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bj88

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Mar 19, 2010, 2:23:22 PM3/19/10
to
Hello People,im doing research on virtual particles for my project and
was wondering if virtual particles travel backwards in time without
travelin violating causality as i understand relativity says if you
travel faster than light you will end up in the past.what im Asking if
you can travel backwards in time without traveling backwards in time?


[[Mod. note -- The physics newsgroup FAQ,
http://www.edu-observatory.org/physics-faq/index.html
has a good discussion of virtual particles (section "Quantum Physics",
entry "Some frequently asked questions about virtual particles").

Alas, many -- perhaps most -- popular accounts of virtual particles
have serious factual and/or conceptual mistakes. :(
-- jt]]

Arnold Neumaier

unread,
Mar 20, 2010, 2:00:39 AM3/20/10
to
bj88 wrote:
> Hello People,im doing research on virtual particles for my project and
> was wondering if virtual particles travel backwards in time without
> travelin violating causality as i understand relativity says if you
> travel faster than light you will end up in the past.what im Asking if
> you can travel backwards in time without traveling backwards in time?
>
>
> [[Mod. note -- The physics newsgroup FAQ,
> http://www.edu-observatory.org/physics-faq/index.html
> has a good discussion of virtual particles (section "Quantum Physics",
> entry "Some frequently asked questions about virtual particles").

Although the discussion given there is in terms of perturbation theory,
it is not emphasized enough there that virtual particles are an
artifact of perturbation theory, rather than objects that exist in
any reasonable physical sense of the word. (Individual Feynman diagrams
do not make formal sense - they come all out infinite, even after
renormalization. Only the sum of all Feynman diagrams of a given order
can be given a finite meaning; but then their interpretation in space
and time is lost!)

> Alas, many -- perhaps most -- popular accounts of virtual particles
> have serious factual and/or conceptual mistakes. :(
> -- jt]]

My theoretical Physics FAQ at
http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html
has in Chapter A7 several articles related to this topic:

Virtual particles and Coulomb interaction
Are virtual particles and decaying particles the same?
How real are 'virtual particles'?
Can particles go backward in time?
What about particles faster than light (tachyons)?
Superluminal light pulses

Arnold Neumaier

Oh No

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Mar 21, 2010, 3:19:05 AM3/21/10
to
Thus spake bj88 <bolan...@yahoo.com>

>Hello People,im doing research on virtual particles for my project and
>was wondering if virtual particles travel backwards in time without
>travelin violating causality as i understand relativity says if you
>travel faster than light you will end up in the past.what im Asking if
>you can travel backwards in time without traveling backwards in time?
>
In the Stuckelberg-Feynman interpretation antiparticles are particles
travelling backwards in time. More on this at
http://rqgravity.net/TheDiracEquation#Antiparticles

Relativity does not allow you to travel backwards in time - faster than
light means space-like, which is different from the negative light cone.
http://rqgravity.net/IntroductionToVectorSpace#TheLightCone

Antimatter obeys the same kind of causality as matter. Individual
particles can be thought of as "turning round in time" but what we see
is particle-antiparticle annihilation, and the release of energy. It
would not be possible for macroscopic matter (such as yourself) to
participate in this kind of interaction.

[ Mod. note: As pointed out in the example above, any situation which
may be thought of as involving propagation backward in time may
also be thought of as one where all particles propagated in the
same direction, but special events like particle creation or
annihilation occur. Thus this kind of interpretation is a matter
of personal preference. -ik ]

Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.rqgravity.net


FrediFizzx

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Mar 21, 2010, 4:35:46 AM3/21/10
to
"Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
news:4BA3D4F...@univie.ac.at...

> bj88 wrote:
>> Hello People,im doing research on virtual particles for my project
>> and
>> was wondering if virtual particles travel backwards in time without
>> travelin violating causality as i understand relativity says if you
>> travel faster than light you will end up in the past.what im Asking
>> if
>> you can travel backwards in time without traveling backwards in time?

Tricky question but not sure it even should be applying to your research
on virtual particles. IMHO, virtual particles do not travel backward in
time any more than real particles travel backward in time.

>>
>> [[Mod. note -- The physics newsgroup FAQ,
>> http://www.edu-observatory.org/physics-faq/index.html
>> has a good discussion of virtual particles (section "Quantum
>> Physics",
>> entry "Some frequently asked questions about virtual particles").

I would add to that FAQ that "virtual" in particle physics simply means
"off-mass-shell" which is allowed by the Uncertainty Principle.

> Although the discussion given there is in terms of perturbation
> theory,
> it is not emphasized enough there that virtual particles are an
> artifact of perturbation theory, rather than objects that exist in
> any reasonable physical sense of the word.

How are we to explain static electric and magnetic fields if virtual
photons are just a mathematical artifact? If I stick a steel plate to a
magnet, are the photons holding the plate to the magnet, real or
virtual? If I rub a balloon on my hair and stick it to the wall, are
the photons holding it to the wall, real or virtual?

>> Alas, many -- perhaps most -- popular accounts of virtual particles
>> have serious factual and/or conceptual mistakes. :(
>> -- jt]]
>
> My theoretical Physics FAQ at
> http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html
> has in Chapter A7 several articles related to this topic:
>
> Virtual particles and Coulomb interaction
> Are virtual particles and decaying particles the same?
> How real are 'virtual particles'?
> Can particles go backward in time?
> What about particles faster than light (tachyons)?
> Superluminal light pulses

I read through all of the articles in your Chapter A7 and I am not
satisfied that virtual particles are just mathematical artifacts. You
seem to be missing something to make your explanation complete (or I
missed it somewhere). And Gordon Kane (famous particle physicist) does
not agree with you in this article,

http://www.scientificamerican.com/article.cfm?id=are-virtual-particles-rea

"Virtual particles are indeed real particles. Quantum theory predicts
that every particle spends some time as a combination of other particles
in all possible ways. These predictions are very well understood and
tested."

Modern particle physics doesn't make complete sense unless we allow
virtual particles to exist as Kane points out in the rest of the
article.

Best,

Fred Diether

Arnold Neumaier

unread,
Mar 21, 2010, 6:05:52 PM3/21/10
to

Coulomb forces, Casimir forces, and the like (i.e., electromagnetic
fields) are real, the photons used to describe them in perturbation
theory cannot be. This can be seen from the fact that different
approximation schemes (e.g., the standard textbook perturbation theory
and light front calculations) lead to completely different properties
of the virtual particles involved.

Photons are transverse excitations of the e/m field, i.e.,
deviations from the mean value of the field. To consider the
Coulomb field, say, as being composed of virtual photons is a
superficial misunderstanding of quantum field theory.
None of the many very successful calculations comparing theory with
experiment depends on the reality of such virtual particles.


>>> Alas, many -- perhaps most -- popular accounts of virtual particles
>>> have serious factual and/or conceptual mistakes. :(
>>> -- jt]]
>> My theoretical Physics FAQ at
>> http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html
>> has in Chapter A7 several articles related to this topic:
>>
>> Virtual particles and Coulomb interaction
>> Are virtual particles and decaying particles the same?
>> How real are 'virtual particles'?
>> Can particles go backward in time?
>> What about particles faster than light (tachyons)?
>> Superluminal light pulses
>
> I read through all of the articles in your Chapter A7 and I am not
> satisfied that virtual particles are just mathematical artifacts.

Why then do they behave so differently depending on the perturbation
scheme used?

Why then is the mass of virtual electrons infinite?
This has never been observed.

Why then do the values of all diagrams in QED involving virtual
particles come out infinite?

All three questions need a convincing answer in order to regard
virtual particles as more than artifacts.

The only place where one can assign virtual particles some
physical meaning is
- (i) in nonrelativistic quantum mechanics, where electromagnetic
fields are represented either by explicit Coulomb terms or by
external e/m potentials, and there are no photons,
- (ii) in a tree approximation to relativistic QFT; but this is
not enough for any of the great successes of QED.

> You
> seem to be missing something to make your explanation complete (or I
> missed it somewhere). And Gordon Kane (famous particle physicist) does
> not agree with you in this article,
>
> http://www.scientificamerican.com/article.cfm?id=are-virtual-particles-rea
>
> "Virtual particles are indeed real particles. Quantum theory predicts
> that every particle spends some time as a combination of other particles
> in all possible ways. These predictions are very well understood and
> tested."

Unfortunately, the article fails to give references to support
the claim that virtual particles are indeed real particles.

There is not a single verified measurement of a virtual particle.
Measured particles are always on-shell.

The tests mentioned are not tests of the reality of virtual
particles, but tests of QED and the standard model in general,
together will talk about the behavior of virtual particles that
may be a common fantasy of many physicists, but has no support at
all by experiments.

The Lamb shift and the Casimir effect can be calculated in a number
of different ways, not all requiring the concept of virtual particles.
How can it then be experimental proof of the latter?

On the Z-boson, the article writes, ''the value agreed with that
obtained from the virtual particle analysis, providing a dramatic
test of our understanding of virtual particles.''
This virtual particle analysis (when followed through - no
references are given) consists in perturbative calculations
that start with Feynman diagrams involving virtual particles.
But these are unmeasurable things with an infinite mass and
infinite interactions, both physically meaningless. They are turned
into predictions only by a perturbative renormalization process
that restores physics to the ill-conceived bare interpretation.

After renormalization, when everything is finite, there is not
the smallest trace left of virtual particles. Moreover, modern
nonperturbatives renormalization techniques via lattice gauge
theory or similarity flows give comparable predictions for quantum
field theories, without encountering virtual particles anywhere.


Arnold Neumaier

FrediFizzx

unread,
Mar 22, 2010, 2:27:00 AM3/22/10
to
"Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
news:4BA62CD1...@univie.ac.at...

> FrediFizzx wrote:
>> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
>> news:4BA3D4F...@univie.ac.at...

>>> Although the discussion given there is in terms of perturbation
>>> theory,
>>> it is not emphasized enough there that virtual particles are an
>>> artifact of perturbation theory, rather than objects that exist in
>>> any reasonable physical sense of the word.
>>
>> How are we to explain static electric and magnetic fields if virtual
>> photons are just a mathematical artifact? If I stick a steel plate
>> to a magnet, are the photons holding the plate to the magnet, real or
>> virtual? If I rub a balloon on my hair and stick it to the wall, are
>> the photons holding it to the wall, real or virtual?
>
> Coulomb forces, Casimir forces, and the like (i.e., electromagnetic
> fields) are real, the photons used to describe them in perturbation
> theory cannot be. This can be seen from the fact that different
> approximation schemes (e.g., the standard textbook perturbation theory
> and light front calculations) lead to completely different properties
> of the virtual particles involved.

Well, that is not surprising that different schemes might lead to
somewhat different properties.

> Photons are transverse excitations of the e/m field, i.e.,
> deviations from the mean value of the field. To consider the
> Coulomb field, say, as being composed of virtual photons is a
> superficial misunderstanding of quantum field theory.
> None of the many very successful calculations comparing theory with
> experiment depends on the reality of such virtual particles.

Are you saying that the Coulomb field is composed of real photons then?
If not virtual photons nor virtual fermionic pairs, then there is
nothing else to explain it within the context of QED other than real
photons.

>>>> Alas, many -- perhaps most -- popular accounts of virtual particles
>>>> have serious factual and/or conceptual mistakes. :(
>>>> -- jt]]
>>> My theoretical Physics FAQ at
>>> http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html
>>> has in Chapter A7 several articles related to this topic:
>>>
>>> Virtual particles and Coulomb interaction
>>> Are virtual particles and decaying particles the same?
>>> How real are 'virtual particles'?
>>> Can particles go backward in time?
>>> What about particles faster than light (tachyons)?
>>> Superluminal light pulses
>>
>> I read through all of the articles in your Chapter A7 and I am not
>> satisfied that virtual particles are just mathematical artifacts.
>
> Why then do they behave so differently depending on the perturbation
> scheme used?

I didn't know that they did behave differently. What is a simple
example?

> Why then is the mass of virtual electrons infinite?
> This has never been observed.

What makes you think it would be infinite? All virtual processes
involved in real processes must still abide by conservation laws in the
total scheme of the whole process.

> Why then do the values of all diagrams in QED involving virtual
> particles come out infinite?

I was under the impression when you add them all up for a particular
real process, that most of them cancel out.

> All three questions need a convincing answer in order to regard
> virtual particles as more than artifacts.
>
> The only place where one can assign virtual particles some
> physical meaning is
> - (i) in nonrelativistic quantum mechanics, where electromagnetic
> fields are represented either by explicit Coulomb terms or by
> external e/m potentials, and there are no photons,
> - (ii) in a tree approximation to relativistic QFT; but this is
> not enough for any of the great successes of QED.

You lost me here. As in, what you are saying doesn't make sense. What
is the "physical meaning" you are trying to show?

>> You seem to be missing something to make your explanation complete
>> (or I missed it somewhere). And Gordon Kane (famous particle
>> physicist) does not agree with you in this article,
>>
>> http://www.scientificamerican.com/article.cfm?id=are-virtual-particles-rea
>>
>> "Virtual particles are indeed real particles. Quantum theory predicts
>> that every particle spends some time as a combination of other
>> particles in all possible ways. These predictions are very well
>> understood and tested."
>
> Unfortunately, the article fails to give references to support
> the claim that virtual particles are indeed real particles.

That article was just a response to a question that a reader sent.
However, I do believe that Kane teaches or has taught a university
graduate level particle physics course(s) so I expect he knows what he
is talking about.

> There is not a single verified measurement of a virtual particle.
> Measured particles are always on-shell.

:-) Of course not; that is part of the definition of a virtual
particle. It is not directly observable. And your second sentence is
part of the definition of real particles. I do believe that they are
indirectly observable as I seem to recall seeing bubble chamber photos
of particle interactions that can only be explained by a virtual
particle that is not showing in the photo (didn't leave track). In
fact, on page 35 in Griffiths' "Introduction to Elementary Particles"
there is a bubble chamber photo of the discovery of the Omega minus
particle that shows there must have been two virtual gamma photons that
each decayed to real electron-positron pairs. They had to be virtual as
a real gamma photon can't do that. I would imagine they see this same
sort of phenomena in modern particle detectors also; real
electron-positron pairs (and other fermionic pairs) just appearing with
no trace of virtual photons (or other virtual particles) showing.

[ Mod. note: The above paragraph confuses the meaning of "directly
observable". Lack of a vapor trail in a bubble chamber indicates
lack of electric charge, not "virtuality". Neutral particles can
also be directly observed in bubble chambers, through their decay
products, the same way your eyes observe the computer screen in
front of you. The statement from Griffiths sounds dubious, but
I have no copy on hand to check it. -ik ]

> The tests mentioned are not tests of the reality of virtual
> particles, but tests of QED and the standard model in general,
> together will talk about the behavior of virtual particles that
> may be a common fantasy of many physicists, but has no support at
> all by experiments.

Sorry, but I have to disagree since without virtual W bosons, the decay
of muons makes no sense.

> The Lamb shift and the Casimir effect can be calculated in a number
> of different ways, not all requiring the concept of virtual particles.
> How can it then be experimental proof of the latter?

1/40th of the Lamb shift (~27MHz) is explained by virtual particles
(vacuum polarization) in the calculations I have seen. Do you have a
reference that shows a calculation of the Lamb shift without resorting
to vacuum polarization? Better yet; how about a reference for the Lamb
shift calculation of muonic atoms? There, vacuum polarization effects
are more pronounced.

> On the Z-boson, the article writes, ''the value agreed with that
> obtained from the virtual particle analysis, providing a dramatic
> test of our understanding of virtual particles.''
> This virtual particle analysis (when followed through - no
> references are given) consists in perturbative calculations
> that start with Feynman diagrams involving virtual particles.
> But these are unmeasurable things with an infinite mass and
> infinite interactions, both physically meaningless. They are turned
> into predictions only by a perturbative renormalization process
> that restores physics to the ill-conceived bare interpretation.

I don't agree with that. They are immeasurable but they still have to
abide by conservation laws for the entire process so don't think they
would have infinite mass-energy nor an infinite amount of interactions.

> After renormalization, when everything is finite, there is not
> the smallest trace left of virtual particles. Moreover, modern
> nonperturbatives renormalization techniques via lattice gauge
> theory or similarity flows give comparable predictions for quantum
> field theories, without encountering virtual particles anywhere.

Is there any online references for those techniques? How about for
decay of a charged pion to anti-neutrino and lepton that involves a
virtual W minus boson? How does any of those techniques make the
virtual W boson disappear from the decay process?

Best,

Fred Diether

bj88

unread,
Mar 22, 2010, 3:10:38 AM3/22/10
to
=============== Moderator's note ========================================
I shortened the full quote of the previous posting to the necessary small
piece needed to understand the follow-up question of this posting. Please
help to keep the newsgroup better readable by shortening quotes to a
sensible length!
HvH.
=========================================================================

On Mar 21, 2:05 pm, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
wrote:

> The only place where one can assign virtual particles some
> physical meaning is
> - (i) in nonrelativistic quantum mechanics, where electromagnetic
> fields are represented either by explicit Coulomb terms or by
> external e/m potentials, and there are no photons,
> - (ii) in a tree approximation to relativistic QFT; but this is
> not enough for any of the great successes of QED.

so virtual particles don't travel faster than light becuase they would
violate casuality?

Hendrik van Hees

unread,
Mar 22, 2010, 11:31:08 AM3/22/10
to
Oh No wrote:

> In the Stuckelberg-Feynman interpretation antiparticles are particles
> travelling backwards in time. More on this at
> http://rqgravity.net/TheDiracEquation#Antiparticles

That's only half of the story of the Stueckelberg interpretation. The
complete argument goes as follows:

When looking for the modes of definite momentum and helicity of free
quantum fields, one finds always solutions with positive and negative
frequency, and for the interacting theory one necessarily has to
superpose both classes of solutions in order to obtain a local
realization of the Poincare group, i.e., field operators which transform
under (proper orthochronous) Poincare transformation

x'=L x+a

in the same way as their classical counterparts:

(1) psi'(x',lambda')=\sum_{\lambda} D_{lambda',lambda}(L)
psi(L^{-1} (x'-a),lambda)

This admits to build local interactions leading to a Poincare covariant
causal S matrix.

The Feynman-Stueckelberg trick now consists in two steps: First one
realizes that the propagation of a field mode with negative frequencey
*would* look like a particle running backwards in time *if* one would
quantize that modes with an annihhilation operator, but it would look
like a particle of another kind, running *forwards* in time with an
opposite momentum. This other particles have the same properties as the
particles (i.e. the same mass and spin) but opposite currents (if their
exist any). It can also be that the anti-particles are indistinguishable
from the particles, i.e., then one has strictly neutral particles (like
photons in the standard model or Majorana fermions which might be the
right realization of neutrinos if those turn out to be strictly neutral
and not Dirac particles as assumed in the standard model).

Thus, the great merit of the Stueckelberg-Feynman trick is that there
are in fact no particles running backward in time but only particles and
anti particles running forward in time, thus establishing the causality
of relativistic quantum field theory of massive and massless particles.

Another merit is, that one has got rid of the ugly construct of the
Dirac sea, i.e. the picture of the vacuum as a sea of filled single-
particle states. If you take the original application of electrons and
protons in QED, for which Dirac invented his famous equation, then a
Dirac sea would imply that one has an infinite background of negative
charges which are not observed. In the Feynman-Stueckelberg picture
there is just charge renormalization by normal ordering, and
everything's fine and causal.

>
> Relativity does not allow you to travel backwards in time - faster
> than light means space-like, which is different from the negative
> light cone.
> http://rqgravity.net/IntroductionToVectorSpace#TheLightCone
>
> Antimatter obeys the same kind of causality as matter. Individual
> particles can be thought of as "turning round in time" but what we see
> is particle-antiparticle annihilation, and the release of energy. It
> would not be possible for macroscopic matter (such as yourself) to
> participate in this kind of interaction.

That's the point! Nothing is running backwards in time!


>
> [ Mod. note: As pointed out in the example above, any situation which
> may be thought of as involving propagation backward in time may
> also be thought of as one where all particles propagated in the
> same direction, but special events like particle creation or
> annihilation occur. Thus this kind of interpretation is a matter
> of personal preference. -ik ]

I'd put it stronger: There's no merit of the esoteric interpretation of
backward-in-time-running particles but only confusion!

--
Hendrik van Hees
Institut f?r Theoretische Physik
Justus-Liebig-Universit?t Gie?en
http://theorie.physik.uni-giessen.de/~hees/

Bob_for_short

unread,
Mar 22, 2010, 11:31:57 AM3/22/10
to
On 21 mar, 09:35, "FrediFizzx" <fredifi...@hotmail.com> wrote:

>
> I would add to that FAQ that "virtual" in particle physics simply means
> "off-mass-shell" which is allowed by the Uncertainty Principle.
>

>


> How are we to explain static electric and magnetic fields if virtual

> photons are just a mathematical artefact? If I stick a steel plate to a


> magnet, are the photons holding the plate to the magnet, real or
> virtual? If I rub a balloon on my hair and stick it to the wall, are
> the photons holding it to the wall, real or virtual?
>

>


> "Virtual particles are indeed real particles. Quantum theory predicts
> that every particle spends some time as a combination of other particles
> in all possible ways. These predictions are very well understood and
> tested."
>
> Modern particle physics doesn't make complete sense unless we allow
> virtual particles to exist as Kane points out in the rest of the
> article.
>

Quasi-stationary Coulomb and magnetic interaction terms are not
virtual particles since they are not independent and are not
propagating. It is charge interaction terms (inter-charge action on
each other).

