Charles Francis
Apr7-04, 08:56 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=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 <xQY6c.5174$PY1.123189@bgtnsc05-news.ops.worldnet.att.net>,\nCalvin Ritchie <DonRitchie870@csWebmail.com> writes\n>Is there any reason, other than the "required" Lorentz gauge (equivalent in\n>QFT to \'conservation\'), why a massless vector particle must be transverse? I\n>am interested in both the classical and quantum fields.\n\nOnly the observable part of the field is transverse. Lorentz gauge does\nnot enforce this, btw.\n\n> I have read (and re-read, and re-read, ....) Weinberg\'s Vol. 1\n>discussions of massless fields (pgs. 69-74, and 246-255) and *think* that I\n>understand all of it pretty well. In particular, I\'ve found why the problem\n>of constructing the transverse field, i.e. the inconsistency that Weinberg\n>points out in Eqs. (5.9.19) - (5.9.21), stems directly from the exclusion of\n>the spin zero field and why we must impose Eq. (2.5.38) to retain Lorentz\n>gauge.\n\nAnd in another post you state:\n\n>Weinberg (QTFT, Vol. 1, 250) states it\n>clearly:\n>"We have thus come to\n>the conclusion that no four-vector field\n>can be constructed from the annihilation\n>and creation operators for a particle of\n>mass zero and helicity +/- 1."\n>\n>Do you disagree with that?\n\nYes I do disagree with it. I don\'t have Weinberg. He was ridiculed by my\nlecturers in Cambridge, both for his ideas on gravity and for doing qed\nwith only two polarisation states, and I have never had occasion to\nrevise that opinion. If you do qed with only two polarisation states you\ncannot derive the Coulomb force (as Arnold remarked elsewhere) and you\nhave yourself commented that if you discard the lightlike states they\ncome back in as soon as you transform.\n\nThis can be straightened out by using four creation and annihilation\noperators, longitudinal and timelike as well as two transverse, in\nLorentz gauge. A time like creation operator would create a state with\nnegative probability, but this is fine if the time like and longitudinal\nstates are always created with the same amplitude. That gives zero\nprobability for observing such states. So we find that there are not\nonly two polarisation states, but there is also an unobservable\ntime-like state.\n\nAs you have remarked elsewhere simply discarding or "modding out" this\nunobservable state leads to inconsistency. It also makes it impossible\nto derive Maxwell\'s equations in the classical correspondence, which to\nmy mind proves it wrong. The calculation of Maxwell\'s equations is quite\nstraightforward, and is done for quite general electron and photon\nstates by calculating the expectation of the photon field A and the\nelectron current\nj = psi_hat gamma psi\nMaxwell\'s equations are recovered in the Lorentz gauge,.\n\nI sketch the argument below in ASCII. This argument was known in the\nseventies, but I have not seen it in text books and seems to have since\nbeen forgotten.\n\nStarting from the defining equations\n\nd.A(x) = 0 [1]\n\n\nBox(A) = 0 [2]\n\nand with the interaction\n\nH = e A j [3]\n\nwe want\n\nBox(<A>) = i d_a { [H, d_a A] + <d_a A> }\n\n= i [H, Box(A)] + i [H, d_a A] + <Box(A)>\n\n= i [H, d_a A] [4]\n\nby [2]. Now because of the equal time commutator, [A(x),A(y)] = 0, [4]\nreduces at once to\n\nBox(<A>) = i [H, d_0 A] [5]\n\nSo for an interaction density H = e A j we obtain the form of Maxwell\'s\nequations in Lorentz gauge i.e.\n\nBox(<A>) = - e <j>\n\nIt seems to me important that we should do this calculation more often,\nbecause quantisation as it is usually carried out is not a rigorous\ndeduction, and also because the bare charge appearing in [3] is\nunmodified in Maxwell\'s equations, which is important for fixing the\ndivergent terms via renormalisation.\n\n\n\n\nRegards\n\n--\nCharles Francis\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 article <xQY6c.5174$PY1.123189@bgtnsc05-news.ops.worldnet.att.net>,
Calvin Ritchie <DonRitchie870@csWebmail.com> writes
>Is there any reason, other than the "required" Lorentz gauge (equivalent in
>QFT to 'conservation'), why a massless vector particle must be transverse? I
>am interested in both the classical and quantum fields.
Only the observable part of the field is transverse. Lorentz gauge does
not enforce this, btw.
> I have read (and re-read, and re-read, ....) Weinberg's Vol. 1
>discussions of massless fields (pgs. 69-74, and 246-255) and *think* that I
>understand all of it pretty well. In particular, I've found why the problem
>of constructing the transverse field, i.e. the inconsistency that Weinberg
>points out in Eqs. (5.9.19) - (5.9.21), stems directly from the exclusion of
>the spin zero field and why we must impose Eq. (2.5.38) to retain Lorentz
>gauge.
And in another post you state:
>Weinberg (QTFT, Vol. 1, 250) states it
>clearly:
>"We have thus come to
>the conclusion that no four-vector field
>can be constructed from the annihilation
>and creation operators for a particle of
>mass zero and helicity +/- 1."
>
>Do you disagree with that?
Yes I do disagree with it. I don't have Weinberg. He was ridiculed by my
lecturers in Cambridge, both for his ideas on gravity and for doing qed
with only two polarisation states, and I have never had occasion to
revise that opinion. If you do qed with only two polarisation states you
cannot derive the Coulomb force (as Arnold remarked elsewhere) and you
have yourself commented that if you discard the lightlike states they
come back in as soon as you transform.
