Gauge Transformations in Momentum Space?

mikem@despammed.com

<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>Most textbook treatments of gauge transformations do it in\nposition space. So far, I haven\'t found any that discuss\nin detail what they look like in momentum space, and what\nissues arise in the QFT Fock space.\n\nFor example, consider the U(1) group for electromagnetism\nacting on Dirac electrons in QED. Textbooks write it as\nsomething like:\n\nPsi(x) -> exp(i theta(x)) Psi(x)\n\nwhere theta(x) is a real scalar function of x (in 3+1\nspacetime of course).\n\nTo pass to momentum space, we need to assume that\nexp(i theta(x)) has a reasonable Fourier transform,\nsuch that the transformation can be represented\nin momentum space as an integral operator whose\nkernel is a distribution.\n\nTake a simple case: theta(x) = wt, where \'w\' is a real constant.\nI.e: in position space we have\n\nPsi(x) -> exp(iwt) Psi(x)\n\nIn momentum space, this just shifts the energy by an amount \'w\'\nI.e: E -> E - w.\n\nSo old positive-energy modes in the energy range 0 to w get\ntransformed into negative-energy modes. In the 2nd-quantized Fock\nspace this means we\'re mixing some of the annihilation and creation\noperators - because they were defined in terms of the original +ve\nand -ve energy modes. Such mixing usually means that we\'re mapping\nbetween unitarily inequivalent representations, i.e: between\northogonal Fock spaces.\n\nI\'m interested in finding explicit operators which are form-invariant\nin both representations. I tried Google-Scholar but didn\'t have much\nsuccess.\n\nSo my question is:\n\nDo any textbooks or review papers discuss this stuff at length?\n(I don\'t mean just the usual Bogoliubov transformations from\ncondensed matter physics which map between inequivalent reps,\nbut specifically for standard model gauge transformations\nin momentum space, and hence Fock space(s).)\n\nTIA.\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>Most textbook treatments of gauge transformations do it in
position space. So far, I haven't found any that discuss
in detail what they look like in momentum space, and what
issues arise in the QFT Fock space.

For example, consider the U(1) group for electromagnetism
acting on Dirac electrons in QED. Textbooks write it as
something like:

[tex]\Psi(x) -> \exp(i \theta(x)) \Psi(x)[/tex]

where [itex]\theta(x)[/itex] is a real scalar function of x (in 3+1
spacetime of course).

To pass to momentum space, we need to assume that
[itex]\exp(i \theta(x))[/itex] has a reasonable Fourier transform,
such that the transformation can be represented
in momentum space as an integral operator whose
kernel is a distribution.

Take a simple case: [itex]\theta(x) =[/itex] wt, where 'w' is a real constant.
I.e: in position space we have

[tex]\Psi(x) -> \exp(iwt) \Psi(x)[/tex]

In momentum space, this just shifts the energy by an amount 'w'
I.e: [itex]E -> E - w[/itex].

So old positive-energy modes in the energy range to w get
transformed into negative-energy modes. In the 2nd-quantized Fock
space this means we're mixing some of the annihilation and creation
operators - because they were defined in terms of the original +ve
and -ve energy modes. Such mixing usually means that we're mapping
between unitarily inequivalent representations, i.e: between
orthogonal Fock spaces.

I'm interested in finding explicit operators which are form-invariant
in both representations. I tried Google-Scholar but didn't have much
success.

So my question is:

Do any textbooks or review papers discuss this stuff at length?
(I don't mean just the usual Bogoliubov transformations from
condensed matter physics which map between inequivalent reps,
but specifically for standard model gauge transformations
in momentum space, and hence Fock space(s).)

TIA.

Eugene Stefanovich

<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>mikem@despammed.com wrote:\n> Most textbook treatments of gauge transformations do it in\n> position space. So far, I haven\'t found any that discuss\n> in detail what they look like in momentum space, and what\n> issues arise in the QFT Fock space.\n>\n> For example, consider the U(1) group for electromagnetism\n> acting on Dirac electrons in QED. Textbooks write it as\n> something like:\n>\n> Psi(x) -> exp(i theta(x)) Psi(x)\n>\n> where theta(x) is a real scalar function of x (in 3+1\n> spacetime of course).\n>\n> To pass to momentum space, we need to assume that\n> exp(i theta(x)) has a reasonable Fourier transform,\n> such that the transformation can be represented\n> in momentum space as an integral operator whose\n> kernel is a distribution.\n>\n> Take a simple case: theta(x) = wt, where \'w\' is a real constant.\n> I.e: in position space we have\n>\n> Psi(x) -> exp(iwt) Psi(x)\n>\n> In momentum space, this just shifts the energy by an amount \'w\'\n> I.e: E -> E - w.\n>\n> So old positive-energy modes in the energy range 0 to w get\n> transformed into negative-energy modes. In the 2nd-quantized Fock\n> space this means we\'re mixing some of the annihilation and creation\n> operators - because they were defined in terms of the original +ve\n> and -ve energy modes. Such mixing usually means that we\'re mapping\n> between unitarily inequivalent representations, i.e: between\n> orthogonal Fock spaces.\n>\n> I\'m interested in finding explicit operators which are form-invariant\n> in both representations. I tried Google-Scholar but didn\'t have much\n> success.\n>\n> So my question is:\n>\n> Do any textbooks or review papers discuss this stuff at length?\n> (I don\'t mean just the usual Bogoliubov transformations from\n> condensed matter physics which map between inequivalent reps,\n> but specifically for standard model gauge transformations\n> in momentum space, and hence Fock space(s).)\n>\n> TIA.\n\nI think quantum fields and their gauge transformations in the "momentum"\nrepresentation have no meaning at all. You can switch between position\nand momentum representations of wave functions, but quantum fields are\ncompletely different beasts (see recent thread "Why no tensors in\nquantum mechanics?").\n\nThe only point to introduce quantum fields Psi(x,t) in QFT is to have\nconvenient "building blocks" for the interacting Hamiltonian.\nThe gauge invariance of Psi(x,t) is a heuristic aid in this\nconstruction. All this works only when (x,t) are coordinates on an\nabstract Minkowski space. Then you can explicitly ensure that\nPsi(x,t) transform by linear Lorentz formulas wrt the non-interacting\nrepresentation of the Poincare group and that Psi(x,t) (anti)commute\nat "space-like" separation. This, in turn, guarantees (see Weinberg,\nvol. 1) that the interaction operator in the Hamiltonian constructed\nfrom Psi(x,t) is relativistically invariant.\n\nI have no idea how and why would you want to use "momentum-space"\nPsi(p,t) for this purpose.\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>mikem@despammed.com wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> [itex]\Psi(x) -> \exp(i \theta(x)) \Psi(x)[/itex]
>
> where [itex]\theta(x)[/itex] is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> [itex]\exp(i \theta(x))[/itex] has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.
>
> Take a simple case: [itex]\theta(x) =[/itex] wt, where 'w' is a real constant.
> I.e: in position space we have
>
> [itex]\Psi(x) -> \exp(iwt) \Psi(x)[/itex]
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: [itex]E -> E - w[/itex].
>
> So old positive-energy modes in the energy range to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>
> I'm interested in finding explicit operators which are form-invariant
> in both representations. I tried Google-Scholar but didn't have much
> success.
>
> So my question is:
>
> Do any textbooks or review papers discuss this stuff at length?
> (I don't mean just the usual Bogoliubov transformations from
> condensed matter physics which map between inequivalent reps,
> but specifically for standard model gauge transformations
> in momentum space, and hence Fock space(s).)
>
> TIA.