Real photons, on the other hand, are independent and propagating. They
carry energy-momentum, E = hv, for example, and no mention of their
source is here necessary. Also their polarizations are quite different
from that of quasi-stationary fields.

At far distances we can observe a neutral and non magnetic body with
help of photons - real particles since the quasi-stationary fields
compensate each other. Photons cannot compensate each other.

In QED we often consider the quasi-stationary interaction by the
perturbation theory - together with interaction with the quantized
electromagnetic field. This treatment created a confusion of these two
interaction terms and led to the notion of virtual particles. In fact,
it is quite misleading since the quasi-stationary interaction terms
can be taken into account exactly (see Hydrogen atom description).
Besides, even in the perturbation theory one can work in the
configuration space where no "off-shell" explanation (distinction) is
possible.

Juan R. González-Álvarez

unread,
Mar 22, 2010, 11:32:22 AM3/22/10
to
Oh No wrote on Sun, 21 Mar 2010 07:19:05 +0000:

> In the Stuckelberg-Feynman interpretation antiparticles are particles
> travelling backwards in time. More on this at
> http://rqgravity.net/TheDiracEquation#Antiparticles

More exactly negative energy particles travelling backward in time are
*reinterpreted* as positive energy antiparticles travelling forward.

The *reinterpretation* is needed because neither negative energy
particles nor particles travelling backward are detected/observed in
the lab.


--
http://www.canonicalscience.org/

BLOG:
http://www.canonicalscience.org/publications/canonicalsciencetoday/canonicalsciencetoday.html

Oh No

unread,
Mar 22, 2010, 1:23:32 PM3/22/10
to
Thus spake Arnold Neumaier <Arnold....@univie.ac.at>
>FrediFizzx wrote:

>Coulomb forces, Casimir forces, and the like (i.e., electromagnetic
>fields) are real, the photons used to describe them in perturbation
>theory cannot be. This can be seen from the fact that different
>approximation schemes (e.g., the standard textbook perturbation theory
>and light front calculations) lead to completely different properties
>of the virtual particles involved.

Actually photons, virtual or otherwise, can be perfectly well defined

>
>Why then do they behave so differently depending on the perturbation
>scheme used?

That depends on whatever misunderstandings have been built into
whichever schemes you refer to


>
>Why then is the mass of virtual electrons infinite?

it isn't

>This has never been observed.

even if it were, the definition of virtual particles is that they are
non-observable states.


>
>Why then do the values of all diagrams in QED involving virtual
>particles come out infinite?

Firstly, they don't. e.g. the simple diagram of a single virtual photon
exchange is not infinite.

Secondly even for those diagrams which contain ultraviolet divergences,
the infinity is due to the mistreatment of limits and our lack of
understanding of very short scale behaviour.


>
>All three questions need a convincing answer in order to regard
>virtual particles as more than artifacts.

What is really necessary is to give qed a proper mathematical treatment,
as distinct from a typical text book treatment. Then all three questions
do have answers.

>There is not a single verified measurement of a virtual particle.
>Measured particles are always on-shell.

I think you need to study the derivation of Feynman rules from the Dyson
expansion, and to get a better idea of what "virtual" means in this
context.

Message has been deleted

Bob_for_short

unread,
Mar 22, 2010, 2:54:55 PM3/22/10
to
On 22 mar, 07:27, "FrediFizzx" <fredifi...@hotmail.com> wrote:

>
> Are you saying that the Coulomb field is composed of real photons then?
> If not virtual photons nor virtual fermionic pairs, then there is
> nothing else to explain it within the context of QED other than real
> photons.
>

Why we should "explain" the Coulomb interaction term via particles?

>
> 1/40th of the Lamb shift (~27MHz) is explained by virtual particles
> (vacuum polarization) in the calculations I have seen.  Do you have a
> reference that shows a calculation of the Lamb shift without resorting
> to vacuum polarization?  Better yet; how about a reference for the Lamb
> shift calculation of muonic atoms?  There, vacuum polarization effects
> are more pronounced.

The *entire* Coulomb interaction is often called a "virtual photon".
It is not corrections to it but the entire quasi-stationary
interaction which is a "virtual particle". But this interaction term
is expressed via charge coordinates and velocities directly, no need
in intermediary particles for that. It is quite different from real,
propagating particles.

Radiative decay of an excited atom does not make it "virtual".

Arnold Neumaier

unread,
Mar 22, 2010, 2:54:56 PM3/22/10
to
FrediFizzx wrote:
> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
> news:4BA62CD1...@univie.ac.at...
>> FrediFizzx wrote:
>>> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
>>> news:4BA3D4F...@univie.ac.at...
>
>
>>>> Although the discussion given there is in terms of perturbation theory,
>>>> it is not emphasized enough there that virtual particles are an
>>>> artifact of perturbation theory, rather than objects that exist in
>>>> any reasonable physical sense of the word.
>>>
>>> How are we to explain static electric and magnetic fields if virtual
>>> photons are just a mathematical artifact? If I stick a steel plate
>>> to a magnet, are the photons holding the plate to the magnet, real or
>>> virtual? If I rub a balloon on my hair and stick it to the wall, are
>>> the photons holding it to the wall, real or virtual?
>>
>> Coulomb forces, Casimir forces, and the like (i.e., electromagnetic
>> fields) are real, the photons used to describe them in perturbation
>> theory cannot be. This can be seen from the fact that different
>> approximation schemes (e.g., the standard textbook perturbation theory
>> and light front calculations) lead to completely different properties
>> of the virtual particles involved.
>
> Well, that is not surprising that different schemes might lead to
> somewhat different properties.

But not to completely different properties.
Different approximations should lead to close results for physical
quantities. But already the Feynman diagrams are completely different.

>> Photons are transverse excitations of the e/m field, i.e.,
>> deviations from the mean value of the field. To consider the
>> Coulomb field, say, as being composed of virtual photons is a
>> superficial misunderstanding of quantum field theory.
>> None of the many very successful calculations comparing theory with
>> experiment depends on the reality of such virtual particles.
>
> Are you saying that the Coulomb field is composed of real photons then?

No. The Coulomb field has nothig to do with photons. It arises as an
interaction term in the QED Hamiltonian in the
Coulomb gauge.


> If not virtual photons nor virtual fermionic pairs, then there is
> nothing else to explain it within the context of QED other than real
> photons.

The standard Lagrangian --> Hamiltonian procedure is enough to
explain it.


>>> I read through all of the articles in your Chapter A7 and I am not
>>> satisfied that virtual particles are just mathematical artifacts.
>>
>> Why then do they behave so differently depending on the perturbation
>> scheme used?
>
> I didn't know that they did behave differently. What is a simple example?

Look at arbitrary articles on light front quantization, and you'll
find hat the rules are completely different from those in the
textbook. Nothing looks comparable.


>> Why then is the mass of virtual electrons infinite?
>> This has never been observed.
>
> What makes you think it would be infinite? All virtual processes
> involved in real processes must still abide by conservation laws in the
> total scheme of the whole process.

Look into any book on quantum field theory to see that the bare
particles (and only these arise in internal lines of Feynman diagrams)
have infinite mass.


>> Why then do the values of all diagrams in QED involving virtual
>> particles come out infinite?
>
> I was under the impression when you add them all up for a particular
> real process, that most of them cancel out.

Yes, but virtual particles are associated with individual Feynman
diagrams, not with their sum. In the sum, there is no recognizable
space-time behavior of anything that could travel.


>> All three questions need a convincing answer in order to regard
>> virtual particles as more than artifacts.
>>
>> The only place where one can assign virtual particles some
>> physical meaning is
>> - (i) in nonrelativistic quantum mechanics, where electromagnetic
>> fields are represented either by explicit Coulomb terms or by
>> external e/m potentials, and there are no photons,
>> - (ii) in a tree approximation to relativistic QFT; but this is
>> not enough for any of the great successes of QED.
>
> You lost me here. As in, what you are saying doesn't make sense. What
> is the "physical meaning" you are trying to show?

OK, then forget it. This was only a guarding remark in case you could
have made sense of it.

>>> You seem to be missing something to make your explanation complete
>>> (or I missed it somewhere). And Gordon Kane (famous particle
>>> physicist) does not agree with you in this article,
>>>
>>> http://www.scientificamerican.com/article.cfm?id=are-virtual-particles-rea
>>>
>>>
>>> "Virtual particles are indeed real particles. Quantum theory predicts
>>> that every particle spends some time as a combination of other
>>> particles in all possible ways. These predictions are very well
>>> understood and tested."
>>
>> Unfortunately, the article fails to give references to support
>> the claim that virtual particles are indeed real particles.
>
> That article was just a response to a question that a reader sent.
> However, I do believe that Kane teaches or has taught a university
> graduate level particle physics course(s) so I expect he knows what he
> is talking about.

He knows quantum field theory. but he does not know more than anyone
else about virtual particles, since these are unobservable.
He simply equates quantum field theory with the use of virtual
particles, without sufficient reasons.


>> There is not a single verified measurement of a virtual particle.
>> Measured particles are always on-shell.
>
> :-) Of course not; that is part of the definition of a virtual
> particle. It is not directly observable. And your second sentence is
> part of the definition of real particles. I do believe that they are
> indirectly observable as I seem to recall seeing bubble chamber photos
> of particle interactions that can only be explained by a virtual
> particle that is not showing in the photo (didn't leave track). In
> fact, on page 35 in Griffiths' "Introduction to Elementary Particles"
> there is a bubble chamber photo of the discovery of the Omega minus
> particle that shows there must have been two virtual gamma photons that
> each decayed to real electron-positron pairs. They had to be virtual as
> a real gamma photon can't do that.

Two real gamma photons can decay into a real electron-positron pair.
How do you distinguish experimentally between an invisible ghost
called a virtual photon from two real photons that can be observed on
numerous other occasions?


> I would imagine they see this same

Not your imagination is called for, but facts supporting claaims.


> sort of phenomena in modern particle detectors also; real
> electron-positron pairs (and other fermionic pairs) just appearing with
> no trace of virtual photons (or other virtual particles) showing.

The correct creation/annihilation process is
2 gamma <---> e+ + e-
and involves real particles only, on both sides of the process.
The corresponding cross section can be calculated and measured.


> [ Mod. note: The above paragraph confuses the meaning of "directly
> observable". Lack of a vapor trail in a bubble chamber indicates
> lack of electric charge, not "virtuality". Neutral particles can
> also be directly observed in bubble chambers, through their decay
> products, the same way your eyes observe the computer screen in
> front of you. The statement from Griffiths sounds dubious, but
> I have no copy on hand to check it. -ik ]
>
>> The tests mentioned are not tests of the reality of virtual
>> particles, but tests of QED and the standard model in general,
>> together will talk about the behavior of virtual particles that
>> may be a common fantasy of many physicists, but has no support at
>> all by experiments.
>
> Sorry, but I have to disagree since without virtual W bosons, the decay
> of muons makes no sense.

As with all other particles, observable W-bosons are real, not virtual.


>> The Lamb shift and the Casimir effect can be calculated in a number
>> of different ways, not all requiring the concept of virtual particles.
>> How can it then be experimental proof of the latter?
>
> 1/40th of the Lamb shift (~27MHz) is explained by virtual particles
> (vacuum polarization) in the calculations I have seen. Do you have a
> reference that shows a calculation of the Lamb shift without resorting
> to vacuum polarization? Better yet; how about a reference for the Lamb
> shift calculation of muonic atoms? There, vacuum polarization effects
> are more pronounced.

Vacuum polarization is usually calculated by means of virtual photons,
but does not have to. High precision calculations of the Lamb shift
using NRQED are not based on virtual photons.


>> On the Z-boson, the article writes, ''the value agreed with that
>> obtained from the virtual particle analysis, providing a dramatic
>> test of our understanding of virtual particles.''
>> This virtual particle analysis (when followed through - no
>> references are given) consists in perturbative calculations
>> that start with Feynman diagrams involving virtual particles.
>> But these are unmeasurable things with an infinite mass and
>> infinite interactions, both physically meaningless. They are turned
>> into predictions only by a perturbative renormalization process
>> that restores physics to the ill-conceived bare interpretation.
>
> I don't agree with that. They are immeasurable but they still have to
> abide by conservation laws for the entire process so don't think they
> would have infinite mass-energy nor an infinite amount of interactions.

The formal conservation laws become meaningless since inf+inf=inf+inf,
no matter what you do. Bare particles have no physical properties,
but only bare particles figure in Feynman diagrams and give rise to
talk about virtual particles. Real particles are renormalized
versions of the bare particles and have finite, physical properties,
but (at least in vacuum) you cannot find them off-shell.


>> After renormalization, when everything is finite, there is not
>> the smallest trace left of virtual particles. Moreover, modern
>> nonperturbatives renormalization techniques via lattice gauge
>> theory or similarity flows give comparable predictions for quantum
>> field theories, without encountering virtual particles anywhere.
>
> Is there any online references for those techniques?

scholar.google.com on "lattice gauge theory",
"similarity renormalization", "light front", or NRQED gives articles
on four techniques very different from each other and from the
typical textbook treatment.

You may wish to read the Nobel lecture by Frank Wilczek at
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1150826/
where he first talks about the traditiona virtual particle picture:
''Loosely speaking, energy can be borrowed to make evanescent
virtual particles''. Note his qualification that indicates that
this cannot be taken seriously. He also says why - because one
encounters divergernces by taking them seriously. Then he gets
more serious and shows how renormalization fixes the problems,
though he does not say that this comes at the cost of making the
virtual particles infinitely heavy (and hence again physically
meaningless). But this can be read in any textbook on QFT.
Later, he slips back into the traditional jargon since it
provides a vivid intuition about Feynman diagrams -- especially
for the many nonexperts in his audience, but again he does so
with a careful, explicit caveat:
''(I'm being a little sloppy in my terminology; instead of saying
the number of virtual particles, it would be more proper to speak
of the number of internal loops in a Feynman graph.)''

Towards the middle he mentions lattice discretizations, and how they
cope with the problem in a nonperturbative way by not invoking virtual
particles (i.e., formally correct, a loop expansion) at all.


> How about for
> decay of a charged pion to anti-neutrino and lepton that involves a
> virtual W minus boson? How does any of those techniques make the
> virtual W boson disappear from the decay process?

As already said, in the tree approximation - where renormalization is
not needed - one can think meaningfully of virtual particles (if one
is inclined to mystery). This covers the case you talk about.
But not beyond. This just shows that virtual particles are an artifact
of perturbation theory.

Thjerefore, it is physically correct to not use this terminology,
except (as Frank Wilczek did) as a figure of speech to interpret
Feynman diagrams (where it is harmless if one knows what one does).

The correct way -- which is in full accord with renormalized
quantum field theory and remains valid nonperturbatively -- to think
about particle creation, annihilation, or scattering processes is
that particles are excitations of fields. The latter create and absorb
particles in a similar way as flowing water creates and absorbs
water waves.

Thus in your case, the charged pion is surrounded by (and in fact
part of) a weak-force field, which absorbs the pion and then creates
two leptons. During the interaction there is neither pion nor W-boson
nor leptons, since the identification of individual excitations
requires the latter to be well enough separated so that the
asymptotic S-matrix regime applies. Only then the particle nature
of the field is apparent.


Arnold Neumaier


Oh No

unread,
Mar 22, 2010, 4:29:01 PM3/22/10
to
Thus spake Hendrik van Hees <Hendrik...@physik.uni-giessen.de>

>I'd put it stronger: There's no merit of the esoteric interpretation of
>backward-in-time-running particles but only confusion!

Seems to me you gave two great merits. Those, plus the fact that the
explanation gives a straightforward way of explaining the mathematical
form of the theory. For example the fact that a field operator creates
an antiparticle or annihilates a particle, which thus appear in the
theory as aspects of the same process.

Oh No

unread,
Mar 22, 2010, 4:29:05 PM3/22/10
to
Thus spake Juan R. González-Álvarez <email....@not.supplied>

>Oh No wrote on Sun, 21 Mar 2010 07:19:05 +0000:
>
>> In the Stuckelberg-Feynman interpretation antiparticles are particles
>> travelling backwards in time. More on this at
>> http://rqgravity.net/TheDiracEquation#Antiparticles
>
>More exactly negative energy particles travelling backward in time are
>*reinterpreted* as positive energy antiparticles travelling forward.
>
>The *reinterpretation* is needed because neither negative energy
>particles nor particles travelling backward are detected/observed in
>the lab.

In fact one cannot say this is not observed, or even that
reinterpretation is needed, because the observation of a particle
travelling backwards in time is identical to the observation of an
antiparticle travelling forwards. The point is that the mathematical
structure of the theory arises naturally in the interpretation of
particles travelling backwards in time, whereas it requires explanation
if one thinks of antiparticles travelling forwards.

Juan R.

unread,
Mar 22, 2010, 5:33:34 PM3/22/10
to
Oh No wrote on Mon, 22 Mar 2010 21:29:05 +0100:

> Thus spake Juan R. González-Álvarez <email....@not.supplied>
>>Oh No wrote on Sun, 21 Mar 2010 07:19:05 +0000:
>>
>>> In the Stuckelberg-Feynman interpretation antiparticles are
>>> particles travelling backwards in time. More on this at
>>> http://rqgravity.net/TheDiracEquation#Antiparticles
>>
>> More exactly negative energy particles travelling backward in time
>> are *reinterpreted* as positive energy antiparticles travelling
>> forward.
>>
>> The *reinterpretation* is needed because neither negative energy
>> particles nor particles travelling backward are detected/observed
>> in the lab.
>
> In fact one cannot say this is not observed, or even that
> reinterpretation is needed, because the observation of a particle
> travelling backwards in time is identical to the observation of an
> antiparticle travelling forwards.

We do not detect negative energy particles becoming from future.
We detect positive energy particles going to future, or if you prefer
we detect at the present time particles which were prepared or
generated at earlier times.

This is the reason which Stuckelberg and Feynman were forced to
*reinterpret* the negative energy states, not the inverse.

bj88

unread,
Mar 23, 2010, 3:08:11 AM3/23/10
to
On Mar 22, 10:54 am, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
wrote:
> > so virtual particles don't travelfasterthan light becuase they would
> > violate casuality?
>
> They don't travelfasterthan light because they don't exist,
> hence don't travel at all. Excepti o=in Science Fiction.
>
> Even if one considers virtual particles as existing in case (i) or (ii),
> both apply to reality only at speeds much lower than the speed of light.
> It leads only to misunderstandings if one takes pieces of theory out of
> the context in which they match reality.- Hide quoted text -
>
> - Show quoted text -

what about this article that says they do?
http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html

Oh No

unread,
Mar 23, 2010, 11:52:53 AM3/23/10
to
Thus spake Juan R. González-Álvarez <now...@canonicalscience.com>

>Oh No wrote on Mon, 22 Mar 2010 21:29:05 +0100:
>
>> Thus spake Juan R. González-à lvarez <email....@not.supplied>

>>>Oh No wrote on Sun, 21 Mar 2010 07:19:05 +0000:
>>>
>>>> In the Stuckelberg-Feynman interpretation antiparticles are
>>>> particles travelling backwards in time. More on this at
>>>> http://rqgravity.net/TheDiracEquation#Antiparticles
>>>
>>> More exactly negative energy particles travelling backward in time
>>> are *reinterpreted* as positive energy antiparticles travelling
>>> forward.
>>>
>>> The *reinterpretation* is needed because neither negative energy
>>> particles nor particles travelling backward are detected/observed
>>> in the lab.
>>
>> In fact one cannot say this is not observed, or even that
>> reinterpretation is needed, because the observation of a particle
>> travelling backwards in time is identical to the observation of an
>> antiparticle travelling forwards.
>
>We do not detect negative energy particles becoming from future.
>We detect positive energy particles going to future,
>.
>
What we detect is a track in a bubble chamber, wire chamber or
equivalent. Both interpretations are identical.

Oh No

unread,
Mar 23, 2010, 11:53:01 AM3/23/10
to
Thus spake Arnold Neumaier <Arnold....@univie.ac.at>
>FrediFizzx wrote:
>> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
>>news:4BA62CD1...@univie.ac.at...
>>> FrediFizzx wrote:

>> Are you saying that the Coulomb field is composed of real photons
>>then?
>
>No. The Coulomb field has nothig to do with photons. It arises as an
>interaction term in the QED Hamiltonian in the
>Coulomb gauge.

It arises from the interchange of photons, which is allowed from the
interaction term in the QED Hamiltonian (which creates or annihilates
photons) irrespective of gauge (owing to gauge invariance).

>> If not virtual photons nor virtual fermionic pairs, then there is
>>nothing else to explain it within the context of QED other than real
>>photons.
>
>The standard Lagrangian --> Hamiltonian procedure is enough to
>explain it.

The standard Lagrangian --> Hamiltonian procedure is a procedure, not an
explanation.

