please help my confusion about particles and irreps.

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>Igor Khavkine wrote:\n> On 2005-06-03, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n>\n>>The representation of the Lorentz group used in the Dirac\'s construction\n>>is 4-dimensional, and therefore not unitary. Lorentz group is not\n>>compact, so any non-trivial unitary representation must be infinite-\n>>dimensional. Thus, there cannot be a unitary mapping between Dirac\'s\n>>and Wigner\'s wavefunctions. The condition of unitarity is essential,\n>>because it guarantees the preservation of probabilities in different\n>>inertial reference frames, and therefore only unitary representations\n>>satisfy the relativity principle. Only Wigner\'s construction\n>>satisfies this condition. Dirac\'s functions cannot be regarded as\n>>probability amplitudes.\n>\n>\n> This argument shows a deep misunderstanding of how the Dirac wave\n> function represents the Poincare group. The wave function *itself* is an\n> element of the (infinite dimensional) unitary representation of the\n> Poincare group. The transformations act as\n>\n> [(Lambda,a)psi](x) = psi(Lambda^{-1}x - a).\n\n>\n> These transformations are unitary, since the action of the Poincare\n> group preserve the inner product\n>\n> (psi,phi) = int psi(x)* phi(x) dx.\n\nYou forgot the transformation of the field components\n\n[(Lambda,a)psi](x) = D(Lambda) psi(Lambda^{-1}x - a). (1)\n\nwhere D(Lambda) are 4-by-4 matrices representing the Lorentz subgroup\nof the Poincare group.\nThe action (1) is not unitary, because matrices D(Lambda) are not\nunitary.\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>Igor Khavkine wrote:
> On [itex]2005-06-03,[/itex] Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>
>>The representation of the Lorentz group used in the Dirac's construction
>>is 4-dimensional, and therefore not unitary. Lorentz group is not
>>compact, so any non-trivial unitary representation must be infinite-
>>dimensional. Thus, there cannot be a unitary mapping between Dirac's
>>and Wigner's wavefunctions. The condition of unitarity is essential,
>>because it guarantees the preservation of probabilities in different
>>inertial reference frames, and therefore only unitary representations
>>satisfy the relativity principle. Only Wigner's construction
>>satisfies this condition. Dirac's functions cannot be regarded as
>>probability amplitudes.
>
>
> This argument shows a deep misunderstanding of how the Dirac wave
> function represents the Poincare group. The wave function *itself* is an
> element of the (infinite dimensional) unitary representation of the
> Poincare group. The transformations act as
>
> [itex][(\Lambda,a)\psi](x) = \psi(\Lambda^{-1}x - a)[/itex].

>
> These transformations are unitary, since the action of the Poincare
> group preserve the inner product
>
> [itex](\psi,\phi) = \int \psi(x)* \phi(x) dx[/itex].

You forgot the transformation of the field components

[itex][(\Lambda,a)\psi](x) = D(\Lambda) \psi(\Lambda^{-1}x - a)[/itex]. (1)

where [itex]D(\Lambda)[/itex] are [itex]4-by-4[/itex] matrices representing the Lorentz subgroup
of the Poincare group.
The action (1) is not unitary, because matrices [itex]D(\Lambda)[/itex] are not
unitary.

Eugene.

Igor Khavkine

<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>On 2005-06-03, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n\n> Yes, you can obtain numerical functions satisfying the Dirac equation\n> using Weinberg\'s formulas (14.1.4) - (14.1.5). However, these functions\n> have nothing to do with probability amplitudes describing states.\n> I learned in quantum mechanics that wave functions are expansion\n> coefficients of the state vectors in certain orthonormal basis.\n> I haven\'t seen the definition of a wave function as a matrix element\n> of quantum field, like in (14.1.4) - (14.1.5). This looks like a\n> completely new approach to quantum mechanics. How would you write\n> the wave function of a 2-particle system in this notation?\n\nNo, this is not how quantum mechanics is usually taught, but that\nshouldn\'t stop you from learning to look at it this way when you have\nthe opportunity. This approach is not new at all. Have you been exposed\nto second quantization outside the treatment of QED? Schroedinger\'s wave\nfunction can be obtained in exactly the same way, but as matrix elements\nof the Schroedinger field operators. Have you read Dirac\'s book\n_Principles of Quantum Mechanics_? He deals with this kind of\nequivalence issue in the chapter on identical particles. The\nprobabilistic interpretation only needs a positive definite inner\nproduct. In both the Schroedinger and Dirac wave pictures, one is\nreadily available.\n\nIt is easy to write multiplarticle states as well. They correspond to\nmultivariable wave functions. For instance:\n\npsi++(x,y) = <0|PSI(x) PSI(y) |2-particle>\npsi-+(x,y) = <0|PSI(x)* PSI(y) |2-particle>\npsi+-(x,y) = <0|PSI(x) PSI(y)*|2-particle>\npsi--(x,y) = <0|PSI(x)* PSI(y)*|2-particle>\n\nWill give you, respectively, the pos-pos, neg-pos, pos-neg, and neg-neg\nfrequency components of a 2-particle antisymmetric wave function.\n\n> I am surprised that Weinberg took this route in his chapter 14.\n> There is a more consistent way to the stationary states of\n> the hydrogen atom and Lamb shifts.\n\nThis is exactly the kind of statement that I criticised earlier. You\nhave failed to point out any inconsistency in the equivalence between\nthe Dirac wave function and Fock space state points of view. Yet you\ncontinue to make unsubstantiated claims about its consistency, despite\nan explicit demonstration of the equivalence. If you make this statement\nbecause you are confident of your opinion, this casts a shadow on the\nvalidity of other statements that you believe to be true.\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 [itex]2005-06-03,[/itex] Eugene Stefanovich <eugenev@synopsys.com> wrote:

> Yes, you can obtain numerical functions satisfying the Dirac equation
> using Weinberg's formulas (14.1.[itex]4) - (14.1.5).[/itex] However, these functions
> have nothing to do with probability amplitudes describing states.
> I learned in quantum mechanics that wave functions are expansion
> coefficients of the state vectors in certain orthonormal basis.
> I haven't seen the definition of a wave function as a matrix element
> of quantum field, like in (14.1.[itex]4) - (14.1.5).[/itex] This looks like a
> completely new approach to quantum mechanics. How would you write
> the wave function of a 2-particle system in this notation?

No, this is not how quantum mechanics is usually taught, but that
shouldn't stop you from learning to look at it this way when you have
the opportunity. This approach is not new at all. Have you been exposed
to second quantization outside the treatment of QED? Schroedinger's wave
function can be obtained in exactly the same way, but as matrix elements
of the Schroedinger field operators. Have you read Dirac's book
_Principles of Quantum Mechanics_? He deals with this kind of
equivalence issue in the chapter on identical particles. The
probabilistic interpretation only needs a positive definite inner
product. In both the Schroedinger and Dirac wave pictures, one is
readily available.

It is easy to write multiplarticle states as well. They correspond to
multivariable wave functions. For instance:

[tex]\psi++(x,y) = <0|\PSI(x) \PSI(y)[/itex] |2-particle>
[itex]\psi-+(x,y) = <0|\PSI(x)* \PSI(y)[/itex] |2-particle>
[itex]\psi+-(x,y) = <0|\PSI(x) \PSI(y)*|2-particle>[/itex]
\psi--(x,y) [itex]= <0|\PSI(x)* \PSI(y)*|2-particle>[/tex]

Will give you, respectively, the pos-pos, neg-pos, pos-neg, and neg-neg
frequency components of a 2-particle antisymmetric wave function.

> I am surprised that Weinberg took this route in his chapter 14.
> There is a more consistent way to the stationary states of
> the hydrogen atom and Lamb shifts.

This is exactly the kind of statement that I criticised earlier. You
have failed to point out any inconsistency in the equivalence between
the Dirac wave function and Fock space state points of view. Yet you
continue to make unsubstantiated claims about its consistency, despite
an explicit demonstration of the equivalence. If you make this statement
because you are confident of your opinion, this casts a shadow on the
validity of other statements that you believe to be true.