I think quantum fields and their gauge transformations in the "momentum"
representation have no meaning at all. You can switch between position
and momentum representations of wave functions, but quantum fields are
completely different beasts (see recent thread "Why no tensors in
quantum mechanics?").

The only point to introduce quantum fields [itex]\Psi(x,t)[/itex] in QFT is to have
convenient "building blocks" for the interacting Hamiltonian.
The gauge invariance of [itex]\Psi(x,t)[/itex] is a heuristic aid in this
construction. All this works only when (x,t) are coordinates on an
abstract Minkowski space. Then you can explicitly ensure that
[itex]\Psi(x,t)[/itex] transform by linear Lorentz formulas wrt the non-interacting
representation of the Poincare group and that [itex]\Psi(x,t)[/itex] (anti)commute
at "space-like" separation. This, in turn, guarantees (see Weinberg,
vol. 1) that the interaction operator in the Hamiltonian constructed
from [itex]\Psi(x,t)[/itex] is relativistically invariant.

I have no idea how and why would you want to use "momentum-space"
[itex]\Psi(p,t)[/itex] for this purpose.

Eugene.

mikem@despammed.com

<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 think quantum fields and their gauge transformations\n> in the "momentum" representation have no meaning at all.\n> [...]\n\nCertainly they\'re not of much use in standard treatments of\nQFT and the standard model. But "no meaning at all" seems\na bit strong.\n\n\n> [...]\n> I have no idea how and why would you want to use\n> "momentum-space" Psi(p,t) for this purpose.\n\nI\'m trying to find out whether standard model gauge\ngroups (acting on fermions) correspond to\nBogoliubov transformations mapping between\ndisjoint Fock spaces, i.e: between unitarily\ninequivalent representations. The textbooks I\'m aware\nof give various calculation techniques using momentum-space\nannihilation/creation operators exclusively. So in the\nhope of leveraging those techniques I need to learn\nmore about gauge transformations in momentum space.\n\n\ntechniques for investigating such Bogoliubov transformations\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 think quantum fields and their gauge transformations
> in the "momentum" representation have no meaning at all.
> [...]

Certainly they're not of much use in standard treatments of
QFT and the standard model. But "no meaning at all" seems
a bit strong.

> [...]
> I have no idea how and why would you want to use
> "momentum-space[itex]" \Psi(p,t)[/itex] for this purpose.

I'm trying to find out whether standard model gauge
groups (acting on fermions) correspond to
Bogoliubov transformations mapping between
disjoint Fock spaces, i.e: between unitarily
inequivalent representations. The textbooks I'm aware
of give various calculation techniques using momentum-space
annihilation/creation operators exclusively. So in the
hope of leveraging those techniques I need to learn
more about gauge transformations in momentum space.

techniques for investigating such Bogoliubov transformations

mikem@despammed.com

<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> Unitarily inequivalent representations and disjoint Fock spaces\n> is something I couldn\'t understand for a long time. This looks\n> like infamous "parallel universes" to me.\n\nNot at all. They arise because the Fock space construction\nmust arbitrarily restrict to finite particle numbers in order for\nan inner product to exist. This is explained in Umezawa (below).\n\n\n> Could you give one good example where these things are\n> necessary for understanding real physical phenomena.\n\nI\'ll give you several...\n\nThey arise in condensed matter physics, i.e: bose-einstein\ncondensation, superfluidity and superconductivity. In textbooks\non these subjects, look for "Bogoliubov transformation" and\nin most cases you\'ll find UIRs lurking, although the textbooks\ndon\'t always bring out this point explicitly. Umezawa\'s text on\n"Thermofield Dynamics and Condensed States" gives the\nmost understandable presentation I\'ve seen so far.\n\nA different, more recent, area is neutrino oscillations.\nBlasone et al have shown that the Fock space of\ndefinite flavour states is unitarily inequivalent to\nthat definite mass states. See, for example,\nhep-ph/9501263, and also the review article by\nCapolupo: hep-th/0408228. This means that to\nunderstand the QFT of neutrino oscillations fully,\nwe need to understand UIRs and disjoint Fock\nspaces.\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:

> Unitarily inequivalent representations and disjoint Fock spaces
> is something I couldn't understand for a long time. This looks
> like infamous "parallel universes" to me.

Not at all. They arise because the Fock space construction
must arbitrarily restrict to finite particle numbers in order for
an inner product to exist. This is explained in Umezawa (below).

> Could you give one good example where these things are
> necessary for understanding real physical phenomena.

I'll give you several...

They arise in condensed matter physics, i.e: bose-einstein
condensation, superfluidity and superconductivity. In textbooks
on these subjects, look for "Bogoliubov transformation" and
in most cases you'll find UIRs lurking, although the textbooks
don't always bring out this point explicitly. Umezawa's text on
"Thermofield Dynamics and Condensed States" gives the
most understandable presentation I've seen so far.

A different, more recent, area is neutrino oscillations.
Blasone et al have shown that the Fock space of
definite flavour states is unitarily inequivalent to
that definite mass states. See, for example,
http://www.arxiv.org/abs/hep-ph/9501263, and also the review article by
Capolupo: http://www.arxiv.org/abs/hep-th/0408228. This means that to
understand the QFT of neutrino oscillations fully,
we need to understand UIRs and disjoint Fock
spaces.

