanovich wrote:\n\n> Another advantage of having finite well-behaving Hamiltonian is that\n> now you can find bound states (both energies and wave functions)\n> by simple diagonalization, and you can study time-dependent dynamics.\n\nYes, this is true; I like this about your approach.\n\nUnfortunately, you don\'t manage to carry out even\nthe simplest bound state calculation that would show 1-loop corrections\nfrom the electromagnetic field. To be competitive with the traditional\napproach using Bethe-Salpeter or Schwinger-Dyson equations for bound states,\nyou need to show that you can at least match their results.\n\nSo there is more work ahead before getting recognition...\n\n\nArnold Neumaier\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> Another advantage of having finite well-behaving Hamiltonian is that
> now you can find bound states (both energies and wave functions)
> by simple diagonalization, and you can study time-dependent dynamics.

Yes, this is true; I like this about your approach.

Unfortunately, you don't manage to carry out even
the simplest bound state calculation that would show 1-loop corrections
from the electromagnetic field. To be competitive with the traditional
approach using Bethe-Salpeter or Schwinger-Dyson equations for bound states,
you need to show that you can at least match their results.

So there is more work ahead before getting recognition...

Arnold Neumaier

Arnold Neumaier

Oct25-04, 08:09 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\nChris Oakley wrote:\n\n>>I gave you instead an exercise. Physicists are too lazy to do things as\n>>carefully as you wish, so you have to do it yourself. I gave you the\n>>outline, so that you can fill in the details.\n\n>\n> I have not done it because I do not believe that it can be done. Products of\n> field operators in the interaction Hamiltonian density must refer to the\n> same time if we are to be able to transform to the Interaction Picture.\n\nIf the issue is so important to you that you keep it up over many years,\nyou\'d be willing to spend a little time on it. I had to go through\nmany, many wrong tracks to arrive at my present understanding; who only\ntakes the paths trodden by many only gets to know what everyone knows,\nand this is the sloppy traditional formalism.\n\n\n> I might just add that I am not the first person to be bothered by the sloppy\n> mathematics in QFT. People have been working on this, on and off, since\n> 1928. If there was a way of making the procedure rigorous, it would have\n> found its way into the text books by now. However the best QFT text book *I*\n> know of - Weinberg - makes no mention of the possibility of defeating\n> infinite subtractions via non-local field equations. Neither does the\n> supposedly more rigorous Scharf.\n\nWeinberg\'s treatise does _not_ attempt to do things with rigor, nowhere.\nScharf is rigorous, but just one particular way. There are many ways to\ndo renormalization, but few who both have the energy to write a textbook\nand find it worth spelling out detais that require independent thinking\nto find.\n\n\n> Limiting the integral in p^0 space results in a\n> smearing in the time integral. This is just basic Fourier analysis. It is\n> the latter which prevents a transformation between the Heisenberg and\n> Interaction pictures for the field operators.\n\nEverything needs the interaction picture only.\nIf you follow my recipe of writing the time-ordered exponential\nin the momentum representation, you can extract a formula for how to\nidentify the interaction term in the Hamiltonian corresponding to the\nregularized action.\n\nIf you are not even willing to try this, I find it useless to\ncontinue the discussion.\n\n>>Should you ever want to visit Vienna, I\'d show you\n>>on the blackboard what to do.\n>\n> A kind offer, but be careful - I might take you up on it.\n\nYou are welcome!\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:

>>I gave you instead an exercise. Physicists are too lazy to do things as
>>carefully as you wish, so you have to do it yourself. I gave you the
>>outline, so that you can fill in the details.

>
> I have not done it because I do not believe that it can be done. Products of
> field operators in the interaction Hamiltonian density must refer to the
> same time if we are to be able to transform to the Interaction Picture.

If the issue is so important to you that you keep it up over many years,
you'd be willing to spend a little time on it. I had to go through
many, many wrong tracks to arrive at my present understanding; who only
takes the paths trodden by many only gets to know what everyone knows,
and this is the sloppy traditional formalism.

> I might just add that I am not the first person to be bothered by the sloppy
> mathematics in QFT. People have been working on this, on and off, since
> 1928. If there was a way of making the procedure rigorous, it would have
> found its way into the text books by now. However the best QFT text book *I*
> know of - Weinberg - makes no mention of the possibility of defeating
> infinite subtractions via non-local field equations. Neither does the
> supposedly more rigorous Scharf.

Weinberg's treatise does _not_ attempt to do things with rigor, nowhere.
Scharf is rigorous, but just one particular way. There are many ways to
do renormalization, but few who both have the energy to write a textbook
and find it worth spelling out detais that require independent thinking
to find.

> Limiting the integral in p^0 space results in a
> smearing in the time integral. This is just basic Fourier analysis. It is
> the latter which prevents a transformation between the Heisenberg and
> Interaction pictures for the field operators.

Everything needs the interaction picture only.
If you follow my recipe of writing the time-ordered exponential
in the momentum representation, you can extract a formula for how to
identify the interaction term in the Hamiltonian corresponding to the
regularized action.

If you are not even willing to try this, I find it useless to
continue the discussion.

>>Should you ever want to visit Vienna, I'd show you
>>on the blackboard what to do.
>
> A kind offer, but be careful - I might take you up on it.

You are welcome!

Arnold Neumaier

Eugene Stefanovich

Oct25-04, 08:09 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\n\n\nChris Oakley wrote:\n> However the best QFT text book *I*\n> know of - Weinberg - makes no mention of the possibility of defeating\n> infinite subtractions via non-local field equations. Neither does the\n> supposedly more rigorous Scharf.\n>\n\nLocality is not required for relativistic invariance. One can build a\nnon-local (and finite) quantum field theory which is\nrelativistically invariant\n(Poincare commutation relations are satisfied). See\n\nM.I. Shirokov "On relativistic nonlocal quantum field theory"\nInt. J. Theor. Phys. 41 (2002), 1027\n\nH. Kita, "A non-trivial example of a relativistic\nquantum theory of particles without divergence difficulties",\nProgr. Theor. Phys. 35 (1966), 934.\n\nI don\'t think, however, that this is the way to go in\nformulating a consistent approach to QED. "Dressing transformation"\nis the way.\n\nEugene\nwww.geocities.com/meopemuk\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
> However the best QFT text book *I*
> know of - Weinberg - makes no mention of the possibility of defeating
> infinite subtractions via non-local field equations. Neither does the
> supposedly more rigorous Scharf.
>

Locality is not required for relativistic invariance. One can build a
non-local (and finite) quantum field theory which is
relativistically invariant
(Poincare commutation relations are satisfied). See

M.I. Shirokov "On relativistic nonlocal quantum field theory"
\Int. J. Theor. Phys. 41 (2002), 1027

H. Kita, "A non-trivial example of a relativistic
quantum theory of particles without divergence difficulties",
Progr. Theor. Phys. 35 (1966), 934.

I don't think, however, that this is the way to go in
formulating a consistent approach to QED. "Dressing transformation"
is the way.

Eugene
www.geocities.com/meopemuk

Chris Oakley

Oct25-04, 08:09 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n> The less serious problem is revealed by Haag\'s theorem. In a nutshell,\n> it says that interacting fields cannot have Lorentz transformation\n> properties. In my approach I do not have interacting fields at all.\n> The interaction in my Hamiltonian is written in terms of particle\n> creation and annihilation operators. From this Hamiltonian I can\n> (in principle) calculate S-matrix, bound states, time evolution,\n> whatever I like, without ever mentioning "interacting quantum fields".\n> So, the Haag\'s theorem does not apply.\n\nA corollary of this is that if you force the fields to have Lorentz\ntransformation properties then the Hamiltonian must be the free one. I don\'t\nhave a problem with this. If an interacting scalar field \\phi(x) is composed\nfrom a free field \\phi_0(x) through (e.g.)\n\n\\phi(x) = \\phi_0(x) + \\int d^4x\'d^4x\'\' f(x-x\',x-x\'\') \\phi_0(x\')\\phi_0(x\'\')\n\nfor some invariant function f - the kind of approach I advocate - then you\nfind that the free Hamiltonian works just as well as the time displacement\noperator as the interacting one (just apply it directly & you will see). You\ndon\'t therefore need to build an interacting Hamiltonian at all.\n\n> The more serious problem is related to renormalization. You are right\n> that if we take usual QED Hamiltonian (e.g., (8.4.22) - (8.4.25) in\n> Weinberg\'s book) and calculate elements of the S-matrix, we\'ll get\n> infinite results. My idea is that this Hamiltonian H is simply wrong.\n> It is wrong, but not completely wrong. It contain\'s a grain of truth.\n> In order to find the correct Hamiltonian H\'\', one should perform\n> two steps:\n>\n> 1. The first step is renormalization. We add infinite (mass and\n> charge renormalization) counterterms to H, so that in the process of\n> calculation of the S-matrix all infinities nicely cancel each other\n> and we get agreement with experiment to 10^{-11}%. I agree that\n> resulting Hamiltonian H\' is infinite and ugly, but this is the only\n> way to calculate the S-matrix we have so far. We swept infinities\n> "under the rug", this rug is Hamiltonian. That\'s were QED stands today.\n>\n> 2. The second step is "dressing transformation". That\'s what I suggest\n> in my book. We can apply a unitary transformation U to the Hamiltonian\n> H\'\n>\n> H\'\' = U H\' U^{-1}\n>\n> This transformation can be chosen so that\n> 1) the good S-matrix obtained in p. 1. is preserved\n> 2) H\'\' is finite.\n>\n> All infinities from H\' "transfer" to U. We swept infinities under\n> another "rug". This time the rug is operator U. Since (in contrast\n> to Hamiltonian) U has no physical meaning whatsoever, we can now\n> throw this rug away and forget the problems with ultraviolet infinities:\n> We have finite Hamiltonian H\'\' with interactions written in terms\n> of particle creation and annihilation operators. We can choose H\'\',\n> so that there are no more self-interaction effects in vacuum and\n> one-particle subspaces. So, one can work with H\'\' just as we work\n> with non-relativistic Hamiltonians (though, everything is relativistic,\n> of course). The only difference is that H\'\' does not conserve the\n> number of particles.\n>\n> The Hamiltonian H\'\' is rather complex. It has interactions in\n> all perturbation orders starting from 2nd. It has Coulomb,\n> magnetic, spin-orbit, etc. interactions between charged particles.\n> It has interaction responsible for the Lamb\'s shifts in 4th and\n> higher orders. It has interactions responsible for the Compton\n> scattering, bremsstrahlung, pair creation, etc.\n>\n> It would be great if we could write operator H\'\' somehow from\n> "general principles". Then we could avoid the two bad-smelling\n> steps I wrote above. However, I simply don\'t know how it is possible.\n>\n> I think the original problem of QED dates way back to the end of 1920\'s\n> when Dirac, Born, Heisenberg, and others guessed the form of QED\n> Hamiltonian (from canonical quantization of Maxwell\'s theory).\n> Their guess was fortunate and unfortunate at the same\n> time. It was unfortunate that their interaction had self-interaction\n> effects, and we pay now with all these tricks (regularization,\n> renormalization, dressing, etc.) to get rid of these effects.\n> It was fortunate that their guess contained a grain of truth,\n> which gives us accurate S-matrix and a hope to finally find\n> the correct Hamiltonian H\'\'.\n\nIt may be that no-one has found a "clean" way to construct this interacting\nHamiltonian simply because it cannot be done ... indeed, it was the belief\nthat I personally could not do any better that led me to try something\ncompletely different.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>The less serious problem is revealed by Haag's theorem. In a nutshell,
> it says that interacting fields cannot have Lorentz transformation
> properties. In my approach I do not have interacting fields at all.
> The interaction in my Hamiltonian is written in terms of particle
> creation and annihilation operators. From this Hamiltonian I can
> (in principle) calculate S-matrix, bound states, time evolution,
> whatever I like, without ever mentioning "interacting quantum fields".
> So, the Haag's theorem does not apply.

A corollary of this is that if you force the fields to have Lorentz
transformation properties then the Hamiltonian must be the free one. I don't
have a problem with this. If an interacting scalar field \phi(x) is composed
from a free field \phi_0(x) through (e.g.)

\phi(x) = \phi_0(x) + \int d^{4x}'d^{4x}'' f(x-x',x-x'') \phi_0(x')\phi_0(x'')

for some invariant function f - the kind of approach I advocate - then you
find that the free Hamiltonian works just as well as the time displacement
operator as the interacting one (just apply it directly & you will see). You
don't therefore need to build an interacting Hamiltonian at all.

> The more serious problem is related to renormalization. You are right
> that if we take usual QED Hamiltonian (e.g., (8.4.22) - (8.4.25) in
> Weinberg's book) and calculate elements of the S-matrix, we'll get
> infinite results. My idea is that this Hamiltonian H is simply wrong.
> It is wrong, but not completely wrong. It contain's a grain of truth.
> In order to find the correct Hamiltonian H'', one should perform
> two steps:
>
> 1. The first step is renormalization. We add infinite (mass and
> charge renormalization) counterterms to H, so that in the process of
> calculation of the S-matrix all infinities nicely cancel each other
> and we get agreement with experiment to 10^{-11}%. I agree that
> resulting Hamiltonian H' is infinite and ugly, but this is the only
> way to calculate the S-matrix we have so far. We swept infinities
> "under the rug", this rug is Hamiltonian. That's were QED stands today.
>
> 2. The second step is "dressing transformation". That's what I suggest
> in my book. We can apply a unitary transformation U to the Hamiltonian
> H'
>
> H'' = U H' U^{-1}
>
> This transformation can be chosen so that
> 1) the good S-matrix obtained in p. 1. is preserved
> 2) H'' is finite.
>
> All infinities from H' "transfer" to U. We swept infinities under
> another "rug". This time the rug is operator U. Since (in contrast
> to Hamiltonian) U has no physical meaning whatsoever, we can now
> throw this rug away and forget the problems with ultraviolet infinities:
> We have finite Hamiltonian H'' with interactions written in terms
> of particle creation and annihilation operators. We can choose H'',
> so that there are no more self-interaction effects in vacuum and
> one-particle subspaces. So, one can work with H'' just as we work
> with non-relativistic Hamiltonians (though, everything is relativistic,
> of course). The only difference is that H'' does not conserve the
> number of particles.
>
> The Hamiltonian H'' is rather complex. It has interactions in
> all perturbation orders starting from 2nd. It has Coulomb,
> magnetic, spin-orbit, etc. interactions between charged particles.
> It has interaction responsible for the Lamb's shifts in 4th and
> higher orders. It has interactions responsible for the Compton
> scattering, bremsstrahlung, pair creation, etc.
>
> It would be great if we could write operator H'' somehow from
> "general principles". Then we could avoid the two bad-smelling
> steps I wrote above. However, I simply don't know how it is possible.
>
> I think the original problem of QED dates way back to the end of 1920's
> when Dirac, Born, Heisenberg, and others guessed the form of QED
> Hamiltonian (from canonical quantization of Maxwell's theory).
> Their guess was fortunate and unfortunate at the same
> time. It was unfortunate that their interaction had self-interaction
> effects, and we pay now with all these tricks (regularization,
> renormalization, dressing, etc.) to get rid of these effects.
> It was fortunate that their guess contained a grain of truth,
> which gives us accurate S-matrix and a hope to finally find
> the correct Hamiltonian H''.

It may be that no-one has found a "clean" way to construct this interacting
Hamiltonian simply because it cannot be done ... indeed, it was the belief
that I personally could not do any better that led me to try something
completely different.

Arnold Neumaier

Oct25-04, 08:13 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nEugene Stefanovich wrote:\n>\n> There is a version of perturbation theory which allows you\n> to write the S-operator in manifestly unitary form in each perturbation\n> order. This version was suggested by Magnus.\n> It is not well known. You can find references in my Ann. Phys. paper.\n\nThis is not yet the same as having S as a unitary operator, since the\nperturbation series probably still diverges.\n\nAs long as perturbation theory _defines_ the S-matrix (or, in your approach,\nthe Hamiltonian), something important is missing unless one could show that\nthe series converges. Perturbation theory in nonrelativistic QM is different\nsince the objects exists mathematically without recourse to perturbation\ntheory, and the latter is only used to find numerical approximations.\n\n\nArnold Neumaier\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> There is a version of perturbation theory which allows you
> to write the S-operator in manifestly unitary form in each perturbation
> order. This version was suggested by Magnus.
> It is not well known. You can find references in my Ann. Phys. paper.

This is not yet the same as having S as a unitary operator, since the
perturbation series probably still diverges.

As long as perturbation theory _defines_ the S-matrix (or, in your approach,
the Hamiltonian), something important is missing unless one could show that
the series converges. Perturbation theory in nonrelativistic QM is different
since the objects exists mathematically without recourse to perturbation
theory, and the latter is only used to find numerical approximations.

Arnold Neumaier

Thomas Larsson

Oct26-04, 12:56 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<415856B6.8050907@univie.ac.at>...\n\n> This is one of a series of papers by Connes and coauthors which recast\n> the renormalization procedure of QFT in algebraic terms, making\n> high loop calculations and renormalizability proofs (beyond just\n> superficial counting) more accessible to those who understand Hopf\n> algebras, and giving some insight into the structure of renormalization.\n> On this level, the renormalization group is embedded into a much larger\n> (in fact huge) group. The authors equate group = symmetries, but these\n> symmetries are far, far away from those which - say - give rise to\n> Noether currents.\n>\n> The above paper deepens the relations between physics and abstract\n> mathematics, but it does not seem to have any consequences for working\n> with the standard model, say.\n\nSome of Dirk Kreimer\'s papers have really impressive titles. He claims\nthat he has done perturbation theory to order 30 loops on a computer or\nsomething like that. I would like to understand exactly what that means.\nWhat it definitely cannot mean is the naive thing, that he is able to\ncompute say the anomalous magnetic moment of the electron to order\nalpha^30.\n\nWhat I think it means is that once you know all 1PI (one-particle\nirreducible) diagrams, you can get all Feynmann diagrams by pure\ncombinatorics. While it is nice to have algorithms for this, the\nproblem of evaluating infinitely many 1PI diagrams still remains.\nCan Connes address this problem?\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<415856B6.8050907@univie.ac.at>...

> This is one of a series of papers by Connes and coauthors which recast
> the renormalization procedure of QFT in algebraic terms, making
> high loop calculations and renormalizability proofs (beyond just
> superficial counting) more accessible to those who understand Hopf
> algebras, and giving some insight into the structure of renormalization.
> On this level, the renormalization group is embedded into a much larger
> (in fact huge) group. The authors equate group = symmetries, but these
> symmetries are far, far away from those which - say - give rise to
> Noether currents.
>
> The above paper deepens the relations between physics and abstract
> mathematics, but it does not seem to have any consequences for working
> with the standard model, say.

Some of Dirk Kreimer's papers have really impressive titles. He claims
that he has done perturbation theory to order 30 loops on a computer or
something like that. I would like to understand exactly what that means.
What it definitely cannot mean is the naive thing, that he is able to
compute say the anomalous magnetic moment of the electron to order
\alpha^30.

What I think it means is that once you know all 1PI (one-particle
irreducible) diagrams, you can get all Feynmann diagrams by pure
combinatorics. While it is nice to have algorithms for this, the
problem of evaluating infinitely many 1PI diagrams still remains.
Can Connes address this problem?

Arnold Neumaier

Oct26-04, 12:56 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Frank Hellmann wrote:\n> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<415856B6.8050907@univie.ac.at>...\n>\n>>pind are_poet wrote:\n>>\n>>>in math.NT/0409306, Connes and Marcolli investigate the occurence of\n>>>\'the action of a certain universal "Motivic Galois Group" on the set\n>>>of physical theories\' and wrote:\n>>>\n>>>"...these facts altogether indicate that the divergences of Quantum\n>>>Field Theory, far from just being an unwanted nuisance, are a clear\n>>>sign of totally unexpected symmetries of geometrical origin."\n>>\n>>This is one of a series of papers by Connes and coauthors which recast\n>>the renormalization procedure of QFT in algebraic terms, making\n>>high loop calculations and renormalizability proofs (beyond just\n>>superficial counting) more accessible to those who understand Hopf\n>>algebras, and giving some insight into the structure of renormalization.\n>>On this level, the renormalization group is embedded into a much larger\n>>(in fact huge) group. The authors equate group = symmetries, but these\n>>symmetries are far, far away from those which - say - give rise to\n>>Noether currents.\n>>\n>>The above paper deepens the relations between physics and abstract\n>>mathematics, but it does not seem to have any consequences for working\n>>with the standard model, say.\n>\n> Are they in fact thinking about a symmetry on the "space of physical\n> theories" here? What do they define as a physical theory then? And how\n> does that space acquire geometry?\n\nAs far as I understand, a \'physical theory\' should be taken here as\nsynonymous with \'an action defined by a Lagrangian\' (and the class\nof theories defined by them through renormalization); so the space is\nthe space of all Lagrangians for a given family of fields and symmetries.\n\nBut I am not very familiar with that stuff, and since (at my present\nunderstanding) it is unlikely to have impact on the practice of QFT,\nI have no intention of going into it more deeply.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Frank Hellmann wrote:
> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<415856B6.8050907@univie.ac.at>...
>
>>pindare_poet wrote:
>>
>>>in math.NT/0409306, Connes and Marcolli investigate the occurence of
>>>'the action of a certain universal "Motivic Galois Group" on the set
>>>of physical theories' and wrote:
>>>
>>>"...these facts altogether indicate that the divergences of Quantum
>>>Field Theory, far from just being an unwanted nuisance, are a clear
>>>sign of totally unexpected symmetries of geometrical origin."
>>
>>This is one of a series of papers by Connes and coauthors which recast
>>the renormalization procedure of QFT in algebraic terms, making
>>high loop calculations and renormalizability proofs (beyond just
>>superficial counting) more accessible to those who understand Hopf
>>algebras, and giving some insight into the structure of renormalization.
>>On this level, the renormalization group is embedded into a much larger
>>(in fact huge) group. The authors equate group = symmetries, but these
>>symmetries are far, far away from those which - say - give rise to
>>Noether currents.
>>
>>The above paper deepens the relations between physics and abstract
>>mathematics, but it does not seem to have any consequences for working
>>with the standard model, say.
>
> Are they in fact thinking about a symmetry on the "space of physical
> theories" here? What do they define as a physical theory then? And how
> does that space acquire geometry?

As far as I understand, a 'physical theory' should be taken here as
synonymous with 'an action defined by a Lagrangian' (and the class
of theories defined by them through renormalization); so the space is
the space of all Lagrangians for a given family of fields and symmetries.

But I am not very familiar with that stuff, and since (at my present
understanding) it is unlikely to have impact on the practice of QFT,
I have no intention of going into it more deeply.

Arnold Neumaier

Chris Oakley

Oct27-04, 10:55 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n> Everything needs the interaction picture only.\n> If you follow my recipe of writing the time-ordered exponential\n> in the momentum representation, you can extract a formula for how to\n> identify the interaction term in the Hamiltonian corresponding to the\n> regularized action.\n>\n> If you are not even willing to try this, I find it useless to\n> continue the discussion.\n\nThis is your argument, not mine. Post it on ArXiv or your web site & I will\nstudy it.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Everything needs the interaction picture only.
> If you follow my recipe of writing the time-ordered exponential
> in the momentum representation, you can extract a formula for how to
> identify the interaction term in the Hamiltonian corresponding to the
> regularized action.
>
> If you are not even willing to try this, I find it useless to
> continue the discussion.

This is your argument, not mine. Post it on ArXiv or your web site & I will
study it.

Arnold Neumaier

Oct27-04, 10:55 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\nThomas Larsson wrote:\n>\n> Some of Dirk Kreimer\'s papers have really impressive titles. He claims\n> that he has done perturbation theory to order 30 loops on a computer or\n> something like that. I would like to understand exactly what that means.\n> What it definitely cannot mean is the naive thing, that he is able to\n> compute say the anomalous magnetic moment of the electron to order\n> alpha^30.\n>\n> What I think it means is that once you know all 1PI (one-particle\n> irreducible) diagrams, you can get all Feynmann diagrams by pure\n> combinatorics.\n\nI don\'t think it can mean this; this has been known for a long time.\n\nIt sounds as if they can actually reorganize the computations in such\na way that they reexpress the sum of the myriads of individual Feynman\ndiagram contributions such that a 30 loop calculation becomes tractable.\n\nI imagine they are doing something like writing\nsum_{k=1:inf} g^k/k = sum_{l=1:inf} (g^{2l-1}/(2l-1)+g^{2l}/(2l))\n= sum_{l=1:inf} g^{2l-1}(2l(1+g)-g)/(4l^2-2l)\nwhich doubles the order, but applied to the loop expansion in place\nof this trivial example. QFTists use all kinds of resummations to reduce\nthe combinatorial explosion of terms.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Thomas Larsson wrote:
>
> Some of Dirk Kreimer's papers have really impressive titles. He claims
> that he has done perturbation theory to order 30 loops on a computer or
> something like that. I would like to understand exactly what that means.
> What it definitely cannot mean is the naive thing, that he is able to
> compute say the anomalous magnetic moment of the electron to order
> \alpha^30.
>
> What I think it means is that once you know all 1PI (one-particle
> irreducible) diagrams, you can get all Feynmann diagrams by pure
> combinatorics.

I don't think it can mean this; this has been known for a long time.

It sounds as if they can actually reorganize the computations in such
a way that they reexpress the sum of the myriads of individual Feynman
diagram contributions such that a 30 loop calculation becomes tractable.

I imagine they are doing something like writing
sum_{k=1:inf} g^k/k = sum_{l=1:inf} (g^{2l-1}/(2l-1)+g^{2l}/(2l))= sum_{l=1:inf} g^{2l-1}(2l(1+g)-g)/(4l^2-2l)
which doubles the order, but applied to the loop expansion in place
of this trivial example. QFTists use all kinds of resummations to reduce
the combinatorial explosion of terms.

Arnold Neumaier

Arnold Neumaier

Oct27-04, 10:55 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\nEugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>Eugene Stefanovich wrote:\n>>\n>>>(vi) Apply unitary dressing transformation U to the Hamiltonian H\',\n>>> so that\n>>>\n>>>1) the new Hamiltonian\n>>>\n>>> H\'\' = U H\' U^{-1}\n>>>\n>>>is finite\n>>>\n>>>2) The S-matrices obtained with H\' and H\'\' are exactly the same.\n>>>\n>>>Now you can forget the nightmares of QFT, canonical quantization,\n>>>Haags theorem, etc. You can use H\'\' in all usual quantum mechanical\n>>>formulas (obtain time evolution operator, S-matrix, bound states via\n>>>diagonalization, etc) without any need of regularization,\n>>>renormalization, and other tricks.\n>>\n>>Well, this is a research program - the question is whether you can\n>>actually carry it out. How do you construct U? It is there that all\n>>the renormalization issues will recur. Since H\' is an ill-defined object\n>>one must regularize it, then construct U for the regularized theory,\n>>and then go to the limit and show that it really exists. This poses\n>>the same problems as the original renormalization program.\n>>\n>\n> You are absolutely right. That\'s one way. This is a long way, but it\n> reaches the goal. There is, however, a shortcut: we can simply fit\n> a finite interaction Hamiltonian H\'\' which yields the same (finite and\n> accurate S-matrix of QED). The key condition on H\'\' is that interaction\n> must be "physical" - that\'s my terminology for saying that operator\n> yields zero when acting on vacuum and one-particle states. This\n> guarantees that the theory with H\'\' does not have self-interaction\n> effects.\n\nI understand what you are doing - but you don\'t find in fact H\'\',\nbut you find a sequence of approximations H_j=U_j H\' U_j^{-1},\nwhere the U_j and H\' is ill-defined but the H_j make sense to order j.\nAnd there is no guarantee that the H_j will converge to the desired H\'\',\nor that the infinite series that agrees with H_j to order j is more\nthan asymptotic.\n\n\n> You are probably right that QED is a sick theory: there is Landau pole,\n> the perturbation series does not converge, etc. My approach does not\n> cure all deseases. It achieves one important goal, though. It makes\n> the Hamiltonian of QED finite and well-behaving.\n\nThus while traditional QFT renormalizes the S-matrix you renormalize the\nHamiltonian. This is a useful achievement, and may have important\nconsequences for bound state calculations.\n\nBut in both cases, renormalization is needed to get rid of\nill-defined terms, and in both cases, the convergence questions remain\nopen.\n\n\n> Few low order terms converge to extremely accurate QED\n> results for electron\'s magnetic moment, Lamb\'s shifts, etc.\n\nYou did not show that you get the Lamb shift, only made it plausible.\nComputing it and showing that to order O(e^2) you get the same value\nas QFT would be an important step in convincing people that your\ntheory is indeed equivalent to the traditional approach.\n\n\n> This is guaranteed, because my approach does not change the S-matrix by\n> construction.\n\nWell, the initial theory is ill-defined and you make it defined in\na different way than tradition. This does not mean that one knows that\nboth give the same results. In principle one can get from inf-inf\ndifferent finite answers by interpreting it in different ways.\n\nTo be convincing, one needs an argument showing that the two finite,\nwell-defined versions give the same results by relating one to the\nother in terms of well-defined operations or limits.\n\n\n> Another advantage of having finite well-behaving Hamiltonian is that\n> now you can find bound states (both energies and wave functions)\n> by simple diagonalization, and you can study time-dependent dynamics.\n\nOn the other hand, a disatvantage is that all calculations are done\nnoncovariantly, which makes them somewhat messy...\nIn spite of that, I think that the dressing approach is useful.\nIf your Hamiltonian QED is shown equivalent with QED (which it is on a\nformal level, but this is not necessarily inherited on the renormalized\nlevel), it would give QED a dynamical content that is difficult to discern\nin the traditional formulation.\n\nHow does your work go beyond that of Shebeko and Shirokov in\nnucl-th/0102037 ?\n\nWhy don\'t you put your papers on the arxiv?\nIt is never too late, and you reach more people.\n\n\nArnold Neumaier\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>Eugene Stefanovich wrote:
>>
>>>(vi) Apply unitary dressing transformation U to the Hamiltonian H',
>>> so that
>>>
>>>1) the new Hamiltonian
>>>
>>> H'' = U H' U^{-1}
>>>
>>>is finite
>>>
>>>2) The S-matrices obtained with H' and H'' are exactly the same.
>>>
>>>Now you can forget the nightmares of QFT, canonical quantization,
>>>Haags theorem, etc. You can use H'' in all usual quantum mechanical
>>>formulas (obtain time evolution operator, S-matrix, bound states via
>>>diagonalization, etc) without any need of regularization,
>>>renormalization, and other tricks.
>>
>>Well, this is a research program - the question is whether you can
>>actually carry it out. How do you construct U? It is there that all
>>the renormalization issues will recur. Since H' is an ill-defined object
>>one must regularize it, then construct U for the regularized theory,
>>and then go to the limit and show that it really exists. This poses
>>the same problems as the original renormalization program.
>>
>
> You are absolutely right. That's one way. This is a long way, but it
> reaches the goal. There is, however, a shortcut: we can simply fit
> a finite interaction Hamiltonian H'' which yields the same (finite and
> accurate S-matrix of QED). The key condition on H'' is that interaction
> must be "physical" - that's my terminology for saying that operator
> yields zero when acting on vacuum and one-particle states. This
> guarantees that the theory with H'' does not have self-interaction
> effects.

I understand what you are doing - but you don't find in fact H'',
but you find a sequence of approximations H_j=U_j H' U_j^{-1},
where the U_j and H' is ill-defined but the H_j make sense to order j.
And there is no guarantee that the H_j will converge to the desired H'',
or that the infinite series that agrees with H_j to order j is more
than asymptotic.

> You are probably right that QED is a sick theory: there is Landau pole,
> the perturbation series does not converge, etc. My approach does not
> cure all deseases. It achieves one important goal, though. It makes
> the Hamiltonian of QED finite and well-behaving.

Thus while traditional QFT renormalizes the S-matrix you renormalize the
Hamiltonian. This is a useful achievement, and may have important
consequences for bound state calculations.

But in both cases, renormalization is needed to get rid of
ill-defined terms, and in both cases, the convergence questions remain
open.

> Few low order terms converge to extremely accurate QED
> results for electron's magnetic moment, Lamb's shifts, etc.

You did not show that you get the Lamb shift, only made it plausible.
Computing it and showing that to order O(e^2) you get the same value
as QFT would be an important step in convincing people that your
theory is indeed equivalent to the traditional approach.

> This is guaranteed, because my approach does not change the S-matrix by
> construction.

Well, the initial theory is ill-defined and you make it defined in
a different way than tradition. This does not mean that one knows that
both give the same results. In principle one can get from inf-inf
different finite answers by interpreting it in different ways.

To be convincing, one needs an argument showing that the two finite,
well-defined versions give the same results by relating one to the
other in terms of well-defined operations or limits.

> Another advantage of having finite well-behaving Hamiltonian is that
> now you can find bound states (both energies and wave functions)
> by simple diagonalization, and you can study time-dependent dynamics.

On the other hand, a disatvantage is that all calculations are done
noncovariantly, which makes them somewhat messy...
In spite of that, I think that the dressing approach is useful.
If your Hamiltonian QED is shown equivalent with QED (which it is on a
formal level, but this is not necessarily inherited on the renormalized
level), it would give QED a dynamical content that is difficult to discern
in the traditional formulation.

How does your work go beyond that of Shebeko and Shirokov in
http://www.arxiv.org/abs/nucl-th/0102037 ?

Why don't you put your papers on the arxiv?
It is never too late, and you reach more people.

Arnold Neumaier

Chris Oakley

Oct27-04, 10:55 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n> Locality is not required for relativistic invariance. One can build a\n> non-local (and finite) quantum field theory which is\n> relativistically invariant\n> (Poincare commutation relations are satisfied). See\n>\n> M.I. Shirokov "On relativistic nonlocal quantum field theory"\n> Int. J. Theor. Phys. 41 (2002), 1027\n>\n> H. Kita, "A non-trivial example of a relativistic\n> quantum theory of particles without divergence difficulties",\n> Progr. Theor. Phys. 35 (1966), 934.\n\nI\'ll chase these up ... thank-you. The first one does not seem to have been\nposted on arXiv (I would have seen it if it had).\n\n> I don\'t think, however, that this is the way to go in\n> formulating a consistent approach to QED. "Dressing transformation"\n> is the way.\n\n.... and if you manage to (i) make it covariant through and through & (ii)\nhave no infinite subtractions I will gladly follow.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Locality is not required for relativistic invariance. One can build a
> non-local (and finite) quantum field theory which is
> relativistically invariant
> (Poincare commutation relations are satisfied). See
>
> M.I. Shirokov "On relativistic nonlocal quantum field theory"
> \Int. J. Theor. Phys. 41 (2002), 1027
>
> H. Kita, "A non-trivial example of a relativistic
> quantum theory of particles without divergence difficulties",
> Progr. Theor. Phys. 35 (1966), 934.

I'll chase these up ... thank-you. The first one does not seem to have been
posted on arXiv (I would have seen it if it had).

> I don't think, however, that this is the way to go in
> formulating a consistent approach to QED. "Dressing transformation"
> is the way.

.... and if you manage to (i) make it covariant through and through & (ii)
have no infinite subtractions I will gladly follow.

Eugene Stefanovich

Oct27-04, 10:56 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\n\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>There is a version of perturbation theory which allows you\n>>to write the S-operator in manifestly unitary form in each perturbation\n>>order. This version was suggested by Magnus.\n>>It is not well known. You can find references in my Ann. Phys. paper.\n>\n>\n> This is not yet the same as having S as a unitary operator, since the\n> perturbation series probably still diverges.\n>\n> As long as perturbation theory _defines_ the S-matrix (or, in your approach,\n> the Hamiltonian), something important is missing unless one could show that\n> the series converges. Perturbation theory in nonrelativistic QM is different\n> since the objects exists mathematically without recourse to perturbation\n> theory, and the latter is only used to find numerical approximations.\n>\n>\n> Arnold Neumaier\n>\n>\n\n\nYou are right, I was concerned only to make finite each term\nin the perturbation expansion. I have no idea whether the full\nseries converge or diverge. The only thing I know is that few\nlower-order terms converge nicely to experimental numbers.\nIt may happen that somewhere around 137th term the things will go wild.\nI don\'t know how to solve that yet.\n\nEugene.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>There is a version of perturbation theory which allows you
>>to write the S-operator in manifestly unitary form in each perturbation
>>order. This version was suggested by Magnus.
>>It is not well known. You can find references in my Ann. Phys. paper.
>
>
> This is not yet the same as having S as a unitary operator, since the
> perturbation series probably still diverges.
>
> As long as perturbation theory _defines_ the S-matrix (or, in your approach,
> the Hamiltonian), something important is missing unless one could show that
> the series converges. Perturbation theory in nonrelativistic QM is different
> since the objects exists mathematically without recourse to perturbation
> theory, and the latter is only used to find numerical approximations.
>
>
> Arnold Neumaier
>
>

You are right, I was concerned only to make finite each term
in the perturbation expansion. I have no idea whether the full
series converge or diverge. The only thing I know is that few
lower-order terms converge nicely to experimental numbers.
It may happen that somewhere around 137th term the things will go wild.
I don't know how to solve that yet.

Eugene.

Eugene Stefanovich

Oct27-04, 10:56 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\n\n\nChris Oakley wrote:\n>>The less serious problem is revealed by Haag\'s theorem. In a nutshell,\n>>it says that interacting fields cannot have Lorentz transformation\n>>properties. In my approach I do not have interacting fields at all.\n>>The interaction in my Hamiltonian is written in terms of particle\n>>creation and annihilation operators. From this Hamiltonian I can\n>>(in principle) calculate S-matrix, bound states, time evolution,\n>>whatever I like, without ever mentioning "interacting quantum fields".\n>>So, the Haag\'s theorem does not apply.\n>\n>\n> A corollary of this is that if you force the fields to have Lorentz\n> transformation properties then the Hamiltonian must be the free one. I don\'t\n> have a problem with this. If an interacting scalar field \\phi(x) is composed\n> from a free field \\phi_0(x) through (e.g.)\n>\n> \\phi(x) = \\phi_0(x) + \\int d^4x\'d^4x\'\' f(x-x\',x-x\'\') \\phi_0(x\')\\phi_0(x\'\')\n>\n> for some invariant function f - the kind of approach I advocate - then you\n> find that the free Hamiltonian works just as well as the time displacement\n> operator as the interacting one (just apply it directly & you will see). You\n> don\'t therefore need to build an interacting Hamiltonian at all.\n\nThat\'s not the way I would like to do things. In my view, fields (either\ninteracting or non-interacting) do not have any physical meaning. I\nhaven\'t seen any experiment measuring fields. I formulate my approach\nin terms of particles. I express the Hamiltonian (generator of time\ntranslations) through particle creation and annihilation operators. Once\nI have interacting Hamiltonian, I can do all kinds of physical\ncalculations: bound states, time evolution, S-matrix, etc. Fields\nare not needed.\n\n>\n>\n>>The more serious problem is related to renormalization. You are right\n>>that if we take usual QED Hamiltonian (e.g., (8.4.22) - (8.4.25) in\n>>Weinberg\'s book) and calculate elements of the S-matrix, we\'ll get\n>>infinite results. My idea is that this Hamiltonian H is simply wrong.\n>>It is wrong, but not completely wrong. It contain\'s a grain of truth.\n>>In order to find the correct Hamiltonian H\'\', one should perform\n>>two steps:\n>>\n>>1. The first step is renormalization. We add infinite (mass and\n>>charge renormalization) counterterms to H, so that in the process of\n>>calculation of the S-matrix all infinities nicely cancel each other\n>>and we get agreement with experiment to 10^{-11}%. I agree that\n>>resulting Hamiltonian H\' is infinite and ugly, but this is the only\n>>way to calculate the S-matrix we have so far. We swept infinities\n>>"under the rug", this rug is Hamiltonian. That\'s were QED stands today.\n>>\n>>2. The second step is "dressing transformation". That\'s what I suggest\n>>in my book. We can apply a unitary transformation U to the Hamiltonian\n>> H\'\n>>\n>>H\'\' = U H\' U^{-1}\n>>\n>>This transformation can be chosen so that\n>>1) the good S-matrix obtained in p. 1. is preserved\n>>2) H\'\' is finite.\n>>\n>>All infinities from H\' "transfer" to U. We swept infinities under\n>>another "rug". This time the rug is operator U. Since (in contrast\n>>to Hamiltonian) U has no physical meaning whatsoever, we can now\n>>throw this rug away and forget the problems with ultraviolet infinities:\n>>We have finite Hamiltonian H\'\' with interactions written in terms\n>>of particle creation and annihilation operators. We can choose H\'\',\n>>so that there are no more self-interaction effects in vacuum and\n>>one-particle subspaces. So, one can work with H\'\' just as we work\n>>with non-relativistic Hamiltonians (though, everything is relativistic,\n>>of course). The only difference is that H\'\' does not conserve the\n>>number of particles.\n>>\n>>The Hamiltonian H\'\' is rather complex. It has interactions in\n>>all perturbation orders starting from 2nd. It has Coulomb,\n>>magnetic, spin-orbit, etc. interactions between charged particles.\n>>It has interaction responsible for the Lamb\'s shifts in 4th and\n>>higher orders. It has interactions responsible for the Compton\n>>scattering, bremsstrahlung, pair creation, etc.\n>>\n>>It would be great if we could write operator H\'\' somehow from\n>>"general principles". Then we could avoid the two bad-smelling\n>>steps I wrote above. However, I simply don\'t know how it is possible.\n>>\n>>I think the original problem of QED dates way back to the end of 1920\'s\n>>when Dirac, Born, Heisenberg, and others guessed the form of QED\n>>Hamiltonian (from canonical quantization of Maxwell\'s theory).\n>>Their guess was fortunate and unfortunate at the same\n>>time. It was unfortunate that their interaction had self-interaction\n>>effects, and we pay now with all these tricks (regularization,\n>>renormalization, dressing, etc.) to get rid of these effects.\n>>It was fortunate that their guess contained a grain of truth,\n>>which gives us accurate S-matrix and a hope to finally find\n>>the correct Hamiltonian H\'\'.\n>\n>\n> It may be that no-one has found a "clean" way to construct this interacting\n> Hamiltonian simply because it cannot be done ...\n\nI have done that. You are probably right that the road to the\nHamiltonian H\'\' is somewhat untidy (moving infinities around\nis not a rigorous mathematical exercise). However, after H\'\'\nis obtained (I have clear recipes how to do that) we can forever\nforget such terms as "regularization", "renormalization",\n"ultraviolet divergences". We can just postulate H\'\' as the\ncorrect Hamiltonian of the theory. Then doing calculations in\nQED becomes not more difficult than in non-relativistic QM.\n\nEugene\nwww.geocities.com/meopemuk\n\n> indeed, it was the belief\n> that I personally could not do any better that led me to try something\n> completely different.\n>\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
>>The less serious problem is revealed by Haag's theorem. In a nutshell,
>>it says that interacting fields cannot have Lorentz transformation
>>properties. In my approach I do not have interacting fields at all.
>>The interaction in my Hamiltonian is written in terms of particle
>>creation and annihilation operators. From this Hamiltonian I can
>>(in principle) calculate S-matrix, bound states, time evolution,
>>whatever I like, without ever mentioning "interacting quantum fields".
>>So, the Haag's theorem does not apply.
>
>
> A corollary of this is that if you force the fields to have Lorentz
> transformation properties then the Hamiltonian must be the free one. I don't
> have a problem with this. If an interacting scalar field \phi(x) is composed
> from a free field \phi_0(x) through (e.g.)
>
> \phi(x) = \phi_0(x) + \int d^{4x}'d^{4x}'' f(x-x',x-x'') \phi_0(x')\phi_0(x'')
>
> for some invariant function f - the kind of approach I advocate - then you
> find that the free Hamiltonian works just as well as the time displacement
> operator as the interacting one (just apply it directly & you will see). You
> don't therefore need to build an interacting Hamiltonian at all.

That's not the way I would like to do things. In my view, fields (either
interacting or non-interacting) do not have any physical meaning. I
haven't seen any experiment measuring fields. I formulate my approach
in terms of particles. I express the Hamiltonian (generator of time
translations) through particle creation and annihilation operators. Once
I have interacting Hamiltonian, I can do all kinds of physical
calculations: bound states, time evolution, S-matrix, etc. Fields
are not needed.

>
>
>>The more serious problem is related to renormalization. You are right
>>that if we take usual QED Hamiltonian (e.g., (8.4.22) - (8.4.25) in
>>Weinberg's book) and calculate elements of the S-matrix, we'll get
>>infinite results. My idea is that this Hamiltonian H is simply wrong.
>>It is wrong, but not completely wrong. It contain's a grain of truth.
>>In order to find the correct Hamiltonian H'', one should perform
>>two steps:
>>
>>1. The first step is renormalization. We add infinite (mass and
>>charge renormalization) counterterms to H, so that in the process of
>>calculation of the S-matrix all infinities nicely cancel each other
>>and we get agreement with experiment to 10^{-11}%. I agree that
>>resulting Hamiltonian H' is infinite and ugly, but this is the only
>>way to calculate the S-matrix we have so far. We swept infinities
>>"under the rug", this rug is Hamiltonian. That's were QED stands today.
>>
>>2. The second step is "dressing transformation". That's what I suggest
>>in my book. We can apply a unitary transformation U to the Hamiltonian
>> H'
>>
>>H'' = U H' U^{-1}
>>
>>This transformation can be chosen so that
>>1) the good S-matrix obtained in p. 1. is preserved
>>2) H'' is finite.
>>
>>All infinities from H' "transfer" to U. We swept infinities under
>>another "rug". This time the rug is operator U. Since (in contrast
>>to Hamiltonian) U has no physical meaning whatsoever, we can now
>>throw this rug away and forget the problems with ultraviolet infinities:
>>We have finite Hamiltonian H'' with interactions written in terms
>>of particle creation and annihilation operators. We can choose H'',
>>so that there are no more self-interaction effects in vacuum and
>>one-particle subspaces. So, one can work with H'' just as we work
>>with non-relativistic Hamiltonians (though, everything is relativistic,
>>of course). The only difference is that H'' does not conserve the
>>number of particles.
>>
>>The Hamiltonian H'' is rather complex. It has interactions in
>>all perturbation orders starting from 2nd. It has Coulomb,
>>magnetic, spin-orbit, etc. interactions between charged particles.
>>It has interaction responsible for the Lamb's shifts in 4th and
>>higher orders. It has interactions responsible for the Compton
>>scattering, bremsstrahlung, pair creation, etc.
>>
>>It would be great if we could write operator H'' somehow from
>>"general principles". Then we could avoid the two bad-smelling
>>steps I wrote above. However, I simply don't know how it is possible.
>>
>>I think the original problem of QED dates way back to the end of 1920's
>>when Dirac, Born, Heisenberg, and others guessed the form of QED
>>Hamiltonian (from canonical quantization of Maxwell's theory).
>>Their guess was fortunate and unfortunate at the same
>>time. It was unfortunate that their interaction had self-interaction
>>effects, and we pay now with all these tricks (regularization,
>>renormalization, dressing, etc.) to get rid of these effects.
>>It was fortunate that their guess contained a grain of truth,
>>which gives us accurate S-matrix and a hope to finally find
>>the correct Hamiltonian H''.
>
>
> It may be that no-one has found a "clean" way to construct this interacting
> Hamiltonian simply because it cannot be done ...

I have done that. You are probably right that the road to the
Hamiltonian H'' is somewhat untidy (moving infinities around
is not a rigorous mathematical exercise). However, after H''
is obtained (I have clear recipes how to do that) we can forever
forget such terms as "regularization", "renormalization",
"ultraviolet divergences". We can just postulate H'' as the
correct Hamiltonian of the theory. Then doing calculations in
QED becomes not more difficult than in non-relativistic QM.

Eugene
www.geocities.com/meopemuk

> indeed, it was the belief
> that I personally could not do any better that led me to try something
> completely different.
>

Eugene Stefanovich

Oct27-04, 10:56 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\n\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>Another advantage of having finite well-behaving Hamiltonian is that\n>>now you can find bound states (both energies and wave functions)\n>>by simple diagonalization, and you can study time-dependent dynamics.\n>\n>\n> Yes, this is true; I like this about your approach.\n>\n> Unfortunately, you don\'t manage to carry out even\n> the simplest bound state calculation that would show 1-loop corrections\n> from the electromagnetic field. To be competitive with the traditional\n> approach using Bethe-Salpeter or Schwinger-Dyson equations for bound states,\n> you need to show that you can at least match their results.\n>\n> So there is more work ahead before getting recognition...\n>\n>\n> Arnold Neumaier\n\nI do have a proof that the\nS-matrix of my approach is exactly the same as in QED. This should\ncover the Lamb\'s shifts as well. However, I agree with you, I need to\nshow that explicitly. There are so many things to do and so little time.\n\nEugene\nwww.geocities.com/meopemuk\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>Another advantage of having finite well-behaving Hamiltonian is that
>>now you can find bound states (both energies and wave functions)
>>by simple diagonalization, and you can study time-dependent dynamics.
>
>
> Yes, this is true; I like this about your approach.
>
> Unfortunately, you don't manage to carry out even
> the simplest bound state calculation that would show 1-loop corrections
> from the electromagnetic field. To be competitive with the traditional
> approach using Bethe-Salpeter or Schwinger-Dyson equations for bound states,
> you need to show that you can at least match their results.
>
> So there is more work ahead before getting recognition...
>
>
> Arnold Neumaier

I do have a proof that the
S-matrix of my approach is exactly the same as in QED. This should
cover the Lamb's shifts as well. However, I agree with you, I need to
show that explicitly. There are so many things to do and so little time.

Eugene
www.geocities.com/meopemuk

Arnold Neumaier

Oct28-04, 01:28 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\nEugene Stefanovich wrote:\n\n> I do have a proof that the\n> S-matrix of my approach is exactly the same as in QED.\n\nWhere, without relying on intermedate infinities?\nThat is the question.\n\nJust showing that both reduce to the ill-defined Dyson series\nfor the Lagrangian with infinite counterterms, as I understand you did,\nis not enough.\n\nIt is well-known that a \'proof\' that has an intermediate step\namounting to \'inf minus inf\' can show two arbitrary different things\nto be the same.\n\n\nArnold Neumaier\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> I do have a proof that the
> S-matrix of my approach is exactly the same as in QED.

Where, without relying on intermedate infinities?
That is the question.

Just showing that both reduce to the ill-defined Dyson series
for the Lagrangian with infinite counterterms, as I understand you did,
is not enough.

It is well-known that a 'proof' that has an intermediate step
amounting to 'inf minus inf' can show two arbitrary different things
to be the same.

Arnold Neumaier

Eugene Stefanovich

Oct28-04, 01:29 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>Eugene Stefanovich wrote:\n>>>\n>>>\n>>>>(vi) Apply unitary dressing transformation U to the Hamiltonian H\',\n>>>> so that\n>>>>\n>>>>1) the new Hamiltonian\n>>>>\n>>>> H\'\' = U H\' U^{-1}\n>>>>\n>>>>is finite\n>>>>\n>>>>2) The S-matrices obtained with H\' and H\'\' are exactly the same.\n>>>>\n>>>>Now you can forget the nightmares of QFT, canonical quantization,\n>>>>Haags theorem, etc. You can use H\'\' in all usual quantum mechanical\n>>>>formulas (obtain time evolution operator, S-matrix, bound states via\n>>>>diagonalization, etc) without any need of regularization,\n>>>>renormalization, and other tricks.\n>>>\n>>>Well, this is a research program - the question is whether you can\n>>>actually carry it out. How do you construct U? It is there that all\n>>>the renormalization issues will recur. Since H\' is an ill-defined object\n>>>one must regularize it, then construct U for the regularized theory,\n>>>and then go to the limit and show that it really exists. This poses\n>>>the same problems as the original renormalization program.\n>>>\n>>\n>>You are absolutely right. That\'s one way. This is a long way, but it\n>>reaches the goal. There is, however, a shortcut: we can simply fit\n>>a finite interaction Hamiltonian H\'\' which yields the same (finite and\n>>accurate S-matrix of QED). The key condition on H\'\' is that interaction\n>>must be "physical" - that\'s my terminology for saying that operator\n>>yields zero when acting on vacuum and one-particle states. This\n>>guarantees that the theory with H\'\' does not have self-interaction\n>>effects.\n>\n>\n> I understand what you are doing - but you don\'t find in fact H\'\',\n> but you find a sequence of approximations H_j=U_j H\' U_j^{-1},\n> where the U_j and H\' is ill-defined but the H_j make sense to order j.\n> And there is no guarantee that the H_j will converge to the desired H\'\',\n> or that the infinite series that agrees with H_j to order j is more\n> than asymptotic.\n>\n\nI work under assumption (maybe incorrect?) that QED series for the\nS-operator is convergent. If there is no well-defined S-operator in\nrenormalized QED, then, of course, one cannot fit a well-defined\nHamiltonian, and my approach doesn\'t work. It would probably break\nafter some high perturbation order. If this is true, then the problem\nis not with my approach, but with the original QED Hamiltonian.\n\n>\n>\n>>You are probably right that QED is a sick theory: there is Landau pole,\n>>the perturbation series does not converge, etc. My approach does not\n>>cure all deseases. It achieves one important goal, though. It makes\n>>the Hamiltonian of QED finite and well-behaving.\n>\n>\n> Thus while traditional QFT renormalizes the S-matrix you renormalize the\n> Hamiltonian. This is a useful achievement, and may have important\n> consequences for bound state calculations.\n>\n> But in both cases, renormalization is needed to get rid of\n> ill-defined terms, and in both cases, the convergence questions remain\n> open.\n>\n\n\nI have not addressed the convergence question yet. I heard many times\nthat QED series for the S-matrix is asymptotic, but I have never seen\nthe proof. Could you recommend some reading on this issue? Thanks.\n\n>\n>>Few low order terms converge to extremely accurate QED\n>>results for electron\'s magnetic moment, Lamb\'s shifts, etc.\n>\n>\n> You did not show that you get the Lamb shift, only made it plausible.\n> Computing it and showing that to order O(e^2) you get the same value\n> as QFT would be an important step in convincing people that your\n> theory is indeed equivalent to the traditional approach.\n\nI agree, more work needs to be done,\nthough I believe the Lamb shift appears in the 4th order, so\nit is proportional to e^4.\n\n\n>\n>\n>\n>>This is guaranteed, because my approach does not change the S-matrix by\n>>construction.\n>\n>\n> Well, the initial theory is ill-defined and you make it defined in\n> a different way than tradition. This does not mean that one knows that\n> both give the same results. In principle one can get from inf-inf\n> different finite answers by interpreting it in different ways.\n>\n> To be convincing, one needs an argument showing that the two finite,\n> well-defined versions give the same results by relating one to the\n> other in terms of well-defined operations or limits.\n\nAgreed.\n\n>\n>\n>\n>>Another advantage of having finite well-behaving Hamiltonian is that\n>>now you can find bound states (both energies and wave functions)\n>>by simple diagonalization, and you can study time-dependent dynamics.\n>\n>\n> On the other hand, a disatvantage is that all calculations are done\n> noncovariantly, which makes them somewhat messy...\n> In spite of that, I think that the dressing approach is useful.\n> If your Hamiltonian QED is shown equivalent with QED (which it is on a\n> formal level, but this is not necessarily inherited on the renormalized\n> level), it would give QED a dynamical content that is difficult to discern\n> in the traditional formulation.\n>\n> How does your work go beyond that of Shebeko and Shirokov in\n> nucl-th/0102037 ?\n\nI am in contact with Shebeko and Shirokov. I met them personally in\nDubna in 2002. There are few differences in our approach to dressing:\n\n1. They prefer to talk about "dressing" of particles, but I don\'t change\nthe definition of particles and modify the Hamiltonian. These two\napproaches are philosophically different, but numerically equivalent\n(see subsection 12.1.8 in my book)\n\n2. I think my biggest achievement is the (formal) proof that elimination\nof infinities from the Hamiltonian works in each perturbation order.\n\nI noticed from your posts that you are well informed in current\nliterature.\nFor example, you often mentioned works by Coester, Klink, and Polyzou.\nThey are an important source of inspiration for my work as well.\n\n>\n> Why don\'t you put your papers on the arxiv?\n> It is never too late, and you reach more people.\n\nI think it\'s a good idea, though I have a couple of questions:\n\n1. I am not affiliated with any academic institution, so direct\nuploading doesn\'t work.\n2. Eventually, I may want to publish my book. I think posting\non the arxiv will complicate the publication.\n\nThanks for your good comments.\nEugene Stefanovich.\n\n>\n>\n> Arnold Neumaier\n>\n>\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>Eugene Stefanovich wrote:
>>>
>>>
>>>>(vi) Apply unitary dressing transformation U to the Hamiltonian H',
>>>> so that
>>>>
>>>>1) the new Hamiltonian
>>>>
>>>> H'' = U H' U^{-1}
>>>>
>>>>is finite
>>>>
>>>>2) The S-matrices obtained with H' and H'' are exactly the same.
>>>>
>>>>Now you can forget the nightmares of QFT, canonical quantization,
>>>>Haags theorem, etc. You can use H'' in all usual quantum mechanical
>>>>formulas (obtain time evolution operator, S-matrix, bound states via
>>>>diagonalization, etc) without any need of regularization,
>>>>renormalization, and other tricks.
>>>
>>>Well, this is a research program - the question is whether you can
>>>actually carry it out. How do you construct U? It is there that all
>>>the renormalization issues will recur. Since H' is an ill-defined object
>>>one must regularize it, then construct U for the regularized theory,
>>>and then go to the limit and show that it really exists. This poses
>>>the same problems as the original renormalization program.
>>>
>>
>>You are absolutely right. That's one way. This is a long way, but it
>>reaches the goal. There is, however, a shortcut: we can simply fit
>>a finite interaction Hamiltonian H'' which yields the same (finite and
>>accurate S-matrix of QED). The key condition on H'' is that interaction
>>must be "physical" - that's my terminology for saying that operator
>>yields zero when acting on vacuum and one-particle states. This
>>guarantees that the theory with H'' does not have self-interaction
>>effects.
>
>
> I understand what you are doing - but you don't find in fact H'',
> but you find a sequence of approximations H_j=U_j H' U_j^{-1},
> where the U_j and H' is ill-defined but the H_j make sense to order j.
> And there is no guarantee that the H_j will converge to the desired H'',
> or that the infinite series that agrees with H_j to order j is more
> than asymptotic.
>

I work under assumption (maybe incorrect?) that QED series for the
S-operator is convergent. If there is no well-defined S-operator in
renormalized QED, then, of course, one cannot fit a well-defined
Hamiltonian, and my approach doesn't work. It would probably break
after some high perturbation order. If this is true, then the problem
is not with my approach, but with the original QED Hamiltonian.

>
>
>>You are probably right that QED is a sick theory: there is Landau pole,
>>the perturbation series does not converge, etc. My approach does not
>>cure all deseases. It achieves one important goal, though. It makes
>>the Hamiltonian of QED finite and well-behaving.
>
>
> Thus while traditional QFT renormalizes the S-matrix you renormalize the
> Hamiltonian. This is a useful achievement, and may have important
> consequences for bound state calculations.
>
> But in both cases, renormalization is needed to get rid of
> ill-defined terms, and in both cases, the convergence questions remain
> open.
>

I have not addressed the convergence question yet. I heard many times
that QED series for the S-matrix is asymptotic, but I have never seen
the proof. Could you recommend some reading on this issue? Thanks.

>
>>Few low order terms converge to extremely accurate QED
>>results for electron's magnetic moment, Lamb's shifts, etc.
>
>
> You did not show that you get the Lamb shift, only made it plausible.
> Computing it and showing that to order O(e^2) you get the same value
> as QFT would be an important step in convincing people that your
> theory is indeed equivalent to the traditional approach.

I agree, more work needs to be done,
though I believe the Lamb shift appears in the 4th order, so
it is proportional to e^4.

>
>
>
>>This is guaranteed, because my approach does not change the S-matrix by
>>construction.
>
>
> Well, the initial theory is ill-defined and you make it defined in
> a different way than tradition. This does not mean that one knows that
> both give the same results. In principle one can get from inf-inf
> different finite answers by interpreting it in different ways.
>
> To be convincing, one needs an argument showing that the two finite,
> well-defined versions give the same results by relating one to the
> other in terms of well-defined operations or limits.

Agreed.

>
>
>
>>Another advantage of having finite well-behaving Hamiltonian is that
>>now you can find bound states (both energies and wave functions)
>>by simple diagonalization, and you can study time-dependent dynamics.
>
>
> On the other hand, a disatvantage is that all calculations are done
> noncovariantly, which makes them somewhat messy...
> In spite of that, I think that the dressing approach is useful.
> If your Hamiltonian QED is shown equivalent with QED (which it is on a
> formal level, but this is not necessarily inherited on the renormalized
> level), it would give QED a dynamical content that is difficult to discern
> in the traditional formulation.
>
> How does your work go beyond that of Shebeko and Shirokov in
> http://www.arxiv.org/abs/nucl-th/0102037 ?

I am in contact with Shebeko and Shirokov. I met them personally in
Dubna in 2002. There are few differences in our approach to dressing:

1. They prefer to talk about "dressing" of particles, but I don't change
the definition of particles and modify the Hamiltonian. These two
approaches are philosophically different, but numerically equivalent
(see subsection 12.1.8 in my book)

2. I think my biggest achievement is the (formal) proof that elimination
of infinities from the Hamiltonian works in each perturbation order.

I noticed from your posts that you are well informed in current
literature.
For example, you often mentioned works by Coester, Klink, and Polyzou.
They are an important source of inspiration for my work as well.