Igor

Igor Khavkine

<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>On 2005-06-05, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n>\n>> If you agree that fields are a good starting point for the description\n>> of quasiparticles in a metal (as you seem to in the above quote), you\n>> have to agree that fields are also a good starting point for the\n>> description of electrons in a vacuum. Because if you don\'t, you\'ll have\n>> proverbial egg on your face if the vacuum is discovered to have some\n>> "more fundamental" structure.\n>\n> I don\'t think that vacuum has some fundamental structure.\n> There is no experimental evidence to suggest that, and your favorite\n> Occam\'s razor cuts out any remaining doubts.\n\nIt is true that we have not observed any non-trivial structure in the\nvacuum. In this case Occam\'s razor dismisses theories that do postulate\nsome fundamental structure to the vacuum in order to describe\nobservations. And I completely agree with that. For the purposes of\nexplaining observations, the Standard Model is plenty good enough and\nit\'s vacuum has no structure.\n\nOccam\'s razor is irrelevant to the present discussion because we are\ntalking about which mathematical starting point (local fields or\nPoincare invariant particle states) is advantageous in formulating field\ntheory. In QED, or any Poincare invariant theory, the two formulations\nare equivalent. But this is no longer so when looking at other\napplications of QFT. The particle interpretation changes wildly from\ntheory to theory, while the field interpretation provides a common\nground for all of them (given examples were curved space times,\naccelerated frames of reference, emergent systems like metals and\nsuperconductors).\n\nTaking insight gained elsewhere and coming back to QED, fields provide a\nbetter formulation of the theory. In addition, fields provide their\nusers with an additional layer of security agains getting proverbial egg\non their faces *if* the vacuum of QED is discovered to have some\nstructure.\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 [itex]2005-06-05,[/itex] Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:
>
>> If you agree that fields are a good starting point for the description
>> of quasiparticles in a metal (as you seem to in the above quote), you
>> have to agree that fields are also a good starting point for the
>> description of electrons in a vacuum. Because if you don't, you'll have
>> proverbial egg on your face if the vacuum is discovered to have some
>> "more fundamental" structure.
>
> I don't think that vacuum has some fundamental structure.
> There is no experimental evidence to suggest that, and your favorite
> Occam's razor cuts out any remaining doubts.

It is true that we have not observed any non-trivial structure in the
vacuum. In this case Occam's razor dismisses theories that do postulate
some fundamental structure to the vacuum in order to describe
observations. And I completely agree with that. For the purposes of
explaining observations, the Standard Model is plenty good enough and
it's vacuum has no structure.

Occam's razor is irrelevant to the present discussion because we are
talking about which mathematical starting point (local fields or
Poincare invariant particle states) is advantageous in formulating field
theory. In QED, or any Poincare invariant theory, the two formulations
are equivalent. But this is no longer so when looking at other
applications of QFT. The particle interpretation changes wildly from
theory to theory, while the field interpretation provides a common
ground for all of them (given examples were curved space times,
accelerated frames of reference, emergent systems like metals and
superconductors).

Taking insight gained elsewhere and coming back to QED, fields provide a
better formulation of the theory. In addition, fields provide their
users with an additional layer of security agains getting proverbial egg
on their faces *if* the vacuum of QED is discovered to have some
structure.

Igor

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\nIgor Khavkine wrote:\n\n> Occam\'s razor is irrelevant to the present discussion because we are\n> talking about which mathematical starting point (local fields or\n> Poincare invariant particle states) is advantageous in formulating field\n> theory. In QED, or any Poincare invariant theory, the two formulations\n> are equivalent. But this is no longer so when looking at other\n> applications of QFT. The particle interpretation changes wildly from\n> theory to theory, while the field interpretation provides a common\n> ground for all of them (given examples were curved space times,\n> accelerated frames of reference, emergent systems like metals and\n> superconductors).\n\nI can\'t tell anything about curved space-times or accelerated frames.\nThough, I would like to mention that there is a significant\ndifference between QFT applications to solids and QED.\n\nFor example, the electron-phonon interaction operator in QFT treatment\nof solids formally resembles the electron-photon trilinear interaction\nin QED. The difference is that an electron in a solid *really* can\nspontaneously emit a phonon. Electrons in a solid are *really* dressed\nwith a coat of virtual phonons, thus forming a polaron. The mass and\ncharge of the polaron are *really* different from those of the bare\nparticle. In this case, the renormalization describes a *real* physical\nphenomenon of interaction between electron and physical medium (crystal\nlattice).\n\nThis is not the case in QED. You cannot take the physical electron\nout of the medium (vacuum). Unlike in solid state, bare electrons in\nQED are just mathematical artifacts. The "dressed particle" formalism\ndescribes all experimental data (e.g., the S-matrix) without involving\nthe concept of renormalization.\n\nMy point is that you cannot invoke analogy with solid state\napplications to justify you statements about QED.\nThere are similarities, but there are importanrt differences as well.\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>Igor Khavkine wrote:

> Occam's razor is irrelevant to the present discussion because we are
> talking about which mathematical starting point (local fields or
> Poincare invariant particle states) is advantageous in formulating field
> theory. In QED, or any Poincare invariant theory, the two formulations
> are equivalent. But this is no longer so when looking at other
> applications of QFT. The particle interpretation changes wildly from
> theory to theory, while the field interpretation provides a common
> ground for all of them (given examples were curved space times,
> accelerated frames of reference, emergent systems like metals and
> superconductors).

I can't tell anything about curved space-times or accelerated frames.
Though, I would like to mention that there is a significant
difference between QFT applications to solids and QED.

For example, the electron-phonon interaction operator in QFT treatment
of solids formally resembles the electron-photon trilinear interaction
in QED. The difference is that an electron in a solid *really* can
spontaneously emit a phonon. Electrons in a solid are *really* dressed
with a coat of virtual phonons, thus forming a polaron. The mass and
charge of the polaron are *really* different from those of the bare
particle. In this case, the renormalization describes [itex]a *real*[/itex] physical
phenomenon of interaction between electron and physical medium (crystal
lattice).

This is not the case in QED. You cannot take the physical electron
out of the medium (vacuum). Unlike in solid state, bare electrons in
QED are just mathematical artifacts. The "dressed particle" formalism
describes all experimental data (e.g., the S-matrix) without involving
the concept of renormalization.

My point is that you cannot invoke analogy with solid state
applications to justify you statements about QED.
There are similarities, but there are importanrt differences as well.

Eugene.

Igor Khavkine

<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>On 2005-06-05, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n\n>> This argument shows a deep misunderstanding of how the Dirac wave\n>> function represents the Poincare group. The wave function *itself* is an\n>> element of the (infinite dimensional) unitary representation of the\n>> Poincare group. The transformations act as\n>>\n>> [(Lambda,a)psi](x) = psi(Lambda^{-1}x - a).\n>\n>>\n>> These transformations are unitary, since the action of the Poincare\n>> group preserve the inner product\n>>\n>> (psi,phi) = int psi(x)* phi(x) dx.\n>\n> You forgot the transformation of the field components\n>\n> [(Lambda,a)psi](x) = D(Lambda) psi(Lambda^{-1}x - a). (1)\n>\n> where D(Lambda) are 4-by-4 matrices representing the Lorentz subgroup\n> of the Poincare group.\n> The action (1) is not unitary, because matrices D(Lambda) are not\n> unitary.\n\nNo I have not forgotten this. To be explicit:\n\n(psi,phi) = sum_s int d^3x psi_s(x,t)* phi_s(x,t).\n\nThe integral is over the spacial slice at time t. Clearly, translations\nare represented unitarily. Since spacial rotations are represented\nunitarily over the spinor indices, the are also represented unitarily\nwith respect to the above inner product. Time translations are\nrepresented unitarily as a consequence of the equation of motion (the\nDirac equation).\n\nThe only non-trivial case to check are the boost transformations. Boost\ntransformations mix x and t coordinates in the argument of psi(x,t). But\nthe integral in the inner product is only over the values of of psi(x,t)\nat a fixed time. However, the equation of motion comes to the rescue\nagain and keeps the above inner product invariant. This can be checked\nexplicitly by verifying that the infinitesimal boost generator\n(combining the parts acting on the spinor indices s and the space-time\ncoordinates (x,t)) is hermitian with respect to the above inner product.\nThis covers the full algebra of Poincare generators.\n\nThe important lesson here is that the above inner product is defined on\nthe set of solutions of the Dirac equation, not for arbitrary functions\npsi(x,t).\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 [itex]2005-06-05,[/itex] Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:

>> This argument shows a deep misunderstanding of how the Dirac wave
>> function represents the Poincare group. The wave function *itself* is an
>> element of the (infinite dimensional) unitary representation of the
>> Poincare group. The transformations act as
>>
>> [itex][(\Lambda,a)\psi](x) = \psi(\Lambda^{-1}x - a)[/itex].
>
>>
>> These transformations are unitary, since the action of the Poincare
>> group preserve the inner product
>>
>> [itex](\psi,\phi) = \int \psi(x)* \phi(x) dx[/itex].
>
> You forgot the transformation of the field components
>
> [itex][(\Lambda,a)\psi](x) = D(\Lambda) \psi(\Lambda^{-1}x - a)[/itex]. (1)
>
> where [itex]D(\Lambda)[/itex] are [itex]4-by-4[/itex] matrices representing the Lorentz subgroup
> of the Poincare group.
> The action (1) is not unitary, because matrices [itex]D(\Lambda)[/itex] are not
> unitary.

No I have not forgotten this. To be explicit:

[itex](\psi,\phi) = sum_s \int d^{3x} \psi_s(x,t)* \phi_s(x,t)[/itex].