Arnold Neumaier

<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> <mikem@despammed.com> wrote in message\n> news:1126839691.917142.163680@g43g2000cwa.googlegroups.com...\n>\n&g t;>I\'m trying to find out whether standard model gauge\n>>groups (acting on fermions) correspond to\n>>Bogoliubov transformations mapping between\n>>disjoint Fock spaces, i.e: between unitarily\n>>inequivalent representations.\n>\n>\n> Unitarily inequivalent representations and disjoint Fock spaces\n> is something I couldn\'t understand for a long time. This looks like\n> infamous "parallel universes" to me. Could you give one\n> good example where these things are\n> necessary for understanding real physical phenomena.\n\nSuperconductivity is the most conspicuous example.\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:
> <mikem@despammed.com> wrote in message
> news:1126839691.917142.163680@g43g20...egroups.com...
>
>>I'm trying to find out whether standard model gauge
>>groups (acting on fermions) correspond to
>>Bogoliubov transformations mapping between
>>disjoint Fock spaces, i.e: between unitarily
>>inequivalent representations.
>
>
> Unitarily inequivalent representations and disjoint Fock spaces
> is something I couldn't understand for a long time. This looks like
> infamous "parallel universes" to me. Could you give one
> good example where these things are
> necessary for understanding real physical phenomena.

Superconductivity is the most conspicuous example.

Arnold Neumaier

mikem@despammed.com

<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> [...]\n> OK, let\'s skip superconductivity and talk about neutrinos.\n> I looked at Blasone-Vitiello paper. This is a good example\n> of what seems confusing about UIR for me.\n>\n> They find a unitary transformation which makes flavor\n> eigenstates (or creation-annihilation operators) from\n> mass eigenstates (or creation-annihilation operators).\n> This transformation also changes the vacuum vector.\n> In particular, it makes the new vacuum |0\'> orthogonal\n> to the old vacuum |0>.\n>\n> I have two questions:\n>\n> 1. In my opinion this construction does not mean that the\n> new vacuum lies in a different Fock state.\n\nI presume you meant Fock "space". :-)\n\n> This wouldn\'t be the case even if all components of |0\'>\n> in the old basis were "zero" in the limit of infinite volume.\n\nActually, it would. I.e: it means that the new vacuum\ncannot be expressed as a mathematically meaningful\nsuperposition of vectors in the old Fock space.\n\nUnfortunately, I have to go away for a couple of days, so\nI can\'t respond more thoroughly right now. But I promise\nto do so when I return. For now, I\'ll just make a few general\nremarks:\n\nIt is probably true that *all* the vectors in the new Fock space\nare orthogonal to *every* vector in the old Fock space,\n(though I haven\'t explicitly checked this for the\nBlasone-Vitiello - I should remedy that). However, Umezawa\ncontains the essence of such a calculation, though he\nleaves it to the reader to fill in a fair bit of detail. Do you\nhave access to a copy of Umezawa? It explains a lot of\nthe theory of UIR much better than I can. The basic idea\nis that every new vector is orthogonal to every old vector.\nTherefore, none of the new vectors can be expressed as\nsuperpositions of the old vectors. That\'s essentially what\ndefines a UIR.\n\nI\'ll follow up on this, and your 2nd question, more fully\nin a few days time.\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's skip superconductivity and talk about neutrinos.
> I looked at Blasone-Vitiello paper. This is a good example
> of what seems confusing about UIR for me.
>
> They find a unitary transformation which makes flavor
> eigenstates (or creation-annihilation operators) from
> mass eigenstates (or creation-annihilation operators).
> This transformation also changes the vacuum vector.
> In particular, it makes the new vacuum [itex]|0'>[/itex] orthogonal
> to the old vacuum |0>.
>
> I have two questions:
>
> 1. In my opinion this construction does not mean that the
> new vacuum lies in a different Fock state.

I presume you meant Fock "space". :-)

> This wouldn't be the case even if all components of [itex]|0'>[/itex]
> in the old basis were "zero" in the limit of infinite volume.

Actually, it would. I.e: it means that the new vacuum
cannot be expressed as a mathematically meaningful
superposition of vectors in the old Fock space.

Unfortunately, I have to go away for a couple of days, so
I can't respond more thoroughly right now. But I promise
to do so when I return. For now, I'll just make a few general
remarks:

It is probably true that *all* the vectors in the new Fock space
are orthogonal to *every* vector in the old Fock space,
(though I haven't explicitly checked this for the
Blasone-Vitiello - I should remedy that). However, Umezawa
contains the essence of such a calculation, though he
leaves it to the reader to fill in a fair bit of detail. Do you
have access to a copy of Umezawa? It explains a lot of
the theory of UIR much better than I can. The basic idea
is that every new vector is orthogonal to every old vector.
Therefore, none of the new vectors can be expressed as
superpositions of the old vectors. That's essentially what
defines a UIR.

I'll follow up on this, and your 2nd question, more fully
in a few days time.