This can be straightened out by using four creation and annihilation
operators, longitudinal and timelike as well as two transverse, in
Lorentz gauge. A time like creation operator would create a state with
negative probability, but this is fine if the time like and longitudinal
states are always created with the same amplitude. That gives zero
probability for observing such states. So we find that there are not
only two polarisation states, but there is also an unobservable
time-like state.
As you have remarked elsewhere simply discarding or "modding out" this
unobservable state leads to inconsistency. It also makes it impossible
to derive Maxwell's equations in the classical correspondence, which to
my mind proves it wrong. The calculation of Maxwell's equations is quite
straightforward, and is done for quite general electron and photon
states by calculating the expectation of the photon field A and the
electron current
j = \psi_hat \gamma \psi
Maxwell's equations are recovered in the Lorentz gauge,.
I sketch the argument below in ASCII. This argument was known in the
seventies, but I have not seen it in text books and seems to have since
been forgotten.
Starting from the defining equations
d.A(x) = [1]
Box(A) = [2]
and with the interaction
H = e A j[/itex] [3]
we want
Box(<A>) = i d_a { [H, [itex]d_a A] + <d_a A> }
= i [H, Box(A)] + i [H, d_a A] + <Box(A)>
= i [H, d_a A] [4]
by [2]. Now because of the equal time commutator, [A(x),A(y)] = 0, [4]
reduces at once to
Box(<A>) = i [H, d_0 A] [5]
So for an interaction density H = e A j we obtain the form of Maxwell's
equations in Lorentz gauge i.e.
Box(<A>) = - e <j>
It seems to me important that we should do this calculation more often,
because quantisation as it is usually carried out is not a rigorous
deduction, and also because the bare charge appearing in [3] is
unmodified in Maxwell's equations, which is important for fixing the
divergent terms via renormalisation.
Regards
--
Charles Francis
Calvin Ritchie <DonRitchie870@csWebmail.com> writes
>Is there any reason, other than the "required" Lorentz gauge (equivalent in
>QFT to 'conservation'), why a massless vector particle must be transverse? I
>am interested in both the classical and quantum fields.
Only the observable part of the field is transverse. Lorentz gauge does
not enforce this, btw.
> I have read (and re-read, and re-read, ....) Weinberg's Vol. 1
>discussions of massless fields (pgs. 69-74, and 246-255) and *think* that I
>understand all of it pretty well. In particular, I've found why the problem
>of constructing the transverse field, i.e. the inconsistency that Weinberg
>points out in Eqs. (5.9.19) - (5.9.21), stems directly from the exclusion of
>the spin zero field and why we must impose Eq. (2.5.38) to retain Lorentz
>gauge.
And in another post you state:
>Weinberg (QTFT, Vol. 1, 250) states it
>clearly:
>"We have thus come to
>the conclusion that no four-vector field
>can be constructed from the annihilation
>and creation operators for a particle of
>mass zero and helicity +/- 1."
>
>Do you disagree with that?
Yes I do disagree with it. I don't have Weinberg. He was ridiculed by my
lecturers in Cambridge, both for his ideas on gravity and for doing qed
with only two polarisation states, and I have never had occasion to
revise that opinion. If you do qed with only two polarisation states you
cannot derive the Coulomb force (as Arnold remarked elsewhere) and you
have yourself commented that if you discard the lightlike states they
come back in as soon as you transform.
This can be straightened out by using four creation and annihilation
operators, longitudinal and timelike as well as two transverse, in
Lorentz gauge. A time like creation operator would create a state with
negative probability, but this is fine if the time like and longitudinal
states are always created with the same amplitude. That gives zero
probability for observing such states. So we find that there are not
only two polarisation states, but there is also an unobservable
time-like state.
As you have remarked elsewhere simply discarding or "modding out" this
unobservable state leads to inconsistency. It also makes it impossible
to derive Maxwell's equations in the classical correspondence, which to
my mind proves it wrong. The calculation of Maxwell's equations is quite
straightforward, and is done for quite general electron and photon
states by calculating the expectation of the photon field A and the
electron current
j = \psi_hat \gamma \psi
Maxwell's equations are recovered in the Lorentz gauge,.
I sketch the argument below in ASCII. This argument was known in the
seventies, but I have not seen it in text books and seems to have since
been forgotten.
Starting from the defining equations
d.A(x) = [1]
Box(A) = [2]
and with the interaction
H = e A j[/itex] [3]
we want
Box(<A>) = i d_a { [H, [itex]d_a A] + <d_a A> }
= i [H, Box(A)] + i [H, d_a A] + <Box(A)>
= i [H, d_a A] [4]
by [2]. Now because of the equal time commutator, [A(x),A(y)] = 0, [4]
reduces at once to
Box(<A>) = i [H, d_0 A] [5]
So for an interaction density H = e A j we obtain the form of Maxwell's
equations in Lorentz gauge i.e.
Box(<A>) = - e <j>
It seems to me important that we should do this calculation more often,
because quantisation as it is usually carried out is not a rigorous
deduction, and also because the bare charge appearing in [3] is
unmodified in Maxwell's equations, which is important for fixing the
divergent terms via renormalisation.
Regards
--
Charles Francis