The integral is over the spacial slice at time t. Clearly, translations
are represented unitarily. Since spacial rotations are represented
unitarily over the spinor indices, the are also represented unitarily
with respect to the above inner product. Time translations are
represented unitarily as a consequence of the equation of motion (the
Dirac equation).

The only non-trivial case to check are the boost transformations. Boost
transformations mix x and t coordinates in the argument of [itex]\psi(x,t)[/itex]. But
the integral in the inner product is only over the values of of [itex]\psi(x,t)at a[/itex] fixed time. However, the equation of motion comes to the rescue
again and keeps the above inner product invariant. This can be checked
explicitly by verifying that the infinitesimal boost generator
(combining the parts acting on the spinor indices s and the space-time
coordinates (x,t)) is hermitian with respect to the above inner product.
This covers the full algebra of Poincare generators.

The important lesson here is that the above inner product is defined on
the set of solutions of the Dirac equation, not for arbitrary functions
[itex]\psi(x,t)[/itex].

Igor

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>Aaron Bergman wrote:\n\n>>Could we stay within QED, please?\n>\n>\n> No. If you want to talk about QFT, you have to talk about QFT.\n\n\nI want to talk only about QED. That\'s the subject I know and understand\npretty well. Let me then say that the quantum theory of electromagnetic\ninteractions can be formulated as a theory of particles rather than\nfields. This "dressed particle" approach has important advantages:\nexistence of a finite Hamiltonian with clear physical meaning;\nrenormalization is not needed; bound states and time evolution can\nbe calculated directly by standard formulas of quantum mechanics,\netc.\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>Aaron Bergman wrote:

>>Could we stay within QED, please?
>
>
> No. If you want to talk about QFT, you have to talk about QFT.

I want to talk only about QED. That's the subject I know and understand
pretty well. Let me then say that the quantum theory of electromagnetic
interactions can be formulated as a theory of particles rather than
fields. This "dressed particle" approach has important advantages:
existence of a finite Hamiltonian with clear physical meaning;
renormalization is not needed; bound states and time evolution can
be calculated directly by standard formulas of quantum mechanics,
etc.

Eugene

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>Igor Khavkine wrote:\n> On 2005-06-03, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n>\n>>Yes, you can obtain numerical functions satisfying the Dirac equation\n>>using Weinberg\'s formulas (14.1.4) - (14.1.5). However, these functions\n>>have nothing to do with probability amplitudes describing states.\n>>I learned in quantum mechanics that wave functions are expansion\n>>coefficients of the state vectors in certain orthonormal basis.\n>>I haven\'t seen the definition of a wave function as a matrix element\n>>of quantum field, like in (14.1.4) - (14.1.5). This looks like a\n>>completely new approach to quantum mechanics. How would you write\n>>the wave function of a 2-particle system in this notation?\n>\n>\n> No, this is not how quantum mechanics is usually taught, but that\n> shouldn\'t stop you from learning to look at it this way when you have\n> the opportunity. This approach is not new at all. Have you been exposed\n> to second quantization outside the treatment of QED? Schroedinger\'s wave\n> function can be obtained in exactly the same way, but as matrix elements\n> of the Schroedinger field operators. Have you read Dirac\'s book\n> _Principles of Quantum Mechanics_? He deals with this kind of\n> equivalence issue in the chapter on identical particles. The\n> probabilistic interpretation only needs a positive definite inner\n> product. In both the Schroedinger and Dirac wave pictures, one is\n> readily available.\n\nI may agree with you if you can prove that the representation of\nthe Poincare group built on Dirac\'s functions\n\nU(Lambda, a) psi(x) = D(Lambda^{-1}) psi(Lambda x + a) (1)\n\nis unitarily equivalent to the Wigner\'s representation. But this is not\ntrue. Representation (1) is not unitary. All irreducible unitary\nrepresentation were built in Wigner\'s 1939 paper, and (1) is not\none of them.\n\nI know exactly what is the meaning of 2 components of the Wigner\'s\nwave function. Their squares give the probabilities of measuring spin\nup and spin down, respectively. These are the probabilities\nthat the particle will go up (or down) in the Stern-Gerlach apparatus.\nCould you explain me what is\nthe physical meaning of 4 components of the Dirac\'s wave function?\n\n>\n> It is easy to write multiplarticle states as well. They correspond to\n> multivariable wave functions. For instance:\n>\n> psi++(x,y) = <0|PSI(x) PSI(y) |2-particle>\n> psi-+(x,y) = <0|PSI(x)* PSI(y) |2-particle>\n> psi+-(x,y) = <0|PSI(x) PSI(y)*|2-particle>\n> psi--(x,y) = <0|PSI(x)* PSI(y)*|2-particle>\n>\n> Will give you, respectively, the pos-pos, neg-pos, pos-neg, and neg-neg\n> frequency components of a 2-particle antisymmetric wave function.\n\nAgain, the physical meaning of positive and negative frequency\ncomponents is rather obscure. What about pos-pos, etc. components?\nIs there an experimental setup that can measure these components?\n\n>\n>\n>>I am surprised that Weinberg took this route in his chapter 14.\n>>There is a more consistent way to the stationary states of\n>>the hydrogen atom and Lamb shifts.\n>\n>\n> This is exactly the kind of statement that I criticised earlier. You\n> have failed to point out any inconsistency in the equivalence between\n> the Dirac wave function and Fock space state points of view. Yet you\n> continue to make unsubstantiated claims about its consistency, despite\n> an explicit demonstration of the equivalence.\n\nYou haven\'t demonstrated the equivalence yet. I don\'t think there is\nan unitary equivalence.\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>Igor Khavkine wrote:
> On [itex]2005-06-03,[/itex] Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>
>>Yes, you can obtain numerical functions satisfying the Dirac equation
>>using Weinberg's formulas (14.1.[itex]4) - (14.1.5).[/itex] However, these functions
>>have nothing to do with probability amplitudes describing states.
>>I learned in quantum mechanics that wave functions are expansion
>>coefficients of the state vectors in certain orthonormal basis.
>>I haven't seen the definition of a wave function as a matrix element
>>of quantum field, like in (14.1.[itex]4) - (14.1.5).[/itex] This looks like a
>>completely new approach to quantum mechanics. How would you write
>>the wave function of a 2-particle system in this notation?
>
>
> No, this is not how quantum mechanics is usually taught, but that
> shouldn't stop you from learning to look at it this way when you have
> the opportunity. This approach is not new at all. Have you been exposed
> to second quantization outside the treatment of QED? Schroedinger's wave
> function can be obtained in exactly the same way, but as matrix elements
> of the Schroedinger field operators. Have you read Dirac's book
> _Principles of Quantum Mechanics_? He deals with this kind of
> equivalence issue in the chapter on identical particles. The
> probabilistic interpretation only needs a positive definite inner
> product. In both the Schroedinger and Dirac wave pictures, one is
> readily available.

I may agree with you if you can prove that the representation of
the Poincare group built on Dirac's functions

[itex]U(\Lambda, a) \psi(x) = D(\Lambda^{-1}) \psi(\Lambda x + a)[/itex] (1)

is unitarily equivalent to the Wigner's representation. But this is not
true. Representation (1) is not unitary. All irreducible unitary
representation were built in Wigner's 1939 paper, and (1) is not
one of them.

I know exactly what is the meaning of 2 components of the Wigner's
wave function. Their squares give the probabilities of measuring spin
up and spin down, respectively. These are the probabilities
that the particle will go up (or down) in the Stern-Gerlach apparatus.
Could you explain me what is
the physical meaning of 4 components of the Dirac's wave function?

>
> It is easy to write multiplarticle states as well. They correspond to
> multivariable wave functions. For instance:
>
> [itex]\psi++(x,y) = <0|\PSI(x) \PSI(y)[/itex] |2-particle>
> [itex]\psi-+(x,y) = <0|\PSI(x)* \PSI(y)[/itex] |2-particle>
> [itex]\psi+-(x,y) = <0|\PSI(x) \PSI(y)*|2-particle>[/itex]
> \psi--(x,y) [itex]= <0|\PSI(x)* \PSI(y)*|2-particle>[/itex]
>
> Will give you, respectively, the pos-pos, neg-pos, pos-neg, and neg-neg
> frequency components of a 2-particle antisymmetric wave function.

Again, the physical meaning of positive and negative frequency
components is rather obscure. What about pos-pos, etc. components?
Is there an experimental setup that can measure these components?

>
>
>>I am surprised that Weinberg took this route in his chapter 14.
>>There is a more consistent way to the stationary states of
>>the hydrogen atom and Lamb shifts.
>
>
> This is exactly the kind of statement that I criticised earlier. You
> have failed to point out any inconsistency in the equivalence between
> the Dirac wave function and Fock space state points of view. Yet you
> continue to make unsubstantiated claims about its consistency, despite
> an explicit demonstration of the equivalence.