>
> Why don't you put your papers on the arxiv?
> It is never too late, and you reach more people.

I think it's a good idea, though I have a couple of questions:

1. I am not affiliated with any academic institution, so direct
uploading doesn't work.
2. Eventually, I may want to publish my book. I think posting
on the arxiv will complicate the publication.

Thanks for your good comments.
Eugene Stefanovich.

>
>
> Arnold Neumaier
>
>

Eugene Stefanovich

Oct29-04, 02:07 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>I do have a proof that the\n>>S-matrix of my approach is exactly the same as in QED.\n>\n>\n> Where, without relying on intermedate infinities?\n> That is the question.\n>\n> Just showing that both reduce to the ill-defined Dyson series\n> for the Lagrangian with infinite counterterms, as I understand you did,\n> is not enough.\n>\n> It is well-known that a \'proof\' that has an intermediate step\n> amounting to \'inf minus inf\' can show two arbitrary different things\n> to be the same.\n>\n>\n> Arnold Neumaier\n>\n>\n>\nOne way to proceed without intermediate infinities is the following:\n\n1. Take usual QED Hamiltonian\n\n2. Perform regularization. This can be simple momentum cutoff (\\Lambda)\nin loop integrals, or you can choose to use Kita-Shirokov\nregularization\nwhich preserves the relativistic invariance:\n\nM.I. Shirokov "On relativistic nonlocal quantum field theory"\nInt. J. Theor. Phys. 41 (2002), 1027\n\nH. Kita, "A non-trivial example of a relativistic\nquantum theory of particles without divergence difficulties",\nProgr. Theor. Phys. 35 (1966), 934.\n\n3. Introduce mass and charge renormalization counterterms in the\nHamiltonian. Everything is finite so far. But the S-matrix\nS(\\Lambda) is not accurate yet.\n\n[One way to get the accurate S-matrix is to follow the\nusual recipe - remove the regularization (e.g., set the cutoff momentum\nto infinity). Then you get the S-matrix S(inf) as a sum of finite terms.\nJust few low-order terms are enough to get astonishing agreement with\nexperiment. I have no idea about the convergence of the series for the\nS-matrix. I am not going to discuss it here. I assume that this S-matrix\nis the best we can get. The problem with this approach is that the\nHamiltonian H(inf) is infinite. To solve this problem I suggest step 4.]\n\n4. Perform unitary "dressing" transformation H(\\Lambda) -> H\'(\\Lambda)\nwhich removes from\nH(\\Lambda) "renorm" and "unphysical" terms and makes "physical"\nterms in H\'(\\Lambda) rapidly decaying outside the energy shell which\nguarantees convergence of loop integrals even after regularization\nis removed. This "dressing" transformation, in addition, does not\nchange the S-matrix S(\\Lambda)\n\n5. Now we can remove the regularization (e.g., set the cutoff momentum\nto infinity). In the limit we get the same S-matrix S(inf) as\ntraditional QED. We also get the new finite Hamiltonian H\'(inf).\n\nOne can look at this problem from another angle as well.\nRenormalized QED gives us S-operator which is very accurate\n(again, I am not discussing here the convergence of the perturbation\nseries). I do not\ncare much how this operator was obtained. The calculation of S involved\nvarious tricks, like regularization, renormalization, cancellation of\ninfinities, etc. But at the end we got S-operator which describes\nscattering events and bound states with amazing accuracy. In spite of\nthe bad smell of the derivation procedure, let us assume that the\nresult is correct and S is the exact S-operator fo charged particles and\nphotons. Then we can just\nfit a "physical" finite Hamiltonian H\' which produces this\nS-operator upon substitution to the usual perturbation theory formula\nexpressing S through H\' (this could be Dyson formula, or old-fashioned\nformula, or Magnus formula, it doesn\'t matter much). This is what I did\nin section 12.1. of the book.\n\nEugene.\nwww.geocities.com/meopemuk\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>I do have a proof that the
>>S-matrix of my approach is exactly the same as in QED.
>
>
> Where, without relying on intermedate infinities?
> That is the question.
>
> Just showing that both reduce to the ill-defined Dyson series
> for the Lagrangian with infinite counterterms, as I understand you did,
> is not enough.
>
> It is well-known that a 'proof' that has an intermediate step
> amounting to 'inf minus inf' can show two arbitrary different things
> to be the same.
>
>
> Arnold Neumaier
>
>
>
One way to proceed without intermediate infinities is the following:

1. Take usual QED Hamiltonian

2. Perform regularization. This can be simple momentum cutoff (\Lambda)
in loop integrals, or you can choose to use Kita-Shirokov
regularization
which preserves the relativistic invariance:

M.I. Shirokov "On relativistic nonlocal quantum field theory"
\Int. J. Theor. Phys. 41 (2002), 1027

H. Kita, "A non-trivial example of a relativistic
quantum theory of particles without divergence difficulties",
Progr. Theor. Phys. 35 (1966), 934.

3. Introduce mass and charge renormalization counterterms in the
Hamiltonian. Everything is finite so far. But the S-matrix
S(\Lambda) is not accurate yet.

[One way to get the accurate S-matrix is to follow the
usual recipe - remove the regularization (e.g., set the cutoff momentum
to infinity). Then you get the S-matrix S(inf) as a sum of finite terms.
Just few low-order terms are enough to get astonishing agreement with
experiment. I have no idea about the convergence of the series for the
S-matrix. I am not going to discuss it here. I assume that this S-matrix
is the best we can get. The problem with this approach is that the
Hamiltonian H(inf) is infinite. To solve this problem I suggest step 4.]

4. Perform unitary "dressing" transformation H(\Lambda) -> H'(\Lambda)
which removes from
H(\Lambda) "renorm" and "unphysical" terms and makes "physical"
terms in H'(\Lambda) rapidly decaying outside the energy shell which
guarantees convergence of loop integrals even after regularization
is removed. This "dressing" transformation, in addition, does not
change the S-matrix S(\Lambda)

5. Now we can remove the regularization (e.g., set the cutoff momentum
to infinity). In the limit we get the same S-matrix S(inf) as
traditional QED. We also get the new finite Hamiltonian H'(inf).

One can look at this problem from another angle as well.
Renormalized QED gives us S-operator which is very accurate
(again, I am not discussing here the convergence of the perturbation
series). I do not
care much how this operator was obtained. The calculation of S involved
various tricks, like regularization, renormalization, cancellation of
infinities, etc. But at the end we got S-operator which describes
scattering events and bound states with amazing accuracy. In spite of
the bad smell of the derivation procedure, let us assume that the
result is correct and S is the exact S-operator fo charged particles and
photons. Then we can just
fit a "physical" finite Hamiltonian H' which produces this
S-operator upon substitution to the usual perturbation theory formula
expressing S through H' (this could be Dyson formula, or old-fashioned
formula, or Magnus formula, it doesn't matter much). This is what I did
in section 12.1. of the book.

Eugene.
www.geocities.com/meopemuk

Arnold Neumaier

Oct29-04, 02:07 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\nEugene Stefanovich wrote:\n\n> I have not addressed the convergence question yet. I heard many times\n> that QED series for the S-matrix is asymptotic, but I have never seen\n> the proof. Could you recommend some reading on this issue? Thanks.\n\nThere is no proof, but some arguments. See my theoretical physics FAQ at\nhttp://www.mat.univie.ac.at/~neum/physics-faq.txt\nunder \'Summing divergent series\'.\n\n\n> 1. I am not affiliated with any academic institution, so direct\n> uploading doesn\'t work.\n\nI\'d endorse those of your papers that concentrate on your math\n(which is interesting) rather than your philosophy (which is speculative).\n\n\nArnold Neumaier\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> I have not addressed the convergence question yet. I heard many times
> that QED series for the S-matrix is asymptotic, but I have never seen
> the proof. Could you recommend some reading on this issue? Thanks.

There is no proof, but some arguments. See my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
under 'Summing divergent series'.

> 1. I am not affiliated with any academic institution, so direct
> uploading doesn't work.

I'd endorse those of your papers that concentrate on your math
(which is interesting) rather than your philosophy (which is speculative).

Arnold Neumaier

Arnold Neumaier

Nov2-04, 12:38 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\n\nEugene Stefanovich wrote:\n>\n> One way to proceed without intermediate infinities is the following:\n>\n> 1. Take usual QED Hamiltonian\n>\n> 2. Perform regularization. This can be simple momentum cutoff (\\Lambda)\n> in loop integrals, or you can choose to use Kita-Shirokov\n> regularization\n> which preserves the relativistic invariance:\n>\n> M.I. Shirokov "On relativistic nonlocal quantum field theory"\n> Int. J. Theor. Phys. 41 (2002), 1027\n\nPlease point me to the page where they do regularization.\n\n\n> 3. Introduce mass and charge renormalization counterterms in the\n> Hamiltonian. Everything is finite so far. But the S-matrix\n> S(\\Lambda) is not accurate yet.\n>\n> [One way to get the accurate S-matrix is to follow the\n> usual recipe - remove the regularization (e.g., set the cutoff momentum\n> to infinity). Then you get the S-matrix S(inf) as a sum of finite terms.\n> Just few low-order terms are enough to get astonishing agreement with\n> experiment. I have no idea about the convergence of the series for the\n> S-matrix. I am not going to discuss it here. I assume that this S-matrix\n> is the best we can get. The problem with this approach is that the\n> Hamiltonian H(inf) is infinite. To solve this problem I suggest step 4.]\n>\n> 4. Perform unitary "dressing" transformation H(\\Lambda) -> H\'(\\Lambda)\n> which removes from\n> H(\\Lambda) "renorm" and "unphysical" terms and makes "physical"\n> terms in H\'(\\Lambda) rapidly decaying outside the energy shell which\n> guarantees convergence of loop integrals even after regularization\n> is removed. This "dressing" transformation, in addition, does not\n> change the S-matrix S(\\Lambda)\n>\n> 5. Now we can remove the regularization (e.g., set the cutoff momentum\n> to infinity). In the limit we get the same S-matrix S(inf) as\n> traditional QED. We also get the new finite Hamiltonian H\'(inf).\n\nOK; this sounds good. So it looks like you found a finite, dynamical\nformulation of QED at any finite loop order. It would be very interesting\nif you could prove convergence of your procedure as the number of loops\ngets large. Perhaps first for phi^4 theory, which should be simpler\nin this respect...\n\n\nArnold Neumaier\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> One way to proceed without intermediate infinities is the following:
>
> 1. Take usual QED Hamiltonian
>
> 2. Perform regularization. This can be simple momentum cutoff (\Lambda)
> in loop integrals, or you can choose to use Kita-Shirokov
> regularization
> which preserves the relativistic invariance:
>
> M.I. Shirokov "On relativistic nonlocal quantum field theory"
> \Int. J. Theor. Phys. 41 (2002), 1027

Please point me to the page where they do regularization.

> 3. Introduce mass and charge renormalization counterterms in the
> Hamiltonian. Everything is finite so far. But the S-matrix
> S(\Lambda) is not accurate yet.
>
> [One way to get the accurate S-matrix is to follow the
> usual recipe - remove the regularization (e.g., set the cutoff momentum
> to infinity). Then you get the S-matrix S(inf) as a sum of finite terms.
> Just few low-order terms are enough to get astonishing agreement with
> experiment. I have no idea about the convergence of the series for the
> S-matrix. I am not going to discuss it here. I assume that this S-matrix
> is the best we can get. The problem with this approach is that the
> Hamiltonian H(inf) is infinite. To solve this problem I suggest step 4.]
>
> 4. Perform unitary "dressing" transformation H(\Lambda) -> H'(\Lambda)
> which removes from
> H(\Lambda) "renorm" and "unphysical" terms and makes "physical"
> terms in H'(\Lambda) rapidly decaying outside the energy shell which
> guarantees convergence of loop integrals even after regularization
> is removed. This "dressing" transformation, in addition, does not
> change the S-matrix S(\Lambda)
>
> 5. Now we can remove the regularization (e.g., set the cutoff momentum
> to infinity). In the limit we get the same S-matrix S(inf) as
> traditional QED. We also get the new finite Hamiltonian H'(inf).

OK; this sounds good. So it looks like you found a finite, dynamical
formulation of QED at any finite loop order. It would be very interesting
if you could prove convergence of your procedure as the number of loops
gets large. Perhaps first for \phi^4 theory, which should be simpler
in this respect...

Arnold Neumaier

Chris Oakley

Nov3-04, 09:38 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>> Chris Oakley wrote:\n> >>The less serious problem is revealed by Haag\'s theorem. In a nutshell,\n> >>it says that interacting fields cannot have Lorentz transformation\n> >>properties. In my approach I do not have interacting fields at all.\n> >>The interaction in my Hamiltonian is written in terms of particle\n> >>creation and annihilation operators. From this Hamiltonian I can\n> >>(in principle) calculate S-matrix, bound states, time evolution,\n> >>whatever I like, without ever mentioning "interacting quantum fields".\n> >>So, the Haag\'s theorem does not apply.\n> >\n> >\n> > A corollary of this is that if you force the fields to have Lorentz\n> > transformation properties then the Hamiltonian must be the free one. I\ndon\'t\n> > have a problem with this. If an interacting scalar field \\phi(x) is\ncomposed\n> > from a free field \\phi_0(x) through (e.g.)\n> >\n> > \\phi(x) = \\phi_0(x) + \\int d^4x\'d^4x\'\' f(x-x\',x-x\'\')\n\\phi_0(x\')\\phi_0(x\'\')\n> >\n> > for some invariant function f - the kind of approach I advocate - then\nyou\n> > find that the free Hamiltonian works just as well as the time\ndisplacement\n> > operator as the interacting one (just apply it directly & you will see).\nYou\n> > don\'t therefore need to build an interacting Hamiltonian at all.\n>\n> That\'s not the way I would like to do things. In my view, fields (either\n> interacting or non-interacting) do not have any physical meaning. I\n> haven\'t seen any experiment measuring fields. I formulate my approach\n> in terms of particles. I express the Hamiltonian (generator of time\n> translations) through particle creation and annihilation operators. Once\n> I have interacting Hamiltonian, I can do all kinds of physical\n> calculations: bound states, time evolution, S-matrix, etc. Fields\n> are not needed.\n\nYour creation/annihilation operators have definite 3-momentum, so (e.g.)\n\n[P_i, a^\\dagger({\\bf p})] = p_i a^\\dagger({\\bf p})\n\nIt follows from this that the 3D Fourier transform \\phi satisfies\n\n[P_i, \\phi({\\bf x})] = -i\\partial_i \\phi({\\bf x})\n\n.... as required for an operator associated with a particular point in space.\nSo if you have got creation/annihilation operators, you can also get field\noperators, whether you want them or not.\n\nWhilst I am not arguing that operators of definite 3-momentum are generally\nmore useful in doing calculations than operators associated with a point in\nspace, if you choose to be explicitly relativistic you have no choice simply\nbecause you need time to be explicit in your matrix elements, and in a\nrelativistic environment, that necessitates space being explicit as well. So\ntwo particle elastic scattering is going to involve a matrix element\n\n<0|\\phi(x^1)\\phi(x^2)\\phi(x^3)\\phi( x^4)|0>\n\nwith x^1_0 = x^2_0 = t (the final state time) & x^3_0 = x^4_0 = 0 (the\ninitial state time)\n\n.... even though a more useful representation is obtained by a F.T. of the\nspace components.\n\n<0|\\phi(t;{bf q}_1)\\phi(t;{bf q}_2)\\phi(0;{bf p}_1)\\phi(0;{bf p}_1)|0>\n\nSee for example http://www.cgoakley.demon.co.uk/qft/dipe.pdf - section 6.\n\nAs far as I know, all the attempts to build a theory by using\nannihilation/creation operators that work at a particular time and which are\ntime-evolved using a Hamiltonian expressed in terms of these do not work\nbecause of loops. Your approach may have tidied certain things up, but it\nstill has infinite subtractions.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>> Chris Oakley wrote:
> >>The less serious problem is revealed by Haag's theorem. In a nutshell,
> >>it says that interacting fields cannot have Lorentz transformation
> >>properties. In my approach I do not have interacting fields at all.
> >>The interaction in my Hamiltonian is written in terms of particle
> >>creation and annihilation operators. From this Hamiltonian I can
> >>(in principle) calculate S-matrix, bound states, time evolution,
> >>whatever I like, without ever mentioning "interacting quantum fields".
> >>So, the Haag's theorem does not apply.
> >
> >
> > A corollary of this is that if you force the fields to have Lorentz
> > transformation properties then the Hamiltonian must be the free one. I
don't
> > have a problem with this. If an interacting scalar field \phi(x) is
composed
> > from a free field \phi_0(x) through (e.g.)
> >
> > \phi(x) = \phi_0(x) + \int d^{4x}'d^{4x}'' f(x-x',x-x'')\phi_0(x')\phi_0(x'')
> >
> > for some invariant function f - the kind of approach I advocate - then
you
> > find that the free Hamiltonian works just as well as the time
displacement
> > operator as the interacting one (just apply it directly & you will see).
You
> > don't therefore need to build an interacting Hamiltonian at all.
>
> That's not the way I would like to do things. In my view, fields (either
> interacting or non-interacting) do not have any physical meaning. I
> haven't seen any experiment measuring fields. I formulate my approach
> in terms of particles. I express the Hamiltonian (generator of time
> translations) through particle creation and annihilation operators. Once
> I have interacting Hamiltonian, I can do all kinds of physical
> calculations: bound states, time evolution, S-matrix, etc. Fields
> are not needed.

Your creation/annihilation operators have definite 3-momentum, so (e.g.)

[P_i, a^\dagger({\bf p})] = p_i a^\dagger({\bf p})

It follows from this that the 3D Fourier transform \phi satisfies

[P_i, \phi({\bf x})] = -i\partial_i \phi({\bf x})

.... as required for an operator associated with a particular point in space.
So if you have got creation/annihilation operators, you can also get field
operators, whether you want them or not.

Whilst I am not arguing that operators of definite 3-momentum are generally
more useful in doing calculations than operators associated with a point in
space, if you choose to be explicitly relativistic you have no choice simply
because you need time to be explicit in your matrix elements, and in a
relativistic environment, that necessitates space being explicit as well. So
two particle elastic scattering is going to involve a matrix element

<0|\phi(x^1)\phi(x^2)\phi(x^3)\phi(x^4)|0>

with x^{1_0} = x^{2_0} = t (the final state time) & x^{3_0} = x^{4_0} = (the
initial state time)

.... even though a more useful representation is obtained by a F.T. of the
space components.

<0|\phi(t;{bf q}_1)\phi(t;{bf q}_2)\phi(0;{bf p}_1)\phi(0;{bf p}_1)|0>

See for example http://www.cgoakley.demon.co.uk/qft/dipe.pdf - section 6.

As far as I know, all the attempts to build a theory by using
annihilation/creation operators that work at a particular time and which are
time-evolved using a Hamiltonian expressed in terms of these do not work
because of loops. Your approach may have tidied certain things up, but it
still has infinite subtractions.

Eugene Stefanovich

Nov3-04, 09:38 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Chris Oakley wrote:\n>>Locality is not required for relativistic invariance. One can build a\n>>non-local (and finite) quantum field theory which is\n>>relativistically invariant\n>>(Poincare commutation relations are satisfied). See\n>>\n>>M.I. Shirokov "On relativistic nonlocal quantum field theory"\n>>Int. J. Theor. Phys. 41 (2002), 1027\n>>\n>>H. Kita, "A non-trivial example of a relativistic\n>>quantum theory of particles without divergence difficulties",\n>>Progr. Theor. Phys. 35 (1966), 934.\n>\n>\n> I\'ll chase these up ... thank-you. The first one does not seem to have been\n> posted on arXiv (I would have seen it if it had).\n>\n>\n>>I don\'t think, however, that this is the way to go in\n>>formulating a consistent approach to QED. "Dressing transformation"\n>>is the way.\n>\n>\n> ... and if you manage to (i) make it covariant through and through & (ii)\n> have no infinite subtractions I will gladly follow.\n>\n\nOne way to achieve that, which may satisfy you, could be:\n1. Apply Kita-Shirokov regularization (this preserves Poincare\ninvariance of the regularized theory)\n2. Apply usual renormalization approach (add mass and charge\nrenormalization counterterms to the Hamiltonian) and obtain\nfinite S-operator\n3. Apply "dressing" transformation which removes "bad" or\n"unphysical" terms from the Hamiltonian while preserving\nthe S-operator (see my paper in Ann. Phys. or\nShebeko-Shirokov review)\n4. Remove regularization. As a result you\'ll obtain a finite\ntheory (both Hamiltonian and S-operator are finite)\nwithout self-interaction.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
>>Locality is not required for relativistic invariance. One can build a
>>non-local (and finite) quantum field theory which is
>>relativistically invariant
>>(Poincare commutation relations are satisfied). See
>>
>>M.I. Shirokov "On relativistic nonlocal quantum field theory"
>>\Int. J. Theor. Phys. 41 (2002), 1027
>>
>>H. Kita, "A non-trivial example of a relativistic
>>quantum theory of particles without divergence difficulties",
>>Progr. Theor. Phys. 35 (1966), 934.
>
>
> I'll chase these up ... thank-you. The first one does not seem to have been
> posted on arXiv (I would have seen it if it had).
>
>
>>I don't think, however, that this is the way to go in
>>formulating a consistent approach to QED. "Dressing transformation"
>>is the way.
>
>
> ... and if you manage to (i) make it covariant through and through & (ii)
> have no infinite subtractions I will gladly follow.
>

One way to achieve that, which may satisfy you, could be:
1. Apply Kita-Shirokov regularization (this preserves Poincare
invariance of the regularized theory)
2. Apply usual renormalization approach (add mass and charge
renormalization counterterms to the Hamiltonian) and obtain
finite S-operator
3. Apply "dressing" transformation which removes "bad" or
"unphysical" terms from the Hamiltonian while preserving
the S-operator (see my paper in Ann. Phys. or
Shebeko-Shirokov review)
4. Remove regularization. As a result you'll obtain a finite
theory (both Hamiltonian and S-operator are finite)
without self-interaction.

Eugene.

Arnold Neumaier

Nov3-04, 09:43 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> That\'s not the way I would like to do things. In my view, fields (either\n> interacting or non-interacting) do not have any physical meaning. I\n> haven\'t seen any experiment measuring fields.\n\nPeople measure the electromagnetic field routinely in daily life.\nIt is the basis of all our modern technical culture. So at least\nthe electromagnetic part of QED cannot be discussed away...\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> That's not the way I would like to do things. In my view, fields (either
> interacting or non-interacting) do not have any physical meaning. I
> haven't seen any experiment measuring fields.

People measure the electromagnetic field routinely in daily life.
It is the basis of all our modern technical culture. So at least
the electromagnetic part of QED cannot be discussed away...

Arnold Neumaier

Eugene Stefanovich

Nov3-04, 09:59 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>I have not addressed the convergence question yet. I heard many times\n>>that QED series for the S-matrix is asymptotic, but I have never seen\n>>the proof. Could you recommend some reading on this issue? Thanks.\n>\n>\n> There is no proof, but some arguments. See my theoretical physics FAQ at\n> http://www.mat.univie.ac.at/~neum/physics-faq.txt\n> under \'Summing divergent series\'.\n\nThanks.\n>\n>\n>\n>>1. I am not affiliated with any academic institution, so direct\n>> uploading doesn\'t work.\n>\n>\n> I\'d endorse those of your papers that concentrate on your math\n> (which is interesting) rather than your philosophy (which is speculative).\n>\n>\n\nI am ready to discuss (and defend) both math and phil aspects of my work.\nBest regards.\nEugene.\n\n> Arnold Neumaier\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>I have not addressed the convergence question yet. I heard many times
>>that QED series for the S-matrix is asymptotic, but I have never seen
>>the proof. Could you recommend some reading on this issue? Thanks.
>
>
> There is no proof, but some arguments. See my theoretical physics FAQ at
> http://www.mat.univie.ac.at/~neum/physics-faq.txt
> under 'Summing divergent series'.

Thanks.
>
>
>
>>1. I am not affiliated with any academic institution, so direct
>> uploading doesn't work.
>
>
> I'd endorse those of your papers that concentrate on your math
> (which is interesting) rather than your philosophy (which is speculative).
>
>

I am ready to discuss (and defend) both math and phil aspects of my work.
Best regards.
Eugene.

> Arnold Neumaier

Eugene Stefanovich

Nov3-04, 10:07 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>One way to proceed without intermediate infinities is the following:\n>>\n>>1. Take usual QED Hamiltonian\n>>\n>>2. Perform regularization. This can be simple momentum cutoff (\\Lambda)\n>> in loop integrals, or you can choose to use Kita-Shirokov\n>> regularization\n>> which preserves the relativistic invariance:\n>>\n>>M.I. Shirokov "On relativistic nonlocal quantum field theory"\n>>Int. J. Theor. Phys. 41 (2002), 1027\n>\n>\n> Please point me to the page where they do regularization.\n\nHe doesn\'t write specifically about regularization. He shows how\none can introduce "relativistic formfactors" in usual local\ntrilinear interactions, so that resulting theory remains Poincare\ninvariant and becomes finite. There is a very small step from this\nobservation to regularization. In my copy of this paper\n(I only have a preprint version, but I believe the journal version\nis close) I can read at the end of CONCLUSION:\n\n"This example shows that this nonlocal theory can be used as a way\nof relativistic regularization of the local one."\n\n>\n>\n>\n>>3. Introduce mass and charge renormalization counterterms in the\n>>Hamiltonian. Everything is finite so far. But the S-matrix\n>>S(\\Lambda) is not accurate yet.\n>>\n>>[One way to get the accurate S-matrix is to follow the\n>>usual recipe - remove the regularization (e.g., set the cutoff momentum\n>>to infinity). Then you get the S-matrix S(inf) as a sum of finite terms.\n>>Just few low-order terms are enough to get astonishing agreement with\n>>experiment. I have no idea about the convergence of the series for the\n>>S-matrix. I am not going to discuss it here. I assume that this S-matrix\n>>is the best we can get. The problem with this approach is that the\n>>Hamiltonian H(inf) is infinite. To solve this problem I suggest step 4.]\n>>\n>>4. Perform unitary "dressing" transformation H(\\Lambda) -> H\'(\\Lambda)\n>> which removes from\n>> H(\\Lambda) "renorm" and "unphysical" terms and makes "physical"\n>> terms in H\'(\\Lambda) rapidly decaying outside the energy shell which\n>> guarantees convergence of loop integrals even after regularization\n>> is removed. This "dressing" transformation, in addition, does not\n>> change the S-matrix S(\\Lambda)\n>>\n>>5. Now we can remove the regularization (e.g., set the cutoff momentum\n>> to infinity). In the limit we get the same S-matrix S(inf) as\n>> traditional QED. We also get the new finite Hamiltonian H\'(inf).\n>\n>\n> OK; this sounds good. So it looks like you found a finite, dynamical\n> formulation of QED at any finite loop order. It would be very interesting\n> if you could prove convergence of your procedure as the number of loops\n> gets large. Perhaps first for phi^4 theory, which should be simpler\n> in this respect...\n\nSorry, I don\'t have an explicit proof at the moment.\nThe only thing I can say\nis that convergence properties of my approach are exactly the same as\nof the usual renormalization approach: If usual perturbation series for\nthe S-matrix converges, then my perturbation series converges too.\nThey are guaranteed to be the same in each perturbation order.\n\nEugene.\n\n>\n>\n> Arnold Neumaier\n>\n>\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>One way to proceed without intermediate infinities is the following:
>>
>>1. Take usual QED Hamiltonian
>>
>>2. Perform regularization. This can be simple momentum cutoff (\Lambda)
>> in loop integrals, or you can choose to use Kita-Shirokov
>> regularization
>> which preserves the relativistic invariance:
>>
>>M.I. Shirokov "On relativistic nonlocal quantum field theory"
>>\Int. J. Theor. Phys. 41 (2002), 1027
>
>
> Please point me to the page where they do regularization.

He doesn't write specifically about regularization. He shows how
one can introduce "relativistic formfactors" in usual local
trilinear interactions, so that resulting theory remains Poincare
invariant and becomes finite. There is a very small step from this
observation to regularization. In my copy of this paper
(I only have a preprint version, but I believe the journal version
is close) I can read at the end of CONCLUSION:

"This example shows that this nonlocal theory can be used as a way
of relativistic regularization of the local one."

>
>
>
>>3. Introduce mass and charge renormalization counterterms in the
>>Hamiltonian. Everything is finite so far. But the S-matrix
>>S(\Lambda) is not accurate yet.
>>
>>[One way to get the accurate S-matrix is to follow the
>>usual recipe - remove the regularization (e.g., set the cutoff momentum
>>to infinity). Then you get the S-matrix S(inf) as a sum of finite terms.
>>Just few low-order terms are enough to get astonishing agreement with
>>experiment. I have no idea about the convergence of the series for the
>>S-matrix. I am not going to discuss it here. I assume that this S-matrix
>>is the best we can get. The problem with this approach is that the
>>Hamiltonian H(inf) is infinite. To solve this problem I suggest step 4.]
>>
>>4. Perform unitary "dressing" transformation H(\Lambda) -> H'(\Lambda)
>> which removes from
>> H(\Lambda) "renorm" and "unphysical" terms and makes "physical"
>> terms in H'(\Lambda) rapidly decaying outside the energy shell which
>> guarantees convergence of loop integrals even after regularization
>> is removed. This "dressing" transformation, in addition, does not
>> change the S-matrix S(\Lambda)
>>
>>5. Now we can remove the regularization (e.g., set the cutoff momentum
>> to infinity). In the limit we get the same S-matrix S(inf) as
>> traditional QED. We also get the new finite Hamiltonian H'(inf).
>
>
> OK; this sounds good. So it looks like you found a finite, dynamical
> formulation of QED at any finite loop order. It would be very interesting
> if you could prove convergence of your procedure as the number of loops
> gets large. Perhaps first for \phi^4 theory, which should be simpler
> in this respect...

Sorry, I don't have an explicit proof at the moment.
The only thing I can say
is that convergence properties of my approach are exactly the same as
of the usual renormalization approach: If usual perturbation series for
the S-matrix converges, then my perturbation series converges too.
They are guaranteed to be the same in each perturbation order.

Eugene.

>
>
> Arnold Neumaier
>
>
>

Arnold Neumaier

Nov4-04, 03:38 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Chris Oakley wrote:\n\n> As far as I know, all the attempts to build a theory by using\n> annihilation/creation operators that work at a particular time and which are\n> time-evolved using a Hamiltonian expressed in terms of these do not work\n> because of loops. Your approach may have tidied certain things up, but it\n> still has infinite subtractions.\n\nThis does not make sense as _any_ multiparticle QM can be written in\nsecond quantized form in terms of creation and annihilation operators,\nincluding standard nonrelativistic QM. There one does not get any\ndivergences.\n\nThe divergences come only from the fact that standard relativistic QFTs\nhave singular interactions. Even in nonrelativistic QM of single particles,\nsingular interactions produce divergences and the need for divergent\ncounterterms. But there it is well-understood how it works, even on a\nmathematical level.\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:

> As far as I know, all the attempts to build a theory by using
> annihilation/creation operators that work at a particular time and which are
> time-evolved using a Hamiltonian expressed in terms of these do not work
> because of loops. Your approach may have tidied certain things up, but it
> still has infinite subtractions.

This does not make sense as _any_ multiparticle QM can be written in
second quantized form in terms of creation and annihilation operators,
including standard nonrelativistic QM. There one does not get any
divergences.

The divergences come only from the fact that standard relativistic QFTs
have singular interactions. Even in nonrelativistic QM of single particles,
singular interactions produce divergences and the need for divergent
counterterms. But there it is well-understood how it works, even on a
mathematical level.

Arnold Neumaier

Arnold Neumaier

Nov4-04, 03:40 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>OK; this sounds good. So it looks like you found a finite, dynamical\n>>formulation of QED at any finite loop order. It would be very interesting\n>>if you could prove convergence of your procedure as the number of loops\n>>gets large. Perhaps first for phi^4 theory, which should be simpler\n>>in this respect...\n>\n> Sorry, I don\'t have an explicit proof at the moment.\n> The only thing I can say\n> is that convergence properties of my approach are exactly the same as\n> of the usual renormalization approach: If usual perturbation series for\n> the S-matrix converges, then my perturbation series converges too.\n\nAccording to\nF.J. Dyson,\nDivergence of perturbation theory in quantum electrodynamics,\nPhys. Rev. 85 (1952), 613--632.\nit is very likely that the S-matrix expansion does _not_ converge\nbut is only asymptotic. On the other hand, the Hamiltonian expansion\nis different and might well converge; Dyson\'s arguments no longer apply.\n\nNot that it would be easy to prove convergence, since it would\nimply an existence proof of QED. So far, no interacting Lorentz invariant\ntheory has been proved to exist nonperturbatively. But since your approach\nto QED is different, the existence question might be easier from that\nperspective...\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>OK; this sounds good. So it looks like you found a finite, dynamical
>>formulation of QED at any finite loop order. It would be very interesting
>>if you could prove convergence of your procedure as the number of loops
>>gets large. Perhaps first for \phi^4 theory, which should be simpler
>>in this respect...
>
> Sorry, I don't have an explicit proof at the moment.
> The only thing I can say
> is that convergence properties of my approach are exactly the same as
> of the usual renormalization approach: If usual perturbation series for
> the S-matrix converges, then my perturbation series converges too.

According to
F.J. Dyson,
Divergence of perturbation theory in quantum electrodynamics,
Phys. Rev. 85 (1952), 613--632.
it is very likely that the S-matrix expansion does _not_ converge
but is only asymptotic. On the other hand, the Hamiltonian expansion
is different and might well converge; Dyson's arguments no longer apply.

Not that it would be easy to prove convergence, since it would
imply an existence proof of QED. So far, no interacting Lorentz invariant
theory has been proved to exist nonperturbatively. But since your approach
to QED is different, the existence question might be easier from that
perspective...

Arnold Neumaier

Eugene Stefanovich

Nov5-04, 08:20 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>One way to proceed without intermediate infinities is the following:\n>>\n>>1. Take usual QED Hamiltonian\n>>\n>>2. Perform regularization. This can be simple momentum cutoff (\\Lambda)\n>> in loop integrals, or you can choose to use Kita-Shirokov\n>> regularization\n>> which preserves the relativistic invariance:\n>>\n>>M.I. Shirokov "On relativistic nonlocal quantum field theory"\n>>Int. J. Theor. Phys. 41 (2002), 1027\n>\n>\n> Please point me to the page where they do regularization.\n>\n>\n>\n>>3. Introduce mass and charge renormalization counterterms in the\n>>Hamiltonian. Everything is finite so far. But the S-matrix\n>>S(\\Lambda) is not accurate yet.\n>>\n>>[One way to get the accurate S-matrix is to follow the\n>>usual recipe - remove the regularization (e.g., set the cutoff momentum\n>>to infinity). Then you get the S-matrix S(inf) as a sum of finite terms.\n>>Just few low-order terms are enough to get astonishing agreement with\n>>experiment. I have no idea about the convergence of the series for the\n>>S-matrix. I am not going to discuss it here. I assume that this S-matrix\n>>is the best we can get. The problem with this approach is that the\n>>Hamiltonian H(inf) is infinite. To solve this problem I suggest step 4.]\n>>\n>>4. Perform unitary "dressing" transformation H(\\Lambda) -> H\'(\\Lambda)\n>> which removes from\n>> H(\\Lambda) "renorm" and "unphysical" terms and makes "physical"\n>> terms in H\'(\\Lambda) rapidly decaying outside the energy shell which\n>> guarantees convergence of loop integrals even after regularization\n>> is removed. This "dressing" transformation, in addition, does not\n>> change the S-matrix S(\\Lambda)\n>>\n>>5. Now we can remove the regularization (e.g., set the cutoff momentum\n>> to infinity). In the limit we get the same S-matrix S(inf) as\n>> traditional QED. We also get the new finite Hamiltonian H\'(inf).\n>\n>\n> OK; this sounds good. So it looks like you found a finite, dynamical\n> formulation of QED at any finite loop order. It would be very interesting\n> if you could prove convergence of your procedure as the number of loops\n> gets large. Perhaps first for phi^4 theory, which should be simpler\n> in this respect...\n>\n>\n> Arnold Neumaier\n>\nSuppose for a moment that the series for the S-matrix and/or the\nHamiltonian\ndo not converge (though I hope this is not true).\nThis is not the end of the world. This would simply\nmean that our guess for the QED interaction Hamiltonian was wrong, and\nwe need to find a better guess. I am saying that currently accepted QED\ninteraction is just a guess (though this guess turns out to be\nexceptionally good). The quantization of the Maxwell\'s theory cannot\nbe considered as a rigorous derivation of QED. Maxwell\'s\ntheory can be at best regarded as a very rough approximation to the full\nquantum theory (it is probably enough to keep 2nd and 3rd perturbation\norders and c^{-2} approximation to deduce Maxwell\'s theory from QED).\nWe cannot rigorously derive a quantum theory from its approximate\nversion.\n\nAlso, I wouldn\'t waste time on phi^4 theory. This is a toy model\nhaving little relevance to real physics. If a toy doesn\'t work it\nusualy ends up in the garbage can.\n\nEugene Stefanovich.\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>One way to proceed without intermediate infinities is the following:
>>
>>1. Take usual QED Hamiltonian
>>
>>2. Perform regularization. This can be simple momentum cutoff (\Lambda)
>> in loop integrals, or you can choose to use Kita-Shirokov
>> regularization
>> which preserves the relativistic invariance:
>>
>>M.I. Shirokov "On relativistic nonlocal quantum field theory"
>>\Int. J. Theor. Phys. 41 (2002), 1027
>
>
> Please point me to the page where they do regularization.
>
>
>
>>3. Introduce mass and charge renormalization counterterms in the
>>Hamiltonian. Everything is finite so far. But the S-matrix
>>S(\Lambda) is not accurate yet.
>>
>>[One way to get the accurate S-matrix is to follow the
>>usual recipe - remove the regularization (e.g., set the cutoff momentum
>>to infinity). Then you get the S-matrix S(inf) as a sum of finite terms.
>>Just few low-order terms are enough to get astonishing agreement with
>>experiment. I have no idea about the convergence of the series for the
>>S-matrix. I am not going to discuss it here. I assume that this S-matrix
>>is the best we can get. The problem with this approach is that the
>>Hamiltonian H(inf) is infinite. To solve this problem I suggest step 4.]
>>
>>4. Perform unitary "dressing" transformation H(\Lambda) -> H'(\Lambda)
>> which removes from
>> H(\Lambda) "renorm" and "unphysical" terms and makes "physical"
>> terms in H'(\Lambda) rapidly decaying outside the energy shell which
>> guarantees convergence of loop integrals even after regularization
>> is removed. This "dressing" transformation, in addition, does not
>> change the S-matrix S(\Lambda)
>>
>>5. Now we can remove the regularization (e.g., set the cutoff momentum
>> to infinity). In the limit we get the same S-matrix S(inf) as
>> traditional QED. We also get the new finite Hamiltonian H'(inf).
>
>
> OK; this sounds good. So it looks like you found a finite, dynamical
> formulation of QED at any finite loop order. It would be very interesting
> if you could prove convergence of your procedure as the number of loops
> gets large. Perhaps first for \phi^4 theory, which should be simpler
> in this respect...
>
>
> Arnold Neumaier
>
Suppose for a moment that the series for the S-matrix and/or the
Hamiltonian
do not converge (though I hope this is not true).
This is not the end of the world. This would simply
mean that our guess for the QED interaction Hamiltonian was wrong, and
we need to find a better guess. I am saying that currently accepted QED
interaction is just a guess (though this guess turns out to be
exceptionally good). The quantization of the Maxwell's theory cannot
be considered as a rigorous derivation of QED. Maxwell's
theory can be at best regarded as a very rough approximation to the full
quantum theory (it is probably enough to keep 2nd and 3rd perturbation
orders and c^{-2} approximation to deduce Maxwell's theory from QED).
We cannot rigorously derive a quantum theory from its approximate
version.

Also, I wouldn't waste time on \phi^4 theory. This is a toy model
having little relevance to real physics. If a toy doesn't work it
usualy ends up in the garbage can.

Eugene Stefanovich.

Eugene Stefanovich

Nov5-04, 08:20 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>That\'s not the way I would like to do things. In my view, fields (either\n>>interacting or non-interacting) do not have any physical meaning. I\n>>haven\'t seen any experiment measuring fields.\n>\n>\n> People measure the electromagnetic field routinely in daily life.\n> It is the basis of all our modern technical culture. So at least\n> the electromagnetic part of QED cannot be discussed away...\n>\n>\n> Arnold Neumaier\n>\n\nI disagree. When people "measure electromagnetic field" they actually\nput a test charge at a point in space and measure a force exerted on\nthe charge by surrounding charges. This can be described in terms of\nparticles and direct interactions between them. No fields are needed.\nWhen people speak of "transversal electromagnetic wave" or light,\nthey actually speak about a flow of particles - photons. No fields\nare needed. This is the main point of my approach.\n\nEugene.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>That's not the way I would like to do things. In my view, fields (either
>>interacting or non-interacting) do not have any physical meaning. I
>>haven't seen any experiment measuring fields.
>
>
> People measure the electromagnetic field routinely in daily life.
> It is the basis of all our modern technical culture. So at least
> the electromagnetic part of QED cannot be discussed away...
>
>
> Arnold Neumaier
>

I disagree. When people "measure electromagnetic field" they actually
put a test charge at a point in space and measure a force exerted on
the charge by surrounding charges. This can be described in terms of
particles and direct interactions between them. No fields are needed.
When people speak of "transversal electromagnetic wave" or light,
they actually speak about a flow of particles - photons. No fields
are needed. This is the main point of my approach.

Eugene.

Eugene Stefanovich

Nov5-04, 08:20 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nChris Oakley wrote:\n\n>\n> As far as I know, all the attempts to build a theory by using\n> annihilation/creation operators that work at a particular time and which are\n> time-evolved using a Hamiltonian expressed in terms of these do not work\n> because of loops. Your approach may have tidied certain things up, but it\n> still has infinite subtractions.\n>\n\nIn my book I succeeded in building interacting theory where all\nquantities are expressed in terms of annihilation and creation\noperators, and fields are not used. Loops are not a problem, because\ninteraction terms rapidly fall off outside the energy shell, which\nguarantees convergence of all loop integrals. All this can be done\nwithout infinite subtractions. Let me repeat what I wrote a few days\nago in response to similar question from A. Neumaier\n\n\n\nOne way to proceed without intermediate infinities is the following:\n\n1. Take usual QED Hamiltonian\n\n2. Perform regularization. This can be simple momentum cutoff (\\Lambda)\nin loop integrals, or you can choose to use Kita-Shirokov\nregularization\nwhich preserves the relativistic invariance:\n\nM.I. Shirokov "On relativistic nonlocal quantum field theory"\nInt. J. Theor. Phys. 41 (2002), 1027\n\nH. Kita, "A non-trivial example of a relativistic\nquantum theory of particles without divergence difficulties",\nProgr. Theor. Phys. 35 (1966), 934.\n\n3. Introduce mass and charge renormalization counterterms in the\nHamiltonian. Everything is finite so far. But the S-matrix\nS(\\Lambda) is not accurate yet.\n\n[One way to get the accurate S-matrix is to follow the\nusual recipe - remove the regularization (e.g., set the cutoff momentum\nto infinity). Then you get the S-matrix S(inf) as a sum of finite terms.\nJust few low-order terms are enough to get astonishing agreement with\nexperiment. I have no idea about the convergence of the series for the\nS-matrix. I am not going to discuss it here. I assume that this S-matrix\nis the best we can get. The problem with this approach is that the\nHamiltonian H(inf) is infinite. To solve this problem I suggest step 4.]\n\n4. Perform unitary "dressing" transformation H(\\Lambda) -> H\'(\\Lambda)\nwhich removes from\nH(\\Lambda) "renorm" and "unphysical" terms and makes "physical"\nterms in H\'(\\Lambda) rapidly decaying outside the energy shell which\nguarantees convergence of loop integrals even after regularization\nis removed. This "dressing" transformation, in addition, does not\nchange the S-matrix S(\\Lambda)\n\n5. Now we can remove the regularization (e.g., set the cutoff momentum\nto infinity). In the limit we get the same S-matrix S(inf) as\ntraditional QED. We also get the new finite Hamiltonian H\'(inf).\n\nEugene.\nwww.geocities.com/meopemuk\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:

>
> As far as I know, all the attempts to build a theory by using
> annihilation/creation operators that work at a particular time and which are
> time-evolved using a Hamiltonian expressed in terms of these do not work
> because of loops. Your approach may have tidied certain things up, but it
> still has infinite subtractions.
>

In my book I succeeded in building interacting theory where all
quantities are expressed in terms of annihilation and creation
operators, and fields are not used. Loops are not a problem, because
interaction terms rapidly fall off outside the energy shell, which
guarantees convergence of all loop integrals. All this can be done
without infinite subtractions. Let me repeat what I wrote a few days
ago in response to similar question from A. Neumaier

One way to proceed without intermediate infinities is the following:

1. Take usual QED Hamiltonian

2. Perform regularization. This can be simple momentum cutoff (\Lambda)
in loop integrals, or you can choose to use Kita-Shirokov
regularization
which preserves the relativistic invariance:

M.I. Shirokov "On relativistic nonlocal quantum field theory"
\Int. J. Theor. Phys. 41 (2002), 1027

H. Kita, "A non-trivial example of a relativistic
quantum theory of particles without divergence difficulties",
Progr. Theor. Phys. 35 (1966), 934.

3. Introduce mass and charge renormalization counterterms in the
Hamiltonian. Everything is finite so far. But the S-matrix
S(\Lambda) is not accurate yet.

[One way to get the accurate S-matrix is to follow the
usual recipe - remove the regularization (e.g., set the cutoff momentum
to infinity). Then you get the S-matrix S(inf) as a sum of finite terms.
Just few low-order terms are enough to get astonishing agreement with
experiment. I have no idea about the convergence of the series for the
S-matrix. I am not going to discuss it here. I assume that this S-matrix
is the best we can get. The problem with this approach is that the
Hamiltonian H(inf) is infinite. To solve this problem I suggest step 4.]

4. Perform unitary "dressing" transformation H(\Lambda) -> H'(\Lambda)
which removes from
H(\Lambda) "renorm" and "unphysical" terms and makes "physical"
terms in H'(\Lambda) rapidly decaying outside the energy shell which
guarantees convergence of loop integrals even after regularization
is removed. This "dressing" transformation, in addition, does not
change the S-matrix S(\Lambda)

5. Now we can remove the regularization (e.g., set the cutoff momentum
to infinity). In the limit we get the same S-matrix S(inf) as
traditional QED. We also get the new finite Hamiltonian H'(inf).

Eugene.
www.geocities.com/meopemuk

Chris Oakley

Nov5-04, 08:20 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\nnews:41890343.3090703@univie.ac.at...\n> Chris Oakley wrote:\n>\n> > As far as I know, all the attempts to build a theory by using\n> > annihilation/creation operators that work at a particular time and which\nare\n> > time-evolved using a Hamiltonian expressed in terms of these do not work\n> > because of loops. Your approach may have tidied certain things up, but\nit\n> > still has infinite subtractions.\n>\n> This does not make sense as _any_ multiparticle QM can be written in\n> second quantized form in terms of creation and annihilation operators,\n> including standard nonrelativistic QM. There one does not get any\n> divergences.\n\nOne can certainly write multi-particle quantum mechanics with interactions\nbut no additional creation or annihilation of particles in second-quantised\nform, i.e. in terms of annihilation and creation operators, and I agree that\nthis can be done without pathological divergences, since all one is doing is\npresenting the same theory in a different way. The problem is that one\nnormally *does* want creation & annihilation of particles, e.g. a photon\nbeing created when an atom goes from an excited to a de-excited state. The\ntrilinear operator that one writes to create photons as desired causes a\nvery undesirable infinity in the second order perturbation expansion. But\nfor this mathematical inconvenience, quantum electrodynamics would have been\nwrapped up by 1930.\n\n> The divergences come only from the fact that standard relativistic QFTs\n> have singular interactions. Even in nonrelativistic QM of single\nparticles,\n> singular interactions produce divergences and the need for divergent\n> counterterms. But there it is well-understood how it works, even on a\n> mathematical level.\n\nI would not put it like this. Familiarity breeds contempt, as they say.\nPeople have gotten so used to lowering their standards of mathematical\nrigour in QFT that they forget now that they are doing it at all.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message
news:41890343.3090703@univie.ac.at...
> Chris Oakley wrote:
>
> > As far as I know, all the attempts to build a theory by using
> > annihilation/creation operators that work at a particular time and which
are
> > time-evolved using a Hamiltonian expressed in terms of these do not work
> > because of loops. Your approach may have tidied certain things up, but
it
> > still has infinite subtractions.
>
> This does not make sense as _any_ multiparticle QM can be written in
> second quantized form in terms of creation and annihilation operators,
> including standard nonrelativistic QM. There one does not get any
> divergences.

One can certainly write multi-particle quantum mechanics with interactions
but no additional creation or annihilation of particles in second-quantised
form, i.e. in terms of annihilation and creation operators, and I agree that
this can be done without pathological divergences, since all one is doing is
presenting the same theory in a different way. The problem is that one
normally *does* want creation & annihilation of particles, e.g. a photon
being created when an atom goes from an excited to a de-excited state. The
trilinear operator that one writes to create photons as desired causes a
very undesirable infinity in the second order perturbation expansion. But
for this mathematical inconvenience, quantum electrodynamics would have been
wrapped up by 1930.

> The divergences come only from the fact that standard relativistic QFTs
> have singular interactions. Even in nonrelativistic QM of single
particles,
> singular interactions produce divergences and the need for divergent
> counterterms. But there it is well-understood how it works, even on a
> mathematical level.

I would not put it like this. Familiarity breeds contempt, as they say.
People have gotten so used to lowering their standards of mathematical
rigour in QFT that they forget now that they are doing it at all.

Arnold Neumaier

Nov6-04, 11:06 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> Suppose for a moment that the series for the S-matrix and/or the\n> Hamiltonian\n> do not converge (though I hope this is not true).\n> This is not the end of the world. This would simply\n> mean that our guess for the QED interaction Hamiltonian was wrong, and\n> we need to find a better guess.\n\nNo; there are many functions f(x) which make perfect sense for x>=0 but if\nyou expand it around x=0 the resulting power series has convergence radius\nzero, so that the expansion does not determine the function. This is the\ncase for the S-matrix elements of field theories in 2 and 3 dimensions.\nThere the interactions are the right ones but the way of extracting\ninformation by means of perturbation theory is faulty. Probably the\nsame holds for QED. Even more probably, the same holds for QCD,\nwhich is asymptotically free but also is unlikely to have a convergent\nperturbation theory.)\n\n\n> I am saying that currently accepted QED\n> interaction is just a guess (though this guess turns out to be\n> exceptionally good). The quantization of the Maxwell\'s theory cannot\n> be considered as a rigorous derivation of QED. Maxwell\'s\n> theory can be at best regarded as a very rough approximation to the full\n> quantum theory (it is probably enough to keep 2nd and 3rd perturbation\n> orders and c^{-2} approximation to deduce Maxwell\'s theory from QED).\n> We cannot rigorously derive a quantum theory from its approximate\n> version.\n>\n> Also, I wouldn\'t waste time on phi^4 theory. This is a toy model\n> having little relevance to real physics. If a toy doesn\'t work it\n> usually ends up in the garbage can.\n\nIf you look at the literature you can see that a lot is done on Phi^4\ntheory. If you can\'t do that, it is most likely that you can\'t do QED.\nIf you can do QED, you\'ll be almost certainly able to do Phi^4,\nand everything is much simpler - and hence easier to understand and\neasier for others to learn.\n\nThat\'s precisely the value of toy models. One doesn\'t give children\nreal cars, but begin with toy versions, so that they can learn...\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> Suppose for a moment that the series for the S-matrix and/or the
> Hamiltonian
> do not converge (though I hope this is not true).
> This is not the end of the world. This would simply
> mean that our guess for the QED interaction Hamiltonian was wrong, and
> we need to find a better guess.

No; there are many functions f(x) which make perfect sense for x>=0 but if
you expand it around x=0 the resulting power series has convergence radius
zero, so that the expansion does not determine the function. This is the
case for the S-matrix elements of field theories in 2 and 3 dimensions.
There the interactions are the right ones but the way of extracting
information by means of perturbation theory is faulty. Probably the
same holds for QED. Even more probably, the same holds for QCD,
which is asymptotically free but also is unlikely to have a convergent
perturbation theory.)

> I am saying that currently accepted QED
> interaction is just a guess (though this guess turns out to be
> exceptionally good). The quantization of the Maxwell's theory cannot
> be considered as a rigorous derivation of QED. Maxwell's
> theory can be at best regarded as a very rough approximation to the full
> quantum theory (it is probably enough to keep 2nd and 3rd perturbation
> orders and c^{-2} approximation to deduce Maxwell's theory from QED).
> We cannot rigorously derive a quantum theory from its approximate
> version.
>
> Also, I wouldn't waste time on \phi^4 theory. This is a toy model
> having little relevance to real physics. If a toy doesn't work it
> usually ends up in the garbage can.

If you look at the literature you can see that a lot is done on \Phi^4
theory. If you can't do that, it is most likely that you can't do QED.
If you can do QED, you'll be almost certainly able to do \Phi^4,
and everything is much simpler - and hence easier to understand and
easier for others to learn.

That's precisely the value of toy models. One doesn't give children
real cars, but begin with toy versions, so that they can learn...

Arnold Neumaier

Arnold Neumaier

Nov6-04, 11:06 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>Eugene Stefanovich wrote:\n>>\n>>>That\'s not the way I would like to do things. In my view, fields (either\n>>>interacting or non-interacting) do not have any physical meaning. I\n>>>haven\'t seen any experiment measuring fields.\n>>\n>>People measure the electromagnetic field routinely in daily life.\n>>It is the basis of all our modern technical culture. So at least\n>>the electromagnetic part of QED cannot be discussed away...\n>\n> I disagree. When people "measure electromagnetic field" they actually\n> put a test charge at a point in space and measure a force exerted on\n> the charge by surrounding charges. This can be described in terms of\n> particles and direct interactions between them. No fields are needed.\n> When people speak of "transversal electromagnetic wave" or light,\n> they actually speak about a flow of particles - photons. No fields\n> are needed. This is the main point of my approach.\n\nBut this is the useless part of your approach; what you consider a step\nforward is in fact a step back. The only valuable part in your papers is\nthe renormalized Hamiltonian.\n\nTo do any realistic calculations in an external field (e.g. that produced\nby large accelerators) you\'ll need the field; making it up by particles\nwill make things hopelessly complicated. So you\'d extend your method to\nwork for QED with an external field. This is needed anyway if you want\nto recover the anomalous magnetic moment of the electron within your\napproach.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>Eugene Stefanovich wrote:
>>
>>>That's not the way I would like to do things. In my view, fields (either
>>>interacting or non-interacting) do not have any physical meaning. I
>>>haven't seen any experiment measuring fields.
>>
>>People measure the electromagnetic field routinely in daily life.
>>It is the basis of all our modern technical culture. So at least
>>the electromagnetic part of QED cannot be discussed away...
>
> I disagree. When people "measure electromagnetic field" they actually
> put a test charge at a point in space and measure a force exerted on
> the charge by surrounding charges. This can be described in terms of
> particles and direct interactions between them. No fields are needed.
> When people speak of "transversal electromagnetic wave" or light,
> they actually speak about a flow of particles - photons. No fields
> are needed. This is the main point of my approach.

But this is the useless part of your approach; what you consider a step
forward is in fact a step back. The only valuable part in your papers is
the renormalized Hamiltonian.

To do any realistic calculations in an external field (e.g. that produced
by large accelerators) you'll need the field; making it up by particles
will make things hopelessly complicated. So you'd extend your method to
work for QED with an external field. This is needed anyway if you want
to recover the anomalous magnetic moment of the electron within your
approach.

Arnold Neumaier

Arnold Neumaier

Nov6-04, 11:06 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Chris Oakley wrote:\n> "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\n> news:41890343.3090703@univie.ac.at...\n>\n>>_any_ multiparticle QM can be written in\n>>second quantized form in terms of creation and annihilation operators,\n>>including standard nonrelativistic QM. There one does not get any\n>>divergences.\n>\n> One can certainly write multi-particle quantum mechanics with interactions\n> but no additional creation or annihilation of particles in second-quantised\n> form, i.e. in terms of annihilation and creation operators, and I agree that\n> this can be done without pathological divergences, since all one is doing is\n> presenting the same theory in a different way. The problem is that one\n> normally *does* want creation & annihilation of particles, e.g. a photon\n> being created when an atom goes from an excited to a de-excited state. The\n> trilinear operator that one writes to create photons as desired causes a\n> very undesirable infinity in the second order perturbation expansion.\n\nOnly because of the very singualr interaction demanded by microlocality.\nThere are perfectly well-behaved nonrelativistic quantum theories that\ndo not conserve particle number, but allow reactions like\n1+1 <-> 1+1+1\nAs soon as single particles are stable and interactions are not pointlike,\nthere are no problems at all with particle creation/annihilation in\nnonrelativistic theories.\n\n\n>>The divergences come only from the fact that standard relativistic QFTs\n>>have singular interactions. Even in nonrelativistic QM of single particles,\n>>singular interactions produce divergences and the need for divergent\n>>counterterms. But there it is well-understood how it works, even on a\n>>mathematical level.\n>\n> I would not put it like this.\n\nI deliberately put it like this; I know what I am talking about.\n\n\n> People have gotten so used to lowering their standards of mathematical\n> rigour in QFT that they forget now that they are doing it at all.\n\nDifferent people have different degrees of rigor.\n\nMathematical physicists generally don\'t compromise at all in their\nrigor; they do proper mathematics, but they don\'t see a problem in\nrenormalization of nonrelativistic theories. It is just singular\nperturbation theory with Hamiltonians that are only defined as\nresolvent limits of Hamiltonians of the form T+V.\n\nYour reasoning is like that of someone who rejects modern analysis\nbecause Euler calculated seeming nonsense with divergent series,\nand who never took the troubles to look at how mathematicians\nhandle similar things today with full rigor.\n\nI suggest that you study the volumes by Reed and Simon to get the\nfunctional analysis background needed for coping with singular limits,\nand then look at modern mathematical work on renormalization, starting\nfor example with Glimm and Jaffe, who do the 2-dimensional case with\ncomplete rigor, without using perturbation theory at all.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
> "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message
> news:41890343.3090703@univie.ac.at...
>
>>_any_ multiparticle QM can be written in
>>second quantized form in terms of creation and annihilation operators,
>>including standard nonrelativistic QM. There one does not get any
>>divergences.
>
> One can certainly write multi-particle quantum mechanics with interactions
> but no additional creation or annihilation of particles in second-quantised
> form, i.e. in terms of annihilation and creation operators, and I agree that
> this can be done without pathological divergences, since all one is doing is
> presenting the same theory in a different way. The problem is that one
> normally *does* want creation & annihilation of particles, e.g. a photon
> being created when an atom goes from an excited to a de-excited state. The
> trilinear operator that one writes to create photons as desired causes a
> very undesirable infinity in the second order perturbation expansion.

Only because of the very singualr interaction demanded by microlocality.
There are perfectly well-behaved nonrelativistic quantum theories that
do not conserve particle number, but allow reactions like
1+1 <-> 1+1+1
As soon as single particles are stable and interactions are not pointlike,
there are no problems at all with particle creation/annihilation in
nonrelativistic theories.

>>The divergences come only from the fact that standard relativistic QFTs
>>have singular interactions. Even in nonrelativistic QM of single particles,
>>singular interactions produce divergences and the need for divergent
>>counterterms. But there it is well-understood how it works, even on a
>>mathematical level.
>
> I would not put it like this.

I deliberately put it like this; I know what I am talking about.

> People have gotten so used to lowering their standards of mathematical
> rigour in QFT that they forget now that they are doing it at all.

Different people have different degrees of rigor.

Mathematical physicists generally don't compromise at all in their
rigor; they do proper mathematics, but they don't see a problem in
renormalization of nonrelativistic theories. It is just singular
perturbation theory with Hamiltonians that are only defined as
resolvent limits of Hamiltonians of the form T+V.

Your reasoning is like that of someone who rejects modern analysis
because Euler calculated seeming nonsense with divergent series,
and who never took the troubles to look at how mathematicians
handle similar things today with full rigor.

I suggest that you study the volumes by Reed and Simon to get the
functional analysis background needed for coping with singular limits,
and then look at modern mathematical work on renormalization, starting
for example with Glimm and Jaffe, who do the 2-dimensional case with
complete rigor, without using perturbation theory at all.