Eugene Stefanovich

<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\nmikem@despammed.com wrote:\n> Eugene Stefanovich wrote:\n>\n> > [...]\n> > OK, let\'s skip superconductivity and talk about neutrinos.\n> > I looked at Blasone-Vitiello paper. This is a good example\n> > of what seems confusing about UIR for me.\n> >\n> > They find a unitary transformation which makes flavor\n> > eigenstates (or creation-annihilation operators) from\n> > mass eigenstates (or creation-annihilation operators).\n> > This transformation also changes the vacuum vector.\n> > In particular, it makes the new vacuum |0\'> orthogonal\n> > to the old vacuum |0>.\n> >\n> > I have two questions:\n> >\n> > 1. In my opinion this construction does not mean that the\n> > new vacuum lies in a different Fock state.\n>\n> I presume you meant Fock "space". :-)\n\nThat\'s right. Sorry.\n\n>\n> > This wouldn\'t be the case even if all components of |0\'>\n> > in the old basis were "zero" in the limit of infinite volume.\n>\n> Actually, it would. I.e: it means that the new vacuum\n> cannot be expressed as a mathematically meaningful\n> superposition of vectors in the old Fock space.\n\nI doubt that.\n\n\n> It is probably true that *all* the vectors in the new Fock space\n> are orthogonal to *every* vector in the old Fock space,\n> (though I haven\'t explicitly checked this for the\n> Blasone-Vitiello - I should remedy that). However, Umezawa\n> contains the essence of such a calculation, though he\n> leaves it to the reader to fill in a fair bit of detail. Do you\n> have access to a copy of Umezawa?\n\nI couldn\'t find this book in my usual library. I\'ll try another\nplace later.\n\n> It explains a lot of\n> the theory of UIR much better than I can. The basic idea\n> is that every new vector is orthogonal to every old vector.\n> Therefore, none of the new vectors can be expressed as\n> superpositions of the old vectors. That\'s essentially what\n> defines a UIR.\n\nI have seen similar discussions in many places. I don\'t have\nthese sources with me now. If I remember well, the idea\nwas to\n\n1) apply a unitary tranformation to the vacuum vector\n|0\'> = U|0>\n2) Find components of |0\'> in a basis\n3) Observe that in some limit all these components tend to zero\n4) Conclude that the vector |0\'> goes outside the original\nHilbert space H in this limit.\n\nThis seems unfair to me. In these examples, even if all components\nof |0\'>\ntend to zero, their number tends to infinity, and the sum of squares\nof the components |0\'> in any basis in H should remain 1.\nThe unitary operator U is explicitly defined within the Hilbert space\nH, so it is beyond me how it can bring any vector outside of H.\n\nIn my opinion, mathematicians overemphasise the subtle differences\nbetween finite-dimensional and infinite-dimensional Hilbert spaces.\nI am not convinced that these differences have any physical meaning.\nThough, this could be just my ignorance.\n\nEugene.\n\n>\n> I\'ll follow up on this, and your 2nd question, more fully\n> in a few days time.\n\nHave a nice trip. I look to hear from you soon.\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>mikem@despammed.com wrote:
> Eugene Stefanovich wrote:
>
> > [...]
> > OK, let's skip superconductivity and talk about neutrinos.
> > I looked at Blasone-Vitiello paper. This is a good example
> > of what seems confusing about UIR for me.
> >
> > They find a unitary transformation which makes flavor
> > eigenstates (or creation-annihilation operators) from
> > mass eigenstates (or creation-annihilation operators).
> > This transformation also changes the vacuum vector.
> > In particular, it makes the new vacuum [itex]|0'>[/itex] orthogonal
> > to the old vacuum |0>.
> >
> > I have two questions:
> >
> > 1. In my opinion this construction does not mean that the
> > new vacuum lies in a different Fock state.
>
> I presume you meant Fock "space". :-)

That's right. Sorry.

>
> > This wouldn't be the case even if all components of [itex]|0'>[/itex]
> > in the old basis were "zero" in the limit of infinite volume.
>
> Actually, it would. I.e: it means that the new vacuum
> cannot be expressed as a mathematically meaningful
> superposition of vectors in the old Fock space.

I doubt that.

> It is probably true that *all* the vectors in the new Fock space
> are orthogonal to *every* vector in the old Fock space,
> (though I haven't explicitly checked this for the
> Blasone-Vitiello - I should remedy that). However, Umezawa
> contains the essence of such a calculation, though he
> leaves it to the reader to fill in a fair bit of detail. Do you
> have access to a copy of Umezawa?

I couldn't find this book in my usual library. I'll try another
place later.

> It explains a lot of
> the theory of UIR much better than I can. The basic idea
> is that every new vector is orthogonal to every old vector.
> Therefore, none of the new vectors can be expressed as
> superpositions of the old vectors. That's essentially what
> defines a UIR.

I have seen similar discussions in many places. I don't have
these sources with me now. If I remember well, the idea
was to

1) apply a unitary tranformation to the vacuum vector
[itex]|0'> = U|0>[/itex]
2) Find components of [itex]|0'>[/itex] in a basis
3) Observe that in some limit all these components tend to zero
4) Conclude that the vector [itex]|0'>[/itex] goes outside the original
Hilbert space H in this limit.

This seems unfair to me. In these examples, even if all components
of [itex]|0'>[/itex]
tend to zero, their number tends to infinity, and the sum of squares
of the components [itex]|0'>[/itex] in any basis in H should remain 1.
The unitary operator U is explicitly defined within the Hilbert space
H, so it is beyond me how it can bring any vector outside of H.

In my opinion, mathematicians overemphasise the subtle differences
between finite-dimensional and infinite-dimensional Hilbert spaces.
I am not convinced that these differences have any physical meaning.
Though, this could be just my ignorance.

Eugene.

>
> I'll follow up on this, and your 2nd question, more fully
> in a few days time.

Have a nice trip. I look to hear from you soon.

Eugene.