>Look into any book on quantum field theory to see that the bare
>particles (and only these arise in internal lines of Feynman diagrams)
>have infinite mass.

Look into Scharf, Finite QED, to see that this an error because of
incorrect mathematical treatments.

>
>He knows quantum field theory. but he does not know more than anyone
>else about virtual particles,

I hope you include yourself.

>>> There is not a single verified measurement of a virtual particle.
>>> Measured particles are always on-shell.
>>
>> :-) Of course not; that is part of the definition of a virtual
>> particle. It is not directly observable. And your second sentence
>>is part of the definition of real particles. I do believe that they
>>are indirectly observable as I seem to recall seeing bubble chamber
>>photos of particle interactions that can only be explained by a
>>virtual particle that is not showing in the photo (didn't leave
>>track). In fact, on page 35 in Griffiths' "Introduction to
>>Elementary Particles" there is a bubble chamber photo of the
>>discovery of the Omega minus particle that shows there must have been
>>two virtual gamma photons that each decayed to real electron-positron
>>pairs. They had to be virtual as a real gamma photon can't do that.
>
>Two real gamma photons can decay into a real electron-positron pair.
>How do you distinguish experimentally between an invisible ghost
>called a virtual photon from two real photons that can be observed on
>numerous other occasions?

You distinguish because the process is mathematically described by a
single photon, which is necessarily virtual.


>
>
>> [ Mod. note: The above paragraph confuses the meaning of "directly
>> observable". Lack of a vapor trail in a bubble chamber indicates
>> lack of electric charge, not "virtuality". Neutral particles can
>> also be directly observed in bubble chambers, through their decay
>> products, the same way your eyes observe the computer screen in
>> front of you. The statement from Griffiths sounds dubious, but
>> I have no copy on hand to check it. -ik ]

It corresponds precisely to my own recollection from other bubble
chamber photographs, which I can't precisely place (being some time
ago), but I know was not Griffiths because I have not read it.


>>> The tests mentioned are not tests of the reality of virtual
>>> particles, but tests of QED and the standard model in general,
>>> together will talk about the behavior of virtual particles that
>>> may be a common fantasy of many physicists, but has no support at
>>> all by experiments.
>> Sorry, but I have to disagree since without virtual W bosons, the
>>decay of muons makes no sense.
>
>As with all other particles, observable W-bosons are real, not virtual.

I agree with Fred. I am not sure that one can even legitimately say that
a W-boson is observable, since its life is 3x10^-25 secs. In any case,
what about Z bosons, which are neutral, and cannot be directly
observable.

>>> The Lamb shift and the Casimir effect can be calculated in a number
>>> of different ways, not all requiring the concept of virtual particles.
>>> How can it then be experimental proof of the latter?
>> 1/40th of the Lamb shift (~27MHz) is explained by virtual particles
>>(vacuum polarization) in the calculations I have seen. Do you have a
>>reference that shows a calculation of the Lamb shift without resorting
>>to vacuum polarization? Better yet; how about a reference for the
>>Lamb shift calculation of muonic atoms? There, vacuum polarization
>>effects are more pronounced.
>
>Vacuum polarization is usually calculated by means of virtual photons,
>but does not have to. High precision calculations of the Lamb shift
>using NRQED are not based on virtual photons.

Those calculations do not incorporate the effect Fred is talking about,
which provides a precision test of qed.

>>> On the Z-boson, the article writes, ''the value agreed with that
>>> obtained from the virtual particle analysis, providing a dramatic
>>> test of our understanding of virtual particles.''
>>> This virtual particle analysis (when followed through - no
>>> references are given) consists in perturbative calculations
>>> that start with Feynman diagrams involving virtual particles.
>>> But these are unmeasurable things with an infinite mass and
>>> infinite interactions, both physically meaningless. They are turned
>>> into predictions only by a perturbative renormalization process
>>> that restores physics to the ill-conceived bare interpretation.
>> I don't agree with that. They are immeasurable but they still have
>>to abide by conservation laws for the entire process so don't think
>>they would have infinite mass-energy nor an infinite amount of
>>interactions.
>
>The formal conservation laws become meaningless since inf+inf=inf+inf,
>no matter what you do.

This is not true. See finite QED by Scharf, or my own paper

http://papers.rqgravity.net/RQGQED.pdf

>Bare particles have no physical properties,
>but only bare particles figure in Feynman diagrams and give rise to
>talk about virtual particles. Real particles are renormalized
>versions of the bare particles and have finite, physical properties,
>but (at least in vacuum) you cannot find them off-shell.

This is also not true. See same refs.

Arnold Neumaier

unread,
Mar 23, 2010, 5:23:38 PM3/23/10
to
Oh No wrote:
> Thus spake Arnold Neumaier <Arnold....@univie.ac.at>
>> FrediFizzx wrote:
>>> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
>>> news:4BA62CD1...@univie.ac.at...
>>>> FrediFizzx wrote:
>
>>> Are you saying that the Coulomb field is composed of real photons
>>> then?
>> No. The Coulomb field has nothig to do with photons. It arises as an
>> interaction term in the QED Hamiltonian in the
>> Coulomb gauge.
>
> It arises from the interchange of photons, which is allowed from the
> interaction term in the QED Hamiltonian (which creates or annihilates
> photons) irrespective of gauge (owing to gauge invariance).

Nothin in the standard approach to get a Hamiltonian from a Lagrangian
depends on talk about virtual particles. The latter arise only when one
considers a perturbative expansion around a free dynamics.


>>> If not virtual photons nor virtual fermionic pairs, then there is
>>> nothing else to explain it within the context of QED other than real
>>> photons.
>> The standard Lagrangian --> Hamiltonian procedure is enough to
>> explain it.
>
> The standard Lagrangian --> Hamiltonian procedure is a procedure, not an
> explanation.

This procedure explains perfectly, without any approximation, how the
Coloumb force arises from the action. Thus it is far superior to any
attempted explanation by means of approximate, nonconvergent
perturbation expansions that give rise to virtual particles with
extremely strange, unobservable properties.


>> Look into any book on quantum field theory to see that the bare
>> particles (and only these arise in internal lines of Feynman diagrams)
>> have infinite mass.
>
> Look into Scharf, Finite QED, to see that this an error because of
> incorrect mathematical treatments.

The correct mathematical treatment in Scharf has no room left
for virtual photons. It computes time-ordered expectation values
without any reference to virtual particles.


>>> In fact, on page 35 in Griffiths' "Introduction to
>>> Elementary Particles" there is a bubble chamber photo of the
>>> discovery of the Omega minus particle that shows there must have been
>>> two virtual gamma photons that each decayed to real electron-positron
>>> pairs. They had to be virtual as a real gamma photon can't do that.
>> Two real gamma photons can decay into a real electron-positron pair.
>> How do you distinguish experimentally between an invisible ghost
>> called a virtual photon from two real photons that can be observed on
>> numerous other occasions?
>
> You distinguish because the process is mathematically described by a
> single photon, which is necessarily virtual.

It is a mistake to assume without proof that the process is described by
a single photon, since the latter is unobservable. So ne needs an
argument that can be verified experimentally.

Now the cross section can be calculated and verified. Itcomes from an
S-matrix calculation involving two real photons (all S-matrix elements
describe scattering of real particles). Thus there is no need to invoke
ghosts called virtual particles.


>>>> The Lamb shift and the Casimir effect can be calculated in a number
>>>> of different ways, not all requiring the concept of virtual particles.
>>>> How can it then be experimental proof of the latter?
>>> 1/40th of the Lamb shift (~27MHz) is explained by virtual particles
>>> (vacuum polarization) in the calculations I have seen. Do you have a
>>> reference that shows a calculation of the Lamb shift without resorting
>>> to vacuum polarization? Better yet; how about a reference for the
>>> Lamb shift calculation of muonic atoms? There, vacuum polarization
>>> effects are more pronounced.
>> Vacuum polarization is usually calculated by means of virtual photons,
>> but does not have to. High precision calculations of the Lamb shift
>> using NRQED are not based on virtual photons.
>
> Those calculations do not incorporate the effect Fred is talking about,
> which provides a precision test of qed.

NRQED gives the highest known precision calculation of Lamb shifts,
correct ot O(alpha^6).


Arnold Neumaier

Oh No

unread,
Mar 24, 2010, 4:04:42 AM3/24/10
to
Thus spake Arnold Neumaier <Arnold....@univie.ac.at>
>Oh No wrote:
>> Thus spake Arnold Neumaier <Arnold....@univie.ac.at>
>>> FrediFizzx wrote:
>>>> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
>>>> news:4BA62CD1...@univie.ac.at...
>>>>> FrediFizzx wrote:
>>
>>>> Are you saying that the Coulomb field is composed of real photons
>>>> then?
>>> No. The Coulomb field has nothig to do with photons. It arises as an
>>> interaction term in the QED Hamiltonian in the
>>> Coulomb gauge.
>> It arises from the interchange of photons, which is allowed from the
>> interaction term in the QED Hamiltonian (which creates or annihilates
>> photons) irrespective of gauge (owing to gauge invariance).
>
>Nothin in the standard approach to get a Hamiltonian from a Lagrangian
>depends on talk about virtual particles.

That is hardly relevant. I would rather start with the Hamiltonian than
the Lagrangian anyway.

>The latter arise only when one
>considers a perturbative expansion around a free dynamics.


>>>> If not virtual photons nor virtual fermionic pairs, then there is
>>>> nothing else to explain it within the context of QED other than real
>>>> photons.
>>> The standard Lagrangian --> Hamiltonian procedure is enough to
>>> explain it.
>> The standard Lagrangian --> Hamiltonian procedure is a procedure,
>>not an
>> explanation.
>
>This procedure explains perfectly, without any approximation, how the
>Coloumb force arises from the action.

This does not affect my point. metaphysical concepts like the action do
not constitute an explanation of anything.

>Thus it is far superior to any attempted explanation by means of
>approximate, nonconvergent perturbation expansions that give rise to
>virtual particles with extremely strange, unobservable properties.

I do not find anything strange about them, apart from the word "virtual"
which is a poor choice. Issues like nonconvergence (the expansion is
asymptotic) can be understood by studying the derivation.

>>> Look into any book on quantum field theory to see that the bare
>>> particles (and only these arise in internal lines of Feynman diagrams)
>>> have infinite mass.
>> Look into Scharf, Finite QED, to see that this an error because of
>> incorrect mathematical treatments.
>
>The correct mathematical treatment in Scharf has no room left
>for virtual photons. It computes time-ordered expectation values
>without any reference to virtual particles.

Many treatments do not find it necessary to use the word "virtual",
which is a poor choice.

>>>> In fact, on page 35 in Griffiths' "Introduction to
>>>> Elementary Particles" there is a bubble chamber photo of the
>>>> discovery of the Omega minus particle that shows there must have been
>>>> two virtual gamma photons that each decayed to real electron-positron
>>>> pairs. They had to be virtual as a real gamma photon can't do that.
>>> Two real gamma photons can decay into a real electron-positron pair.
>>> How do you distinguish experimentally between an invisible ghost
>>> called a virtual photon from two real photons that can be observed on
>>> numerous other occasions?
>> You distinguish because the process is mathematically described by a
>> single photon, which is necessarily virtual.
>
>It is a mistake to assume without proof that the process is described by
>a single photon, since the latter is unobservable. So ne needs an
>argument that can be verified experimentally.
>
>Now the cross section can be calculated and verified. Itcomes from an
>S-matrix calculation involving two real photons (all S-matrix elements
>describe scattering of real particles). Thus there is no need to invoke
>ghosts called virtual particles.

I strongly suspect you must have some other definition of "virtual" in
this context from the one I use.

Hendrik van Hees

unread,
Mar 24, 2010, 5:51:51 AM3/24/10
to
Juan R. González-Álvarez wrote:

> We do not detect negative energy particles becoming from future.
> We detect positive energy particles going to future, or if you prefer
> we detect at the present time particles which were prepared or
> generated at earlier times.
>
> This is the reason which Stuckelberg and Feynman were forced to
> *reinterpret* the negative energy states, not the inverse.

I post in reply to this message, but it's meant more as a general
contribution to the whole thread.

I think the trouble with "virtual particles" is again just a
mystification of rather simple facts. The problem is to speak in plain
(English) language about things one cannot express in plain language but
has to use mathematics. Another complication is, that this mathematics
is quite dirty, but we have nothing better in physics than quantum
(gauge) field theory, and in fact it's the best kind of theories we have
so far.

The Stueckelberg-Feynman trick is in fact no trick but a necessary
mathematical construction, given the fact that (by definition!) physics
deals with causal descriptions of nature, i.e., for any observer an
observable fact "now" can depend only on events in his "past". That's
also inherent in the very definition of time. So it doesn't make sense
to have "particles" running backwards in time. That's why I wrote "modes
of negative frequency" and not "plane-wave functions of negative energy"
as is sometimes done even in textbooks, adding to the confusion rather
than helping with the understanding. This forces us to write these modes
with a creation operator instead of an annihilation operator in the
expansion of the field operator. Nothing could be simpler than this!
However, it makes relativistic "vacuum QFT" manifestly different (and
more complicated) than non-relativistic "vacuum QFT". Solving these
difficulties in turn solves a nearly one century old obstacle of
classical field theory (i.e., Maxwell-Lorentz electromagnetics), namely
the question of the back reaction of a particle's own electromagnetic
field on its (accelerated) motion (at least in a perturbation-
theoretical sense, and I only talk about interacting field theories in
the perturbative sense in this posting!). The particle simply gets a
self energy with a diverging mass and a diverging renormalization factor
of the corresponding single-particle state.

That these infinities appear in perturbation theory is no big surprise
anymore nowadays, but was a bit obstacle in the early 30ies when QFT has
been invented by Pauli, Born, Jordan, and Dirac. The solution came in
1948 with the discovery of renormalization theory (which was developed
in full glory only in the mid 60ies or so with the works of
Bogoliubov+Parasiuk, Hepp, and Zimmermann with their famous BPHZ
formalism). Although it's not a very easy task in the concrete
calculations, the physical meaning is simple. In perturbation theory,
everything is expanded in terms of free-particle states. In QED that
means that at 0th order perturbation theory there are non-interacting
electrons and non-interacting photons. This already shows that this is a
men-made construct and not a description of what we observe in nature!
Particularly the electron in this order has no electromagnetic field
around it and in the same way the photon (i.e., the elementary
electromagnetic excitations) have no electron-positron cloud around it.
Thus, one speaks (in the usual funny slang of QFT physicists) about
"bare particles" and/or "bare fields".

Now, taking into account the electromagnetic interaction, we calculate
things with a "point of view" (technically speaking we formulate the
theory in a free-particle Fock basis) which handles fictitious objects,
namely the bare particles and bare fields. In reality, these objects
don't exist. Thus, order by order perturbation theory, we have to
construct the Fock basis of the real-particle/fields states, and this is
done by expressing everything in terms of physical parameters (like
wave-function normalization factors, masses and coupling constants),
approximated in the sense of perturbation theory (which is formally an
expansion in the coupling constant and/or \hbar). These parameters are
perfectly finite and can be adjusted by observable facts. Fortunately
there's a class of QFTs, called Dyson-renormalizable, where there's a
finite number coupling and mass parameters which is sufficient to absorb
all infinities of perturbation theory, i.e., the model has only a finite
number of parameters. I don't speak about the other case, namely
"effective theories" that are not renormalizable in the Dyson sense but
can be used in a certain (mostly low-) energy range and are expansions
with respect to powers in energy/momentum rather than in terms of
coupling constants/hbar (although both expansions are closely related,
which makes these theories very useful, like in chiral perturbation
theory to describe the low-energy behavior of strongly interacting
particles in terms of the hadronic "relevant" degrees of freedom rather
than with quarks and gluons, which makes the issue of hadron
interactions a very complicated matter).

Another fortunate property of the perturbative expansion of interacting
QFTs is that everything can be expressed in a *formal* diagrammatical
way, invented first by Feynman. He presented this diagram technique to
his astonished theory colleagues at the famous Shelter-Island
conference. Nowadays this technique is nearly an own special subject in
theoretical physics and far developed, particularly it's also applied in
a much wider range of applications, particularly in many-body QT (be it
non-relativistic of relativistic) in and out off equilibrium. Here, I
stick to the special case of the "vacuum theory", dealing with a small
number of particles (usually two in the initial and a few in the final
state of a collision process).

The original Feynman diagrams stand for terms in the perturbative
expansion of scattering-matrix elements. This said, it is clear that one
should not put too much physical interpretation into them although they
are very intuitive, if one keeps their abstract meaning in mind, and S-
matrix elements are pretty abstract objects. They are the probability
amplitudes for the transition of an asymptotically free initial state
(usually two particles which are prepared far away from each other with
well-adjusted momenta (and sometimes also spin/polarization)) to an
asymptotically free final state (whatever one decides to detect after
the collision about the then approximately freely streaming reaction
products). These kind of Feynman diagrams lead indeed to observable
quantitities, namely these transition-probability rates which are
usually expressed in terms of a cross section.

The Feynman diagrams themselves are built out of elementary Feynman
diagrams which stand for the bare-particle propagators (lines connecting
two points, where the particle or antiparticle is created and reabsorbed
again) and the interaction points where three or more lines meet at a
space-time point where "a perturbation" occurs in the sense of
perturbation theory. This also shows that the number of vertices in a
diagram counts the number of coupling constants in the mathematical
expression of the corresponding contribution to the S-matrix element.
Unfortunately the internal propagator lines are also called "virtual
particles" in the funny jargon of QFT physicists, and this gives all the
trouble we are discussing about, but the words "virtual particles" in
fact have no other meaning than that given by the mathematical
expressions. It stands for a propagator of bare particles no more no
less.

A bare-particle propagator is of course not observable in principle
since we cannot do experiments with bare particles/fields, because these
are fictitous inventions of the human mind to organize the perturbation
series! What is observable in principle are the properties of single
real particles, meaning in a more accurate language, single-particle
states for real particles. The properties of the single-particle states
can be reconstructed from the full two-point Green's function which can
also be calculated with the Feynman-diagram technique. Again, each
Feynman diagram stands for a certain correction term to the bare
propagator no more no less, and one has to ask what is observable. A
single-particle state is characterized by some set of compatible
observables, e.g., momentum and spin states. If it comes to
asymptotically free (stable or quasi-stable) particles, there's also a
relation between energy and momentum, and this is observable in
principle (in case of a stable particle, it's a strict relation
omega=omeg(\vec{p}), in the case of an unstable but long-lived single-
particle excitation it's the spectral function). The spectral function
is given by the imaginary part of the full propagator. Approximately
(for not too broad spectral functions) the spectral function is
characterized by the possition of the mass peak and its width, which in
turn are given by the poles of the propagators in the complex energy
plane, and this is what is observable in principle (e.g., looking at
elastic pion-nucleon scattering one observes a strong resonance, called
Delta(1232) or lepton scattering giving the W- and Z-boson peaks from
weak interaction etc.).

The notion of virtual particles doesn't play a big role in the
interpretation of QFTs, it's only part of a convenient jargon of
experts. Unfortunately in many attempts to popularize this fascinating
theory by crude simplifications about the meaning of Feynman diagrams
"virtual particles" are made to something intuitively "real", and that's
imho a big didactical mistake. One shouldn't speak about the theory in
these terms. To popularize particle physics it's much wiser to
concentrate on phenomenology, i.e., the observable facts about particles
rather than to chat about highly specialized theories which are not
suitable for popularization since the real meaning of the abstract
objects (including Feynman diagrams, as I hope to have made clear above)
is incomprehensible without the appropriate training in theoretical
physics (that's why relativistic QFT is taught only in higher semesters
for physics majors, and renormalization theory and all that perhaps even
only at graduate-student level). What one can popularize, however, is
the general world view which follows from the application of quantum
theory to real phenomena, but for that one shouldn't (and doesn't need
to) talk about "virtual particles"!

--
Hendrik van Hees
Institut für Theoretische Physik
Justus-Liebig-Universität Gießen
http://theorie.physik.uni-giessen.de/~hees/

Bob_for_short

unread,
Mar 24, 2010, 11:00:03 AM3/24/10
to
A hot body loses its energy by radiating real photons.

Another body may be heated with this radiation.

It is another distinction between the real particles (photons) and
permanently acting forces between charges (Coulomb forces, for
example).

Bob_for_short

unread,
Mar 24, 2010, 9:01:19 PM3/24/10
to
On 24 mar, 10:51, Hendrik van Hees <Hendrik.vanH...@physik.uni-
giessen.de> wrote:


> ...The particle simply gets a


> self energy with a diverging mass and a diverging renormalization
> factor of the corresponding single-particle state.
>

Indeed, it is simple. The question is why electron and photon get
corrections to their masses? I think it is because the interaction
term is of kinetic rather than potential character (see my answers at
http://groups.google.com/group/sci.physics.foundations/browse_thread/thread/0ac9b50e17a05511#).

> That these infinities appear in perturbation theory is no big surprise

> any more nowadays, but was a bit obstacle in the early 30ies when QFT has


> been invented by Pauli, Born, Jordan, and Dirac. The solution came in

> 1948 with the discovery of renormalization theory ...

which is simply a practice of discarding divergent perturbative
corrections to the initial physical parameters, isn't it?