Arnold Neumaier

Eugene Stefanovich

Nov6-04, 11:06 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Chris Oakley wrote:\n> "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\n> news:41890343.3090703@univie.ac.at...\n>\n>>Chris Oakley wrote:\n>>\n>>\n>>>As far as I know, all the attempts to build a theory by using\n>>>annihilation/creation operators that work at a particular time and which\n>>\n> are\n>\n>>>time-evolved using a Hamiltonian expressed in terms of these do not work\n>>>because of loops. Your approach may have tidied certain things up, but\n>>\n> it\n>\n>>>still has infinite subtractions.\n>>\n>>This does not make sense as _any_ multiparticle QM can be written in\n>>second quantized form in terms of creation and annihilation operators,\n>>including standard nonrelativistic QM. There one does not get any\n>>divergences.\n>\n>\n> One can certainly write multi-particle quantum mechanics with interactions\n> but no additional creation or annihilation of particles in second-quantised\n> form, i.e. in terms of annihilation and creation operators, and I agree that\n> this can be done without pathological divergences, since all one is doing is\n> presenting the same theory in a different way. The problem is that one\n> normally *does* want creation & annihilation of particles, e.g. a photon\n> being created when an atom goes from an excited to a de-excited state. The\n> trilinear operator that one writes to create photons as desired causes a\n> very undesirable infinity in the second order perturbation expansion. But\n> for this mathematical inconvenience, quantum electrodynamics would have been\n> wrapped up by 1930.\n\nIn my opinion, "trilinear interactions" are the source of all problems\nin QED. Trilinear interaction allows free electron to emit photons.\nIt also allows creation of particle pairs from vacuum. These effects\nhave never been observed experimentally, and I believe they do not\nexist. I call trilinear operators "unphysical". Another problem is that\na product or commutator of two unphysical operators contain renorm\noperators, like a^{\\dag}a. These renorm operators contribute to the mass\nof free particles, and you get infinite mass renormalization due to them.\n\nThe main idea of the "dressed particle" approach is to remove these\ntrilinear "unphysical" operators from the Hamiltonian. This can be done\nby a unitary transformation which does not change the S-matrix.\nThen, the theory becomes very similar to multi-particle quantum\nmechanics. The only difference is that creation and annihilation of\nparticles is permitted.\nThe interaction has only "physical" terms: interaction is\nnonzero only if there are two or more particles. Single particles and\nvacuum do not have self-interactions.\n\nEugene.\n\n\n\n>\n>\n>>The divergences come only from the fact that standard relativistic QFTs\n>>have singular interactions. Even in nonrelativistic QM of single\n>\n> particles,\n>\n>>singular interactions produce divergences and the need for divergent\n>>counterterms. But there it is well-understood how it works, even on a\n>>mathematical level.\n>\n>\n> I would not put it like this. Familiarity breeds contempt, as they say.\n> People have gotten so used to lowering their standards of mathematical\n> rigour in QFT that they forget now that they are doing it at all.\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:
> "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message
> news:41890343.3090703@univie.ac.at...
>
>>Chris Oakley wrote:
>>
>>
>>>As far as I know, all the attempts to build a theory by using
>>>annihilation/creation operators that work at a particular time and which
>>
> are
>
>>>time-evolved using a Hamiltonian expressed in terms of these do not work
>>>because of loops. Your approach may have tidied certain things up, but
>>
> it
>
>>>still has infinite subtractions.
>>
>>This does not make sense as _any_ multiparticle QM can be written in
>>second quantized form in terms of creation and annihilation operators,
>>including standard nonrelativistic QM. There one does not get any
>>divergences.
>
>
> One can certainly write multi-particle quantum mechanics with interactions
> but no additional creation or annihilation of particles in second-quantised
> form, i.e. in terms of annihilation and creation operators, and I agree that
> this can be done without pathological divergences, since all one is doing is
> presenting the same theory in a different way. The problem is that one
> normally *does* want creation & annihilation of particles, e.g. a photon
> being created when an atom goes from an excited to a de-excited state. The
> trilinear operator that one writes to create photons as desired causes a
> very undesirable infinity in the second order perturbation expansion. But
> for this mathematical inconvenience, quantum electrodynamics would have been
> wrapped up by 1930.

In my opinion, "trilinear interactions" are the source of all problems
in QED. Trilinear interaction allows free electron to emit photons.
It also allows creation of particle pairs from vacuum. These effects
have never been observed experimentally, and I believe they do not
exist. I call trilinear operators "unphysical". Another problem is that
a product or commutator of two unphysical operators contain renorm
operators, like a^{\dag}a. These renorm operators contribute to the mass
of free particles, and you get infinite mass renormalization due to them.

The main idea of the "dressed particle" approach is to remove these
trilinear "unphysical" operators from the Hamiltonian. This can be done
by a unitary transformation which does not change the S-matrix.
Then, the theory becomes very similar to multi-particle quantum
mechanics. The only difference is that creation and annihilation of
particles is permitted.
The interaction has only "physical" terms: interaction is
nonzero only if there are two or more particles. Single particles and
vacuum do not have self-interactions.

Eugene.

>
>
>>The divergences come only from the fact that standard relativistic QFTs
>>have singular interactions. Even in nonrelativistic QM of single
>
> particles,
>
>>singular interactions produce divergences and the need for divergent
>>counterterms. But there it is well-understood how it works, even on a
>>mathematical level.
>
>
> I would not put it like this. Familiarity breeds contempt, as they say.
> People have gotten so used to lowering their standards of mathematical
> rigour in QFT that they forget now that they are doing it at all.
>

Chris Oakley

Nov7-04, 07:44 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\nnews:418BB95D.4090809@univie.ac.at...\n> Chris Oakley wrote:\n> > "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\n> > news:41890343.3090703@univie.ac.at...\n> >\n> >>_any_ multiparticle QM can be written in\n> >>second quantized form in terms of creation and annihilation operators,\n> >>including standard nonrelativistic QM. There one does not get any\n> >>divergences.\n> >\n> > One can certainly write multi-particle quantum mechanics with\ninteractions\n> > but no additional creation or annihilation of particles in\nsecond-quantised\n> > form, i.e. in terms of annihilation and creation operators, and I agree\nthat\n> > this can be done without pathological divergences, since all one is\ndoing is\n> > presenting the same theory in a different way. The problem is that one\n> > normally *does* want creation & annihilation of particles, e.g. a photon\n> > being created when an atom goes from an excited to a de-excited state.\nThe\n> > trilinear operator that one writes to create photons as desired causes a\n> > very undesirable infinity in the second order perturbation expansion.\n>\n> Only because of the very singualr interaction demanded by microlocality.\n> There are perfectly well-behaved nonrelativistic quantum theories that\n> do not conserve particle number, but allow reactions like\n> 1+1 <-> 1+1+1\n> As soon as single particles are stable and interactions are not pointlike,\n> there are no problems at all with particle creation/annihilation in\n> nonrelativistic theories.\n>\n>\n> >>The divergences come only from the fact that standard relativistic QFTs\n> >>have singular interactions. Even in nonrelativistic QM of single\nparticles,\n> >>singular interactions produce divergences and the need for divergent\n> >>counterterms. But there it is well-understood how it works, even on a\n> >>mathematical level.\n> >\n> > I would not put it like this.\n>\n> I deliberately put it like this; I know what I am talking about.\n>\n>\n> > People have gotten so used to lowering their standards of mathematical\n> > rigour in QFT that they forget now that they are doing it at all.\n>\n> Different people have different degrees of rigor.\n>\n> Mathematical physicists generally don\'t compromise at all in their\n> rigor; they do proper mathematics, but they don\'t see a problem in\n> renormalization of nonrelativistic theories. It is just singular\n> perturbation theory with Hamiltonians that are only defined as\n> resolvent limits of Hamiltonians of the form T+V.\n>\n> Your reasoning is like that of someone who rejects modern analysis\n> because Euler calculated seeming nonsense with divergent series,\n> and who never took the troubles to look at how mathematicians\n> handle similar things today with full rigor.\n>\n> I suggest that you study the volumes by Reed and Simon to get the\n> functional analysis background needed for coping with singular limits,\n> and then look at modern mathematical work on renormalization, starting\n> for example with Glimm and Jaffe, who do the 2-dimensional case with\n> complete rigor, without using perturbation theory at all.\n>\n>\n> Arnold Neumaier\n>\n\nMy original statement was:\n\n"As far as I know, all the attempts to build a theory by using\nannihilation/creation operators that work at a particular time and which are\ntime-evolved using a Hamiltonian expressed in terms of these do not work\nbecause of loops."\n\nI should have qualified this with the words "realistic, relativistic". My\napologies.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message
news:418BB95D.4090809@univie.ac.at...
> Chris Oakley wrote:
> > "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message
> > news:41890343.3090703@univie.ac.at...
> >
> >>_any_ multiparticle QM can be written in
> >>second quantized form in terms of creation and annihilation operators,
> >>including standard nonrelativistic QM. There one does not get any
> >>divergences.
> >
> > One can certainly write multi-particle quantum mechanics with
interactions
> > but no additional creation or annihilation of particles in
second-quantised
> > form, i.e. in terms of annihilation and creation operators, and I agree
that
> > this can be done without pathological divergences, since all one is
doing is
> > presenting the same theory in a different way. The problem is that one
> > normally *does* want creation & annihilation of particles, e.g. a photon
> > being created when an atom goes from an excited to a de-excited state.
The
> > trilinear operator that one writes to create photons as desired causes a
> > very undesirable infinity in the second order perturbation expansion.
>
> Only because of the very singualr interaction demanded by microlocality.
> There are perfectly well-behaved nonrelativistic quantum theories that
> do not conserve particle number, but allow reactions like
> 1+1 <-> 1+1+1
> As soon as single particles are stable and interactions are not pointlike,
> there are no problems at all with particle creation/annihilation in
> nonrelativistic theories.
>
>
> >>The divergences come only from the fact that standard relativistic QFTs
> >>have singular interactions. Even in nonrelativistic QM of single
particles,
> >>singular interactions produce divergences and the need for divergent
> >>counterterms. But there it is well-understood how it works, even on a
> >>mathematical level.
> >
> > I would not put it like this.
>
> I deliberately put it like this; I know what I am talking about.
>
>
> > People have gotten so used to lowering their standards of mathematical
> > rigour in QFT that they forget now that they are doing it at all.
>
> Different people have different degrees of rigor.
>
> Mathematical physicists generally don't compromise at all in their
> rigor; they do proper mathematics, but they don't see a problem in
> renormalization of nonrelativistic theories. It is just singular
> perturbation theory with Hamiltonians that are only defined as
> resolvent limits of Hamiltonians of the form T+V.
>
> Your reasoning is like that of someone who rejects modern analysis
> because Euler calculated seeming nonsense with divergent series,
> and who never took the troubles to look at how mathematicians
> handle similar things today with full rigor.
>
> I suggest that you study the volumes by Reed and Simon to get the
> functional analysis background needed for coping with singular limits,
> and then look at modern mathematical work on renormalization, starting
> for example with Glimm and Jaffe, who do the 2-dimensional case with
> complete rigor, without using perturbation theory at all.
>
>
> Arnold Neumaier
>

My original statement was:

"As far as I know, all the attempts to build a theory by using
annihilation/creation operators that work at a particular time and which are
time-evolved using a Hamiltonian expressed in terms of these do not work
because of loops."

I should have qualified this with the words "realistic, relativistic". My
apologies.

Hendrik van Hees

Nov7-04, 07:44 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nEugene Stefanovich wrote:\n\n> In my opinion, "trilinear interactions" are the source of all problems\n> in QED.\n\n> Trilinear interaction allows free electron to emit photons.\n\nThis is not true, as I wrote some days ago. There are simple kinematic\nreasons that this process cannot happen.\n\n> It also allows creation of particle pairs from vacuum.\n\nSince the vacuum is unchanged under the time evolution (by\nconstruction), this is also impossible.\n\n> These effects\n> have never been observed experimentally, and I believe they do not\n> exist.\n\nNeither are they observed nor are they happening within QED.\n\n> I call trilinear operators "unphysical". Another problem is\n> that a product or commutator of two unphysical operators contain\n> renorm operators, like a^{\\dag}a. These renorm operators contribute to\n> the mass of free particles, and you get infinite mass renormalization\n> due to them.\n\nIf you like to stay with renormalisable models (like QED), there is no\nother interaction available than the usual electron-photon vertex\n(simple power counting).\n>\n> The main idea of the "dressed particle" approach is to remove these\n> trilinear "unphysical" operators from the Hamiltonian. This can be\n> done by a unitary transformation which does not change the S-matrix.\n> Then, the theory becomes very similar to multi-particle quantum\n> mechanics. The only difference is that creation and annihilation of\n> particles is permitted.\n\nBut creation and annihilation of particles is an observed fact!\n\n> The interaction has only "physical" terms: interaction is\n> nonzero only if there are two or more particles. Single particles and\n> vacuum do not have self-interactions.\n\nNeither do they in usual QED.\n\n--\nHendrik van Hees Cyclotron Institute\nPhone: +1 979/845-1411 Texas A&M University\nFax: +1 979/845-1899 Cyclotron Institute, MS-3366\nhttp://theory.gsi.de/~vanhees/ College Station, TX 77843-3366\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> In my opinion, "trilinear interactions" are the source of all problems
> in QED.

> Trilinear interaction allows free electron to emit photons.

This is not true, as I wrote some days ago. There are simple kinematic
reasons that this process cannot happen.

> It also allows creation of particle pairs from vacuum.

Since the vacuum is unchanged under the time evolution (by
construction), this is also impossible.

> These effects
> have never been observed experimentally, and I believe they do not
> exist.

Neither are they observed nor are they happening within QED.

> I call trilinear operators "unphysical". Another problem is
> that a product or commutator of two unphysical operators contain
> renorm operators, like a^{\dag}a. These renorm operators contribute to
> the mass of free particles, and you get infinite mass renormalization
> due to them.

If you like to stay with renormalisable models (like QED), there is no
other interaction available than the usual electron-photon vertex
(simple power counting).
>
> The main idea of the "dressed particle" approach is to remove these
> trilinear "unphysical" operators from the Hamiltonian. This can be
> done by a unitary transformation which does not change the S-matrix.
> Then, the theory becomes very similar to multi-particle quantum
> mechanics. The only difference is that creation and annihilation of
> particles is permitted.

But creation and annihilation of particles is an observed fact!

> The interaction has only "physical" terms: interaction is
> nonzero only if there are two or more particles. Single particles and
> vacuum do not have self-interactions.

Neither do they in usual QED.

--
Hendrik van Hees Cyclotron Institute
Phone: +1 979/845-1411 Texas A&M University
Fax: +1 979/845-1899 Cyclotron Institute, MS-3366
http://theory.gsi.de/~vanhees/ College Station, TX 77843-3366

Hendrik van Hees

Nov7-04, 04:05 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n\n> No; there are many functions f(x) which make perfect sense for x>=0\n> but if you expand it around x=0 the resulting power series has\n> convergence radius zero, so that the expansion does not determine the\n> function. This is the case for the S-matrix elements of field theories\n> in 2 and 3 dimensions. There the interactions are the right ones but\n> the way of extracting information by means of perturbation theory is\n> faulty. Probably the same holds for QED. Even more probably, the same\n> holds for QCD, which is asymptotically free but also is unlikely to\n> have a convergent perturbation theory.)\n\nI am not an expert in these questions, but as far as I know the\nliterature, most qft\'s are likely to have 0 convergence radius of the\npert. series. The pert. series has to be understood as an asymptotic\nseries.\n\nI also think to remember that, due to asymptotic freedom, QCD is likely\nto be Borel resummable.\n\nThe "convergence" of the perturbative series is sometimes good,\nsometimes bad. With "convergence" physicists mean that the (n+1st)\norder is smaller than the nth order (usually for small n).\n\nA famous example that this is not always really the case is thermal QCD,\nwhere the pressure (or other bulk thermodynamic quantities) are badly\nconverging. Of course, in thermal QFT, one always has to resum\ninfinitely many diagrams of the naive perturbation series, because due\nto thermal masses the naive power counting breaks down. Also this\nhard-thermal-loop improved perturbation theory is not well convergent\nfor bulk thermodynamical quantities. An improvement is the treatment\nbased on the Baym functional (also known as Cornwall-Tomboulis-Jackiw\nor CJT action), but there one has trouble with gauge invariance. As far\nas I know these issues are not completely solved yet, except by using\nlattice QCD, which also has its limitations of course (mainly we have\nonly the quenched results).\n\n> If you look at the literature you can see that a lot is done on Phi^4\n> theory. If you can\'t do that, it is most likely that you can\'t do QED.\n\nphi^4 theory is a nice pedagogical example to learn renormalisation\ntechniques etc. without being worried by gauge invariance. Nevertheless\nit is a toy model and indeed it\'s physical relevance is quite limited,\nalthough of course in the standard model the Higgs sector can be seen\nas a refined version of phi^4 theory. In effective mesonic models also\nthe linear sigma models are nice to describe medium-energy nuclear\nphysics. Don\'t underestimate the power of simple models!\n\n> If you can do QED, you\'ll be almost certainly able to do Phi^4,\n> and everything is much simpler - and hence easier to understand and\n> easier for others to learn.\n>\n> That\'s precisely the value of toy models. One doesn\'t give children\n> real cars, but begin with toy versions, so that they can learn...\n\nThat\'s right, but when the children grow up, they won\'t be satisfied\nwith the toys anymore and like to deal with the real world. And then\nQED is a nice starting point (don\'t give the kids the big sports car\nbefore they can handle normal ones ;-)).\n\n--\nHendrik van Hees Cyclotron Institute\nPhone: +1 979/845-1411 Texas A&M University\nFax: +1 979/845-1899 Cyclotron Institute, MS-3366\nhttp://theory.gsi.de/~vanhees/ College Station, TX 77843-3366\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:

> No; there are many functions f(x) which make perfect sense for x>=0
> but if you expand it around x=0 the resulting power series has
> convergence radius zero, so that the expansion does not determine the
> function. This is the case for the S-matrix elements of field theories
> in 2 and 3 dimensions. There the interactions are the right ones but
> the way of extracting information by means of perturbation theory is
> faulty. Probably the same holds for QED. Even more probably, the same
> holds for QCD, which is asymptotically free but also is unlikely to
> have a convergent perturbation theory.)

I am not an expert in these questions, but as far as I know the
literature, most qft's are likely to have convergence radius of the
pert. series. The pert. series has to be understood as an asymptotic
series.

I also think to remember that, due to asymptotic freedom, QCD is likely
to be Borel resummable.

The "convergence" of the perturbative series is sometimes good,
sometimes bad. With "convergence" physicists mean that the (n+1st)
order is smaller than the nth order (usually for small n).

A famous example that this is not always really the case is thermal QCD,
where the pressure (or other bulk thermodynamic quantities) are badly
converging. Of course, in thermal QFT, one always has to resum
infinitely many diagrams of the naive perturbation series, because due
to thermal masses the naive power counting breaks down. Also this
hard-thermal-loop improved perturbation theory is not well convergent
for bulk thermodynamical quantities. An improvement is the treatment
based on the Baym functional (also known as Cornwall-Tomboulis-Jackiw
or CJT action), but there one has trouble with gauge invariance. As far
as I know these issues are not completely solved yet, except by using
lattice QCD, which also has its limitations of course (mainly we have
only the quenched results).

> If you look at the literature you can see that a lot is done on \Phi^4
> theory. If you can't do that, it is most likely that you can't do QED.

\phi^4[/itex] theory is a nice pedagogical example to learn renormalisation
techniques etc. without being worried by gauge invariance. Nevertheless
it is a toy model and indeed it's physical relevance is quite limited,
although of course in the standard model the Higgs sector can be seen
as a refined version of \phi^4 theory. In effective mesonic models also
the linear \sigma models are nice to describe medium-energy nuclear
physics. Don't underestimate the power of simple models!

> If you can do QED, you'll be almost certainly able to do \Phi^4,
> and everything is much simpler - and hence easier to understand and
> easier for others to learn.
>
> That's precisely the value of toy models. One doesn't give children
> real cars, but begin with toy versions, so that they can learn...

That's right, but when the children grow up, they won't be satisfied
with the toys anymore and like to deal with the real world. And then
QED is a nice starting point (don't give the kids the big sports car
before they can handle normal ones ;-)).

--
Hendrik van Hees Cyclotron Institute
Phone: +1 979/845-1411 Texas A&M University
Fax: +1 979/845-1899 Cyclotron Institute, MS-3366
http://theory.gsi.de/~vanhees/ College Station, [itex]TX 77843-3366

Arnold Neumaier

Nov8-04, 05:10 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Chris Oakley wrote:\n\n> My original statement was:\n>\n> "As far as I know, all the attempts to build a theory by using\n> annihilation/creation operators that work at a particular time and which are\n> time-evolved using a Hamiltonian expressed in terms of these do not work\n> because of loops."\n>\n> I should have qualified this with the words "realistic, relativistic". My\n> apologies.\n\nIf one is interested, one can still learn from realistic, nonrelativistic\ntheories with singular potentials or from relativistic theories in 2D and 3D\nhow renormalization of formally infinite quantities can be done\nin a mathematically rigorous way. The standard perturbative textbook\ntreatment produces in these cases the same kind of infinities\nas in QCD, say. QCD is harder only because of the difficulty of getting\ntight enough estimates (since the Sobolev embedding theorems in 4D\nare weaker than in 2D and 3D). But in an ultrametric topology which treats\nthe coupling constant as an indeterminate rather than a number, everything\nis fine - this corresponds exactly to renormalized perturbation theory\nat all orders.\n\nSince nonrelativistic and 2D or 3D relativistic theories also have\na rigorous nonperturbative renormalization formulation,\none knows that the perturbative infinities are spurious\nand not worse than computing log 2 = sum (-1)^k/(k+1) by first summing\nall positive contributions and then subtracting the negative\ncontributions.\n\nThus your arguments against renormalization per se are irrelevant since\nthey would apply as well to nonrelativistic or 3D relativistic theories.\n\nIf you insist on not learning these rigorous ways you miss the chance\nof understanding what can be understood.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Chris Oakley wrote:

> My original statement was:
>
> "As far as I know, all the attempts to build a theory by using
> annihilation/creation operators that work at a particular time and which are
> time-evolved using a Hamiltonian expressed in terms of these do not work
> because of loops."
>
> I should have qualified this with the words "realistic, relativistic". My
> apologies.

If one is interested, one can still learn from realistic, nonrelativistic
theories with singular potentials or from relativistic theories in 2D and 3D
how renormalization of formally infinite quantities can be done
in a mathematically rigorous way. The standard perturbative textbook
treatment produces in these cases the same kind of infinities
as in QCD, say. QCD is harder only because of the difficulty of getting
tight enough estimates (since the Sobolev embedding theorems in 4D
are weaker than in 2D and 3D). But in an ultrametric topology which treats
the coupling constant as an indeterminate rather than a number, everything
is fine - this corresponds exactly to renormalized perturbation theory
at all orders.

Since nonrelativistic and 2D or 3D relativistic theories also have
a rigorous nonperturbative renormalization formulation,
one knows that the perturbative infinities are spurious
and not worse than computing log 2 = sum (-1)^k/(k+1) by first summing
all positive contributions and then subtracting the negative
contributions.

Thus your arguments against renormalization per se are irrelevant since
they would apply as well to nonrelativistic or 3D relativistic theories.

If you insist on not learning these rigorous ways you miss the chance
of understanding what can be understood.

Arnold Neumaier

Arnold Neumaier

Nov8-04, 05:10 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hendrik van Hees wrote:\n> Arnold Neumaier wrote:\n>\n>>No; there are many functions f(x) which make perfect sense for x>=0\n>>but if you expand it around x=0 the resulting power series has\n>>convergence radius zero, so that the expansion does not determine the\n>>function. This is the case for the S-matrix elements of field theories\n>>in 2 and 3 dimensions. There the interactions are the right ones but\n>>the way of extracting information by means of perturbation theory is\n>>faulty. Probably the same holds for QED. Even more probably, the same\n>>holds for QCD, which is asymptotically free but also is unlikely to\n>>have a convergent perturbation theory.)\n>\n> I am not an expert in these questions, but as far as I know the\n> literature, most qft\'s are likely to have 0 convergence radius of the\n> pert. series. The pert. series has to be understood as an asymptotic\n> series.\n\nThe perturbation series for the renormalized S-matrix, yes.\nIt may be different for the perturbation series for the renormalized\nHamiltonian which Stefanovich constructed, and which might actually\nconverge.\n\n\n> The "convergence" of the perturbative series is sometimes good,\n> sometimes bad. With "convergence" physicists mean that the (n+1st)\n> order is smaller than the nth order (usually for small n).\n\nOf course _only_ for small n. The asymptotic series for S-matrix elements\ntend to have coefficients that grow like O(n!) for large order n.\n\n\n>>If you look at the literature you can see that a lot is done on Phi^4\n>>theory. If you can\'t do that, it is most likely that you can\'t do QED.\n>\n> phi^4 theory is a nice pedagogical example to learn renormalisation\n> techniques etc. without being worried by gauge invariance. Nevertheless\n> it is a toy model and indeed it\'s physical relevance is quite limited,\n> although of course in the standard model the Higgs sector can be seen\n> as a refined version of phi^4 theory. In effective mesonic models also\n> the linear sigma models are nice to describe medium-energy nuclear\n> physics. Don\'t underestimate the power of simple models!\n>\n>>If you can do QED, you\'ll be almost certainly able to do Phi^4,\n>>and everything is much simpler - and hence easier to understand and\n>>easier for others to learn.\n>>\n>>That\'s precisely the value of toy models. One doesn\'t give children\n>>real cars, but begin with toy versions, so that they can learn...\n>\n> That\'s right, but when the children grow up, they won\'t be satisfied\n> with the toys anymore and like to deal with the real world. And then\n> QED is a nice starting point (don\'t give the kids the big sports car\n> before they can handle normal ones ;-)).\n\nOf course. Stefanovich already constructed a renormalized Hamiltonian for\nQED (and including pointlike protons) with a finite perturbative\nexpansion.\n\nNow I challenged him to investigate the convergence of the expansion.\nThis is a much more difficult question, so I suggested he\'d look first\nat Phi^4 theory since this should be simpler but already show the\nimportant aspects. And since he was thinking little of Phi^4\nI tried to make it more acceptable for him...\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hendrik van Hees wrote:
> Arnold Neumaier wrote:
>
>>No; there are many functions f(x) which make perfect sense for x>=0
>>but if you expand it around x=0 the resulting power series has
>>convergence radius zero, so that the expansion does not determine the
>>function. This is the case for the S-matrix elements of field theories
>>in 2 and 3 dimensions. There the interactions are the right ones but
>>the way of extracting information by means of perturbation theory is
>>faulty. Probably the same holds for QED. Even more probably, the same
>>holds for QCD, which is asymptotically free but also is unlikely to
>>have a convergent perturbation theory.)
>
> I am not an expert in these questions, but as far as I know the
> literature, most qft's are likely to have convergence radius of the
> pert. series. The pert. series has to be understood as an asymptotic
> series.

The perturbation series for the renormalized S-matrix, yes.
It may be different for the perturbation series for the renormalized
Hamiltonian which Stefanovich constructed, and which might actually
converge.

> The "convergence" of the perturbative series is sometimes good,
> sometimes bad. With "convergence" physicists mean that the (n+1st)
> order is smaller than the nth order (usually for small n).

Of course _only_ for small n. The asymptotic series for S-matrix elements
tend to have coefficients that grow like O(n!) for large order n.

>>If you look at the literature you can see that a lot is done on \Phi^4
>>theory. If you can't do that, it is most likely that you can't do QED.
>
> \phi^4 theory is a nice pedagogical example to learn renormalisation
> techniques etc. without being worried by gauge invariance. Nevertheless
> it is a toy model and indeed it's physical relevance is quite limited,
> although of course in the standard model the Higgs sector can be seen
> as a refined version of \phi^4 theory. In effective mesonic models also
> the linear \sigma models are nice to describe medium-energy nuclear
> physics. Don't underestimate the power of simple models!
>
>>If you can do QED, you'll be almost certainly able to do \Phi^4,
>>and everything is much simpler - and hence easier to understand and
>>easier for others to learn.
>>
>>That's precisely the value of toy models. One doesn't give children
>>real cars, but begin with toy versions, so that they can learn...
>
> That's right, but when the children grow up, they won't be satisfied
> with the toys anymore and like to deal with the real world. And then
> QED is a nice starting point (don't give the kids the big sports car
> before they can handle normal ones ;-)).

Of course. Stefanovich already constructed a renormalized Hamiltonian for
QED (and including pointlike protons) with a finite perturbative
expansion.

Now I challenged him to investigate the convergence of the expansion.
This is a much more difficult question, so I suggested he'd look first
at \Phi^4 theory since this should be simpler but already show the
important aspects. And since he was thinking little of \Phi^4
I tried to make it more acceptable for him...

Arnold Neumaier

Eugene Stefanovich

Nov9-04, 03:44 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Hendrik van Hees wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>No; there are many functions f(x) which make perfect sense for x>=0\n>>>but if you expand it around x=0 the resulting power series has\n>>>convergence radius zero, so that the expansion does not determine the\n>>>function. This is the case for the S-matrix elements of field theories\n>>>in 2 and 3 dimensions. There the interactions are the right ones but\n>>>the way of extracting information by means of perturbation theory is\n>>>faulty. Probably the same holds for QED. Even more probably, the same\n>>>holds for QCD, which is asymptotically free but also is unlikely to\n>>>have a convergent perturbation theory.)\n>>\n>>I am not an expert in these questions, but as far as I know the\n>>literature, most qft\'s are likely to have 0 convergence radius of the\n>>pert. series. The pert. series has to be understood as an asymptotic\n>>series.\n>\n>\n> The perturbation series for the renormalized S-matrix, yes.\n> It may be different for the perturbation series for the renormalized\n> Hamiltonian which Stefanovich constructed, and which might actually\n> converge.\n>\n>\n\n\nI don\'t quite understand your logic. Suppose that the perturbation\nseries for the renormalized S-matrix does not converge, then we can\nbe certain that the corresponding "dressed" Hamiltonian is wrong\neven if it\nis finite (convergent). This was the case with the original (finite)\nHamiltonian of QED. This Hamiltonian was finite, but the resulting\nS-matrix was infinite. This is why this Hamiltonian must be\n"renormalized" and "dressed".\n\nEugene.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Hendrik van Hees wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>No; there are many functions f(x) which make perfect sense for x>=0
>>>but if you expand it around x=0 the resulting power series has
>>>convergence radius zero, so that the expansion does not determine the
>>>function. This is the case for the S-matrix elements of field theories
>>>in 2 and 3 dimensions. There the interactions are the right ones but
>>>the way of extracting information by means of perturbation theory is
>>>faulty. Probably the same holds for QED. Even more probably, the same
>>>holds for QCD, which is asymptotically free but also is unlikely to
>>>have a convergent perturbation theory.)
>>
>>I am not an expert in these questions, but as far as I know the
>>literature, most qft's are likely to have convergence radius of the
>>pert. series. The pert. series has to be understood as an asymptotic
>>series.
>
>
> The perturbation series for the renormalized S-matrix, yes.
> It may be different for the perturbation series for the renormalized
> Hamiltonian which Stefanovich constructed, and which might actually
> converge.
>
>

I don't quite understand your logic. Suppose that the perturbation
series for the renormalized S-matrix does not converge, then we can
be certain that the corresponding "dressed" Hamiltonian is wrong
even if it
is finite (convergent). This was the case with the original (finite)
Hamiltonian of QED. This Hamiltonian was finite, but the resulting
S-matrix was infinite. This is why this Hamiltonian must be
"renormalized" and "dressed".

Eugene.

Arnold Neumaier

Nov9-04, 03:44 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nHendrik van Hees wrote:\n> Eugene Stefanovich wrote:\n>\n>>In my opinion, "trilinear interactions" are the source of all problems\n>>in QED.\n>\n>>Trilinear interaction allows free electron to emit photons.\n>\n> This is not true, as I wrote some days ago. There are simple kinematic\n> reasons that this process cannot happen.\n>\n>>It also allows creation of particle pairs from vacuum.\n>\n> Since the vacuum is unchanged under the time evolution (by\n> construction), this is also impossible.\n\n[etc.]\n\nIt sees you two are saying the same in words that sound contradictory.\nEugene Stefanovich argues that bare particles (which are not subject to\nkinematic restrictions) create nonsense and hence do not exist;\nHendrik van Hees argues that they do not exist since they are not\nphysical (\'happening\' or \'observed\').\n\nThe unfortunate thing is that they seem to be needed to _define_ what\nQED is. After that, there are a number of constructive ways of\nrenormalizing the naive definition to turn it into something that\nis at least perturbatively meaningful.\n\nThe fact that these different approaches all end up giving the same\nresults is, for me, an indication that there is something neat but\nundiscovered behind. The smoke tells of the fire, even if the latter\nis not yet discovered...\n\n\nArnold Neumaier\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hendrik van Hees wrote:
> Eugene Stefanovich wrote:
>
>>In my opinion, "trilinear interactions" are the source of all problems
>>in QED.
>
>>Trilinear interaction allows free electron to emit photons.
>
> This is not true, as I wrote some days ago. There are simple kinematic
> reasons that this process cannot happen.
>
>>It also allows creation of particle pairs from vacuum.
>
> Since the vacuum is unchanged under the time evolution (by
> construction), this is also impossible.

[etc.]

It sees you two are saying the same in words that sound contradictory.
Eugene Stefanovich argues that bare particles (which are not subject to
kinematic restrictions) create nonsense and hence do not exist;
Hendrik van Hees argues that they do not exist since they are not
physical ('happening' or 'observed').

The unfortunate thing is that they seem to be needed to _define_ what
QED is. After that, there are a number of constructive ways of
renormalizing the naive definition to turn it into something that
is at least perturbatively meaningful.

The fact that these different approaches all end up giving the same
results is, for me, an indication that there is something neat but
undiscovered behind. The smoke tells of the fire, even if the latter
is not yet discovered...

Arnold Neumaier

Eugene Stefanovich

Nov9-04, 03:45 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nHendrik van Hees wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>In my opinion, "trilinear interactions" are the source of all problems\n>>in QED.\n>\n>\n>>Trilinear interaction allows free electron to emit photons.\n>\n>\n> This is not true, as I wrote some days ago. There are simple kinematic\n> reasons that this process cannot happen.\n\nI agree that terms like\n\ne -> e + \\gamma\n0 -> e + p + \\gamma\n\n(e=electron, p=positron, \\gamma = photon, 0=vacuum) are forbidden\nby the energy-momentum\nconservation law, and they are not present in the S-matrix\n(where everything is integrated in time from -infinity to +infinity).\nHowever,\nthese trilinear terms are present in the Hamiltonian, and\n(according to QED) there should\nbe corresponding physical processes (creation of particles from vacuum,\netc.) at intermediate times. These processes (even short ones) have\nnever been observed.\n\n>\n>\n>>It also allows creation of particle pairs from vacuum.\n>\n>\n> Since the vacuum is unchanged under the time evolution (by\n> construction), this is also impossible.\n\nCould you please show me the Hamiltonian you are using to\ndescribe the time evolution? (reference is OK) How do you prove\nthat the vacuum vector remains unchanged during time evolution?\n\n>\n>\n>>These effects\n>>have never been observed experimentally, and I believe they do not\n>>exist.\n>\n>\n> Neither are they observed nor are they happening within QED.\n\nThey are not present in the S-matrix. True. But they are present in the\nHamiltonian, so they should happen at short times (if we take the QED\nHamiltonian at face value).\n\n>\n>\n>>I call trilinear operators "unphysical". Another problem is\n>>that a product or commutator of two unphysical operators contain\n>>renorm operators, like a^{\\dag}a. These renorm operators contribute to\n>>the mass of free particles, and you get infinite mass renormalization\n>>due to them.\n>\n>\n> If you like to stay with renormalisable models (like QED), there is no\n> other interaction available than the usual electron-photon vertex\n> (simple power counting).\n\nPower counting applies to Hamiltonians (or Lagrangians) written as\nproducts of fields. My Hamiltonian cannot be written in terms of\nfields. It can be only written in terms of creation and annihilation\noperators of particles. The power counting scheme does not apply.\nMoreover, I have a proof that the S-matrix of my approach is exactly\nthe same as the S-matrix of renormalized QED. My approach is\nnot just renormalizable. It is finite. Both Hamiltonian and the S-matrix\nare finite.\n\n>\n>>The main idea of the "dressed particle" approach is to remove these\n>>trilinear "unphysical" operators from the Hamiltonian. This can be\n>>done by a unitary transformation which does not change the S-matrix.\n>>Then, the theory becomes very similar to multi-particle quantum\n>>mechanics. The only difference is that creation and annihilation of\n>>particles is permitted.\n>\n>\n> But creation and annihilation of particles is an observed fact!\n\nThis is my point. Multi-particle quantum mechanics does not\ndescribe creation and annihilation of particles (in disagreement\nwith experiment). The "dressed particle" approach does describe\ncreation and annihilation of particles. In the dressed particle\nHamiltonian there are no terms like\n\ne -> e + \\gamma\n0 -> e + p + \\gamma\n\nwhich have never been observed, but there are (observable) terms like\n\ne + e -> e + e + \\gamma (bremsstrahlung)\ne + p -> \\gamma + \\gamma (annihilation)\n\netc.\n\n\n>\n>\n>>The interaction has only "physical" terms: interaction is\n>>nonzero only if there are two or more particles. Single particles and\n>>vacuum do not have self-interactions.\n>\n>\n> Neither do they in usual QED.\n\nThe self-interaction of single particles and vacuum is present in QED.\nThe renormalization approach cancels the effects of these\nself-interactions on the S-matrix. However, the self-interactions\nremain in the Hamiltonian and, therefore, reveal themselves in the\ntime evolution.\n\nQED conveniently bypasses this problem by refusing to calculate the\ntime evolution. This is the weak point of QED exposed in my book\n(see www.geocities.com/meopemuk). The only way (which I know of) to\nremove the self-interaction effects from the Hamiltonian is\nto use the "dressing transformation".\n\nEugene.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hendrik van Hees wrote:
> Eugene Stefanovich wrote:
>
>
>>In my opinion, "trilinear interactions" are the source of all problems
>>in QED.
>
>
>>Trilinear interaction allows free electron to emit photons.
>
>
> This is not true, as I wrote some days ago. There are simple kinematic
> reasons that this process cannot happen.

I agree that terms like

e -> e + \gamma-> e + p + \gamma

(e=electron, p=positron, \gamma = photon, 0=vacuum) are forbidden
by the energy-momentum
conservation law, and they are not present in the S-matrix
(where everything is integrated in time from -infinity to +infinity).
However,
these trilinear terms are present in the Hamiltonian, and
(according to QED) there should
be corresponding physical processes (creation of particles from vacuum,
etc.) at intermediate times. These processes (even short ones) have
never been observed.

>
>
>>It also allows creation of particle pairs from vacuum.
>
>
> Since the vacuum is unchanged under the time evolution (by
> construction), this is also impossible.

Could you please show me the Hamiltonian you are using to
describe the time evolution? (reference is OK) How do you prove
that the vacuum vector remains unchanged during time evolution?

>
>
>>These effects
>>have never been observed experimentally, and I believe they do not
>>exist.
>
>
> Neither are they observed nor are they happening within QED.

They are not present in the S-matrix. True. But they are present in the
Hamiltonian, so they should happen at short times (if we take the QED
Hamiltonian at face value).

>
>
>>I call trilinear operators "unphysical". Another problem is
>>that a product or commutator of two unphysical operators contain
>>renorm operators, like a^{\dag}a. These renorm operators contribute to
>>the mass of free particles, and you get infinite mass renormalization
>>due to them.
>
>
> If you like to stay with renormalisable models (like QED), there is no
> other interaction available than the usual electron-photon vertex
> (simple power counting).

Power counting applies to Hamiltonians (or Lagrangians) written as
products of fields. My Hamiltonian cannot be written in terms of
fields. It can be only written in terms of creation and annihilation
operators of particles. The power counting scheme does not apply.
Moreover, I have a proof that the S-matrix of my approach is exactly
the same as the S-matrix of renormalized QED. My approach is
not just renormalizable. It is finite. Both Hamiltonian and the S-matrix
are finite.

>
>>The main idea of the "dressed particle" approach is to remove these
>>trilinear "unphysical" operators from the Hamiltonian. This can be
>>done by a unitary transformation which does not change the S-matrix.
>>Then, the theory becomes very similar to multi-particle quantum
>>mechanics. The only difference is that creation and annihilation of
>>particles is permitted.
>
>
> But creation and annihilation of particles is an observed fact!

This is my point. Multi-particle quantum mechanics does not
describe creation and annihilation of particles (in disagreement
with experiment). The "dressed particle" approach does describe
creation and annihilation of particles. In the dressed particle
Hamiltonian there are no terms like

e -> e + \gamma-> e + p + \gamma

which have never been observed, but there are (observable) terms like

e + e -> e + e + \gamma (bremsstrahlung)
e + p -> \gamma + \gamma (annihilation)

etc.

>
>
>>The interaction has only "physical" terms: interaction is
>>nonzero only if there are two or more particles. Single particles and
>>vacuum do not have self-interactions.
>
>
> Neither do they in usual QED.

The self-interaction of single particles and vacuum is present in QED.
The renormalization approach cancels the effects of these
self-interactions on the S-matrix. However, the self-interactions
remain in the Hamiltonian and, therefore, reveal themselves in the
time evolution.

QED conveniently bypasses this problem by refusing to calculate the
time evolution. This is the weak point of QED exposed in my book
(see www.geocities.com/meopemuk). The only way (which I know of) to
remove the self-interaction effects from the Hamiltonian is
to use the "dressing transformation".

Eugene.

Eugene Stefanovich

Nov9-04, 03:45 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>Eugene Stefanovich wrote:\n>>>\n>>>\n>>>>That\'s not the way I would like to do things. In my view, fields (either\n>>>>interacting or non-interacting) do not have any physical meaning. I\n>>>>haven\'t seen any experiment measuring fields.\n>>>\n>>>People measure the electromagnetic field routinely in daily life.\n>>>It is the basis of all our modern technical culture. So at least\n>>>the electromagnetic part of QED cannot be discussed away...\n>>\n>>I disagree. When people "measure electromagnetic field" they actually\n>>put a test charge at a point in space and measure a force exerted on\n>>the charge by surrounding charges. This can be described in terms of\n>>particles and direct interactions between them. No fields are needed.\n>>When people speak of "transversal electromagnetic wave" or light,\n>>they actually speak about a flow of particles - photons. No fields\n>>are needed. This is the main point of my approach.\n>\n>\n> But this is the useless part of your approach; what you consider a step\n> forward is in fact a step back. The only valuable part in your papers is\n> the renormalized Hamiltonian.\n\nThe renormalized Hamiltonian is just one (though important) part of my\napproach. There are other ingredients, which are also important.\nMy theory is about particles and direct interactions between them.\nThere are no fields and there are no degrees of freedom associated with\nfields.\n\n>\n> To do any realistic calculations in an external field (e.g. that produced\n> by large accelerators) you\'ll need the field; making it up by particles\n> will make things hopelessly complicated.\n> So you\'d extend your method to\n> work for QED with an external field. This is needed anyway if you want\n> to recover the anomalous magnetic moment of the electron within your\n> approach.\n>\n>\n> Arnold Neumaier\n>\n\nI agree that fields provide a useful tools for dealing with systems in\nwhich a large number of charged particles is involved (like\naccelerators). I think, I can introduce in my approach an (approximate)\nconcept of field. This could be something like the total force acting\non a test charge from a large collection of charged particles\n(for example, this could be the sum of Coulomb (12.26) and Darwin\'s\n(12.28) potentials). However, it would be rather difficult to make this\nconcept exact. First, interparticle interactions have terms proportional\nto all powers of e, not just e^2 as in Maxwell\'s theory. Second, in\naddition to pair potentials (as in Maxwell\'s theory) there are\n3-particle, 4-particle, etc. potentials. The exact electromagnetic\nfield would be rather nonlinear. Of course, all these higher order\nnonlinear corrections are very small, but they should be taken into\naccount in a rigorous approach.\n\nMost importantly, in contrast to QED, where fields are considered basic\ningredients of the theory, my "field" will be just a\nshorthand approximate notation for the interparticle interactions.\n\nI think that the idea of external field is not necessary for deriving\nthe anomalous magnetic moment of the electron. One can extract this\nquantity from scattering data between electron and other charged\nsingle particles.\n\nEugene.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>Eugene Stefanovich wrote:
>>>
>>>
>>>>That's not the way I would like to do things. In my view, fields (either
>>>>interacting or non-interacting) do not have any physical meaning. I
>>>>haven't seen any experiment measuring fields.
>>>
>>>People measure the electromagnetic field routinely in daily life.
>>>It is the basis of all our modern technical culture. So at least
>>>the electromagnetic part of QED cannot be discussed away...
>>
>>I disagree. When people "measure electromagnetic field" they actually
>>put a test charge at a point in space and measure a force exerted on
>>the charge by surrounding charges. This can be described in terms of
>>particles and direct interactions between them. No fields are needed.
>>When people speak of "transversal electromagnetic wave" or light,
>>they actually speak about a flow of particles - photons. No fields
>>are needed. This is the main point of my approach.
>
>
> But this is the useless part of your approach; what you consider a step
> forward is in fact a step back. The only valuable part in your papers is
> the renormalized Hamiltonian.

The renormalized Hamiltonian is just one (though important) part of my
approach. There are other ingredients, which are also important.
My theory is about particles and direct interactions between them.
There are no fields and there are no degrees of freedom associated with
fields.

>
> To do any realistic calculations in an external field (e.g. that produced
> by large accelerators) you'll need the field; making it up by particles
> will make things hopelessly complicated.
> So you'd extend your method to
> work for QED with an external field. This is needed anyway if you want
> to recover the anomalous magnetic moment of the electron within your
> approach.
>
>
> Arnold Neumaier
>

I agree that fields provide a useful tools for dealing with systems in
which a large number of charged particles is involved (like
accelerators). I think, I can introduce in my approach an (approximate)
concept of field. This could be something like the total force acting
on a test charge from a large collection of charged particles
(for example, this could be the sum of Coulomb (12.26) and Darwin's
(12.28) potentials). However, it would be rather difficult to make this
concept exact. First, interparticle interactions have terms proportional
to all powers of e, not just e^2 as in Maxwell's theory. Second, in
addition to pair potentials (as in Maxwell's theory) there are
3-particle, 4-particle, etc. potentials. The exact electromagnetic
field would be rather nonlinear. Of course, all these higher order
nonlinear corrections are very small, but they should be taken into
account in a rigorous approach.

Most importantly, in contrast to QED, where fields are considered basic
ingredients of the theory, my "field" will be just a
shorthand approximate notation for the interparticle interactions.

I think that the idea of external field is not necessary for deriving
the anomalous magnetic moment of the electron. One can extract this
quantity from scattering data between electron and other charged
single particles.

Eugene.

Eugene Stefanovich

Nov9-04, 03:45 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>Suppose for a moment that the series for the S-matrix and/or the\n>>Hamiltonian\n>>do not converge (though I hope this is not true).\n>>This is not the end of the world. This would simply\n>> mean that our guess for the QED interaction Hamiltonian was wrong, and\n>> we need to find a better guess.\n>\n>\n> No; there are many functions f(x) which make perfect sense for x>=0 but if\n> you expand it around x=0 the resulting power series has convergence radius\n> zero, so that the expansion does not determine the function. This is the\n> case for the S-matrix elements of field theories in 2 and 3 dimensions.\n> There the interactions are the right ones but the way of extracting\n> information by means of perturbation theory is faulty. Probably the\n> same holds for QED. Even more probably, the same holds for QCD,\n> which is asymptotically free but also is unlikely to have a convergent\n> perturbation theory.)\n\nI agree with you that the perturbation approach may be\nfaulty as well.\n\n>\n>\n>\n>>I am saying that currently accepted QED\n>>interaction is just a guess (though this guess turns out to be\n>> exceptionally good). The quantization of the Maxwell\'s theory cannot\n>>be considered as a rigorous derivation of QED. Maxwell\'s\n>>theory can be at best regarded as a very rough approximation to the full\n>>quantum theory (it is probably enough to keep 2nd and 3rd perturbation\n>> orders and c^{-2} approximation to deduce Maxwell\'s theory from QED).\n>>We cannot rigorously derive a quantum theory from its approximate\n>> version.\n>>\n>>Also, I wouldn\'t waste time on phi^4 theory. This is a toy model\n>>having little relevance to real physics. If a toy doesn\'t work it\n>>usually ends up in the garbage can.\n>\n>\n> If you look at the literature you can see that a lot is done on Phi^4\n> theory. If you can\'t do that, it is most likely that you can\'t do QED.\n> If you can do QED, you\'ll be almost certainly able to do Phi^4,\n> and everything is much simpler - and hence easier to understand and\n> easier for others to learn.\n>\n> That\'s precisely the value of toy models. One doesn\'t give children\n> real cars, but begin with toy versions, so that they can learn...\n\nI also agree with you here. My only (rather trivial)\npoint was that if a toy model\ndoes not work for real life applications, this could the problem of\nthe model\nrather than of our general approach. I am not complaining that I cannot\nride a toy car (read phi^4 theory) from home to work. However, if my\nreal car (read QED) can\'t do that, then I have a real problem and I need\nto call a mechanic.\n\nEugene Stefanovich.\n\n\n>\n>\n> Arnold Neumaier\n>\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>Suppose for a moment that the series for the S-matrix and/or the
>>Hamiltonian
>>do not converge (though I hope this is not true).
>>This is not the end of the world. This would simply
>> mean that our guess for the QED interaction Hamiltonian was wrong, and
>> we need to find a better guess.
>
>
> No; there are many functions f(x) which make perfect sense for x>=0 but if
> you expand it around x=0 the resulting power series has convergence radius
> zero, so that the expansion does not determine the function. This is the
> case for the S-matrix elements of field theories in 2 and 3 dimensions.
> There the interactions are the right ones but the way of extracting
> information by means of perturbation theory is faulty. Probably the
> same holds for QED. Even more probably, the same holds for QCD,
> which is asymptotically free but also is unlikely to have a convergent
> perturbation theory.)

I agree with you that the perturbation approach may be
faulty as well.

>
>
>
>>I am saying that currently accepted QED
>>interaction is just a guess (though this guess turns out to be
>> exceptionally good). The quantization of the Maxwell's theory cannot
>>be considered as a rigorous derivation of QED. Maxwell's
>>theory can be at best regarded as a very rough approximation to the full
>>quantum theory (it is probably enough to keep 2nd and 3rd perturbation
>> orders and c^{-2} approximation to deduce Maxwell's theory from QED).
>>We cannot rigorously derive a quantum theory from its approximate
>> version.
>>
>>Also, I wouldn't waste time on \phi^4 theory. This is a toy model
>>having little relevance to real physics. If a toy doesn't work it
>>usually ends up in the garbage can.
>
>
> If you look at the literature you can see that a lot is done on \Phi^4
> theory. If you can't do that, it is most likely that you can't do QED.
> If you can do QED, you'll be almost certainly able to do \Phi^4,
> and everything is much simpler - and hence easier to understand and
> easier for others to learn.
>
> That's precisely the value of toy models. One doesn't give children
> real cars, but begin with toy versions, so that they can learn...

I also agree with you here. My only (rather trivial)
point was that if a toy model
does not work for real life applications, this could the problem of
the model
rather than of our general approach. I am not complaining that I cannot
ride a toy car (read \phi^4 theory) from home to work. However, if my
real car (read QED) can't do that, then I have a real problem and I need
to call a mechanic.

Eugene Stefanovich.

>
>
> Arnold Neumaier
>

Hendrik van Hees

Nov9-04, 02:54 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> Could you please show me the Hamiltonian you are using to\n> describe the time evolution? (reference is OK) How do you prove\n> that the vacuum vector remains unchanged during time evolution?\n\nThe QED-Lagrangian (and thus also the Hamiltonian) can be found in any\ntextbook on quantum field theory.\n\n> They are not present in the S-matrix. True. But they are present in\n> the Hamiltonian, so they should happen at short times (if we take the\n> QED Hamiltonian at face value).\n\nIn the usually choosen representation of the Poincare group, which is a\nunitary one and not a ray representation, the vacuum is an eigenstate\nof the Hamiltonian with eigenvalue 0 and thus the vacuum state stays\nconstant in time as it should be.\n\n>> If you like to stay with renormalisable models (like QED), there is\n>> no other interaction available than the usual electron-photon vertex\n>> (simple power counting).\n>\n> Power counting applies to Hamiltonians (or Lagrangians) written as\n> products of fields. My Hamiltonian cannot be written in terms of\n> fields. It can be only written in terms of creation and annihilation\n> operators of particles.\n\nThen you should check, whether the whole thing is really physically\nmeaningful, especially whether the S-matrix is Lorentz invariant,\nunitary, and fulfills the Cluster-Decomposition property. See\nWeinberg\'s Quantum Theory of fields, Vol I, Cam. Uni. Press for\ndetails, why local QFTs guarantee these important properties of the S\nmatrix.\n\n> The power counting scheme does not apply.\n> Moreover, I have a proof that the S-matrix of my approach is exactly\n> the same as the S-matrix of renormalized QED. My approach is\n> not just renormalizable. It is finite. Both Hamiltonian and the\n> S-matrix are finite.\n\nThen your theory should be identical with QED in some way.\n>\n>>\n>>>The main idea of the "dressed particle" approach is to remove these\n>>>trilinear "unphysical" operators from the Hamiltonian. This can be\n>>>done by a unitary transformation which does not change the S-matrix.\n>>>Then, the theory becomes very similar to multi-particle quantum\n>>>mechanics. The only difference is that creation and annihilation of\n>>>particles is permitted.\n>>\n>>\n>> But creation and annihilation of particles is an observed fact!\n>\n> This is my point. Multi-particle quantum mechanics does not\n> describe creation and annihilation of particles (in disagreement\n> with experiment).\n\nThis I do not understand, since quantum field theory exactly does\ndescribe creation and annihilation processes all the time. Even the\nsimple propagation of a free particle can be understood as creation and\nannihilation of particles ;-).\n\n> The "dressed particle" approach does describe\n> creation and annihilation of particles. In the dressed particle\n> Hamiltonian there are no terms like\n>\n> e -> e + \\gamma\n> 0 -> e + p + \\gamma\n>\n> which have never been observed, but there are (observable) terms like\n>\n> e + e -> e + e + \\gamma (bremsstrahlung)\n> e + p -> \\gamma + \\gamma (annihilation)\n\n> The self-interaction of single particles and vacuum is present in QED.\n> The renormalization approach cancels the effects of these\n> self-interactions on the S-matrix. However, the self-interactions\n> remain in the Hamiltonian and, therefore, reveal themselves in the\n> time evolution.\n\nHow comes? I mean, of course you have to calculate with the physical\nfinite quantities.\n\n--\nHendrik van Hees Cyclotron Institute\nPhone: +1 979/845-1411 Texas A&M University\nFax: +1 979/845-1899 Cyclotron Institute, MS-3366\nhttp://theory.gsi.de/~vanhees/ College Station, TX 77843-3366\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> Could you please show me the Hamiltonian you are using to
> describe the time evolution? (reference is OK) How do you prove
> that the vacuum vector remains unchanged during time evolution?

The QED-Lagrangian (and thus also the Hamiltonian) can be found in any
textbook on quantum field theory.

> They are not present in the S-matrix. True. But they are present in
> the Hamiltonian, so they should happen at short times (if we take the
> QED Hamiltonian at face value).

In the usually choosen representation of the Poincare group, which is a
unitary one and not a ray representation, the vacuum is an eigenstate
of the Hamiltonian with eigenvalue and thus the vacuum state stays
constant in time as it should be.

>> If you like to stay with renormalisable models (like QED), there is
>> no other interaction available than the usual electron-photon vertex
>> (simple power counting).
>
> Power counting applies to Hamiltonians (or Lagrangians) written as
> products of fields. My Hamiltonian cannot be written in terms of
> fields. It can be only written in terms of creation and annihilation
> operators of particles.

Then you should check, whether the whole thing is really physically
meaningful, especially whether the S-matrix is Lorentz invariant,
unitary, and fulfills the Cluster-Decomposition property. See
Weinberg's Quantum Theory of fields, Vol I, Cam. Uni. Press for
details, why local QFTs guarantee these important properties of the S
matrix.

> The power counting scheme does not apply.
> Moreover, I have a proof that the S-matrix of my approach is exactly
> the same as the S-matrix of renormalized QED. My approach is
> not just renormalizable. It is finite. Both Hamiltonian and the
> S-matrix are finite.

Then your theory should be identical with QED in some way.
>
>>
>>>The main idea of the "dressed particle" approach is to remove these
>>>trilinear "unphysical" operators from the Hamiltonian. This can be
>>>done by a unitary transformation which does not change the S-matrix.
>>>Then, the theory becomes very similar to multi-particle quantum
>>>mechanics. The only difference is that creation and annihilation of
>>>particles is permitted.
>>
>>
>> But creation and annihilation of particles is an observed fact!
>
> This is my point. Multi-particle quantum mechanics does not
> describe creation and annihilation of particles (in disagreement
> with experiment).

This I do not understand, since quantum field theory exactly does
describe creation and annihilation processes all the time. Even the
simple propagation of a free particle can be understood as creation and
annihilation of particles ;-).

> The "dressed particle" approach does describe
> creation and annihilation of particles. In the dressed particle
> Hamiltonian there are no terms like
>
> e -> e + \gamma
> -> e + p + \gamma
>
> which have never been observed, but there are (observable) terms like
>
> e + e -> e + e + \gamma (bremsstrahlung)
> e + p -> \gamma + \gamma (annihilation)

> The self-interaction of single particles and vacuum is present in QED.
> The renormalization approach cancels the effects of these
> self-interactions on the S-matrix. However, the self-interactions
> remain in the Hamiltonian and, therefore, reveal themselves in the
> time evolution.

How comes? I mean, of course you have to calculate with the physical
finite quantities.

--
Hendrik van Hees Cyclotron Institute
Phone: +1 979/845-1411 Texas A&M University
Fax: +1 979/845-1899 Cyclotron Institute, MS-3366
http://theory.gsi.de/~vanhees/ College Station, TX 77843-3366

Eugene Stefanovich

Nov14-04, 06:26 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>\n>\n>>>The perturbation series for the renormalized S-matrix, yes.\n>>>It may be different for the perturbation series for the renormalized\n>>>Hamiltonian which Stefanovich constructed, and which might actually\n>>>converge.\n>>>\n>>\n>>I don\'t quite understand your logic. Suppose that the perturbation\n>> series for the renormalized S-matrix does not converge, then we can\n>>be certain that the corresponding "dressed" Hamiltonian is wrong\n>>even if it\n>>is finite (convergent).\n>\n>\n> No. The perturbation series for the renormalized S-matrix does not\n> converge almost certainly, no matter whether you calculate it from\n> standard QED or fron your recipe, which, as you proved, gives identical\n> results.\n>\n> But since the perturbation series often diverges even for nonrelativistic\n> Hamiltonians of the textbook kind, your series for the Hamiltonian might\n> well converge without forcing any conclusion for the S-matrix.\n>\n>\n> Arnold Neumaier\n\nIn this case, I would say that the Hamiltonian is wrong (correct\nHamiltonian must result in finite scattering amplitudes), and studying\nits convergence properties is an academic exercise. We should find\nthe correct Hamiltonian first.\n\nEugene.\n\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>
>
>>>The perturbation series for the renormalized S-matrix, yes.
>>>It may be different for the perturbation series for the renormalized
>>>Hamiltonian which Stefanovich constructed, and which might actually
>>>converge.
>>>
>>
>>I don't quite understand your logic. Suppose that the perturbation
>> series for the renormalized S-matrix does not converge, then we can
>>be certain that the corresponding "dressed" Hamiltonian is wrong
>>even if it
>>is finite (convergent).
>
>
> No. The perturbation series for the renormalized S-matrix does not
> converge almost certainly, no matter whether you calculate it from
> standard QED or fron your recipe, which, as you proved, gives identical
> results.
>
> But since the perturbation series often diverges even for nonrelativistic
> Hamiltonians of the textbook kind, your series for the Hamiltonian might
> well converge without forcing any conclusion for the S-matrix.
>
>
> Arnold Neumaier

In this case, I would say that the Hamiltonian is wrong (correct
Hamiltonian must result in finite scattering amplitudes), and studying
its convergence properties is an academic exercise. We should find
the correct Hamiltonian first.

Eugene.

>

Arnold Neumaier

Nov14-04, 11:59 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nEugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>But since the perturbation series often diverges even for nonrelativistic\n>>Hamiltonians of the textbook kind, your series for the Hamiltonian might\n>>well converge without forcing any conclusion for the S-matrix.\n>\n> In this case, I would say that the Hamiltonian is wrong (correct\n> Hamiltonian must result in finite scattering amplitudes), and studying\n> its convergence properties is an academic exercise. We should find\n> the correct Hamiltonian first.\n\nThis means that you still have to do some work. For your Hamiltonian\nis only defined order by order. To give a well-defined expression for\nthe full Hamiltonian requires showing that the sequence of approximate\nHamiltonians you constructed converges.\n\nOn the other hand, there seems to be a misunderstanding here.\n\nNo one knows at present whether QED as currently formulated predicts\nfinite scattering amplitudes. it only predicts finite contributions at\neach order. Your reformulation of QED does not change this since your\narguments are also only order by order. With your recipe all one can do\nis pick one of your approximate Hamiltonians at a particular order k\nand compute its S-matrix expansion. This gives finite scattering amplitudes,\nbut for the approximate theory only. It will (most likely, though not\neven that seems to be proved at the moment) have a perturbation expansion\nwhose terms up to order k coincide with those of QED.\nIt does _not_ give finite scattering amplitudes for QED itself.\n\nNote:\nIf we have a well-defined Hamiltonian H(g) depending on a parameter g,\nit has a well-defined S-matrix S(g) also depending on g.\nPerturbation theory compute a power series expansion\nS(g)=S_0 + S_1 g + ...\nwhich often diverges for all g although each S_k is finite.\nThis happens already for the anharmonic oscillator with\nH(g)= 1/2 (p^2+q^2) +g q^4.\nThus a correct Hamultonian with a convergent (in the harmonic\noscillator case even finite, hence trivially convergent) expansion\nis quite consistent with a divergent expansion of the S-matrix.\n\nThus while it is clearly not reasonable to demanding a convergent\nS-matrix expansion, your Hamiltonian expansion could still be\nconvergent. If - and only if - you could show that, you\'d indeed\nhave found the correct Hamiltonian.\n\n\nArnold Neumaier\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>But since the perturbation series often diverges even for nonrelativistic
>>Hamiltonians of the textbook kind, your series for the Hamiltonian might
>>well converge without forcing any conclusion for the S-matrix.
>
> In this case, I would say that the Hamiltonian is wrong (correct
> Hamiltonian must result in finite scattering amplitudes), and studying
> its convergence properties is an academic exercise. We should find
> the correct Hamiltonian first.