mikem@despammed.com

<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> 1. In my opinion [the Blasone-Vitiello] construction does not mean\n> that the new vacuum lies in a different Fock state. This wouldn\'t be\n> the case even if all components of |0\'> in the old basis were "zero"\n> in the limit of infinite volume. Each of the components may tend to\n> zero, but the number of components tends to infinity. So that if you\n> correctly sum up the infinite number of "zeros" you should still get\n> a vector of unit norm.\n>\n> In my view, this is not dissimilar to the normalized plane wave. The\n> wavefunction of the state with definite momentum is "zero"\neverywhere\n> in the position space. However, if you integrate its square over the\n> entire universe you should get 1. You wouldn\'t say that momentum\n> eigenstates lie in a separate Hilbert space, wouldn\'t you?\n\nI\'m not quite sure what you mean here. My textbooks say that\n<x|p> = exp(ipx), meaning that a position eigenstate |x> is _not_\northogonal to a momentum eigenstate |p>. But perhaps you meant\nsomething else?\n\n> 2. There is an infinite number of unitary transformations from\nflavor\n> eigenstates to mass eigenstates. Blasone-Vitiello\'s transformation\n> changes vacuum, which seems unphysical to me. I would prefer to have\na\n> unique vacuum without particles of any kind. This is achieved, for\n> example, by the following transformation:\n>\n> U = a_v* a_1 + a_u* a_2\n>\n> where a_1, a_2 are annihilation operators of the mass eigenstates\n> a_v* and a_u* are creation operators of the flavor eigenstates\n> (e.g., a_v* = cos(phi) a_1* + sin(phi) a_2*)\n>\n> It is\n> 1) unitary in the 0-particle and 1-particle sectors\n> 2) transforms a_1, a_2 to a_v and a_u, respectively\n> 3) does not change vacuum.\n\nI stared at this for a while, but I\'m still unsure what you mean.\nDid you omit an "exp" and/or an Integral and/or a "phi" in\nyour definition of U above?\n\n>> [...] The basic idea is that every new vector is orthogonal to\nevery\n>> old vector. Therefore, none of the new vectors can be expressed as\n>> superpositions of the old vectors. That\'s essentially what defines\na\n>> UIR.\n\n> [...] If I remember well, the idea was to\n>\n> 1) apply a unitary tranformation to the vacuum vector |0\'> = U|0>\n> 2) Find components of |0\'> in a basis\n> 3) Observe that in some limit all these components tend to zero\n> 4) Conclude that the vector |0\'> goes outside the original\n> Hilbert space H in this limit.\n>\n> This seems unfair to me. In these examples, even if all components\nof\n> |0\'> tend to zero, their number tends to infinity, and the sum of\n> squares of the components |0\'> in any basis in H should remain 1.\n\nI look at it this way: if we have a (continuously-parametrized,\ninfinite) basis |b> for H, then any other vector |v> can be expressed\nas an integral superposition, whose coefficients are given by taking\nthe inner product between |v> and each respective |b>, i.e:\n\n|v> = Integral db <b|v> |b>\n\nSo if <b|v> is 0 for every b , the Integral must be 0, showing that |v>\ncannot be expressed as a superposition of |b>\'s. This is quite\ndifferent from the position/momentum case where <x|p> = exp(ipx) which\nis non-zero.\n\n> The unitary operator U is explicitly defined within the Hilbert\nspace\n> H, so it is beyond me how it can bring any vector outside of H.\n\nIf one examines the U carefully, it is not really correct to say\nthat it is explicitly defined "within" the Hilbert (Fock) space. A\ncrucial step in the construction of Fock space is to restrict it\nto have only state vectors whose total particle number is finite.\nWithout this restriction, one cannot define an inner product on the\nspace, because the usual Riemann-Lebesgue integral calculus doesn\'t\nwork: we can\'t approximate an arbitrary vector therein by a countable\nsum arbitrarily closely, as is necessary when defining integrals\nrigorously. Umezawa explains (some of) this. But if you can\'t get a\ncopy, part of it appears in a summary I posted to spr back on\n15-Dec-2004 in a thread titled "Degenerate vacua in QFT":\n\nhttp://www.lns.cornell.edu/spr/2004-12/msg0065860.html\n\nmodulo some followup corrections by Arnold Neumaier. :-)\n\nThe "U" maps vectors in the Fock space into other vectors in the\nlarger non-separable space, of which the Fock space is but a\nsubspace. The total particle number of those "other vectors" turns\nout to be infinite, proving that they lie outside Fock space, which\nby construction contains only vectors of *finite* total particle\nnumber.\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:

> 1. In my opinion [the Blasone-Vitiello] construction does not mean
> that the new vacuum lies in a different Fock state. This wouldn't be
> the case even if all components of [itex]|0'>[/itex] in the old basis were "zero"
> in the limit of infinite volume. Each of the components may tend to
> zero, but the number of components tends to infinity. So that if you
> correctly sum up the infinite number of "zeros" you should still get
> a vector of unit norm.
>
> In my view, this is not dissimilar to the normalized plane wave. The
> wavefunction of the state with definite momentum is "zero"
everywhere
> in the position space. However, if you integrate its square over the
> entire universe you should get 1. You wouldn't say that momentum
> eigenstates lie in a separate Hilbert space, wouldn't you?

I'm not quite sure what you mean here. My textbooks say that
[itex]<x|p> = \exp(ipx),[/itex] meaning that a position eigenstate |x> is _not_
orthogonal to a momentum eigenstate |p>. But perhaps you meant
something else?

> 2. There is an infinite number of unitary transformations from
flavor
> eigenstates to mass eigenstates. Blasone-Vitiello's transformation
> changes vacuum, which seems unphysical to me. I would prefer to have
a
> unique vacuum without particles of any kind. This is achieved, for
> example, by the following transformation:
>
> [itex]U = a_v* a_1 + a_u* a_2[/itex]
>
> where [itex]a_1, a_2[/itex] are annihilation operators of the mass eigenstates
> [itex]a_v*[/itex] and [itex]a_u*[/itex] are creation operators of the flavor eigenstates
> (e.g.[itex], a_v* = cos(\phi) a_1* + sin(\phi) a_2*)[/itex]
>
> It is
> 1) unitary in the 0-particle and 1-particle sectors
> 2) transforms [itex]a_1, a_2[/itex] to [itex]a_v[/itex] and [itex]a_u,[/itex] respectively
> 3) does not change vacuum.

I stared at this for a while, but I'm still unsure what you mean.
Did you omit an "[itex]\exp" and/or[/itex] an Integral [itex]and/or a "\phi"[/itex] in
your definition of U above?

>> [...] The basic idea is that every new vector is orthogonal to
every
>> old vector. Therefore, none of the new vectors can be expressed as
>> superpositions of the old vectors. That's essentially what defines
a
>> UIR.

> [...] If I remember well, the idea was to
>
> 1) apply a unitary tranformation to the vacuum vector [itex]|0'> = U|0>[/itex]
> 2) Find components of [itex]|0'>[/itex] in a basis
> 3) Observe that in some limit all these components tend to zero
> 4) Conclude that the vector [itex]|0'>[/itex] goes outside the original
> Hilbert space H in this limit.
>
> This seems unfair to me. In these examples, even if all components
of
> [itex]|0'>[/itex] tend to zero, their number tends to infinity, and the sum of
> squares of the components [itex]|0'>[/itex] in any basis in H should remain 1.

I look at it this way: if we have a (continuously-parametrized,
infinite) basis |b> for H, then any other vector |v> can be expressed
as an integral superposition, whose coefficients are given by taking
the inner product between |v> and each respective [itex]|b>, i[/itex].e:

[tex]|v> =[/itex] Integral [itex]db <b|v> |b>[/tex]

So if [itex]<b|v>[/itex] is for every b , the Integral must be 0, showing that |v>
cannot be expressed as a superposition of [itex]|b>'s[/itex]. This is quite
different from the position/momentum case where [itex]<x|p> = \exp(ipx)[/itex] which
is non-zero.

> The unitary operator U is explicitly defined within the Hilbert
space
> H, so it is beyond me how it can bring any vector outside of H.