> In perturbation theory,
> everything is expanded in terms of free-particle states. In QED that

> means that at 0-th order perturbation theory there are non-interacting


> electrons and non-interacting photons. This already shows that this is a
> men-made construct and not a description of what we observe in nature!
> Particularly the electron in this order has no electromagnetic field
> around it and in the same way the photon (i.e., the elementary
> electromagnetic excitations) have no electron-positron cloud around it.
> Thus, one speaks (in the usual funny slang of QFT physicists) about
> "bare particles" and/or "bare fields".
>

If something is a solution of a free equation, it does not mean that
this something is not observable.

Besides, we start solving the equations of interacting, observable
particles. Another thing is our way of the equation solving with the
perturbation theory which gives approximate rather than exact
solutions.

> Now, taking into account the electromagnetic interaction, we calculate
> things with a "point of view" (technically speaking we formulate the
> theory in a free-particle Fock basis) which handles fictitious objects,
> namely the bare particles and bare fields. In reality, these objects
> don't exist.

When do we discover that the "initial" particles are "bare" with
infinite parameters? When we obtain corrections to the initial
solutions. It is these corrections that are infinite, not the original
physical parameters.

> ... These parameters are


> perfectly finite and can be adjusted by observable facts. Fortunately
> there's a class of QFTs, called Dyson-renormalizable, where there's a
> finite number coupling and mass parameters which is sufficient to absorb
> all infinities of perturbation theory, i.e., the model has only a finite
> number of parameters.
>

Everybody speaks of "absorbing" infinite corrections by the initial
parameters. It simply means *discarding* these corrections and keeping
the initial physical parameters intact.

> A bare-particle propagator is of course not observable in principle

> since we cannot do experiments with bare particles/fields ...

Nevertheless we require the exact propagator to have a pole exactly in
the same form as the bare, non observable propagator. Isn't is funny?

I studies this question and arrived at a quite reasonable explanation.
You may find my results in my web log: http://vladimirkalitvianski.wordpress.com

Regards,

Vladimir.

Juan R. González-Álvarez

unread,
Mar 24, 2010, 11:21:24 PM3/24/10
to
FrediFizzx wrote on Mon, 22 Mar 2010 06:27:00 +0000:

> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
> news:4BA62CD1...@univie.ac.at...
>> FrediFizzx wrote:
>>> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
>>> news:4BA3D4F...@univie.ac.at...

(...)

>> Photons are transverse excitations of the e/m field, i.e.,
>> deviations from the mean value of the field. To consider the
>> Coulomb field, say, as being composed of virtual photons is a
>> superficial misunderstanding of quantum field theory. None of the
>> many very successful calculations comparing theory with experiment
>> depends on the reality of such virtual particles.
>
> Are you saying that the Coulomb field is composed of real photons
> then? If not virtual photons nor virtual fermionic pairs, then
> there is nothing else to explain it within the context of QED other
> than real photons.

It was already proven in [1,2,3] that the usual classical field theory
interpretation of Coulomb interactions is misguided and not valid
anymore. The proof has matematical rigor (indeed Prof. Smirnov-Rueda
is a mathematician)

[[Mod. note -- I think it's fair to say that the references cited by
the author are highly controversial. I believe that most researchers
in classical field theory do not consider these results to be correct.

If people want to debate this, please start a *new* thread with an
appropriate subject line, since such debate would not really be about
virtual particles (the nominal subject of this thread).
-- jt]]

Currently I am working to extend the mathematical proof to quantum
field theory. I can advance now to you that quantum Coulomb
interactions do not arise from interchanging virtual quanta.

[1] Phys. Rev. E 1997, 53, 5373.
[2] Phys. Rev. E 1998, 57, 3683.
[3] http://www.canonicalscience.org/publications/canonicalsciencetoday/20100301.html

FrediFizzx

unread,
Mar 24, 2010, 11:22:47 PM3/24/10
to
"Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
news:4BA7B949...@univie.ac.at...

You need to give me a couple of concrete examples if you say they are
that much different. And don't just give me some general reference.
Just do it right here on the group. The results would be fine.

>>> Photons are transverse excitations of the e/m field, i.e.,
>>> deviations from the mean value of the field. To consider the
>>> Coulomb field, say, as being composed of virtual photons is a
>>> superficial misunderstanding of quantum field theory.
>>> None of the many very successful calculations comparing theory with
>>> experiment depends on the reality of such virtual particles.
>>
>> Are you saying that the Coulomb field is composed of real photons
>> then?
>
> No. The Coulomb field has nothig to do with photons. It arises as an
> interaction term in the QED Hamiltonian in the
> Coulomb gauge.

Sorry, but that is contrary to the standard model of particle physics
and I believe fairly speculative. For QED we have photons and
electrically charged elementary particles such as the electron, etc.
And via the Uncertainty Principle we also have their virtual versions.
If you are saying the Coulomb field has nothing to do with photons
either real or virtual, then we probably should not even continue this
discussion. OK, if not photons, then it has to be virtual fermionic
pairs (my favorite for the Coulomb and magnetic fields) but you don't
allow that either. So what is this new theory called that you are
proposing that allows a magic *real* Coulomb field that is not supported
by elementary standard model particles? You have got to do much better
than "arises as an interaction term". Is that another mathematical
"artifact"? ;-)

>> If not virtual photons nor virtual fermionic pairs, then there is
>> nothing else to explain it within the context of QED other than real
>> photons.
>
> The standard Lagrangian --> Hamiltonian procedure is enough to
> explain it.

No, within QED you have to explain it with the allowed building blocks
of the standard model of particle physics. Which *does* include virtual
particles if you wish to use them.

[big snip; will study what you said later]

> You may wish to read the Nobel lecture by Frank Wilczek at
> http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1150826/
> where he first talks about the traditiona virtual particle picture:
> ''Loosely speaking, energy can be borrowed to make evanescent
> virtual particles''. Note his qualification that indicates that
> this cannot be taken seriously. He also says why - because one
> encounters divergernces by taking them seriously. Then he gets
> more serious and shows how renormalization fixes the problems,
> though he does not say that this comes at the cost of making the
> virtual particles infinitely heavy (and hence again physically
> meaningless). But this can be read in any textbook on QFT.
> Later, he slips back into the traditional jargon since it
> provides a vivid intuition about Feynman diagrams -- especially
> for the many nonexperts in his audience, but again he does so
> with a careful, explicit caveat:
> ''(I'm being a little sloppy in my terminology; instead of saying
> the number of virtual particles, it would be more proper to speak
> of the number of internal loops in a Feynman graph.)''

I prefer the part where he says, "These pictures make it clear and
tangible that the quantum vacuum is a dynamic medium, whose properties
and responses largely determine the behavior of matter". Note the word
"tangible". I don't think he would say that if he thought virtual
particles are mathematical artifacts. In fact, that sentence pretty
much frames my research effort.

Best,

Fred Diether

Bob_for_short

unread,
Mar 24, 2010, 11:23:05 PM3/24/10
to
On 23 mar, 16:53, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
> >No. The Coulomb field has nothig to do with photons. It arises as an
> >interaction term in the QED Hamiltonian in the
> >Coulomb gauge.
>
> It arises from the interchange of photons, which is allowed from the
> interaction term in the QED Hamiltonian (which creates or annihilates
> photons) irrespective of gauge (owing to gauge invariance).
>

No, it is not an interchange of photons. One charge is a source of a
strong quasi-stationary electromagnetic filed for another charge, and
the other charge is a source for the first one. Perturbative treatment
of this interaction term (it depends only on the charge relative
distance and charge velocities) does not make a virtual photon. As I
said, this term is an explicit solution expressed via particle
variables solely so no intermediary agent is necessary.

Note, a free photon cannot be entirely absorbed by a free electron but
scattered. The "virtual" photon, on the other hand cannot be scattered
but only "absorbed". In other words, it does not propagate. It is an
intrinsic property of charge intercation and nothing else. We need not
virtual photons to describe and understant bound states and charge
scattering. let us not fool ourselves.

Oh No

unread,
Mar 24, 2010, 11:23:48 PM3/24/10
to
Thus spake Hendrik van Hees <Hendrik...@physik.uni-giessen.de>
>The Stueckelberg-Feynman trick is in fact no trick but a necessary
>mathematical construction, given the fact that (by definition!) physics
>deals with causal descriptions of nature, i.e., for any observer an
>observable fact "now" can depend only on events in his "past". That's
>also inherent in the very definition of time. So it doesn't make sense
>to have "particles" running backwards in time.

To me this makes perfect sense. The energy momentum vector is a timelike
4-vector. If energy is positive, it is a vector into the future light
cone. If energy is negative, it is a vector into the past light cone. I
don't think we need say more than that to say what we mean by particles
running backwards in time.

>That's why I wrote "modes of negative frequency" and not "plane-wave
>functions of negative energy" as is sometimes done even in textbooks,
>adding to the confusion rather than helping with the understanding.
>This forces us to write these modes with a creation operator instead of
>an annihilation operator in the expansion of the field operator.
>Nothing could be simpler than this!

quite.

> Unfortunately in many attempts to popularize this fascinating theory
>by crude simplifications about the meaning of Feynman diagrams "virtual
>particles" are made to something intuitively "real", and that's imho a
>big didactical mistake.

The answer to this should not be to reject the approach, but rather to
develop a more sophisticated language in which to discuss quantum theory
in general and Feynman diagrams in particular.

Arnold Neumaier

unread,
Mar 25, 2010, 3:49:35 PM3/25/10
to

I don't have the time to locate a suitable reference that has it all in
one place. But it is clear to anyone who learnt more than one approach
to calculations with quantum fields that the intermediate expressions
that arise cannot have any reality since they occur only in a particular
approach.

Giving virtual reality to intermediate expressions in a technique to
calculate numbers with physical meaning is like saying the terms in
the series pi/4=1 - 1/3 + 1/5 - 1/7 + 1/9 - ... represent virtual
pieces and antipieces of the quarter-circle. But someone else computes
pi from pi^2/12=1 - 1/4 + 1/9 - 1/16 + 1/25 - ... and gets virtual
pieces with completely different properties! And of course, we have no
reason to take such virtuality seriously. It lives in quantum fieeld
theory for the sole reason that the formulas are abstract and messy,
and couching them in a pictorial language with comprehensible names
and an illusion of reality makes them easier to remember and to refer
to.

It is the perturbative treatment -- and only that -- that generates
virtual particles as internal lines in diagrams that represent
multidimiensional integrals. Different perturbative approaches
of course give different integral representations and hence virtual
particles with different properties. And nonperturbative approaches do
not see any of these ghosts. Look at an _arbitrary_ book on lattice
gauge theory and try to find virtual particles. You'll not find them,
but I can't give you references to the nonexisting books that tell
you about this lack of occurence.

If you cannot accept that much without reference to particular papers
then you have to find these references yourself.


>>> Are you saying that the Coulomb field is composed of real photons
>>> then?
>> No. The Coulomb field has nothig to do with photons. It arises as an
>> interaction term in the QED Hamiltonian in the
>> Coulomb gauge.
>
> Sorry, but that is contrary to the standard model of particle physics
> and I believe fairly speculative.

This is not speculative. You can read it in the quantum field theory
book of Bjorken and Drell,
http://en.wikipedia.org/wiki/James_Bjorken
which was for many years the textbook from which people learnt the
ropes, before path integral techniques took over.


> If you are saying the Coulomb field has nothing to do with photons
> either real or virtual, then we probably should not even continue this
> discussion.

You may quit the discussion anytime like anyone else.
It doesn't change the truth.


>>> If not virtual photons nor virtual fermionic pairs, then there is
>>> nothing else to explain it within the context of QED other than real
>>> photons.
>> The standard Lagrangian --> Hamiltonian procedure is enough to
>> explain it.
>
> No, within QED you have to explain it with the allowed building blocks
> of the standard model of particle physics.

Look at Bjorken/Drell, and then we can discuss this with more background.


>> You may wish to read the Nobel lecture by Frank Wilczek at
>> http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1150826/
>> where he first talks about the traditiona virtual particle picture:
>> ''Loosely speaking, energy can be borrowed to make evanescent
>> virtual particles''. Note his qualification that indicates that
>> this cannot be taken seriously. He also says why - because one
>> encounters divergernces by taking them seriously. Then he gets
>> more serious and shows how renormalization fixes the problems,
>> though he does not say that this comes at the cost of making the
>> virtual particles infinitely heavy (and hence again physically
>> meaningless). But this can be read in any textbook on QFT.
>> Later, he slips back into the traditional jargon since it
>> provides a vivid intuition about Feynman diagrams -- especially
>> for the many nonexperts in his audience, but again he does so
>> with a careful, explicit caveat:
>> ''(I'm being a little sloppy in my terminology; instead of saying
>> the number of virtual particles, it would be more proper to speak
>> of the number of internal loops in a Feynman graph.)''
>
> I prefer the part where he says, "These pictures make it clear and
> tangible that the quantum vacuum is a dynamic medium, whose properties
> and responses largely determine the behavior of matter". Note the word
> "tangible".

Tangible to the imagination, not to an experiment.
Only the latter defines what is real in science.


Arnold Neumaier

Arnold Neumaier

unread,
Mar 25, 2010, 3:49:36 PM3/25/10
to
bj88 wrote:
> On Mar 22, 10:54 am, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
> wrote:
>> bj88 wrote:
>>> so virtual particles don't travelfasterthan light becuase they would
>>> violate casuality?
>> They don't travelfasterthan light because they don't exist,
>> hence don't travel at all. Except in Science Fiction.

>>
> what about this article that says they do?
> http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html

Does the article say that virtual particles exist?
I couldn't find the word ``exist'' in the article.

If someone tells a nice story to illustrate a possibility,
you should not take it as a proof of existence.

This article begins with:

''Electromagnetic fields can do things other than vibration. For
instance, the electric field produces an attractive or repulsive force
between charged objects, which varies as the inverse square of distance.
The force can change the momenta of the objects.
Can this be understood in terms of photons as well? It turns out that,
in a sense, it can.''

So it can, but (i) only ``in a sense'', and (ii) it doesn't have to.

If you can understand a phenomenon in terms of measurable forces and
also in terms of unobservable ghosts (so that you do not need the
latter for understanding), is this then a proof that ghosts exist?

Common sense will give you the answer.


Arnold Neumaier

Bob_for_short

unread,
Mar 25, 2010, 3:49:33 PM3/25/10
to
On 23 mar, 16:53, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Thus spake Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
...

> >No. The Coulomb field has nothig to do with photons. It arises as an
> >interaction term in the QED Hamiltonian in the
> >Coulomb gauge.
>
> It arises from the interchange of photons, which is allowed from the
> interaction term in the QED Hamiltonian (which creates or annihilates
> photons) irrespective of gauge (owing to gauge invariance).
>
I agree with Arnold here.

The Coulomb (or radiation) gauge is a gauge where the transversal
(propagating) and longitudinal fields are separated explicitly. In a
certain sense it is a gauge-invariant formulation in the so called
Dirac variables. The Coulomb interaction term H_int =
integral_d^3xd^3y J0(x)J0(y)/|x-y| is a quasi-stationary potential
energy of interacting charges. Using another gauge and in the
perturbative approach does not cancel this physics. Again, this term
is explicit and free of ambiguities, no photons or exchanges with
something else are involved here. QED does not cancel the charge
property to interact with another charge via the Coulomb (or more
generally, quasi-stationary) forces. Incident real photons arriving
from far away, on the other hand, can be described without knowing
their source - via boundary conditions that determine the photon flux
properties.

Regards,

Vladimir.

FrediFizzx

unread,
Mar 26, 2010, 7:07:11 PM3/26/10
to
"Hendrik van Hees" <Hendrik...@physik.uni-giessen.de> wrote in
message news:hocl6e$oi0$1...@hoshi.visyn.net...

> The notion of virtual particles doesn't play a big role in the
> interpretation of QFTs, it's only part of a convenient jargon of
> experts. Unfortunately in many attempts to popularize this fascinating
> theory by crude simplifications about the meaning of Feynman diagrams
> "virtual particles" are made to something intuitively "real", and
> that's
> imho a big didactical mistake. One shouldn't speak about the theory in
> these terms. To popularize particle physics it's much wiser to
> concentrate on phenomenology, i.e., the observable facts about
> particles
> rather than to chat about highly specialized theories which are not
> suitable for popularization since the real meaning of the abstract
> objects (including Feynman diagrams, as I hope to have made clear
> above)
> is incomprehensible without the appropriate training in theoretical
> physics (that's why relativistic QFT is taught only in higher
> semesters
> for physics majors, and renormalization theory and all that perhaps
> even
> only at graduate-student level). What one can popularize, however, is
> the general world view which follows from the application of quantum
> theory to real phenomena, but for that one shouldn't (and doesn't need
> to) talk about "virtual particles"!

So what are we doing here? Are we going back over 50 years to Feynman's
"blob" and ignoring the billions of dollars spent on elementary particle
research since then? Forget about Feynman diagrams and all that jazz
for a minute. Quarks are also not directly observable but there are not
many particle physicists that don't think they exist. They produce some
very real jets in particle detectors. Are we going to toss out the
Uncertainty Principle (UP) that allows virtual particles to exist?

Gordon Kane, a very prominent particle physicist, says they are real.
Frank Wilczek also thinks they are real; he is a closet modern etherist
so I know he believes they are real [1]. My former particle physics
tutor, Dr. Andy Inopin - a hadronic particle physics theorist, also
thinks they are real. And that they are just like real particles but
simply "off-mass-shell" that is allowed by the UP.

There are many aspects of particle theory that can't be *properly*
explained without the concept of virtual particles such as muon, pion
and neutron decay. You have to have an "off-mass-shell" W boson to
explain those. Are you going to go back to Fermi's old theory to
explain those? There are many more examples of processes with virtual
particles that produce real detectable phenomena. And then there is the
simple coulomb field around a charged body. Is it just there by magic?
Quantum theory and the standard model of particle physics says there has
to be quantum objects involved. What are they?

[1]
http://www.amazon.com/Lightness-Being-Ether-Unification-Forces/dp/0465003214#noop

Best,

Fred Diether

Arnold Neumaier

unread,
Mar 26, 2010, 7:08:12 PM3/26/10
to
Oh No wrote:
> Thus spake Hendrik van Hees <Hendrik...@physik.uni-giessen.de>
>> The Stueckelberg-Feynman trick is in fact no trick but a necessary
>> mathematical construction, given the fact that (by definition!) physics
>> deals with causal descriptions of nature, i.e., for any observer an
>> observable fact "now" can depend only on events in his "past". That's
>> also inherent in the very definition of time. So it doesn't make sense
>> to have "particles" running backwards in time.
>
> To me this makes perfect sense. The energy momentum vector is a timelike
> 4-vector. If energy is positive, it is a vector into the future light
> cone. If energy is negative, it is a vector into the past light cone. I
> don't think we need say more than that to say what we mean by particles
> running backwards in time.

How do you cause a particle to move backwards in time?
You cannot.

In all scatering experiments that produce antiparticles,
these are caused by the collision of particles created brior to the
collision, and result in the antiparticles that have, according to your
view, allegedely arrived from the future to be present just at the time
where they meet the collision experiment.

Very, very strage, such a view, if taken seriously.

Physicists take causality as the most basic princilple, and this
immediately rules out anything moving backward in time.


Arnold Neumaier

Hendrik van Hees

unread,
Mar 26, 2010, 7:09:24 PM3/26/10
to
Bob_for_short wrote:

> Indeed, it is simple. The question is why electron and photon get
> corrections to their masses? I think it is because the interaction
> term is of kinetic rather than potential character (see my answers at

I tried to convince you that not the electron gets a correction to its
mass but a crude approximation to an elekctron called "bare electron"
becomes an infinite correction to its mass. The mass of the bare
electron is unobservable and thus nobody cares about this infiniteness.

> which is simply a practice of discarding divergent perturbative
> corrections to the initial physical parameters, isn't it?

It is surely not! You have to renormalize perturbative expansions
anyway, no matter whether you have infinities or not. That's most
evident in the stationary perturbation theory, where you always have to
determine the normalization of the "corrected" state separately by the
normalization condition, which makes the pertinent coefficient uniquely
determined up to an imaginary part which translates into the freedom to
choose an (approximate) phase of the exact state in question.

> If something is a solution of a free equation, it does not mean that
> this something is not observable.

A particle which is not interacting is only very indirectly observable,
but in the case of an electron, for sure a free electron is unobservable
since in nature the electron couples to the electromagnetic and the
weak-force (aka photon and weakon) fields.


>
> Besides, we start solving the equations of interacting, observable
> particles. Another thing is our way of the equation solving with the
> perturbation theory which gives approximate rather than exact
> solutions.

We cannot even start to solve the equations of interacting particles
since so far nobody has come up with a clever idea how to do so (except
in interesting but academic cases in 1+1-dimensional toy models).


>
>> Now, taking into account the electromagnetic interaction, we
>> calculate things with a "point of view" (technically speaking we
>> formulate the theory in a free-particle Fock basis) which handles
>> fictitious objects, namely the bare particles and bare fields. In
>> reality, these objects don't exist.
>
> When do we discover that the "initial" particles are "bare" with
> infinite parameters? When we obtain corrections to the initial
> solutions. It is these corrections that are infinite, not the original
> physical parameters.