This means that you still have to do some work. For your Hamiltonian
is only defined order by order. To give a well-defined expression for
the full Hamiltonian requires showing that the sequence of approximate
Hamiltonians you constructed converges.

On the other hand, there seems to be a misunderstanding here.

No one knows at present whether QED as currently formulated predicts
finite scattering amplitudes. it only predicts finite contributions at
each order. Your reformulation of QED does not change this since your
arguments are also only order by order. With your recipe all one can do
is pick one of your approximate Hamiltonians at a particular order k
and compute its S-matrix expansion. This gives finite scattering amplitudes,
but for the approximate theory only. It will (most likely, though not
even that seems to be proved at the moment) have a perturbation expansion
whose terms up to order k coincide with those of QED.
It does _not_ give finite scattering amplitudes for QED itself.

Note:
If we have a well-defined Hamiltonian H(g) depending on a parameter g,
it has a well-defined S-matrix S(g) also depending on g.
Perturbation theory compute a power series expansion
S(g)=S_0 + S_1 g + ...
which often diverges for all g although each S_k is finite.
This happens already for the anharmonic oscillator with
H(g)= 1/2 (p^2+q^2) +g q^4.
Thus a correct Hamultonian with a convergent (in the harmonic
oscillator case even finite, hence trivially convergent) expansion
is quite consistent with a divergent expansion of the S-matrix.

Thus while it is clearly not reasonable to demanding a convergent
S-matrix expansion, your Hamiltonian expansion could still be
convergent. If - and only if - you could show that, you'd indeed
have found the correct Hamiltonian.

Arnold Neumaier

Arnold Neumaier

Nov14-04, 11:59 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nEugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>Eugene Stefanovich wrote:\n>>\n>>>My theory is about particles and direct interactions between them.\n>>>There are no fields and there are no degrees of freedom associated with\n>>>fields.\n>>\n>>There are fields, namely the creation and annihilation operators,\n>>or if you prefer their real and imaginary parts, just like in\n>>the free theory.\n>>\n>>You close your eyes to these fields since they don\'t fit your\n>>personal philosophy. A possible but not a very fruitful way to\n>>face reality.\n>\n> I agree that we can write fields as mathematical objects, and\n> sometimes they are convenient to do math. I disagree that fields\n> have any observable meaning. All we measure in experiments are\n> properties of particles: position, momentum, spin, etc.\n\nFields had an observable meaning already long before the advent of QM,\nand this is not going to change just because you don\'t like them.\n\n\n>>>I think that the idea of external field is not necessary for deriving\n>>>the anomalous magnetic moment of the electron. One can extract this\n>>>quantity from scattering data between electron and other charged\n>>>single particles.\n>>\n>>Well, do it to prove your point! Claims are not enough. To be convincing\n>>you need at least to outline how it can actually be done.\n>\n> I have no proof.\n\nThen better be silent, or at least formulate your speculations more\ncarefully, adding qualifiers like \'probably\' or \'I expect\' instead of\nboldly claiming \'can\'!\n\nIt is _very_ easy to make wrong claims in QM just by guessing on\nthe basis of analogies or untested ideas. The history of particle physics\nis full of that...\n\n\n> I base my statement on the understanding that\n> electron\'s magnetic moment plays a role in interaction\n> electron - electron. So, there should be some indications of\n> the value of this moment in electron-electron scattering data.\n> Moreover, I believe, the electron\'s magnetic moment can be\n> extracted from fine and hyperfine structure of the hydrogen spectrum.\n>\n> My point is that the magnetic moment can be considered in 2-particle\n> problems, without invoking the concept of the external field.\n\nBut the _measured_ anomalous magnetic moment requires a strong\nexternal field. No way to measure it in 2-particle experiments.\n\n\nArnold Neumaier\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>Eugene Stefanovich wrote:
>>
>>>My theory is about particles and direct interactions between them.
>>>There are no fields and there are no degrees of freedom associated with
>>>fields.
>>
>>There are fields, namely the creation and annihilation operators,
>>or if you prefer their real and imaginary parts, just like in
>>the free theory.
>>
>>You close your eyes to these fields since they don't fit your
>>personal philosophy. A possible but not a very fruitful way to
>>face reality.
>
> I agree that we can write fields as mathematical objects, and
> sometimes they are convenient to do math. I disagree that fields
> have any observable meaning. All we measure in experiments are
> properties of particles: position, momentum, spin, etc.

Fields had an observable meaning already long before the advent of QM,
and this is not going to change just because you don't like them.

>>>I think that the idea of external field is not necessary for deriving
>>>the anomalous magnetic moment of the electron. One can extract this
>>>quantity from scattering data between electron and other charged
>>>single particles.
>>
>>Well, do it to prove your point! Claims are not enough. To be convincing
>>you need at least to outline how it can actually be done.
>
> I have no proof.

Then better be silent, or at least formulate your speculations more
carefully, adding qualifiers like 'probably' or 'I expect' instead of
boldly claiming 'can'!

It is _very_ easy to make wrong claims in QM just by guessing on
the basis of analogies or untested ideas. The history of particle physics
is full of that...

> I base my statement on the understanding that
> electron's magnetic moment plays a role in interaction
> electron - electron. So, there should be some indications of
> the value of this moment in electron-electron scattering data.
> Moreover, I believe, the electron's magnetic moment can be
> extracted from fine and hyperfine structure of the hydrogen spectrum.
>
> My point is that the magnetic moment can be considered in 2-particle
> problems, without invoking the concept of the external field.

But the _measured_ anomalous magnetic moment requires a strong
external field. No way to measure it in 2-particle experiments.

Arnold Neumaier

Arnold Neumaier

Nov14-04, 11:59 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nEugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>Actually, creation and annihilation operators _are_ field operators,\n>>and one can reconstruct the standard psi(x) as a linear combination\n>>of the a(x) and a^*(x). Thus Stefanovich constructs a field theory,\n>>but his peculiar philosophy blinds him to this fact.\n>\n> The "dressing transformation" cannot be understood in terms of fields\n> psi(x).\n\nOnly for you. I have no difficulties understanding this in terms of field.\nWhat you are doing is applying canonical transformations that transform\nthe physical vacuum and 1-particle states states into the bare states.\nThe corresponding dual canonical transformations transform the\nphysical fields acting on the physical vacuum into fields acting on the\nbare vacuum.\n\nWhat you are doing is just a coordinate transformation in infinite\ndimensions. People have been doing this within field theory almost\nsince the beginning. The only new and interesting thing about your\nwork is that you showed that you can push this through to all orders\nof perturbation theory, thereby renormalizing the Hamiltonian.\n\n\n> This transformation requires separation of each operator in\n> "physical", "unphysical", and "renorm" part. These parts are easily\n> expressed in terms of momentum-space creation and annihilation operators\n> a(p) and a^*(p). I don\'t know how to do the same in terms of fields.\n> Maybe, this can be done, but then field expressions would be very ugly.\n>\n> My point is that fields are just mathematical constructs.\n> They are sometimes useful for doing math\n> (e.g., for proving the relativistic invariance\n> of the theory). They are useless for the physical interpretation of\n> the theory.\n\n.... except that most people, including all practicioners, find it very\nvaluable, while your proposal that they are useless is that of an\noutsider. As long as you support such views it will be very difficult\nfor you to gain acceptance.\n\n\nArnold Neumaier\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>Actually, creation and annihilation operators _are_ field operators,
>>and one can reconstruct the standard \psi(x) as a linear combination
>>of the a(x) and a^*(x). Thus Stefanovich constructs a field theory,
>>but his peculiar philosophy blinds him to this fact.
>
> The "dressing transformation" cannot be understood in terms of fields
> \psi(x).

Only for you. I have no difficulties understanding this in terms of field.
What you are doing is applying canonical transformations that transform
the physical vacuum and 1-particle states states into the bare states.
The corresponding dual canonical transformations transform the
physical fields acting on the physical vacuum into fields acting on the
bare vacuum.

What you are doing is just a coordinate transformation in infinite
dimensions. People have been doing this within field theory almost
since the beginning. The only new and interesting thing about your
work is that you showed that you can push this through to all orders
of perturbation theory, thereby renormalizing the Hamiltonian.

> This transformation requires separation of each operator in
> "physical", "unphysical", and "renorm" part. These parts are easily
> expressed in terms of momentum-space creation and annihilation operators
> a(p) and a^*(p). I don't know how to do the same in terms of fields.
> Maybe, this can be done, but then field expressions would be very ugly.
>
> My point is that fields are just mathematical constructs.
> They are sometimes useful for doing math
> (e.g., for proving the relativistic invariance
> of the theory). They are useless for the physical interpretation of
> the theory.

.... except that most people, including all practicioners, find it very
valuable, while your proposal that they are useless is that of an
outsider. As long as you support such views it will be very difficult
for you to gain acceptance.

Arnold Neumaier

Eugene Stefanovich

Nov16-04, 02:18 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>But since the perturbation series often diverges even for nonrelativistic\n>>>Hamiltonians of the textbook kind, your series for the Hamiltonian might\n>>>well converge without forcing any conclusion for the S-matrix.\n>>\n>>In this case, I would say that the Hamiltonian is wrong (correct\n>>Hamiltonian must result in finite scattering amplitudes), and studying\n>>its convergence properties is an academic exercise. We should find\n>>the correct Hamiltonian first.\n>\n>\n> This means that you still have to do some work. For your Hamiltonian\n> is only defined order by order. To give a well-defined expression for\n> the full Hamiltonian requires showing that the sequence of approximate\n> Hamiltonians you constructed converges.\n\nAgreed.\n\n>\n> On the other hand, there seems to be a misunderstanding here.\n>\n> No one knows at present whether QED as currently formulated predicts\n> finite scattering amplitudes. it only predicts finite contributions at\n> each order. Your reformulation of QED does not change this since your\n> arguments are also only order by order. With your recipe all one can do\n> is pick one of your approximate Hamiltonians at a particular order k\n> and compute its S-matrix expansion. This gives finite scattering amplitudes,\n> but for the approximate theory only. It will (most likely, though not\n> even that seems to be proved at the moment) have a perturbation expansion\n> whose terms up to order k coincide with those of QED.\n> It does _not_ give finite scattering amplitudes for QED itself.\n>\n> Note:\n> If we have a well-defined Hamiltonian H(g) depending on a parameter g,\n> it has a well-defined S-matrix S(g) also depending on g.\n> Perturbation theory compute a power series expansion\n> S(g)=S_0 + S_1 g + ...\n> which often diverges for all g although each S_k is finite.\n> This happens already for the anharmonic oscillator with\n> H(g)= 1/2 (p^2+q^2) +g q^4.\n> Thus a correct Hamultonian with a convergent (in the harmonic\n> oscillator case even finite, hence trivially convergent) expansion\n> is quite consistent with a divergent expansion of the S-matrix.\n>\n> Thus while it is clearly not reasonable to demanding a convergent\n> S-matrix expansion, your Hamiltonian expansion could still be\n> convergent. If - and only if - you could show that, you\'d indeed\n> have found the correct Hamiltonian.\n>\n>\n> Arnold Neumaier\n>\n\n\nI basically agree with what you said. The convergence of the series\nfor both Hamiltonian and the S-matrix should be investigated.\n\n\nEugene\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>But since the perturbation series often diverges even for nonrelativistic
>>>Hamiltonians of the textbook kind, your series for the Hamiltonian might
>>>well converge without forcing any conclusion for the S-matrix.
>>
>>In this case, I would say that the Hamiltonian is wrong (correct
>>Hamiltonian must result in finite scattering amplitudes), and studying
>>its convergence properties is an academic exercise. We should find
>>the correct Hamiltonian first.
>
>
> This means that you still have to do some work. For your Hamiltonian
> is only defined order by order. To give a well-defined expression for
> the full Hamiltonian requires showing that the sequence of approximate
> Hamiltonians you constructed converges.

Agreed.

>
> On the other hand, there seems to be a misunderstanding here.
>
> No one knows at present whether QED as currently formulated predicts
> finite scattering amplitudes. it only predicts finite contributions at
> each order. Your reformulation of QED does not change this since your
> arguments are also only order by order. With your recipe all one can do
> is pick one of your approximate Hamiltonians at a particular order k
> and compute its S-matrix expansion. This gives finite scattering amplitudes,
> but for the approximate theory only. It will (most likely, though not
> even that seems to be proved at the moment) have a perturbation expansion
> whose terms up to order k coincide with those of QED.
> It does _not_ give finite scattering amplitudes for QED itself.
>
> Note:
> If we have a well-defined Hamiltonian H(g) depending on a parameter g,
> it has a well-defined S-matrix S(g) also depending on g.
> Perturbation theory compute a power series expansion
> S(g)=S_0 + S_1 g + ...
> which often diverges for all g although each S_k is finite.
> This happens already for the anharmonic oscillator with
> H(g)= 1/2 (p^2+q^2) +g q^4.
> Thus a correct Hamultonian with a convergent (in the harmonic
> oscillator case even finite, hence trivially convergent) expansion
> is quite consistent with a divergent expansion of the S-matrix.
>
> Thus while it is clearly not reasonable to demanding a convergent
> S-matrix expansion, your Hamiltonian expansion could still be
> convergent. If - and only if - you could show that, you'd indeed
> have found the correct Hamiltonian.
>
>
> Arnold Neumaier
>

I basically agree with what you said. The convergence of the series
for both Hamiltonian and the S-matrix should be investigated.

Eugene

Eugene Stefanovich

Nov16-04, 02:18 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>Actually, creation and annihilation operators _are_ field operators,\n>>>and one can reconstruct the standard psi(x) as a linear combination\n>>>of the a(x) and a^*(x). Thus Stefanovich constructs a field theory,\n>>>but his peculiar philosophy blinds him to this fact.\n>>\n>>The "dressing transformation" cannot be understood in terms of fields\n>>psi(x).\n>\n>\n> Only for you. I have no difficulties understanding this in terms of field.\n> What you are doing is applying canonical transformations that transform\n> the physical vacuum and 1-particle states states into the bare states.\n> The corresponding dual canonical transformations transform the\n> physical fields acting on the physical vacuum into fields acting on the\n> bare vacuum.\n>\n> What you are doing is just a coordinate transformation in infinite\n> dimensions. People have been doing this within field theory almost\n> since the beginning. The only new and interesting thing about your\n> work is that you showed that you can push this through to all orders\n> of perturbation theory, thereby renormalizing the Hamiltonian.\n\nThe "dressed" Hamiltonian can be easily expressed in terms of particle\ncreation and annihilation operators, and these expressions have very\ntransparent physical meaning, and convenient for calculations.\nProbably, the same Hamiltonian can be expressed in terms of fields,\n(creatio-annihilation operators can be written as linear\ncombinations of fields) but I suspect that the formulas would be ugly\nand not particularly illuminating. I haven\'t done that, and I might\nbe wrong. That\'s a qualifier.\n\n>\n>\n>>This transformation requires separation of each operator in\n>>"physical", "unphysical", and "renorm" part. These parts are easily\n>>expressed in terms of momentum-space creation and annihilation operators\n>>a(p) and a^*(p). I don\'t know how to do the same in terms of fields.\n>>Maybe, this can be done, but then field expressions would be very ugly.\n>>\n>>My point is that fields are just mathematical constructs.\n>>They are sometimes useful for doing math\n>>(e.g., for proving the relativistic invariance\n>>of the theory). They are useless for the physical interpretation of\n>>the theory.\n>\n>\n> ... except that most people, including all practicioners, find it very\n> valuable, while your proposal that they are useless is that of an\n> outsider. As long as you support such views it will be very difficult\n> for you to gain acceptance.\n>\n>\n> Arnold Neumaier\n>\n\nThe final results of any QED calculation are either scattering\namplitudes\n(described in terms of momenta and spins of particles) or energies of\nbound states. So, fields may play some role in the theoretical\nformalism, but they are not needed to connect the results of theory\nwith experiment.\n\nI am fully aware that my views are not going to be accepted so easily.\nAs time goes by, I see it more and more clear,\nhow difficult it would be to gain acceptance. My approach attacks so\nmany "well-established" concepts. My attack on fields is probably not\nthe most controversial. I refuse to accept such "pillar of modern\nphysics" as Minkowski spacetime. That will get me in trouble, for\nsure.\n\nNevertheless, I am confident in my approach, and so far I haven\'t\nheard any convincing argument that could shake my confidence.\n\nBest regards.\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>Actually, creation and annihilation operators _are_ field operators,
>>>and one can reconstruct the standard \psi(x) as a linear combination
>>>of the a(x) and a^*(x). Thus Stefanovich constructs a field theory,
>>>but his peculiar philosophy blinds him to this fact.
>>
>>The "dressing transformation" cannot be understood in terms of fields
>>\psi(x).
>
>
> Only for you. I have no difficulties understanding this in terms of field.
> What you are doing is applying canonical transformations that transform
> the physical vacuum and 1-particle states states into the bare states.
> The corresponding dual canonical transformations transform the
> physical fields acting on the physical vacuum into fields acting on the
> bare vacuum.
>
> What you are doing is just a coordinate transformation in infinite
> dimensions. People have been doing this within field theory almost
> since the beginning. The only new and interesting thing about your
> work is that you showed that you can push this through to all orders
> of perturbation theory, thereby renormalizing the Hamiltonian.

The "dressed" Hamiltonian can be easily expressed in terms of particle
creation and annihilation operators, and these expressions have very
transparent physical meaning, and convenient for calculations.
Probably, the same Hamiltonian can be expressed in terms of fields,
(creatio-annihilation operators can be written as linear
combinations of fields) but I suspect that the formulas would be ugly
and not particularly illuminating. I haven't done that, and I might
be wrong. That's a qualifier.

>
>
>>This transformation requires separation of each operator in
>>"physical", "unphysical", and "renorm" part. These parts are easily
>>expressed in terms of momentum-space creation and annihilation operators
>>a(p) and a^*(p). I don't know how to do the same in terms of fields.
>>Maybe, this can be done, but then field expressions would be very ugly.
>>
>>My point is that fields are just mathematical constructs.
>>They are sometimes useful for doing math
>>(e.g., for proving the relativistic invariance
>>of the theory). They are useless for the physical interpretation of
>>the theory.
>
>
> ... except that most people, including all practicioners, find it very
> valuable, while your proposal that they are useless is that of an
> outsider. As long as you support such views it will be very difficult
> for you to gain acceptance.
>
>
> Arnold Neumaier
>

The final results of any QED calculation are either scattering
amplitudes
(described in terms of momenta and spins of particles) or energies of
bound states. So, fields may play some role in the theoretical
formalism, but they are not needed to connect the results of theory
with experiment.

I am fully aware that my views are not going to be accepted so easily.
As time goes by, I see it more and more clear,
how difficult it would be to gain acceptance. My approach attacks so
many "well-established" concepts. My attack on fields is probably not
the most controversial. I refuse to accept such "pillar of modern
physics" as Minkowski spacetime. That will get me in trouble, for
sure.

Nevertheless, I am confident in my approach, and so far I haven't
heard any convincing argument that could shake my confidence.

Best regards.
Eugene Stefanovich.

Eugene Stefanovich

Nov16-04, 02:18 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>Actually, creation and annihilation operators _are_ field operators,\n>>>and one can reconstruct the standard psi(x) as a linear combination\n>>>of the a(x) and a^*(x). Thus Stefanovich constructs a field theory,\n>>>but his peculiar philosophy blinds him to this fact.\n>>\n>>The "dressing transformation" cannot be understood in terms of fields\n>>psi(x).\n>\n>\n> Only for you. I have no difficulties understanding this in terms of field.\n> What you are doing is applying canonical transformations that transform\n> the physical vacuum and 1-particle states states into the bare states.\n> The corresponding dual canonical transformations transform the\n> physical fields acting on the physical vacuum into fields acting on the\n> bare vacuum.\n>\n> What you are doing is just a coordinate transformation in infinite\n> dimensions. People have been doing this within field theory almost\n> since the beginning. The only new and interesting thing about your\n> work is that you showed that you can push this through to all orders\n> of perturbation theory, thereby renormalizing the Hamiltonian.\n>\n>\n>\n>>This transformation requires separation of each operator in\n>>"physical", "unphysical", and "renorm" part. These parts are easily\n>>expressed in terms of momentum-space creation and annihilation operators\n>>a(p) and a^*(p). I don\'t know how to do the same in terms of fields.\n>>Maybe, this can be done, but then field expressions would be very ugly.\n>>\n>>My point is that fields are just mathematical constructs.\n>>They are sometimes useful for doing math\n>>(e.g., for proving the relativistic invariance\n>>of the theory). They are useless for the physical interpretation of\n>>the theory.\n>\n>\n> ... except that most people, including all practicioners, find it very\n> valuable, while your proposal that they are useless is that of an\n> outsider. As long as you support such views it will be very difficult\n> for you to gain acceptance.\n>\n>\n> Arnold Neumaier\n>\n\nOK, you agree that a finite Hamiltonian of QED can be constructed.\n(the Hamiltonian is finite in each order separately, let us set aside\nthe question about convergence). This Hamiltonian is expressed in\nterms of creation and annihilations operators of real particles\n(electrons, photons, protons, etc.). Therefore, the interacting\npotentials between particles act instantaneously. There are no\ndegrees of freedom associated with fields or virtual particles,\nso, there are no degrees of freedom which can "keep" the momentum\nand energy while they are transferring from one particle to another.\nTherefore, the interaction cannot be retarded. It must be instantaneous.\n\nIf this is true, then the usual approach to relativity (with universal\nlinear Lorentz transformations of observables) cannot be used.\nThis approach cannot reconcile the action-at-a-distance with causality.\nTo do that, one needs to admit that boost transformations are\ninteraction-dependent.\n\nYou cannot accept just one part of my approach and ignore the others.\nEverything must be\nconsistent, and this consistency requires deep changes in our\nfundamental views on particles, fields, space, and time. So far,\nnobody (except myself) have subscribed to such changes.\nUnfortunately, my theory doesn\'t have any spectacular new\nexperimental predictions, whose confirmations would make it\nmore acceptable. But I believe, the acceptance is just a matter of time.\n\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>Actually, creation and annihilation operators _are_ field operators,
>>>and one can reconstruct the standard \psi(x) as a linear combination
>>>of the a(x) and a^*(x). Thus Stefanovich constructs a field theory,
>>>but his peculiar philosophy blinds him to this fact.
>>
>>The "dressing transformation" cannot be understood in terms of fields
>>\psi(x).
>
>
> Only for you. I have no difficulties understanding this in terms of field.
> What you are doing is applying canonical transformations that transform
> the physical vacuum and 1-particle states states into the bare states.
> The corresponding dual canonical transformations transform the
> physical fields acting on the physical vacuum into fields acting on the
> bare vacuum.
>
> What you are doing is just a coordinate transformation in infinite
> dimensions. People have been doing this within field theory almost
> since the beginning. The only new and interesting thing about your
> work is that you showed that you can push this through to all orders
> of perturbation theory, thereby renormalizing the Hamiltonian.
>
>
>
>>This transformation requires separation of each operator in
>>"physical", "unphysical", and "renorm" part. These parts are easily
>>expressed in terms of momentum-space creation and annihilation operators
>>a(p) and a^*(p). I don't know how to do the same in terms of fields.
>>Maybe, this can be done, but then field expressions would be very ugly.
>>
>>My point is that fields are just mathematical constructs.
>>They are sometimes useful for doing math
>>(e.g., for proving the relativistic invariance
>>of the theory). They are useless for the physical interpretation of
>>the theory.
>
>
> ... except that most people, including all practicioners, find it very
> valuable, while your proposal that they are useless is that of an
> outsider. As long as you support such views it will be very difficult
> for you to gain acceptance.
>
>
> Arnold Neumaier
>

OK, you agree that a finite Hamiltonian of QED can be constructed.
(the Hamiltonian is finite in each order separately, let us set aside
the question about convergence). This Hamiltonian is expressed in
terms of creation and annihilations operators of real particles
(electrons, photons, protons, etc.). Therefore, the interacting
potentials between particles act instantaneously. There are no
degrees of freedom associated with fields or virtual particles,
so, there are no degrees of freedom which can "keep" the momentum
and energy while they are transferring from one particle to another.
Therefore, the interaction cannot be retarded. It must be instantaneous.

If this is true, then the usual approach to relativity (with universal
linear Lorentz transformations of observables) cannot be used.
This approach cannot reconcile the action-at-a-distance with causality.
To do that, one needs to admit that boost transformations are
interaction-dependent.

You cannot accept just one part of my approach and ignore the others.
Everything must be
consistent, and this consistency requires deep changes in our
fundamental views on particles, fields, space, and time. So far,
nobody (except myself) have subscribed to such changes.
Unfortunately, my theory doesn't have any spectacular new
experimental predictions, whose confirmations would make it
more acceptable. But I believe, the acceptance is just a matter of time.

Eugene.

Eugene Stefanovich

Nov16-04, 02:56 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>Eugene Stefanovich wrote:\n>>>\n>>>\n>>>>My theory is about particles and direct interactions between them.\n>>>>There are no fields and there are no degrees of freedom associated with\n>>>>fields.\n>>>\n>>>There are fields, namely the creation and annihilation operators,\n>>>or if you prefer their real and imaginary parts, just like in\n>>>the free theory.\n>>>\n>>>You close your eyes to these fields since they don\'t fit your\n>>>personal philosophy. A possible but not a very fruitful way to\n>>>face reality.\n>>\n>>I agree that we can write fields as mathematical objects, and\n>>sometimes they are convenient to do math. I disagree that fields\n>>have any observable meaning. All we measure in experiments are\n>>properties of particles: position, momentum, spin, etc.\n>\n>\n> Fields had an observable meaning already long before the advent of QM,\n> and this is not going to change just because you don\'t like them.\n\nNobody has ever measured electric and magnetic fields. In experiment we\nmeasure forces acting on a test charge and depending on the charge\'s\nposition and velocity. Then we invent the theoretical notion of\n"field" which helps us to visualize these forces. This theoretical\nnotion can be safely omitted, and electromagnetic interactions between\nparticles can be described as astion-at-a-distance. My book shows\nthat this view does not contradict any experimental data.\n\n>\n>\n>\n>>>>I think that the idea of external field is not necessary for deriving\n>>>>the anomalous magnetic moment of the electron. One can extract this\n>>>>quantity from scattering data between electron and other charged\n>>>>single particles.\n>>>\n>>>Well, do it to prove your point! Claims are not enough. To be convincing\n>>>you need at least to outline how it can actually be done.\n>>\n>>I have no proof.\n>\n>\n> Then better be silent, or at least formulate your speculations more\n> carefully, adding qualifiers like \'probably\' or \'I expect\' instead of\n> boldly claiming \'can\'!\n\nI would probably agree with you that I can use more\nqualifiers.\n\n>\n> It is _very_ easy to make wrong claims in QM just by guessing on\n> the basis of analogies or untested ideas. The history of particle physics\n> is full of that...\n>\n>\n>\n>>I base my statement on the understanding that\n>>electron\'s magnetic moment plays a role in interaction\n>>electron - electron. So, there should be some indications of\n>>the value of this moment in electron-electron scattering data.\n>>Moreover, I believe, the electron\'s magnetic moment can be\n>>extracted from fine and hyperfine structure of the hydrogen spectrum.\n>>\n>>My point is that the magnetic moment can be considered in 2-particle\n>>problems, without invoking the concept of the external field.\n>\n>\n> But the _measured_ anomalous magnetic moment requires a strong\n> external field. No way to measure it in 2-particle experiments.\n\nThis is a matter of experimental accuracy/sensitivity.\nNot a matter of principle.\n\nIn my approach I prefer to stay within the context of isolated systems.\nI do not consider systems in external fields. This should be possible,\nat the expense of sacrificing some rigor. I simply haven\'t extended\nmy theory to handle such cases yet.\n\nEugene.\n\n>\n>\n> Arnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>Eugene Stefanovich wrote:
>>>
>>>
>>>>My theory is about particles and direct interactions between them.
>>>>There are no fields and there are no degrees of freedom associated with
>>>>fields.
>>>
>>>There are fields, namely the creation and annihilation operators,
>>>or if you prefer their real and imaginary parts, just like in
>>>the free theory.
>>>
>>>You close your eyes to these fields since they don't fit your
>>>personal philosophy. A possible but not a very fruitful way to
>>>face reality.
>>
>>I agree that we can write fields as mathematical objects, and
>>sometimes they are convenient to do math. I disagree that fields
>>have any observable meaning. All we measure in experiments are
>>properties of particles: position, momentum, spin, etc.
>
>
> Fields had an observable meaning already long before the advent of QM,
> and this is not going to change just because you don't like them.

Nobody has ever measured electric and magnetic fields. In experiment we
measure forces acting on a test charge and depending on the charge's
position and velocity. Then we invent the theoretical notion of
"field" which helps us to visualize these forces. This theoretical
notion can be safely omitted, and electromagnetic interactions between
particles can be described as astion-at-a-distance. My book shows
that this view does not contradict any experimental data.

>
>
>
>>>>I think that the idea of external field is not necessary for deriving
>>>>the anomalous magnetic moment of the electron. One can extract this
>>>>quantity from scattering data between electron and other charged
>>>>single particles.
>>>
>>>Well, do it to prove your point! Claims are not enough. To be convincing
>>>you need at least to outline how it can actually be done.
>>
>>I have no proof.
>
>
> Then better be silent, or at least formulate your speculations more
> carefully, adding qualifiers like 'probably' or 'I expect' instead of
> boldly claiming 'can'!

I would probably agree with you that I can use more
qualifiers.

>
> It is _very_ easy to make wrong claims in QM just by guessing on
> the basis of analogies or untested ideas. The history of particle physics
> is full of that...
>
>
>
>>I base my statement on the understanding that
>>electron's magnetic moment plays a role in interaction
>>electron - electron. So, there should be some indications of
>>the value of this moment in electron-electron scattering data.
>>Moreover, I believe, the electron's magnetic moment can be
>>extracted from fine and hyperfine structure of the hydrogen spectrum.
>>
>>My point is that the magnetic moment can be considered in 2-particle
>>problems, without invoking the concept of the external field.
>
>
> But the _measured_ anomalous magnetic moment requires a strong
> external field. No way to measure it in 2-particle experiments.

This is a matter of experimental accuracy/sensitivity.
Not a matter of principle.

In my approach I prefer to stay within the context of isolated systems.
I do not consider systems in external fields. This should be possible,
at the expense of sacrificing some rigor. I simply haven't extended
my theory to handle such cases yet.

Eugene.

>
>
> Arnold Neumaier

Arnold Neumaier

Nov17-04, 10:54 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>Eugene Stefanovich wrote:\n>>\n>>>My point is that fields are just mathematical constructs.\n>>>They are sometimes useful for doing math\n>>>(e.g., for proving the relativistic invariance\n>>>of the theory). They are useless for the physical interpretation of\n>>>the theory.\n>>\n>>... except that most people, including all practicioners, find it very\n>>valuable, while your proposal that they are useless is that of an\n>>outsider. As long as you support such views it will be very difficult\n>>for you to gain acceptance.\n>>\n> OK, you agree that a finite Hamiltonian of QED can be constructed.\n> (the Hamiltonian is finite in each order separately, let us set aside\n> the question about convergence).\n\nWell, there are other caveats in your theory, namely infrared divergences\n(ignored according to p.235 of your book draft).\n\n\n> This Hamiltonian is expressed in\n> terms of creation and annihilations operators of real particles\n> (electrons, photons, protons, etc.). Therefore, the interacting\n> potentials between particles act instantaneously. There are no\n> degrees of freedom associated with fields or virtual particles,\n\nAs in free field theory, fields and particles use the same degrees\nof freedom in complementary ways (particle-wave duality).\nOnce you have a Fock space, you have fields.\n\nFor example you get from a massless spin 1 field the electromagnetic field\nA(x,t) = exp(itH)(eps(x)a(x)+eps^*(x)a^*(x))exp(-itH)\nwith well-defined expectation values and correlation functions.\n\n\n> so, there are no degrees of freedom which can "keep" the momentum\n> and energy while they are transferring from one particle to another.\n> Therefore, the interaction cannot be retarded. It must be instantaneous.\n\nThis is just because you chose the instant form, where the state\nis - by definition - determined by a wave function on the hyperplane\nx_0=t, and the dynamical parameter is t. Of course this implies by\ndefinition that the temporal dynamics is instantaneous.\n\nIf you work instead in the point form, say, the state is determined by a\nwave function on the hyperboloid x^2=l^2. Now the dynamical parameter\nis the scale l, and the corresponding dilation dynamics is instantaneous.\nBut the temporal dynamics is non-instantaneous.\n\nYour renormalization technique probably works as well for the\ndilation generator in the point form. Then you have another\ndescription of QED, as finite as your instant form description,\nwhich shows that there is nothing special about the instant form.\n\n\n> You cannot accept just one part of my approach and ignore the others.\n\nOf course I can; I don\'t need to share your limitations.\nYour work fits quite well into a more general framework\nthat you choose to ignore.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>Eugene Stefanovich wrote:
>>
>>>My point is that fields are just mathematical constructs.
>>>They are sometimes useful for doing math
>>>(e.g., for proving the relativistic invariance
>>>of the theory). They are useless for the physical interpretation of
>>>the theory.
>>
>>... except that most people, including all practicioners, find it very
>>valuable, while your proposal that they are useless is that of an
>>outsider. As long as you support such views it will be very difficult
>>for you to gain acceptance.
>>
> OK, you agree that a finite Hamiltonian of QED can be constructed.
> (the Hamiltonian is finite in each order separately, let us set aside
> the question about convergence).

Well, there are other caveats in your theory, namely infrared divergences
(ignored according to p.235 of your book draft).

> This Hamiltonian is expressed in
> terms of creation and annihilations operators of real particles
> (electrons, photons, protons, etc.). Therefore, the interacting
> potentials between particles act instantaneously. There are no
> degrees of freedom associated with fields or virtual particles,

As in free field theory, fields and particles use the same degrees
of freedom in complementary ways (particle-wave duality).
Once you have a Fock space, you have fields.

For example you get from a massless spin 1 field the electromagnetic field
A(x,t) = \exp(itH)(eps(x)a(x)+eps^*(x)a^*(x))\exp(-itH)
with well-defined expectation values and correlation functions.

> so, there are no degrees of freedom which can "keep" the momentum
> and energy while they are transferring from one particle to another.
> Therefore, the interaction cannot be retarded. It must be instantaneous.

This is just because you chose the instant form, where the state
is - by definition - determined by a wave function on the hyperplane
x_0=t, and the dynamical parameter is t. Of course this implies by
definition that the temporal dynamics is instantaneous.

If you work instead in the point form, say, the state is determined by a
wave function on the hyperboloid x^2=l^2. Now the dynamical parameter
is the scale l, and the corresponding dilation dynamics is instantaneous.
But the temporal dynamics is non-instantaneous.

Your renormalization technique probably works as well for the
dilation generator in the point form. Then you have another
description of QED, as finite as your instant form description,
which shows that there is nothing special about the instant form.

> You cannot accept just one part of my approach and ignore the others.

Of course I can; I don't need to share your limitations.
Your work fits quite well into a more general framework
that you choose to ignore.

Arnold Neumaier

Arnold Neumaier

Nov17-04, 11:13 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nEugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>Eugene Stefanovich wrote:\n\n>>>My point is that fields are just mathematical constructs.\n>>>They are sometimes useful for doing math\n>>>(e.g., for proving the relativistic invariance\n>>>of the theory). They are useless for the physical interpretation of\n>>>the theory.\n>>\n>>... except that most people, including all practicioners, find it very\n>>valuable, while your proposal that they are useless is that of an\n>>outsider. As long as you support such views it will be very difficult\n>>for you to gain acceptance.\n>>\n>>Arnold Neumaier\n>>\n> The final results of any QED calculation are either scattering\n> amplitudes\n> (described in terms of momenta and spins of particles) or energies of\n> bound states. So, fields may play some role in the theoretical\n> formalism, but they are not needed to connect the results of theory\n> with experiment.\n\nBut the input to QED calculations are usually external fields!\nYou can hardly get away without! In practice, one wants to know\nthe response of small systems to macroscopic, external input.\n\n\n> I am fully aware that my views are not going to be accepted so easily.\n> As time goes by, I see it more and more clear,\n> how difficult it would be to gain acceptance. My approach attacks so\n> many "well-established" concepts. My attack on fields is probably not\n> the most controversial. I refuse to accept such "pillar of modern\n> physics" as Minkowski spacetime. That will get me in trouble, for\n> sure.\n\n.... in needless trouble.\n\nTradition is - for good reasons - not as bad as you may think it is.\nRewriting everything in terms of concepts you approve of would make\nphysics quite messy.\n\nMoreover, you need Minkowski space already to define the Poincare group,\nwhich is one of the ingredients that cannot be dispensed with.\nAnd to define the action from which the QED Hamiltonian derives.\nSo your space-time denying approach is intrinsically contradictory.\n\n\n> Nevertheless, I am confident in my approach, and so far I haven\'t\n> heard any convincing argument that could shake my confidence.\n\nThe problem may be with waht you regard as convincing.\nIt is a pity that you combine technical skill with conceptual\nblindfolds. But I won\'t further argue that...\n\n\nArnold Neumaier\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>Eugene Stefanovich wrote:

>>>My point is that fields are just mathematical constructs.
>>>They are sometimes useful for doing math
>>>(e.g., for proving the relativistic invariance
>>>of the theory). They are useless for the physical interpretation of
>>>the theory.
>>
>>... except that most people, including all practicioners, find it very
>>valuable, while your proposal that they are useless is that of an
>>outsider. As long as you support such views it will be very difficult
>>for you to gain acceptance.
>>
>>Arnold Neumaier
>>
> The final results of any QED calculation are either scattering
> amplitudes
> (described in terms of momenta and spins of particles) or energies of
> bound states. So, fields may play some role in the theoretical
> formalism, but they are not needed to connect the results of theory
> with experiment.

But the input to QED calculations are usually external fields!
You can hardly get away without! In practice, one wants to know
the response of small systems to macroscopic, external input.

> I am fully aware that my views are not going to be accepted so easily.
> As time goes by, I see it more and more clear,
> how difficult it would be to gain acceptance. My approach attacks so
> many "well-established" concepts. My attack on fields is probably not
> the most controversial. I refuse to accept such "pillar of modern
> physics" as Minkowski spacetime. That will get me in trouble, for
> sure.

.... in needless trouble.

Tradition is - for good reasons - not as bad as you may think it is.
Rewriting everything in terms of concepts you approve of would make
physics quite messy.

Moreover, you need Minkowski space already to define the Poincare group,
which is one of the ingredients that cannot be dispensed with.
And to define the action from which the QED Hamiltonian derives.
So your space-time denying approach is intrinsically contradictory.

> Nevertheless, I am confident in my approach, and so far I haven't
> heard any convincing argument that could shake my confidence.

The problem may be with waht you regard as convincing.
It is a pity that you combine technical skill with conceptual
blindfolds. But I won't further argue that...

Arnold Neumaier

Eugene Stefanovich

Nov18-04, 12:52 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>Eugene Stefanovich wrote:\n>>>\n>>>\n>>>>My point is that fields are just mathematical constructs.\n>>>>They are sometimes useful for doing math\n>>>>(e.g., for proving the relativistic invariance\n>>>>of the theory). They are useless for the physical interpretation of\n>>>>the theory.\n>>>\n>>>... except that most people, including all practicioners, find it very\n>>>valuable, while your proposal that they are useless is that of an\n>>>outsider. As long as you support such views it will be very difficult\n>>>for you to gain acceptance.\n>>>\n>>\n>>OK, you agree that a finite Hamiltonian of QED can be constructed.\n>>(the Hamiltonian is finite in each order separately, let us set aside\n>>the question about convergence).\n>\n>\n> Well, there are other caveats in your theory, namely infrared divergences\n> (ignored according to p.235 of your book draft).\n\nThat\'s right. There are well-established methods to deal with IR\ndivergences in the traditional approach. I presume that the same\nmethods will be applicable to my theory as well. I haven\'t investigated\nthis issue.\n\n>\n>\n>\n>>This Hamiltonian is expressed in\n>>terms of creation and annihilations operators of real particles\n>>(electrons, photons, protons, etc.). Therefore, the interacting\n>>potentials between particles act instantaneously. There are no\n>>degrees of freedom associated with fields or virtual particles,\n>\n>\n> As in free field theory, fields and particles use the same degrees\n> of freedom in complementary ways (particle-wave duality).\n> Once you have a Fock space, you have fields.\n>\n> For example you get from a massless spin 1 field the electromagnetic field\n> A(x,t) = exp(itH)(eps(x)a(x)+eps^*(x)a^*(x))exp(-itH)\n> with well-defined expectation values and correlation functions.\n\nMy point was that a full-fledged relativistic quantum\ntheory can be formulated entirely in terms of particles and\ninstantaneous interactions. You might be right, and field (or mixed)\ndescription of dynamics could be possible as well. I am not saying\nthat this can\'t be done. But so far we do not have a field theory which\ngoes beyond the S-matrix and predicts time evolution.\n\n>\n>\n>\n>>so, there are no degrees of freedom which can "keep" the momentum\n>>and energy while they are transferring from one particle to another.\n>>Therefore, the interaction cannot be retarded. It must be instantaneous.\n>\n>\n> This is just because you chose the instant form, where the state\n> is - by definition - determined by a wave function on the hyperplane\n> x_0=t, and the dynamical parameter is t. Of course this implies by\n> definition that the temporal dynamics is instantaneous.\n>\n> If you work instead in the point form, say, the state is determined by a\n> wave function on the hyperboloid x^2=l^2. Now the dynamical parameter\n> is the scale l, and the corresponding dilation dynamics is instantaneous.\n> But the temporal dynamics is non-instantaneous.\n>\n> Your renormalization technique probably works as well for the\n> dilation generator in the point form. Then you have another\n> description of QED, as finite as your instant form description,\n> which shows that there is nothing special about the instant form.\n>\n\nI hope you are not trying to say that instant and point form\ndescriptions are physically equivalent. It is true that they are\nequivalent as far as scattering is concerned (Sokolov\'s theorem).\nBut they are not equivalent if dynamics beyond the S-matrix is\nconsidered. So, in nature just one form of dynamics should be present.\nIn my book I gave some arguments that this should be the instant form.\nYou may disagree and put forward some arguments in favor of the point\n(or other) form. However, I would not agree with (often expressed)\nopinion that the choice of the form of dynamics is just a matter of\nconvenience.\n\nSimilarly, there should not be any ambiguity about the speed of\ninteraction. Interactions should be either instantaneous or retarded.\nThis should be verifiable by experiment (there was no a conclusive\nexperiment yet, but I believe such experiment is possible and\ndesperately needed).\n\nEugene Stefanovich.\n\n>\n>\n>>You cannot accept just one part of my approach and ignore the others.\n>\n>\n> Of course I can; I don\'t need to share your limitations.\n> Your work fits quite well into a more general framework\n> that you choose to ignore.\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>Eugene Stefanovich wrote:
>>>
>>>
>>>>My point is that fields are just mathematical constructs.
>>>>They are sometimes useful for doing math
>>>>(e.g., for proving the relativistic invariance
>>>>of the theory). They are useless for the physical interpretation of
>>>>the theory.
>>>
>>>... except that most people, including all practicioners, find it very
>>>valuable, while your proposal that they are useless is that of an
>>>outsider. As long as you support such views it will be very difficult
>>>for you to gain acceptance.
>>>
>>
>>OK, you agree that a finite Hamiltonian of QED can be constructed.
>>(the Hamiltonian is finite in each order separately, let us set aside
>>the question about convergence).
>
>
> Well, there are other caveats in your theory, namely infrared divergences
> (ignored according to p.235 of your book draft).

That's right. There are well-established methods to deal with IR
divergences in the traditional approach. I presume that the same
methods will be applicable to my theory as well. I haven't investigated
this issue.

>
>
>
>>This Hamiltonian is expressed in
>>terms of creation and annihilations operators of real particles
>>(electrons, photons, protons, etc.). Therefore, the interacting
>>potentials between particles act instantaneously. There are no
>>degrees of freedom associated with fields or virtual particles,
>
>
> As in free field theory, fields and particles use the same degrees
> of freedom in complementary ways (particle-wave duality).
> Once you have a Fock space, you have fields.
>
> For example you get from a massless spin 1 field the electromagnetic field
> A(x,t) = \exp(itH)(eps(x)a(x)+eps^*(x)a^*(x))\exp(-itH)
> with well-defined expectation values and correlation functions.

My point was that a full-fledged relativistic quantum
theory can be formulated entirely in terms of particles and
instantaneous interactions. You might be right, and field (or mixed)
description of dynamics could be possible as well. I am not saying
that this can't be done. But so far we do not have a field theory which
goes beyond the S-matrix and predicts time evolution.

>
>
>
>>so, there are no degrees of freedom which can "keep" the momentum
>>and energy while they are transferring from one particle to another.
>>Therefore, the interaction cannot be retarded. It must be instantaneous.
>
>
> This is just because you chose the instant form, where the state
> is - by definition - determined by a wave function on the hyperplane
> x_0=t, and the dynamical parameter is t. Of course this implies by
> definition that the temporal dynamics is instantaneous.
>
> If you work instead in the point form, say, the state is determined by a
> wave function on the hyperboloid x^2=l^2. Now the dynamical parameter
> is the scale l, and the corresponding dilation dynamics is instantaneous.
> But the temporal dynamics is non-instantaneous.
>
> Your renormalization technique probably works as well for the
> dilation generator in the point form. Then you have another
> description of QED, as finite as your instant form description,
> which shows that there is nothing special about the instant form.
>

I hope you are not trying to say that instant and point form
descriptions are physically equivalent. It is true that they are
equivalent as far as scattering is concerned (Sokolov's theorem).
But they are not equivalent if dynamics beyond the S-matrix is
considered. So, in nature just one form of dynamics should be present.
In my book I gave some arguments that this should be the instant form.
You may disagree and put forward some arguments in favor of the point
(or other) form. However, I would not agree with (often expressed)
opinion that the choice of the form of dynamics is just a matter of
convenience.

Similarly, there should not be any ambiguity about the speed of
interaction. Interactions should be either instantaneous or retarded.
This should be verifiable by experiment (there was no a conclusive
experiment yet, but I believe such experiment is possible and
desperately needed).

Eugene Stefanovich.

>
>
>>You cannot accept just one part of my approach and ignore the others.
>
>
> Of course I can; I don't need to share your limitations.
> Your work fits quite well into a more general framework
> that you choose to ignore.
>
>
> Arnold Neumaier
>

Eugene Stefanovich

Nov19-04, 01:27 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>Eugene Stefanovich wrote:\n>>\n>\n>>>>My point is that fields are just mathematical constructs.\n>>>>They are sometimes useful for doing math\n>>>>(e.g., for proving the relativistic invariance\n>>>>of the theory). They are useless for the physical interpretation of\n>>>>the theory.\n>>>\n>>>... except that most people, including all practicioners, find it very\n>>>valuable, while your proposal that they are useless is that of an\n>>>outsider. As long as you support such views it will be very difficult\n>>>for you to gain acceptance.\n>>>\n>>>Arnold Neumaier\n>>>\n>>\n>>The final results of any QED calculation are either scattering\n>>amplitudes\n>>(described in terms of momenta and spins of particles) or energies of\n>>bound states. So, fields may play some role in the theoretical\n>>formalism, but they are not needed to connect the results of theory\n>>with experiment.\n>\n>\n> But the input to QED calculations are usually external fields!\n> You can hardly get away without! In practice, one wants to know\n> the response of small systems to macroscopic, external input.\n\nI would agree with you if you use external fields as an approximate\nconcept. In this sense, I can also introduce E&M fields in my approach\n(as a sum of potentials of external charges acting on the test charge).\nIn a rigorous description, there is a reaction from the test charge\nto the field-generating charges (rigorously, only isolated systems\nshould be considered). However, in many cases such reaction can be\nsafely neglected.\n\nI wouldn\'t agree with you if you consider fields as a fundamental\nindispensible ingredient of theory.\n\n>\n>\n>\n>>I am fully aware that my views are not going to be accepted so easily.\n>>As time goes by, I see it more and more clear,\n>>how difficult it would be to gain acceptance. My approach attacks so\n>>many "well-established" concepts. My attack on fields is probably not\n>>the most controversial. I refuse to accept such "pillar of modern\n>>physics" as Minkowski spacetime. That will get me in trouble, for\n>>sure.\n>\n>\n> ... in needless trouble.\n>\n> Tradition is - for good reasons - not as bad as you may think it is.\n> Rewriting everything in terms of concepts you approve of would make\n> physics quite messy.\n>\n> Moreover, you need Minkowski space already to define the Poincare group,\n> which is one of the ingredients that cannot be dispensed with.\n\nI define the Poincare group as the group of transformations between\ninertial observers without mentioning the Minkowski spacetime\n(see section 2.2)\n\n> And to define the action from which the QED Hamiltonian derives.\n\nI agree that canonical Lagrangean formalism used to derive the QED\nHamiltonian uses quantum fields defined on 4D Minkowski spacetime.\nThese fields\nsatisfy manifestly covariant transformation laws. I also agree that in\nmy approach (without fields) I cannot derive the QED Hamiltonian from\nany general principle (I do not have the idea of gauge invariance in my\ntheory, for\nexample). In my book I just borrow the QED Hamiltonian from the standard\napproach without discussing how it was derived.\n\nThis can be considered a weak point of my theory. I agree.\n\n\nNevertheless, I was always suspicious about the canonical quantization\n+ gauge invariance approach to deriving the QED Hamiltonian. I still\nthink this is just a mathematical trick giving the "correct" result\nsomewhat mysteriously. Fields, gauges, Lagrangeans, etc. never made\nany physical sense to me.\n\nIn any event, the canonical procedure is not perfect, because\nthe resulting QED Hamiltonian is not entirely correct. We need to\nmake two corrective steps (renormalization and dressing) in order\nto arrive to a finite Hamiltonian which can be used for studying\ntime dynamics.\n\nThis finite (and correct!) Hamiltonian is an infinite series\nof more and more complicated terms. Although, all these terms\nhave very clear physical interpretation, I don\'t see a chance\nof rigorously and directly "deriving" them from any\ngeneral principle (I would pay a lot for getting a clue how to do that).\nThe only approach I know is not very pretty to my taste\n\n1. canonical gauge quantization\n2. renormalization\n3. dressing.\n\n> So your space-time denying approach is intrinsically contradictory.\n\nThere is no contradiction. My theory is relativistically (Poincare)\ninvariant, but Minkowski spacetime is not needed for physical\ninterpretation. This spacetime is sometimes used just as a convenient\nmathematical object (e.g., to write fields - another convenient\nmathematical object void of physical interpretation).\n\nEugene Stefanovich.\n\n>\n>\n>\n>>Nevertheless, I am confident in my approach, and so far I haven\'t\n>>heard any convincing argument that could shake my confidence.\n>\n>\n> The problem may be with waht you regard as convincing.\n> It is a pity that you combine technical skill with conceptual\n> blindfolds. But I won\'t further argue that...\n>\n>\n> Arnold Neumaier\n>\n>\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>Eugene Stefanovich wrote:
>>
>
>>>>My point is that fields are just mathematical constructs.
>>>>They are sometimes useful for doing math
>>>>(e.g., for proving the relativistic invariance
>>>>of the theory). They are useless for the physical interpretation of
>>>>the theory.
>>>
>>>... except that most people, including all practicioners, find it very
>>>valuable, while your proposal that they are useless is that of an
>>>outsider. As long as you support such views it will be very difficult
>>>for you to gain acceptance.
>>>
>>>Arnold Neumaier
>>>
>>
>>The final results of any QED calculation are either scattering
>>amplitudes
>>(described in terms of momenta and spins of particles) or energies of
>>bound states. So, fields may play some role in the theoretical
>>formalism, but they are not needed to connect the results of theory
>>with experiment.
>
>
> But the input to QED calculations are usually external fields!
> You can hardly get away without! In practice, one wants to know
> the response of small systems to macroscopic, external input.

I would agree with you if you use external fields as an approximate
concept. In this sense, I can also introduce E&M fields in my approach
(as a sum of potentials of external charges acting on the test charge).
In a rigorous description, there is a reaction from the test charge
to the field-generating charges (rigorously, only isolated systems
should be considered). However, in many cases such reaction can be
safely neglected.

I wouldn't agree with you if you consider fields as a fundamental
indispensible ingredient of theory.

>
>
>
>>I am fully aware that my views are not going to be accepted so easily.
>>As time goes by, I see it more and more clear,
>>how difficult it would be to gain acceptance. My approach attacks so
>>many "well-established" concepts. My attack on fields is probably not
>>the most controversial. I refuse to accept such "pillar of modern
>>physics" as Minkowski spacetime. That will get me in trouble, for
>>sure.
>
>
> ... in needless trouble.
>
> Tradition is - for good reasons - not as bad as you may think it is.
> Rewriting everything in terms of concepts you approve of would make
> physics quite messy.
>
> Moreover, you need Minkowski space already to define the Poincare group,
> which is one of the ingredients that cannot be dispensed with.

I define the Poincare group as the group of transformations between
inertial observers without mentioning the Minkowski spacetime
(see section 2.2)

> And to define the action from which the QED Hamiltonian derives.

I agree that canonical Lagrangean formalism used to derive the QED
Hamiltonian uses quantum fields defined on 4D Minkowski spacetime.
These fields
satisfy manifestly covariant transformation laws. I also agree that in
my approach (without fields) I cannot derive the QED Hamiltonian from
any general principle (I do not have the idea of gauge invariance in my
theory, for
example). In my book I just borrow the QED Hamiltonian from the standard
approach without discussing how it was derived.

This can be considered a weak point of my theory. I agree.

Nevertheless, I was always suspicious about the canonical quantization
+ gauge invariance approach to deriving the QED Hamiltonian. I still
think this is just a mathematical trick giving the "correct" result
somewhat mysteriously. Fields, gauges, Lagrangeans, etc. never made
any physical sense to me.

In any event, the canonical procedure is not perfect, because
the resulting QED Hamiltonian is not entirely correct. We need to
make two corrective steps (renormalization and dressing) in order
to arrive to a finite Hamiltonian which can be used for studying
time dynamics.

This finite (and correct!) Hamiltonian is an infinite series
of more and more complicated terms. Although, all these terms
have very clear physical interpretation, I don't see a chance
of rigorously and directly "deriving" them from any
general principle (I would pay a lot for getting a clue how to do that).
The only approach I know is not very pretty to my taste

1. canonical gauge quantization
2. renormalization
3. dressing.

> So your space-time denying approach is intrinsically contradictory.

There is no contradiction. My theory is relativistically (Poincare)
invariant, but Minkowski spacetime is not needed for physical
interpretation. This spacetime is sometimes used just as a convenient
mathematical object (e.g., to write fields - another convenient
mathematical object void of physical interpretation).

Eugene Stefanovich.

>
>
>
>>Nevertheless, I am confident in my approach, and so far I haven't
>>heard any convincing argument that could shake my confidence.
>
>
> The problem may be with waht you regard as convincing.
> It is a pity that you combine technical skill with conceptual
> blindfolds. But I won't further argue that...
>
>
> Arnold Neumaier
>
>
>

Arnold Neumaier

Nov19-04, 01:30 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>Eugene Stefanovich wrote:\n\n>>>OK, you agree that a finite Hamiltonian of QED can be constructed.\n>>>(the Hamiltonian is finite in each order separately, let us set aside\n>>>the question about convergence).\n>>\n>>Well, there are other caveats in your theory, namely infrared divergences\n>>(ignored according to p.235 of your book draft).\n>\n> That\'s right. There are well-established methods to deal with IR\n> divergences in the traditional approach. I presume that the same\n> methods will be applicable to my theory as well. I haven\'t investigated\n> this issue.\n\nWell, there are also well-established methods to deal with UV\ndivergences in the traditional approach, but you didn\'t like them\nas they apply only to the S-matrix. If you want to have a perturbatively\nfinite Hamiltonian, you need to cancel both the IR and UV divergences!\n\n\n>>As in free field theory, fields and particles use the same degrees\n>>of freedom in complementary ways (particle-wave duality).\n>>Once you have a Fock space, you have fields.\n>>\n>>For example you get from a massless spin 1 field the electromagnetic field\n>> A(x,t) = exp(itH)(eps(x)a(x)+eps^*(x)a^*(x))exp(-itH)\n>>with well-defined expectation values and correlation functions.\n>\n> My point was that a full-fledged relativistic quantum\n> theory can be formulated entirely in terms of particles and\n> instantaneous interactions. You might be right, and field (or mixed)\n> description of dynamics could be possible as well. I am not saying\n> that this can\'t be done. But so far we do not have a field theory which\n> goes beyond the S-matrix and predicts time evolution.\n\nYour theory _is_ a quantum field theory in instant form.\nInsisting on the contrary only shows an irrational bias that\nmakes your work difficult to accept by others.\n\n\n> I hope you are not trying to say that instant and point form\n> descriptions are physically equivalent. It is true that they are\n> equivalent as far as scattering is concerned (Sokolov\'s theorem).\n\nThey are _completely_ equivalent since there is a unitary transform\nbetween the two by means of space dilations. So if you have finite\nspace dilations in your theory - which you _should_ have,\nyou can go from the instant form to the point form.\nAnd doing what you do directly in the point form should give formulas\nthat are much less messy than yours, since they are manifestly\nLorentz invariant.\n\n\n> Similarly, there should not be any ambiguity about the speed of\n> interaction. Interactions should be either instantaneous or retarded.\n\nNo, it depends on what information you assume to be given.\nIn the instant form, you assume the state on a hyperplane to be given,\nwhich cannot be collected even by an ideal observer because of the finite\nspeed of light. But in the point form you need data on a past hyperboloid,\nwhich is available for an ideal observer.\n\nInstantaneous propagation of information that cannot be locally collected\ndoes not contradict a finite propagation of accessible information.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>Eugene Stefanovich wrote:

>>>OK, you agree that a finite Hamiltonian of QED can be constructed.
>>>(the Hamiltonian is finite in each order separately, let us set aside
>>>the question about convergence).
>>
>>Well, there are other caveats in your theory, namely infrared divergences
>>(ignored according to p.235 of your book draft).
>
> That's right. There are well-established methods to deal with IR
> divergences in the traditional approach. I presume that the same
> methods will be applicable to my theory as well. I haven't investigated
> this issue.

Well, there are also well-established methods to deal with UV
divergences in the traditional approach, but you didn't like them
as they apply only to the S-matrix. If you want to have a perturbatively
finite Hamiltonian, you need to cancel both the IR and UV divergences!

>>As in free field theory, fields and particles use the same degrees
>>of freedom in complementary ways (particle-wave duality).
>>Once you have a Fock space, you have fields.
>>
>>For example you get from a massless spin 1 field the electromagnetic field
>> A(x,t) = \exp(itH)(eps(x)a(x)+eps^*(x)a^*(x))\exp(-itH)
>>with well-defined expectation values and correlation functions.
>
> My point was that a full-fledged relativistic quantum
> theory can be formulated entirely in terms of particles and
> instantaneous interactions. You might be right, and field (or mixed)
> description of dynamics could be possible as well. I am not saying
> that this can't be done. But so far we do not have a field theory which
> goes beyond the S-matrix and predicts time evolution.

Your theory _is_ a quantum field theory in instant form.
Insisting on the contrary only shows an irrational bias that
makes your work difficult to accept by others.

> I hope you are not trying to say that instant and point form
> descriptions are physically equivalent. It is true that they are
> equivalent as far as scattering is concerned (Sokolov's theorem).

They are _completely_ equivalent since there is a unitary transform
between the two by means of space dilations. So if you have finite
space dilations in your theory - which you _should_ have,
you can go from the instant form to the point form.
And doing what you do directly in the point form should give formulas
that are much less messy than yours, since they are manifestly
Lorentz invariant.

> Similarly, there should not be any ambiguity about the speed of
> interaction. Interactions should be either instantaneous or retarded.

No, it depends on what information you assume to be given.
In the instant form, you assume the state on a hyperplane to be given,
which cannot be collected even by an ideal observer because of the finite
speed of light. But in the point form you need data on a past hyperboloid,
which is available for an ideal observer.

Instantaneous propagation of information that cannot be locally collected
does not contradict a finite propagation of accessible information.