If one examines the U carefully, it is not really correct to say
that it is explicitly defined "within" the Hilbert (Fock) space. A
crucial step in the construction of Fock space is to restrict it
to have only state vectors whose total particle number is finite.
Without this restriction, one cannot define an inner product on the
space, because the usual Riemann-Lebesgue integral calculus doesn't
work: we can't approximate an arbitrary vector therein by a countable
sum arbitrarily closely, as is necessary when defining integrals
rigorously. Umezawa explains (some of) this. But if you can't get a
copy, part of it appears in a summary I posted to spr back on
15-Dec-2004 in a thread titled "Degenerate vacua in QFT":

http://www.lns.cornell.edu/spr/2004-12/msg0065860.html

modulo some followup corrections by Arnold Neumaier. :-)

The "U" maps vectors in the Fock space into other vectors in the
larger non-separable space, of which the Fock space is but a
subspace. The total particle number of those "other vectors" turns
out to be infinite, proving that they lie outside Fock space, which
by construction contains only vectors of *finite* total particle
number.

Eugene Stefanovich

<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>mikem@despammed.com wrote:\n> > The\n> > wavefunction of the state with definite momentum is "zero"\n> everywhere\n> > in the position space. However, if you integrate its square over the\n> > entire universe you should get 1. You wouldn\'t say that momentum\n> > eigenstates lie in a separate Hilbert space, wouldn\'t you?\n>\n> I\'m not quite sure what you mean here. My textbooks say that\n> <x|p> = exp(ipx), meaning that a position eigenstate |x> is _not_\n> orthogonal to a momentum eigenstate |p>. But perhaps you meant\n> something else?\n\nIf you require that |p> is a normalized vector, then\n\n<x|p> = N exp(ipx)\n\nwhere the normalization factor N is basically "zero". One can formally\nsay that N = 1/sqrt(V) where V is the "volume of space", i.e., infinity.\nThe probability of finding definite momentum particle in each finite\nvolume W is W/V, which is "zero". However, this does not mean that this\nstate is "outside" of the Hilbert space. The probability of finding\nthe particle "somewhere" (i.e. the intergal over V) is V/V = 1.\n\n>\n> > 2. There is an infinite number of unitary transformations from\n> flavor\n> > eigenstates to mass eigenstates. Blasone-Vitiello\'s transformation\n> > changes vacuum, which seems unphysical to me. I would prefer to have\n> a\n> > unique vacuum without particles of any kind. This is achieved, for\n> > example, by the following transformation:\n> >\n> > U = a_v* a_1 + a_u* a_2\n> >\n> > where a_1, a_2 are annihilation operators of the mass eigenstates\n> > a_v* and a_u* are creation operators of the flavor eigenstates\n> > (e.g., a_v* = cos(phi) a_1* + sin(phi) a_2*)\n> >\n> > It is\n> > 1) unitary in the 0-particle and 1-particle sectors\n> > 2) transforms a_1, a_2 to a_v and a_u, respectively\n> > 3) does not change vacuum.\n>\n> I stared at this for a while, but I\'m still unsure what you mean.\n> Did you omit an "exp" and/or an Integral and/or a "phi" in\n> your definition of U above?\n\nSorry, I should have been more specific. The full definition of U is:\n1) U = 1 on the vacuum vector |0>\n2) U = a_v* a_1 + a_u* a_2 on one-particle subspaces\nH_1 (+) H_2 = H_v (+) H_u\n3) U = whatever on the rest of the Fock space.\n\n>\n> >> [...] The basic idea is that every new vector is orthogonal to\n> every\n> >> old vector. Therefore, none of the new vectors can be expressed as\n> >> superpositions of the old vectors. That\'s essentially what defines\n> a\n> >> UIR.\n>\n> > [...] If I remember well, the idea was to\n> >\n> > 1) apply a unitary tranformation to the vacuum vector |0\'> = U|0>\n> > 2) Find components of |0\'> in a basis\n> > 3) Observe that in some limit all these components tend to zero\n> > 4) Conclude that the vector |0\'> goes outside the original\n> > Hilbert space H in this limit.\n> >\n> > This seems unfair to me. In these examples, even if all components\n> of\n> > |0\'> tend to zero, their number tends to infinity, and the sum of\n> > squares of the components |0\'> in any basis in H should remain 1.\n>\n> I look at it this way: if we have a (continuously-parametrized,\n> infinite) basis |b> for H, then any other vector |v> can be expressed\n> as an integral superposition, whose coefficients are given by taking\n> the inner product between |v> and each respective |b>, i.e:\n>\n> |v> = Integral db <b|v> |b>\n>\n> So if <b|v> is 0 for every b , the Integral must be 0, showing that |v>\n> cannot be expressed as a superposition of |b>\'s. This is quite\n> different from the position/momentum case where <x|p> = exp(ipx) which\n> is non-zero.\n\nThere is full analogy. If we use (as we should) normalized vectors\nfor |x> and |p>, then we obtain\n\n<x|p> = 1/sqrt(V) exp(ipx)\n\n|p> = Integral dx <x|p> |x>\n= 1/sqrt(V) Integral dx exp(ipx) |x>\n\nThe wave function (the density of the probability amplitude)\nof |p> in the position representation must be\n\n1/sqrt(V) exp(ipx)\n\nbecause the volume integral of its square is\n\nIntegral dx |1/sqrt(V) exp(ipx)|^2 = 1/V Integral dx\n= 1\n\nas it should. The function exp(ipx) without the normalization\nfactor 1/sqrt(V) does not have a probabilistic\ninterpretation, because the volume integral of its square is infinite.\n\n> > The unitary operator U is explicitly defined within the Hilbert\n> space\n> > H, so it is beyond me how it can bring any vector outside of H.\n>\n> If one examines the U carefully, it is not really correct to say\n> that it is explicitly defined "within" the Hilbert (Fock) space. A\n> crucial step in the construction of Fock space is to restrict it\n> to have only state vectors whose total particle number is finite.\n> Without this restriction, one cannot define an inner product on the\n> space, because the usual Riemann-Lebesgue integral calculus doesn\'t\n> work: we can\'t approximate an arbitrary vector therein by a countable\n> sum arbitrarily closely, as is necessary when defining integrals\n> rigorously. Umezawa explains (some of) this. But if you can\'t get a\n> copy, part of it appears in a summary I posted to spr back on\n> 15-Dec-2004 in a thread titled "Degenerate vacua in QFT":\n>\n> http://www.lns.cornell.edu/spr/2004-12/msg0065860.html\n>\n> modulo some followup corrections by Arnold Neumaier. :-)\n>\n> The "U" maps vectors in the Fock space into other vectors in the\n> larger non-separable space, of which the Fock space is but a\n> subspace. The total particle number of those "other vectors" turns\n> out to be infinite, proving that they lie outside Fock space, which\n> by construction contains only vectors of *finite* total particle\n> number.\n\nThank you for the reference. I\'ve seen similar arguments in other\nplaces, but they do not make much sense to me. I have a strong feeling\nthat the distinction between separable and non-separable Hilbert spaces\nwas invented by mathematicians to make life of physicists miserable.\nI don\'t think there is anything wrong with regarding the Hilbert space\nof a single particle as non-separable. After all, the number of points\nin 3D space is not countable, and one can associate a distinct basis\nvector (eigenvector of the position operator) with each such point.\n\nYour arguments could be correct if your DEFINE the Fock space as\nhaving not more than a finite number of particles.\nThen, why I am not allowed\nto DEFINE the Fock space as having any number of particles from zero\nto infinity?\nI have thought about these issues and came to a conclusion that there\ncould be some non-trivial math involved, but it has no significance to\nphysics. Again, I am not talking about superconductivity and\nspontaneously broken vacuum symmetries - the issues I\'m not so familiar\nwith. Maybe there is deep physical truth concerning UIR in these fields,\nI just don\'t know.\n\nFor myself, I found another more comfortable attitude to\nthese issues. This attitude is not frequently discussed in physics\nliterature, but I found it rather illuminating.\nThis is based on the so-called\n"non-standard analysis" first developed by A. Robinson in 1960.\nThis is too vast a field to be described in one post, but the basic idea\nis to treat finite, "infinitely small" and "infinitely large" quantities\non the same footing. In this approach, the numbers like 1/sqrt(V),\nwhere V is the volume of the entire space, make perfect sense, and there\nis no trouble to calculate the integral\n\nIntegral dx |1/sqrt(V) exp(ipx)|^2 = 1\n\neven if the integrand is "zero" everywhere.\n\nThere are few papers which try to apply this approach to quantum\nmechanics. See, for example\nA. Friedman, "Non-standard extension of quantum logic and\nDirac\'s bra-ket formalism of quantum mechanics", Int. J. Theor. Phys.\n33 (1994), 307 (he was my student back in old times).\n\nThe non-standard analysis is now a well-established branch of\nmathematics. I think, its use in QM may clarify some conceptual issues,\nbut I don\'t expect any physical discoveries there.\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>mikem@despammed.com wrote:
> > The
> > wavefunction of the state with definite momentum is "zero"
> everywhere
> > in the position space. However, if you integrate its square over the
> > entire universe you should get 1. You wouldn't say that momentum
> > eigenstates lie in a separate Hilbert space, wouldn't you?
>
> I'm not quite sure what you mean here. My textbooks say that
> [itex]<x|p> = \exp(ipx),[/itex] meaning that a position eigenstate |x> is _not_
> orthogonal to a momentum eigenstate |p>. But perhaps you meant
> something else?