As I said many times, we cannot observe fictitious bare particles. They
simply don't exist. We prepare asymptotically free particles (to be
distinguished from bare particles). These are only asymptotically free
but interact at some point with other particles!

> Everybody speaks of "absorbing" infinite corrections by the initial
> parameters. It simply means *discarding* these corrections and keeping
> the initial physical parameters intact.

That's not completely right. Of course, you fix the parameters of the
model to their physical values (at the given order of perturbation
theory) at a certain energy-momentum scale (the renormalization point of
your one-particle irreducible Green's functions). Further, I agree that
the most clear way to do calculations is to reorganize the perturbation
series such that you work with the physical parameters at each order and
then introducing counter terms to meet the renormalization condition.
The observable quantities (i.e. S-matrix elements) are independent of
the choice of the renormalization condition (within the realm of small
couplings, where perturbation theory is applicable). This independence
is described formally by the renormalization-group equations.


>
>> A bare-particle propagator is of course not observable in principle
>> since we cannot do experiments with bare particles/fields ...
>
> Nevertheless we require the exact propagator to have a pole exactly in
> the same form as the bare, non observable propagator. Isn't is funny?

That's the very definition of the mass parameter of the theory. By
definition the single-particle propagator's pole determines its mass,
but it's the mass of the observable particles not of unobservable bare
particles (that's why I emphasized that one has to keep in mind that
notions like "bare particles" or "virtual particles" are HEP people's
slang which is convenient as soon as one becomes familiar with it, but
one has to learn the definitions not to get confused!).

--
Hendrik van Hees
Institut f?r Theoretische Physik
Justus-Liebig-Universit?t Gie?en
http://theorie.physik.uni-giessen.de/~hees/

Arnold Neumaier

unread,
Mar 26, 2010, 10:32:24 PM3/26/10
to
Bob_for_short wrote
[in: ``Virtual particles help'']:

> On 24 mar, 10:51, Hendrik van Hees <Hendrik.vanH...@physik.uni-
> giessen.de> wrote:
>
>> ...The particle simply gets a
>> self energy with a diverging mass and a diverging renormalization
>> factor of the corresponding single-particle state.
>
> Indeed, it is simple. The question is why electron and photon get
> corrections to their masses? I think it is because the interaction
> term is of kinetic rather than potential character (see my answers at
> http://groups.google.com/group/sci.physics.foundations/browse_thread/thread/0ac9b50e17a05511#).

The reason is much simpler:

Interactions _generally_ change the ground state energies of a system.
In the relativistic case, the ground state energies at the lowest
definite other quantum numbers define the particle masses.
No surprise therefore that these are different from their bare values
in the free theory.

What is special in relativistic QFT though is the fact that these
corrections must be infinitely large, which comes from the fact that
real particles are very different in their properties from free
particles. E.g., a free electron has no associated electromagnetic
field although it is charged!

Therefore perturbation theory works only in a renormalization limit
where the free particle properties are taken to diverge.


>> That these infinities appear in perturbation theory is no big surprise
>> any more nowadays, but was a bit obstacle in the early 30ies when QFT has
>> been invented by Pauli, Born, Jordan, and Dirac. The solution came in
>> 1948 with the discovery of renormalization theory ...
>
> which is simply a practice of discarding divergent perturbative
> corrections to the initial physical parameters, isn't it?

No. Discarding would produce uncontrollable results.

Renormalization amounts to taking limits in the correct way so that
finite numbers cancel each other before they can get infinite in the limit.

Unrenormalized theory amounts in a very simplified analogy to
representing a number Z(m_0) depending on a bare parameter m_0 in terms
of a cutoff M in the form
Z_0(m_0) := sum_k lim_{M->inf} A_k(m_0,M)
where the limits are plus or minus infinite, no matter which value
of m_0 is chosen.

Renormalized theory amounts to finding a function m_0(M,m) that depends
on the cutoff M and the renormalized mass m such that m(M,m) diverges
for M->inf but with the property that the rearranged expression
Z(m) = lim_{M->inf} sum_k A_k(m_0(M,m),M)
gives a well-defined, finite answer for all finite m.


> When do we discover that the "initial" particles are "bare" with
> infinite parameters? When we obtain corrections to the initial
> solutions. It is these corrections that are infinite, not the original
> physical parameters.

The parameters arising in the action (and these are the bare mass, etc.)
_must_ tend to infinity as the cutoff is removed, in order that the
final results are finite. There is no way to get a finite theory from
the QED action with finite bare masses.


> Everybody speaks of "absorbing" infinite corrections by the initial
> parameters. It simply means *discarding* these corrections and keeping
> the initial physical parameters intact.

No. You'd be more cautious in making your claims.
One can see that you haven't worked through the details,
since your statement flatly contradicts actual practice.

Discarded are only carefully defined terms that casncel under the
limit rearrangement indicated above.

Arnold Neumaier

Bob_for_short

unread,
Mar 28, 2010, 5:15:00 AM3/28/10
to
On 27 mar, 00:09, Hendrik van Hees <Hendrik.vanH...@physik.uni-
giessen.de> wrote:
> ... The mass of the bare

> electron is unobservable and thus nobody cares about this infiniteness.
>

Yes, everybody cares. It is not only infinite, it is also infinite as
necessary to make ends meet while its renormalizing. In particular, it
is cutoff-dependent.

Normalizing wave functions is not renormalization. It is a routine
normalization.

>
> We cannot even start to solve the equations of interacting particles

> since so far nobody has come up with a clever idea how to do so.

That's a correct answer. That is why we practice the ritual of
renormalizations.

> ... the bare particles and bare fields. In


> >> reality, these objects don't exist.

Why then do they exist in our theory?

>
> > At what stage we discover that the "initial" particles are "bare" with


> > infinite parameters? When we obtain corrections to the initial
> > solutions. It is these corrections that are infinite, not the original
> > physical parameters.
>
> As I said many times, we cannot observe fictitious bare particles. They
> simply don't exist. We prepare asymptotically free particles (to be
> distinguished from bare particles). These are only asymptotically free
> but interact at some point with other particles!

This is called an adiabatic hypothesis. It simply means we work with a
wrong modelling. Better modelling may preserve interaction even in the
asymptotic states.


On 27 mar, 03:32, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
wrote:
...


>
> What is special in relativistic QFT though is the fact that these
> corrections must be infinitely large, which comes from the fact that
> real particles are very different in their properties from free
> particles. E.g., a free electron has no associated electromagnetic
> field although it is charged!

So let us make an electron and its quantized electromagnetic filed
come together by construction. Who does prevent us from a better
construction?

> The parameters arising in the action (and these are the bare mass, etc.)
> _must_ tend to infinity as the cutoff is removed, in order that the
> final results are finite. There is no way to get a finite theory from
> the QED action with finite bare masses.

I agree! Let us construct a better QED action then. Why do we stick to
the old QED?

Regards,

Vladimir.

Arnold Neumaier

unread,
Mar 28, 2010, 7:59:01 AM3/28/10
to
Bob_for_short wrote:
> On 27 mar, 00:09, Hendrik van Hees <Hendrik.vanH...@physik.uni-
> giessen.de> wrote:
>
>> We cannot even start to solve the equations of interacting particles
>> since so far nobody has come up with a clever idea how to do so.
>
> That's a correct answer. That is why we practice the ritual of
> renormalizations.
>
>> ... the bare particles and bare fields. In
>>>> reality, these objects don't exist.
>
> Why then do they exist in our theory?

Maybe the new thread ''Harmonic oscillators and counterterms''
that should appear soon on s.p.r. hepls to explain this.

They exist in the textbook theory since our textbooks present field
theory in terms of perturbation theory of free fields. The altter
are auxiliary fields with bare particles used to approximate the
real fields and real particles.

Because the free field approximation is infinitely bad (e.g., bare
electrons have no electromagnetic field), one needs infinitely
large corrections. Starting with better initial theories would
remove this blemish, but nobody so far has found a way to recast
QED in a form that exactly reproduces its results without using
renormalization theory of some sort.

However, at least it has been shown in a nice book on QED by Scharf
that it is possible to avoid all bare stuff. Then no infinities
appear anywhere (except most likely in summing the perturbation series),
and the only masses, charges, etc. are those of the real particles.

No virtual particles anymore!


> On 27 mar, 03:32, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
> wrote:
> ...
>> What is special in relativistic QFT though is the fact that these
>> corrections must be infinitely large, which comes from the fact that
>> real particles are very different in their properties from free
>> particles. E.g., a free electron has no associated electromagnetic
>> field although it is charged!
>
> So let us make an electron and its quantized electromagnetic filed
> come together by construction. Who does prevent us from a better
> construction?

Go ahead and do it if you think it can be done. What are the equations
of this dressed electrons? What are its field operators and its N-point
expectations? How does one do perturbation theory around it to get
the solutions of the equations defined by standard QED?


>> The parameters arising in the action (and these are the bare mass, etc.)
>> _must_ tend to infinity as the cutoff is removed, in order that the
>> final results are finite. There is no way to get a finite theory from
>> the QED action with finite bare masses.
>
> I agree! Let us construct a better QED action then. Why do we stick to
> the old QED?

We cannot change QED, since it is by far the most accurate theory we
have. Therefore we cannot change the action.

We can only change the methods used to find approximate solutions.


Arnold Neumaier

Arnold Neumaier

unread,
Mar 28, 2010, 9:33:14 AM3/28/10
to
FrediFizzx wrote:
[in: Virtual particles help]

> Quantum theory and the standard model of particle physics says there has
> to be quantum objects involved. What are they?

The standard model is a quantum field theory. These theories are called
so after the basic objects in them: Quantum fields.

In quantum field theory, particles are not the basic objects,
but are derived features inherent in quantum fields:
Particles are the localized excitations of quantum fields with
well-define quantum numbers.

In standard quantum field theory, particles are asymptotic objects,
existing only before and after collisions. Then they have definite
properties, including a mass -- one says they are on-shell.
Decaying excitations can be treated in some approximation as such
asymptotic objects if they are long-living enough; these are the
unstable particles. They are slightly off-shell, since their mass is
uncertain by an amount inversely proportional to their lifetime.
If they do not live long, they still manifest themselves as resonances.


During collisions, there is the quantum field, but there are no
discernible particles. Depending on the ways the quantum field is
analyzed, one may ascribe parts of the fields to certain particles.
In schemes that do so, these particles are thought to be free
(although they cannot be, which results in renormalization problems);
they are called virtual particles, and correspond to the internal
lines of corresponding Feynman diagrams. They are off-shell, and
different schemes for analyzing the quantum field during a collision
assign different portions of the field to different and differently
interacting virtual particles. Nonperturbative schemes such as lattice
gauge theory do not permit such an analysis in terms of virtual
particles. Thus the presence and meaning of virtual particles is
scheme-dependent, and one cannot ascribe any objective reality to them.

In a scattering experiment that does not change the number and type of
particles, the in-going particles become virtual (and off-shell) in
perturbative schemes until the interaction is completed, when they
are recognizable again as a real, on-shell particle.


In a dense medium, collisions are so frequent that the asymptotic regime
needed for the preceding interpretations to make sense, However, using
nonperturbative techniques such as the closed-time path integral, one
can define quasiparticles with position-dependent masses (on-shell in
Boltzmann-type equations, off-shell in Kadanoff-Baym-type equations)
that satisfy a quantum kinetic equation with measurable consequences.
These quasiparticles are real and measurable.

Indeed, from a fundamental point of view, most particles (all apart
from leptons, quarks and gluons), and in particular the proton and
the neutron, must be considered as quasiparticles.


Arnold Neumaier

Bob_for_short

unread,
Mar 28, 2010, 3:58:21 PM3/28/10
to
On Mar 27, 4:32 am, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
wrote:

> Bob_for_short wrote
> [in: ``Virtual particles help'']:
>
> > On 24 mar, 10:51, Hendrik van Hees <Hendrik.vanH...@physik.uni-
> > giessen.de> wrote:
>
> >> ...The particle simply gets a
> >> self energy with a diverging mass and a diverging renormalization
> >> factor of the corresponding single-particle state.
>
> > Indeed, it is simple. The question is why electron and photon get
> > corrections to their masses? I think it is because the interaction
> > term is of kinetic rather than potential character (see my answers at
> >http://groups.google.com/group/sci.physics.foundations/browse_thread/...).

>
> The reason is much simpler:
>
> Interactions _generally_ change the ground state energies of a system.
> In the relativistic case, the ground state energies at the lowest
> definite other quantum numbers define the particle masses.
> No surprise therefore that these are different from their bare values
> in the free theory.

Normally correction to the bound state energy are not considered as
corrections to the mass, even in the relativistic theory.

When I was writing about kinetic perturbation term, I meant non-
relativistic case. Namely, if the non-perturbed Hamiltonian is written
as p^2/2m + V(r) and the perturbation (additional interaction to take
into account) is something like c*p^2, it is clear that it brings
correction(s) to the particle mass m and nothing else. If we use the
observable mass value m in the non-perturbed Hamiltonian, then the
perturbation may only worsen agreement with experiment, especially if
it is large or infinite. The only "reasonable" way to cope with such "
unnecessary corrections" is to discard them. That is why
"renormalizations" may work in my opinion.

> Maybe the new thread ''Harmonic oscillators and counterterms''

> that should appear soon on s.p.r. helps to explain this.
> ...an anharmonic oscillator has no simple distinguished
> expansion frequency, Indeed, approximations made by truncating the
> expansion (and in perturbative field theory this sort of thing _must_
> be done) depend on the choice of omega, and the best approximation
> is usually not obtained with the omega that comes from the expansion of
> the Hamiltonian, but a different one. (Google ''variational perturbation
> theory'' for details.) ....

I did not get the picture. We can expand the exact spectrum in series
of small parameter (anharmonicity). We can truncate the series
somewhere. It will be different from the exact value, of course, no
problem with it. Why should we change here the oscillator frequency in
the truncated series - to fit the exact (observable) eigenvalue? Then
it is not a theory of perturbation but the use of truncated series as
a fitting curve with some free parameters. there is no physics in such
an approach. In particular, it does not make the original oscillator
to be "bare".

Regards,

Vladimir.


FrediFizzx

unread,
Mar 30, 2010, 12:22:32 PM3/30/10
to
My first attempt at responding to this was rejected for repetition and a
personal remark. I thank the moderator for catching the somewhat
personal remark as that was an oversight on my end. However, the
moderator's other comment,

"A major point that has been made in this discussion is that "virtual
particles" are NOT necessary for "fundamental explanations of static
electric and magnetic fields". You cannot simply keep asserting the
opposite without new supporting
evidence."

is completely backwards in my opinion. I am following the Standard
Model of particle physics for the *fundamental* explanation of static
electric and magnetic fields. I don't need to present any more
evidence than that because there is no other evidence that I could
present. It is the people that are making statements contrary to the
standard model that need to provide new evidence or a new theory. That
new theory must also explain electroweak interactions mediated by
*virtual* W and Z bosons. Nobel prizes were given out for the current
explanation within the standard model since real W and Z bosons were
discovered. So since there is no way to "get rid of" the virtual W and
Z bosons and still have a consistent theory of elementary particle
interactions, it is compelling evidence for the existence of other
virtual particles allowed by the Uncertainty Principle.

"Bob_for_short" <vladimir.k...@wanadoo.fr> wrote in message
news:304341b9-64c2-4c60...@g10g2000yqh.googlegroups.com...


> On 23 mar, 16:53, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>>
>> >No. The Coulomb field has nothig to do with photons. It arises as an
>> >interaction term in the QED Hamiltonian in the
>> >Coulomb gauge.

The above statement is clearly contrary to the standard model of
particle physics.

>> It arises from the interchange of photons, which is allowed from the
>> interaction term in the QED Hamiltonian (which creates or annihilates
>> photons) irrespective of gauge (owing to gauge invariance).

This statement above is not contrary to the standard model.

> No, it is not an interchange of photons. One charge is a source of a
> strong quasi-stationary electromagnetic filed for another charge, and
> the other charge is a source for the first one. Perturbative treatment
> of this interaction term (it depends only on the charge relative
> distance and charge velocities) does not make a virtual photon. As I
> said, this term is an explicit solution expressed via particle
> variables solely so no intermediary agent is necessary.

Of course the above statement is also contrary to the standard model
since the electric field of an elementary charge has to be modeled
within the standard model as photons (when I say "photons" here, that
includes both real and virtual). But it does bring up the interesting
aspect, that the electric field of say an electron *is* always attached
to that particle. However, we can immediately see that the electric
field is not going to be an invariant as different observers could see
it differently. But for sure, there are some interactions that you can
just use the Coulomb field as a term.

> Note, a free photon cannot be entirely absorbed by a free electron but
> scattered. The "virtual" photon, on the other hand cannot be scattered
> but only "absorbed". In other words, it does not propagate. It is an
> intrinsic property of charge intercation and nothing else. We need not
> virtual photons to describe and understant bound states and charge
> scattering. let us not fool ourselves.

Technically, photons are never scattered nor absorbed even though that
language is used quite a bit. They can only be created and destroyed
and it is energy-momentum that is transferred to and from. Anyways, why
would it be a problem that they do not propagate if they are virtual
(off-mass-shell)? It is very true that we don't need virtual photons to
*calculate* certain interactions but we do need them for the *ultimate*
fundamental explanation. The coulomb field of elementary charged
particles is not invariant like their rest mass, charge, and spin are.
The implications of that are obvious when trying to formulate a
consistent complete theory of elementary particle interactions.

Best,

Fred Diether

Arnold Neumaier

unread,
Mar 30, 2010, 12:22:50 PM3/30/10
to
FrediFizzx wrote:
> Are we going to toss out the
> Uncertainty Principle (UP) that allows virtual particles to exist?

The uncertainty pinciple does not prove anything about virtual
particles. It also allows all sorts of other ghosts, provided they
have conjugate variables and are fuzzy enough, though nobody would
bet on their existence because of that.


> Gordon Kane, a very prominent particle physicist, says they are real.

... without sufficient reasons, as I had pointed out.


> Frank Wilczek also thinks they are real; he is a closet modern etherist
> so I know he believes they are real [1].

In his Nobel lecture, he uses the term with sound judgment, and uses the
qualification ''loosely speaking'', which speaks for itself. I haven't
read his book. But I wouldn't take loose talk in books intended for a
very wide audience including non-specialists for real science!

Please quote the context that proves that he takes virtual particles
as real.


> My former particle physics
> tutor, Dr. Andy Inopin - a hadronic particle physics theorist, also
> thinks they are real. And that they are just like real particles but
> simply "off-mass-shell" that is allowed by the UP.

So apparently he believes in off-shell particles and calls them virtual.
But this renaming does not do justice to the concept as generally used.
He uses the term in a nonstandard way.


> There are many aspects of particle theory that can't be *properly*
> explained without the concept of virtual particles such as muon, pion
> and neutron decay. You have to have an "off-mass-shell" W boson to
> explain those.

Virtual particles are by definition the particles in internal lines of
Feynman diagrams. At least this is how they are introduced in QFT books,
and this is enough to do all calculations that are done with them.

The more esoteric ''definition'' in Wikipedia is irrelevant to
the practice of QFT. It doesn't make the slightest difference to
the perturbative S-matrix calculations whether virtual particles
''exists for a limited time and space, introducing uncertainty in their
energy and momentum due to the uncertainty principle'' or that
''the consequences of its existence are prolonged to such a degree that
it cannot be virtual''.

In fact, it makes no difference whether or not they exist at all!
The assumptions that go into the standard perturbative formalism
of QFT are quantum fields and their multipoint expectations, not
virtual particles. So quantum fields must be assumed to exist,
and their expectations be observable in priniple. All the other
stuff is just auxiliary technicality.


Off-shell is not equivalent to virtual. Vitual particles also
happen to be off-shell, but this is not their defining property.

Real, stable particles in a medium are usually off-shell because
the interqction with the medium changes the official mass.

Resonances (unstable particles) are also off-shell, but unlike
virtual particles they exist in a good approximation as asymptotic
states and can be measured, with definite, repeatable properties.
And the real W-boson is an example of such a resonance.


To explain muon, pion, and neutron decay, you need a weak force field.
And field operators corresponding to its field strength. And an S-matrix
computed from these. No virtual particles.

The calculation of weak field properties with lattice gauge theory is
completely independent of anything that smells like a virtual particle.
You cannot find a place for them, even if you try.

Virtual W-Bosons are only needed in a particular, perturbative approach
that generates corresponding Feynman graphs. Moreover, their properties
depend on the perturbation scheme -- Feynman graphs in the traditional
form and in a light front scheme have nothing in common except that
they are graphs.

Only those who champion the perturbative approach to QFT and don't
take other approaches seriously can speak of virtual particles as real.


Are you going to go back to Fermi's old theory to
> explain those? There are many more examples of processes with virtual
> particles that produce real detectable phenomena. And then there is the
> simple coulomb field around a charged body. Is it just there by magic?

Yes. Just like the magic that produces all the terms in the standard
model. But -- in contrast to the standard model -- the magic involved
in the Coulomb field is known for more than 100 yeears!