Arnold Neumaier

Eugene Stefanovich

Nov21-04, 02:50 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>Eugene Stefanovich wrote:\n>>\n>\n>>>>OK, you agree that a finite Hamiltonian of QED can be constructed.\n>>>>(the Hamiltonian is finite in each order separately, let us set aside\n>>>>the question about convergence).\n>>>\n>>>Well, there are other caveats in your theory, namely infrared divergences\n>>>(ignored according to p.235 of your book draft).\n>>\n>>That\'s right. There are well-established methods to deal with IR\n>> divergences in the traditional approach. I presume that the same\n>>methods will be applicable to my theory as well. I haven\'t investigated\n>>this issue.\n>\n>\n> Well, there are also well-established methods to deal with UV\n> divergences in the traditional approach, but you didn\'t like them\n> as they apply only to the S-matrix. If you want to have a perturbatively\n> finite Hamiltonian, you need to cancel both the IR and UV divergences!\n\nAgreed.\n\n>\n>\n>\n>>>As in free field theory, fields and particles use the same degrees\n>>>of freedom in complementary ways (particle-wave duality).\n>>>Once you have a Fock space, you have fields.\n>>>\n>>>For example you get from a massless spin 1 field the electromagnetic field\n>>> A(x,t) = exp(itH)(eps(x)a(x)+eps^*(x)a^*(x))exp(-itH)\n>>>with well-defined expectation values and correlation functions.\n>>\n>>My point was that a full-fledged relativistic quantum\n>> theory can be formulated entirely in terms of particles and\n>>instantaneous interactions. You might be right, and field (or mixed)\n>>description of dynamics could be possible as well. I am not saying\n>>that this can\'t be done. But so far we do not have a field theory which\n>>goes beyond the S-matrix and predicts time evolution.\n>\n>\n> Your theory _is_ a quantum field theory in instant form.\n\nDisagreed. I would characterize my approach as action-at-a-distance\ntheory in which there are particles and instantaneous interactions\nbetween them (including interactions changing the number of particles).\nYou may notice that I use _free_ quantum fields only in two places in\nmy book:\n1) I use fields to write down the original Hamiltonian H and boost\noperator K of QED, because this notation is compact. I agree that\ncurrently there is no other way to derive these expressions except\nusing the canonical gauge field formalism. I do not discuss this\nformalism in my book. I just use its final result - the expressions\nfor H and K.\n2) I use field notation to prove (a\'la Weinberg) the relativistic\ninvariance of the theory. All formulas are much simpler in this\nformalism. The proof can be done using particle creation and\nannihilation operators only, without ever mentioning fields.\nThough I wouldn\'t even try to do that.\n\nAfter these two tasks are completed, I express all operators\nin terms of particle creation and annihilation operators and\nuse them exclusively to perform the dressing transformation and\nto calculate all physical properties, like dynamics, S-matrix, etc.\n\n> Insisting on the contrary only shows an irrational bias that\n> makes your work difficult to accept by others.\n\n\n\n>\n>\n>\n>>I hope you are not trying to say that instant and point form\n>>descriptions are physically equivalent. It is true that they are\n>>equivalent as far as scattering is concerned (Sokolov\'s theorem).\n>\n>\n> They are _completely_ equivalent since there is a unitary transform\n> between the two by means of space dilations. So if you have finite\n> space dilations in your theory - which you _should_ have,\n> you can go from the instant form to the point form.\n> And doing what you do directly in the point form should give formulas\n> that are much less messy than yours, since they are manifestly\n> Lorentz invariant.\n\nI strongly disagree.\nI do agree that you can unitarily transform from the instant form to the\npoint form. I also agree that this transformation will preserve the\nS-matrix. However, this transformation will affect dynamical properties.\nIn the point form, space translations are not kinematical.\nThe interacting system will look completely different from two frames\nconnected by a space translation. For example, in the case of\nan unstable\nparticle, the particle may look undecayed in one frame and decayed\nin the translated frame. Such behavior has never been observed\nin experiment.\nSo, I maintain that the instant form provides a better description of\nreality. Different forms of dynamics are scattering equivalent, but\nthey are not physically equivalent.\n\n>\n>\n>\n>>Similarly, there should not be any ambiguity about the speed of\n>>interaction. Interactions should be either instantaneous or retarded.\n>\n>\n> No, it depends on what information you assume to be given.\n> In the instant form, you assume the state on a hyperplane to be given,\n> which cannot be collected even by an ideal observer because of the finite\n> speed of light. But in the point form you need data on a past hyperboloid,\n> which is available for an ideal observer.\n>\n> Instantaneous propagation of information that cannot be locally collected\n> does not contradict a finite propagation of accessible information.\n\nI do not understand your argument about the "collection of information".\nSuppose you have two charged particles separated by the distance R.\nYou shake one particle at time t=0 and measure the time when the other\nparticle starts to move in response. If the time is zero, the\ninteraction is instantaneous. If the time is R/c, the interaction is\nretarded. How you arguments about hypersurfaces and information\ncollection apply here?\n\nEugene Stefanovich.\n\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>Eugene Stefanovich wrote:
>>
>
>>>>OK, you agree that a finite Hamiltonian of QED can be constructed.
>>>>(the Hamiltonian is finite in each order separately, let us set aside
>>>>the question about convergence).
>>>
>>>Well, there are other caveats in your theory, namely infrared divergences
>>>(ignored according to p.235 of your book draft).
>>
>>That's right. There are well-established methods to deal with IR
>> divergences in the traditional approach. I presume that the same
>>methods will be applicable to my theory as well. I haven't investigated
>>this issue.
>
>
> Well, there are also well-established methods to deal with UV
> divergences in the traditional approach, but you didn't like them
> as they apply only to the S-matrix. If you want to have a perturbatively
> finite Hamiltonian, you need to cancel both the IR and UV divergences!

Agreed.

>
>
>
>>>As in free field theory, fields and particles use the same degrees
>>>of freedom in complementary ways (particle-wave duality).
>>>Once you have a Fock space, you have fields.
>>>
>>>For example you get from a massless spin 1 field the electromagnetic field
>>> A(x,t) = \exp(itH)(eps(x)a(x)+eps^*(x)a^*(x))\exp(-itH)
>>>with well-defined expectation values and correlation functions.
>>
>>My point was that a full-fledged relativistic quantum
>> theory can be formulated entirely in terms of particles and
>>instantaneous interactions. You might be right, and field (or mixed)
>>description of dynamics could be possible as well. I am not saying
>>that this can't be done. But so far we do not have a field theory which
>>goes beyond the S-matrix and predicts time evolution.
>
>
> Your theory _is_ a quantum field theory in instant form.

Disagreed. I would characterize my approach as action-at-a-distance
theory in which there are particles and instantaneous interactions
between them (including interactions changing the number of particles).
You may notice that I use _free_ quantum fields only in two places in
my book:
1) I use fields to write down the original Hamiltonian H and boost
operator K of QED, because this notation is compact. I agree that
currently there is no other way to derive these expressions except
using the canonical gauge field formalism. I do not discuss this
formalism in my book. I just use its final result - the expressions
for H and K.
2) I use field notation to prove (a'la Weinberg) the relativistic
invariance of the theory. All formulas are much simpler in this
formalism. The proof can be done using particle creation and
annihilation operators only, without ever mentioning fields.
Though I wouldn't even try to do that.

After these two tasks are completed, I express all operators
in terms of particle creation and annihilation operators and
use them exclusively to perform the dressing transformation and
to calculate all physical properties, like dynamics, S-matrix, etc.

> Insisting on the contrary only shows an irrational bias that
> makes your work difficult to accept by others.

>
>
>
>>I hope you are not trying to say that instant and point form
>>descriptions are physically equivalent. It is true that they are
>>equivalent as far as scattering is concerned (Sokolov's theorem).
>
>
> They are _completely_ equivalent since there is a unitary transform
> between the two by means of space dilations. So if you have finite
> space dilations in your theory - which you _should_ have,
> you can go from the instant form to the point form.
> And doing what you do directly in the point form should give formulas
> that are much less messy than yours, since they are manifestly
> Lorentz invariant.

I strongly disagree.
I do agree that you can unitarily transform from the instant form to the
point form. I also agree that this transformation will preserve the
S-matrix. However, this transformation will affect dynamical properties.
In the point form, space translations are not kinematical.
The interacting system will look completely different from two frames
connected by a space translation. For example, in the case of
an unstable
particle, the particle may look undecayed in one frame and decayed
in the translated frame. Such behavior has never been observed
in experiment.
So, I maintain that the instant form provides a better description of
reality. Different forms of dynamics are scattering equivalent, but
they are not physically equivalent.

>
>
>
>>Similarly, there should not be any ambiguity about the speed of
>>interaction. Interactions should be either instantaneous or retarded.
>
>
> No, it depends on what information you assume to be given.
> In the instant form, you assume the state on a hyperplane to be given,
> which cannot be collected even by an ideal observer because of the finite
> speed of light. But in the point form you need data on a past hyperboloid,
> which is available for an ideal observer.
>
> Instantaneous propagation of information that cannot be locally collected
> does not contradict a finite propagation of accessible information.

I do not understand your argument about the "collection of information".
Suppose you have two charged particles separated by the distance R.
You shake one particle at time t=0 and measure the time when the other
particle starts to move in response. If the time is zero, the
interaction is instantaneous. If the time is R/c, the interaction is
retarded. How you arguments about hypersurfaces and information
collection apply here?

Eugene Stefanovich.

>
>
> Arnold Neumaier
>

Arnold Neumaier

Nov21-04, 02:54 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n>\n> Nevertheless, I was always suspicious about the canonical quantization\n> + gauge invariance approach to deriving the QED Hamiltonian. I still\n> think this is just a mathematical trick giving the "correct" result\n> somewhat mysteriously. Fields, gauges, Lagrangeans, etc. never made\n> any physical sense to me.\n\nBut they make sense to most particle physicists, and in 2D and 3D\neven to mathematicians. So this is a deficiency in your understanding,\nand not in physics.\n\n\n> In any event, the canonical procedure is not perfect, because\n> the resulting QED Hamiltonian is not entirely correct.\n\nAfter renormalization, it is entirely correct (at the typical energies\nfor QED) since it gives entirely correct results. And it is unitarily\nequivalent to your supposedly more correct theory. So if yours is correct,\nthe original one is, too.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> Nevertheless, I was always suspicious about the canonical quantization
> + gauge invariance approach to deriving the QED Hamiltonian. I still
> think this is just a mathematical trick giving the "correct" result
> somewhat mysteriously. Fields, gauges, Lagrangeans, etc. never made
> any physical sense to me.

But they make sense to most particle physicists, and in 2D and 3D
even to mathematicians. So this is a deficiency in your understanding,
and not in physics.

> In any event, the canonical procedure is not perfect, because
> the resulting QED Hamiltonian is not entirely correct.

After renormalization, it is entirely correct (at the typical energies
for QED) since it gives entirely correct results. And it is unitarily
equivalent to your supposedly more correct theory. So if yours is correct,
the original one is, too.

Arnold Neumaier

Arnold Neumaier

Nov22-04, 05:14 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nEugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n\n>>Your theory _is_ a quantum field theory in instant form.\n>\n> Disagreed. I would characterize my approach as action-at-a-distance\n> theory in which there are particles and instantaneous interactions\n> between them (including interactions changing the number of particles).\n> You may notice that I use _free_ quantum fields only in two places in\n> my book:\n> 1) I use fields to write down the original Hamiltonian H and boost\n> operator K of QED, because this notation is compact. I agree that\n> currently there is no other way to derive these expressions except\n> using the canonical gauge field formalism. I do not discuss this\n> formalism in my book. I just use its final result - the expressions\n> for H and K.\n> 2) I use field notation to prove (a\'la Weinberg) the relativistic\n> invariance of the theory. All formulas are much simpler in this\n> formalism. The proof can be done using particle creation and\n> annihilation operators only, without ever mentioning fields.\n> Though I wouldn\'t even try to do that.\n\nThis shows that you need fields in a _very_ essential way!\nWithout field theory no Hamiltonian to start with, and only messy\nmathematics.\n\n\n> After these two tasks are completed, I express all operators\n> in terms of particle creation and annihilation operators and\n> use them exclusively to perform the dressing transformation and\n> to calculate all physical properties, like dynamics, S-matrix, etc.\n\nBut creation and annihilation operators _are_ fields, namely linear\ncombinations of free boson or fermion fields and their derivatives.\nNot calling them fields just amounts to blinding yourself.\n_Every_ operator with a space dependence is a quantum field,\njust like every classical object with a space dependence is a field.\n\n\n[instant and point form]\n>>They are _completely_ equivalent since there is a unitary transform\n>>between the two by means of space dilations. So if you have finite\n>>space dilations in your theory - which you _should_ have,\n>>you can go from the instant form to the point form.\n>>And doing what you do directly in the point form should give formulas\n>>that are much less messy than yours, since they are manifestly\n>>Lorentz invariant.\n>\n> I strongly disagree.\n> I do agree that you can unitarily transform from the instant form to the\n> point form. I also agree that this transformation will preserve the\n> S-matrix. However, this transformation will affect dynamical properties.\n> In the point form, space translations are not kinematical.\n> The interacting system will look completely different from two frames\n> connected by a space translation. For example, in the case of\n> an unstable\n> particle, the particle may look undecayed in one frame and decayed\n> in the translated frame. Such behavior has never been observed\n> in experiment.\n\nThis is because in experiments one observes in the rest frame of the\nexperimental set-up, which provides a preferred coordinate system whose\ninstant form defines the right coordinates. nevertheless the formalism\nis completely frame independent, and invariant under arbitrary unitary\ntransformations.\n\n\n> So, I maintain that the instant form provides a better description of reality.\n\nThis is like insisting that classical mechanics should be always done in\na particular coordinate system. One can do it, but it makes many things\nmuch more difficult than necessary. The best description has always\nbeen the most tractable one, and not the one which formulated closest to\nphenomenology. Compatibility with phenomenology only requires that the\nlatter can be calculated from the model.\n\n\n> Different forms of dynamics are scattering equivalent, but\n> they are not physically equivalent.\n\nBut on pp. 316ff of your book you state explicitly that your theory is\ncompletely equivalent with Shirokov\'s, who uses the original Hamiltonian\nand a different vacuum state. You say that his theory is related to yours\nby a unitary transform. Now apply another unitary transform to put\nhis theory into the point form. This just transforms the vacuum again,\nandf the result is again completely equivalent to your version.\n\nThere is little point in restricting unitary transforms to a subset\nthat fits your too narrow conceptions. Mathematics is most powerful\nwhen it uses all the flexibility that is available.\n\n\n>>>Similarly, there should not be any ambiguity about the speed of\n>>>interaction. Interactions should be either instantaneous or retarded.\n>>\n>>No, it depends on what information you assume to be given.\n>>In the instant form, you assume the state on a hyperplane to be given,\n>>which cannot be collected even by an ideal observer because of the finite\n>>speed of light. But in the point form you need data on a past hyperboloid,\n>>which is available for an ideal observer.\n>>\n>>Instantaneous propagation of information that cannot be locally collected\n>>does not contradict a finite propagation of accessible information.\n>\n> I do not understand your argument about the "collection of information".\n> Suppose you have two charged particles separated by the distance R.\n> You shake one particle at time t=0 and measure the time when the other\n> particle starts to move in response.\n\nYou cannot implement this recipe since you cannot control whether\nthe other particle is at rest at time t=0, unless you know the complete\nfield at t=0 -- which you cannot since all information you have must come\nfrom your past cone. This assumes that \'you\' are pointlike\nand located at the position of the shaken particle. If you are not there,\nthings are even worse since you cannot guarantee shaking the particle\nat time t=0.\n\nThere is no way to cheat causality. Nature is governed by symmetric\nhyperbolic differential equations, and hence forbids instantaneous flow\nof information.\n\nThe instant form dynamics appears to be instantaneous, but only\nbecause it assumes the availability of information which in fact\nis not available. The fact that we can use it nevertheless in\nmany cases is because c is so large so that the instantaneous\napproximation works well.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>

>>Your theory _is_ a quantum field theory in instant form.
>
> Disagreed. I would characterize my approach as action-at-a-distance
> theory in which there are particles and instantaneous interactions
> between them (including interactions changing the number of particles).
> You may notice that I use _free_ quantum fields only in two places in
> my book:
> 1) I use fields to write down the original Hamiltonian H and boost
> operator K of QED, because this notation is compact. I agree that
> currently there is no other way to derive these expressions except
> using the canonical gauge field formalism. I do not discuss this
> formalism in my book. I just use its final result - the expressions
> for H and K.
> 2) I use field notation to prove (a'la Weinberg) the relativistic
> invariance of the theory. All formulas are much simpler in this
> formalism. The proof can be done using particle creation and
> annihilation operators only, without ever mentioning fields.
> Though I wouldn't even try to do that.

This shows that you need fields in a _very_ essential way!
Without field theory no Hamiltonian to start with, and only messy
mathematics.

> After these two tasks are completed, I express all operators
> in terms of particle creation and annihilation operators and
> use them exclusively to perform the dressing transformation and
> to calculate all physical properties, like dynamics, S-matrix, etc.

But creation and annihilation operators _are_ fields, namely linear
combinations of free boson or fermion fields and their derivatives.
Not calling them fields just amounts to blinding yourself.
_Every_ operator with a space dependence is a quantum field,
just like every classical object with a space dependence is a field.

[instant and point form]
>>They are _completely_ equivalent since there is a unitary transform
>>between the two by means of space dilations. So if you have finite
>>space dilations in your theory - which you _should_ have,
>>you can go from the instant form to the point form.
>>And doing what you do directly in the point form should give formulas
>>that are much less messy than yours, since they are manifestly
>>Lorentz invariant.
>
> I strongly disagree.
> I do agree that you can unitarily transform from the instant form to the
> point form. I also agree that this transformation will preserve the
> S-matrix. However, this transformation will affect dynamical properties.
> In the point form, space translations are not kinematical.
> The interacting system will look completely different from two frames
> connected by a space translation. For example, in the case of
> an unstable
> particle, the particle may look undecayed in one frame and decayed
> in the translated frame. Such behavior has never been observed
> in experiment.

This is because in experiments one observes in the rest frame of the
experimental set-up, which provides a preferred coordinate system whose
instant form defines the right coordinates. nevertheless the formalism
is completely frame independent, and invariant under arbitrary unitary
transformations.

> So, I maintain that the instant form provides a better description of reality.

This is like insisting that classical mechanics should be always done in
a particular coordinate system. One can do it, but it makes many things
much more difficult than necessary. The best description has always
been the most tractable one, and not the one which formulated closest to
phenomenology. Compatibility with phenomenology only requires that the
latter can be calculated from the model.

> Different forms of dynamics are scattering equivalent, but
> they are not physically equivalent.

But on pp. 316ff of your book you state explicitly that your theory is
completely equivalent with Shirokov's, who uses the original Hamiltonian
and a different vacuum state. You say that his theory is related to yours
by a unitary transform. Now apply another unitary transform to put
his theory into the point form. This just transforms the vacuum again,
andf the result is again completely equivalent to your version.

There is little point in restricting unitary transforms to a subset
that fits your too narrow conceptions. Mathematics is most powerful
when it uses all the flexibility that is available.

>>>Similarly, there should not be any ambiguity about the speed of
>>>interaction. Interactions should be either instantaneous or retarded.
>>
>>No, it depends on what information you assume to be given.
>>In the instant form, you assume the state on a hyperplane to be given,
>>which cannot be collected even by an ideal observer because of the finite
>>speed of light. But in the point form you need data on a past hyperboloid,
>>which is available for an ideal observer.
>>
>>Instantaneous propagation of information that cannot be locally collected
>>does not contradict a finite propagation of accessible information.
>
> I do not understand your argument about the "collection of information".
> Suppose you have two charged particles separated by the distance R.
> You shake one particle at time t=0 and measure the time when the other
> particle starts to move in response.

You cannot implement this recipe since you cannot control whether
the other particle is at rest at time t=0, unless you know the complete
field at t=0 -- which you cannot since all information you have must come
from your past cone. This assumes that 'you' are pointlike
and located at the position of the shaken particle. If you are not there,
things are even worse since you cannot guarantee shaking the particle
at time t=0.

There is no way to cheat causality. Nature is governed by symmetric
hyperbolic differential equations, and hence forbids instantaneous flow
of information.

The instant form dynamics appears to be instantaneous, but only
because it assumes the availability of information which in fact
is not available. The fact that we can use it nevertheless in
many cases is because c is so large so that the instantaneous
approximation works well.

Arnold Neumaier

Eugene Stefanovich

Nov22-04, 05:14 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Nevertheless, I was always suspicious about the canonical quantization\n>>+ gauge invariance approach to deriving the QED Hamiltonian. I still\n>>think this is just a mathematical trick giving the "correct" result\n>>somewhat mysteriously. Fields, gauges, Lagrangeans, etc. never made\n>>any physical sense to me.\n>\n>\n> But they make sense to most particle physicists, and in 2D and 3D\n> even to mathematicians. So this is a deficiency in your understanding,\n> and not in physics.\n\nI would agree with you on this point. I do not have anything against the\nmath of gauge theory. However, I was never able to understand the\nphysics of it.\n\n>\n>\n>\n>>In any event, the canonical procedure is not perfect, because\n>>the resulting QED Hamiltonian is not entirely correct.\n>\n>\n> After renormalization, it is entirely correct (at the typical energies\n> for QED) since it gives entirely correct results.\n\n....only for the S-matrix...\n\n> And it is unitarily\n> equivalent to your supposedly more correct theory. So if yours is correct,\n> the original one is, too.\n\nUnitary equivalence is not the same as physical equivalence.\n\nThe Hamiltonian of renormalized QED is infinite and\ncontains self-interaction of particles and vacuum. It cannot be\nconsidered as correct time evolution generator. However, it contains\nsome truth. This truth can be revealed by the unitary dressing\ntransformation. There is just one correct Hamiltonian in nature.\nTwo are too many.\n\nEugene.\n\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Nevertheless, I was always suspicious about the canonical quantization
>>+ gauge invariance approach to deriving the QED Hamiltonian. I still
>>think this is just a mathematical trick giving the "correct" result
>>somewhat mysteriously. Fields, gauges, Lagrangeans, etc. never made
>>any physical sense to me.
>
>
> But they make sense to most particle physicists, and in 2D and 3D
> even to mathematicians. So this is a deficiency in your understanding,
> and not in physics.

I would agree with you on this point. I do not have anything against the
math of gauge theory. However, I was never able to understand the
physics of it.

>
>
>
>>In any event, the canonical procedure is not perfect, because
>>the resulting QED Hamiltonian is not entirely correct.
>
>
> After renormalization, it is entirely correct (at the typical energies
> for QED) since it gives entirely correct results.

....only for the S-matrix...

> And it is unitarily
> equivalent to your supposedly more correct theory. So if yours is correct,
> the original one is, too.

Unitary equivalence is not the same as physical equivalence.

The Hamiltonian of renormalized QED is infinite and
contains self-interaction of particles and vacuum. It cannot be
considered as correct time evolution generator. However, it contains
some truth. This truth can be revealed by the unitary dressing
transformation. There is just one correct Hamiltonian in nature.
Two are too many.

Eugene.

>
>
> Arnold Neumaier
>

Arnold Neumaier

Nov22-04, 05:41 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> The Hamiltonian of renormalized QED is infinite\n\nNot necessarily. It is a limit of well-defined Hamiltonians with\nperturbatively defined cutoff-dependent coupling coefficients,\nand this limit might well turn out to exist in the resolvent sense.\nIndeed, this happens for similar problem in 1D quantum mechanics with\nsingular potentials. You\'d at least try to understand the part which is\nknown before you draw your conclusions.\n\n\n> and contains self-interaction of particles and vacuum.\n\nThis does not make it ill-defined. Indeed, this Hamiltonian is exactly\nthe same as that of Shirokov, whose theory you called equivalent with\nyours.\n\nBut I am tired of arguing this topic, and will not continue this thread.\nIf you are happy with your views and with disregarding my advice,\nlet it be so.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> The Hamiltonian of renormalized QED is infinite

Not necessarily. It is a limit of well-defined Hamiltonians with
perturbatively defined cutoff-dependent coupling coefficients,
and this limit might well turn out to exist in the resolvent sense.
Indeed, this happens for similar problem in 1D quantum mechanics with
singular potentials. You'd at least try to understand the part which is
known before you draw your conclusions.

> and contains self-interaction of particles and vacuum.

This does not make it ill-defined. Indeed, this Hamiltonian is exactly
the same as that of Shirokov, whose theory you called equivalent with
yours.

But I am tired of arguing this topic, and will not continue this thread.
If you are happy with your views and with disregarding my advice,
let it be so.

Arnold Neumaier

Eugene Stefanovich

Nov23-04, 03:08 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>The Hamiltonian of renormalized QED is infinite\n>\n>\n> Not necessarily. It is a limit of well-defined Hamiltonians with\n> perturbatively defined cutoff-dependent coupling coefficients,\n> and this limit might well turn out to exist in the resolvent sense.\n> Indeed, this happens for similar problem in 1D quantum mechanics with\n> singular potentials. You\'d at least try to understand the part which is\n> known before you draw your conclusions.\n\nI don\'t see how Hamiltonian can be finite when parameters e and m\nturn out to be infinite after renormalization. Could you give me\na reference to similar problems in QM?\n\n>\n>\n>\n>>and contains self-interaction of particles and vacuum.\n>\n>\n> This does not make it ill-defined. Indeed, this Hamiltonian is exactly\n> the same as that of Shirokov, whose theory you called equivalent with\n> yours.\n\nIn Shirokov\'s approach the dressing transformation redefines states\n(vacuum vector and creation and annihilation operators). So, the\nHamiltonian is kept (mathematically)\nthe same, while definitions of particles change.\nThe redefinition of\nparticles in the Shirokov\'s approach makes it wrong to say that\nhe keeps the old QED Hamiltonian with self-interaction. The self-\ninteraction effects got absorbed in the new definitions of particles,\nand the physical content of the Hamiltonian changes.\n\n>\n> But I am tired of arguing this topic, and will not continue this thread.\n> If you are happy with your views and with disregarding my advice,\n> let it be so.\n\nOK. Thanks for you good comments.\n\nEugene.\n\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>The Hamiltonian of renormalized QED is infinite
>
>
> Not necessarily. It is a limit of well-defined Hamiltonians with
> perturbatively defined cutoff-dependent coupling coefficients,
> and this limit might well turn out to exist in the resolvent sense.
> Indeed, this happens for similar problem in 1D quantum mechanics with
> singular potentials. You'd at least try to understand the part which is
> known before you draw your conclusions.

I don't see how Hamiltonian can be finite when parameters e and m
turn out to be infinite after renormalization. Could you give me
a reference to similar problems in QM?

>
>
>
>>and contains self-interaction of particles and vacuum.
>
>
> This does not make it ill-defined. Indeed, this Hamiltonian is exactly
> the same as that of Shirokov, whose theory you called equivalent with
> yours.

In Shirokov's approach the dressing transformation redefines states
(vacuum vector and creation and annihilation operators). So, the
Hamiltonian is kept (mathematically)
the same, while definitions of particles change.
The redefinition of
particles in the Shirokov's approach makes it wrong to say that
he keeps the old QED Hamiltonian with self-interaction. The self-
interaction effects got absorbed in the new definitions of particles,
and the physical content of the Hamiltonian changes.

>
> But I am tired of arguing this topic, and will not continue this thread.
> If you are happy with your views and with disregarding my advice,
> let it be so.

OK. Thanks for you good comments.

Eugene.

>
>
> Arnold Neumaier
>

Eugene Stefanovich

Nov24-04, 01:46 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>\n>\n>>>Your theory _is_ a quantum field theory in instant form.\n>>\n>>Disagreed. I would characterize my approach as action-at-a-distance\n>>theory in which there are particles and instantaneous interactions\n>>between them (including interactions changing the number of particles).\n>>You may notice that I use _free_ quantum fields only in two places in\n>>my book:\n>>1) I use fields to write down the original Hamiltonian H and boost\n>>operator K of QED, because this notation is compact. I agree that\n>>currently there is no other way to derive these expressions except\n>>using the canonical gauge field formalism. I do not discuss this\n>>formalism in my book. I just use its final result - the expressions\n>>for H and K.\n>>2) I use field notation to prove (a\'la Weinberg) the relativistic\n>>invariance of the theory. All formulas are much simpler in this\n>>formalism. The proof can be done using particle creation and\n>>annihilation operators only, without ever mentioning fields.\n>>Though I wouldn\'t even try to do that.\n>\n>\n> This shows that you need fields in a _very_ essential way!\n> Without field theory no Hamiltonian to start with, and only messy\n> mathematics.\n\nMy formalism does not require fields. It does require explicit\nexpressions of the interacting hamiltonian and boost operator\nin the Fock space. I do not know how to derive these operators\n(while preserving the Poincare commutators) myself. So, I take\nready expressions from the field theory. I expand fields in terms\nof creation and annihilation operators and perform all\ncalculations in this notation.\n\nI cannot accept physical relevance of fields. If I do that, then\nI should accept physical relevance of the Minkowski spacetime.\nThis would mean that boosts are kinematical. This, in turn,\nis rejected by\nthe Currie-Jordan-Sudarshan theorem. The CJS is a very important\nstatement which is often underestimated, or even ignored, in QFT.\n\n\n>\n>\n>\n>>After these two tasks are completed, I express all operators\n>>in terms of particle creation and annihilation operators and\n>>use them exclusively to perform the dressing transformation and\n>>to calculate all physical properties, like dynamics, S-matrix, etc.\n>\n>\n> But creation and annihilation operators _are_ fields, namely linear\n> combinations of free boson or fermion fields and their derivatives.\n> Not calling them fields just amounts to blinding yourself.\n\nI am not saying that fields are not useful. They are quite useful\nas abstract mathematical operators. You can express fields as linear\ncombinations of creation and annihilation operators, and you can\nalso express creation and annihilation operators back in terms of\nfields. I maintain that both creation and annihilation operators\nand fields are nothing more as mathematical objects which are\nconvenient for writing down important physical stuff -\ninteraction operators in the Hamiltonian and boost.\n\nI wouldn\'t say that creation operator a^{\\dag}(p) has any physical\nmeaning, because there is no physical process in which electron gets\ncreated out of vacuum. However, Hermitian operators like\na^{\\dag}(p_1)a^{\\dag}(p_2)a(p_3)a(p_4) make perfect sense as\ninteraction terms in the Hamiltonian.\n\nThe fields \\psi(x,t) have no other meaning except as convenient objects\nfor compact writing of the interaction Hamiltonian and proving its\nrelativistic invariance. I insist on this interpretation, most\nimportantly because I reject interpretation of x and t as physical\nposition and time. Fields live in the abstract Minkowski spacetime,\nwhich has no relationship to the real world. The manifest covariance\nimplicit in the notion of field is rejected by the CJS theorem.\n\n> _Every_ operator with a space dependence is a quantum field,\n> just like every classical object with a space dependence is a field.\n>\n>\n> [instant and point form]\n>\n>>>They are _completely_ equivalent since there is a unitary transform\n>>>between the two by means of space dilations. So if you have finite\n>>>space dilations in your theory - which you _should_ have,\n>>>you can go from the instant form to the point form.\n>>>And doing what you do directly in the point form should give formulas\n>>>that are much less messy than yours, since they are manifestly\n>>>Lorentz invariant.\n>>\n>>I strongly disagree.\n>>I do agree that you can unitarily transform from the instant form to the\n>>point form. I also agree that this transformation will preserve the\n>>S-matrix. However, this transformation will affect dynamical properties.\n>>In the point form, space translations are not kinematical.\n>>The interacting system will look completely different from two frames\n>>connected by a space translation. For example, in the case of\n>>an unstable\n>>particle, the particle may look undecayed in one frame and decayed\n>>in the translated frame. Such behavior has never been observed\n>>in experiment.\n>\n>\n> This is because in experiments one observes in the rest frame of the\n> experimental set-up, which provides a preferred coordinate system whose\n> instant form defines the right coordinates. nevertheless the formalism\n> is completely frame independent, and invariant under arbitrary unitary\n> transformations.\n\nI agree that the formalism is invariant with respect to the\nunitary transformations associated with inertial transformations\nof observers (Poincare group transformations). I do not agree\nthat different forms of dynamics (e.g., the instant and point forms),\nwhose generators can be connected by unitary transformations conserving\nthe S-matrix, are physically equivalent.\n\nOf course, everything will be the same if you apply any unitary\ntransformation to everything: total generators of the Poincare group,\nobservables of individual particles, and wavefunctions.\nHowever, by going from the instant form dynamics to the point form\ndynamics you apply a unitary transformation only to the generators\nof the Poincare group. This may preserve the S-matrix, but changes\nother aspects of physics: time dependence, position dependence,\nand velocity dependence of observables. There should be an experiment\nwhich can distinguish whether boost or momentum operators are\ninteraction-dependent.\n\n>\n>\n>\n>>So, I maintain that the instant form provides a better description of reality.\n>\n>\n> This is like insisting that classical mechanics should be always done in\n> a particular coordinate system. One can do it, but it makes many things\n> much more difficult than necessary. The best description has always\n> been the most tractable one, and not the one which formulated closest to\n> phenomenology. Compatibility with phenomenology only requires that the\n> latter can be calculated from the model.\n\nNo, I am not limited to one coordinate system. Once an instant form\nrepresentation of the Poincare group is constructed in the Hilbert space\nof the system, one can transfer between descriptions of different\ninertial observers. A different representation of the Poincare group\n(e.g., the point form representation) would describe a different set\nof transformations between observations of various inertial observers.\n\nThere is just one set of transformations which is correct and fully\nagreeable with experiment. My point is that this set of transformations\n(parameterized by 10 parameters of the Poincare group) must belong to\nthe instant form.\n\n>\n>\n>\n>>Different forms of dynamics are scattering equivalent, but\n>>they are not physically equivalent.\n>\n>\n> But on pp. 316ff of your book you state explicitly that your theory is\n> completely equivalent with Shirokov\'s, who uses the original Hamiltonian\n> and a different vacuum state. You say that his theory is related to yours\n> by a unitary transform.\n\nThat\'s right. This is a kind of transformation in which everything is\nunitarily transformed at once: vacuum, creation and annihilation\noperators, particle observables, generators of the Poincare group.\nYou can just unitarily move entire Hilbert space with everything it\ncontains. Then you obtain just another mathematical description of\nexactly the same physical theory. All physical results remain the\nsame. That\'s the trivial difference between my approach and Shirokov\'s\napproach.\n\n> Now apply another unitary transform to put\n> his theory into the point form. This just transforms the vacuum again,\n> andf the result is again completely equivalent to your version.\n\nThe difference between instant form dynamics and point form dynamics is\nof another sort. Let me remind you how different forms of dynamics are\nconstructed:\n1. We start from Hilbert spaces T_i of unitary irreducible\nrepresentations of the Poincare group corresponding to individual\nparticles. There are operators of particle observables p_i (momentum),\nr_i (position), h_i (energy) etc. defined in these Hilbert spaces.\n\n2. We build a Hilbert space T of the compound system as a tensor\nproduct of particle Hilbert spaces. In the general case we need to\nbuild\na direct sum of 0-particle, 1-particle, 2-particle, etc. spaces\nwhich is called Fock space, but for simplicity, I\'ll stick with\na simple 2-particle example\n\nT = T_1 (x) T_2\n\n3. This tensor product construction uniquely defines operators\nof 1-particle observables in T. I will use the same symbols p_i,\nr_i, h_i etc. to identify these operators in both T_i and in T.\nWe are not allowed to change these operators anymore. This means\nthat the non-interacting representation of the Poincare group in\nT is also uniquely defined\n\nH_0 = h_1 + h_2\nP_0 = p_1 + p_2\nJ_0 = j_1 + j_2\nK_0 = k_1 + k_2\n\n4. Now we need to choose a representation of the Poincare group in\nT which correctly describes interacting dynamics. Apparently, the\nrepresentation (H_0, P_0, J_0, K_0) is not a good choice, because\nit does not have any interaction. Let us consider two other choices:\nthe instant form\n\nH = H_0 + V\nP = P_0\nJ = J_0\nK = K_0 + Z\n\nand the point form\n\nH = H_0 + V\nP = P_0 + Y\nJ = J_0\nK = K_0\n\nIn both cases (H,P,J,K) satisfy Poincare commutators.\n\n5. Suppose that we calculated the S-matrix S with the instant form\ninteraction. Then, Sokolov\'s theorem tells us that there is a unitary\ntransformation U such that by applying U to instant form generators\n\nH\' = UHU^{-1}\nP\' = UPU^{-1}\nJ\' = UJU^{-1}\nK\' = UKU^{-1}\n\nthe transformed (H\',P\',J\',K\') belong to the point form and the S-matrix\nS is not changed. Note, that the unitary transformation U is applied\nONLY\nto the Poincare group generators. It is not applied to vacuum,\none-particle observables, creation and annihilation operators, or\nanything else.\n\n6. Although the S-matrix is the same with both instant and point\nform interactions, the dynamical behavior of particle observables isn\'t.\nTake, for example, momentum of particle 1 and its transformation wrt\nto space translations of the observer. In the general case\n\np_1(a) = exp(-iP.a) p_1 exp(iP.a)\n\nIn the instant form we can write\n\np_1(a) = exp(-iP_0.a) p_1 exp(iP_0.a) = p_1\n\ni.e., the momentum p_1 is not affected by translation, as we know from\nexperiment. On the other hand, in the point form dynamics\n\np_1(a) = exp(-i(P_0 + Y).a) p_1 exp(i(P_0 + Y).a) /= p_1\n\nthe observed momentum of particle 1 depends on the spatial position\nof observer, because generally Y does not commute with p_1. There are\nobservable physical differences between the instant and point forms\nof dynamics.\n\n\n\n>\n> There is little point in restricting unitary transforms to a subset\n> that fits your too narrow conceptions. Mathematics is most powerful\n> when it uses all the flexibility that is available.\n>\n>\n>\n>>>>Similarly, there should not be any ambiguity about the speed of\n>>>>interaction. Interactions should be either instantaneous or retarded.\n>>>\n>>>No, it depends on what information you assume to be given.\n>>>In the instant form, you assume the state on a hyperplane to be given,\n>>>which cannot be collected even by an ideal observer because of the finite\n>>>speed of light. But in the point form you need data on a past hyperboloid,\n>>>which is available for an ideal observer.\n>>>\n>>>Instantaneous propagation of information that cannot be locally collected\n>>>does not contradict a finite propagation of accessible information.\n>>\n>>I do not understand your argument about the "collection of information".\n>>Suppose you have two charged particles separated by the distance R.\n>>You shake one particle at time t=0 and measure the time when the other\n>>particle starts to move in response.\n>\n>\n> You cannot implement this recipe since you cannot control whether\n> the other particle is at rest at time t=0, unless you know the complete\n> field at t=0 -- which you cannot since all information you have must come\n> from your past cone. This assumes that \'you\' are pointlike\n> and located at the position of the shaken particle. If you are not there,\n> things are even worse since you cannot guarantee shaking the particle\n> at time t=0.\n\nI disagree. I don\'t know much about experiment, but I am sure that\none can place two charges (or charged antennas)\n10 cm from each other in a desktop\nexperiment and monitor their positions synchonously with time\nresolution much better than 0.3 ns (that\'s the time in which light\npasses 10 cm distance).\n\n>\n> There is no way to cheat causality.\n\nI disagree. There is a way to send a superluminal (in fact,\ninstantaneous) signal between two charged particles. This does not\ncontradict causality. The key for understanding that is in the\ndynamical character of boosts. If boosts are dynamical (depend on\ninteraction), then even if interaction propagates instantaneously,\nyou do not get the "grandfather paradox" in the moving frame of\nreference.\nIt appears, that the interaction remains instantaneous in all frames\nof reference. This is discussed in the last subsection of 12.3.\n\n>Nature is governed by symmetric\n> hyperbolic differential equations, and hence forbids instantaneous\n>flow of information.\n\nI do not use hyperbolic differential equations in my approach.\n\n>\n> The instant form dynamics appears to be instantaneous, but only\n> because it assumes the availability of information which in fact\n> is not available. The fact that we can use it nevertheless in\n> many cases is because c is so large so that the instantaneous\n> approximation works well.\n\nAs I said earlier, we disagree here.\n\nEugene Stefanovich.\n\n>\n>\n> Arnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>
>
>>>Your theory _is_ a quantum field theory in instant form.
>>
>>Disagreed. I would characterize my approach as action-at-a-distance
>>theory in which there are particles and instantaneous interactions
>>between them (including interactions changing the number of particles).
>>You may notice that I use _free_ quantum fields only in two places in
>>my book:
>>1) I use fields to write down the original Hamiltonian H and boost
>>operator K of QED, because this notation is compact. I agree that
>>currently there is no other way to derive these expressions except
>>using the canonical gauge field formalism. I do not discuss this
>>formalism in my book. I just use its final result - the expressions
>>for H and K.
>>2) I use field notation to prove (a'la Weinberg) the relativistic
>>invariance of the theory. All formulas are much simpler in this
>>formalism. The proof can be done using particle creation and
>>annihilation operators only, without ever mentioning fields.
>>Though I wouldn't even try to do that.
>
>
> This shows that you need fields in a _very_ essential way!
> Without field theory no Hamiltonian to start with, and only messy
> mathematics.

My formalism does not require fields. It does require explicit
expressions of the interacting hamiltonian and boost operator
in the Fock space. I do not know how to derive these operators
(while preserving the Poincare commutators) myself. So, I take
ready expressions from the field theory. I expand fields in terms
of creation and annihilation operators and perform all
calculations in this notation.

I cannot accept physical relevance of fields. If I do that, then
I should accept physical relevance of the Minkowski spacetime.
This would mean that boosts are kinematical. This, in turn,
is rejected by
the Currie-Jordan-Sudarshan theorem. The CJS is a very important
statement which is often underestimated, or even ignored, in QFT.

>
>
>
>>After these two tasks are completed, I express all operators
>>in terms of particle creation and annihilation operators and
>>use them exclusively to perform the dressing transformation and
>>to calculate all physical properties, like dynamics, S-matrix, etc.
>
>
> But creation and annihilation operators _are_ fields, namely linear
> combinations of free boson or fermion fields and their derivatives.
> Not calling them fields just amounts to blinding yourself.

I am not saying that fields are not useful. They are quite useful
as abstract mathematical operators. You can express fields as linear
combinations of creation and annihilation operators, and you can
also express creation and annihilation operators back in terms of
fields. I maintain that both creation and annihilation operators
and fields are nothing more as mathematical objects which are
convenient for writing down important physical stuff -
interaction operators in the Hamiltonian and boost.

I wouldn't say that creation operator a^{\dag}(p) has any physical
meaning, because there is no physical process in which electron gets
created out of vacuum. However, Hermitian operators like
a^{\dag}(p_1)a^{\dag}(p_2)a(p_3)a(p_4) make perfect sense as
interaction terms in the Hamiltonian.

The fields \psi(x,t) have no other meaning except as convenient objects
for compact writing of the interaction Hamiltonian and proving its
relativistic invariance. I insist on this interpretation, most
importantly because I reject interpretation of x and t as physical
position and time. Fields live in the abstract Minkowski spacetime,
which has no relationship to the real world. The manifest covariance
implicit in the notion of field is rejected by the CJS theorem.

> _Every_ operator with a space dependence is a quantum field,
> just like every classical object with a space dependence is a field.
>
>
> [instant and point form]
>
>>>They are _completely_ equivalent since there is a unitary transform
>>>between the two by means of space dilations. So if you have finite
>>>space dilations in your theory - which you _should_ have,
>>>you can go from the instant form to the point form.
>>>And doing what you do directly in the point form should give formulas
>>>that are much less messy than yours, since they are manifestly
>>>Lorentz invariant.
>>
>>I strongly disagree.
>>I do agree that you can unitarily transform from the instant form to the
>>point form. I also agree that this transformation will preserve the
>>S-matrix. However, this transformation will affect dynamical properties.
>>In the point form, space translations are not kinematical.
>>The interacting system will look completely different from two frames
>>connected by a space translation. For example, in the case of
>>an unstable
>>particle, the particle may look undecayed in one frame and decayed
>>in the translated frame. Such behavior has never been observed
>>in experiment.
>
>
> This is because in experiments one observes in the rest frame of the
> experimental set-up, which provides a preferred coordinate system whose
> instant form defines the right coordinates. nevertheless the formalism
> is completely frame independent, and invariant under arbitrary unitary
> transformations.

I agree that the formalism is invariant with respect to the
unitary transformations associated with inertial transformations
of observers (Poincare group transformations). I do not agree
that different forms of dynamics (e.g., the instant and point forms),
whose generators can be connected by unitary transformations conserving
the S-matrix, are physically equivalent.

Of course, everything will be the same if you apply any unitary
transformation to everything: total generators of the Poincare group,
observables of individual particles, and wavefunctions.
However, by going from the instant form dynamics to the point form
dynamics you apply a unitary transformation only to the generators
of the Poincare group. This may preserve the S-matrix, but changes
other aspects of physics: time dependence, position dependence,
and velocity dependence of observables. There should be an experiment
which can distinguish whether boost or momentum operators are
interaction-dependent.

>
>
>
>>So, I maintain that the instant form provides a better description of reality.
>
>
> This is like insisting that classical mechanics should be always done in
> a particular coordinate system. One can do it, but it makes many things
> much more difficult than necessary. The best description has always
> been the most tractable one, and not the one which formulated closest to
> phenomenology. Compatibility with phenomenology only requires that the
> latter can be calculated from the model.

No, I am not limited to one coordinate system. Once an instant form
representation of the Poincare group is constructed in the Hilbert space
of the system, one can transfer between descriptions of different
inertial observers. A different representation of the Poincare group
(e.g., the point form representation) would describe a different set
of transformations between observations of various inertial observers.

There is just one set of transformations which is correct and fully
agreeable with experiment. My point is that this set of transformations
(parameterized by 10 parameters of the Poincare group) must belong to
the instant form.

>
>
>
>>Different forms of dynamics are scattering equivalent, but
>>they are not physically equivalent.
>
>
> But on pp. 316ff of your book you state explicitly that your theory is
> completely equivalent with Shirokov's, who uses the original Hamiltonian
> and a different vacuum state. You say that his theory is related to yours
> by a unitary transform.

That's right. This is a kind of transformation in which everything is
unitarily transformed at once: vacuum, creation and annihilation
operators, particle observables, generators of the Poincare group.
You can just unitarily move entire Hilbert space with everything it
contains. Then you obtain just another mathematical description of
exactly the same physical theory. All physical results remain the
same. That's the trivial difference between my approach and Shirokov's
approach.

> Now apply another unitary transform to put
> his theory into the point form. This just transforms the vacuum again,
> andf the result is again completely equivalent to your version.

The difference between instant form dynamics and point form dynamics is
of another sort. Let me remind you how different forms of dynamics are
constructed:
1. We start from Hilbert spaces T_i of unitary irreducible
representations of the Poincare group corresponding to individual
particles. There are operators of particle observables p_i (momentum),
r_i (position), h_i (energy) etc. defined in these Hilbert spaces.

2. We build a Hilbert space T of the compound system as a tensor
product of particle Hilbert spaces. In the general case we need to
build
a direct sum of 0-particle, 1-particle, 2-particle, etc. spaces
which is called Fock space, but for simplicity, I'll stick with
a simple 2-particle example

T = T_1 (x) T_2

3. This tensor product construction uniquely defines operators
of 1-particle observables in T. I will use the same symbols p_i,r_i, h_i etc. to identify these operators in both T_i and in T.
We are not allowed to change these operators anymore. This means
that the non-interacting representation of the Poincare group in
T is also uniquely defined

H_0 = h_1 + h_2P_0 = p_1 + p_2J_0 = j_1 + j_2K_0 = k_1 + k_2

4. Now we need to choose a representation of the Poincare group in
T which correctly describes interacting dynamics. Apparently, the
representation (H_0, P_0, J_0, K_0) is not a good choice, because
it does not have any interaction. Let us consider two other choices:
the instant form

H = H_0 + VP = P_0J = J_0K = K_0 + Z

and the point form

H = H_0 + VP = P_0 + YJ = J_0K = K_0

In both cases (H,P,J,K) satisfy Poincare commutators.

5. Suppose that we calculated the S-matrix S with the instant form
interaction. Then, Sokolov's theorem tells us that there is a unitary
transformation U such that by applying U to instant form generators

H' = UHU^{-1}P' = UPU^{-1}J' = UJU^{-1}K' = UKU^{-1}

the transformed (H',P',J',K') belong to the point form and the S-matrix
S is not changed. Note, that the unitary transformation U is applied
ONLY
to the Poincare group generators. It is not applied to vacuum,
one-particle observables, creation and annihilation operators, or
anything else.

6. Although the S-matrix is the same with both instant and point
form interactions, the dynamical behavior of particle observables isn't.
Take, for example, momentum of particle 1 and its transformation wrt
to space translations of the observer. In the general case

p_1(a) = \exp(-iP.a) p_1 \exp(iP.a)

In the instant form we can write

p_1(a) = \exp(-iP_0.a) p_1 \exp(iP_0.a) = p_1

i.e., the momentum p_1 is not affected by translation, as we know from
experiment. On the other hand, in the point form dynamics

p_1(a) = \exp(-i(P_0 + Y)[/itex].a) p_1 \exp(i(P_0 + Y).[itex]a) /= p_1

the observed momentum of particle 1 depends on the spatial position
of observer, because generally Y does not commute with p_1. There are
observable physical differences between the instant and point forms
of dynamics.

>
> There is little point in restricting unitary transforms to a subset
> that fits your too narrow conceptions. Mathematics is most powerful
> when it uses all the flexibility that is available.
>
>
>
>>>>Similarly, there should not be any ambiguity about the speed of
>>>>interaction. Interactions should be either instantaneous or retarded.
>>>
>>>No, it depends on what information you assume to be given.
>>>In the instant form, you assume the state on a hyperplane to be given,
>>>which cannot be collected even by an ideal observer because of the finite
>>>speed of light. But in the point form you need data on a past hyperboloid,
>>>which is available for an ideal observer.
>>>
>>>Instantaneous propagation of information that cannot be locally collected
>>>does not contradict a finite propagation of accessible information.
>>
>>I do not understand your argument about the "collection of information".
>>Suppose you have two charged particles separated by the distance R.
>>You shake one particle at time t=0 and measure the time when the other
>>particle starts to move in response.
>
>
> You cannot implement this recipe since you cannot control whether
> the other particle is at rest at time t=0, unless you know the complete
> field at t=0 -- which you cannot since all information you have must come
> from your past cone. This assumes that 'you' are pointlike
> and located at the position of the shaken particle. If you are not there,
> things are even worse since you cannot guarantee shaking the particle
> at time t=0.

I disagree. I don't know much about experiment, but I am sure that
one can place two charges (or charged antennas)
10 cm from each other in a desktop
experiment and monitor their positions synchonously with time
resolution much better than .3 ns (that's the time in which light
passes 10 cm distance).

>
> There is no way to cheat causality.

I disagree. There is a way to send a superluminal (in fact,
instantaneous) signal between two charged particles. This does not
contradict causality. The key for understanding that is in the
dynamical character of boosts. If boosts are dynamical (depend on
interaction), then even if interaction propagates instantaneously,
you do not get the "grandfather paradox" in the moving frame of
reference.
It appears, that the interaction remains instantaneous in all frames
of reference. This is discussed in the last subsection of 12.3.

>Nature is governed by symmetric
> hyperbolic differential equations, and hence forbids instantaneous
>flow of information.

I do not use hyperbolic differential equations in my approach.

>
> The instant form dynamics appears to be instantaneous, but only
> because it assumes the availability of information which in fact
> is not available. The fact that we can use it nevertheless in
> many cases is because c is so large so that the instantaneous
> approximation works well.

As I said earlier, we disagree here.

Eugene Stefanovich.

>
>
> Arnold Neumaier

Arnold Neumaier

Nov24-04, 01:47 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>Eugene Stefanovich wrote:\n>>\n>>>The Hamiltonian of renormalized QED is infinite\n>>\n>>Not necessarily. It is a limit of well-defined Hamiltonians with\n>>perturbatively defined cutoff-dependent coupling coefficients,\n>>and this limit might well turn out to exist in the resolvent sense.\n>>Indeed, this happens for similar problem in 1D quantum mechanics with\n>>singular potentials. You\'d at least try to understand the part which is\n>>known before you draw your conclusions.\n>\n> I don\'t see how Hamiltonian can be finite when parameters e and m\n> turn out to be infinite after renormalization. Could you give me\n> a reference to similar problems in QM?\n\nI had given you one (delta-potential) quite early in our discussions,\nbut you chose to ignore it.\n\n\n>>>and contains self-interaction of particles and vacuum.\n>>\n>>This does not make it ill-defined. Indeed, this Hamiltonian is exactly\n>>the same as that of Shirokov, whose theory you called equivalent with\n>>yours.\n>\n> In Shirokov\'s approach the dressing transformation redefines states\n> (vacuum vector and creation and annihilation operators). So, the\n> Hamiltonian is kept (mathematically)\n> the same, while definitions of particles change.\n\nThese particles are just quasiparticles in the standard sense.\nThis is quite in line with the standard nonrelativisitc case,\nwhere the vacuum is the ground state and the quasi-particles are the\nexcitations of the vacuum. Except that no infinities occur in the\nnonrelativistic case.\n\n\n> The redefinition of\n> particles in the Shirokov\'s approach makes it wrong to say that\n> he keeps the old QED Hamiltonian with self-interaction. The self-\n> interaction effects got absorbed in the new definitions of particles,\n> and the physical content of the Hamiltonian changes.\n\nThis is the usual dressing defining quasiparticles for any Hamiltonian.\nHe shows that one just used the standard Hamiltonian and then\nlooks at excitations of the physical vacuum instead of the bare ones.\nThen everything is finite (In the UV, perturbatively).\nThus what you say does not contradict my conclusions.\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>Eugene Stefanovich wrote:
>>
>>>The Hamiltonian of renormalized QED is infinite
>>
>>Not necessarily. It is a limit of well-defined Hamiltonians with
>>perturbatively defined cutoff-dependent coupling coefficients,
>>and this limit might well turn out to exist in the resolvent sense.
>>Indeed, this happens for similar problem in 1D quantum mechanics with
>>singular potentials. You'd at least try to understand the part which is
>>known before you draw your conclusions.
>
> I don't see how Hamiltonian can be finite when parameters e and m
> turn out to be infinite after renormalization. Could you give me
> a reference to similar problems in QM?

I had given you one (\delta-potential) quite early in our discussions,
but you chose to ignore it.

>>>and contains self-interaction of particles and vacuum.
>>
>>This does not make it ill-defined. Indeed, this Hamiltonian is exactly
>>the same as that of Shirokov, whose theory you called equivalent with
>>yours.
>
> In Shirokov's approach the dressing transformation redefines states
> (vacuum vector and creation and annihilation operators). So, the
> Hamiltonian is kept (mathematically)
> the same, while definitions of particles change.

These particles are just quasiparticles in the standard sense.
This is quite in line with the standard nonrelativisitc case,
where the vacuum is the ground state and the quasi-particles are the
excitations of the vacuum. Except that no infinities occur in the
nonrelativistic case.

> The redefinition of
> particles in the Shirokov's approach makes it wrong to say that
> he keeps the old QED Hamiltonian with self-interaction. The self-
> interaction effects got absorbed in the new definitions of particles,
> and the physical content of the Hamiltonian changes.

This is the usual dressing defining quasiparticles for any Hamiltonian.
He shows that one just used the standard Hamiltonian and then
looks at excitations of the physical vacuum instead of the bare ones.
Then everything is finite (In the UV, perturbatively).
Thus what you say does not contradict my conclusions.

Arnold Neumaier

Arnold Neumaier

Nov25-04, 03:36 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> I cannot accept physical relevance of fields. If I do that, then\n> I should accept physical relevance of the Minkowski spacetime.\n> This would mean that boosts are kinematical. This, in turn,\n> is rejected by\n> the Currie-Jordan-Sudarshan theorem. The CJS is a very important\n> statement which is often underestimated, or even ignored, in QFT.\n\nAccording to Dirac, what is kinematical and what is dynamical is a\nfree choice. The CJS theorem is physically irrelevant if its assumptions\nare not satisfied in reality.\n\n\n> The fields \\psi(x,t) have no other meaning except as convenient objects\n> for compact writing of the interaction Hamiltonian and proving its\n> relativistic invariance.\n\nThey are hermitian linear operators and hence observables according to\nthe standard interpretation of QM. Certainly <dA(x)> is measurabble\nand gives the classical e/m field.\n\n\n> As I said earlier, we disagree here.\n\nYes, we disagree in a lot. I said what I could say, and do not\nwish to repeat myself. But am not convinced by your arguments.\nThe future will reveal that you chose a dead end.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> I cannot accept physical relevance of fields. If I do that, then
> I should accept physical relevance of the Minkowski spacetime.
> This would mean that boosts are kinematical. This, in turn,
> is rejected by
> the Currie-Jordan-Sudarshan theorem. The CJS is a very important
> statement which is often underestimated, or even ignored, in QFT.

According to Dirac, what is kinematical and what is dynamical is a
free choice. The CJS theorem is physically irrelevant if its assumptions
are not satisfied in reality.

> The fields \psi(x,t) have no other meaning except as convenient objects
> for compact writing of the interaction Hamiltonian and proving its
> relativistic invariance.

They are hermitian linear operators and hence observables according to
the standard interpretation of QM. Certainly <dA(x)> is measurabble
and gives the classical e/m field.

> As I said earlier, we disagree here.

Yes, we disagree in a lot. I said what I could say, and do not
wish to repeat myself. But am not convinced by your arguments.
The future will reveal that you chose a dead end.

Arnold Neumaier

Eugene Stefanovich

Nov25-04, 03:37 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>\n>>>>and contains self-interaction of particles and vacuum.\n>>>\n>>>This does not make it ill-defined. Indeed, this Hamiltonian is exactly\n>>>the same as that of Shirokov, whose theory you called equivalent with\n>>>yours.\n>>\n>>In Shirokov\'s approach the dressing transformation redefines states\n>>(vacuum vector and creation and annihilation operators). So, the\n>> Hamiltonian is kept (mathematically)\n>>the same, while definitions of particles change.\n>\n>\n> These particles are just quasiparticles in the standard sense.\n> This is quite in line with the standard nonrelativisitc case,\n> where the vacuum is the ground state and the quasi-particles are the\n> excitations of the vacuum. Except that no infinities occur in the\n> nonrelativistic case.\n>\n>\n>\n>>The redefinition of\n>>particles in the Shirokov\'s approach makes it wrong to say that\n>>he keeps the old QED Hamiltonian with self-interaction. The self-\n>>interaction effects got absorbed in the new definitions of particles,\n>>and the physical content of the Hamiltonian changes.\n>\n>\n> This is the usual dressing defining quasiparticles for any Hamiltonian.\n> He shows that one just used the standard Hamiltonian and then\n> looks at excitations of the physical vacuum instead of the bare ones.\n> Then everything is finite (In the UV, perturbatively).\n> Thus what you say does not contradict my conclusions.\n>\n> Arnold Neumaier\n\nOK, let us agree on this point and move on. My major claim is that the\ndressed Hamiltonian (either in my representation or in Shirokov\'s\nrepresentation) describes instantaneous electromagnetic interactions\nbetween charged particles. I suspect that you want to dispute this\nclaim. Let us discuss this.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>
>>>>and contains self-interaction of particles and vacuum.
>>>
>>>This does not make it ill-defined. Indeed, this Hamiltonian is exactly
>>>the same as that of Shirokov, whose theory you called equivalent with
>>>yours.
>>
>>In Shirokov's approach the dressing transformation redefines states
>>(vacuum vector and creation and annihilation operators). So, the
>> Hamiltonian is kept (mathematically)
>>the same, while definitions of particles change.
>
>
> These particles are just quasiparticles in the standard sense.
> This is quite in line with the standard nonrelativisitc case,
> where the vacuum is the ground state and the quasi-particles are the
> excitations of the vacuum. Except that no infinities occur in the
> nonrelativistic case.
>
>
>
>>The redefinition of
>>particles in the Shirokov's approach makes it wrong to say that
>>he keeps the old QED Hamiltonian with self-interaction. The self-
>>interaction effects got absorbed in the new definitions of particles,
>>and the physical content of the Hamiltonian changes.
>
>
> This is the usual dressing defining quasiparticles for any Hamiltonian.
> He shows that one just used the standard Hamiltonian and then
> looks at excitations of the physical vacuum instead of the bare ones.
> Then everything is finite (In the UV, perturbatively).
> Thus what you say does not contradict my conclusions.
>
> Arnold Neumaier

OK, let us agree on this point and move on. My major claim is that the
dressed Hamiltonian (either in my representation or in Shirokov's
representation) describes instantaneous electromagnetic interactions
between charged particles. I suspect that you want to dispute this
claim. Let us discuss this.

Eugene.

Arnold Neumaier

Nov26-04, 01:17 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> OK, let us agree on this point and move on. My major claim is that the\n> dressed Hamiltonian (either in my representation or in Shirokov\'s\n> representation) describes instantaneous electromagnetic interactions\n> between charged particles. I suspect that you want to dispute this\n> claim. Let us discuss this.\n\nNo; I don\'t want to discuss this. Causality is already classically\nwell established experimentally, and there are no good reasons why it\nshould not be valid in the quantum case. All of QFT relies on it, and is\nin agreement with experiment. Transforming the problem into one with\napparent instantaneous interactions does not affect this. If you want\nto keep your belief, do it, but without me.\n\nI do not discuss things for their own sake, but either to teach\nor to learn. Neither of this is happening anymore in this thread.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> OK, let us agree on this point and move on. My major claim is that the
> dressed Hamiltonian (either in my representation or in Shirokov's
> representation) describes instantaneous electromagnetic interactions
> between charged particles. I suspect that you want to dispute this
> claim. Let us discuss this.

No; I don't want to discuss this. Causality is already classically
well established experimentally, and there are no good reasons why it
should not be valid in the quantum case. All of QFT relies on it, and is
in agreement with experiment. Transforming the problem into one with
apparent instantaneous interactions does not affect this. If you want
to keep your belief, do it, but without me.