If you require that |p> is a normalized vector, then

[tex]<x|p> = N \exp(ipx)[/tex]

where the normalization factor N is basically "zero". One can formally
say that [itex]N = 1/\sqrt(V)[/itex] where V is the "volume of space", i.e., infinity.
The probability of finding definite momentum particle in each finite
volume W is [itex]W/V,[/itex] which is "zero". However, this does not mean that this
state is "outside" of the Hilbert space. The probability of finding
the particle "somewhere" (i.e. the intergal over V) is [itex]V/V = 1[/itex].

>
> > 2. There is an infinite number of unitary transformations from
> flavor
> > eigenstates to mass eigenstates. Blasone-Vitiello's transformation
> > changes vacuum, which seems unphysical to me. I would prefer to have
> a
> > unique vacuum without particles of any kind. This is achieved, for
> > example, by the following transformation:
> >
> > [itex]U = a_v* a_1 + a_u* a_2[/itex]
> >
> > where [itex]a_1, a_2[/itex] are annihilation operators of the mass eigenstates
> > [itex]a_v*[/itex] and [itex]a_u*[/itex] are creation operators of the flavor eigenstates
> > (e.g.[itex], a_v* = cos(\phi) a_1* + sin(\phi) a_2*)[/itex]
> >
> > It is
> > 1) unitary in the 0-particle and 1-particle sectors
> > 2) transforms [itex]a_1, a_2[/itex] to [itex]a_v[/itex] and [itex]a_u,[/itex] respectively
> > 3) does not change vacuum.
>
> I stared at this for a while, but I'm still unsure what you mean.
> Did you omit an "[itex]\exp" and/or[/itex] an Integral [itex]and/or a "\phi"[/itex] in
> your definition of U above?

Sorry, I should have been more specific. The full definition of U is:
1) [itex]U = 1[/itex] on the vacuum vector |0>
[itex]2) U = a_v* a_1 + a_u* a_2[/itex] on one-particle subspaces
[itex]H_1 (+) H_2 = H_v (+) H_u3) U =[/itex] whatever on the rest of the Fock space.

>
> >> [...] The basic idea is that every new vector is orthogonal to
> every
> >> old vector. Therefore, none of the new vectors can be expressed as
> >> superpositions of the old vectors. That's essentially what defines
> a
> >> UIR.
>
> > [...] If I remember well, the idea was to
> >
> > 1) apply a unitary tranformation to the vacuum vector [itex]|0'> = U|0>[/itex]
> > 2) Find components of [itex]|0'>[/itex] in a basis
> > 3) Observe that in some limit all these components tend to zero
> > 4) Conclude that the vector [itex]|0'>[/itex] goes outside the original
> > Hilbert space H in this limit.
> >
> > This seems unfair to me. In these examples, even if all components
> of
> > [itex]|0'>[/itex] tend to zero, their number tends to infinity, and the sum of
> > squares of the components [itex]|0'>[/itex] in any basis in H should remain 1.
>
> I look at it this way: if we have a (continuously-parametrized,
> infinite) basis |b> for H, then any other vector |v> can be expressed
> as an integral superposition, whose coefficients are given by taking
> the inner product between |v> and each respective [itex]|b>, i[/itex].e:
>
> [itex]|v> =[/itex] Integral [itex]db <b|v> |b>[/itex]
>
> So if [itex]<b|v>[/itex] is for every b , the Integral must be 0, showing that |v>
> cannot be expressed as a superposition of [itex]|b>'s[/itex]. This is quite
> different from the position/momentum case where [itex]<x|p> = \exp(ipx)[/itex] which
> is non-zero.