You haven't demonstrated the equivalence yet. I don't think there is
an unitary equivalence.

Eugene.

Igor Khavkine

<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>On 2005-06-06, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n\n> I can\'t tell anything about curved space-times or accelerated frames.\n> Though, I would like to mention that there is a significant\n> difference between QFT applications to solids and QED.\n\nYes, there is. It\'s called Poincare invariance. All the rest is pretty\nmuch the same.\n\n> For example, the electron-phonon interaction operator in QFT treatment\n> of solids formally resembles the electron-photon trilinear interaction\n> in QED. The difference is that an electron in a solid *really* can\n> spontaneously emit a phonon. Electrons in a solid are *really* dressed\n> with a coat of virtual phonons, thus forming a polaron. The mass and\n> charge of the polaron are *really* different from those of the bare\n> particle. In this case, the renormalization describes a *real* physical\n> phenomenon of interaction between electron and physical medium (crystal\n> lattice).\n\nYou are right about one thing. Physical electron states in a crystal\n(those with definite crystal momentum) can spontaneously emit a phonon.\nBut lets examine the reasons:\n\n* Finite temperature. There is a probability of either absorbing from or\nemitting into the phonon thermal bath.\n* Finite size. Electrons are scattered off boundaries of the sample.\n* Impurities. Electrons are scattered off impurities.\n\nNow, lets try to remove each of these obstacles one by one. First\nsynthesize a perfectly pure crystal. No foreign substances, dislocations\nor the like. This is actually possible to a high degree of precision\nwith modern experimental techniques. Even if the crystal is not 100%\npure, it\'s safe to make this assumption formulating a theory.\n\nAssume that the crystal is infinite. Obviously, this is impossible in\nreality. But infinity is a relative term, at least for physicists.\nAtomic spacing in a crystal is of the order of Angstroms while\nmacroscopic samples are of the order of centimeters. That\'s a 10^8 ratio\nof scales, for many purposes that\'s as good as infinity. Again, a safe\ntheoretical assumption to make is that the unerlying crystal lattice is\nunbounded.\n\nLast but not least, for physicists, like infinity, zero is relative\nnumber. The relevant temperature scale in metals is the Fermi energy.\nIts typical values are in the 10^4\'s of Kelvin. Modern liquid Helium\ncryostats can easily get below the 1K mark. Again, an easy theoretical\nassumption to make is that we work at zero temperature.\n\nWhat you get as a result is a zero temperature theory (just like QED)\nwith space and time translation invariance (just like QED). But without\nrotational or boost symmetry (unlike QED). Due to translational\ninvariance, physical electron states (described in this approximation)\nare indeed stable and do not spontaneously emit phonons (just like QED).\n\nCalculations in this theory are performed in exactly the same way as\nany other QFT (including QED). Feynman diagrams, propagators,\nrenormalization, and all the other field theory tools play as much of a\nrole there as anywhere else. As I\'ve claimed previously, mathematically\nthese theories are if not identical at least very close to each other.\n\nEven more differences can be swept away when considering only low energy\nexcitations above the Fermi surface. Since they only involve long\nwavelength modes, the underlying crystal structure could be neglected\nand replaced by a continuum. In this case, renormalization is\nindispensable because ultraviolet infinities can make an appearance as\nwell. However, we always have a recourse to the underlying discrete\ntheory that tells us to which finite values should diverging transitions\namplitudes must be renormalized. This is not possible, because at the\nmoment we have no underlying theory capable of making these predictions.\nBut an appeal to underlying theory need not even be made since these\ntransition amplitudes can be measured experimentally, and the resulting\nvalues could be used instead.\n\nThus I\'m quite justified in saying that QED and the theory of electrons\nin a metal are very similar. The fact that one is assumed to be\nfundamental while the other is not is irrelevant. As you can see, both\ncan be formulated on their own terms without reference to any underlying\ntheory, simply by appealing to experiment. Neither theory looks inside\nits own vacuum, thus its composition is quite irrelevant to calculations.\n\n> This is not the case in QED. You cannot take the physical electron\n> out of the medium (vacuum). Unlike in solid state, bare electrons in\n> QED are just mathematical artifacts. The "dressed particle" formalism\n> describes all experimental data (e.g., the S-matrix) without involving\n> the concept of renormalization.\n\nThe distinction between the vacuua of QED and metals is a matter of\nphysics, not mathematics. It has not once been claimed in this or\nprevious conversations that bare particles are assigned physical\nmeaning. Thus you are not contradicting anyone. Let me remind you that\nthe point of contention was your claim that a mathematical formulation\nof the QED in terms of bare particles is impotent when it comes to\nvarious kinds of calculations. This claim is what I\'ve been\ncontradicting and falsifying with examples not only from QED but from\nother applications of QFT, including condensed matter systems.\n\nAs to the issue of renormalization, you are once again lapsing into the\nhabit of repeating statements that have been shown to be incorrect. No,\nyou cannot do without renormalization. You start with the standard QED\nHamiltonian and use renormalization to evaluate coefficients in your own\nHamiltonian. Thus until you calculate your Hamiltonian to all orders in\nperturbation theory, you will still need to refer to the original\nHamiltonian and apply renormalization. If you do not wish to refer to\nthe original formulation of QED, you have to use experimental input.\nHowever this approach is not much different from writing down a\nphenomenological effective field theory, which can be done from scratch\nwithout any extra work.\n\n> My point is that you cannot invoke analogy with solid state\n> applications to justify you statements about QED.\n> There are similarities, but there are importanrt differences as well.\n\nAnd what statements would that be? I\'ve said the following. QED:\nparticle and field formulations are equivalent. Other QFT\'s, including\nthose from condensed matter physics: sometimes the particle description\nbreaks down, but the field one doesn\'t. Coming back to QED: insight from\nother QFT\'s tells us that fields are a better starting point for the\nformulation.\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 [itex]2005-06-06,[/itex] Eugene Stefanovich <eugenev@synopsys.com> wrote:

> I can't tell anything about curved space-times or accelerated frames.
> Though, I would like to mention that there is a significant
> difference between QFT applications to solids and QED.

Yes, there is. It's called Poincare invariance. All the rest is pretty
much the same.

> For example, the electron-phonon interaction operator in QFT treatment
> of solids formally resembles the electron-photon trilinear interaction
> in QED. The difference is that an electron in a solid *really* can
> spontaneously emit a phonon. Electrons in a solid are *really* dressed
> with a coat of virtual phonons, thus forming a polaron. The mass and
> charge of the polaron are *really* different from those of the bare
> particle. In this case, the renormalization describes [itex]a *real*[/itex] physical
> phenomenon of interaction between electron and physical medium (crystal
> lattice).

You are right about one thing. Physical electron states in a crystal
(those with definite crystal momentum) can spontaneously emit a phonon.
But lets examine the reasons:

* Finite temperature. There is a probability of either absorbing from or
emitting into the phonon thermal bath.
* Finite size. Electrons are scattered off boundaries of the sample.
* Impurities. Electrons are scattered off impurities.

Now, lets try to remove each of these obstacles one by one. First
synthesize a perfectly pure crystal. No foreign substances, dislocations
or the like. This is actually possible to a high degree of precision
with modern experimental techniques. Even if the crystal is not 100%
pure, it's safe to make this assumption formulating a theory.

Assume that the crystal is infinite. Obviously, this is impossible in
reality. But infinity is a relative term, at least for physicists.
Atomic spacing in a crystal is of the order of Angstroms while
macroscopic samples are of the order of centimeters. That's [itex]a 10^8[/itex] ratio
of scales, for many purposes that's as good as infinity. Again, a safe
theoretical assumption to make is that the unerlying crystal lattice is
unbounded.

Last but not least, for physicists, like infinity, zero is relative
number. The relevant temperature scale in metals is the Fermi energy.
Its typical values are in the [itex]10^4's[/itex] of Kelvin. Modern liquid Helium
cryostats can easily get below the 1K mark. Again, an easy theoretical
assumption to make is that we work at zero temperature.

What you get as a result is a zero temperature theory (just like QED)
with space and time translation invariance (just like QED). But without
rotational or boost symmetry (unlike QED). Due to translational
invariance, physical electron states (described in this approximation)
are indeed stable and do not spontaneously emit phonons (just like QED).

Calculations in this theory are performed in exactly the same way as
any other QFT (including QED). Feynman diagrams, propagators,
renormalization, and all the other field theory tools play as much of a
role there as anywhere else. As I've claimed previously, mathematically
these theories are if not identical at least very close to each other.

Even more differences can be swept away when considering only low energy
excitations above the Fermi surface. Since they only involve long
wavelength modes, the underlying crystal structure could be neglected
and replaced by a continuum. In this case, renormalization is
indispensable because ultraviolet infinities can make an appearance as
well. However, we always have a recourse to the underlying discrete
theory that tells us to which finite values should diverging transitions
amplitudes must be renormalized. This is not possible, because at the
moment we have no underlying theory capable of making these predictions.
But an appeal to underlying theory need not even be made since these
transition amplitudes can be measured experimentally, and the resulting
values could be used instead.