I do not discuss things for their own sake, but either to teach
or to learn. Neither of this is happening anymore in this thread.

Arnold Neumaier

Arnold Neumaier

Nov27-04, 06:38 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> This would mean that boosts are kinematical. This, in turn,\n> is rejected by\n> the Currie-Jordan-Sudarshan theorem. The CJS is a very important\n> statement which is often underestimated, or even ignored, in QFT.\n\nThe CJS theorem only forbids a dynamics with observer-independent worldlines\nfor each particle. Field theories do not make this assumption, hence are\nnot affected by the theorem. So it can be safely ignored in QED.\n\n\n>>This is like insisting that classical mechanics should be always done in\n>>a particular coordinate system. One can do it, but it makes many things\n>>much more difficult than necessary. The best description has always\n>>been the most tractable one, and not the one which formulated closest to\n>>phenomenology. Compatibility with phenomenology only requires that the\n>>latter can be calculated from the model.\n>\n> No, I am not limited to one coordinate system. Once an instant form\n> representation of the Poincare group is constructed in the Hilbert space\n> of the system, one can transfer between descriptions of different\n> inertial observers. A different representation of the Poincare group\n> (e.g., the point form representation) would describe a different set\n> of transformations between observations of various inertial observers.\n\nI was not referring to coordinate systems in 4-space but to those in\nthe Hilbert space (or, in the classical analogue, phase space).\nMechanics is invariant under canonical transformations classically,\nand under unitary transformations quantum mechanically. You try to\nenforce a particular representation, although any equivalent representation\nis as good.\n\n\n>>There is no way to cheat causality.\n>\n> I disagree. There is a way to send a superluminal (in fact,\n> instantaneous) signal between two charged particles.\n\nWell, try to convince an experimentalist to check it!\nI doubt you\'ll find one prepared to work seriously on such a project.\n\n\n> >Nature is governed by symmetric\n> > hyperbolic differential equations, and hence forbids instantaneous\n> >flow of information.\n>\n> I do not use hyperbolic differential equations in my approach.\n\nThis is why I wrote this. The world is governed by hyperbolic\ndifferential equations, and you try to do without them, contradicting\nall we know on the classical level. To be convincing, you\'d have to\nrewrite all of physics and show that your view gives a more consistent\npicture.\n\nBut in fact you have a 448 page book on QED which\n1. does not even go far enough to derive the Lamb shift and the anomalous\nmagnetic moment of the electron, with which QED gained its credibility.\n2. claims to have the true Hamiltonian, which is however only a formal\npower series, still with infrared problems, whereas a self-adjoint\noperator on a Hilbert space would be needed.\n3. claims that the whole fabric of physics must be changed, removing\nall traces of hyperbolicity (which is essential in all more complex\nproblems to prove existence and uniqueness of the dynamics)\n4. claims that superluminal information exchange is possible between\nlocal observers, which has never been found expermentally.\nWhy should you be credible??? Having done one good thing (UV renormalization\nof the Hamiltonian) does not give licence for your other claims!\n\nScience proceeds not by speculation but by making sure that one always\nstands on safe ground and is correspondingly cautious with great claims.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> This would mean that boosts are kinematical. This, in turn,
> is rejected by
> the Currie-Jordan-Sudarshan theorem. The CJS is a very important
> statement which is often underestimated, or even ignored, in QFT.

The CJS theorem only forbids a dynamics with observer-independent worldlines
for each particle. Field theories do not make this assumption, hence are
not affected by the theorem. So it can be safely ignored in QED.

>>This is like insisting that classical mechanics should be always done in
>>a particular coordinate system. One can do it, but it makes many things
>>much more difficult than necessary. The best description has always
>>been the most tractable one, and not the one which formulated closest to
>>phenomenology. Compatibility with phenomenology only requires that the
>>latter can be calculated from the model.
>
> No, I am not limited to one coordinate system. Once an instant form
> representation of the Poincare group is constructed in the Hilbert space
> of the system, one can transfer between descriptions of different
> inertial observers. A different representation of the Poincare group
> (e.g., the point form representation) would describe a different set
> of transformations between observations of various inertial observers.

I was not referring to coordinate systems in 4-space but to those in
the Hilbert space (or, in the classical analogue, phase space).
Mechanics is invariant under canonical transformations classically,
and under unitary transformations quantum mechanically. You try to
enforce a particular representation, although any equivalent representation
is as good.

>>There is no way to cheat causality.
>
> I disagree. There is a way to send a superluminal (in fact,
> instantaneous) signal between two charged particles.

Well, try to convince an experimentalist to check it!
I doubt you'll find one prepared to work seriously on such a project.

> >Nature is governed by symmetric
> > hyperbolic differential equations, and hence forbids instantaneous
> >flow of information.
>
> I do not use hyperbolic differential equations in my approach.

This is why I wrote this. The world is governed by hyperbolic
differential equations, and you try to do without them, contradicting
all we know on the classical level. To be convincing, you'd have to
rewrite all of physics and show that your view gives a more consistent
picture.

But in fact you have a 448 page book on QED which
1. does not even go far enough to derive the Lamb shift and the anomalous
magnetic moment of the electron, with which QED gained its credibility.
2. claims to have the true Hamiltonian, which is however only a formal
power series, still with infrared problems, whereas a self-adjoint
operator on a Hilbert space would be needed.
3. claims that the whole fabric of physics must be changed, removing
all traces of hyperbolicity (which is essential in all more complex
problems to prove existence and uniqueness of the dynamics)
4. claims that superluminal information exchange is possible between
local observers, which has never been found expermentally.
Why should you be credible??? Having done one good thing (UV renormalization
of the Hamiltonian) does not give licence for your other claims!

Science proceeds not by speculation but by making sure that one always
stands on safe ground and is correspondingly cautious with great claims.

Arnold Neumaier

Eugene Stefanovich

Nov27-04, 06:40 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>I cannot accept physical relevance of fields. If I do that, then\n>>I should accept physical relevance of the Minkowski spacetime.\n>>This would mean that boosts are kinematical. This, in turn,\n>>is rejected by\n>>the Currie-Jordan-Sudarshan theorem. The CJS is a very important\n>>statement which is often underestimated, or even ignored, in QFT.\n>\n>\n> According to Dirac, what is kinematical and what is dynamical is a\n> free choice.\n\nIf Dirac said that, then I disagree with him.\nTake the point form generator of space translations in a 2-particle\nsystem\n\nP = p_1 + p_2 + Y\n\nwhere interaction operator Y generally does not commute with p_1 and\np_2. Let us find how momenta of particles transform to the translated\nframe of reference\n\np_1(a) = exp(-iPa) p_1 exp(iPa) = p_1 -i [Y,p_1]a + ...\np_2(a) = exp(-iPa) p_2 exp(iPa) = p_2 -i [Y,p_2]a + ...\n\nContrary to observations, these transformations are non-trivial and\ninteraction-dependent. This dependence should be observable in\nexperiment. On the other hand, in the instant form dynamics Y = 0, and\n\np_1(a) = p_1\np_2(a) = p_2\n\nas it should be. The instant and point forms are physically different.\n\n> The CJS theorem is physically irrelevant if its assumptions\n> are not satisfied in reality.\n\nThe assumptions of the CJS theorem are\n1) the Hamiltonian description of dynamics\n2) measurability of particle observables, like p_i and r_i,\nin all frames of reference.\nI think, these assumptions are well justified.\n\n>\n>\n>\n>>The fields \\psi(x,t) have no other meaning except as convenient objects\n>>for compact writing of the interaction Hamiltonian and proving its\n>>relativistic invariance.\n>\n>\n> They are hermitian linear operators and hence observables according to\n> the standard interpretation of QM. Certainly <dA(x)> is measurabble\n> and gives the classical e/m field.\n\nI have a few counterarguments:\n1) the electron-positron field \\psi(x,t) is not Hermitian, so it is\nnot observable according to your criterion\n2) It is true that every observable is described by a Hermitian\noperator, but it is not generally true that each Hermitian operator\ncorresponds to an observable.\n3) What is the meaning of x? It would be natural to assume that x are\neigenvalues of the position operator. Then is it true that t is\nan eigenvalue of the time operator? Nobody was able to introduce\na satisfactory time operator in quantum mechanics.\nThe bottomline of my approach\nis that unification of space and time in the Einsteinian version of\nrelativity theory is in deep contradiction with basic principles of\nquantum mechanics. Time is just not an observable in the quantum-\nmechanical sense. There are now many discussions of this "paradox"\namong people trying to unify quantum mechanics with general\nrelativity. This is considered to be THE main obstacle for\ndevelopment of a consistent "quantum gravity" theory. I don\'t think\nthey are going to succeed until they realize that space and time\ncannot be unified in one 4D continuum.\n\n>\n>\n>\n>>As I said earlier, we disagree here.\n>\n>\n> Yes, we disagree in a lot. I said what I could say, and do not\n> wish to repeat myself. But am not convinced by your arguments.\n> The future will reveal that you chose a dead end.\n\nMy feeling is that theoretical physics is already in the dead end.\nThe main reason for that is the blind acceptance of the 4D space-time\ncontinuum. I have shown that the space-time description does not follow\nfrom Einstein\'s postulates, and is, in fact, an additional hypothesis.\nEverything becomes simple and consistent, if this hypothesis is\ndropped.\n\n\nEugene Stefanovich.\n\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>I cannot accept physical relevance of fields. If I do that, then
>>I should accept physical relevance of the Minkowski spacetime.
>>This would mean that boosts are kinematical. This, in turn,
>>is rejected by
>>the Currie-Jordan-Sudarshan theorem. The CJS is a very important
>>statement which is often underestimated, or even ignored, in QFT.
>
>
> According to Dirac, what is kinematical and what is dynamical is a
> free choice.

If Dirac said that, then I disagree with him.
Take the point form generator of space translations in a 2-particle
system

P = p_1 + p_2 + Y

where interaction operator Y generally does not commute with p_1 and
p_2. Let us find how momenta of particles transform to the translated
frame of reference

p_1(a) = \exp(-iPa) p_1 \exp(iPa) = p_1 -i [Y,p_1]a + .[/itex]..
p_2(a) = \exp(-iPa) p_2 \exp(iPa) = p_2 -i [Y,p_2]a + ...

Contrary to observations, these transformations are non-trivial and
interaction-dependent. This dependence should be observable in
experiment. On the other hand, in the instant form dynamics Y = 0, and

[itex]p_1(a) = p_1p_2(a) = p_2

as it should be. The instant and point forms are physically different.

> The CJS theorem is physically irrelevant if its assumptions
> are not satisfied in reality.

The assumptions of the CJS theorem are
1) the Hamiltonian description of dynamics
2) measurability of particle observables, like p_i and r_i,
in all frames of reference.
I think, these assumptions are well justified.

>
>
>
>>The fields \psi(x,t) have no other meaning except as convenient objects
>>for compact writing of the interaction Hamiltonian and proving its
>>relativistic invariance.
>
>
> They are hermitian linear operators and hence observables according to
> the standard interpretation of QM. Certainly <dA(x)> is measurabble
> and gives the classical e/m field.

I have a few counterarguments:
1) the electron-positron field \psi(x,t) is not Hermitian, so it is
not observable according to your criterion
2) It is true that every observable is described by a Hermitian
operator, but it is not generally true that each Hermitian operator
corresponds to an observable.
3) What is the meaning of x? It would be natural to assume that x are
eigenvalues of the position operator. Then is it true that t is
an eigenvalue of the time operator? Nobody was able to introduce
a satisfactory time operator in quantum mechanics.
The bottomline of my approach
is that unification of space and time in the Einsteinian version of
relativity theory is in deep contradiction with basic principles of
quantum mechanics. Time is just not an observable in the quantum-
mechanical sense. There are now many discussions of this "paradox"
among people trying to unify quantum mechanics with general
relativity. This is considered to be THE main obstacle for
development of a consistent "quantum gravity" theory. I don't think
they are going to succeed until they realize that space and time
cannot be unified in one 4D continuum.

>
>
>
>>As I said earlier, we disagree here.
>
>
> Yes, we disagree in a lot. I said what I could say, and do not
> wish to repeat myself. But am not convinced by your arguments.
> The future will reveal that you chose a dead end.

My feeling is that theoretical physics is already in the dead end.
The main reason for that is the blind acceptance of the 4D space-time
continuum. I have shown that the space-time description does not follow
from Einstein's postulates, and is, in fact, an additional hypothesis.
Everything becomes simple and consistent, if this hypothesis is
dropped.

Eugene Stefanovich.

>
>
> Arnold Neumaier
>

Eugene Stefanovich

Nov27-04, 06:41 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>OK, let us agree on this point and move on. My major claim is that the\n>>dressed Hamiltonian (either in my representation or in Shirokov\'s\n>>representation) describes instantaneous electromagnetic interactions\n>>between charged particles. I suspect that you want to dispute this\n>>claim. Let us discuss this.\n>\n>\n> No; I don\'t want to discuss this. Causality is already classically\n> well established experimentally,\n\n1) My approach does not contradict causality. The effect never occurs\nbefore the cause in all frames of reference.\n2) As far as I know, nobody have measured experimentally the speed of\npropagation of the Coulomb interaction between two charged particles.\n(I am not talking, of course, about the propagation of the\ntransverse electromagnetic wave = photons).\n\n\nEugene Stefanovich.\n\n> and there are no good reasons why it\n> should not be valid in the quantum case. All of QFT relies on it, and is\n> in agreement with experiment. Transforming the problem into one with\n> apparent instantaneous interactions does not affect this.\n\nYou agree that theory can be cast into form with instantaneous\ninteractions. I think you also agree that only one choice is realized\nin nature (either instantaneous or retarded). Then your statement leads\nto a contradiction: QFT cannot have it both ways.\n\nEugene Stefanovich.\n\n\n> If you want\n> to keep your belief, do it, but without me.\n>\n> I do not discuss things for their own sake, but either to teach\n> or to learn. Neither of this is happening anymore in this thread.\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>OK, let us agree on this point and move on. My major claim is that the
>>dressed Hamiltonian (either in my representation or in Shirokov's
>>representation) describes instantaneous electromagnetic interactions
>>between charged particles. I suspect that you want to dispute this
>>claim. Let us discuss this.
>
>
> No; I don't want to discuss this. Causality is already classically
> well established experimentally,

1) My approach does not contradict causality. The effect never occurs
before the cause in all frames of reference.
2) As far as I know, nobody have measured experimentally the speed of
propagation of the Coulomb interaction between two charged particles.
(I am not talking, of course, about the propagation of the
transverse electromagnetic wave = photons).

Eugene Stefanovich.

> and there are no good reasons why it
> should not be valid in the quantum case. All of QFT relies on it, and is
> in agreement with experiment. Transforming the problem into one with
> apparent instantaneous interactions does not affect this.

You agree that theory can be cast into form with instantaneous
interactions. I think you also agree that only one choice is realized
in nature (either instantaneous or retarded). Then your statement leads
to a contradiction: QFT cannot have it both ways.

Eugene Stefanovich.

> If you want
> to keep your belief, do it, but without me.
>
> I do not discuss things for their own sake, but either to teach
> or to learn. Neither of this is happening anymore in this thread.
>
>
> Arnold Neumaier
>

Eugene Stefanovich

Nov27-04, 12:43 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>output from its instantaneous tracking system is displayed for the\n: brass.\n:\n: In 1993, however, the FBI allowed a reporter who was working on what the\n: bureau expected would be a friendly article to visit the inner sanctum.\n:\n: The command center, she later wrote, "looks not unlike the Starship\n: Enterprise, of \'Star Trek.\' On the rear wall of the room are three giant\n: screens on which neighborhood maps, live field surveillance, and graphs\n: charting the progress of a manhunt can be projected.\n:\n: Law enforcement officials, at stations in three semicircular tiers of\n: desks, can watch---and direct---as criminals are caught in the act.\n:\n: Their computer mouse screen pointers are a gun icon.\n\nOH MY GAWD!!!!\n\nWHAT\'S NEXT, THE WHOLE DAMN COUNTRY???\n\n\nWhat is this?\n\n* Subject: Air Force News Service 01oct96\n* From: webmaster@vnis.com (Veterans News & Information Service)\n* Date: 1996/10/01\n* Newsgroups: soc.veterans\n*\n* Night vision lasers go to court\n*\n* KIRTLAND AIR FORCE BASE, N.M. (AFNS) -- More drug and\n* smuggling convictions may soon result from a laser optics research\n* agreement signed here Sept. 25 between the Air Force Phillips\n* Laboratory and FLIR Systems, Inc.\n*\n* "Current sensors cannot read a license plate, ship registration, or\n* aircraft tail number," said 1st Lieutenant Robert J. Ireland of\n* Phillips\'s Lasers and Imaging Directorate.\n*\n* "But an operator with special eyewear, using a laser spotlight having a\n* wavelength invisible to the unaided eye, may be able to," he said.\n*\n* ----------------------------------------------------------------------\n* Air Force News Agency : DSN: 945-1281\n* AFNEWS/IICT : (210) 925-1281\n* 203 Norton\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>output from its instantaneous tracking system is displayed for the
: brass.
:
: In 1993, however, the FBI allowed a reporter who was working on what the
: bureau expected would be a friendly article to visit the inner sanctum.
:
: The command center, she later wrote, "looks not unlike the Starship
: Enterprise, of 'Star Trek.' On the rear wall of the room are three giant
: screens on which neighborhood maps, live field surveillance, and graphs
: charting the progress of a manhunt can be projected.
:
: Law enforcement officials, at stations in three semicircular tiers of
: desks, can watch---and direct---as criminals are caught in the act.
:
: Their computer mouse screen pointers are a gun icon.

OH MY GAWD!!!!

WHAT'S NEXT, THE WHOLE DAMN COUNTRY???

What is this?

* Subject: Air Force News Service 01oct96
* From: webmaster@vnis.com (Veterans News & Information Service)
* Date: 1996/10/01
* Newsgroups: soc.veterans
*
* Night vision lasers go to court
*
* KIRTLAND AIR FORCE BASE, N.M. (AFNS) -- More drug and
* smuggling convictions may soon result from a laser optics research
* agreement signed here Sept. 25 between the Air Force Phillips
* Laboratory and FLIR Systems, Inc.
*
* "Current sensors cannot read a license plate, ship registration, or
* aircraft tail number," said 1st Lieutenant Robert J. Ireland of
* Phillips's Lasers and Imaging Directorate.
*
* "But an operator with special eyewear, using a laser spotlight having a
* wavelength invisible to the unaided eye, may be able to," he said.
*
* ----------------------------------------------------------------------
* Air Force News Agency : DSN: 945-1281
* AFNEWS/IICT : (210) 925-1281
* 203 Norton

Eugene Stefanovich

Nov27-04, 02:23 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>: * "Unless the encryption issue is RESOLVED soon, criminal conversations\n: * over the telephone and other communications devices will become\n: * indecipherable by law enforcement. This, as much as any issue,\n: * jeopardizes the public safety and national security of this country."\n\nLouis Freeh, banging the Drums of War.\n\n\nIt\'s official:\n\n* http://epic.org/crypto/ban/fbi_dox/impact_text.gif\n*\n* SECRET FBI report\n*\n* NEED FOR A NATIONAL POLICY\n*\n* A national policy embodied in legislation is needed which insures\n* that cryptography use in the United States should be forced to be\n* crackable by law enforcement, so such communications can be monitored\n* with real-time decryption.\n*\n* All cryptography that cannot meet this standard should be prohibited.\n\n\n\nThe U.S. asked the OECD to agree to internationally required Key Recovery.\n\n* What Is The OECD\n*\n* The Organization for Economic Co-operation and Development, based in\n* Paris, France, is a unique forum permitting governments of the\n* industrialized democracies to study and formulate the best policies\n* possible in all economic and social spheres.\n\n: From owner-firewalls-outgoing@GreatCircle.COM Wed May 14 18:54:15 1997\n: Received: from osiris (osiris.nso.org [207.30.58.40]) by ra.nso.org\n: (post.office MTA v1.9.3 ID# 0-13592) with SMTP id AAA322\n: for <firewalls@GreatCircle.COM>; Wed, 14 May 1997 12:56:13 -0400\n: Date: Wed, 14 May 1997 12:58:46 -0400\n: To: firewalls@GreatCircle.COM\n: From: research@isr.net (Research Unit I)\n: Subject: Re: Encryption Outside US\n:\n:\n: I was part of that OECD Expert Group, and believe I may shine at least\n: some light on what exactly was said and happened at the meetings.\n:\n: The main conflict during all sessions was the demand of the US to be\n: able to decrypt anything, anywhere at any time versus t\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>: * "Unless the encryption issue is RESOLVED soon, criminal conversations
: * over the telephone and other communications devices will become
: * indecipherable by law enforcement. This, as much as any issue,
: * jeopardizes the public safety and national security of this country."

Louis Freeh, banging the Drums of War.

It's official:

* http://epic.org/crypto/ban/fbi_dox/impact_text.gif
*
* SECRET FBI report
*
* NEED FOR A NATIONAL POLICY
*
* A national policy embodied in legislation is needed which insures
* that cryptography use in the United States should be forced to be
* crackable by law enforcement, so such communications can be monitored
* with real-time decryption.
*
* All cryptography that cannot meet this standard should be prohibited.

The U.S. asked the OECD to agree to internationally required Key Recovery.

* What Is The OECD
*
* The Organization for Economic Co-operation and Development, based in
* Paris, France, is a unique forum permitting governments of the
* industrialized democracies to study and formulate the best policies
* possible in all economic and social spheres.

: From owner-firewalls-outgoing@GreatCircle.COM Wed May 14 18:54:15 1997
: Received: from osiris (osiris.nso.org [207.30.58.40]) by ra.nso.org
: (post.office MTA v1.9.3 ID# 0-13592) with SMTP id AAA322
: for <firewalls@GreatCircle.COM>; Wed, 14 May 1997 12:56:13 -0400
: Date: Wed, 14 May 1997 12:58:46 -0400
: To: firewalls@GreatCircle.COM
: From: research@isr.net (Research Unit I)
: Subject: Re: Encryption Outside US
:
:
: I was part of that OECD Expert Group, and believe I may shine at least
: some light on what exactly was said and happened at the meetings.
:
: The main conflict during all sessions was the demand of the US to be
: able to decrypt anything, anywhere at any time versus t

Eugene Stefanovich

Nov27-04, 02:45 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>So, "statistical filtration for all homeostatic loops" means one is checking\non the health of the monitored system.\n\nThe cybernetician uses the same language for feedback of weapons systems\n(picking out a submarine from the background noise of the ocean) as they\ndo for describing human life, as they do for the political organization\nof a country.\n\nLike I said, an awesome scope.\n\nNorbert Wiener even came up with a physics-based\ndescription of how life is formed by information.\n\nCheck it out. Hang in there too, it\'s worth it.\n\n* "Platform for Change", by Stafford Beer, 1978, ISBN 0 471 06189 1\n*\n* The term \'entropy\' began life as a subtle measure of energy flow.\n*\n* When something hotter is systemically bound to something cooler, the\n* greater energy of the hotter stuff migrates---inexorably migrates---\n* into the cooler stuff. This is one manifestation of the Second Law of\n* Thermodynamics, which everyone of education has encountered.\n*\n* This is sometimes referred to as \'the universe is running down\'.\n\nOkay, \'entropy\', yeah I remember that kinda. Keep going:\n\nOur solar system is a lot of matter that is NOT sitting in a situation of\nentropy: the sun is radiating heat at the planets. Instead of just matter\nsmoothing out to a common low-energy state, a burning fireball is at work.\n\nCybernetics states that under conditions\nlike this, matter does something special.\n\n* "Platform for Change", by Stafford Beer, 1978, ISBN 0 471 06189 1\n*\n* If we have a universe, which is improbable though it exists, it is\n* because the Second Law of Thermodynamics has two forms. One is concerned\n* with the pressure to even out energy; that is the form which belongs to\n* our stereotyped conception of the u\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>So, "statistical filtration for all homeostatic loops" means one is checking
on the health of the monitored system.

The cybernetician uses the same language for feedback of weapons systems
(picking out a submarine from the background noise of the ocean) as they
do for describing human life, as they do for the political organization
of a country.

Like I said, an awesome scope.

Norbert Wiener even came up with a physics-based
description of how life is formed by information.

Check it out. Hang in there too, it's worth it.

* "Platform for Change", by Stafford Beer, 1978, ISBN 471 06189 1
*
* The term 'entropy' began life as a subtle measure of energy flow.
*
* When something hotter is systemically bound to something cooler, the
* greater energy of the hotter stuff migrates---inexorably migrates---
* into the cooler stuff. This is one manifestation of the Second Law of
* Thermodynamics, which everyone of education has encountered.
*
* This is sometimes referred to as 'the universe is running down'.

Okay, 'entropy', yeah I remember that kinda. Keep going:

Our solar system is a lot of matter that is NOT sitting in a situation of
entropy: the sun is radiating heat at the planets. Instead of just matter
smoothing out to a common low-energy state, a burning fireball is at work.

Cybernetics states that under conditions
like this, matter does something special.

* "Platform for Change", by Stafford Beer, 1978, ISBN 471 06189 1
*
* If we have a universe, which is improbable though it exists, it is
* because the Second Law of Thermodynamics has two forms. One is concerned
* with the pressure to even out energy; that is the form which belongs to
* our stereotyped conception of the u

Eugene Stefanovich

Nov27-04, 03:55 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>pick up the telephone? Are all Internet or fax\nmessages being pored over continuously by shadowy figures somewhere in a\nwindowless building? There is almost never any solid information with which\nto judge what is realistic concern and what is silly paranoia.\n\nWhat follows explains as precisely as possible - and for the first time in\npublic - how the worldwide system works, just how immense and powerful it is\nand what it can and cannot do. The electronic spies are not ubiquitous, but\nthe paranoia is not unfounded.\n\nThe global system has a highly secret codename - ECHELON.\n\nThe intelligence agencies will be shocked to see it named and described for\nthe first time in print.\n\nEach station in the ECHELON network has computers that automatically search\nthrough millions of intercepted messages for ones containing pre-programmed\nkeywords or fax, telex and email addresses. Every word of every message is\nautomatically searched: they do not need your specific telephone number or\nInternet address on the list.\n\nAll the different computers in the network are known, within the UKUSA\nagencies, as the ECHELON Dictionaries.\n\nComputers that can search for keywords have existed since at least the 1970s,\nbut the ECHELON system has been designed to interconnect all these computers\nand allow the stations to function as components of an integrated whole.\n\nUnder the ECHELON system, a particular station\'s\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>pick up the telephone? Are all Internet or fax
messages being pored over continuously by shadowy figures somewhere in a
windowless building? There is almost never any solid information with which
to judge what is realistic concern and what is silly paranoia.

What follows explains as precisely as possible - and for the first time in
public - how the worldwide system works, just how immense and powerful it is
and what it can and cannot do. The electronic spies are not ubiquitous, but
the paranoia is not unfounded.

The global system has a highly secret codename - ECHELON.

The intelligence agencies will be shocked to see it named and described for
the first time in print.

Each station in the ECHELON network has computers that automatically search
through millions of intercepted messages for ones containing pre-programmed
keywords or fax, telex and email addresses. Every word of every message is
automatically searched: they do not need your specific telephone number or
Internet address on the list.

All the different computers in the network are known, within the UKUSA
agencies, as the ECHELON Dictionaries.

Computers that can search for keywords have existed since at least the 1970s,
but the ECHELON system has been designed to interconnect all these computers
and allow the stations to function as components of an integrated whole.

Under the ECHELON system, a particular station's

Eugene Stefanovich

Nov27-04, 04:02 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>*for* the people, by the people\'.\n\nWe have slowly reached a state of McCarthyism against any elected\nofficial who shows ANY "SIGNS OF SOFTNESS" in the War against Crime.\n\nThe constant state of War against imaginary enemies must end.\n\nBy imaginary, I mean crime was going down the whole War time.\n\nAll we are saying, is give peace a chance.\n\n\n----\n\nI repeat: Civil war would have broken out.\n\n----\n\n\nDire suspension of Constitutional protections happens during War:\n\nAbraham Lincoln ordered thousands of people detained without hearings,\nand opposition newspapers shut down during the Civil War. During\nWorld War II: the president orders Japanese and such to be held in\ninternment camps.\n\nSo why do we have all these loss of freedoms during peacetime?\n\nAnswer:\n\nBecause the Military has never stopped fighting World War II.\n\n# "Spy Agencies Faulted for War Focus"\n# By Tim Wiener, The New York Times, June 28, 1996\n#\n# American intelligence agencies devote too much time and money to supporting\n# the Military, and FAR too little to understanding the problems of peace, a\n# new and authoritative critique concludes.\n#\n# The report is one of FOUR MAJOR STUDIES to cite the "alarming imbalance"\n# of spending more than \\$26 billion a year on machines, and less tha\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>*for* the people, by the people'.

We have slowly reached a state of McCarthyism against any elected
official who shows ANY "SIGNS OF SOFTNESS" in the War against Crime.

The constant state of War against imaginary enemies must end.

By imaginary, I mean crime was going down the whole War time.

All we are saying, is give peace a chance.

----

I repeat: Civil war would have broken out.

----

Dire suspension of Constitutional protections happens during War:

Abraham Lincoln ordered thousands of people detained without hearings,
and opposition newspapers shut down during the Civil War. During
World War II: the president orders Japanese and such to be held in
internment camps.

So why do we have all these loss of freedoms during peacetime?

Answer:

Because the Military has never stopped fighting World War II.

# "Spy Agencies Faulted for War Focus"
# By Tim Wiener, The New York Times, June 28, 1996
#
# American intelligence agencies devote too much time and money to supporting
# the Military, and FAR too little to understanding the problems of peace, a
# new and authoritative critique concludes.
#
# The report is one of FOUR MAJOR STUDIES to cite the "alarming imbalance"
# of spending more than $26 billion a year on machines, and less tha

Eugene Stefanovich

Nov27-04, 04:13 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>and Congressional legislation.\n\nTo create a secret agency and a secret sham court.\n\n\nUsed repeatedly to control lawful domestic political protest.\n\nThe Soviet Union and China we were told to fear.\n\n\n*************************************** ***************************************\n********* ************************************************** *******************\n***************************** *************************************************\ n\n\nPart 2: On Monitoring and Being Monitored\n---- - -- ---------- --- ----- ---------\n\no On Monitoring\n- Driver\'s Seat\n- Five Months Statistics\n- The FBI Investigations\n- I Can See What You Are Thinking\n- Why I Monitor\no On Being Monitored\n\n\nIf you are an ordinary citizen, you may not notice the\nINVISIBLE massive spy apparatus until it is too late.\n\nAs has already happened repeatedly: the government will use\nthis National Spying Apparatus to crush political protests,\nand monitor the politically incorrect.\n\nIn the 1960s, 1970s, 1980s... and the 1990s.\n\n\nQuestion:\n\nWhy argue against something that would catch crime?\n\nAnswer:\n\nECHELON is so invasive we lose all privacy.\nIt is infinitely abusable.\nIt has been abused repeatedly.\nCALEA takes us into the abyss.\n\n\nWould monitoring really turn up that many violations?\nMeaning: is it really that effective a mechanism?\n\n\n\n******************************** **********************************************\n\n \n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>and Congressional legislation.

To create a secret agency and a secret sham court.

Used repeatedly to control lawful domestic political protest.

The Soviet Union and China we were told to fear.

************************************************** ************************************************** ************************************************** ************************************************** **********************************

Part 2: On Monitoring and Being Monitored
---- - -- ---------- --- ----- ---------

o On Monitoring
- Driver's Seat
- Five Months Statistics
- The FBI Investigations
- I Can See What You Are Thinking
- Why I Monitor
o On Being Monitored

If you are an ordinary citizen, you may not notice the
INVISIBLE massive spy apparatus until it is too late.

As has already happened repeatedly: the government will use
this National Spying Apparatus to crush political protests,
and monitor the politically incorrect.

In the 1960s, 1970s, 1980s... and the 1990s.

Question:

Why argue against something that would catch crime?

Answer:

ECHELON is so invasive we lose all privacy.
It is infinitely abusable.
It has been abused repeatedly.
CALEA takes us into the abyss.

Would monitoring really turn up that many violations?
Meaning: is it really that effective a mechanism?

************************************************** ****************************

Eugene Stefanovich

Nov28-04, 06:03 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>This would mean that boosts are kinematical. This, in turn,\n>>is rejected by\n>>the Currie-Jordan-Sudarshan theorem. The CJS is a very important\n>>statement which is often underestimated, or even ignored, in QFT.\n>\n>\n> The CJS theorem only forbids a dynamics with observer-independent worldlines\n> for each particle. Field theories do not make this assumption, hence are\n> not affected by the theorem. So it can be safely ignored in QED.\n\nYou are right that CJS theorem has no relevance to the traditional QED,\nbecause this approach only cares about the S-matrix, and does not care\nabout the time-dependence of particle observables (this time dependence\nis represented by trajectories in the classical limit). In particular,\nQED cannot even talk about the speed of propagation of interaction,\nbecause it can say only about the correlation between infinite past\nand infinite future, and doesn\'t know what happens in between.\n\nThe CJS theorem\nbecomes very important in the case of dynamical description of the time\nevolution. Such a description is built in my book, and in agreement with\nthe CJS theorem,\nI found that particle trajectories (or, in general, particle\nobservables) do not transform in a manifestly covariant fashion between\ndifferent reference frames. The dynamical character of the boost\ntransformations is consistent with causality and instantaneous\npropagation of interactions.\n\n>\n>\n>\n>>>This is like insisting that classical mechanics should be always done in\n>>>a particular coordinate system. One can do it, but it makes many things\n>>>much more difficult than necessary. The best description has always\n>>>been the most tractable one, and not the one which formulated closest to\n>>>phenomenology. Compatibility with phenomenology only requires that the\n>>>latter can be calculated from the model.\n>>\n>>No, I am not limited to one coordinate system. Once an instant form\n>>representation of the Poincare group is constructed in the Hilbert space\n>> of the system, one can transfer between descriptions of different\n>>inertial observers. A different representation of the Poincare group\n>>(e.g., the point form representation) would describe a different set\n>>of transformations between observations of various inertial observers.\n>\n>\n> I was not referring to coordinate systems in 4-space but to those in\n> the Hilbert space (or, in the classical analogue, phase space).\n> Mechanics is invariant under canonical transformations classically,\n> and under unitary transformations quantum mechanically. You try to\n> enforce a particular representation, although any equivalent representation\n> is as good.\n\nI disagree that unitary equivalent representations of the Poincare group\nare always physically equivalent. Take two Hamiltonians connected by a\nunitary transformation\n\nH\' = U H U^{-1}\n\nThese two Hamiltonians may have the same S-matrix and energy spectrum of\nbound states. However, the wave functions of bound states are\ndifferent, and the time evolutions described by these two Hamiltonians\nare different too.\n\n>\n>\n>\n>>>There is no way to cheat causality.\n>>\n>>I disagree. There is a way to send a superluminal (in fact,\n>>instantaneous) signal between two charged particles.\n>\n>\n> Well, try to convince an experimentalist to check it!\n> I doubt you\'ll find one prepared to work seriously on such a project.\n\nI would love to talk to an experimentalist, and I hope that I can\nconvince somebody. The question about the speed of propagation of the\nCoulomb and magnetic interactions remains unresolved experimentally\nuntil this day. I think you would agree with me that this\nis the fundamental issue for our understanding of nature, and the\nabsence of experimental data is rather disturbing.\nI hope that there are experimentalists which are curious\nenough to check this for themselves, regardless of\ntheoretical speculations.\n\n>\n>\n>\n>> >Nature is governed by symmetric\n>> > hyperbolic differential equations, and hence forbids instantaneous\n>> >flow of information.\n>>\n>>I do not use hyperbolic differential equations in my approach.\n>\n>\n> This is why I wrote this. The world is governed by hyperbolic\n> differential equations, and you try to do without them, contradicting\n> all we know on the classical level. To be convincing, you\'d have to\n> rewrite all of physics and show that your view gives a more consistent\n> picture.\n>\n> But in fact you have a 448 page book on QED which\n> 1. does not even go far enough to derive the Lamb shift and the anomalous\n> magnetic moment of the electron, with which QED gained its credibility.\n\nI have a proof that in my approach the S-matrix and the energies of\nbound states are the same as in renormalized QED. This indirectly\nsuggests that the Lamb shift and the magnetic moment of the electron\nshould come out correct. Though, I agree with you that to be more\nconvincing, I should perform a direct calculation. That\'s what I am\ndoing now.\n\n> 2. claims to have the true Hamiltonian, which is however only a formal\n> power series, still with infrared problems, whereas a self-adjoint\n> operator on a Hilbert space would be needed.\n\nMore work is needed here as well, I agree.\n\n> 3. claims that the whole fabric of physics must be changed, removing\n> all traces of hyperbolicity (which is essential in all more complex\n> problems to prove existence and uniqueness of the dynamics)\n> 4. claims that superluminal information exchange is possible between\n> local observers, which has never been found expermentally.\n\nThe retarded character of the electromagnetic interaction has never been\nproved experimentally either.\n\n> Why should you be credible??? Having done one good thing (UV renormalization\n> of the Hamiltonian) does not give licence for your other claims!\n>\n> Science proceeds not by speculation but by making sure that one always\n> stands on safe ground and is correspondingly cautious with great claims.\n\nI believe, speculation plays a great role in the development of\nscience. Actually, the universally accepted Minkowski spacetime is also\na great speculation. It doesn\'t follow from two Einstein\'s postulates,\nand it does not have experimental support. I consider my approach more\nconservative and cautious than traditional special relativity.\nI do not accept the dubious postulate of Minkowski spacetime, though I\nkeep all other well-established postulates of relativity and quantum\nmechanics.\n\nEugene Stefanovich.\n\n\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>This would mean that boosts are kinematical. This, in turn,
>>is rejected by
>>the Currie-Jordan-Sudarshan theorem. The CJS is a very important
>>statement which is often underestimated, or even ignored, in QFT.
>
>
> The CJS theorem only forbids a dynamics with observer-independent worldlines
> for each particle. Field theories do not make this assumption, hence are
> not affected by the theorem. So it can be safely ignored in QED.

You are right that CJS theorem has no relevance to the traditional QED,
because this approach only cares about the S-matrix, and does not care
about the time-dependence of particle observables (this time dependence
is represented by trajectories in the classical limit). In particular,
QED cannot even talk about the speed of propagation of interaction,
because it can say only about the correlation between infinite past
and infinite future, and doesn't know what happens in between.

The CJS theorem
becomes very important in the case of dynamical description of the time
evolution. Such a description is built in my book, and in agreement with
the CJS theorem,
I found that particle trajectories (or, in general, particle
observables) do not transform in a manifestly covariant fashion between
different reference frames. The dynamical character of the boost
transformations is consistent with causality and instantaneous
propagation of interactions.

>
>
>
>>>This is like insisting that classical mechanics should be always done in
>>>a particular coordinate system. One can do it, but it makes many things
>>>much more difficult than necessary. The best description has always
>>>been the most tractable one, and not the one which formulated closest to
>>>phenomenology. Compatibility with phenomenology only requires that the
>>>latter can be calculated from the model.
>>
>>No, I am not limited to one coordinate system. Once an instant form
>>representation of the Poincare group is constructed in the Hilbert space
>> of the system, one can transfer between descriptions of different
>>inertial observers. A different representation of the Poincare group
>>(e.g., the point form representation) would describe a different set
>>of transformations between observations of various inertial observers.
>
>
> I was not referring to coordinate systems in 4-space but to those in
> the Hilbert space (or, in the classical analogue, phase space).
> Mechanics is invariant under canonical transformations classically,
> and under unitary transformations quantum mechanically. You try to
> enforce a particular representation, although any equivalent representation
> is as good.

I disagree that unitary equivalent representations of the Poincare group
are always physically equivalent. Take two Hamiltonians connected by a
unitary transformation

H' = U H U^{-1}

These two Hamiltonians may have the same S-matrix and energy spectrum of
bound states. However, the wave functions of bound states are
different, and the time evolutions described by these two Hamiltonians
are different too.

>
>
>
>>>There is no way to cheat causality.
>>
>>I disagree. There is a way to send a superluminal (in fact,
>>instantaneous) signal between two charged particles.
>
>
> Well, try to convince an experimentalist to check it!
> I doubt you'll find one prepared to work seriously on such a project.

I would love to talk to an experimentalist, and I hope that I can
convince somebody. The question about the speed of propagation of the
Coulomb and magnetic interactions remains unresolved experimentally
until this day. I think you would agree with me that this
is the fundamental issue for our understanding of nature, and the
absence of experimental data is rather disturbing.
I hope that there are experimentalists which are curious
enough to check this for themselves, regardless of
theoretical speculations.

>
>
>
>> >Nature is governed by symmetric
>> > hyperbolic differential equations, and hence forbids instantaneous
>> >flow of information.
>>
>>I do not use hyperbolic differential equations in my approach.
>
>
> This is why I wrote this. The world is governed by hyperbolic
> differential equations, and you try to do without them, contradicting
> all we know on the classical level. To be convincing, you'd have to
> rewrite all of physics and show that your view gives a more consistent
> picture.
>
> But in fact you have a 448 page book on QED which
> 1. does not even go far enough to derive the Lamb shift and the anomalous
> magnetic moment of the electron, with which QED gained its credibility.

I have a proof that in my approach the S-matrix and the energies of
bound states are the same as in renormalized QED. This indirectly
suggests that the Lamb shift and the magnetic moment of the electron
should come out correct. Though, I agree with you that to be more
convincing, I should perform a direct calculation. That's what I am
doing now.

> 2. claims to have the true Hamiltonian, which is however only a formal
> power series, still with infrared problems, whereas a self-adjoint
> operator on a Hilbert space would be needed.

More work is needed here as well, I agree.

> 3. claims that the whole fabric of physics must be changed, removing
> all traces of hyperbolicity (which is essential in all more complex
> problems to prove existence and uniqueness of the dynamics)
> 4. claims that superluminal information exchange is possible between
> local observers, which has never been found expermentally.

The retarded character of the electromagnetic interaction has never been
proved experimentally either.

> Why should you be credible??? Having done one good thing (UV renormalization
> of the Hamiltonian) does not give licence for your other claims!
>
> Science proceeds not by speculation but by making sure that one always
> stands on safe ground and is correspondingly cautious with great claims.

I believe, speculation plays a great role in the development of
science. Actually, the universally accepted Minkowski spacetime is also
a great speculation. It doesn't follow from two Einstein's postulates,
and it does not have experimental support. I consider my approach more
conservative and cautious than traditional special relativity.
I do not accept the dubious postulate of Minkowski spacetime, though I
keep all other well-established postulates of relativity and quantum
mechanics.

Eugene Stefanovich.

>
>
> Arnold Neumaier
>

Arnold Neumaier

Nov29-04, 02:49 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>=20\n>>According to Dirac, what is kinematical and what is dynamical is a\n>>free choice.=20\n>=20\n> If Dirac said that, then I disagree with him.\n\nI find it difficult to disagree with well-established mathematical facts.=\n\nRev. Mod. Phys. 21, 392=96399 (1949).\n\n\n> Take the point form generator of space translations in a 2-particle\n> system\n> P =3D p_1 + p_2 + Y\n> where interaction operator Y generally does not commute with p_1 and\n> p_2. Let us find how momenta of particles transform to the translated\n> frame of reference\n> p_1(a) =3D exp(-iPa) p_1 exp(iPa) =3D p_1 -i [Y,p_1]a + ...\n> p_2(a) =3D exp(-iPa) p_2 exp(iPa) =3D p_2 -i [Y,p_2]a + ...\n> Contrary to observations, these transformations are non-trivial and\n> interaction-dependent. This dependence should be observable in\n> experiment.=20\n\nNo; it is - and has to be - as unobservable as coordinate transforms in\nordinary space. It just says that in the point form not only time but\nalso space moves. What remains fixed instead is the past hyperboloid.\n\nYou simply insist on describing things from the perspective of a\nparticular observer who knows the full present (relative to the time\naxis defined by its momentum); then, of course, manifest covariance\nis gone by definition, and you are left with the instant form.\nBut apart from your personal desire, there is no need at all to use\nsuch a restrictive point of view.\n\nAnd indeed, it is not fully adequate since, according to the finite\nbound of c on the speed of communication, established over 100 years ago,=\n\nit is impossible for a physical observer to know the present anywhere\nexcept at its own position.\n\nOf course, since you start from irrealistic assumptions, it is no surpris=\ne\nthat you end up with apparent superluminal behavior.\n\n\n> The assumptions of the CJS theorem are\n> 1) the Hamiltonian description of dynamics\n> 2) measurability of particle observables, like p_i and r_i,\n> in all frames of reference.\n> I think, these assumptions are well justified.\n\nNo. It is a very restrictive assumption that different observers\nmeasure identical p_i and r_i. The CJS theorem simply asserts that\nthis assumption leads to ridiculous consequences. The correct\nconclusion is (as in case of von Veumann\'s theorem versus Bell\'s\ntheorem about hidden variables) that these assumptions are inadequate.\n\n\n>>>The fields \\psi(x,t) have no other meaning except as convenient object=\ns\n>>>for compact writing of the interaction Hamiltonian and proving its\n>>>relativistic invariance.=20\n>>\n>>They are hermitian linear operators and hence observables according to\n>>the standard interpretation of QM. Certainly <dA(x)> is measurable\n>>and gives the classical e/m field.\n\n> I have a few counterarguments:\n\nThese are no counterarguments to my statement that\n\'Certainly <dA(x)> is measurable and gives the classical e/m field.\'\nThis was the important point. Look at how QED is used by practicioners,\nand you can see the truth of my statement.\n\n\n> 1) the electron-positron field \\psi(x,t) is not Hermitian, so it is\n> not observable according to your criterion\n\nTheir expectation is zero anyway, since they are odd operators.\nSo in this case there is nothing to measure.\nBut certain Hermitian quadratic expressions in \\psi(x,t) are well-known\nobservables, and define the measurable mass density, the charge density,\nand the electromagnetic currents. And every quadratic expression in the\n\\psi(x,t) can be given a measurable meaning, though in a more intricate\nway. Simply couple QED to a classical apparatus in a semiclassical\napproximation, and work out the response of the pointer to a fairly gener=\nal\ninteraction...\n\n\n> 2) It is true that every observable is described by a Hermitian\n> operator, but it is not generally true that each Hermitian operator=\n\n> corresponds to an observable.\n\nThis is probably wrong. It is provably wrong in small enough systems.\nIn general, one can probably measure the expectation value of any\nlocal Hermitian operator to arbitrary accuracy, assuming the system to\nbe embedded into a large enough observing system. Of course, in practice,=\n\nthe accuracy is limited; but there is no restriction in principle.\n\n\n> 3) What is the meaning of x? It would be natural to assume that x are\n> eigenvalues of the position operator.=20\n\nYou could ask the same question about the meaning of x in general relativ=\nity.\nIt has no direct physical meaning, but nevertheless general relativity\ndescribes nature, and in a very elegant way. So a negative answer to your=\n\nstatement does not mean that anything is wrong with Minkowski space.\n\nIf you accept general relativity in the large you have to accept Minkowsk=\ni\nspace in the small. Indeed, the latter is just the linear approximation\n(tangent space) to the mainfold of general relativity, and for microscopi=\nc\ndynamics such as collision processes, the curvature of space can be fully=\n\nneglected.\n\n\n> My feeling is that theoretical physics is already in the dead end.\n> The main reason for that is the blind acceptance of the 4D space-time\n> continuum.=20\n\nOf course, feelings are subjective, so I can\'t argue your statement.\nBut most theoretical physicists find theoretical physics very alive\nand progressing.\n\n\n> I have shown that the space-time description does not follow\n> from Einstein\'s postulates, and is, in fact, an additional hypothesis.=\n\n> Everything becomes simple and consistent, if this hypothesis is\n> dropped.\n\nWhat??? I haven\'t seen anything simple in your approach.\nYour Hamiltonian power series expansion is much more messy\nthan covariant QED...\n\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>=20
>>According to Dirac, what is kinematical and what is dynamical is a
>>free choice.=20
>=20
> If Dirac said that, then I disagree with him.

I find it difficult to disagree with well-established mathematical facts.=

Rev. Mod. Phys. 21, 392=96399 (1949).

> Take the point form generator of space translations in a 2-particle
> system
> P =3D p_1 + p_2 + Y
> where interaction operator Y generally does not commute with p_1 and
> p_2. Let us find how momenta of particles transform to the translated
> frame of reference
> p_1(a) =3D \exp(-iPa) p_1 \exp(iPa) =3D p_1 -i [Y,p_1]a + ...
> p_2(a) =3D \exp(-iPa) p_2 \exp(iPa) =3D p_2 -i [Y,p_2]a + ...
> Contrary to observations, these transformations are non-trivial and
> interaction-dependent. This dependence should be observable in
> experiment.=20

No; it is - and has to be - as unobservable as coordinate transforms in
ordinary space. It just says that in the point form not only time but
also space moves. What remains fixed instead is the past hyperboloid.

You simply insist on describing things from the perspective of a
particular observer who knows the full present (relative to the time
axis defined by its momentum); then, of course, manifest covariance
is gone by definition, and you are left with the instant form.
But apart from your personal desire, there is no need at all to use
such a restrictive point of view.

And indeed, it is not fully adequate since, according to the finite
bound of c on the speed of communication, established over 100 years ago,=

it is impossible for a physical observer to know the present anywhere
except at its own position.

Of course, since you start from irrealistic assumptions, it is no surpris=
e
that you end up with apparent superluminal behavior.

> The assumptions of the CJS theorem are
> 1) the Hamiltonian description of dynamics
> 2) measurability of particle observables, like p_i and r_i,
> in all frames of reference.
> I think, these assumptions are well justified.

No. It is a very restrictive assumption that different observers
measure identical p_i and r_i. The CJS theorem simply asserts that
this assumption leads to ridiculous consequences. The correct
conclusion is (as in case of von Veumann's theorem versus Bell's
theorem about hidden variables) that these assumptions are inadequate.

>>>The fields \psi(x,t) have no other meaning except as convenient object=
s
>>>for compact writing of the interaction Hamiltonian and proving its
>>>relativistic invariance.=20
>>
>>They are hermitian linear operators and hence observables according to
>>the standard interpretation of QM. Certainly <dA(x)> is measurable
>>and gives the classical e/m field.

> I have a few counterarguments:

These are no counterarguments to my statement that
'Certainly <dA(x)> is measurable and gives the classical e/m field.'
This was the important point. Look at how QED is used by practicioners,
and you can see the truth of my statement.

> 1) the electron-positron field \psi(x,t) is not Hermitian, so it is
> not observable according to your criterion

Their expectation is zero anyway, since they are odd operators.
So in this case there is nothing to measure.
But certain Hermitian quadratic expressions in \psi(x,t) are well-known
observables, and define the measurable mass density, the charge density,
and the electromagnetic currents. And every quadratic expression in the
\psi(x,t) can be given a measurable meaning, though in a more intricate
way. Simply couple QED to a classical apparatus in a semiclassical
approximation, and work out the response of the pointer to a fairly gener=
al
interaction...

> 2) It is true that every observable is described by a Hermitian
> operator, but it is not generally true that each Hermitian operator=

> corresponds to an observable.

This is probably wrong. It is provably wrong in small enough systems.
In general, one can probably measure the expectation value of any
local Hermitian operator to arbitrary accuracy, assuming the system to
be embedded into a large enough observing system. Of course, in practice,=

the accuracy is limited; but there is no restriction in principle.

> 3) What is the meaning of x? It would be natural to assume that x are
> eigenvalues of the position operator.=20

You could ask the same question about the meaning of x in general relativ=
ity.
It has no direct physical meaning, but nevertheless general relativity
describes nature, and in a very elegant way. So a negative answer to your=

statement does not mean that anything is wrong with Minkowski space.

If you accept general relativity in the large you have to accept Minkowsk=
i
space in the small. Indeed, the latter is just the linear approximation
(tangent space) to the mainfold of general relativity, and for microscopi=
c
dynamics such as collision processes, the curvature of space can be fully=

neglected.

> My feeling is that theoretical physics is already in the dead end.
> The main reason for that is the blind acceptance of the 4D space-time
> continuum.=20

Of course, feelings are subjective, so I can't argue your statement.
But most theoretical physicists find theoretical physics very alive
and progressing.

> I have shown that the space-time description does not follow
> from Einstein's postulates, and is, in fact, an additional hypothesis.=

> Everything becomes simple and consistent, if this hypothesis is
> dropped.

What??? I haven't seen anything simple in your approach.
Your Hamiltonian power series expansion is much more messy
than covariant QED...

Arnold Neumaier

Arnold Neumaier

Nov29-04, 02:49 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>The CJS theorem only forbids a dynamics with observer-independent worldlines\n>>for each particle. Field theories do not make this assumption, hence are\n>>not affected by the theorem. So it can be safely ignored in QED.\n>\n> You are right that CJS theorem has no relevance to the traditional QED,\n> because this approach only cares about the S-matrix, and does not care\n> about the time-dependence of particle observables (this time dependence\n> is represented by trajectories in the classical limit).\n\nBut QED cares about the time-dependence of field observables, which is what\none has in practice. Read about the closed time path formalism.\n\n\n> The CJS theorem\n> becomes very important in the case of dynamical description of the time\n> evolution. Such a description is built in my book, and in agreement with\n> the CJS theorem,\n> I found that particle trajectories (or, in general, particle\n> observables) do not transform in a manifestly covariant fashion between\n> different reference frames.\n\nIn standard terminology, this just says that particle trajectories are\n(and by CJS have to be) observer dependent.\nBut nevertheless, one can translate form one observer\nto another one, if one uses fields instead of trajectories. Then everything\nis covariant. Trajectories are fuzzy anyway, already in nonrelativistic QM,\nso giving them up is not a big deal.\n\n\n\n> I disagree that unitary equivalent representations of the Poincare group\n> are always physically equivalent. Take two Hamiltonians connected by a\n> unitary transformation\n>\n> H\' = U H U^{-1}\n>\n> These two Hamiltonians may have the same S-matrix and energy spectrum of\n> bound states. However, the wave functions of bound states are\n> different, and the time evolutions described by these two Hamiltonians\n> are different too.\n\nBut the physics is exactly the same, since expectations are identical,\nif states and operators are both correctly transformed.\nOnly that counts for observable consequences.\n\n\n>>>I disagree. There is a way to send a superluminal (in fact,\n>>>instantaneous) signal between two charged particles.\n>>\n>>Well, try to convince an experimentalist to check it!\n>>I doubt you\'ll find one prepared to work seriously on such a project.\n>\n> I would love to talk to an experimentalist, and I hope that I can\n> convince somebody. The question about the speed of propagation of the\n> Coulomb and magnetic interactions remains unresolved experimentally\n> until this day. I think you would agree with me that this\n> is the fundamental issue for our understanding of nature, and the\n> absence of experimental data is rather disturbing.\n\nI think this issue is already resolved in favor of an absolute bound\nof c for the speed of any material body or information flow.\nThat\'s what causality is about.\n\n\n\n>>Science proceeds not by speculation but by making sure that one always\n>>stands on safe ground and is correspondingly cautious with great claims.\n>\n> I believe, speculation plays a great role in the development of\n> science.\n\nOnly if it succeeds in better predictivity. Otherwise it is the biggest\nwaste of efforts. Most speculations remain infertile; very, very few\nsucceed. And those that do succeed, do so because they back up their\nspeculations not with a few general arguments but with invincible\npredictions.\n\n\n> Actually, the universally accepted Minkowski spacetime is also\n> a great speculation.\n\nIt has manifold experimental verifications, both in the classical\nand in the quantum domain. If you call that a great speculation\nthen all of physics is a great speculation.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>The CJS theorem only forbids a dynamics with observer-independent worldlines
>>for each particle. Field theories do not make this assumption, hence are
>>not affected by the theorem. So it can be safely ignored in QED.
>
> You are right that CJS theorem has no relevance to the traditional QED,
> because this approach only cares about the S-matrix, and does not care
> about the time-dependence of particle observables (this time dependence
> is represented by trajectories in the classical limit).

But QED cares about the time-dependence of field observables, which is what
one has in practice. Read about the closed time path formalism.

> The CJS theorem
> becomes very important in the case of dynamical description of the time
> evolution. Such a description is built in my book, and in agreement with
> the CJS theorem,
> I found that particle trajectories (or, in general, particle
> observables) do not transform in a manifestly covariant fashion between
> different reference frames.

In standard terminology, this just says that particle trajectories are
(and by CJS have to be) observer dependent.
But nevertheless, one can translate form one observer
to another one, if one uses fields instead of trajectories. Then everything
is covariant. Trajectories are fuzzy anyway, already in nonrelativistic QM,
so giving them up is not a big deal.

> I disagree that unitary equivalent representations of the Poincare group
> are always physically equivalent. Take two Hamiltonians connected by a
> unitary transformation
>
> H' = U H U^{-1}
>
> These two Hamiltonians may have the same S-matrix and energy spectrum of
> bound states. However, the wave functions of bound states are
> different, and the time evolutions described by these two Hamiltonians
> are different too.

But the physics is exactly the same, since expectations are identical,
if states and operators are both correctly transformed.
Only that counts for observable consequences.

>>>I disagree. There is a way to send a superluminal (in fact,
>>>instantaneous) signal between two charged particles.
>>
>>Well, try to convince an experimentalist to check it!
>>I doubt you'll find one prepared to work seriously on such a project.
>
> I would love to talk to an experimentalist, and I hope that I can
> convince somebody. The question about the speed of propagation of the
> Coulomb and magnetic interactions remains unresolved experimentally
> until this day. I think you would agree with me that this
> is the fundamental issue for our understanding of nature, and the
> absence of experimental data is rather disturbing.

I think this issue is already resolved in favor of an absolute bound
of c for the speed of any material body or information flow.
That's what causality is about.

>>Science proceeds not by speculation but by making sure that one always
>>stands on safe ground and is correspondingly cautious with great claims.
>
> I believe, speculation plays a great role in the development of
> science.

Only if it succeeds in better predictivity. Otherwise it is the biggest
waste of efforts. Most speculations remain infertile; very, very few
succeed. And those that do succeed, do so because they back up their
speculations not with a few general arguments but with invincible
predictions.

> Actually, the universally accepted Minkowski spacetime is also
> a great speculation.

It has manifold experimental verifications, both in the classical
and in the quantum domain. If you call that a great speculation
then all of physics is a great speculation.

Arnold Neumaier

Oz

Nov29-04, 02:52 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich <eugenev@synopsys.com> writes\n\n> do not accept the dubious postulate of Minkowski spacetime,\n\nEh? I was under the impression that minkowski spacetime was just a space\nthat conveniently illustrated SR. What postulate is added?\n\n>though I\n>keep all other well-established postulates of relativity and quantum\n>mechanics.\n\nYou probably ought to itemise them.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich <eugenev@synopsys.com> writes

> do not accept the dubious postulate of Minkowski spacetime,

Eh? I was under the impression that minkowski spacetime was just a space
that conveniently illustrated SR. What postulate is added?

>though I
>keep all other well-established postulates of relativity and quantum
>mechanics.

You probably ought to itemise them.

--
Oz
This post is worth absolutely nothing and is probably fallacious.

Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.