There is full analogy. If we use (as we should) normalized vectors
for |x> and [itex]|p>,[/itex] then we obtain

[tex]<x|p> = 1/\sqrt(V) \exp(ipx)|p> =[/itex] Integral [itex]dx <x|p> |x>= 1/\sqrt(V)[/itex] Integral [itex]dx \exp(ipx) |x>[/tex]

The wave function (the density of the probability amplitude)
of |p> in the position representation must be

[tex]1/\sqrt(V) \exp(ipx)[/tex]

because the volume integral of its square is

Integral [itex]dx |1/\sqrt(V) \exp(ipx)|^2 = 1/V[/itex] Integral dx
= 1

as it should. The function [itex]\exp(ipx)[/itex] without the normalization
factor [itex]1/\sqrt(V)[/itex] does not have a probabilistic
interpretation, because the volume integral of its square is infinite.

> > The unitary operator U is explicitly defined within the Hilbert
> space
> > H, so it is beyond me how it can bring any vector outside of H.
>
> If one examines the U carefully, it is not really correct to say
> that it is explicitly defined "within" the Hilbert (Fock) space. A
> crucial step in the construction of Fock space is to restrict it
> to have only state vectors whose total particle number is finite.
> Without this restriction, one cannot define an inner product on the
> space, because the usual Riemann-Lebesgue integral calculus doesn't
> work: we can't approximate an arbitrary vector therein by a countable
> sum arbitrarily closely, as is necessary when defining integrals
> rigorously. Umezawa explains (some of) this. But if you can't get a
> copy, part of it appears in a summary I posted to spr back on
> 15-Dec-2004 in a thread titled "Degenerate vacua in QFT":
>
> http://www.lns.cornell.edu/spr/2004-12/msg0065860.html
>
> modulo some followup corrections by Arnold Neumaier. :-)
>
> The "U" maps vectors in the Fock space into other vectors in the
> larger non-separable space, of which the Fock space is but a
> subspace. The total particle number of those "other vectors" turns
> out to be infinite, proving that they lie outside Fock space, which
> by construction contains only vectors of *finite* total particle
> number.

Thank you for the reference. I've seen similar arguments in other
places, but they do not make much sense to me. I have a strong feeling
that the distinction between separable and non-separable Hilbert spaces
was invented by mathematicians to make life of physicists miserable.
I don't think there is anything wrong with regarding the Hilbert space
of a single particle as non-separable. After all, the number of points
in 3D space is not countable, and one can associate a distinct basis
vector (eigenvector of the position operator) with each such point.

Your arguments could be correct if your DEFINE the Fock space as
having not more than a finite number of particles.
Then, why I am not allowed
to DEFINE the Fock space as having any number of particles from zero
to infinity?
I have thought about these issues and came to a conclusion that there
could be some non-trivial math involved, but it has no significance to
physics. Again, I am not talking about superconductivity and
spontaneously broken vacuum symmetries - the issues I'm not so familiar
with. Maybe there is deep physical truth concerning UIR in these fields,
I just don't know.

For myself, I found another more comfortable attitude to
these issues. This attitude is not frequently discussed in physics
literature, but I found it rather illuminating.
This is based on the so-called
"non-standard analysis" first developed by A. Robinson in 1960.
This is too vast a field to be described in one post, but the basic idea
is to treat finite, "infinitely small" and "infinitely large" quantities
on the same footing. In this approach, the numbers like [itex]1/\sqrt(V),[/itex]
where V is the volume of the entire space, make perfect sense, and there
is no trouble to calculate the integral

Integral [itex]dx |1/\sqrt(V) \exp(ipx)|^2 = 1[/itex]

even if the integrand is "zero" everywhere.

There are few papers which try to apply this approach to quantum
mechanics. See, for example
A. Friedman, "Non-standard extension of quantum logic and
Dirac's bra-ket formalism of quantum mechanics"[itex], \Int[/itex]. J. Theor. Phys.
33 (1994), 307 (he was my student back in old times).

The non-standard analysis is now a well-established branch of
mathematics. I think, its use in QM may clarify some conceptual issues,
but I don't expect any physical discoveries there.

Eugene.

mikem@despammed.com

<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 in part:\n\n> [...] I don\'t think there is anything wrong with\n> regarding the Hilbert space of a single particle\n> as non-separable. After all, the number of points\n> in 3D space is not countable, and one can associate\n> a distinct basis vector (eigenvector of the position\n> operator) with each such point.\n\nThat\'s because we can integrate over a 3D space.\nBut, (at least with standard integration), we can\'t\nintegrate over an infinite dimensional space in the\nsame way. But about here, my detailed knowledge\nstarts to dry up so I can\'t say much more.\n\n> Your arguments could be correct if your DEFINE\n> the Fock space as having not more than a finite\n> number of particles. Then, why I am not allowed\n> to DEFINE the Fock space as having any number\n> of particles from zero to infinity?\n\nOnly because of the difficulty with performing\nstandard integration over uncountably-infinite\ndimensional spaces.\n\n> [...] See, for example: A. Friedman, "Non-standard\n> extension of quantum logic and Dirac\'s bra-ket\n> formalism of quantum mechanics", Int. J. Theor. Phys.\n> 33 (1994), 307 [...]\n\nIs this paper on the archive, or somewhere else online?\n(It\'s a pain for me to travel to university libraries these\ndays.)\n\n\nRegarding the other items in your post, I need to think\nabout them for a while before replying.\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 in part:

> [...] I don't think there is anything wrong with
> regarding the Hilbert space of a single particle
> as non-separable. After all, the number of points
> in 3D space is not countable, and one can associate
> a distinct basis vector (eigenvector of the position
> operator) with each such point.

That's because we can integrate over a 3D space.
But, (at least with standard integration), we can't
integrate over an infinite dimensional space in the
same way. But about here, my detailed knowledge
starts to dry up so I can't say much more.

> Your arguments could be correct if your DEFINE
> the Fock space as having not more than a finite
> number of particles. Then, why I am not allowed
> to DEFINE the Fock space as having any number
> of particles from zero to infinity?

Only because of the difficulty with performing
standard integration over uncountably-infinite
dimensional spaces.

> [...] See, for example: A. Friedman, "Non-standard
> extension of quantum logic and Dirac's bra-ket
> formalism of quantum mechanics"[itex], \Int[/itex]. J. Theor. Phys.
> 33 (1994), 307 [...]

Is this paper on the archive, or somewhere else online?
(It's a pain for me to travel to university libraries these
days.)

Regarding the other items in your post, I need to think
about them for a while before replying.

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