Thus I'm quite justified in saying that QED and the theory of electrons
in a metal are very similar. The fact that one is assumed to be
fundamental while the other is not is irrelevant. As you can see, both
can be formulated on their own terms without reference to any underlying
theory, simply by appealing to experiment. Neither theory looks inside
its own vacuum, thus its composition is quite irrelevant to calculations.

> This is not the case in QED. You cannot take the physical electron
> out of the medium (vacuum). Unlike in solid state, bare electrons in
> QED are just mathematical artifacts. The "dressed particle" formalism
> describes all experimental data (e.g., the S-matrix) without involving
> the concept of renormalization.

The distinction between the vacuua of QED and metals is a matter of
physics, not mathematics. It has not once been claimed in this or
previous conversations that bare particles are assigned physical
meaning. Thus you are not contradicting anyone. Let me remind you that
the point of contention was your claim that a mathematical formulation
of the QED in terms of bare particles is impotent when it comes to
various kinds of calculations. This claim is what I've been
contradicting and falsifying with examples not only from QED but from
other applications of QFT, including condensed matter systems.

As to the issue of renormalization, you are once again lapsing into the
habit of repeating statements that have been shown to be incorrect. No,
you cannot do without renormalization. You start with the standard QED
Hamiltonian and use renormalization to evaluate coefficients in your own
Hamiltonian. Thus until you calculate your Hamiltonian to all orders in
perturbation theory, you will still need to refer to the original
Hamiltonian and apply renormalization. If you do not wish to refer to
the original formulation of QED, you have to use experimental input.
However this approach is not much different from writing down a
phenomenological effective field theory, which can be done from scratch
without any extra work.

> My point is that you cannot invoke analogy with solid state
> applications to justify you statements about QED.
> There are similarities, but there are importanrt differences as well.

And what statements would that be? I've said the following. QED:
particle and field formulations are equivalent. Other QFT's, including
those from condensed matter physics: sometimes the particle description
breaks down, but the field one doesn't. Coming back to QED: insight from
other QFT's tells us that fields are a better starting point for the
formulation.

Igor

Aaron Bergman

<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 <42A3F65C.8040401@synopsys.com>,\nEugene Stefanovich <eugenev@synopsys.com> wrote:\n\n> Aaron Bergman wrote:\n>\n> >>Could we stay within QED, please?\n> >\n> >\n> > No. If you want to talk about QFT, you have to talk about QFT.\n>\n> I want to talk only about QED. That\'s the subject I know and understand\n> pretty well. Let me then say that the quantum theory of electromagnetic\n> interactions can be formulated as a theory of particles rather than\n> fields. This "dressed particle" approach has important advantages:\n> existence of a finite Hamiltonian with clear physical meaning;\n> renormalization is not needed; bound states and time evolution can\n> be calculated directly by standard formulas of quantum mechanics,\n> etc.\n\nThe disadvantages to QED are, of course, that it probably does not exist\nexcept as an effective field theory. I have not followed this thread\n(and don\'t have the energy to get too involved in it), but it sounds\nlike you\'re exactly doing renormalization using different words.\n\nThat you can do QED in flat space in a first quantized form isn\'t a\nsurprise to anyone, but I tend to think that quarks are important, too.\n\nAaron\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 <42A3F65C.8040401@synopsys.com>,
Eugene Stefanovich <eugenev@synopsys.com> wrote:

> Aaron Bergman wrote:
>
> >>Could we stay within QED, please?
> >
> >
> > No. If you want to talk about QFT, you have to talk about QFT.
>
> I want to talk only about QED. That's the subject I know and understand
> pretty well. Let me then say that the quantum theory of electromagnetic
> interactions can be formulated as a theory of particles rather than
> fields. This "dressed particle" approach has important advantages:
> existence of a finite Hamiltonian with clear physical meaning;
> renormalization is not needed; bound states and time evolution can
> be calculated directly by standard formulas of quantum mechanics,
> etc.

The disadvantages to QED are, of course, that it probably does not exist
except as an effective field theory. I have not followed this thread
(and don't have the energy to get too involved in it), but it sounds
like you're exactly doing renormalization using different words.

That you can do QED in flat space in a first quantized form isn't a
surprise to anyone, but I tend to think that quarks are important, too.

Aaron

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\nIgor Khavkine wrote:\n\n> You are right about one thing. Physical electron states in a crystal\n> (those with definite crystal momentum) can spontaneously emit a phonon.\n> But lets examine the reasons:\n>\n> * Finite temperature. There is a probability of either absorbing from or\n> emitting into the phonon thermal bath.\n> * Finite size. Electrons are scattered off boundaries of the sample.\n> * Impurities. Electrons are scattered off impurities.\n>\n\nNo, these are not the only reasons. Electrons are coupled to phonons\neven in the perfect infinite crystal at zero temperatures. Physically,\nthe coupling occurs because the electron has charge and crystal\nlattice is\npolarizable (this applies mostly to dielectrics, and at much lesser\nextent to metals). Extra electron in the conduction band polarizes\nthe lattice around itself. Atoms (or ions) in the lattice become\ndisplaced from their normal positions. In QFT language this is described\nas a virtual phonon "coat" that follows the electron wherever it goes.\nAnother name for this object is "polaron". Polaron is the analogue of\nthe "dressed electron" in QED. The difference is that\n\n1) Polaron dressing does not not produce infinite corrections to the\nmass and charge (as in QED), due to the underlying granular structure\nof the lattice.\n\n2) Polaron dressing is perfectly physical and\neven observable, unlike in QED.\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>Igor Khavkine wrote:

> You are right about one thing. Physical electron states in a crystal
> (those with definite crystal momentum) can spontaneously emit a phonon.
> But lets examine the reasons:
>
> * Finite temperature. There is a probability of either absorbing from or
> emitting into the phonon thermal bath.
> * Finite size. Electrons are scattered off boundaries of the sample.
> * Impurities. Electrons are scattered off impurities.
>

No, these are not the only reasons. Electrons are coupled to phonons
even in the perfect infinite crystal at zero temperatures. Physically,
the coupling occurs because the electron has charge and crystal
lattice is
polarizable (this applies mostly to dielectrics, and at much lesser
extent to metals). Extra electron in the conduction band polarizes
the lattice around itself. Atoms (or ions) in the lattice become
displaced from their normal positions. In QFT language this is described
as a virtual phonon "coat" that follows the electron wherever it goes.
Another name for this object is "polaron". Polaron is the analogue of
the "dressed electron" in QED. The difference is that

1) Polaron dressing does not not produce infinite corrections to the
mass and charge (as in QED), due to the underlying granular structure
of the lattice.

2) Polaron dressing is perfectly physical and
even observable, unlike in QED.

Eugene.

Igor Khavkine

<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>On 2005-06-07, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n>\n> Igor Khavkine wrote:\n>\n>> You are right about one thing. Physical electron states in a crystal\n>> (those with definite crystal momentum) can spontaneously emit a phonon.\n>> But lets examine the reasons:\n>>\n>> * Finite temperature. There is a probability of either absorbing from or\n>> emitting into the phonon thermal bath.\n>> * Finite size. Electrons are scattered off boundaries of the sample.\n>> * Impurities. Electrons are scattered off impurities.\n>\n> No, these are not the only reasons. Electrons are coupled to phonons\n> even in the perfect infinite crystal at zero temperatures.\n\nAs I described, a single particle electron state in this situation will\nnever emit a phonon, simply because it is an eigenstate of the\nHamiltonian.\n\n> Physically,\n> the coupling occurs because the electron has charge and crystal\n> lattice is\n> polarizable (this applies mostly to dielectrics, and at much lesser\n> extent to metals). Extra electron in the conduction band polarizes\n> the lattice around itself. Atoms (or ions) in the lattice become\n> displaced from their normal positions. In QFT language this is described\n> as a virtual phonon "coat" that follows the electron wherever it goes.\n> Another name for this object is "polaron". Polaron is the analogue of\n> the "dressed electron" in QED. The difference is that\n>\n> 1) Polaron dressing does not not produce infinite corrections to the\n> mass and charge (as in QED), due to the underlying granular structure\n> of the lattice.\n>\n> 2) Polaron dressing is perfectly physical and\n> even observable, unlike in QED.\n\nOnce you remove the lattice all together (take the continuum limit and\nrestrict yourself to long wavelength measurements only), you run into\nthe same ultraviolet difficulties as you would expect in a continuum\ntheory. Then you get infinite renormalizing corrections. They are\ninfinite compared to what would be the mass and charge of a bare\nelectron. In this case a bare electron is not the same as an electron\noutside the crystal. But it doesn\'t matter, because at the assumed level\nof experimental precision you can\'t observe an electron outside the\nlattice anyway. Formulating the theory in terms of bare electrons is\nperfectly reasonable since they are not observed.\n\nThis is the same\nsituation as in QED. Our instruments tell us that we can\'t take an\nelectron out of the vacuum and look at its properties without\ninteraction. Hence the theory need not make any sensible statments about\nwhat, say, the mass and charge of a bare electron are. So a mathematical\nformulation of the theory where the bare mass and charge are infnite is\nacceptable (with all the fine print provided by renormalization).\n\nBut we are getting farther and farther from the topic. This subthread is\nabout the relative merits of the field and particle formulations. So on\ntopic, let me throw someting else into the mix. The mathematical\nformulation of QFT can handle a large variety of theories, with or\nwithout symmetries. Take for instance a quantum field interacting with a\ntime and space varying external field. There is neither space nor time\ntranslation symmetry in this theory. So particle states cannot be\nidentified as eigenstates of any sort of symmetry generators. However, a\nsatisfactory quantization can be constructed by solving partial\ndifferential equations for the wave functions obtained as matrix\nelements of field operators. And yes, such a model has numerous physical\napplications.\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 [itex]2005-06-07,[/itex] Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>
> Igor Khavkine wrote:
>
>> You are right about one thing. Physical electron states in a crystal
>> (those with definite crystal momentum) can spontaneously emit a phonon.
>> But lets examine the reasons:
>>
[itex]>> *[/itex] Finite temperature. There is a probability of either absorbing from or
>> emitting into the phonon thermal bath.
[itex]>> *[/itex] Finite size. Electrons are scattered off boundaries of the sample.
[itex]>> *[/itex] Impurities. Electrons are scattered off impurities.
>
> No, these are not the only reasons. Electrons are coupled to phonons
> even in the perfect infinite crystal at zero temperatures.