Eugene Stefanovich

Nov30-04, 12:49 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>=20\n>>\n>>>According to Dirac, what is kinematical and what is dynamical is a\n>>>free choice.=20\n>>\n>>=20\n>>If Dirac said that, then I disagree with him.\n>\n>\n> I find it difficult to disagree with well-established mathematical facts.=\n>\n> Rev. Mod. Phys. 21, 392=96399 (1949).\n\nI fully agree with Dirac\'s math in this paper. I don\'t agree with\nyour interpretation of this math.\n\n>\n>\n>\n>>Take the point form generator of space translations in a 2-particle\n>>system\n>> P =3D p_1 + p_2 + Y\n>>where interaction operator Y generally does not commute with p_1 and\n>>p_2. Let us find how momenta of particles transform to the translated\n>>frame of reference\n>> p_1(a) =3D exp(-iPa) p_1 exp(iPa) =3D p_1 -i [Y,p_1]a + ...\n>> p_2(a) =3D exp(-iPa) p_2 exp(iPa) =3D p_2 -i [Y,p_2]a + ...\n>>Contrary to observations, these transformations are non-trivial and\n>>interaction-dependent. This dependence should be observable in\n>>experiment.=20\n>\n>\n> No; it is - and has to be - as unobservable as coordinate transforms in\n> ordinary space.\n\nAre you saying that momenta of particles are not observable in different\nframes of reference? What IS observable then?\n\n\n> It just says that in the point form not only time but\n> also space moves. What remains fixed instead is the past hyperboloid.\n>\n> You simply insist on describing things from the perspective of a\n> particular observer who knows the full present (relative to the time\n> axis defined by its momentum); then, of course, manifest covariance\n> is gone by definition, and you are left with the instant form.\n> But apart from your personal desire, there is no need at all to use\n> such a restrictive point of view.\n\nSuppose there are two particles 1 and 2. Particle 1 is on my desk,\nand particle 2 is on Mars.\nI can measure observables f_1(t) and f_2(t) for both particles, where\nt is time measured by my local clock. Then I can\nintroduce a correction due to the finite speed of the signal and\nfind out that f_1(t-R/c) and f_2(t) were simultaneous values.\nHere R is the distance Earth-Mars, and c is the speed of light.\n\n>\n> And indeed, it is not fully adequate since, according to the finite\n> bound of c on the speed of communication, established over 100 years ago,=\n\nI am not arguing that particles (photons, electrons, etc.) cannot\npropagate faster than light. Surely, they can\'t.\nI am saying that electromagnetic\ninteraction (Coulomb and magnetic) between charged particles is\ninstantaneous. If you know an experiment which demonstrated otherwise,\nplease let me know. The theoretical "proof" offered by special\nrelativity ("grandfather paradox") is not adequate because it\nassumes linear interaction-independent Lorentz transformations,\nwhich are not applicable in the case of interacting particles.\n\n>\n> it is impossible for a physical observer to know the present anywhere\n> except at its own position.\n>\n> Of course, since you start from irrealistic assumptions, it is no surpris=\n> e\n> that you end up with apparent superluminal behavior.\n>\n>\n>\n>>The assumptions of the CJS theorem are\n>>1) the Hamiltonian description of dynamics\n>>2) measurability of particle observables, like p_i and r_i,\n>> in all frames of reference.\n>>I think, these assumptions are well justified.\n>\n>\n> No. It is a very restrictive assumption that different observers\n> measure identical p_i and r_i.\n\nOne observer can measure p_i(t) and r_i(t), where t is time measured\nby his clock. Another observer can also measure p\'_i(t\') and r\'_i(t\')\nwhere t\' is time measured by her clock. I hope you are not arguing\nwith that. The CJS theorem say that values of (p,r,t) and (p\',r\',t\')\nare not connected by linear Lorentz transformations.\n\n\n> The CJS theorem simply asserts that\n> this assumption leads to ridiculous consequences. The correct\n> conclusion is (as in case of von Veumann\'s theorem versus Bell\'s\n> theorem about hidden variables) that these assumptions are inadequate.\n>\n>\n>\n>>>>The fields \\psi(x,t) have no other meaning except as convenient object=\n>>>\n> s\n>\n>>>>for compact writing of the interaction Hamiltonian and proving its\n>>>>relativistic invariance.=20\n>>>\n>>>They are hermitian linear operators and hence observables according to\n>>>the standard interpretation of QM. Certainly <dA(x)> is measurable\n>>>and gives the classical e/m field.\n>>\n>\n>>I have a few counterarguments:\n>\n>\n> These are no counterarguments to my statement that\n> \'Certainly <dA(x)> is measurable and gives the classical e/m field.\'\n> This was the important point. Look at how QED is used by practicioners,\n> and you can see the truth of my statement.\n\nAs far as I know, QED is used to calculate scattering cross-sections\n(between particles) and energies of bound states (of particles).\nMaybe I\'m missing something, but I\'ve never seen QED calculations of\nthe E and B fields and their comparison with experiment.\nThese fields do appear at intermediate stages of calculations,\nbut never as a final result. Or sometimes these fields are considered\nas external (fixed) classical fields. This is not a rigorous approach,\nof course.\n\n>\n>\n>\n>>1) the electron-positron field \\psi(x,t) is not Hermitian, so it is\n>> not observable according to your criterion\n>\n>\n> Their expectation is zero anyway, since they are odd operators.\n> So in this case there is nothing to measure.\n> But certain Hermitian quadratic expressions in \\psi(x,t) are well-known\n> observables, and define the measurable mass density, the charge density,\n> and the electromagnetic currents.\n\nThere are other operators (i.e., observables of position and momentum\n(or velocity) of\nindividual particles) which allow us to define these measurable\nquantities, . In a system with fixed number of particles it is\nstraightforward to define positions and momenta of all constituents, and\nthus derive such things as charge density, etc. I am wondering if the\nresult will be the same when using expressions like\n\n\\rho(x,t) = -e \\psi^{\\dag}(x,t) \\psi(x,t)\n\n\n> And every quadratic expression in the\n> \\psi(x,t) can be given a measurable meaning, though in a more intricate\n> way. Simply couple QED to a classical apparatus in a semiclassical\n> approximation, and work out the response of the pointer to a fairly gener=\n> al\n> interaction...\n>\n>\n[...]\n>\n>\n>>3) What is the meaning of x? It would be natural to assume that x are\n>> eigenvalues of the position operator.=20\n>\n>\n> You could ask the same question about the meaning of x in general relativ=\n> ity.\n> It has no direct physical meaning, but nevertheless general relativity\n> describes nature, and in a very elegant way.\n> So a negative answer to your=\n>\n> statement does not mean that anything is wrong with Minkowski space.\n>\n> If you accept general relativity in the large you have to accept Minkowsk=\n> i\n> space in the small. Indeed, the latter is just the linear approximation\n> (tangent space) to the mainfold of general relativity, and for microscopi=\n> c\n> dynamics such as collision processes, the curvature of space can be fully=\n>\n> neglected.\n\nI agree that general relativity is mathematically elegant. However, let\nus not take the mathematical elegance as a proof of physical soundness.\nI do not accept GR\'s fundamental premise: the common universal\nbackground continuum, which degenerates to the flat Minkowski spacetime\nin the absence of gravity. I am not impressed by special relativity and\nGR, because the Minkowski spacetime idea was pulled out of thin air by\nMinkowski and Einstein. This idea does not have solid basis neither in\nEinstein\'s postulates nor in experiment. I agree that so far there were\nno experiments contradicting this idea. However, this does not look like\na strong argument to me.\n\nI have constructed a relativistic quantum theory\nof electromagnetic interactions, which does not use the idea of\nMinkowski spacetime and contradicts some basic results of special\nrelativity (e.g., the retarded character of interactions). Nevertheless,\nthis approach has no less predictive power than QED. Of course, you\nmay argue that I haven\'t done explicit calculations of the Lamb shift,\nand I haven\'t cleaned up the infrared sector of the theory, and I\ndo not have any approach to gravity yet, etc, etc. You may dismiss\nmy approach altogether based on these deficiencies.\n\nHowever, I urge you to pay attention to one observation which seems\nimportant to me: there is no proof that the equivalence of inertial\nobservers and the invariance of the speed of light implies the universality\nof Lorentz transformations and the unification\nof space and time in the 4D spacetime. These remain unproved\nassumptions, which, nevertheless, are the basis of all relativistic\nphysics as we know it.\nIsn\'t it dangerous to build all our theories on such a shaky foundation?\n\n\n\n>\n>\n>\n>>My feeling is that theoretical physics is already in the dead end.\n>>The main reason for that is the blind acceptance of the 4D space-time\n>>continuum.=20\n>\n>\n> Of course, feelings are subjective, so I can\'t argue your statement.\n> But most theoretical physicists find theoretical physics very alive\n> and progressing.\n>\n>\n> > I have shown that the space-time description does not follow\n> > from Einstein\'s postulates, and is, in fact, an additional hypothesis.=\n>\n> > Everything becomes simple and consistent, if this hypothesis is\n> > dropped.\n>\n> What??? I haven\'t seen anything simple in your approach.\n> Your Hamiltonian power series expansion is much more messy\n> than covariant QED...\n\nThe simplicity is not in short expressions for the Hamiltonian and\nother quantities. The simplicity is in adhering to well-tested\nphysical postulates, in clear physical meaning of all theoretical\ningredients, and in the absence of logical contradictions.\n\nEugene Stefanovich.\n\n>\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>=20
>>
>>>According to Dirac, what is kinematical and what is dynamical is a
>>>free choice.=20
>>
>>=20
>>If Dirac said that, then I disagree with him.
>
>
> I find it difficult to disagree with well-established mathematical facts.=
>
> Rev. Mod. Phys. 21, 392=96399 (1949).

I fully agree with Dirac's math in this paper. I don't agree with
your interpretation of this math.

>
>
>
>>Take the point form generator of space translations in a 2-particle
>>system
>> P =3D p_1 + p_2 + Y
>>where interaction operator Y generally does not commute with p_1 and
>>p_2. Let us find how momenta of particles transform to the translated
>>frame of reference
>> p_1(a) =3D \exp(-iPa) p_1 \exp(iPa) =3D p_1 -i [Y,p_1]a + ...
>> p_2(a) =3D \exp(-iPa) p_2 \exp(iPa) =3D p_2 -i [Y,p_2]a + ...
>>Contrary to observations, these transformations are non-trivial and
>>interaction-dependent. This dependence should be observable in
>>experiment.=20
>
>
> No; it is - and has to be - as unobservable as coordinate transforms in
> ordinary space.

Are you saying that momenta of particles are not observable in different
frames of reference? What IS observable then?

> It just says that in the point form not only time but
> also space moves. What remains fixed instead is the past hyperboloid.
>
> You simply insist on describing things from the perspective of a
> particular observer who knows the full present (relative to the time
> axis defined by its momentum); then, of course, manifest covariance
> is gone by definition, and you are left with the instant form.
> But apart from your personal desire, there is no need at all to use
> such a restrictive point of view.

Suppose there are two particles 1 and 2. Particle 1 is on my desk,
and particle 2 is on Mars.
I can measure observables f_1(t) and f_2(t) for both particles, where
t is time measured by my local clock. Then I can
introduce a correction due to the finite speed of the signal and
find out that f_1(t-R/c) and f_2(t) were simultaneous values.
Here R is the distance Earth-Mars, and c is the speed of light.

>
> And indeed, it is not fully adequate since, according to the finite
> bound of c on the speed of communication, established over 100 years ago,=

I am not arguing that particles (photons, electrons, etc.) cannot
propagate faster than light. Surely, they can't.
I am saying that electromagnetic
interaction (Coulomb and magnetic) between charged particles is
instantaneous. If you know an experiment which demonstrated otherwise,
please let me know. The theoretical "proof" offered by special
relativity ("grandfather paradox") is not adequate because it
assumes linear interaction-independent Lorentz transformations,
which are not applicable in the case of interacting particles.

>
> it is impossible for a physical observer to know the present anywhere
> except at its own position.
>
> Of course, since you start from irrealistic assumptions, it is no surpris=
> e
> that you end up with apparent superluminal behavior.
>
>
>
>>The assumptions of the CJS theorem are
>>1) the Hamiltonian description of dynamics
>>2) measurability of particle observables, like p_i and r_i,
>> in all frames of reference.
>>I think, these assumptions are well justified.
>
>
> No. It is a very restrictive assumption that different observers
> measure identical p_i and r_i.

One observer can measure p_i(t) and r_i(t), where t is time measured
by his clock. Another observer can also measure p'_i(t') and r'_i(t')
where t' is time measured by her clock. I hope you are not arguing
with that. The CJS theorem say that values of (p,r,t) and (p',r',t')
are not connected by linear Lorentz transformations.

> The CJS theorem simply asserts that
> this assumption leads to ridiculous consequences. The correct
> conclusion is (as in case of von Veumann's theorem versus Bell's
> theorem about hidden variables) that these assumptions are inadequate.
>
>
>
>>>>The fields \psi(x,t) have no other meaning except as convenient object=
>>>
> s
>
>>>>for compact writing of the interaction Hamiltonian and proving its
>>>>relativistic invariance.=20
>>>
>>>They are hermitian linear operators and hence observables according to
>>>the standard interpretation of QM. Certainly <dA(x)> is measurable
>>>and gives the classical e/m field.
>>
>
>>I have a few counterarguments:
>
>
> These are no counterarguments to my statement that
> 'Certainly <dA(x)> is measurable and gives the classical e/m field.'
> This was the important point. Look at how QED is used by practicioners,
> and you can see the truth of my statement.

As far as I know, QED is used to calculate scattering cross-sections
(between particles) and energies of bound states (of particles).
Maybe I'm missing something, but I've never seen QED calculations of
the E and B fields and their comparison with experiment.
These fields do appear at intermediate stages of calculations,
but never as a final result. Or sometimes these fields are considered
as external (fixed) classical fields. This is not a rigorous approach,
of course.

>
>
>
>>1) the electron-positron field \psi(x,t) is not Hermitian, so it is
>> not observable according to your criterion
>
>
> Their expectation is zero anyway, since they are odd operators.
> So in this case there is nothing to measure.
> But certain Hermitian quadratic expressions in \psi(x,t) are well-known
> observables, and define the measurable mass density, the charge density,
> and the electromagnetic currents.

There are other operators (i.e., observables of position and momentum
(or velocity) of
individual particles) which allow us to define these measurable
quantities, . In a system with fixed number of particles it is
straightforward to define positions and momenta of all constituents, and
thus derive such things as charge density, etc. I am wondering if the
result will be the same when using expressions like

\rho(x,t) = -e \psi^{\dag}(x,t) \psi(x,t)

> And every quadratic expression in the
> \psi(x,t) can be given a measurable meaning, though in a more intricate
> way. Simply couple QED to a classical apparatus in a semiclassical
> approximation, and work out the response of the pointer to a fairly gener=
> al
> interaction...
>
>
[...]
>
>
>>3) What is the meaning of x? It would be natural to assume that x are
>> eigenvalues of the position operator.=20
>
>
> You could ask the same question about the meaning of x in general relativ=
> ity.
> It has no direct physical meaning, but nevertheless general relativity
> describes nature, and in a very elegant way.
> So a negative answer to your=
>
> statement does not mean that anything is wrong with Minkowski space.
>
> If you accept general relativity in the large you have to accept Minkowsk=
> i
> space in the small. Indeed, the latter is just the linear approximation
> (tangent space) to the mainfold of general relativity, and for microscopi=
> c
> dynamics such as collision processes, the curvature of space can be fully=
>
> neglected.

I agree that general relativity is mathematically elegant. However, let
us not take the mathematical elegance as a proof of physical soundness.
I do not accept GR's fundamental premise: the common universal
background continuum, which degenerates to the flat Minkowski spacetime
in the absence of gravity. I am not impressed by special relativity and
GR, because the Minkowski spacetime idea was pulled out of thin air by
Minkowski and Einstein. This idea does not have solid basis neither in
Einstein's postulates nor in experiment. I agree that so far there were
no experiments contradicting this idea. However, this does not look like
a strong argument to me.

I have constructed a relativistic quantum theory
of electromagnetic interactions, which does not use the idea of
Minkowski spacetime and contradicts some basic results of special
relativity (e.g., the retarded character of interactions). Nevertheless,
this approach has no less predictive power than QED. Of course, you
may argue that I haven't done explicit calculations of the Lamb shift,
and I haven't cleaned up the infrared sector of the theory, and I
do not have any approach to gravity yet, etc, etc. You may dismiss
my approach altogether based on these deficiencies.

However, I urge you to pay attention to one observation which seems
important to me: there is no proof that the equivalence of inertial
observers and the invariance of the speed of light implies the universality
of Lorentz transformations and the unification
of space and time in the 4D spacetime. These remain unproved
assumptions, which, nevertheless, are the basis of all relativistic
physics as we know it.
Isn't it dangerous to build all our theories on such a shaky foundation?

>
>
>
>>My feeling is that theoretical physics is already in the dead end.
>>The main reason for that is the blind acceptance of the 4D space-time
>>continuum.=20
>
>
> Of course, feelings are subjective, so I can't argue your statement.
> But most theoretical physicists find theoretical physics very alive
> and progressing.
>
>
> > I have shown that the space-time description does not follow
> > from Einstein's postulates, and is, in fact, an additional hypothesis.=
>
> > Everything becomes simple and consistent, if this hypothesis is
> > dropped.
>
> What??? I haven't seen anything simple in your approach.
> Your Hamiltonian power series expansion is much more messy
> than covariant QED...

The simplicity is not in short expressions for the Hamiltonian and
other quantities. The simplicity is in adhering to well-tested
physical postulates, in clear physical meaning of all theoretical
ingredients, and in the absence of logical contradictions.

Eugene Stefanovich.

>
>
>
> Arnold Neumaier
>

Eugene Stefanovich

Nov30-04, 12:49 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>The CJS theorem only forbids a dynamics with observer-independent worldlines\n>>>for each particle. Field theories do not make this assumption, hence are\n>>>not affected by the theorem. So it can be safely ignored in QED.\n>>\n>>You are right that CJS theorem has no relevance to the traditional QED,\n>>because this approach only cares about the S-matrix, and does not care\n>>about the time-dependence of particle observables (this time dependence\n>>is represented by trajectories in the classical limit).\n>\n>\n> But QED cares about the time-dependence of field observables, which is what\n> one has in practice. Read about the closed time path formalism.\n\nThanks for the suggestion. I\'ll try to find something on the "closed\ntime path formalism". It would be interesting to see how they can manage\nto predict any kind of time evolution without having finite well-defined\nHamiltonian.\n\n>\n>\n>\n>>The CJS theorem\n>>becomes very important in the case of dynamical description of the time\n>>evolution. Such a description is built in my book, and in agreement with\n>>the CJS theorem,\n>>I found that particle trajectories (or, in general, particle\n>>observables) do not transform in a manifestly covariant fashion between\n>>different reference frames.\n>\n>\n> In standard terminology, this just says that particle trajectories are\n> (and by CJS have to be) observer dependent.\n> But nevertheless, one can translate form one observer\n> to another one, if one uses fields instead of trajectories. Then everything\n> is covariant. Trajectories are fuzzy anyway, already in nonrelativistic QM,\n> so giving them up is not a big deal.\n\nI do not agree. Any quantum theory should have a classical limit.\nIn this limit, trajectories are well defined. If the manifest covariance\nis true, then these trajectories must transform by Lorentz (or, as you\nsay, they must be observer independent).\n\n>\n>\n>\n>\n>>I disagree that unitary equivalent representations of the Poincare group\n>>are always physically equivalent. Take two Hamiltonians connected by a\n>>unitary transformation\n>>\n>>H\' = U H U^{-1}\n>>\n>>These two Hamiltonians may have the same S-matrix and energy spectrum of\n>>bound states. However, the wave functions of bound states are\n>>different, and the time evolutions described by these two Hamiltonians\n>>are different too.\n>\n>\n> But the physics is exactly the same, since expectations are identical,\n> if states and operators are both correctly transformed.\n\nI didn\'t say that states are transformed. Only the Hamiltonian\n(and 9 other generators of the Poincare group) is transformed.\nWhen the unitary equivalence of different forms of dynamics is proved,\nthe states are not subject to transformation.\n\nIf you unitarily transform\nboth states and operators, then, of course, the physical content of\nthe theory does not change. That\'s rather trivial.\n\n> Only that counts for observable consequences.\n>\n>\n>\n>>>>I disagree. There is a way to send a superluminal (in fact,\n>>>>instantaneous) signal between two charged particles.\n>>>\n>>>Well, try to convince an experimentalist to check it!\n>>>I doubt you\'ll find one prepared to work seriously on such a project.\n>>\n>>I would love to talk to an experimentalist, and I hope that I can\n>>convince somebody. The question about the speed of propagation of the\n>>Coulomb and magnetic interactions remains unresolved experimentally\n>>until this day. I think you would agree with me that this\n>>is the fundamental issue for our understanding of nature, and the\n>>absence of experimental data is rather disturbing.\n>\n>\n> I think this issue is already resolved in favor of an absolute bound\n> of c for the speed of any material body or information flow.\n> That\'s what causality is about.\n\nThis is "proved" in special relativity by assuming the interaction\nindependence of boost transformations. This proof is not valid in\nmy approach, where boost transformations are interaction dependent.\nIn my approach, interactions are superluminal, and causality is\nrespected (the effect never happens earlier than the cause).\n\n>\n>\n>\n>\n>>>Science proceeds not by speculation but by making sure that one always\n>>>stands on safe ground and is correspondingly cautious with great claims.\n>>\n>>I believe, speculation plays a great role in the development of\n>>science.\n>\n>\n> Only if it succeeds in better predictivity. Otherwise it is the biggest\n> waste of efforts. Most speculations remain infertile; very, very few\n> succeed. And those that do succeed, do so because they back up their\n> speculations not with a few general arguments but with invincible\n> predictions.\n\nI do have experimental predictions: the instantaneous propagation of\nthe Coulomb force, and small corrections to the decay laws of moving\nparticles. If experiment will say that I was wrong, I\'ll shut up.\n\n>\n>\n>\n>>Actually, the universally accepted Minkowski spacetime is also\n>>a great speculation.\n>\n>\n> It has manifold experimental verifications, both in the classical\n> and in the quantum domain. If you call that a great speculation\n> then all of physics is a great speculation.\n\nLet us set aside the effects of gravity and focus on predictions of\nspecial relativity. I agree that there are lot of experimental data\nreferring to properties of free particles (the Doppler effect,\nthe relativistic relationship between mass, momentum, and energy,\nthe Lorentz invariance of the S-matrix, etc.)\nIn such cases, my approach predicts exactly the same properties, because\nno interaction is involved.\n\nThere are much much fewer relativistic\neffects studied in isolated systems with interactions. I actually know\nonly one such effect: the slowing down of the decay of moving unstable\nparticles. Most accurate experiments with muons\nconfirm Einstein\'s time dilation formula with the accuracy of 10^{-5}.\nMy approach predicts corrections 10 orders of magnitude smaller for this\nsystem. So, from the experimental point of view, my speculation is as\ngood as Einstein\'s speculation.\n\nEugene Stefanovich.\n\n\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>The CJS theorem only forbids a dynamics with observer-independent worldlines
>>>for each particle. Field theories do not make this assumption, hence are
>>>not affected by the theorem. So it can be safely ignored in QED.
>>
>>You are right that CJS theorem has no relevance to the traditional QED,
>>because this approach only cares about the S-matrix, and does not care
>>about the time-dependence of particle observables (this time dependence
>>is represented by trajectories in the classical limit).
>
>
> But QED cares about the time-dependence of field observables, which is what
> one has in practice. Read about the closed time path formalism.

Thanks for the suggestion. I'll try to find something on the "closed
time path formalism". It would be interesting to see how they can manage
to predict any kind of time evolution without having finite well-defined
Hamiltonian.

>
>
>
>>The CJS theorem
>>becomes very important in the case of dynamical description of the time
>>evolution. Such a description is built in my book, and in agreement with
>>the CJS theorem,
>>I found that particle trajectories (or, in general, particle
>>observables) do not transform in a manifestly covariant fashion between
>>different reference frames.
>
>
> In standard terminology, this just says that particle trajectories are
> (and by CJS have to be) observer dependent.
> But nevertheless, one can translate form one observer
> to another one, if one uses fields instead of trajectories. Then everything
> is covariant. Trajectories are fuzzy anyway, already in nonrelativistic QM,
> so giving them up is not a big deal.

I do not agree. Any quantum theory should have a classical limit.
In this limit, trajectories are well defined. If the manifest covariance
is true, then these trajectories must transform by Lorentz (or, as you
say, they must be observer independent).

>
>
>
>
>>I disagree that unitary equivalent representations of the Poincare group
>>are always physically equivalent. Take two Hamiltonians connected by a
>>unitary transformation
>>
>>H' = U H U^{-1}
>>
>>These two Hamiltonians may have the same S-matrix and energy spectrum of
>>bound states. However, the wave functions of bound states are
>>different, and the time evolutions described by these two Hamiltonians
>>are different too.
>
>
> But the physics is exactly the same, since expectations are identical,
> if states and operators are both correctly transformed.

I didn't say that states are transformed. Only the Hamiltonian
(and 9 other generators of the Poincare group) is transformed.
When the unitary equivalence of different forms of dynamics is proved,
the states are not subject to transformation.

If you unitarily transform
both states and operators, then, of course, the physical content of
the theory does not change. That's rather trivial.

> Only that counts for observable consequences.
>
>
>
>>>>I disagree. There is a way to send a superluminal (in fact,
>>>>instantaneous) signal between two charged particles.
>>>
>>>Well, try to convince an experimentalist to check it!
>>>I doubt you'll find one prepared to work seriously on such a project.
>>
>>I would love to talk to an experimentalist, and I hope that I can
>>convince somebody. The question about the speed of propagation of the
>>Coulomb and magnetic interactions remains unresolved experimentally
>>until this day. I think you would agree with me that this
>>is the fundamental issue for our understanding of nature, and the
>>absence of experimental data is rather disturbing.
>
>
> I think this issue is already resolved in favor of an absolute bound
> of c for the speed of any material body or information flow.
> That's what causality is about.

This is "proved" in special relativity by assuming the interaction
independence of boost transformations. This proof is not valid in
my approach, where boost transformations are interaction dependent.
In my approach, interactions are superluminal, and causality is
respected (the effect never happens earlier than the cause).

>
>
>
>
>>>Science proceeds not by speculation but by making sure that one always
>>>stands on safe ground and is correspondingly cautious with great claims.
>>
>>I believe, speculation plays a great role in the development of
>>science.
>
>
> Only if it succeeds in better predictivity. Otherwise it is the biggest
> waste of efforts. Most speculations remain infertile; very, very few
> succeed. And those that do succeed, do so because they back up their
> speculations not with a few general arguments but with invincible
> predictions.

I do have experimental predictions: the instantaneous propagation of
the Coulomb force, and small corrections to the decay laws of moving
particles. If experiment will say that I was wrong, I'll shut up.

>
>
>
>>Actually, the universally accepted Minkowski spacetime is also
>>a great speculation.
>
>
> It has manifold experimental verifications, both in the classical
> and in the quantum domain. If you call that a great speculation
> then all of physics is a great speculation.

Let us set aside the effects of gravity and focus on predictions of
special relativity. I agree that there are lot of experimental data
referring to properties of free particles (the Doppler effect,
the relativistic relationship between mass, momentum, and energy,
the Lorentz invariance of the S-matrix, etc.)
In such cases, my approach predicts exactly the same properties, because
no interaction is involved.

There are much much fewer relativistic
effects studied in isolated systems with interactions. I actually know
only one such effect: the slowing down of the decay of moving unstable
particles. Most accurate experiments with muons
confirm Einstein's time dilation formula with the accuracy of 10^{-5}.
My approach predicts corrections 10 orders of magnitude smaller for this
system. So, from the experimental point of view, my speculation is as
good as Einstein's speculation.

Eugene Stefanovich.

>
>
> Arnold Neumaier
>

Eugene Stefanovich

Nov30-04, 12:50 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Oz wrote:\n> Eugene Stefanovich <eugenev@synopsys.com> writes\n>\n>\n>>do not accept the dubious postulate of Minkowski spacetime,\n>\n>\n> Eh? I was under the impression that minkowski spacetime was just a space\n> that conveniently illustrated SR. What postulate is added?\n\nThis is a common misconception that two Einstein\'s postulates\n(the equivalence of all inertial observers and the invariance of the\nspeed of light) immediately imply the unification of space and time.\nIf you think for a moment about these two postulates you\'ll realize\nthat they can tell you only about a very limited class of physical\nsystems, i.e., freely propagated light pulses and events associated\nwith them. It\'s not a surprise, that for such systems (as well as\nfor systems of non-interacting particles) the linear\nand universal Lorentz transformations for time and position can\nbe rigorously derived. What about systems of massive interacting\nparticles? The Einstein\'s postulates can tell nothing about them.\nEinstein and Minkowski simply assumed that the same linear Lorentz\ntransformations are valid for all physical systems. This allowed them\nto say that boost transformations are kinematical, and to introduce\nthe Minkowski spacetime as a mathematical realization of this\nassumption.\n\nIn my approach, I do not make the assumption of the kinematical\n(interaction independent) character of boosts. Actually, I prove that\nthis assumption is inconsistent and should not be used.\n\n\n>\n>\n>>though I\n>>keep all other well-established postulates of relativity and quantum\n>>mechanics.\n>\n>\n> You probably ought to itemise them.\n>\n\nSure. You can find all the details in my online book\nwww.meopemuk.com/book.pdf (see also www.geocities.com/meopemuk\nfor other publications)\n\nPostulate A - the principle of relativity\nPostulate H - the Poincare group properties of transformations\nbetween inertial observers.\nPostulates I-M - usual postulates of quantum mechanics\nPostulate P - quantum description of compound systems\nPostulate R - the kinematical character of time translations\nand rotations (=assumption of the Dirac\'s instant form\nof dynamics)\nPostulate S - cluster separability of interactions\nPostulate V - charge renormalization condition\nPostulate W - stability of vacuum and 1-particle states wrt interaction.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz wrote:
> Eugene Stefanovich <eugenev@synopsys.com> writes
>
>
>>do not accept the dubious postulate of Minkowski spacetime,
>
>
> Eh? I was under the impression that minkowski spacetime was just a space
> that conveniently illustrated SR. What postulate is added?

This is a common misconception that two Einstein's postulates
(the equivalence of all inertial observers and the invariance of the
speed of light) immediately imply the unification of space and time.
If you think for a moment about these two postulates you'll realize
that they can tell you only about a very limited class of physical
systems, i.e., freely propagated light pulses and events associated
with them. It's not a surprise, that for such systems (as well as
for systems of non-interacting particles) the linear
and universal Lorentz transformations for time and position can
be rigorously derived. What about systems of massive interacting
particles? The Einstein's postulates can tell nothing about them.
Einstein and Minkowski simply assumed that the same linear Lorentz
transformations are valid for all physical systems. This allowed them
to say that boost transformations are kinematical, and to introduce
the Minkowski spacetime as a mathematical realization of this
assumption.

In my approach, I do not make the assumption of the kinematical
(interaction independent) character of boosts. Actually, I prove that
this assumption is inconsistent and should not be used.

>
>
>>though I
>>keep all other well-established postulates of relativity and quantum
>>mechanics.
>
>
> You probably ought to itemise them.
>

Sure. You can find all the details in my online book
www.meopemuk.com/book.pdf (see also www.geocities.com/meopemuk
for other publications)

Postulate A - the principle of relativity
Postulate H - the Poincare group properties of transformations
between inertial observers.
Postulates I-M - usual postulates of quantum mechanics
Postulate P - quantum description of compound systems
Postulate R - the kinematical character of time translations
and rotations (=assumption of the Dirac's instant form
of dynamics)
Postulate S - cluster separability of interactions
Postulate V - charge renormalization condition
Postulate W - stability of vacuum and 1-particle states wrt interaction.

Eugene Stefanovich.

Arnold Neumaier

Dec1-04, 11:04 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>But QED cares about the time-dependence of field observables, which is what\n>>one has in practice. Read about the closed time path formalism.\n>\n> Thanks for the suggestion. I\'ll try to find something on the "closed\n> time path formalism". It would be interesting to see how they can manage\n> to predict any kind of time evolution without having finite well-defined\n> Hamiltonian.\n\nThey only need computable formulas for time correlations. This gives\nin principle (disregarding questions of interpretation of the functional\nintegrals) all dynamical information needed, and gives in practice\nreasonable approximations. How do you think people derive kinetic equations\nif there is no trace of dynamics. It is just somewhat implicit, but not\ncompletely absent.\n\n\n> I do not agree. Any quantum theory should have a classical limit.\n\nWell, what is the classical limit of QED, in your opinion?\nHow do you cope with particle creation and destruction in you classical\nlimit? You fail to discuss this in your book.\n\n> In this limit, trajectories are well defined.\n\nWhy? It depends on the theory, and there is very little.\n\n> If the manifest covariance\n> is true, then these trajectories must transform by Lorentz (or, as you\n> say, they must be observer independent).\n\nNo. There must only be a well-defined recipe how the set of trajectories\nof one observer translates into that of another. And this is what covariance\nis about.\n\n\n>>But the physics is exactly the same, since expectations are identical,\n>>if states and operators are both correctly transformed.\n>\n> I didn\'t say that states are transformed.\n\nBut I did.\n\n\n> Only the Hamiltonian\n> (and 9 other generators of the Poincare group) is transformed.\n> When the unitary equivalence of different forms of dynamics is proved,\n> the states are not subject to transformation.\n\nOf course the states must transform, too. You can see this already\nin the nonrelativistic case when you go from the Schroedinger representation\nto the interaction representation. Same physics, different states and\nobservables, related by a unitary transformation.\n\n\n> If you unitarily transform\n> both states and operators, then, of course, the physical content of\n> the theory does not change. That\'s rather trivial.\n\nThis is also what happens in the equivalence of Shirikov\'s version\nwith yours. You keep the vacuum, hee keeps the Hamiltonian; to relate the\ntwo you must transform both.\n\n\n\n\n\n> My approach predicts corrections 10 orders of magnitude smaller for this\n> system. So, from the experimental point of view, my speculation is as\n> good as Einstein\'s speculation.\n\nHe predicted corrections for the Mercury trajectory that were unexpleained\nbefore, and _significant_ bending of light which was observed a few years\nlater. That\'s why his theory is viewed _only_ by outsiders such as you\nas speculations. The GPS confirms his views every day.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>But QED cares about the time-dependence of field observables, which is what
>>one has in practice. Read about the closed time path formalism.
>
> Thanks for the suggestion. I'll try to find something on the "closed
> time path formalism". It would be interesting to see how they can manage
> to predict any kind of time evolution without having finite well-defined
> Hamiltonian.

They only need computable formulas for time correlations. This gives
in principle (disregarding questions of interpretation of the functional
integrals) all dynamical information needed, and gives in practice
reasonable approximations. How do you think people derive kinetic equations
if there is no trace of dynamics. It is just somewhat implicit, but not
completely absent.

> I do not agree. Any quantum theory should have a classical limit.

Well, what is the classical limit of QED, in your opinion?
How do you cope with particle creation and destruction in you classical
limit? You fail to discuss this in your book.

> In this limit, trajectories are well defined.

Why? It depends on the theory, and there is very little.

> If the manifest covariance
> is true, then these trajectories must transform by Lorentz (or, as you
> say, they must be observer independent).

No. There must only be a well-defined recipe how the set of trajectories
of one observer translates into that of another. And this is what covariance
is about.

>>But the physics is exactly the same, since expectations are identical,
>>if states and operators are both correctly transformed.
>
> I didn't say that states are transformed.

But I did.

> Only the Hamiltonian
> (and 9 other generators of the Poincare group) is transformed.
> When the unitary equivalence of different forms of dynamics is proved,
> the states are not subject to transformation.

Of course the states must transform, too. You can see this already
in the nonrelativistic case when you go from the Schroedinger representation
to the interaction representation. Same physics, different states and
observables, related by a unitary transformation.

> If you unitarily transform
> both states and operators, then, of course, the physical content of
> the theory does not change. That's rather trivial.

This is also what happens in the equivalence of Shirikov's version
with yours. You keep the vacuum, hee keeps the Hamiltonian; to relate the
two you must transform both.

> My approach predicts corrections 10 orders of magnitude smaller for this
> system. So, from the experimental point of view, my speculation is as
> good as Einstein's speculation.

He predicted corrections for the Mercury trajectory that were unexpleained
before, and _significant_ bending of light which was observed a few years
later. That's why his theory is viewed _only_ by outsiders such as you
as speculations. The GPS confirms his views every day.

Arnold Neumaier

Arnold Neumaier

Dec1-04, 11:04 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>>p_2. Let us find how momenta of particles transform to the translated\n>>>frame of reference\n>>> p_1(a) =3D exp(-iPa) p_1 exp(iPa) =3D p_1 -i [Y,p_1]a + ...\n>>> p_2(a) =3D exp(-iPa) p_2 exp(iPa) =3D p_2 -i [Y,p_2]a + ...\n>>>Contrary to observations, these transformations are non-trivial and\n>>>interaction-dependent. This dependence should be observable in\n>>>experiment.=20\n>>\n>>No; it is - and has to be - as unobservable as coordinate transforms in\n>>ordinary space.\n>\n> Are you saying that momenta of particles are not observable in different\n> frames of reference? What IS observable then?\n\nNo; I am saying that the transformation dependence is unobservable.\nTransformations are just represent the same physics in different\ncoordinate systems.\n\n\n>>>The assumptions of the CJS theorem are\n>>>1) the Hamiltonian description of dynamics\n>>>2) measurability of particle observables, like p_i and r_i,\n>>> in all frames of reference.\n>>>I think, these assumptions are well justified.\n>>\n>>No. It is a very restrictive assumption that different observers\n>>measure identical p_i and r_i.\n>\n> One observer can measure p_i(t) and r_i(t), where t is time measured\n> by his clock. Another observer can also measure p\'_i(t\') and r\'_i(t\')\n> where t\' is time measured by her clock. I hope you are not arguing\n> with that. The CJS theorem say that values of (p,r,t) and (p\',r\',t\')\n> are not connected by linear Lorentz transformations.\n\nOnly under assumptions which must be rejected.\n\n\n>>These are no counterarguments to my statement that\n>> \'Certainly <dA(x)> is measurable and gives the classical e/m field.\'\n>>This was the important point. Look at how QED is used by practicioners,\n>>and you can see the truth of my statement.\n>\n>\n> As far as I know, QED is used to calculate scattering cross-sections\n> (between particles) and energies of bound states (of particles).\n> Maybe I\'m missing something, but I\'ve never seen QED calculations of\n> the E and B fields and their comparison with experiment.\n\nThere is a lot done with QED that is not in the QFT textbooks.\nPlasma physics people use QED for deriving their kinetic equations\nwhich are about electromagnetic fields changing in time.\n\nQuantum optics people use coherent states made from electromagnetic\nfields discussed within QED.\n\nQuantum chemists use QED to calculate the response of molecules to\nelectromagnetic fields and laser impulses.\n\nAnd probably there is much more that I haven\'t seen.\n\n\n\n> These fields do appear at intermediate stages of calculations,\n> but never as a final result. Or sometimes these fields are considered\n> as external (fixed) classical fields. This is not a rigorous approach,\n> of course.\n\nYour approach (formal perturbation theory) is not rigorous either.\n\n\n>>>1) the electron-positron field \\psi(x,t) is not Hermitian, so it is\n>>> not observable according to your criterion\n>>\n>>Their expectation is zero anyway, since they are odd operators.\n>>So in this case there is nothing to measure.\n>>But certain Hermitian quadratic expressions in \\psi(x,t) are well-known\n>>observables, and define the measurable mass density, the charge density,\n>>and the electromagnetic currents.\n>\n> There are other operators (i.e., observables of position and momentum\n> (or velocity) of\n> individual particles) which allow us to define these measurable\n> quantities, . In a system with fixed number of particles it is\n> straightforward to define positions and momenta of all constituents, and\n> thus derive such things as charge density, etc. I am wondering if the\n> result will be the same when using expressions like\n>\n> \\rho(x,t) = -e \\psi^{\\dag}(x,t) \\psi(x,t)\n\nYou better show that they are, instead of wondering only, if your\napproach is to replace the standard approach, as you want.\n\n\n\n> I agree that general relativity is mathematically elegant. However, let\n> us not take the mathematical elegance as a proof of physical soundness.\n\nIt is elegant _and_ sound. Your theory is not; but QED is also elegant\nand sound; it agrees with all known facts below a certain energy and\ndistance.\n\n\n> I do not accept GR\'s fundamental premise: the common universal\n> background continuum, which degenerates to the flat Minkowski spacetime\n> in the absence of gravity. I am not impressed by special relativity and\n> GR, because the Minkowski spacetime idea was pulled out of thin air by\n> Minkowski and Einstein.\n\nIt was pulled out of the vacuum, or at least the speed of light in vacuum.\nThin air is also physics.\n\n> This idea does not have solid basis neither in\n> Einstein\'s postulates nor in experiment.\n\nWell, insisting on this isolates you completely.\n\n\n> I agree that so far there were\n> no experiments contradicting this idea. However, this does not look like\n> a strong argument to me.\n\nIt is the _only_ argument needed to make good physics.\nThat you find it weak only shows your weak position.\n\n\n>\n> I have constructed a relativistic quantum theory\n> of electromagnetic interactions, which does not use the idea of\n> Minkowski spacetime and contradicts some basic results of special\n> relativity (e.g., the retarded character of interactions). Nevertheless,\n> this approach has no less predictive power than QED. Of course, you\n> may argue that I haven\'t done explicit calculations of the Lamb shift,\n> and I haven\'t cleaned up the infrared sector of the theory, and I\n> do not have any approach to gravity yet, etc, etc. You may dismiss\n> my approach altogether based on these deficiencies.\n\nI only dismiss your strange philosophy.\n\n\n>> > I have shown that the space-time description does not follow\n>> > from Einstein\'s postulates, and is, in fact, an additional hypothesis.=\n>>\n>> > Everything becomes simple and consistent, if this hypothesis is\n>> > dropped.\n>>\n>>What??? I haven\'t seen anything simple in your approach.\n>>Your Hamiltonian power series expansion is much more messy\n>>than covariant QED...\n>\n> The simplicity is not in short expressions for the Hamiltonian and\n> other quantities. The simplicity is in adhering to well-tested\n> physical postulates, in clear physical meaning of all theoretical\n> ingredients, and in the absence of logical contradictions.\n\nHave you _anyone_ besides yourself convinced of that?\n\nThere are exactly the same logical contradictions as in QED, namely the\nmissing mathematical foundations that make sense nonperturbatively.\nWithout that, there is no logical basis to decide about consistency.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>>p_2. Let us find how momenta of particles transform to the translated
>>>frame of reference
>>> p_1(a) =3D \exp(-iPa) p_1 \exp(iPa) =3D p_1 -i [Y,p_1]a + ...
>>> p_2(a) =3D \exp(-iPa) p_2 \exp(iPa) =3D p_2 -i [Y,p_2]a + ...
>>>Contrary to observations, these transformations are non-trivial and
>>>interaction-dependent. This dependence should be observable in
>>>experiment.=20
>>
>>No; it is - and has to be - as unobservable as coordinate transforms in
>>ordinary space.
>
> Are you saying that momenta of particles are not observable in different
> frames of reference? What IS observable then?

No; I am saying that the transformation dependence is unobservable.
Transformations are just represent the same physics in different
coordinate systems.

>>>The assumptions of the CJS theorem are
>>>1) the Hamiltonian description of dynamics
>>>2) measurability of particle observables, like p_i and r_i,
>>> in all frames of reference.
>>>I think, these assumptions are well justified.
>>
>>No. It is a very restrictive assumption that different observers
>>measure identical p_i and r_i.
>
> One observer can measure p_i(t) and r_i(t), where t is time measured
> by his clock. Another observer can also measure p'_i(t') and r'_i(t')
> where t' is time measured by her clock. I hope you are not arguing
> with that. The CJS theorem say that values of (p,r,t) and (p',r',t')
> are not connected by linear Lorentz transformations.

Only under assumptions which must be rejected.

>>These are no counterarguments to my statement that
>> 'Certainly <dA(x)> is measurable and gives the classical e/m field.'
>>This was the important point. Look at how QED is used by practicioners,
>>and you can see the truth of my statement.
>
>
> As far as I know, QED is used to calculate scattering cross-sections
> (between particles) and energies of bound states (of particles).
> Maybe I'm missing something, but I've never seen QED calculations of
> the E and B fields and their comparison with experiment.

There is a lot done with QED that is not in the QFT textbooks.
Plasma physics people use QED for deriving their kinetic equations
which are about electromagnetic fields changing in time.

Quantum optics people use coherent states made from electromagnetic
fields discussed within QED.

Quantum chemists use QED to calculate the response of molecules to
electromagnetic fields and laser impulses.

And probably there is much more that I haven't seen.

> These fields do appear at intermediate stages of calculations,
> but never as a final result. Or sometimes these fields are considered
> as external (fixed) classical fields. This is not a rigorous approach,
> of course.

Your approach (formal perturbation theory) is not rigorous either.

>>>1) the electron-positron field \psi(x,t) is not Hermitian, so it is
>>> not observable according to your criterion
>>
>>Their expectation is zero anyway, since they are odd operators.
>>So in this case there is nothing to measure.
>>But certain Hermitian quadratic expressions in \psi(x,t) are well-known
>>observables, and define the measurable mass density, the charge density,
>>and the electromagnetic currents.
>
> There are other operators (i.e., observables of position and momentum
> (or velocity) of
> individual particles) which allow us to define these measurable
> quantities, . In a system with fixed number of particles it is
> straightforward to define positions and momenta of all constituents, and
> thus derive such things as charge density, etc. I am wondering if the
> result will be the same when using expressions like
>
> \rho(x,t) = -e \psi^{\dag}(x,t) \psi(x,t)

You better show that they are, instead of wondering only, if your
approach is to replace the standard approach, as you want.

> I agree that general relativity is mathematically elegant. However, let
> us not take the mathematical elegance as a proof of physical soundness.

It is elegant _and_ sound. Your theory is not; but QED is also elegant
and sound; it agrees with all known facts below a certain energy and
distance.

> I do not accept GR's fundamental premise: the common universal
> background continuum, which degenerates to the flat Minkowski spacetime
> in the absence of gravity. I am not impressed by special relativity and
> GR, because the Minkowski spacetime idea was pulled out of thin air by
> Minkowski and Einstein.

It was pulled out of the vacuum, or at least the speed of light in vacuum.
Thin air is also physics.

> This idea does not have solid basis neither in
> Einstein's postulates nor in experiment.

Well, insisting on this isolates you completely.

> I agree that so far there were
> no experiments contradicting this idea. However, this does not look like
> a strong argument to me.

It is the _only_ argument needed to make good physics.
That you find it weak only shows your weak position.

>
> I have constructed a relativistic quantum theory
> of electromagnetic interactions, which does not use the idea of
> Minkowski spacetime and contradicts some basic results of special
> relativity (e.g., the retarded character of interactions). Nevertheless,
> this approach has no less predictive power than QED. Of course, you
> may argue that I haven't done explicit calculations of the Lamb shift,
> and I haven't cleaned up the infrared sector of the theory, and I
> do not have any approach to gravity yet, etc, etc. You may dismiss
> my approach altogether based on these deficiencies.

I only dismiss your strange philosophy.

>> > I have shown that the space-time description does not follow
>> > from Einstein's postulates, and is, in fact, an additional hypothesis.=
>>
>> > Everything becomes simple and consistent, if this hypothesis is
>> > dropped.
>>
>>What??? I haven't seen anything simple in your approach.
>>Your Hamiltonian power series expansion is much more messy
>>than covariant QED...
>
> The simplicity is not in short expressions for the Hamiltonian and
> other quantities. The simplicity is in adhering to well-tested
> physical postulates, in clear physical meaning of all theoretical
> ingredients, and in the absence of logical contradictions.

Have you _anyone_ besides yourself convinced of that?

There are exactly the same logical contradictions as in QED, namely the
missing mathematical foundations that make sense nonperturbatively.
Without that, there is no logical basis to decide about consistency.

Arnold Neumaier

Eugene Stefanovich

Dec2-04, 06:14 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>But QED cares about the time-dependence of field observables, which is what\n>>>one has in practice. Read about the closed time path formalism.\n>>\n>>Thanks for the suggestion. I\'ll try to find something on the "closed\n>>time path formalism". It would be interesting to see how they can manage\n>>to predict any kind of time evolution without having finite well-defined\n>>Hamiltonian.\n>\n>\n> They only need computable formulas for time correlations. This gives\n> in principle (disregarding questions of interpretation of the functional\n> integrals) all dynamical information needed, and gives in practice\n> reasonable approximations. How do you think people derive kinetic equations\n> if there is no trace of dynamics. It is just somewhat implicit, but not\n> completely absent.\n\nI briefly looked at this theory, and apparently I need more time to\nunderstand what is going on there. However, the complexity of this\napproach made me suspicious. In a quantum theory time evolution should\nbe described simply by applying the time evolution operator exp(iHt)\nto the\nstate vector. This simple picture is lacking in the "closed time path\nformalism". Do you know if this approach can solve a simple time\nevolution task, such as time dependence of the state of two interacting\nelectrons? I found some applications to statistical systems with\ntemperature, but they do not impress me until the method is demonstrated\nto work for simple models.\n\n\n>\n>\n>\n>>I do not agree. Any quantum theory should have a classical limit.\n>\n>\n> Well, what is the classical limit of QED, in your opinion?\n> How do you cope with particle creation and destruction in you classical\n> limit? You fail to discuss this in your book.\n\nIn the classical limit of the "dressed particle" QED, there are\nparticles interacting with each other via (instantaneous) Coulomb and\nmagnetic forces. They are described by the terms of the type\na^{\\dag}a^{\\dag}aa. In addition, acceleration of charges causes emission\nof photons (bremsstrahlung). In the dressed particle Hamiltonian, this\neffect is represented by the terms\na^{\\dag}a^{\\dag}c^{\\dag}aa. With these simplest contributions, the\nclassical limit of RQD should be very similar (though not identical)\nto the Maxwell\'s theory.\n\nI agree that I should have written more about the classical limit of my\napproach.\n\n>\n>\n>>In this limit, trajectories are well defined.\n>\n>\n> Why? It depends on the theory, and there is very little.\n\nSee section 7.3 of the book\n\n>\n>\n>>If the manifest covariance\n>>is true, then these trajectories must transform by Lorentz (or, as you\n>> say, they must be observer independent).\n>\n>\n> No. There must only be a well-defined recipe how the set of trajectories\n> of one observer translates into that of another. And this is what covariance\n> is about.\n\nThe whole point of the "manifest covariance" is the this well-defined\nrecipe must be nothing else but linear Lorentz formulas. If you allow\nother recipes (as I do), then you deviate from the strict special\nrelativity.\n\n>\n>\n>\n>>>But the physics is exactly the same, since expectations are identical,\n>>>if states and operators are both correctly transformed.\n>>\n>>I didn\'t say that states are transformed.\n>\n>\n> But I did.\n>\n>\n>\n>>Only the Hamiltonian\n>>(and 9 other generators of the Poincare group) is transformed.\n>>When the unitary equivalence of different forms of dynamics is proved,\n>>the states are not subject to transformation.\n>\n>\n> Of course the states must transform, too. You can see this already\n> in the nonrelativistic case when you go from the Schroedinger representation\n> to the interaction representation. Same physics, different states and\n> observables, related by a unitary transformation.\n\nThis is not correct. The connection between the instant and point forms\nof dynamics is not of the type as connection between the Schroedinger,\nHeisenberg, and interaction representation. The definition of states\ndoes not change when a unitary transformation between different\nforms is made(see below).\n\n>\n>\n>\n>>If you unitarily transform\n>>both states and operators, then, of course, the physical content of\n>>the theory does not change. That\'s rather trivial.\n>\n>\n> This is also what happens in the equivalence of Shirikov\'s version\n> with yours. You keep the vacuum, hee keeps the Hamiltonian; to relate the\n> two you must transform both.\n\nThat\'s why I said that my approach and Shirokov\'s approach are trivially\nidentical. This is not the case with instant and point form dynamics.\nThe unitary transformations between them does not touch the states.\nOtherwise the proof of the Sokolov\'s theorem could be done in just one\nsentence. In order to define which form of dynamics you have\n(i.e., which generators contain interaction) you need to have free\ngenerators as a reference point. The definition of free generators\ndepends on the definition of states of free particles. When you\ntransform states, you also transform free generators, and you change\nyour reference point. That\'s not the way to transform between different\nforms of dynamics.\n\n>\n>\n>\n>\n>\n>\n>>My approach predicts corrections 10 orders of magnitude smaller for this\n>>system. So, from the experimental point of view, my speculation is as\n>>good as Einstein\'s speculation.\n>\n>\n> He predicted corrections for the Mercury trajectory that were unexpleained\n> before, and _significant_ bending of light which was observed a few years\n> later. That\'s why his theory is viewed _only_ by outsiders such as you\n> as speculations. The GPS confirms his views every day.\n\n1. I am not trying to diminish successes of GR in explaining and\npredicting various experimental facts.\n2. I believe that the same (or even better) predictions can be made\nwithout the concept of the space-time curvature. Though, I do not\nhave a proof for that.\n3. The reason for my belief is that unified spacetime is in\ncontradiction with quantum mechanics, and was never properly\njustified in special relativity. GR is just using the same\nunproven concept.\n4. The agreement with experiment is a great plus, but any consistent\ntheory should be also free of internal contradictions.\nThis is not the case with SR and GR.\n\nEugene Stefanovich.\n\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>But QED cares about the time-dependence of field observables, which is what
>>>one has in practice. Read about the closed time path formalism.
>>
>>Thanks for the suggestion. I'll try to find something on the "closed
>>time path formalism". It would be interesting to see how they can manage
>>to predict any kind of time evolution without having finite well-defined
>>Hamiltonian.
>
>
> They only need computable formulas for time correlations. This gives
> in principle (disregarding questions of interpretation of the functional
> integrals) all dynamical information needed, and gives in practice
> reasonable approximations. How do you think people derive kinetic equations
> if there is no trace of dynamics. It is just somewhat implicit, but not
> completely absent.

I briefly looked at this theory, and apparently I need more time to
understand what is going on there. However, the complexity of this
approach made me suspicious. In a quantum theory time evolution should
be described simply by applying the time evolution operator \exp(iHt)
to the
state vector. This simple picture is lacking in the "closed time path
formalism". Do you know if this approach can solve a simple time
evolution task, such as time dependence of the state of two interacting
electrons? I found some applications to statistical systems with
temperature, but they do not impress me until the method is demonstrated
to work for simple models.

>
>
>
>>I do not agree. Any quantum theory should have a classical limit.
>
>
> Well, what is the classical limit of QED, in your opinion?
> How do you cope with particle creation and destruction in you classical
> limit? You fail to discuss this in your book.

In the classical limit of the "dressed particle" QED, there are
particles interacting with each other via (instantaneous) Coulomb and
magnetic forces. They are described by the terms of the type
a^{\dag}a^{\dag}aa. In addition, acceleration of charges causes emission
of photons (bremsstrahlung). In the dressed particle Hamiltonian, this
effect is represented by the terms
a^{\dag}a^{\dag}c^{\dag}aa. With these simplest contributions, the
classical limit of RQD should be very similar (though not identical)
to the Maxwell's theory.

I agree that I should have written more about the classical limit of my
approach.

>
>
>>In this limit, trajectories are well defined.
>
>
> Why? It depends on the theory, and there is very little.

See section 7.3 of the book

>
>
>>If the manifest covariance
>>is true, then these trajectories must transform by Lorentz (or, as you
>> say, they must be observer independent).
>
>
> No. There must only be a well-defined recipe how the set of trajectories
> of one observer translates into that of another. And this is what covariance
> is about.

The whole point of the "manifest covariance" is the this well-defined
recipe must be nothing else but linear Lorentz formulas. If you allow
other recipes (as I do), then you deviate from the strict special
relativity.

>
>
>
>>>But the physics is exactly the same, since expectations are identical,
>>>if states and operators are both correctly transformed.
>>
>>I didn't say that states are transformed.
>
>
> But I did.
>
>
>
>>Only the Hamiltonian
>>(and 9 other generators of the Poincare group) is transformed.
>>When the unitary equivalence of different forms of dynamics is proved,
>>the states are not subject to transformation.
>
>
> Of course the states must transform, too. You can see this already
> in the nonrelativistic case when you go from the Schroedinger representation
> to the interaction representation. Same physics, different states and
> observables, related by a unitary transformation.

This is not correct. The connection between the instant and point forms
of dynamics is not of the type as connection between the Schroedinger,
Heisenberg, and interaction representation. The definition of states
does not change when a unitary transformation between different
forms is made(see below).

>
>
>
>>If you unitarily transform
>>both states and operators, then, of course, the physical content of
>>the theory does not change. That's rather trivial.
>
>
> This is also what happens in the equivalence of Shirikov's version
> with yours. You keep the vacuum, hee keeps the Hamiltonian; to relate the
> two you must transform both.

That's why I said that my approach and Shirokov's approach are trivially
identical. This is not the case with instant and point form dynamics.
The unitary transformations between them does not touch the states.
Otherwise the proof of the Sokolov's theorem could be done in just one
sentence. In order to define which form of dynamics you have
(i.e., which generators contain interaction) you need to have free
generators as a reference point. The definition of free generators
depends on the definition of states of free particles. When you
transform states, you also transform free generators, and you change
your reference point. That's not the way to transform between different
forms of dynamics.

>
>
>
>
>
>
>>My approach predicts corrections 10 orders of magnitude smaller for this
>>system. So, from the experimental point of view, my speculation is as
>>good as Einstein's speculation.
>
>
> He predicted corrections for the Mercury trajectory that were unexpleained
> before, and _significant_ bending of light which was observed a few years
> later. That's why his theory is viewed _only_ by outsiders such as you
> as speculations. The GPS confirms his views every day.

1. I am not trying to diminish successes of GR in explaining and
predicting various experimental facts.
2. I believe that the same (or even better) predictions can be made
without the concept of the space-time curvature. Though, I do not
have a proof for that.
3. The reason for my belief is that unified spacetime is in
contradiction with quantum mechanics, and was never properly
justified in special relativity. GR is just using the same
unproven concept.
4. The agreement with experiment is a great plus, but any consistent
theory should be also free of internal contradictions.
This is not the case with SR and GR.

Eugene Stefanovich.

>
>
> Arnold Neumaier
>

Eugene Stefanovich

Dec2-04, 06:15 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>>p_2. Let us find how momenta of particles transform to the translated\n>>>>frame of reference\n>>>> p_1(a) =3D exp(-iPa) p_1 exp(iPa) =3D p_1 -i [Y,p_1]a + ...\n>>>> p_2(a) =3D exp(-iPa) p_2 exp(iPa) =3D p_2 -i [Y,p_2]a + ...\n>>>>Contrary to observations, these transformations are non-trivial and\n>>>>interaction-dependent. This dependence should be observable in\n>>>>experiment.=20\n>>>\n>>>No; it is - and has to be - as unobservable as coordinate transforms in\n>>>ordinary space.\n>>\n>>Are you saying that momenta of particles are not observable in different\n>>frames of reference? What IS observable then?\n>\n>\n> No; I am saying that the transformation dependence is unobservable.\n> Transformations are just represent the same physics in different\n> coordinate systems.\n\nI believe you agree that time evolution is measurable.\nIt is also clear from observations which time evolution is free\n(without interaction) and which is interacting. Time evolution is\nnothing but transformation of observables to the reference frame\nshifted in time. There is no any fundamental difference between\ntime translations and space translations. So, transformations of\nobservables wrt space translations should be observable as well,\nand the dependence of these transformations on interaction should\nbe also easily recognizable.\n\nFor observer shifted by distance a, all positions of particles\nappear shifted by -a. In the case of point form dynamics, this\nsimple rule does not work.\n\n>\n>\n>\n>>>>The assumptions of the CJS theorem are\n>>>>1) the Hamiltonian description of dynamics\n>>>>2) measurability of particle observables, like p_i and r_i,\n>>>> in all frames of reference.\n>>>>I think, these assumptions are well justified.\n>>>\n>>>No. It is a very restrictive assumption that different observers\n>>>measure identical p_i and r_i.\n>>\n>>One observer can measure p_i(t) and r_i(t), where t is time measured\n>>by his clock. Another observer can also measure p\'_i(t\') and r\'_i(t\')\n>>where t\' is time measured by her clock. I hope you are not arguing\n>>with that. The CJS theorem say that values of (p,r,t) and (p\',r\',t\')\n>>are not connected by linear Lorentz transformations.\n>\n>\n> Only under assumptions which must be rejected.\n\nWhich are these assumptions? If you reject the observability of\nparticle\'s positions and momenta, then no physics is left.\n\n>\n>\n>\n>>>These are no counterarguments to my statement that\n>>> \'Certainly <dA(x)> is measurable and gives the classical e/m field.\'\n>>>This was the important point. Look at how QED is used by practicioners,\n>>>and you can see the truth of my statement.\n>>\n>>\n>>As far as I know, QED is used to calculate scattering cross-sections\n>>(between particles) and energies of bound states (of particles).\n>>Maybe I\'m missing something, but I\'ve never seen QED calculations of\n>>the E and B fields and their comparison with experiment.\n>\n>\n> There is a lot done with QED that is not in the QFT textbooks.\n> Plasma physics people use QED for deriving their kinetic equations\n> which are about electromagnetic fields changing in time.\n>\n> Quantum optics people use coherent states made from electromagnetic\n> fields discussed within QED.\n>\n> Quantum chemists use QED to calculate the response of molecules to\n> electromagnetic fields and laser impulses.\n\nI worked in quantum chemistry for many years, and I haven\'t seen\napplications of full-fledged QED. Only some approximate models\n"derived" from QED. Of course, there are many things which I do\nnot know, and there is a good chance that I\'m wrong.\n\n>\n> And probably there is much more that I haven\'t seen.\n\n\n\n\n>\n>\n>\n>\n>>These fields do appear at intermediate stages of calculations,\n>>but never as a final result. Or sometimes these fields are considered\n>>as external (fixed) classical fields. This is not a rigorous approach,\n>>of course.\n>\n>\n> Your approach (formal perturbation theory) is not rigorous either.\n>\n>\n>\n>>>>1) the electron-positron field \\psi(x,t) is not Hermitian, so it is\n>>>> not observable according to your criterion\n>>>\n>>>Their expectation is zero anyway, since they are odd operators.\n>>>So in this case there is nothing to measure.\n>>>But certain Hermitian quadratic expressions in \\psi(x,t) are well-known\n>>>observables, and define the measurable mass density, the charge density,\n>>>and the electromagnetic currents.\n>>\n>>There are other operators (i.e., observables of position and momentum\n>> (or velocity) of\n>>individual particles) which allow us to define these measurable\n>>quantities, . In a system with fixed number of particles it is\n>>straightforward to define positions and momenta of all constituents, and\n>>thus derive such things as charge density, etc. I am wondering if the\n>>result will be the same when using expressions like\n>>\n>>\\rho(x,t) = -e \\psi^{\\dag}(x,t) \\psi(x,t)\n>\n>\n> You better show that they are, instead of wondering only, if your\n> approach is to replace the standard approach, as you want.\n\nI will think about the connection between these two charge densities.\nThanks for the suggestions.\nI agree that my approach is not mature enough to be a viable\nsubstitute for existing theories. There are lot of things to be done,\nlot of questions to be answered.\n\n>\n>\n>\n>\n>>I agree that general relativity is mathematically elegant. However, let\n>>us not take the mathematical elegance as a proof of physical soundness.\n>\n>\n> It is elegant _and_ sound. Your theory is not; but QED is also elegant\n> and sound; it agrees with all known facts below a certain energy and\n> distance.\n>\n>\n>\n>>I do not accept GR\'s fundamental premise: the common universal\n>>background continuum, which degenerates to the flat Minkowski spacetime\n>>in the absence of gravity. I am not impressed by special relativity and\n>>GR, because the Minkowski spacetime idea was pulled out of thin air by\n>>Minkowski and Einstein.\n>\n>\n> It was pulled out of the vacuum, or at least the speed of light in vacuum.\n> Thin air is also physics.\n>\n>\n>>This idea does not have solid basis neither in\n>>Einstein\'s postulates nor in experiment.\n>\n>\n> Well, insisting on this isolates you completely.\n>\n>\n>\n>>I agree that so far there were\n>>no experiments contradicting this idea. However, this does not look like\n>>a strong argument to me.\n>\n>\n> It is the _only_ argument needed to make good physics.\n> That you find it weak only shows your weak position.\n>\n>\n>\n>>I have constructed a relativistic quantum theory\n>>of electromagnetic interactions, which does not use the idea of\n>>Minkowski spacetime and contradicts some basic results of special\n>>relativity (e.g., the retarded character of interactions). Nevertheless,\n>>this approach has no less predictive power than QED. Of course, you\n>>may argue that I haven\'t done explicit calculations of the Lamb shift,\n>>and I haven\'t cleaned up the infrared sector of the theory, and I\n>>do not have any approach to gravity yet, etc, etc. You may dismiss\n>>my approach altogether based on these deficiencies.\n>\n>\n> I only dismiss your strange philosophy.\n\nI challenge you to prove that boost transformations of\nphysical observables are given by universal linear Lorentz formulas.\n(This is the basis for introducing the Minkowski spacetime, isn\'t\nit?) You can use the relativity postulate, the invariance of the\nspeed of light, and any other well-tested postulates.\nMy bet is that you cannot do that. So, you need to introduce\nthe Minkowski spacetime and the unification of space and time\nas an additional independent postulate in your theory.\n\nIf so, there is always a legitimate question how the theory will\nlook like if this postulate is dropped. That\'s what I did.\nAnd I found that all predictions of the new theory are very close\nto the old one (In the new theory boost transformations are\ninteraction-dependent, but in most cases this dependence is too\nweak to be seen in experiment). Moreover, I found that\nthe spacetime postulate contradicts other parts of the theory\n(quantum mechanics), so dropping it just makes the theory more\nconsistent.\n\nThat\'s basically the justification of my approach. You may not\nlike my philosophy, but so far you wasn\'t able to point any\nlogical contradiction in my arguments, or any disagreement with\nexperimental data.\n\n>\n>\n>\n>>>>I have shown that the space-time description does not follow\n>>>>from Einstein\'s postulates, and is, in fact, an additional hypothesis.=\n>>>\n>>>>Everything becomes simple and consistent, if this hypothesis is\n>>>>dropped.\n>>>\n>>>What??? I haven\'t seen anything simple in your approach.\n>>>Your Hamiltonian power series expansion is much more messy\n>>>than covariant QED...\n>>\n>>The simplicity is not in short expressions for the Hamiltonian and\n>>other quantities. The simplicity is in adhering to well-tested\n>>physical postulates, in clear physical meaning of all theoretical\n>>ingredients, and in the absence of logical contradictions.\n>\n>\n> Have you _anyone_ besides yourself convinced of that?\n>\n> There are exactly the same logical contradictions as in QED, namely the\n> missing mathematical foundations that make sense nonperturbatively.\n> Without that, there is no logical basis to decide about consistency.\n\nI agree that this problem is not solved neither in QED nor in my\napproach. I don\'t think you expect me to solve all problems in\ntheoretical physics.\n\nThe simplicity of my approach, as I see it, is in formulating\nQFT in the language of ordinary quantum mechanics, where states are\ndescribed by wave functions, the time evolution is described by\na finite unitary operator, the bound states are calculated via\ndiagonalization of the Hamiltonian, etc. The only significant\ndifference is that creation and annihilation of particles becomes\nallowed. I don\'t see this simple and intuitive picture in\nthe traditional formulation of QFT. That\'s why I think my\napproach is better.\n\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>>p_2. Let us find how momenta of particles transform to the translated
>>>>frame of reference
>>>> p_1(a) =3D \exp(-iPa) p_1 \exp(iPa) =3D p_1 -i [Y,p_1]a + ...
>>>> p_2(a) =3D \exp(-iPa) p_2 \exp(iPa) =3D p_2 -i [Y,p_2]a + ...
>>>>Contrary to observations, these transformations are non-trivial and
>>>>interaction-dependent. This dependence should be observable in
>>>>experiment.=20
>>>
>>>No; it is - and has to be - as unobservable as coordinate transforms in
>>>ordinary space.
>>
>>Are you saying that momenta of particles are not observable in different
>>frames of reference? What IS observable then?
>
>
> No; I am saying that the transformation dependence is unobservable.
> Transformations are just represent the same physics in different
> coordinate systems.