It is there because the Hamiltonian for the QED action in Coloumb gauge
(appropriate for low energies) has a Coulomb potential.
No virtual particles are involved in the transformation from the action
to the Hamiltonian, but the Coulomb term is there.


Quantum field theory is magical (with many astonishing surprises
revealed and probably to be revealed).

But it is not magical enough to create really existing ghosts called
virtual particles.


Arnold Neumaier

Bob_for_short

unread,
Mar 30, 2010, 12:39:16 PM3/30/10
to
On 28 mar, 13:59, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
wrote:

> > So let us make an electron and its quantized electromagnetic filed
> > come together by construction. Who does prevent us from a better
> > construction?
>
> Go ahead and do it if you think it can be done. What are the equations
> of this dressed electrons? What are its field operators and its N-point
> expectations? How does one do perturbation theory around it to get
> the solutions of the equations defined by standard QED?
>

I presented my physical and technical constructions in my papers:
http://arxiv.org/abs/0806.2635 and http://arxiv.org/abs/0811.4416. I
admit they are not perfect but I think they are reasonable. It is just
difficult to accept or to get used to the dressed electron solution at
first. We all think of a separated, free, and point-like electron
althoutgh we recongize it is in interaction with the quantized EMF.
What I propose is a solution of coupled charge and its quantized EMF.
The key idea can be formulated now in a simple way: as soon as the
electron acceleration excites the field oscillators, then the electron
is just a part of them and they all represent a compound system. We
know well how to describe compound systems: with its center of inertia
and relative motion variables. Advancing such a construction for the
dressed electron ("electronium") is the answer to your first question.
In other words, I propose another, "potential" coupling. The electron
excites the oscillators because it is a part of them. The electron in
electronium is a point of the external force application. Doing so, we
excite the relative (internal) motion of a compound system in a
natural way. Such a decription is as natural as description of an atom
(as a compound system). I think this idea merits studying before
getting rejected.

What I wanted to say is that with the old QED we are bound to do
renormalizations and in order to get out of this situation we have to
advance some better physical solutions, as we feel them, and then
develop the corresponding math. I did what I felt reasonable and what
was based on my own experience. It is still an open question for
researchers and everybody is wellcome to participate.

Regards,

Vladimir.

Arnold Neumaier

unread,
Mar 30, 2010, 7:41:12 PM3/30/10
to
FrediFizzx wrote:

>>>> No. The Coulomb field has nothing to do with photons. It arises as an


>>>> interaction term in the QED Hamiltonian in the
>>>> Coulomb gauge.
>
> The above statement is clearly contrary to the standard model of
> particle physics.

Please prove this claim. A mere assertion is not enough.

To all who know the subject, the opposite is clearly the case.
The Coulomb field is one of the simplest conclusion one can draw
from the QED action (which is part of the standard model action
after it is written in the right form), and one can do so without
any of the machinery needed for perturbation theory (whic hdefines
virtual particles), and even without the introduction of photons
(excited states). This can be seen from many textbooks, for example
from that by Bjorcken and Drell.


Arnold Neumaier

Arnold Neumaier

unread,
Mar 30, 2010, 7:41:13 PM3/30/10
to
Bob_for_short wrote:
> On 28 mar, 13:59, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
> wrote:
>
>>> So let us make an electron and its quantized electromagnetic filed
>>> come together by construction. Who does prevent us from a better
>>> construction?
>> Go ahead and do it if you think it can be done. What are the equations
>> of this dressed electrons? What are its field operators and its N-point
>> expectations? How does one do perturbation theory around it to get
>> the solutions of the equations defined by standard QED?
>
> I presented my physical and technical constructions in my papers:
> http://arxiv.org/abs/0806.2635 and http://arxiv.org/abs/0811.4416. I
> admit they are not perfect but I think they are reasonable.

They are completely unrelated to QED, apart from lip service paid.

You do not show that your solutions have anything to do with the action
defining QED. You just invent a different Hamiltonian and declare that
this should be used in place of QED since you can solve it approximately
without renormalization.

Any Hamiltonian that is not too symmetric will produce a hyperfine
splitting and hence a Lamb shift. But only the right one (namely that
coming from QED and a good renormalization procedure) gives the correct
Lamb shift to a high order in the fine structure constant. So getting a
Lamb shift is no proof of having an alternative to QED.


Things are simply not interesting as long as you speculate about QED
but do not connect your alleged solutions with the standard definition
of QED and show that you reproduce the relevant effects to the high
accuracy that is the state of the art.


> I think this idea merits studying before
> getting rejected.

So please study it to the point where you can show that it is relevant
to true QED. Then others will become interested.

> What I wanted to say is that with the old QED we are bound to do
> renormalizations

With the old QED we get excellent agreement with experiment.

With your bogus replacement we don't know what we get.
You ask us to buy a pig in a poke and throw away the excellent tools
that we have, simply because they are not to your taste.

> in order to get out of this situation we have to
> advance some better physical solutions, as we feel them, and then
> develop the corresponding math. I did what I felt reasonable and what
> was based on my own experience.

Unfortunately, what you feel is reasonable is not shared by many.

If you wnat the recognition of the scientific community for your claims
you must support your claims in a way that meets their standards.
You can't get it cheaper.


Arnold Neumaier

Oh No

unread,
Mar 31, 2010, 2:06:55 AM3/31/10
to
Thus spake Arnold Neumaier <Arnold....@univie.ac.at>

I have read Bjorken and Drell, and I do not recognise your reading of
it. Irrespective of perturbation theory, photons are created and
annihilated by the interaction term in the QED hamiltonian. They are
there whether you mention them by name or not. I find it very peculiar
to describe photons as excited states. Certainly it is straightforward
enough to derive the coulomb field but since this is essentially found
from an expectation of photon states and has nothing whatsoever to do
with perturbation theory, what you say makes very little sense to me.

FrediFizzx

unread,
Mar 31, 2010, 4:27:42 AM3/31/10
to
"Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
news:4BB25FA2...@univie.ac.at...

> FrediFizzx wrote:
>
>>>>> No. The Coulomb field has nothing to do with photons. It arises as
>>>>> an
>>>>> interaction term in the QED Hamiltonian in the
>>>>> Coulomb gauge.
>>
>> The above statement is clearly contrary to the standard model of
>> particle physics.
>
> Please prove this claim. A mere assertion is not enough.

It looks like my "proof" (Coulomb fields are not invariant) got snipped
out here but OK..., let's try something different and use some of what
you posted in another message about this;

"The standard model is a quantum field theory. These theories are called
so after the basic objects in them: Quantum fields.

In quantum field theory, particles are not the basic objects,
but are derived features inherent in quantum fields:
Particles are the localized excitations of quantum fields with
well-define quantum numbers."

A Coulomb field is a classical field. I seems to me that you are trying
to assert that it is a quantum field directly or something like that.
You are the one that must prove your claim that photons are not
involved. I am pretty sure the standard model asserts that classical
fields are constructed from "localized excitations of quantum fields
with well-defined quantum numbers". Now, the underlying quantum fields
are not observable, but we must take them to be real as they produce
real matter and real classical fields. So what is the problem with
having excitations that are not observable of this unobservable
underlying quantum field that we must assume is real? They have all the
same quantum numbers as the observable excitations the only difference
being they don't strictly obey the Einstein relation due to the
Uncertainty Principle.

> To all who know the subject, the opposite is clearly the case.
> The Coulomb field is one of the simplest conclusion one can draw
> from the QED action (which is part of the standard model action
> after it is written in the right form), and one can do so without
> any of the machinery needed for perturbation theory (whic hdefines
> virtual particles), and even without the introduction of photons
> (excited states). This can be seen from many textbooks, for example
> from that by Bjorcken and Drell.

Please see Weinberg "The Quantum Theory of Fields, Vol I", Section 8.5
"The Photon Propagator" where he shows how the Coulomb interaction
drops out in the case of an internal photon line. So you can work that
all backwards and recreate the Coulomb interaction field showing how it
is composed of the more fundamental excitations of the quantum field,
photons in this case. I think I will believe what Weinberg has to say
about this. Sorry, your reference is pretty out-dated and I am not
going to bother trying to get one. A lot has happened in particle
physics since Bjorken and Drell.

[ Mod. note: An implicit flaw in the above reasoning is the assumptions
that only one of Weinberg or Bjorken & Drell can be right. In fact, they
are both right. And, for reference, the Coulomb potential makes an
explicit appearance in the last equation of section 89 of B&Dv2. Since
the Coulomb potential appears both in covariant perturbation theory and
in canonical quantization in Coulomb gauge, it cannot be "fundamentally"
linked to calculational artifacts of either approach. -ik ]

Best,

Fred Diether

Arnold Neumaier

unread,
Mar 31, 2010, 6:56:37 AM3/31/10
to
============= Moderator's note ============================================

I disagree with the notion of mass used below. The modern use of this word
is that mass is to be taken as the invariant mass of the particle which is
(in relativistic physics) a Casimir operator of the Poincare group. It's a
scalar quantity, while the energy is the time component of the energy-momentum
four vector. Of course the energy of a particle changes (because its momentum
changes and particles have E=sqrt(m^2+p^2), while its mass stays the same:
E^2-p^2=m^2 is an invariant!

HvH.
===========================================================================

Bob_for_short wrote:
> On Mar 27, 4:32 am, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
> wrote:
>> Bob_for_short wrote
>> [in: ``Virtual particles help'']:
>>
>>> On 24 mar, 10:51, Hendrik van Hees <Hendrik.vanH...@physik.uni-
>>> giessen.de> wrote:
>>>> ...The particle simply gets a
>>>> self energy with a diverging mass and a diverging renormalization
>>>> factor of the corresponding single-particle state.
>>> Indeed, it is simple. The question is why electron and photon get
>>> corrections to their masses? I think it is because the interaction
>>> term is of kinetic rather than potential character (see my answers at
>>> http://groups.google.com/group/sci.physics.foundations/browse_thread/...).
>> The reason is much simpler:
>>
>> Interactions _generally_ change the ground state energies of a system.
>> In the relativistic case, the ground state energies at the lowest
>> definite other quantum numbers define the particle masses.
>> No surprise therefore that these are different from their bare values
>> in the free theory.
>
> Normally correction to the bound state energy are not considered as
> corrections to the mass, even in the relativistic theory.

This is not correct. In relativistic treatments, E=mc^2, so one cannot
change the energy unless one changes the mass. In fact, nuclear physics
talks of the mass spectrum which replaces the energy spectrum of
nonrelativistic quantum mechanics.


> When I was writing about kinetic perturbation term, I meant non-
> relativistic case.

But you talk about QED which is relativistic, so that your
nonrelativiastic results are irrelevant. There are no renormalization
difficulties in the usual nonrelativistic approximations to QED.


>> Maybe the new thread ''Harmonic oscillators and counterterms''
>> that should appear soon on s.p.r. helps to explain this.
>> ...an anharmonic oscillator has no simple distinguished
>> expansion frequency, Indeed, approximations made by truncating the
>> expansion (and in perturbative field theory this sort of thing _must_
>> be done) depend on the choice of omega, and the best approximation
>> is usually not obtained with the omega that comes from the expansion of
>> the Hamiltonian, but a different one. (Google ''variational perturbation
>> theory'' for details.) ....
>
> I did not get the picture. We can expand the exact spectrum in series
> of small parameter (anharmonicity). We can truncate the series
> somewhere. It will be different from the exact value, of course, no
> problem with it. Why should we change here the oscillator frequency in
> the truncated series - to fit the exact (observable) eigenvalue?

Since any perturbation series must be truncated, the truncations at
different bare (i.e., harmonic) frequencies give different results.
One wants to choose the frequency such that few terms of the
perturbation series suffice to get a meaningful accuracy.

In nonrelativistic problems, this is done using variational perturbation
theory, which finds an optimal frequency in a self-consistent manner.
In relativistic problems, the optimial frequency is ininite, and one
needs the usual renormalization techniques to get finite results.

But the procedure can be motivated by considering the approximate theory
with a large cutoff mass M. In this case, the optimal frequency
omega(M.m) depends on the cutoff M and the coefficient m of phi^2/2
in the action (usuall called the bare mass, but this is just a name).
The results (typically S-matrix elements accurate to some power of
the fine structure constant alpha) are finite and also depend on M
and m, Now one wants to get a finite limit as M-->inf, in order to
restore the Poincare covariance that was broken by the cutoff.

It turns out that in order to get finite results in this limit, one
also needs to take m as a function of M, m=m(M), in a way that m
diverges as M-->inf, but that the results are finite and Poincare
covariant.

The latter is the essential point. If you want to replace QED by
something else, nobody will take you seriously unless you can produce
results that are finte and Poincare covariant. In this direction you
have currently nothing to offer but empty words. This is why your owrk
is disregarded. Your laims do not match your results and make your work
incredible.


Arnold Neumaier

Igor Khavkine

unread,
Mar 31, 2010, 5:04:39 PM3/31/10
to
On Mar 31, 8:06 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Thus spake Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
[...]

> >To all who know the subject, the opposite is clearly the case.
> >The Coulomb field is one of the simplest conclusion one can draw
> >from the QED action (which is part of the standard model action
> >after it is written in the right form), and one can do so without
> >any of the machinery needed for perturbation theory (whic hdefines
> >virtual particles), and even without the introduction of photons
> >(excited states). This can be seen from many textbooks, for example
> >from that by Bjorcken and Drell.
>
> I have read Bjorken and Drell, and I do not recognise your reading of
> it. [...]

Please take another look. The Coulomb field, in the way that Arnold is
referring to, makes an explicit appearance in the last equation of
section 89 of Bjorken & Drell's volume 2.

Igor

Igor Khavkine

unread,
Apr 1, 2010, 4:15:08 AM4/1/10
to
On Mar 30, 6:22�pm, "FrediFizzx" <fredifi...@hotmail.com> wrote:
> My first attempt at responding to this was rejected for repetition and a
> personal remark. �I thank the moderator for catching the somewhat
> personal remark as that was an oversight on my end. �However, the
> moderator's other comment,
>
> "A major point that has been made in this discussion is that "virtual
> particles" are NOT necessary for "fundamental explanations of static
> electric and magnetic fields". �You cannot simply keep asserting the
> opposite without new supporting
> evidence."
>
> is completely backwards in my opinion. �I am following the Standard
> Model of particle physics for the *fundamental* explanation of static
> electric �and magnetic fields.

And yet you appear to do so once more. :-(

> �I don't need to present any more


> evidence than that because there is no other evidence that I could
> present. �It is the people that are making statements contrary to the
> standard model that need to provide new evidence or a new theory. �That
> new theory must also explain electroweak interactions mediated by
> *virtual* W and Z bosons. �Nobel prizes were given out for the current
> explanation within the standard model since real W and Z bosons were
> discovered. �So since there is no way to "get rid of" the virtual W and
> Z bosons and still have a consistent theory of elementary particle
> interactions, it is compelling evidence for the existence of other
> virtual particles allowed by the Uncertainty Principle.

Fred, you are confusing "the Standard Model" with "perturbative
calculations in the Standard Model". You are also confusing sufficient
with necessary. And that is were the crux of the argument lies.

No-one is arguing with the *sufficiency* of perturbative QFT
calculations (which is the only context in which virtual particles
make an appearance, as a calculation artifact) to reproduce the many
successes of the Standard Model, which earned numerous Nobel prizes
for its developers.

However, sufficiency does not imply *necessity*. You, on the other
hand keep insisting on the necessity of perturbative QFT. That is the
statement that you must demonstrate, if you insist on it. If you
assume such a position, you also assume the burden of proof that comes
with it.

On the other hand, Arnold has pointed out several other calculation
schemes that extract physical results from the Standard Model (same
model, different way of doing approximate calculations), which have
their own calculation artifacts that do not include virtual particles.

Let me summarize and repeat once more:

* QFT is not the same as perturbative calculations in QFT

* sufficient is not the same as necessary

Igor

Arnold Neumaier

unread,
Apr 1, 2010, 5:42:22 AM4/1/10
to
FrediFizzx wrote:
> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
> news:4BB25FA2...@univie.ac.at...
>> FrediFizzx wrote:
>>
>>>>>> No. The Coulomb field has nothing to do with photons. It arises as
>>>>>> an
>>>>>> interaction term in the QED Hamiltonian in the
>>>>>> Coulomb gauge.
>>>
>>> The above statement is clearly contrary to the standard model of
>>> particle physics.
>>
>> Please prove this claim. A mere assertion is not enough.
>
> It looks like my "proof" (Coulomb fields are not invariant) got snipped
> out here

It is enough that the Hamiltonian with the Coulomb terms in it is
the time translation generator of a representation of the Pojncare
group, and this is proved in many books on quantum field theory.

Since you object to old textbooks like those of Bjorcken and Drell
(without valid reasons, since most of what they say is still
considered valid today), let me cite instead Weinberg, whom you accept
as authoritative. He derives the Coulomb interaction (83.11-12)
at the top of page p.350 of his quantum field theory book, Volume I,
directly from the action and before he does anything else with the QED
action. This shows the fundamental character of the Coulomb force,
completely independent of any particle interpretation of QED.

The particle picture of QED is introduced only in the next section via
the interaction picture, where he defines in (8.4.15-21) bare photon
creation and annihilation operators. And as you can see in (8.4.24-25),
the Coulomb interacten is still present explicitly in this particle
formulation of QED. Not the slightest hint is given that there should
anywhere be virtual particles involved. They are completely irrelevant
for the Coulomb interaction.

But of course, Feynman diagrams in covariant perturbation theory where
no explicit Coulomb interaction term is present must still reproduce the
effect of the Coulomb interaction, which Weinberg shows in Section 8.5,
again without using any virtual photon language. But of course, one can
reinterpret his argument there as the tree contribution to electron
scattering, in which a virtual photon appears in the Feynman diagram.

Thus the relation between the Coulomb interaction and virtual photons
is quite indirect. Weinberg makes no use of it; he never claims that the
internal photon line represents an existing virtual photon.

Therefore your claimed contradiction does not exist.


> but OK..., let's try something different and use some of what
> you posted in another message about this;
>
> "The standard model is a quantum field theory. These theories are called
> so after the basic objects in them: Quantum fields.
>
> In quantum field theory, particles are not the basic objects,
> but are derived features inherent in quantum fields:
> Particles are the localized excitations of quantum fields with
> well-define quantum numbers."
>
> A Coulomb field is a classical field.

The Coulomb interaction is an interaction term in the Hamiltonian of
QED, written in the instant form. There is nothing intrinsically
classical about it except for the name. The whole QED Hamiltonian
can be interpreted classically (using anticommuting fermion variables)
or in a quantum fashion.


> I seems to me that you are trying
> to assert that it is a quantum field directly or something like that.

> You are the one that must prove your claim that photons are not
> involved.

One can write down the QED Hamiltonian and prove that it gives a
representation of the Poincare group and hence a satisfying
relativistic quantum field theory. This Hamiltonian has clearly
separated the effects of the Coulomb field and those of the photon
creation and annihilation terms. They have nothing to do with each
other, as one could drop either term and get a meaningful
noncovariant approximation of QED, though of course not matching
reality. But this is enough to see that there is no contradiction
and that the concepts are independent.

They are related only by the fact that both parts of the Hamiltonian
are needed for a covariant theory, and this is the content of
Weinberg's demonstration that you cite.


>> To all who know the subject, the opposite is clearly the case.
>> The Coulomb field is one of the simplest conclusion one can draw
>> from the QED action (which is part of the standard model action
>> after it is written in the right form), and one can do so without
>> any of the machinery needed for perturbation theory (whic hdefines
>> virtual particles), and even without the introduction of photons
>> (excited states). This can be seen from many textbooks, for example
>> from that by Bjorcken and Drell.
>
> Please see Weinberg "The Quantum Theory of Fields, Vol I", Section 8.5
> "The Photon Propagator" where he shows how the Coulomb interaction
> drops out in the case of an internal photon line.

He says there that the term _cancels_ the Coulomb term that was already
present in the Hamiltonian before any calculations were done.
This is quite different from your claim that the virtual particles
are the origin and explanation of the Coulomb force.


> So you can work that
> all backwards and recreate the Coulomb interaction field showing how it
> is composed of the more fundamental excitations of the quantum field,
> photons in this case. I think I will believe what Weinberg has to say
> about this. Sorry, your reference is pretty out-dated and I am not
> going to bother trying to get one. A lot has happened in particle
> physics since Bjorken and Drell.

There is a big difference between contradicting (that you claimed in
your previous mail but could not substantiate) and being unfashionable
(what you retreat to now with this argument).

But you are mistaken in thinking that the Bjorcken-Drell approach is
obsolete. Although less fashionable than covariant quantization, the
Hamiltonian approach in Coulomb gauge is still alive and healthy in
the standard model literature, and is being used to study,
e.g., confinement. See http://arxiv.org/pdf/0710.0316.
(2.2) and (2.3) give in case of an abelian gauge group the Coulomb
Hamiltonian directly from the action of the quantum fields, without any
particles in the derivation.