As I described, a single particle electron state in this situation will
never emit a phonon, simply because it is an eigenstate of the
Hamiltonian.

> Physically,
> the coupling occurs because the electron has charge and crystal
> lattice is
> polarizable (this applies mostly to dielectrics, and at much lesser
> extent to metals). Extra electron in the conduction band polarizes
> the lattice around itself. Atoms (or ions) in the lattice become
> displaced from their normal positions. In QFT language this is described
> as a virtual phonon "coat" that follows the electron wherever it goes.
> Another name for this object is "polaron". Polaron is the analogue of
> the "dressed electron" in QED. The difference is that
>
> 1) Polaron dressing does not not produce infinite corrections to the
> mass and charge (as in QED), due to the underlying granular structure
> of the lattice.
>
> 2) Polaron dressing is perfectly physical and
> even observable, unlike in QED.

Once you remove the lattice all together (take the continuum limit and
restrict yourself to long wavelength measurements only), you run into
the same ultraviolet difficulties as you would expect in a continuum
theory. Then you get infinite renormalizing corrections. They are
infinite compared to what would be the mass and charge of a bare
electron. In this case a bare electron is not the same as an electron
outside the crystal. But it doesn't matter, because at the assumed level
of experimental precision you can't observe an electron outside the
lattice anyway. Formulating the theory in terms of bare electrons is
perfectly reasonable since they are not observed.

This is the same
situation as in QED. Our instruments tell us that we can't take an
electron out of the vacuum and look at its properties without
interaction. Hence the theory need not make any sensible statments about
what, say, the mass and charge of a bare electron are. So a mathematical
formulation of the theory where the bare mass and charge are infnite is
acceptable (with all the fine print provided by renormalization).

But we are getting farther and farther from the topic. This subthread is
about the relative merits of the field and particle formulations. So on
topic, let me throw someting else into the mix. The mathematical
formulation of QFT can handle a large variety of theories, with or
without symmetries. Take for instance a quantum field interacting with a
time and space varying external field. There is neither space nor time
translation symmetry in this theory. So particle states cannot be
identified as eigenstates of any sort of symmetry generators. However, a
satisfactory quantization can be constructed by solving partial
differential equations for the wave functions obtained as matrix
elements of field operators. And yes, such a model has numerous physical
applications.

Igor

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>Igor Khavkine wrote:\n> On 2005-06-05, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n>>Igor Khavkine wrote:\n>\n>\n>>>This argument shows a deep misunderstanding of how the Dirac wave\n>>>function represents the Poincare group. The wave function *itself* is an\n>>>element of the (infinite dimensional) unitary representation of the\n>>>Poincare group. The transformations act as\n>>>\n>>> [(Lambda,a)psi](x) = psi(Lambda^{-1}x - a).\n>>\n>>>These transformations are unitary, since the action of the Poincare\n>>>group preserve the inner product\n>>>\n>>> (psi,phi) = int psi(x)* phi(x) dx.\n>>\n>>You forgot the transformation of the field components\n>>\n>>[(Lambda,a)psi](x) = D(Lambda) psi(Lambda^{-1}x - a). (1)\n>>\n>>where D(Lambda) are 4-by-4 matrices representing the Lorentz subgroup\n>>of the Poincare group.\n>>The action (1) is not unitary, because matrices D(Lambda) are not\n>>unitary.\n>\n>\n> No I have not forgotten this. To be explicit:\n>\n> (psi,phi) = sum_s int d^3x psi_s(x,t)* phi_s(x,t).\n>\n> The integral is over the spacial slice at time t. Clearly, translations\n> are represented unitarily. Since spacial rotations are represented\n> unitarily over the spinor indices, the are also represented unitarily\n> with respect to the above inner product. Time translations are\n> represented unitarily as a consequence of the equation of motion (the\n> Dirac equation).\n>\n> The only non-trivial case to check are the boost transformations. Boost\n> transformations mix x and t coordinates in the argument of psi(x,t). But\n> the integral in the inner product is only over the values of of psi(x,t)\n> at a fixed time. However, the equation of motion comes to the rescue\n> again and keeps the above inner product invariant. This can be checked\n> explicitly by verifying that the infinitesimal boost generator\n> (combining the parts acting on the spinor indices s and the space-time\n> coordinates (x,t)) is hermitian with respect to the above inner product.\n> This covers the full algebra of Poincare generators.\n>\n> The important lesson here is that the above inner product is defined on\n> the set of solutions of the Dirac equation, not for arbitrary functions\n> psi(x,t).\n\nLet me first agree with you. Yes, you can project out 2 extra\ncomponents, and restrict the space of functions using Dirac\'s\nequation, and show that resulting functions afford an unitary\nirreducible representation of the Poincare group, just as\nWigner\'s functions do. You can also apply the Foldy-Wouthuysen\ntransformation to Dirac\'s functions and arrive to the operators\nof velocity and position that make certain physical sense.\nYes, you can do all that. My question is whether all these non-trivial\nmanipulations have any physical meaning? It seems to me that\nwe have started from something (solutions of Dirac\'s equation)\nthat didn\'t have much physical sense and then applied a series of\npatches (like Foldy-Wouthuysen transformations) to put our creation\nin a better shape.\n\nWhy should we go through all this hassle when we have a perfect\nsolution for the 1-particle problem given by the Wigner\'s prescription?\nThe physical foundations for this prescription are impeccable:\n1. The Hilbert space of the system must contain an unitary\n(= preservation of probabilities) representation of the Poincare\ngroup (= principle of relativity).\n2. Since the system is elementary, this should be the "smallest"\n(= irreducible) Hilbert space.\n\nFrom these two postulate one immediately obtains the classification\nof elementary particles by mass and spin, wave functions without\nextra unphysical components, the transformation laws\n(wrt all inertial transformations) preserving the inner product.\nExplicit expressions for all operators of observables (momentum, energy,\nvelocity, spin, position, etc.) are obtained in a straightforward\nfashion without any adjustments.\n\nUnfortunately, most textbooks prefer to talk about Dirac\'s wave\nfunctions, and don\'t pay any attention to the brilliant Wigner\'s\napproach. I think, this is just for historical reasons.\nSimply, Dirac was first. I remember reading somewhere that Wigner\'s\npaper was not accepted in the major physical journal of that time\n(Phys. Rev.?) and he was forced to publish in a mathematical\njournal. I find it weird, because to me his article\n(E. Wigner, On Unitary Representations of the Inhomogeneous Lorentz\nGroup, Ann. of Math., 40, 149, (1939)) represents one of the few\nfinest works of theoretical physics in the 20th century.\n\n\nPlease understand me right. I don\'t\nwant to diminish the hugh contribution that Dirac\'s equation made\nto the development of theoretical physics. However, I think, in\n2005, almost 80 years after its discovery, it is naive to think\nthat Dirac\'s equation is a relativistic equivalent of the Schroedinger\nequation. Now we know that in relativistic physics the energy of a\nfree particle with mass m is related to the momentum by formula\nE = sqrt(m^2 + p^2). Therefore, the Hamiltonian in the momentum\nrepresentation is simply H = sqrt(M^2 + P^2), and not the 4x4 matrix\nwritten by Dirac.\n\nIf it is not a good idea to describe 1-electron states by Dirac\'s\n"wavefunctions", then maybe these functions have no place in\nrelativistic quantum physics? No, not at all. There is a place\nwhere you simply cannot do without Dirac\'s functions.\nAs Weinberg showed in his Chapter 5, Dirac\'s functions\n(this time you need to consider operator functions, or quantum fields,\nbuilt as linear combinations of creation and annihilation operators)\nare absolutely essential for construction of Poincare invariant\ncluster separable interactions in the Fock space.\n\nSo, my suggestion is\n1) use Wigner\'s approach for description of 1-particle states,\n2) build n-particle Hilbert spaces as tensor product\nof Wigner\'s 1-particle spaces\n3) Build the Fock space as a direct sum of n-particle spaces.\n4) Use Dirac\'s (and other) quantum fields to construct non-trivial\ninterparticle interactions in the Fock space.\n\nEugene.\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>Igor Khavkine wrote:
> On [itex]2005-06-05,[/itex] Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>>Igor Khavkine wrote:
>
>
>>>This argument shows a deep misunderstanding of how the Dirac wave
>>>function represents the Poincare group. The wave function *itself* is an
>>>element of the (infinite dimensional) unitary representation of the
>>>Poincare group. The transformations act as
>>>
>>> [itex][(\Lambda,a)\psi](x) = \psi(\Lambda^{-1}x - a)[/itex].
>>
>>>These transformations are unitary, since the action of the Poincare
>>>group preserve the inner product
>>>
>>> [itex](\psi,\phi) = \int \psi(x)* \phi(x) dx[/itex].
>>
>>You forgot the transformation of the field components
>>
[itex]>>[(\Lambda,a)\psi](x) = D(\Lambda) \psi(\Lambda^{-1}x - a).[/itex] (1)
>>
>>where [itex]D(\Lambda)[/itex] are [itex]4-by-4[/itex] matrices representing the Lorentz subgroup
>>of the Poincare group.
>>The action (1) is not unitary, because matrices [itex]D(\Lambda)[/itex] are not
>>unitary.
>
>
> No I have not forgotten this. To be explicit:
>
> [itex](\psi,\phi) = sum_s \int d^{3x} \psi_s(x,t)* \phi_s(x,t)[/itex].
>
> The integral is over the spacial slice at time t. Clearly, translations
> are represented unitarily. Since spacial rotations are represented
> unitarily over the spinor indices, the are also represented unitarily
> with respect to the above inner product. Time translations are
> represented unitarily as a consequence of the equation of motion (the
> Dirac equation).
>
> The only non-trivial case to check are the boost transformations. Boost
> transformations mix x and t coordinates in the argument of [itex]\psi(x,t)[/itex]. But
> the integral in the inner product is only over the values of of [itex]\psi(x,t)[/itex]
> at a fixed time. However, the equation of motion comes to the rescue
> again and keeps the above inner product invariant. This can be checked
> explicitly by verifying that the infinitesimal boost generator
> (combining the parts acting on the spinor indices s and the space-time
> coordinates (x,t)) is hermitian with respect to the above inner product.
> This covers the full algebra of Poincare generators.
>
> The important lesson here is that the above inner product is defined on
> the set of solutions of the Dirac equation, not for arbitrary functions
> [itex]\psi(x,t)[/itex].