I believe you agree that time evolution is measurable.
It is also clear from observations which time evolution is free
(without interaction) and which is interacting. Time evolution is
nothing but transformation of observables to the reference frame
shifted in time. There is no any fundamental difference between
time translations and space translations. So, transformations of
observables wrt space translations should be observable as well,
and the dependence of these transformations on interaction should
be also easily recognizable.

For observer shifted by distance a, all positions of particles
appear shifted by -a. In the case of point form dynamics, this
simple rule does not work.

>
>
>
>>>>The assumptions of the CJS theorem are
>>>>1) the Hamiltonian description of dynamics
>>>>2) measurability of particle observables, like p_i and r_i,
>>>> in all frames of reference.
>>>>I think, these assumptions are well justified.
>>>
>>>No. It is a very restrictive assumption that different observers
>>>measure identical p_i and r_i.
>>
>>One observer can measure p_i(t) and r_i(t), where t is time measured
>>by his clock. Another observer can also measure p'_i(t') and r'_i(t')
>>where t' is time measured by her clock. I hope you are not arguing
>>with that. The CJS theorem say that values of (p,r,t) and (p',r',t')
>>are not connected by linear Lorentz transformations.
>
>
> Only under assumptions which must be rejected.

Which are these assumptions? If you reject the observability of
particle's positions and momenta, then no physics is left.

>
>
>
>>>These are no counterarguments to my statement that
>>> 'Certainly <dA(x)> is measurable and gives the classical e/m field.'
>>>This was the important point. Look at how QED is used by practicioners,
>>>and you can see the truth of my statement.
>>
>>
>>As far as I know, QED is used to calculate scattering cross-sections
>>(between particles) and energies of bound states (of particles).
>>Maybe I'm missing something, but I've never seen QED calculations of
>>the E and B fields and their comparison with experiment.
>
>
> There is a lot done with QED that is not in the QFT textbooks.
> Plasma physics people use QED for deriving their kinetic equations
> which are about electromagnetic fields changing in time.
>
> Quantum optics people use coherent states made from electromagnetic
> fields discussed within QED.
>
> Quantum chemists use QED to calculate the response of molecules to
> electromagnetic fields and laser impulses.

I worked in quantum chemistry for many years, and I haven't seen
applications of full-fledged QED. Only some approximate models
"derived" from QED. Of course, there are many things which I do
not know, and there is a good chance that I'm wrong.

>
> And probably there is much more that I haven't seen.

>
>
>
>
>>These fields do appear at intermediate stages of calculations,
>>but never as a final result. Or sometimes these fields are considered
>>as external (fixed) classical fields. This is not a rigorous approach,
>>of course.
>
>
> Your approach (formal perturbation theory) is not rigorous either.
>
>
>
>>>>1) the electron-positron field \psi(x,t) is not Hermitian, so it is
>>>> not observable according to your criterion
>>>
>>>Their expectation is zero anyway, since they are odd operators.
>>>So in this case there is nothing to measure.
>>>But certain Hermitian quadratic expressions in \psi(x,t) are well-known
>>>observables, and define the measurable mass density, the charge density,
>>>and the electromagnetic currents.
>>
>>There are other operators (i.e., observables of position and momentum
>> (or velocity) of
>>individual particles) which allow us to define these measurable
>>quantities, . In a system with fixed number of particles it is
>>straightforward to define positions and momenta of all constituents, and
>>thus derive such things as charge density, etc. I am wondering if the
>>result will be the same when using expressions like
>>
>>\rho(x,t) = -e \psi^{\dag}(x,t) \psi(x,t)
>
>
> You better show that they are, instead of wondering only, if your
> approach is to replace the standard approach, as you want.

I will think about the connection between these two charge densities.
Thanks for the suggestions.
I agree that my approach is not mature enough to be a viable
substitute for existing theories. There are lot of things to be done,
lot of questions to be answered.

>
>
>
>
>>I agree that general relativity is mathematically elegant. However, let
>>us not take the mathematical elegance as a proof of physical soundness.
>
>
> It is elegant _and_ sound. Your theory is not; but QED is also elegant
> and sound; it agrees with all known facts below a certain energy and
> distance.
>
>
>
>>I do not accept GR's fundamental premise: the common universal
>>background continuum, which degenerates to the flat Minkowski spacetime
>>in the absence of gravity. I am not impressed by special relativity and
>>GR, because the Minkowski spacetime idea was pulled out of thin air by
>>Minkowski and Einstein.
>
>
> It was pulled out of the vacuum, or at least the speed of light in vacuum.
> Thin air is also physics.
>
>
>>This idea does not have solid basis neither in
>>Einstein's postulates nor in experiment.
>
>
> Well, insisting on this isolates you completely.
>
>
>
>>I agree that so far there were
>>no experiments contradicting this idea. However, this does not look like
>>a strong argument to me.
>
>
> It is the _only_ argument needed to make good physics.
> That you find it weak only shows your weak position.
>
>
>
>>I have constructed a relativistic quantum theory
>>of electromagnetic interactions, which does not use the idea of
>>Minkowski spacetime and contradicts some basic results of special
>>relativity (e.g., the retarded character of interactions). Nevertheless,
>>this approach has no less predictive power than QED. Of course, you
>>may argue that I haven't done explicit calculations of the Lamb shift,
>>and I haven't cleaned up the infrared sector of the theory, and I
>>do not have any approach to gravity yet, etc, etc. You may dismiss
>>my approach altogether based on these deficiencies.
>
>
> I only dismiss your strange philosophy.

I challenge you to prove that boost transformations of
physical observables are given by universal linear Lorentz formulas.
(This is the basis for introducing the Minkowski spacetime, isn't
it?) You can use the relativity postulate, the invariance of the
speed of light, and any other well-tested postulates.
My bet is that you cannot do that. So, you need to introduce
the Minkowski spacetime and the unification of space and time
as an additional independent postulate in your theory.

If so, there is always a legitimate question how the theory will
look like if this postulate is dropped. That's what I did.
And I found that all predictions of the new theory are very close
to the old one (In the new theory boost transformations are
interaction-dependent, but in most cases this dependence is too
weak to be seen in experiment). Moreover, I found that
the spacetime postulate contradicts other parts of the theory
(quantum mechanics), so dropping it just makes the theory more
consistent.

That's basically the justification of my approach. You may not
like my philosophy, but so far you wasn't able to point any
logical contradiction in my arguments, or any disagreement with
experimental data.

>
>
>
>>>>I have shown that the space-time description does not follow
>>>>from Einstein's postulates, and is, in fact, an additional hypothesis.=
>>>
>>>>Everything becomes simple and consistent, if this hypothesis is
>>>>dropped.
>>>
>>>What??? I haven't seen anything simple in your approach.
>>>Your Hamiltonian power series expansion is much more messy
>>>than covariant QED...
>>
>>The simplicity is not in short expressions for the Hamiltonian and
>>other quantities. The simplicity is in adhering to well-tested
>>physical postulates, in clear physical meaning of all theoretical
>>ingredients, and in the absence of logical contradictions.
>
>
> Have you _anyone_ besides yourself convinced of that?
>
> There are exactly the same logical contradictions as in QED, namely the
> missing mathematical foundations that make sense nonperturbatively.
> Without that, there is no logical basis to decide about consistency.

I agree that this problem is not solved neither in QED nor in my
approach. I don't think you expect me to solve all problems in
theoretical physics.

The simplicity of my approach, as I see it, is in formulating
QFT in the language of ordinary quantum mechanics, where states are
described by wave functions, the time evolution is described by
a finite unitary operator, the bound states are calculated via
diagonalization of the Hamiltonian, etc. The only significant
difference is that creation and annihilation of particles becomes
allowed. I don't see this simple and intuitive picture in
the traditional formulation of QFT. That's why I think my
approach is better.

>
>
> Arnold Neumaier
>

Eugene Stefanovich

Dec2-04, 06:15 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n\n>\n>>>>I have shown that the space-time description does not follow\n>>>>from Einstein\'s postulates, and is, in fact, an additional hypothesis.=\n>>>\n>>>>Everything becomes simple and consistent, if this hypothesis is\n>>>>dropped.\n>>>\n>>>What??? I haven\'t seen anything simple in your approach.\n>>>Your Hamiltonian power series expansion is much more messy\n>>>than covariant QED...\n>>\n>>The simplicity is not in short expressions for the Hamiltonian and\n>>other quantities. The simplicity is in adhering to well-tested\n>>physical postulates, in clear physical meaning of all theoretical\n>>ingredients, and in the absence of logical contradictions.\n>\n>\n> Have you _anyone_ besides yourself convinced of that?\n>\n\n\nThere are two papers (preprints) which confirm my results on\nthe decay law of moving particles:\n\nL. A. Khalfin, Quantum theory of unstable particles and\nrelativity,\nPreprint of Steklov Mathematical Institute, St. Petersburg Department,\nPDMI-6/1997 (1997). http://www.pdmi.ras.ru/preprint/1997/97-06.html\n\n\nM.I. Shirokov, Decay law of moving unstable\nparticle, Preprint JINR, E2-2004-4, Dubna (2004)\n(to appear in Int. J. Theor. Phys.)\nhttp://www.jinr.ru/publish/Preprints/2004/004(E2-2004-4).pdf\n\nThese authors do not discuss much the philosophical implications\nof this result, but if they did they would arrive to the same\nconclusions that the space-time unification and the\nLorentz transformations are approximate concepts.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:

>
>>>>I have shown that the space-time description does not follow
>>>>from Einstein's postulates, and is, in fact, an additional hypothesis.=
>>>
>>>>Everything becomes simple and consistent, if this hypothesis is
>>>>dropped.
>>>
>>>What??? I haven't seen anything simple in your approach.
>>>Your Hamiltonian power series expansion is much more messy
>>>than covariant QED...
>>
>>The simplicity is not in short expressions for the Hamiltonian and
>>other quantities. The simplicity is in adhering to well-tested
>>physical postulates, in clear physical meaning of all theoretical
>>ingredients, and in the absence of logical contradictions.
>
>
> Have you _anyone_ besides yourself convinced of that?
>

There are two papers (preprints) which confirm my results on
the decay law of moving particles:

L. A. Khalfin, Quantum theory of unstable particles and
relativity,
Preprint of Steklov Mathematical Institute, St. Petersburg Department,
PDMI-6/1997 (1997). http://www.pdmi.ras.ru/preprint/1997/97-06.html

M.I. Shirokov, Decay law of moving unstable
particle, Preprint JINR, E2-2004-4, Dubna (2004)
(to appear in \Int. J. Theor. Phys.)
http://www.jinr.ru/publish/Preprints/2004/004(E2-2004-4).pdf

These authors do not discuss much the philosophical implications
of this result, but if they did they would arrive to the same
conclusions that the space-time unification and the
Lorentz transformations are approximate concepts.

Eugene.

Arnold Neumaier

Dec3-04, 04:48 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>=20\n>>Eugene Stefanovich wrote:\n>>\n>>>Arnold Neumaier wrote:\n>>>\n>>>>But QED cares about the time-dependence of field observables, which i=\ns what\n>>>>one has in practice. Read about the closed time path formalism.\n>>>\n> I briefly looked at this theory, and apparently I need more time to\n> understand what is going on there. However, the complexity of this\n> approach made me suspicious. In a quantum theory time evolution should\n> be described simply by applying the time evolution operator exp(iHt)\n> to the\n> state vector. This simple picture is lacking in the "closed time path\n> formalism".=20\n\nexp(iHt) is applied by shifting all arguments in the correlation function=\ns\nby t. That there is no explicit rule for it doesn\'t mean that there is no=\n\nsuch operator.\n\n\n> Do you know if this approach can solve a simple time\n> evolution task, such as time dependence of the state of two interacting=\n\n> electrons?=20\n\nWhat\'s the use of such a time dependence? People do complicated calculati=\nons\nonly if something useful comes out. In the present case, reduced equation=\ns\nfor the dynamics of a macroscopic number of particles, an important\nproblem in plasma physics.\n\n\n> I found some applications to statistical systems with\n> temperature, but they do not impress me until the method is demonstrate=\nd\n> to work for simple models.\n\nThose people are not impressed unless a method works many-particle models=\n=2E\n\n\n>>>I do not agree. Any quantum theory should have a classical limit.\n>>\n>>Well, what is the classical limit of QED, in your opinion?\n>>How do you cope with particle creation and destruction in you classical=\n\n>>limit? You fail to discuss this in your book.\n>=20\n> In the classical limit of the "dressed particle" QED, there are\n> particles interacting with each other via (instantaneous) Coulomb and\n> magnetic forces. They are described by the terms of the type\n> a^{\\dag}a^{\\dag}aa. In addition, acceleration of charges causes emissio=\nn\n> of photons (bremsstrahlung). In the dressed particle Hamiltonian, this\n> effect is represented by the terms\n> a^{\\dag}a^{\\dag}c^{\\dag}aa. With these simplest contributions, the\n> classical limit of RQD should be very similar (though not identical)\n> to the Maxwell\'s theory.\n>=20\n> I agree that I should have written more about the classical limit of my=\n\n> approach.\n\nSince you write in the subjunctive, I suspect that you don\'t have studied=\n\nthe classical limit. Thus you should not demand more from your competitio=\nn.\nIt is trivial to write a classical limit in NRQED. But no one has found a=\n\nworking classical limit of QED that is covariant. I doubt that you have.\n\n\n>>>In this limit, trajectories are well defined.=20\n>>\n>>Why? It depends on the theory, and there is very little.\n>=20\n> See section 7.3 of the book\n\nI haven\'t seen there any covariant dynamical equation for classical\nparticles. There is a brief remark on top of p.167 that could count for i=\nt,\nbut lacking details nothing can be said. You\'d have to give an explicit\nclassical Hamiltonian on phase space and prove that the classical replace=\nment you get\nis compatible with the Poincare group and satisfies some sort of cluster\nseparability.\n\n\n>>>If the manifest covariance\n>>>is true, then these trajectories must transform by Lorentz (or, as yo=\nu\n>>> say, they must be observer independent).\n>>\n>>No. There must only be a well-defined recipe how the set of trajectorie=\ns\n>>of one observer translates into that of another. And this is what covar=\niance\n>>is about.\n>=20\n> The whole point of the "manifest covariance" is the this well-defined\n> recipe must be nothing else but linear Lorentz formulas. If you allow\n> other recipes (as I do), then you deviate from the strict special\n> relativity.\n\nManifest or not is irrelevant. What counts is only that there is a\ncompatible unitary representation of the Poincare group. This _is_\nspecial relativity.\n\n\n>>>When the unitary equivalence of different forms of dynamics is proved,=\n\n>>>the states are not subject to transformation.\n>>\n>>Of course the states must transform, too. You can see this already\n>>in the nonrelativistic case when you go from the Schroedinger represent=\nation\n>>to the interaction representation. Same physics, different states and\n>>observables, related by a unitary transformation.\n>=20\n> This is not correct. The connection between the instant and point forms=\n=20\n> of dynamics is not of the type as connection between the Schroedinger,\n> Heisenberg, and interaction representation. The definition of states\n> does not change when a unitary transformation between different\n> forms is made(see below).\n\nThe definition of states is independent of the representation but the\nstates themselves _must_ be transformed unitarily. And they are, in spite=\nof\nwhat you say below.\n\n\n>>>If you unitarily transform\n>>>both states and operators, then, of course, the physical content of\n>>>the theory does not change. That\'s rather trivial.\n>>\n>>This is also what happens in the equivalence of Shirikov\'s version\n>>with yours. You keep the vacuum, hee keeps the Hamiltonian; to relate t=\nhe\n>>two you must transform both.\n>=20\n> That\'s why I said that my approach and Shirokov\'s approach are triviall=\ny\n> identical. This is not the case with instant and point form dynamics.\n> The unitary transformations between them does not touch the states.\n> Otherwise the proof of the Sokolov\'s theorem could be done in just one\n> sentence.=20\n\nThe only nontrivial thing in the equivalence is to exhibit the right\nunitary transform to do that. I haven\'t read Sokolov\'s proof,\nbut the classical version of this transform was exhibited by Bakamjian in=\n\nPhys. Rev. 121, 1849=961851 (1961),\nand the quantum version is similar.\n\n\n> 1. I am not trying to diminish successes of GR in explaining and\n> predicting various experimental facts.\n> 2. I believe that the same (or even better) predictions can be made\n> without the concept of the space-time curvature. Though, I do not\n> have a proof for that.\n\nWhich makes you incredible. Statements of that weight should not be made\nwithout proof.\n\n\n> 3. The reason for my belief is that unified spacetime is in\n> contradiction with quantum mechanics, and was never properly\n> justified in special relativity. GR is just using the same\n> unproven concept.\n\nNo. There exists relativistic quantum field theory at least as\nconsistent as your theory. The difficulty is not relativity but\nintegrating curvature into this picture.\n\n\n> 4. The agreement with experiment is a great plus, but any consistent\n> theory should be also free of internal contradictions.\n> This is not the case with SR and GR.\n\nThere are no internal contradictions between quantum field theory\nand special relativity. And it is not even clear that there are any\nwith general relativity. See the topic\n\'Is quantum mechanics compatible with general relativity?\'\nin my theoretical physics FAQ at\nhttp://www.mat.univie.ac.at/~neum/physics-faq.txt\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>=20
>>Eugene Stefanovich wrote:
>>
>>>Arnold Neumaier wrote:
>>>
>>>>But QED cares about the time-dependence of field observables, which i=
s what
>>>>one has in practice. Read about the closed time path formalism.
>>>
> I briefly looked at this theory, and apparently I need more time to
> understand what is going on there. However, the complexity of this
> approach made me suspicious. In a quantum theory time evolution should
> be described simply by applying the time evolution operator \exp(iHt)
> to the
> state vector. This simple picture is lacking in the "closed time path
> formalism".=20

\exp(iHt) is applied by shifting all arguments in the correlation function=
s
by t. That there is no explicit rule for it doesn't mean that there is no=

such operator.

> Do you know if this approach can solve a simple time
> evolution task, such as time dependence of the state of two interacting=

> electrons?=20

What's the use of such a time dependence? People do complicated calculati=
ons
only if something useful comes out. In the present case, reduced equation=
s
for the dynamics of a macroscopic number of particles, an important
problem in plasma physics.

> I found some applications to statistical systems with
> temperature, but they do not impress me until the method is demonstrate=
d
> to work for simple models.

Those people are not impressed unless a method works many-particle models=
=2E

>>>I do not agree. Any quantum theory should have a classical limit.
>>
>>Well, what is the classical limit of QED, in your opinion?
>>How do you cope with particle creation and destruction in you classical=

>>limit? You fail to discuss this in your book.
>=20
> In the classical limit of the "dressed particle" QED, there are
> particles interacting with each other via (instantaneous) Coulomb and
> magnetic forces. They are described by the terms of the type
> a^{\dag}a^{\dag}aa. In addition, acceleration of charges causes emissio=
n
> of photons (bremsstrahlung). In the dressed particle Hamiltonian, this
> effect is represented by the terms
> a^{\dag}a^{\dag}c^{\dag}aa. With these simplest contributions, the
> classical limit of RQD should be very similar (though not identical)
> to the Maxwell's theory.
>=20
> I agree that I should have written more about the classical limit of my=

> approach.

Since you write in the subjunctive, I suspect that you don't have studied=

the classical limit. Thus you should not demand more from your competitio=
n.
It is trivial to write a classical limit in NRQED. But no one has found a=

working classical limit of QED that is covariant. I doubt that you have.

>>>In this limit, trajectories are well defined.=20
>>
>>Why? It depends on the theory, and there is very little.
>=20
> See section 7.3 of the book

I haven't seen there any covariant dynamical equation for classical
particles. There is a brief remark on top of p.167 that could count for i=
t,
but lacking details nothing can be said. You'd have to give an explicit
classical Hamiltonian on phase space and prove that the classical replace=
ment you get
is compatible with the Poincare group and satisfies some sort of cluster
separability.

>>>If the manifest covariance
>>>is true, then these trajectories must transform by Lorentz (or, as yo=
u
>>> say, they must be observer independent).
>>
>>No. There must only be a well-defined recipe how the set of trajectorie=
s
>>of one observer translates into that of another. And this is what covar=
iance
>>is about.
>=20
> The whole point of the "manifest covariance" is the this well-defined
> recipe must be nothing else but linear Lorentz formulas. If you allow
> other recipes (as I do), then you deviate from the strict special
> relativity.

Manifest or not is irrelevant. What counts is only that there is a
compatible unitary representation of the Poincare group. This _is_
special relativity.

>>>When the unitary equivalence of different forms of dynamics is proved,=

>>>the states are not subject to transformation.
>>
>>Of course the states must transform, too. You can see this already
>>in the nonrelativistic case when you go from the Schroedinger represent=
ation
>>to the interaction representation. Same physics, different states and
>>observables, related by a unitary transformation.
>=20
> This is not correct. The connection between the instant and point forms=
=20
> of dynamics is not of the type as connection between the Schroedinger,
> Heisenberg, and interaction representation. The definition of states
> does not change when a unitary transformation between different
> forms is made(see below).

The definition of states is independent of the representation but the
states themselves _must_ be transformed unitarily. And they are, in spite=
of
what you say below.

>>>If you unitarily transform
>>>both states and operators, then, of course, the physical content of
>>>the theory does not change. That's rather trivial.
>>
>>This is also what happens in the equivalence of Shirikov's version
>>with yours. You keep the vacuum, hee keeps the Hamiltonian; to relate t=
he
>>two you must transform both.
>=20
> That's why I said that my approach and Shirokov's approach are triviall=
y
> identical. This is not the case with instant and point form dynamics.
> The unitary transformations between them does not touch the states.
> Otherwise the proof of the Sokolov's theorem could be done in just one
> sentence.=20

The only nontrivial thing in the equivalence is to exhibit the right
unitary transform to do that. I haven't read Sokolov's proof,
but the classical version of this transform was exhibited by Bakamjian in=

Phys. Rev. 121, 1849=961851 (1961),
and the quantum version is similar.

> 1. I am not trying to diminish successes of GR in explaining and
> predicting various experimental facts.
> 2. I believe that the same (or even better) predictions can be made
> without the concept of the space-time curvature. Though, I do not
> have a proof for that.

Which makes you incredible. Statements of that weight should not be made
without proof.

> 3. The reason for my belief is that unified spacetime is in
> contradiction with quantum mechanics, and was never properly
> justified in special relativity. GR is just using the same
> unproven concept.

No. There exists relativistic quantum field theory at least as
consistent as your theory. The difficulty is not relativity but
integrating curvature into this picture.

> 4. The agreement with experiment is a great plus, but any consistent
> theory should be also free of internal contradictions.
> This is not the case with SR and GR.

There are no internal contradictions between quantum field theory
and special relativity. And it is not even clear that there are any
with general relativity. See the topic
'Is quantum mechanics compatible with general relativity?'
in my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt

Arnold Neumaier

Arnold Neumaier

Dec3-04, 04:48 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>Eugene Stefanovich wrote:\n>\n> There are two papers (preprints) which confirm my results on\n> the decay law of moving particles:\n>\n> L. A. Khalfin, Quantum theory of unstable particles and\n> relativity,\n> Preprint of Steklov Mathematical Institute, St. Petersburg Department,\n> PDMI-6/1997 (1997). http://www.pdmi.ras.ru/preprint/1997/97-06.html\n\nHe argues on the basis of classical physics and an assumed exponential\nlaw (3). It is clear that (3) cannot be exact since it is ridiculous\nfar in the past.\n\nI have seen papers on corrections calculated from QM,\nthough I don\'t remember where. To get exponential decay, one needs to\nneglect memory, a typical approximation step in statistical mechanics.\nOnce one considers corrections induced by the memory, deviations from\nthe exponential behavior appear. This is already the case in a\nnonrelativistic treatment of decay, hence says nothing about Minkowski\nspace.\n\nIn any case, such deviations do not need your theory\nto be explained, as this paper shows. He doesn\'t even mention you.\n\n\n> M.I. Shirokov, Decay law of moving unstable\n> particle, Preprint JINR, E2-2004-4, Dubna (2004)\n> (to appear in Int. J. Theor. Phys.)\n> http://www.jinr.ru/publish/Preprints/2004/004(E2-2004-4).pdf\n\nThis paper, although quoting you, derives a deviation within the\ntraditional relativistic framework. Thus there is no need to\nrenounce Minkowski space to accomodate that.\n\n\n> These authors do not discuss much the philosophical implications\n> of this result,\n\nThis just means that they are wiser than you, and know how easy it\nis to draw unfounded philosophical consequences from some formuals.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>Eugene Stefanovich wrote:
>
> There are two papers (preprints) which confirm my results on
> the decay law of moving particles:
>
> L. A. Khalfin, Quantum theory of unstable particles and
> relativity,
> Preprint of Steklov Mathematical Institute, St. Petersburg Department,
> PDMI-6/1997 (1997). http://www.pdmi.ras.ru/preprint/1997/97-06.html

He argues on the basis of classical physics and an assumed exponential
law (3). It is clear that (3) cannot be exact since it is ridiculous
far in the past.

I have seen papers on corrections calculated from QM,
though I don't remember where. To get exponential decay, one needs to
neglect memory, a typical approximation step in statistical mechanics.
Once one considers corrections induced by the memory, deviations from
the exponential behavior appear. This is already the case in a
nonrelativistic treatment of decay, hence says nothing about Minkowski
space.

In any case, such deviations do not need your theory
to be explained, as this paper shows. He doesn't even mention you.

> M.I. Shirokov, Decay law of moving unstable
> particle, Preprint JINR, E2-2004-4, Dubna (2004)
> (to appear in \Int. J. Theor. Phys.)
> http://www.jinr.ru/publish/Preprints/2004/004(E2-2004-4).pdf

This paper, although quoting you, derives a deviation within the
traditional relativistic framework. Thus there is no need to
renounce Minkowski space to accomodate that.

> These authors do not discuss much the philosophical implications
> of this result,

This just means that they are wiser than you, and know how easy it
is to draw unfounded philosophical consequences from some formuals.

Arnold Neumaier

Eugene Stefanovich

Dec4-04, 03:14 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n\n>\n>>>>I do not agree. Any quantum theory should have a classical limit.\n>>>\n>>>Well, what is the classical limit of QED, in your opinion?\n>>>How do you cope with particle creation and destruction in you classical=\n>>\n>\n>>>limit? You fail to discuss this in your book.\n>>\n>>=20\n>>In the classical limit of the "dressed particle" QED, there are\n>>particles interacting with each other via (instantaneous) Coulomb and\n>>magnetic forces. They are described by the terms of the type\n>>a^{\\dag}a^{\\dag}aa. In addition, acceleration of charges causes emissio=\n>\n> n\n>\n>>of photons (bremsstrahlung). In the dressed particle Hamiltonian, this\n>>effect is represented by the terms\n>>a^{\\dag}a^{\\dag}c^{\\dag}aa. With these simplest contributions, the\n>>classical limit of RQD should be very similar (though not identical)\n>>to the Maxwell\'s theory.\n>>=20\n>>I agree that I should have written more about the classical limit of my=\n>\n>\n>>approach.\n>\n>\n> Since you write in the subjunctive, I suspect that you don\'t have studied=\n>\n> the classical limit. Thus you should not demand more from your competitio=\n> n.\n> It is trivial to write a classical limit in NRQED. But no one has found a=\n>\n> working classical limit of QED that is covariant. I doubt that you have.\n\nNo, I haven\'t. However, I see clearly how to do it if light emission by\naccelerated charges is neglected: I have ready expressions for\ninterparticle instantaneous potentials which can be used directly in\nclassical equations of motion.\n\nThe situation with light emission and\nabsorption is trickier. In my approach, light is a flow of particles -\nphotons, however for photons quantum effects (diffraction,\ninterference, etc.) are apparent even in the macroscopic classical\nworld. So, description of photons as classical point particles is not\ngood. Even in the classical limit, photons should be treated in the\nwave picture.\n\nI need to think more about that, but I feel that some mixed\nparticle-wave "classical" limit of the "dressed particle" theory\nis possible. This limit should be close to Maxwell\'s theory, but\nnot identical. Some of the differences are:\nthe photon field has no direct connection with interparticle\nCoulomb and magnetic potentials, magnetic forces are not described\nby the Biot-Savart law but rather by the Darwin potential,\nthe interaction between charged particles is instantaneous...\n\n\n\n>\n>\n>\n>>>>In this limit, trajectories are well defined.=20\n>>>\n>>>Why? It depends on the theory, and there is very little.\n>>\n>>=20\n>>See section 7.3 of the book\n>\n>\n> I haven\'t seen there any covariant dynamical equation for classical\n> particles. There is a brief remark on top of p.167 that could count for i=\n> t,\n> but lacking details nothing can be said. You\'d have to give an explicit\n> classical Hamiltonian on phase space and prove that the classical replace=\n> ment you get\n> is compatible with the Poincare group and satisfies some sort of cluster\n> separability.\n\nI have quantum Hamiltonian satisfying relativistic invariance.\nI started to write it down in subsection 12.2.3.\nIf we assume that I can continue to write the terms of higher orders,\nand that the series converge, then the classical Hamiltonian\ncan be obtained from its quantum counterpart\nby simply neglecting the non-commutativity of r and p.\n\nAs I said above, the tricky part is how to interpret\nterms responsible for light emission and absorption (bremsstrahlung) in\nthe classical\nlimit. Photons cannot be described accurately as classical particles\nmoving along trajectories.\n\n>\n>\n>\n>>>>If the manifest covariance\n>>>>is true, then these trajectories must transform by Lorentz (or, as yo=\n>>>\n> u\n>\n>>>>say, they must be observer independent).\n>>>\n>>>No. There must only be a well-defined recipe how the set of trajectorie=\n>>\n> s\n>\n>>>of one observer translates into that of another. And this is what covar=\n>>\n> iance\n>\n>>>is about.\n>>\n>>=20\n>>The whole point of the "manifest covariance" is the this well-defined\n>>recipe must be nothing else but linear Lorentz formulas. If you allow\n>>other recipes (as I do), then you deviate from the strict special\n>>relativity.\n>\n>\n> Manifest or not is irrelevant. What counts is only that there is a\n> compatible unitary representation of the Poincare group. This _is_\n> special relativity.\n\nGreat! That\'s what I was saying all along. Poincare invariance - yes;\nmanifest covariance - no. However, to be fair, I wouldn\'t call "special\nrelativity" a theory which does not respect manifest covariance. It is\nrelativity, but neither "special" nor "general". Universal Lorentz\ntransformations, Minkowski spacetime, and manifest covariance are\ningredients of the special relativity, which are not present in my\n(still relativistic) approach.\n\n\n>\n>\n>\n>>>>When the unitary equivalence of different forms of dynamics is proved,=\n>>>\n>\n>>>>the states are not subject to transformation.\n>>>\n>>>Of course the states must transform, too. You can see this already\n>>>in the nonrelativistic case when you go from the Schroedinger represent=\n>>\n> ation\n>\n>>>to the interaction representation. Same physics, different states and\n>>>observables, related by a unitary transformation.\n>>\n>>=20\n>>This is not correct. The connection between the instant and point forms=\n>\n> =20\n>\n>>of dynamics is not of the type as connection between the Schroedinger,\n>>Heisenberg, and interaction representation. The definition of states\n>>does not change when a unitary transformation between different\n>>forms is made(see below).\n>\n>\n> The definition of states is independent of the representation but the\n> states themselves _must_ be transformed unitarily. And they are, in spite=\n> of\n> what you say below.\n>\n>\n>\n>>>>If you unitarily transform\n>>>>both states and operators, then, of course, the physical content of\n>>>>the theory does not change. That\'s rather trivial.\n>>>\n>>>This is also what happens in the equivalence of Shirikov\'s version\n>>>with yours. You keep the vacuum, hee keeps the Hamiltonian; to relate t=\n>>\n> he\n>\n>>>two you must transform both.\n>>\n>>=20\n>>That\'s why I said that my approach and Shirokov\'s approach are triviall=\n>\n> y\n>\n>>identical. This is not the case with instant and point form dynamics.\n>>The unitary transformations between them does not touch the states.\n>>Otherwise the proof of the Sokolov\'s theorem could be done in just one\n>>sentence.=20\n>\n>\n> The only nontrivial thing in the equivalence is to exhibit the right\n> unitary transform to do that. I haven\'t read Sokolov\'s proof,\n> but the classical version of this transform was exhibited by Bakamjian in=\n>\n> Phys. Rev. 121, 1849=961851 (1961),\n> and the quantum version is similar.\n\nThe Sokolov\'s proof is not much different from the Bakamjian\'s version.\nThe significant addition made by Sokolov is demonstration that the\nunitary operator connecting the generators in the two forms satisfies\nCoester\'s condition which is necessary for the preservation of the\nS-matrix.\n\nYou can find Sokolov\'s arguments in\n\nS.N.Sokolov, "Physical equivalence of the point and instantaneous\nforms of relativistic dynamics", Theor. Math. Phys. 24 (1975), 799\n\nS.N.Sokolov, A.N.Shatnii, "Physical equivalence of the three forms\nof relativistic dynamics and addition of interactions in the front and\ninstant forms", Theor. Math. Phys. 37 (1978), 1029.\n\nYou may notice that neither Bakamjian nor Sokolov apply the\nunitary transformation to the observables of individual particles.\nThe definitions of particles and their free states do not change\nby this transformation. Only 10 generators of the Poincare group\nare transformed.\n\nBy rereading Sokolov\'s papers I found a couple of sentences which\nexplain how he\nunderstood the "physical equivalence" and vindicate my point of view.\nThese are the first two sentences in the 1978 paper:\n\n"If one adopts the point of view, first expressed by Heisenberg, that\nall experimental information about the physical world is ultimately\ndeduced from scattering experiments and reduces to knowledge of\ncertain elements of the scattering matrix (or the analogous classical\nquantity), then different dynamical theories which lead to the same S\nmatrix must be regarded as physically equivalent. It is a simple\nmatter to construct both nonrelativistic and relativistic examples of\ndifferent Hamiltonians that give the same S matrix."\n\nI do not adopt Heisenberg\'s point of view. For me, different\nHamiltonians and forms of dynamics are physically distinguishable,\neven though they give the same S matrix, and even though\nexperimental observations\nof time dynamics are extremely difficult.\n\n[...]\n>\n>\n>>4. The agreement with experiment is a great plus, but any consistent\n>>theory should be also free of internal contradictions.\n>>This is not the case with SR and GR.\n>\n>\n> There are no internal contradictions between quantum field theory\n> and special relativity.\n\nIf you define special relativity as a theory requiring the manifest\ncovariance of all physical laws and if you cast QFT in the form\nof Hamiltonian dynamical theory (as I did) the contradiction\nbecomes apparent - this is the CJS theorem which says that\ntrajectories (in the classical limit, of course) do not transform\ncovariantly.\n\nHowever, if you drop the requirement of manifest covariance and\nrequire only Poincare invariance (as I did), then the contradiction\ndisappears. That\'s the main point of my book.\n\nThe contradiction does not show up also if you keep the manifest\ncovariance and pay attention only to the S matrix in QFT. That\'s\nthe way QFT is currently presented (I leave aside the time dependent\napproaches you mentioned in previous posts. I haven\'t understood\nthese approaches yet.) In this formulation, QFT is not a complete\ntheory.\n\n> And it is not even clear that there are any\n> with general relativity. See the topic\n> \'Is quantum mechanics compatible with general relativity?\'\n> in my theoretical physics FAQ at\n> http://www.mat.univie.ac.at/~neum/physics-faq.txt\n\nI think the main contradiction is the following: General relativity\n(just as the special relativity) implies that position and time\nare equivalent quantities (different components of the same 4-vector).\nOn the other hand,\nin quantum mechanics, position and time play very different roles.\nPosition is an observable (Hermitian operator), and time is just a\nclassical\nparameter labeling different reference frames. This controversy was\ndiscussed from different points of view by Isham in gr-qc/9210011.\nPersonally, I don\'t see any prospect of resolving this problem unless\nthe GR approach to gravity (the curvature of the 4D spacetime) is\nabandoned.\n\nEugene Stefanovich.\n\n\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:

>
>>>>I do not agree. Any quantum theory should have a classical limit.
>>>
>>>Well, what is the classical limit of QED, in your opinion?
>>>How do you cope with particle creation and destruction in you classical=
>>
>
>>>limit? You fail to discuss this in your book.
>>
>>=20
>>In the classical limit of the "dressed particle" QED, there are
>>particles interacting with each other via (instantaneous) Coulomb and
>>magnetic forces. They are described by the terms of the type
>>a^{\dag}a^{\dag}aa. In addition, acceleration of charges causes emissio=
>
> n
>
>>of photons (bremsstrahlung). In the dressed particle Hamiltonian, this
>>effect is represented by the terms
>>a^{\dag}a^{\dag}c^{\dag}aa. With these simplest contributions, the
>>classical limit of RQD should be very similar (though not identical)
>>to the Maxwell's theory.
>>=20
>>I agree that I should have written more about the classical limit of my=
>
>
>>approach.
>
>
> Since you write in the subjunctive, I suspect that you don't have studied=
>
> the classical limit. Thus you should not demand more from your competitio=
> n.
> It is trivial to write a classical limit in NRQED. But no one has found a=
>
> working classical limit of QED that is covariant. I doubt that you have.

No, I haven't. However, I see clearly how to do it if light emission by
accelerated charges is neglected: I have ready expressions for
interparticle instantaneous potentials which can be used directly in
classical equations of motion.

The situation with light emission and
absorption is trickier. In my approach, light is a flow of particles -
photons, however for photons quantum effects (diffraction,
interference, etc.) are apparent even in the macroscopic classical
world. So, description of photons as classical point particles is not
good. Even in the classical limit, photons should be treated in the
wave picture.

I need to think more about that, but I feel that some mixed
particle-wave "classical" limit of the "dressed particle" theory
is possible. This limit should be close to Maxwell's theory, but
not identical. Some of the differences are:
the photon field has no direct connection with interparticle
Coulomb and magnetic potentials, magnetic forces are not described
by the Biot-Savart law but rather by the Darwin potential,
the interaction between charged particles is instantaneous...

>
>
>
>>>>In this limit, trajectories are well defined.=20
>>>
>>>Why? It depends on the theory, and there is very little.
>>
>>=20
>>See section 7.3 of the book
>
>
> I haven't seen there any covariant dynamical equation for classical
> particles. There is a brief remark on top of p.167 that could count for i=
> t,
> but lacking details nothing can be said. You'd have to give an explicit
> classical Hamiltonian on phase space and prove that the classical replace=
> ment you get
> is compatible with the Poincare group and satisfies some sort of cluster
> separability.

I have quantum Hamiltonian satisfying relativistic invariance.
I started to write it down in subsection 12.2.3.
If we assume that I can continue to write the terms of higher orders,
and that the series converge, then the classical Hamiltonian
can be obtained from its quantum counterpart
by simply neglecting the non-commutativity of r and p.

As I said above, the tricky part is how to interpret
terms responsible for light emission and absorption (bremsstrahlung) in
the classical
limit. Photons cannot be described accurately as classical particles
moving along trajectories.

>
>
>
>>>>If the manifest covariance
>>>>is true, then these trajectories must transform by Lorentz (or, as yo=
>>>
> u
>
>>>>say, they must be observer independent).
>>>
>>>No. There must only be a well-defined recipe how the set of trajectorie=
>>
> s
>
>>>of one observer translates into that of another. And this is what covar=
>>
> iance
>
>>>is about.
>>
>>=20
>>The whole point of the "manifest covariance" is the this well-defined
>>recipe must be nothing else but linear Lorentz formulas. If you allow
>>other recipes (as I do), then you deviate from the strict special
>>relativity.
>
>
> Manifest or not is irrelevant. What counts is only that there is a
> compatible unitary representation of the Poincare group. This _is_
> special relativity.

Great! That's what I was saying all along. Poincare invariance - yes;
manifest covariance - no. However, to be fair, I wouldn't call "special
relativity" a theory which does not respect manifest covariance. It is
relativity, but neither "special" nor "general". Universal Lorentz
transformations, Minkowski spacetime, and manifest covariance are
ingredients of the special relativity, which are not present in my
(still relativistic) approach.

>
>
>
>>>>When the unitary equivalence of different forms of dynamics is proved,=
>>>
>
>>>>the states are not subject to transformation.
>>>
>>>Of course the states must transform, too. You can see this already
>>>in the nonrelativistic case when you go from the Schroedinger represent=
>>
> ation
>
>>>to the interaction representation. Same physics, different states and
>>>observables, related by a unitary transformation.
>>
>>=20
>>This is not correct. The connection between the instant and point forms=
>
> =20
>
>>of dynamics is not of the type as connection between the Schroedinger,
>>Heisenberg, and interaction representation. The definition of states
>>does not change when a unitary transformation between different
>>forms is made(see below).
>
>
> The definition of states is independent of the representation but the
> states themselves _must_ be transformed unitarily. And they are, in spite=
> of
> what you say below.
>
>
>
>>>>If you unitarily transform
>>>>both states and operators, then, of course, the physical content of
>>>>the theory does not change. That's rather trivial.
>>>
>>>This is also what happens in the equivalence of Shirikov's version
>>>with yours. You keep the vacuum, hee keeps the Hamiltonian; to relate t=
>>
> he
>
>>>two you must transform both.
>>
>>=20
>>That's why I said that my approach and Shirokov's approach are triviall=
>
> y
>
>>identical. This is not the case with instant and point form dynamics.
>>The unitary transformations between them does not touch the states.
>>Otherwise the proof of the Sokolov's theorem could be done in just one
>>sentence.=20
>
>
> The only nontrivial thing in the equivalence is to exhibit the right
> unitary transform to do that. I haven't read Sokolov's proof,
> but the classical version of this transform was exhibited by Bakamjian in=
>
> Phys. Rev. 121, 1849=961851 (1961),
> and the quantum version is similar.

The Sokolov's proof is not much different from the Bakamjian's version.
The significant addition made by Sokolov is demonstration that the
unitary operator connecting the generators in the two forms satisfies
Coester's condition which is necessary for the preservation of the
S-matrix.

You can find Sokolov's arguments in

S.N.Sokolov, "Physical equivalence of the point and instantaneous
forms of relativistic dynamics", Theor. Math. Phys. 24 (1975), 799

S.N.Sokolov, A.N.Shatnii, "Physical equivalence of the three forms
of relativistic dynamics and addition of interactions in the front and
instant forms", Theor. Math. Phys. 37 (1978), 1029.

You may notice that neither Bakamjian nor Sokolov apply the
unitary transformation to the observables of individual particles.
The definitions of particles and their free states do not change
by this transformation. Only 10 generators of the Poincare group
are transformed.

By rereading Sokolov's papers I found a couple of sentences which
explain how he
understood the "physical equivalence" and vindicate my point of view.
These are the first two sentences in the 1978 paper:

"If one adopts the point of view, first expressed by Heisenberg, that
all experimental information about the physical world is ultimately
deduced from scattering experiments and reduces to knowledge of
certain elements of the scattering matrix (or the analogous classical
quantity), then different dynamical theories which lead to the same S
matrix must be regarded as physically equivalent. It is a simple
matter to construct both nonrelativistic and relativistic examples of
different Hamiltonians that give the same S matrix."

I do not adopt Heisenberg's point of view. For me, different
Hamiltonians and forms of dynamics are physically distinguishable,
even though they give the same S matrix, and even though
experimental observations
of time dynamics are extremely difficult.

[...]
>
>
>>4. The agreement with experiment is a great plus, but any consistent
>>theory should be also free of internal contradictions.
>>This is not the case with SR and GR.
>
>
> There are no internal contradictions between quantum field theory
> and special relativity.

If you define special relativity as a theory requiring the manifest
covariance of all physical laws and if you cast QFT in the form
of Hamiltonian dynamical theory (as I did) the contradiction
becomes apparent - this is the CJS theorem which says that
trajectories (in the classical limit, of course) do not transform
covariantly.

However, if you drop the requirement of manifest covariance and
require only Poincare invariance (as I did), then the contradiction
disappears. That's the main point of my book.

The contradiction does not show up also if you keep the manifest
covariance and pay attention only to the S matrix in QFT. That's
the way QFT is currently presented (I leave aside the time dependent
approaches you mentioned in previous posts. I haven't understood
these approaches yet.) In this formulation, QFT is not a complete
theory.

> And it is not even clear that there are any
> with general relativity. See the topic
> 'Is quantum mechanics compatible with general relativity?'
> in my theoretical physics FAQ at
> http://www.mat.univie.ac.at/~neum/physics-faq.txt

I think the main contradiction is the following: General relativity
(just as the special relativity) implies that position and time
are equivalent quantities (different components of the same 4-vector).
On the other hand,
in quantum mechanics, position and time play very different roles.
Position is an observable (Hermitian operator), and time is just a
classical
parameter labeling different reference frames. This controversy was
discussed from different points of view by Isham in http://www.arxiv.org/abs/gr-qc/9210011.
Personally, I don't see any prospect of resolving this problem unless
the GR approach to gravity (the curvature of the 4D spacetime) is
abandoned.

Eugene Stefanovich.

>
>
> Arnold Neumaier
>

Eugene Stefanovich

Dec4-04, 03:16 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>Eugene Stefanovich wrote:\n>>\n>>There are two papers (preprints) which confirm my results on\n>>the decay law of moving particles:\n>>\n>> L. A. Khalfin, Quantum theory of unstable particles and\n>>relativity,\n>>Preprint of Steklov Mathematical Institute, St. Petersburg Department,\n>>PDMI-6/1997 (1997). http://www.pdmi.ras.ru/preprint/1997/97-06.html\n>\n>\n> He argues on the basis of classical physics and an assumed exponential\n> law (3). It is clear that (3) cannot be exact since it is ridiculous\n> far in the past.\n\nYou probably missed the point. The exponential character of decay is\nknown to be approximate, and the argument of the paper does not use\nthe exponentiality of the decay in any substantial way. The point is\nthat Einstein\'s theory demands that if the decay law of a particle at\nrest is \\omega(t), then the decay law of the moving particle is\n\\omega(t/\\gamma), i.e. \\gamma times slower. This is a universal result\nof special relativity which does not depend on the particular form\n(exponential or not) of the function \\omega.\n\nKhalfin shows that this universal result does not hold. From his paper\nit is not immediately clear why special relativity becomes violated.\nThis is explained in my paper\n\nE. V. Stefanovich, Quantum effects in\nrelativistic decays, Int. J. Theor. Phys. 35 (1996), 2539.\n\nand in section 13.2 of the book as the result of the presence of\ninteraction terms in the boost operator, i.e., the dynamic character of\nboosts. Note that his eq. (8) is the same as my eq. (13.28) in the book.\n\n\n\n>\n> I have seen papers on corrections calculated from QM,\n> though I don\'t remember where. To get exponential decay, one needs to\n> neglect memory, a typical approximation step in statistical mechanics.\n> Once one considers corrections induced by the memory, deviations from\n> the exponential behavior appear. This is already the case in a\n> nonrelativistic treatment of decay, hence says nothing about Minkowski\n> space.\n>\n> In any case, such deviations do not need your theory\n> to be explained, as this paper shows. He doesn\'t even mention you.\n\nApparently, he was not aware of my 1996 paper. I didn\'t know Khalfin\npersonally.\n\n>\n>\n>\n>>M.I. Shirokov, Decay law of moving unstable\n>>particle, Preprint JINR, E2-2004-4, Dubna (2004)\n>>(to appear in Int. J. Theor. Phys.)\n>>http://www.jinr.ru/publish/Preprints/2004/004(E2-2004-4).pdf\n>\n>\n> This paper, although quoting you, derives a deviation within the\n> traditional relativistic framework. Thus there is no need to\n> renounce Minkowski space to accomodate that.\n\nAgain, Shirokov skips some important steps in derivation.\nSee, e.g., after eq. (4) "I suppose that c(u) in eq. (4) does\nnot depend on p\'". A consistent derivation of his formula\n(9), which is equivalent to Khalfin\'s eq. (8) and my eq. (13.28)\nrequires working in the instant form dynamics where boost\noperator has interaction terms, i.e., is dynamical. This is\nreflected in the "decays caused by boost" (see subsection 13.2.1\nin my book) which entirely contradicts the manifest covariance.\n\n>\n>\n>\n>>These authors do not discuss much the philosophical implications\n>>of this result,\n>\n>\n> This just means that they are wiser than you, and know how easy it\n> is to draw unfounded philosophical consequences from some formuals.\n\nThis is not "some formula", this is important Einstein\'s time\ndilation formula\nwhich is assumed to be valid exactly in special relativity.\nIf you allow this formula to\nbe approximate, then you should allow that Lorentz transformations\nmay be approximate as well, and that Minkowski spacetime picture\nis also just an approximation suitable only for weakly interacting\nsystems.\n\nEugene Stefanovich.\n\n\n>\n>\n> Arnold Neumaier\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>Eugene Stefanovich wrote:
>>
>>There are two papers (preprints) which confirm my results on
>>the decay law of moving particles:
>>
>> L. A. Khalfin, Quantum theory of unstable particles and
>>relativity,
>>Preprint of Steklov Mathematical Institute, St. Petersburg Department,
>>PDMI-6/1997 (1997). http://www.pdmi.ras.ru/preprint/1997/97-06.html
>
>
> He argues on the basis of classical physics and an assumed exponential
> law (3). It is clear that (3) cannot be exact since it is ridiculous
> far in the past.

You probably missed the point. The exponential character of decay is
known to be approximate, and the argument of the paper does not use
the exponentiality of the decay in any substantial way. The point is
that Einstein's theory demands that if the decay law of a particle at
rest is \omega(t), then the decay law of the moving particle is
\omega(t/\gamma), i.e. \gamma times slower. This is a universal result
of special relativity which does not depend on the particular form
(exponential or not) of the function \omega.

Khalfin shows that this universal result does not hold. From his paper
it is not immediately clear why special relativity becomes violated.
This is explained in my paper

E. V. Stefanovich, Quantum effects in
relativistic decays, \Int. J. Theor. Phys. 35 (1996), 2539.

and in section 13.2 of the book as the result of the presence of
interaction terms in the boost operator, i.e., the dynamic character of
boosts. Note that his eq. (8) is the same as my eq. (13.28) in the book.

>
> I have seen papers on corrections calculated from QM,
> though I don't remember where. To get exponential decay, one needs to
> neglect memory, a typical approximation step in statistical mechanics.
> Once one considers corrections induced by the memory, deviations from
> the exponential behavior appear. This is already the case in a
> nonrelativistic treatment of decay, hence says nothing about Minkowski
> space.
>
> In any case, such deviations do not need your theory
> to be explained, as this paper shows. He doesn't even mention you.

Apparently, he was not aware of my 1996 paper. I didn't know Khalfin
personally.

>
>
>
>>M.I. Shirokov, Decay law of moving unstable
>>particle, Preprint JINR, E2-2004-4, Dubna (2004)
>>(to appear in \Int. J. Theor. Phys.)
>>http://www.jinr.ru/publish/Preprints/2004/004(E2-2004-4).pdf
>
>
> This paper, although quoting you, derives a deviation within the
> traditional relativistic framework. Thus there is no need to
> renounce Minkowski space to accomodate that.

Again, Shirokov skips some important steps in derivation.
See, e.g., after eq. (4) "I suppose that c(u) in eq. (4) does
not depend on p'". A consistent derivation of his formula
(9), which is equivalent to Khalfin's eq. (8) and my eq. (13.28)
requires working in the instant form dynamics where boost
operator has interaction terms, i.e., is dynamical. This is
reflected in the "decays caused by boost" (see subsection 13.2.1
in my book) which entirely contradicts the manifest covariance.

>
>
>
>>These authors do not discuss much the philosophical implications
>>of this result,
>
>
> This just means that they are wiser than you, and know how easy it
> is to draw unfounded philosophical consequences from some formuals.

This is not "some formula", this is important Einstein's time
dilation formula
which is assumed to be valid exactly in special relativity.
If you allow this formula to
be approximate, then you should allow that Lorentz transformations
may be approximate as well, and that Minkowski spacetime picture
is also just an approximation suitable only for weakly interacting
systems.

Eugene Stefanovich.

>
>
> Arnold Neumaier
>

Eugene Stefanovich

Dec4-04, 03:16 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>Eugene Stefanovich wrote:\n>>\n>>There are two papers (preprints) which confirm my results on\n>>the decay law of moving particles:\n>>\n>> L. A. Khalfin, Quantum theory of unstable particles and\n>>relativity,\n>>Preprint of Steklov Mathematical Institute, St. Petersburg Department,\n>>PDMI-6/1997 (1997). http://www.pdmi.ras.ru/preprint/1997/97-06.html\n>\n>\n> He argues on the basis of classical physics and an assumed exponential\n> law (3). It is clear that (3) cannot be exact since it is ridiculous\n> far in the past.\n>\n> I have seen papers on corrections calculated from QM,\n> though I don\'t remember where. To get exponential decay, one needs to\n> neglect memory, a typical approximation step in statistical mechanics.\n> Once one considers corrections induced by the memory, deviations from\n> the exponential behavior appear. This is already the case in a\n> nonrelativistic treatment of decay, hence says nothing about Minkowski\n> space.\n>\n> In any case, such deviations do not need your theory\n> to be explained, as this paper shows. He doesn\'t even mention you.\n>\n>\n>\n>>M.I. Shirokov, Decay law of moving unstable\n>>particle, Preprint JINR, E2-2004-4, Dubna (2004)\n>>(to appear in Int. J. Theor. Phys.)\n>>http://www.jinr.ru/publish/Preprints/2004/004(E2-2004-4).pdf\n>\n>\n> This paper, although quoting you, derives a deviation within the\n> traditional relativistic framework. Thus there is no need to\n> renounce Minkowski space to accomodate that.\n>\n>\n>\n>>These authors do not discuss much the philosophical implications\n>>of this result,\n>\n>\n> This just means that they are wiser than you, and know how easy it\n> is to draw unfounded philosophical consequences from some formuals.\n>\n>\n> Arnold Neumaier\n\nThe formulas obtained by Khalfin, Shirokov, and myself for the decay law\nof moving unstable particles consistent with experiment\n(slowing of the decay) can be derived only in the instant form dynamics.\nIn the last eq. of subsection 13.2.1 I derive the analogous formula for\nthe point form dynamics which results in acceleration of the\ndecay of\nmoving particles in clear contradiction to experiment. This\n(in addition to arguments about the dynamical character of space\ntranslations) allows me to rule out the point form interactions. This\nalso demonstrates the physical inequivalence of different forms of\ndynamics which I discussed in other posts.\n\nEugene Stefanovich.\n\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>Eugene Stefanovich wrote:
>>
>>There are two papers (preprints) which confirm my results on
>>the decay law of moving particles:
>>
>> L. A. Khalfin, Quantum theory of unstable particles and
>>relativity,
>>Preprint of Steklov Mathematical Institute, St. Petersburg Department,
>>PDMI-6/1997 (1997). http://www.pdmi.ras.ru/preprint/1997/97-06.html
>
>
> He argues on the basis of classical physics and an assumed exponential
> law (3). It is clear that (3) cannot be exact since it is ridiculous
> far in the past.
>
> I have seen papers on corrections calculated from QM,
> though I don't remember where. To get exponential decay, one needs to
> neglect memory, a typical approximation step in statistical mechanics.
> Once one considers corrections induced by the memory, deviations from
> the exponential behavior appear. This is already the case in a
> nonrelativistic treatment of decay, hence says nothing about Minkowski
> space.
>
> In any case, such deviations do not need your theory
> to be explained, as this paper shows. He doesn't even mention you.
>
>
>
>>M.I. Shirokov, Decay law of moving unstable
>>particle, Preprint JINR, E2-2004-4, Dubna (2004)
>>(to appear in \Int. J. Theor. Phys.)
>>http://www.jinr.ru/publish/Preprints/2004/004(E2-2004-4).pdf
>
>
> This paper, although quoting you, derives a deviation within the
> traditional relativistic framework. Thus there is no need to
> renounce Minkowski space to accomodate that.
>
>
>
>>These authors do not discuss much the philosophical implications
>>of this result,
>
>
> This just means that they are wiser than you, and know how easy it
> is to draw unfounded philosophical consequences from some formuals.
>
>
> Arnold Neumaier

The formulas obtained by Khalfin, Shirokov, and myself for the decay law
of moving unstable particles consistent with experiment
(slowing of the decay) can be derived only in the instant form dynamics.
In the last eq. of subsection 13.2.1 I derive the analogous formula for
the point form dynamics which results in acceleration of the
decay of
moving particles in clear contradiction to experiment. This
(in addition to arguments about the dynamical character of space
translations) allows me to rule out the point form interactions. This
also demonstrates the physical inequivalence of different forms of
dynamics which I discussed in other posts.

Eugene Stefanovich.

>