It is also very useful in the study of bound states in a QED framework;
see
M. Piatek, V.N. Pervushin, A.B. Arbuzov
Generating Functional for Bound States in QED
Fizika B17:189-196,2008
http://arxiv.org/pdf/0810.5199
or mathematically rigorous approaches to QED, see, e.g.,
http://arxiv.org/pdf/0706.1486


Arnold Neumaier

Arnold Neumaier

unread,
Apr 1, 2010, 6:16:39 AM4/1/10
to
Arnold Neumaier wrote:
> ============= Moderator's note ============================================
>
> I disagree with the notion of mass used below. The modern use of this word
> is that mass is to be taken as the invariant mass of the particle which is
> (in relativistic physics) a Casimir operator of the Poincare group. It's a
> scalar quantity, while the energy is the time component of the energy-momentum
> four vector. Of course the energy of a particle changes (because its momentum
> changes and particles have E=sqrt(m^2+p^2), while its mass stays the same:
> E^2-p^2=m^2 is an invariant!
>
> HvH.
> ===========================================================================


Of course.

But I was implicitly assuming that calculations are done in the rest
frame, where the center of mass motion is separated from the system.
Then p=0, and E=m (in units where c=1).


Arnold Neumaier

Bob_for_short

unread,
Apr 1, 2010, 8:26:55 AM4/1/10
to
> FrediFizzx wrote:

> A Coulomb field is a classical field.

The Coulomb potential dependence 1/|r2 - r1| is simple but in QM and
QED it is not a "classical" any more. The force acts between waves,
not particles. This leads to bound rather than collapsing states.
Moreover, each charge is in permanent "interaction" (or "coupling")
with the quantized EMF. This makes the charge smear quantum
mechanically. So real (or dressed) charges are described with elastic
and inelastic form-factors. And only inclusive picture "corresponds"
to a simple 1/r law.

> Please see Weinberg "The Quantum Theory of Fields, Vol I", Section 8.5
> "The Photon Propagator" where he shows how the Coulomb interaction

> drops out in the case of an internal photon line. So you can work that


> all backwards and recreate the Coulomb interaction field showing how it
> is composed of the more fundamental excitations of the quantum field,
> photons in this case.

In the Coulomb (or radiation) gauge the Hamiltonian contains only
physical degrees of freedom - those that carry real energy-momentum.
In this gauge one real charge interacts with another real charge via
Coulomb term and with a real photon via jA_rad term. The photon
propagator in this gauge is not relativistically invariant.
Relativistically invariant should be S-matrix, which is, of course,
the case. "Compensation" of the Coulomb terms from the propagator and
from the Hamiltonian "happens" only in the first order of perturbation
theory. In higher orders there are some other corrections to the
photon propagator. So you cannot say that the Coulomb term disappears
in the exact formulation. Moreover, the "compensation can be cast in
the another way: the "relativistic part" of the photon propagator is
not much different from the Coulomb interaction term. It looks more
relativistic but in the non-relativistic case it reduced to the
Coulomb one so the photon propagator contribution effectively
disappears and the potential scattering occurs due to the Coulomb
interaction coming from the Hamiltonian. And in the relativistic (or
exact) case we need anyway to take into account radiative corrections
too so we are obliged to work with an inclusive picture. This picture
is different from a purely potential, elastic one.

And one more remark. Some think that 1/r "surrounds" a charge. It is a
mistake. 1/r is an interaction term for at least two charges q2*q1/|r2
- r1|. When we speak of the Coulomb potential we mean exactly the
latter case and nothing else. In other words, this term is always
involved in two equations.

Regards,

Vladimir.

Tom Roberts

unread,
Apr 1, 2010, 12:44:08 PM4/1/10
to
FrediFizzx wrote:
> It is very true that we don't need virtual photons to
> *calculate* certain interactions but we do need them for the *ultimate*
> fundamental explanation.

I disagree.

While I have no idea what the "*ultimate* fundamental explanation" is or will
be, I do know that photons are not part of QED at the fundamental level. Since
they are not fundamental in QED, I see no reason to expect that they will
somehow become fundamental in some future "ultimate explanation".

In QED, photons are not fundamental because they arise in the perturbation
APPROXIMATION to the theory. Photons are the assumed-free quantum excitations of
the EM field, and the fact that they are not truly free indicates how artificial
photons actually are [#]. That is, perturbation theory inherently ASSUMES that
electromagnetic interactions can be modeled as free photons occasionally
interacting with charges; the photon lines in diagrams represent the fields
satisfying the FREE Hamiltonian, and the vertexes are the ASSUMED SMALL
interactions (i.e. the interaction part of the Hamiltonian << the free part of
the Hamiltonian) [$]. While in QED this happens to work and gives excellent
accuracy for modeling many experimental results, it is QUITE CLEARLY an ad hoc
procedure, and the APPROXIMATE nature of this approach is obvious [@].

[#] In the perturbation approximation to QED, there are not
very many diagrams of interest that have no vertexes.

[$] Once one understands this, one instantly realizes that the
typical reification of the photon propagator as a "physical particle
propagating from interaction to interaction" is unwarranted. Yes,
that is how the diagrams look, but the diagrams are not the theory,
they are merely a computational tool for the APPROXIMATION to the
theory.

[@] For QED we are lucky that the values of the physical constants
in the theory happen to make the perturbation series converge
rapidly; for QCD we are not so lucky.


Tom Roberts

Oh No

unread,
Apr 1, 2010, 12:52:42 PM4/1/10
to
Thus spake Igor Khavkine <igo...@gmail.com>

I have taken another look. My copy does not even have a section 89. In
any case, I was not saying that the Coulomb field cannot be derived from
the interaction Hamiltonian, but that even when it is derived from the
interaction Hamiltonian the interpretation that it is a force
transmitted by photons remains.

Oh No

unread,
Apr 1, 2010, 12:56:08 PM4/1/10
to
Thus spake Arnold Neumaier <Arnold....@univie.ac.at>

Strange is not a criterion we can use to assess the correctness of
fundamental theory, or indeed to assess any quantum phenomenon.


>Physicists take causality as the most basic princilple, and this
>immediately rules out anything moving backward in time.

If so they are not acting as physicists should, for they are allowing
preconceptions based on the perceptions of the human mind to take
precedence over observation of physical law. The universe has past
present and future, and all fundamental observed laws are time
reversible. The asymmetry of the mind, that we remember the past but do
not know the future, is ultimately down to the law of entropy. Entropy
is not a fundamental law but is derived statistically from the
application of time-symmetrical laws together with time-asymmetric
boundary conditions. Clearly entropy does not dictate individual
processes in the quantum domain.

[[Mod. note -- To anyone wishing to further discuss the issues raised
in the immediately preceding paragraph: Please change the subject
line to something more appropriate than "Re: virtual particles help"!
Maybe "The arrow of time" or something along those lines??
-- jt]]

Bob_for_short

unread,
Apr 1, 2010, 12:56:54 PM4/1/10
to
> On Mar 31, 12:56 pm, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>

> wrote:
>> Since any perturbation series must be truncated, the truncations at
>> different bare (i.e., harmonic) frequencies give different results.
>> One wants to choose the frequency such that few terms of the
>> perturbation series suffice to get a meaningful accuracy.

I did not know that truncation can be used as a pretext to vary the
original frequencies in order to fit some data, especially if the
series converges rapidly (due to too small expansion parameter).

In my practice I encountered once a perturbative series that was just
wrong starting from the third order, and if one wanted to use such a
wrong series to "determine" the small parameter value, one would
definitely fall in error. So, in order to use the truncated series as
a fitting curve, one has to be sure that the series has at least the
right signs at each power.

Regards,

Vladimir.

Arnold Neumaier

unread,
Apr 1, 2010, 4:32:24 PM4/1/10
to
Tom Roberts wrote:
>
> In QED, photons are not fundamental because they arise in the perturbation
> APPROXIMATION to the theory. Photons are the assumed-free quantum excitations of
> the EM field, and the fact that they are not truly free indicates how artificial
> photons actually are [#].

It is not that bad. Quantum excitations exist for any quantum system,
and need neither be free nor harmonic to exist. They just need to
be separable enough that one can regard them as asymptotic states at
the time scale of interest.

Photons are elementary quantum excitations of the e/m field. They
exist as such in very good approximation and are abundantly used
in quantum optics. In quantum optics, one cannot dispense with
1-photon and 2-photon states. Whether one treats them as particles
since they are not exactly localizable is a different matter that
has nothing to do with whether virtual photons exist.

But for the present discussion, the crucial point is that they neither
_constitute_ nor _cause_ the electromagnetic field, as some erroneously
interpret the derivability of the Coulomb force from the tree diagram
for electron scattering with an internal photon line.


Arnold Neumaier

Arnold Neumaier

unread,
Apr 1, 2010, 4:32:24 PM4/1/10
to
Bob_for_short wrote:
>> On Mar 31, 12:56 pm, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
>> wrote:
>>> Since any perturbation series must be truncated, the truncations at
>>> different bare (i.e., harmonic) frequencies give different results.
>>> One wants to choose the frequency such that few terms of the
>>> perturbation series suffice to get a meaningful accuracy.
>
> I did not know that truncation can be used as a pretext to vary the
> original frequencies in order to fit some data, especially if the
> series converges rapidly (due to too small expansion parameter).

If the series converges rapidly with the bare frequency/mass, there
is no reason for changing the frequency. But for strongly nonlinear
problems and for quantum field theories the bare frequency is a very
poor choice.


> In my practice I encountered once a perturbative series that was just
> wrong starting from the third order, and if one wanted to use such a
> wrong series to "determine" the small parameter value, one would
> definitely fall in error. So, in order to use the truncated series as
> a fitting curve, one has to be sure that the series has at least the
> right signs at each power.

But how to know the right signs?

The traditional recipe for choosing the parameters that can be freely
chosen is by means of a stationarity test, which gives variational
perturbation theory.

Arnold Neuamier

Juan R.

unread,
Apr 1, 2010, 5:05:05 PM4/1/10
to
Oh No wrote on Thu, 01 Apr 2010 12:56:08 -0400:

> Thus spake Arnold Neumaier <Arnold....@univie.ac.at>
>>Oh No wrote:

(...)

>>How do you cause a particle to move backwards in time? You cannot.
>>
>>In all scatering experiments that produce antiparticles, these are
>>caused by the collision of particles created brior to the collision,
>>and result in the antiparticles that have, according to your view,
>>allegedely arrived from the future to be present just at the time
>>where they meet the collision experiment.
>>
>>Very, very strage, such a view, if taken seriously.
>
> Strange is not a criterion we can use to assess the correctness of
> fundamental theory, or indeed to assess any quantum phenomenon.

The criteria are internal consistency and experimental verification.

Your view that particles were generated in future and travelled
backward in time to arrive at the experiment is not tested. As is well
explained in the chapter 3 of Weinberg textbook in QFT (Volume 1),
experimental physicists prepare particles in the *past*, collide them,
and then study the outcome in *future*.

Both particles and antiparticles travel forward in time. Your model of
electrons travelling backard in time is based in a naive interpretation
of the Dirac wave function equation, which is no more a valid equation
in the modern field formulation of QED. The Hamiltonian of QED is
*positive* definite and there is nothing as negative energy particles
travelling backward in time (check Weinberg, Mandl & Shaw...).

>>Physicists take causality as the most basic princilple, and this
>>immediately rules out anything moving backward in time.
>
> If so they are not acting as physicists should, for they are
> allowing preconceptions based on the perceptions of the human mind
> to take precedence over observation of physical law.

Evidently, physicists are acting as physicists...

> The universe has past
> present and future, and all fundamental observed laws are time
> reversible.

This is rather debatable but irrelevant for the topic of causality.

> The asymmetry of the mind, that we remember the past but do
> not know the future, is ultimately down to the law of entropy.
> Entropy is not a fundamental law but is derived statistically from
> the application of time-symmetrical laws together with
> time-asymmetric boundary conditions. Clearly entropy does not
> dictate individual processes in the quantum domain.

This is rather wrong, but again irrelevant for the topic.

Causality, as defined in QFT, is not related to entropy or
time-assymmetry.

The fundamental equation of motion in QFT is time-symmetric, and it
conserves entropy but of course causality implies that time flows
from past to future. This is the reason which we define the S-matrix
as

S = U(+oo, -oo)

and not otherwise.


--
http://www.canonicalscience.org/

BLOG:
http://www.canonicalscience.org/publications/canonicalsciencetoday/canonicalsciencetoday.html

Bob_for_short

unread,
Apr 2, 2010, 3:39:06 AM4/2/10
to
On 25 mar, 05:22, "FrediFizzx" <fredifi...@hotmail.com> wrote:

> >> Are you saying that the Coulomb field is composed of real photons
> >> then?
>
> > No. The Coulomb field has nothig to do with photons. It arises as an


> > interaction term in the QED Hamiltonian in the
> > Coulomb gauge.
>

> Sorry, but that is contrary to the standard model of particle physics
> and I believe fairly speculative.  For QED we have photons and
> electrically charged elementary particles such as the electron, etc. ...

I used to work with a non stationary heat conduction (or diffusion)
equations. There the non stationary Green's function may also be
expressed via eigenfunctions and eigenvalues but no physical particles
are involved, as you understand. The Green's functions may have
explicit analytical expressions without mentioning the eigenstates.
Solving a non stationary heat conduction problem by the perturbation
theory with help of non stationary Green's function is quite similar
to solving QFT problems with help of propagators. That is why many
insist that no "particle" meaning should be given to the internal
lines in Feynman diagrams.

Igor Khavkine

unread,
Apr 2, 2010, 3:39:07 AM4/2/10
to
On Apr 1, 6:52 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Thus spake Igor Khavkine <igor...@gmail.com>

> >On Mar 31, 8:06 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:

> >> I have read Bjorken and Drell, and I do not recognise your reading of
> >> it. [...]
>
> >Please take another look. The Coulomb field, in the way that Arnold is
> >referring to, makes an explicit appearance in the last equation of
> >section 89 of Bjorken & Drell's volume 2.
>
> I have taken another look. My copy does not even have a section 89.

I can only speak for the Russian edition I have access to. It has
sections numbered as follows:

Vol.2, Ch.15: Interacting fields
Sec.88: Introduction
Sec.89: Electromagnetic interaction
Sec.90: Lorentz and translation invariance
...

> In
> any case, I was not saying that the Coulomb field cannot be derived from
> the interaction Hamiltonian, but that even when it is derived from the
> interaction Hamiltonian the interpretation that it is a force
> transmitted by photons remains.

I would call that a stretching of the definition of a photon, as it
occurs in perturbative QFT.

Igor

Oh No

unread,
Apr 2, 2010, 4:00:32 AM4/2/10
to
Thus spake Arnold Neumaier <Arnold....@univie.ac.at>
>However, at least it has been shown in a nice book on QED by Scharf
>that it is possible to avoid all bare stuff. Then no infinities appear
>anywhere (except most likely in summing the perturbation series), and
>the only masses, charges, etc. are those of the real particles.

It would be more accurate to say that Scharf shows how to avoid all the
"dressed" stuff.

>No virtual particles anymore!

Not so. They appear in Scharf as in any other book on field theory

Oh No

unread,
Apr 2, 2010, 1:24:04 PM4/2/10
to
Thus spake Igor Khavkine <igo...@gmail.com>
>On Apr 1, 6:52 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake Igor Khavkine <igor...@gmail.com>
>> >On Mar 31, 8:06 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
>> >> I have read Bjorken and Drell, and I do not recognise your reading of
>> >> it. [...]
>>
>> >Please take another look. The Coulomb field, in the way that Arnold is
>> >referring to, makes an explicit appearance in the last equation of
>> >section 89 of Bjorken & Drell's volume 2.
>>
>> I have taken another look. My copy does not even have a section 89.
>
>I can only speak for the Russian edition I have access to. It has
>sections numbered as follows:
>
>Vol.2, Ch.15: Interacting fields
>Sec.88: Introduction
>Sec.89: Electromagnetic interaction
>Sec.90: Lorentz and translation invariance

ok, a change in numbering. in my edition, these are 15.1, 15.2, 15.3.

>> In
>> any case, I was not saying that the Coulomb field cannot be derived from
>> the interaction Hamiltonian, but that even when it is derived from the
>> interaction Hamiltonian the interpretation that it is a force
>> transmitted by photons remains.
>
>I would call that a stretching of the definition of a photon, as it
>occurs in perturbative QFT.

I don't think so. In my copy of B&D in sec15.2 A is described as the
photon field, which is how I also would describe it. In common with
other field operators, A has both a creation and annihilation part.
These create or annihilate the "excitations of the quantum field" which
have been mentioned in this thread, though I rather think that
"excitation of the quantum field" is just a gobbledygook way of saying
"particle", or, in this case "photon". Elsewhere in either B&D or P&S I
seem to recall that the photon wave function is described as <|A(x)|f>,
though I wouldn't expect to find it easily.

So, what we should see is that the interaction Hamiltonian couples a
charged particle to the Coulomb field through the mechanism or creating
or annihilating a photon. Similarly another charged particle generates
the coulomb field by the same mechanism. Hence the coulomb force is
described in qed quite generally by the transmission of photons,
irrespective of any perturbation theoretic treatment. Considering both
the constraints on a legitimate relativistic quantum theory, and the
empirical success of the theory, I don't have any problem considering
that these photons are real entities, which have real consequences (i.e.
the force), notwithstanding the fact that they are sometimes
misleadingly called "virtual" by dint of the fact that they are only
observable through their effect, not as individual objects.

I have given a more general derivation of Maxwell's equations and the
Lorentz force, without any use of perturbation theory, in
http://papers.rqgravity.net/RQGQED.pdf
(under review).

Igor Khavkine

unread,
Apr 2, 2010, 6:34:31 PM4/2/10
to
On Apr 2, 7:24 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Thus spake Igor Khavkine <igor...@gmail.com>
> >On Apr 1, 6:52 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:

> >> In
> >> any case, I was not saying that the Coulomb field cannot be derived from
> >> the interaction Hamiltonian, but that even when it is derived from the
> >> interaction Hamiltonian the interpretation that it is a force
> >> transmitted by photons remains.
>
> >I would call that a stretching of the definition of a photon, as it
> >occurs in perturbative QFT.
>
> I don't think so. In my copy of B&D in sec15.2 A is described as the
> photon field, which is how I also would describe it.

This nomenclature is standard and is not part of the disagreement.

> In common with
> other field operators, A has both a creation and annihilation part.
> These create or annihilate the "excitations of the quantum field" which

> have been mentioned in this thread, [...]

This is all fine, except that it does not apply to the A_0 component
of the electromagnetic vector potential in Coulomb gauge and that is
the component responsible for the Coulomb field.

> So, what we should see is that the interaction Hamiltonian couples a
> charged particle to the Coulomb field through the mechanism or creating
> or annihilating a photon.

That's the part that is false. No photon creation or annihilation
operators appear in the last equation of B&D's section 15.2.

Igor

Igor Khavkine

unread,
Apr 3, 2010, 10:21:43 AM4/3/10
to
On Apr 2, 10:00 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Thus spake Arnold Neumaier <Arnold.Neuma...@univie.ac.at>

>
> >However, at least it has been shown in a nice book on QED by Scharf
> >that it is possible to avoid all bare stuff. Then no infinities appear
> >anywhere (except most likely in summing the perturbation series), and
> >the only masses, charges, etc. are those of the real particles.
>
> It would be more accurate to say that Scharf shows how to avoid all the
> "dressed" stuff.
>
> >No virtual particles anymore!
>
> Not so. They appear in Scharf as in any other book on field theory

Then, please, if you do not mind, would you share with us an example
of where Scharf uses the terminology of virtual particles?

Igor

Arnold Neumaier

unread,
Apr 3, 2010, 11:04:15 AM4/3/10
to
Oh No wrote:
> Thus spake Arnold Neumaier <Arnold....@univie.ac.at>
>> However, at least it has been shown in a nice book on QED by Scharf
>> that it is possible to avoid all bare stuff. Then no infinities appear
>> anywhere (except most likely in summing the perturbation series), and
>> the only masses, charges, etc. are those of the real particles.
>
> It would be more accurate to say that Scharf shows how to avoid all the
> "dressed" stuff.

Everything in his treatement of QED is already dressed (= physical).
No bare (= infinite) masses or charges appear.


>> No virtual particles anymore!
>
> Not so. They appear in Scharf as in any other book on field theory

Where?

Oh No

unread,
Apr 3, 2010, 11:07:02 AM4/3/10
to
Thus spake Igor Khavkine <igo...@gmail.com>
>On Apr 2, 7:24 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake Igor Khavkine <igor...@gmail.com>
>> >On Apr 1, 6:52 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
>> >> In
>> >> any case, I was not saying that the Coulomb field cannot be derived from
>> >> the interaction Hamiltonian, but that even when it is derived from the
>> >> interaction Hamiltonian the interpretation that it is a force
>> >> transmitted by photons remains.
>>
>> >I would call that a stretching of the definition of a photon, as it
>> >occurs in perturbative QFT.
>>
>> I don't think so. In my copy of B&D in sec15.2 A is described as the
>> photon field, which is how I also would describe it.
>
>This nomenclature is standard and is not part of the disagreement.
>
>> In common with
>> other field operators, A has both a creation and annihilation part.
>> These create or annihilate the "excitations of the quantum field" which
>> have been mentioned in this thread, [...]
>
>This is all fine, except that it does not apply to the A_0 component
>of the electromagnetic vector potential in Coulomb gauge and that is
>the component responsible for the Coulomb field.