Let me first agree with you. Yes, you can project out 2 extra
components, and restrict the space of functions using Dirac's
equation, and show that resulting functions afford an unitary
irreducible representation of the Poincare group, just as
Wigner's functions do. You can also apply the Foldy-Wouthuysen
transformation to Dirac's functions and arrive to the operators
of velocity and position that make certain physical sense.
Yes, you can do all that. My question is whether all these non-trivial
manipulations have any physical meaning? It seems to me that
we have started from something (solutions of Dirac's equation)
that didn't have much physical sense and then applied a series of
patches (like Foldy-Wouthuysen transformations) to put our creation
in a better shape.

Why should we go through all this hassle when we have a perfect
solution for the 1-particle problem given by the Wigner's prescription?
The physical foundations for this prescription are impeccable:
1. The Hilbert space of the system must contain an unitary
(= preservation of probabilities) representation of the Poincare
group (= principle of relativity).
2. Since the system is elementary, this should be the "smallest"
(= irreducible) Hilbert space.

From these two postulate one immediately obtains the classification
of elementary particles by mass and spin, wave functions without
extra unphysical components, the transformation laws
(wrt all inertial transformations) preserving the inner product.
Explicit expressions for all operators of observables (momentum, energy,
velocity, spin, position, etc.) are obtained in a straightforward
fashion without any adjustments.

Unfortunately, most textbooks prefer to talk about Dirac's wave
functions, and don't pay any attention to the brilliant Wigner's
approach. I think, this is just for historical reasons.
Simply, Dirac was first. I remember reading somewhere that Wigner's
paper was not accepted in the major physical journal of that time
(Phys. Rev.?) and he was forced to publish in a mathematical
journal. I find it weird, because to me his article
(E. Wigner, On Unitary Representations of the Inhomogeneous Lorentz
Group, Ann. of Math., 40, 149, (1939)) represents one of the few
finest works of theoretical physics in the 20th century.

Please understand me right. I don't
want to diminish the hugh contribution that Dirac's equation made
to the development of theoretical physics. However, I think, in
2005, almost 80 years after its discovery, it is naive to think
that Dirac's equation is a relativistic equivalent of the Schroedinger
equation. Now we know that in relativistic physics the energy of a
free particle with mass m is related to the momentum by formula
[itex]E = \sqrt(m^2 + p^2)[/itex]. Therefore, the Hamiltonian in the momentum
representation is simply [itex]H = \sqrt(M^2 + P^2),[/itex] and not the 4x4 matrix
written by Dirac.

If it is not a good idea to describe 1-electron states by Dirac's
"wavefunctions", then maybe these functions have no place in
relativistic quantum physics? No, not at all. There is a place
where you simply cannot do without Dirac's functions.
As Weinberg showed in his Chapter 5, Dirac's functions
(this time you need to consider operator functions, or quantum fields,
built as linear combinations of creation and annihilation operators)
are absolutely essential for construction of Poincare invariant
cluster separable interactions in the Fock space.

So, my suggestion is
1) use Wigner's approach for description of 1-particle states,
2) build n-particle Hilbert spaces as tensor product
of Wigner's 1-particle spaces
3) Build the Fock space as a direct sum of n-particle spaces.
4) Use Dirac's (and other) quantum fields to construct non-trivial
interparticle interactions in the Fock space.

Eugene.

Eugene.

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>Aaron Bergman wrote:\n\n> The disadvantages to QED are, of course, that it probably does not exist\n> except as an effective field theory. I have not followed this thread\n> (and don\'t have the energy to get too involved in it), but it sounds\n> like you\'re exactly doing renormalization using different words.\n\nThere were other threads, like "How real are the virtual particles?"\nand others during last half-year, where these issues were discussed at\nlength. But I\'m afraid you\'ll lose all your energy if you try to follow\nthem. You can get dizzy by following soundbites and counter-soundbites\nin those threads. If you want to form your own opinion, I would\nrecommend to read my book "Relativistic quantum dynamics"\nphysics/0504062. I tried to present there in a logical fashion (as best\nas I can) the development of ideas that lead from the principle of\nrelativity and quantum postulates to the physical\nparticle interactions in quantum electrodynamics.\n\nRenormalization is one important part of this story, but not the\nwhole story. There is another big part called "dressing" which\nis less known. This part clarifies the description of physical\nparticles and clears up numerous paradoxes left by the renormalization\napproach.\n\nI haven\'t invented "dressing" in my book. This theory was initiated by\nvan Hove, Greenberg & Schweber in the late 50\'s and continued, most\nnotably, by Faddeev and M. I. Shirokov.\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>Aaron Bergman wrote:

> The disadvantages to QED are, of course, that it probably does not exist
> except as an effective field theory. I have not followed this thread
> (and don't have the energy to get too involved in it), but it sounds
> like you're exactly doing renormalization using different words.

There were other threads, like "How real are the virtual particles?"
and others during last half-year, where these issues were discussed at
length. But I'm afraid you'll lose all your energy if you try to follow
them. You can get dizzy by following soundbites and counter-soundbites
in those threads. If you want to form your own opinion, I would
recommend to read my book "Relativistic quantum dynamics"
http://www.arxiv.org/abs/physics/0504062. I tried to present there in a logical fashion (as best
as I can) the development of ideas that lead from the principle of
relativity and quantum postulates to the physical
particle interactions in quantum electrodynamics.

Renormalization is one important part of this story, but not the
whole story. There is another big part called "dressing" which
is less known. This part clarifies the description of physical
particles and clears up numerous paradoxes left by the renormalization
approach.

I haven't invented "dressing" in my book. This theory was initiated by
van Hove, Greenberg & Schweber in the late 50's and continued, most
notably, by Faddeev and M. I. Shirokov.

Eugene.

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>Aaron Bergman wrote:\n\n> The disadvantages to QED are, of course, that it probably does not exist\n> except as an effective field theory.\n\nI often hear this statement, but I don\'t understand what it means\nexactly. What are your grievances about QED?\n\n1. renormalization?\n2. high-energy behavior (Landau pole)?\n3. non-convergence of the perturbative series?\n4. ???\n\nI can\'t help you much with 2. - 3. (I am not convinced that these are\nreal problems), but I am pretty sure that 1. is not a problem at all.\nWith the help of the "dressed particle" approach, QED can be\nmade a fully logical and self-consistent theory in the range\nof experimentally accessible energies, giving predictions\nagreeing with experiment within error bounds. What else do you expect\nfrom a successful physical theory?\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>Aaron Bergman wrote:

> The disadvantages to QED are, of course, that it probably does not exist
> except as an effective field theory.

I often hear this statement, but I don't understand what it means
exactly. What are your grievances about QED?

1. renormalization?
2. high-energy behavior (Landau pole)?
3. non-convergence of the perturbative series?
4. ???

I can't help you much with 2[itex]. - 3[/itex]. (I am not convinced that these are
real problems), but I am pretty sure that 1. is not a problem at all.
With the help of the "dressed particle" approach, QED can be
made a fully logical and self-consistent theory in the range
of experimentally accessible energies, giving predictions
agreeing with experiment within error bounds. What else do you expect
from a successful physical theory?

Eugene.

Aaron Bergman

<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 <42A60AD5.9090005@synopsys.com>,\nEugene Stefanovich <eugenev@synopsys.com> wrote:\n\n> Aaron Bergman wrote:\n>\n> > The disadvantages to QED are, of course, that it probably does not exist\n> > except as an effective field theory.\n>\n> I often hear this statement, but I don\'t understand what it means\n> exactly. What are your grievances about QED?\n>\n> 1. renormalization?\n\nNeither regularization nor renormalization is a mystery.\n\n> 2. high-energy behavior (Landau pole)?\n\nBingo\n\n> 3. non-convergence of the perturbative series?\n\nNot an issue.\n\nAaron\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 <42A60AD5.9090005@synopsys.com>,
Eugene Stefanovich <eugenev@synopsys.com> wrote:

> Aaron Bergman wrote:
>
> > The disadvantages to QED are, of course, that it probably does not exist
> > except as an effective field theory.
>
> I often hear this statement, but I don't understand what it means
> exactly. What are your grievances about QED?
>
> 1. renormalization?

Neither regularization nor renormalization is a mystery.

> 2. high-energy behavior (Landau pole)?

Bingo

> 3. non-convergence of the perturbative series?

Not an issue.

Aaron

Aaron Bergman

<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 <42A5F921.5020408@synopsys.com>,\nEugene Stefanovich <eugenev@synopsys.com> wrote:\n\n> Renormalization is one important part of this story, but not the\n> whole story. There is another big part called "dressing" which\n> is less known. This part clarifies the description of physical\n> particles and clears up numerous paradoxes left by the renormalization\n> approach.\n\nThe description of physical particles doesn\'t need clearing up. And, as\nyou say, dressing isn\'t new (I thought it was in Weinberg, even, but I\nmust have picked up the term elsewhere.) What is in Weinberg is\nsections 10.2-10.3 (among others) where he discussed the physical poles\nvs. the bare fields.\n\nAaron\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 <42A5F921.5020408@synopsys.com>,
Eugene Stefanovich <eugenev@synopsys.com> wrote:

> Renormalization is one important part of this story, but not the
> whole story. There is another big part called "dressing" which
> is less known. This part clarifies the description of physical
> particles and clears up numerous paradoxes left by the renormalization
> approach.

The description of physical particles doesn't need clearing up. And, as
you say, dressing isn't new (I thought it was in Weinberg, even, but I
must have picked up the term elsewhere.) What is in Weinberg is
sections 10.[itex]2-10[/itex].3 (among others) where he discussed the physical poles
vs. the bare fields.

Aaron

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\nIgor Khavkine wrote:\n\n>\n> But we are getting farther and farther from the topic. This subthread is\n> about the relative merits of the field and particle formulations. So on\n> topic, let me throw someting else into the mix. The mathematical\n> formulation of QFT can handle a large variety of theories, with or\n> without symmetries. Take for instance a quantum field interacting with a\n> time and space varying external field. There is neither space nor time\n> translation symmetry in this theory. So particle states cannot be\n> identified as eigenstates of any sort of symmetry generators. However, a\n> satisfactory quantization can be constructed by solving partial\n> differential equations for the wave functions obtained as matrix\n> elements of field operators. And yes, such a model has numerous physical\n> applications.\n\nAgain, I would prefer to limit our discussion to straight QED.\nIf we remove philosophical noise, then the question boils down to this:\nis it better to represent operators in the Fock space as functions of\nquantum fields or as functions of creation/annihilation operators of\nparticles? Strictly mathematically, these two representations are\ninterchangeable: fields can be uniquely obtained from particle operators\nand vice versa. Then we can ask which way is more convenient and\nphysically transparent?\n\nIf we speak about traditional QED which is interested, almost\nexclusively, in the S-matrix, then fields are certainly more\nconvenient. Scattering amplitudes can be obtained\nfrom Feynman rules that can be read directly from the Lagrangian.\nThese formulas involve propagators that are expectation values of\nfield products. For these calculations one even does not need an\nexplicit expression for the Hamiltonian.\nOf course, the same results could be obtained by using interaction\nterms in the Hamiltonian expressed through creation and annihilation\noperators. But this involves much more labor and, understandably,\nis not popular.\n\nThe story is different if we try to calculate the time evolution.\nFor that, we need the Hamiltonian in the "dressed particle"\nrepresentation. The "dressing transformation" requires sorting\nof different operators into three groups: renorm, phys, and unphys.\nOperators in these groups are most easily identified by their\ncomposition in terms of creation and annihilation operators.\nFor example\n\na^*a is renorm\na^*ac is unphys\na^*a^*aa is phys.\n\nSo, for time evolution and bound states calculations in RQD,\nthe representation in terms of creation and annihilation operators\nis certainly more preferable.\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>Igor Khavkine wrote:

>
> But we are getting farther and farther from the topic. This subthread is
> about the relative merits of the field and particle formulations. So on
> topic, let me throw someting else into the mix. The mathematical
> formulation of QFT can handle a large variety of theories, with or
> without symmetries. Take for instance a quantum field interacting with a
> time and space varying external field. There is neither space nor time
> translation symmetry in this theory. So particle states cannot be
> identified as eigenstates of any sort of symmetry generators. However, a
> satisfactory quantization can be constructed by solving partial
> differential equations for the wave functions obtained as matrix
> elements of field operators. And yes, such a model has numerous physical
> applications.

Again, I would prefer to limit our discussion to straight QED.
If we remove philosophical noise, then the question boils down to this:
is it better to represent operators in the Fock space as functions of
quantum fields or as functions of creation/annihilation operators of
particles? Strictly mathematically, these two representations are
interchangeable: fields can be uniquely obtained from particle operators
and vice versa. Then we can ask which way is more convenient and
physically transparent?

If we speak about traditional QED which is interested, almost
exclusively, in the S-matrix, then fields are certainly more
convenient. Scattering amplitudes can be obtained
from Feynman rules that can be read directly from the Lagrangian.
These formulas involve propagators that are expectation values of
field products. For these calculations one even does not need an
explicit expression for the Hamiltonian.
Of course, the same results could be obtained by using interaction
terms in the Hamiltonian expressed through creation and annihilation
operators. But this involves much more labor and, understandably,
is not popular.

The story is different if we try to calculate the time evolution.
For that, we need the Hamiltonian in the "dressed particle"
representation. The "dressing transformation" requires sorting
of different operators into three groups: renorm, phys, and unphys.
Operators in these groups are most easily identified by their
composition in terms of creation and annihilation operators.
For example

[itex]a^*a[/itex] is renorm
[itex]a^*ac[/itex] is unphys
[itex]a^*a^*aa[/itex] is phys.

So, for time evolution and bound states calculations in RQD,
the representation in terms of creation and annihilation operators
is certainly more preferable.

Eugene.

Similar Threads for: please help my confusion about particles and irreps.
Thread	Forum	Replies
re: Irreps and gauge fields	General Physics	4
re: particles and irreps	General Physics	0
SO(2n),SO(2n +1) Irreps Dimension Formulas	Beyond the Standard Model	0
Global particles, local particles (Colosi, Rovelli)	Beyond the Standard Model	0