Ah, you have put your finger on a fundamental disagreement which I have
with quantisation in the Coulomb gauge. Coulomb gauge is not covariant,
so it doesn't tell the whole story. One can ignore quantisation of the
A_0 component, but one cannot actually eliminate it because it reappears
with Lorentz transformation. There is a big difference between ignoring
something and eliminating it.

>> So, what we should see is that the interaction Hamiltonian couples a
>> charged particle to the Coulomb field through the mechanism or creating
>> or annihilating a photon.
>
>That's the part that is false. No photon creation or annihilation
>operators appear in the last equation of B&D's section 15.2.

Reason seems to be that Coulomb gauge is not a full quantisation. If one
works in Lorenz gauge then everything I said holds up.

Bob_for_short

unread,
Apr 3, 2010, 11:39:22 AM4/3/10
to

======== Moderator's note ===============================================

QED is a gauge invariant theory. Thus there's no change in the physical
meaning of the theory with changing the gauge fixing when quantizing. In
Coulomb gauge you have a static Coulomb-interaction term, and that makes it
convenient to get approximations for bound states. The full field-theoretical
treatment of the bound-state problem is via the Bethe-Salpeter equation and
(non-relativistic) reduction schemes. A fully consistent non-perturbative
treatment of relativistic bound states has not been achieved yet (as far as
I know). The corresponding energy levels are given by the poles of the 2-body
(4-point) Green's function (scattering amplitude) and is thus a gauge-independent
quantity (as it must be for observables!).

For (perturbative) S-matrix amplitudes the Coulomb term is always cancelled order
by order by certain other non-covariant parts of the photon propagator, and you end
up with the correct causal propagators (time ordered propagators to be precise)
within the S-matrix as it must be for any field theory. In other gauges the same
cancellations due to local gauge symmetries takes place. E.g., in Feynman gauge you
have time-like and longitudinal photons which after all do not contribute to the
S-matrix element.

================================================================================

On 3 avr, 17:07, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
> Ah, you have put your finger on a fundamental disagreement which I have
> with quantisation in the Coulomb gauge. Coulomb gauge is not covariant,
> so it doesn't tell the whole story.

...


> One can ignore quantisation of the
> A_0 component, but one cannot actually eliminate it because it reappears
> with Lorentz transformation. There is a big difference between ignoring
> something and eliminating it.

...


> Reason seems to be that Coulomb gauge is not a full quantisation. If one
> works in Lorenz gauge then everything I said holds up.
>

Coulomb gauge is not explicitly covariant but it is covariant
implicitly. Namely extra "degrees of freedom" appearing after Lorentz
transformation can be eliminated with a suitable gauge transformation,
i.e., they are not physical and the Coulomb form of the new
Hamiltonian can always be restored.

Potential energy of charge interaction q2q1/|r2 - r1| leads to bound
states. It is a part of a body energy. Photons carry away some energy-
momentum if emitted. No more entities that "contain" energy exists. So
no additional physical variables (degrees of freedom) are necessary.

Canonic quantization in the Coulomb gauge was named "fundamental
quantization" by J. Schwinger. It is distinguished from the path
integral (heuristic) quantization, the latter being more suitable for
scattering rather than for bound state calculations.

Bob_for_short

unread,
Apr 3, 2010, 1:58:12 PM4/3/10
to
On 3 avr, 17:39, Bob_for_short <vladimir.kalitvian...@wanadoo.fr>
wrote:

> ======== Moderator's note ===============================================
>
> QED is a gauge invariant theory. ....
>
> ================================================================================
>

I did not say the contrary.

Besides, with speaking of the fundamental quantization I meant also
that quantization in the Coulomb gauge is complete rather than "not a
full".

Tom Roberts

unread,
Apr 3, 2010, 2:03:52 PM4/3/10
to
Arnold Neumaier wrote:
> Tom Roberts wrote:
>> In QED, photons are not fundamental because they arise in the
>> perturbation APPROXIMATION to the theory. Photons are the assumed-free
>> quantum excitations of the EM field, and the fact that they are not
>> truly free indicates how artificial photons actually are [#].
>
> It is not that bad.

It's not "bad", but as I understand perturbation theory, it is how things are.


> Quantum excitations exist for any quantum system,
> and need neither be free nor harmonic to exist. They just need to
> be separable enough that one can regard them as asymptotic states at
> the time scale of interest.

Sure.


> Photons are elementary quantum excitations of the e/m field.

Yes. But not just any excitations, they are the specific excitations that
correspond to the squiggly lines we draw in Feynman diagrams. Those excitations
are solutions of the FREE Hamiltonian. The essence of perturbation theory is
approximating the theory as free excitations (photons) propagating from one
interaction (vertex) to the next.

[Plus fermions, which we are not discussing.]

While the squiggly lines of a diagram are solutions to the free Hamiltonian,
they are not really free -- they have interaction vertexes. That's why I said
that photons are "artificial". And they are not fundamental in QED, because they
really only arise in the perturbation approximation to QED -- that approximation
is essential to permit one to separate the problem into photons (excitations of
the field satisfying the free Hamiltonian) and interactions.


> They
> exist as such in very good approximation and are abundantly used
> in quantum optics. In quantum optics, one cannot dispense with
> 1-photon and 2-photon states. Whether one treats them as particles
> since they are not exactly localizable is a different matter that
> has nothing to do with whether virtual photons exist.

Yes.

The distinction between "virtual" and "real" photons is just another, quite
different approximation: for a typical scattering experiment it is assumed that
the physical experiment can be separated into three parts connected by "real"
(on mass shell) particles:
1) creation of the incident particles
2) the interaction of interest
3) detection of the outgoing particles
The QED computation of the interaction (2) is the usual computation that
textbooks cover. It uses external legs for the incident and outgoing particles,
requiring them to be on mass shell. But a more inclusive analysis would not make
this approximation, and those "real particle" external legs of (2) will merely
be "virtual" legs in a larger diagram that includes the creation (1) and
detection (3) processes that were ignored in the first computation. In that
larger diagram, such particles will not be exactly on mass shell, but their
deviations will be incredibly small for macroscopic distances between their
interaction vertexes (constraints corresponding to a real experiment).

This larger analysis is hopelessly complicated, because the
constraints are quite difficult (tantamount to being impossible),
because enormous numbers of diagrams usually contribute, and
because one must (anti-)symmetrize over macroscopic numbers of
particles in sources and detectors.


> But for the present discussion, the crucial point is that they neither
> _constitute_ nor _cause_ the electromagnetic field, as some erroneously
> interpret the derivability of the Coulomb force from the tree diagram
> for electron scattering with an internal photon line.

Yes.


Tom Roberts

Arnold Neumaier

unread,
Apr 3, 2010, 4:16:46 PM4/3/10
to
Tom Roberts wrote:
> Arnold Neumaier wrote:
>> Tom Roberts wrote:
>>> In QED, photons are not fundamental because they arise in the
>>> perturbation APPROXIMATION to the theory. Photons are the
>>> assumed-free quantum excitations of the EM field, and the fact that
>>> they are not truly free indicates how artificial photons actually are
>>> [#].
>>
>> It is not that bad.
>
> It's not "bad", but as I understand perturbation theory, it is how
> things are.

Photons are not artificial; they are quite natural.

Virtual photons, however are nonexistent, except for the imagination
to render abstract multiple integrals more intuitive.


>> Photons are elementary quantum excitations of the e/m field.
>
> Yes. But not just any excitations, they are the specific excitations
> that correspond to the squiggly lines we draw in Feynman diagrams.

Real photons correspond to external lines in Feynman diagrams, which
define real, measurable cross sections.


> Those
> excitations are solutions of the FREE Hamiltonian.

No. This is the bare, unrenormalized version, which is only the formal
foreplay to the real, renormalized thing. Real photons also have form
factors associated with them.


> While the squiggly lines of a diagram are solutions to the free
> Hamiltonian, they are not really free -- they have interaction vertexes.
> That's why I said that photons are "artificial". And they are not
> fundamental in QED, because they really only arise in the perturbation
> approximation to QED

No. The S-matrix is the essence of QED, perturbative or not.
And its input states are real particle states - those of real photons,
real electrons, and real positrons. If it were not so, quantum optics
would not exist as a subject.


> The distinction between "virtual" and "real" photons is just another,
> quite different approximation:

This is not an approximation. It is the distinction between internal and
external lines in a Feynamn diagram.


> The QED computation of the interaction (2) is the usual computation that
> textbooks cover. It uses external legs for the incident and outgoing
> particles, requiring them to be on mass shell. But a more inclusive
> analysis would not make this approximation, and those "real particle"
> external legs of (2) will merely be "virtual" legs in a larger diagram

With such an interpretation, all physics would be artificial, since we
do not have anywhere exact models but only approximations. But with such
an inclusive use of the word ''artificial'' it would become completely
meaningless.


Arnold Neumaier

Juan R.

unread,
Apr 5, 2010, 8:57:03 PM4/5/10
to
Bob_for_short wrote on Thu, 01 Apr 2010 12:26:55 +0000:

>> FrediFizzx wrote:
>
>> A Coulomb field is a classical field.
>
> The Coulomb potential dependence 1/|r2 - r1| is simple but in QM and
> QED it is not a "classical" any more.

Of course!

(...)

> And one more remark. Some think that 1/r "surrounds" a charge. It is
> a mistake. 1/r is an interaction term for at least two charges
> q2*q1/|r2 - r1|. When we speak of the Coulomb potential we mean
> exactly the latter case and nothing else. In other words, this term
> is always involved in two equations.

In my previous message of day 24 March to FrediFizzx (archived here) I
already gave him three rigorous references were his mistakes (and
those done by Charles Francis) are corrected.

For instance, the reference [3] already reported what you are now
saying about the mistake of believing that 1/r surrounds a charge.
Indeed explicit expressions (more detailed than your own of above) are
given and analized in [3].

An instantaneous scalar potential in field theory is given by
phi_F = phi_F(x,t) and the distance used in the potential by

r = |x - y(t)|

For the Coulomb potential in AAAD theory phi_C = phi_C(R(t)) and

R = |X(T) - Y(T)|

phi_F, r, x, y, t and phi_C, R, X, Y, T are different quantities with
completely different mathematics and physics, as rigorously showed.

The reference [3] also explains why it is a mistake to believe that
Coulomb interaction can be thought as interchange of quanta.

There is also a section specifically devoted to self-interactions,
renormalization, and bare and dressed particles, which explains with
detail why field theory needs renormalization to eliminate the
unphysical modes from the interaction and how renormalization is no
more needed when one uses a better model of interactions build over
real particles (not over bare particles).

Maybe it is time to read before continue doing the same mistakes
already corrected before.

Note: Moderator jt wrote,

[[Mod. note -- I think it's fair to say that the references cited by
the author are highly controversial. I believe that most researchers
in classical field theory do not consider these results to be correct.

If people want to debate this, please start a *new* thread with an
appropriate subject line, since such debate would not really be about
virtual particles (the nominal subject of this thread).
-- jt]]

I see absolutely no problem with moderators writting warnings when the
content of a message is non-standard, revolutionary, or controversial.

Another very different issue is when a moderator does invalid comments
about the content of references, or when state his personal *beliefs*
about others' opinions.

I kindly invite everyone to give references showing that the rigorous
proofs given in references [1-3] are misguided :-D


[1] Phys. Rev. E 1997, 53, 5373.
[2] Phys. Rev. E 1998, 57, 3683.
[3] http://www.canonicalscience.org/publications/canonicalsciencetoday/20100301.html

Arnold Neumaier

unread,
Apr 5, 2010, 9:07:25 PM4/5/10
to
Oh No wrote:
> Ah, you have put your finger on a fundamental disagreement which I have
> with quantisation in the Coulomb gauge. Coulomb gauge is not covariant,
> so it doesn't tell the whole story.

It tells the whole story. Bjorcken and Drell prove covariace of their
QED setting in the Coulomb gauge, and derive all the main results that
are nowadays derived in a more manifestly covariant setting.

That Bjorcken and Drell's setting is not manifestly covariant
is immaterial. Not even the 2 x 2 spinor matrices are manifestly
covariant, since they presuppose a choice of coordinates to select
p_0 and p_3.


> Reason seems to be that Coulomb gauge is not a full quantisation.

Strange reasons, since Bjorcken and Drell derived in the Coulomb gauge
everything of interest.

Igor Khavkine

unread,
Apr 5, 2010, 9:07:50 PM4/5/10
to
On Apr 3, 5:07 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Thus spake Igor Khavkine <igor...@gmail.com>
> >On Apr 2, 7:24 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:

> >> So, what we should see is that the interaction Hamiltonian couples a
> >> charged particle to the Coulomb field through the mechanism or creating
> >> or annihilating a photon.
>
> >That's the part that is false. No photon creation or annihilation
> >operators appear in the last equation of B&D's section 15.2.
>
> Reason seems to be that Coulomb gauge is not a full quantisation.

I'm sorry, but that is patently false. The widespread use of the
Coulomb gauge as well as many fully detailed accounts of it stand as a
testament to its validity. Now is a good time to concede the point
under discussion above, rather than making dramatic unsubstantiated
claims about standard QED methods.

> If one
> works in Lorenz gauge then everything I said holds up.

I hope you've been paying attention to the rest of the discussion in
this thread. One major point that has been made again and again is
that there are many theoretical calculational approaches to QED, all
of which ultimately yield the same physical results, yet involve
significantly different intermediate concepts and mathematical steps.
What you say above is a case in point.

Igor

Oh No

unread,
Apr 5, 2010, 9:12:59 PM4/5/10
to
>======== Moderator's note =============================================

>==
>
>QED is a gauge invariant theory. Thus there's no change in the physical
>meaning of the theory with changing the gauge fixing when quantizing.

There is no change in the predictions of the theory, due to gauge
invariance. But if Lorenz gauge means that we have four polarisation
states (two of which vanish in observations), whereas Coulomb gauge
means there are only two, then I consider that a rather important
change in meaning.

>In Coulomb gauge you have a static Coulomb-interaction term, and that
>makes it convenient to get approximations for bound states. The full
>field-theoretical treatment of the bound-state problem is via the
>Bethe-Salpeter equation and (non-relativistic) reduction schemes. A
>fully consistent non-perturbative treatment of relativistic bound
>states has not been achieved yet (as far as I know).

If you are talking about derivation of Maxwell's equations, the Lorentz
force law and the interacting Dirac equation, I find this a very
surprising assertion, as these seem to me rather straightforward
exercises which I would expect many a dutiful PhD student to perform (as
I did without raising the eyebrows of either examiner). If you are
talking about something else, please clarify. Certainly I only see how
to do this in Lorenz gauge, since the problem requires covariance.
Details are given in

http://papers.rqgravity.net/RQGQED.pdf

If you think there is something wrong with the derivations, I would very
much appreciate knowing what it is.

Oh No

unread,
Apr 6, 2010, 3:43:49 AM4/6/10
to
Thus spake Juan R. González-Álvarez <now...@canonicalscience.com>

>Oh No wrote on Thu, 01 Apr 2010 12:56:08 -0400:
>
>> Thus spake Arnold Neumaier <Arnold....@univie.ac.at>
>>>Oh No wrote:
>
>(...)
>
>>>How do you cause a particle to move backwards in time? You cannot.
>>>
>>>In all scatering experiments that produce antiparticles, these are
>>>caused by the collision of particles created brior to the collision,
>>>and result in the antiparticles that have, according to your view,
>>>allegedely arrived from the future to be present just at the time
>>>where they meet the collision experiment.
>>>
>>>Very, very strage, such a view, if taken seriously.
>>
>> Strange is not a criterion we can use to assess the correctness of
>> fundamental theory, or indeed to assess any quantum phenomenon.
>
>The criteria are internal consistency and experimental verification.

Then you should be happy that the view of causality implicit in the
Feynman-Stuckelberg interpretation gains strong support from the Bell
tests and the delayed quantum choice quantum eraser experiment.


>
>Your view that particles were generated in future and travelled
>backward in time to arrive at the experiment is not tested. As is well
>explained in the chapter 3 of Weinberg textbook in QFT (Volume 1),
>experimental physicists prepare particles in the *past*, collide them,
>and then study the outcome in *future*.

As I have explained, the whole operation of preparing initial states and
studying outcomes is dictated by the macroscopic arrow of time, and is
dependent on entropy, which is a statistical law and does not apply to


individual processes in the quantum domain.

>> The universe has past


>> present and future, and all fundamental observed laws are time
>> reversible.
>
>This is rather debatable but irrelevant for the topic of causality.

you mean you have not seen its relevance.


>
>> The asymmetry of the mind, that we remember the past but do
>> not know the future, is ultimately down to the law of entropy.
>> Entropy is not a fundamental law but is derived statistically from
>> the application of time-symmetrical laws together with
>> time-asymmetric boundary conditions. Clearly entropy does not
>> dictate individual processes in the quantum domain.
>
>This is rather wrong, but again irrelevant for the topic.

you mean you have not seen its relevance.

>Causality, as defined in QFT, is not related to entropy or
>time-assymmetry.
>
>The fundamental equation of motion in QFT is time-symmetric, and it
>conserves entropy

Entropy is a statistical law and has no meaning for individual particles
.

>but of course causality implies that time flows
>from past to future.

You are going in circles. Before you can say that causality implies that
time flows from past to future you must have a correct expression of
causality, consistent with the Bell tests and the delayed quantum choice
quantum eraser experiment.

>This is the reason which we define the S-matrix
>as
>
>S = U(+oo, -oo)
>
>and not otherwise.

even that is just a human convention

Juan R. Gonzàlez-Àlvarez

unread,
Apr 7, 2010, 12:46:16 PM4/7/10
to
Oh No wrote on Tue, 06 Apr 2010 09:43:49 +0200:

> Thus spake Juan R. Juan R. Gonzàlez-Àlvarez <now...@canonicalscience.com>


>>Oh No wrote on Thu, 01 Apr 2010 12:56:08 -0400:
>>
>>> Thus spake Arnold Neumaier <Arnold....@univie.ac.at>
>>>>Oh No wrote:
>>
>>(...)
>>
>>>>How do you cause a particle to move backwards in time? You cannot.
>>>>
>>>>In all scatering experiments that produce antiparticles, these are
>>>>caused by the collision of particles created brior to the
>>>>collision, and result in the antiparticles that have, according to
>>>>your view, allegedely arrived from the future to be present just
>>>>at the time where they meet the collision experiment.
>>>>
>>>>Very, very strage, such a view, if taken seriously.
>>>
>>> Strange is not a criterion we can use to assess the correctness of
>>> fundamental theory, or indeed to assess any quantum phenomenon.
>>
>>The criteria are internal consistency and experimental verification.
>
> Then you should be happy that the view of causality implicit in the
> Feynman-Stuckelberg interpretation gains strong support from the
> Bell tests and the delayed quantum choice quantum eraser experiment.

*Nothing* of that invalidates the modern field formulation of QED or
eliminates the known weakness of the original Dirac wave equation,
which is the basis of your misguided approach to causality and
your old approach to electrodynamics.

Modern textbooks in QED explain why the Dirac equation is no more a
wave equation.

>>Your view that particles were generated in future and travelled
>>backward in time to arrive at the experiment is not tested. As is
>>well explained in the chapter 3 of Weinberg textbook in QFT (Volume
>>1), experimental physicists prepare particles in the *past*, collide
>>them, and then study the outcome in *future*.
>
> As I have explained, the whole operation of preparing initial states
> and studying outcomes is dictated by the macroscopic arrow of time,
> and is dependent on entropy, which is a statistical law and does not
> apply to individual processes in the quantum domain.

And as was remarked in previous messages:

"This is rather debatable but irrelevant for the topic of causality."

The emphasis here was on that (i) in QED particles travel forward
in time (ii) there is not negative energy electrons and (iii) all
experiments and observations confirm this.

If you can cite some HEP experiment where some particles were prepared
in future, travelled backward in time and arrived at present to be
detected by experimental physicists, please do it.

Just the contrary is verified in experiments, and this was the reason
which both Stuckleberg and Feynman reinterpreted their negative energy
particles travelling backward in time as positive energy antiparticles
travelling forward in time. Never otherwise.

The modern field formulation of QED avoids gimmnastics as this. And it
is this modern formulation which is the basis for the rest of
interactions in the Standard Model.

>>> The universe has past
>>> present and future, and all fundamental observed laws are time
>>> reversible.
>>
>>This is rather debatable but irrelevant for the topic of causality.
>
> you mean you have not seen its relevance.

No, I mean what I wrote.

>>> The asymmetry of the mind, that we remember the past but do not
>>> know the future, is ultimately down to the law of entropy. Entropy
>>> is not a fundamental law but is derived statistically from the
>>> application of time-symmetrical laws together with time-asymmetric
>>> boundary conditions. Clearly entropy does not dictate individual
>>> processes in the quantum domain.
>>
>>This is rather wrong, but again irrelevant for the topic.
>
> you mean you have not seen its relevance.

No, I mean what I wrote.

>>Causality, as defined in QFT, is not related to entropy or
>>time-assymmetry.
>>
>>The fundamental equation of motion in QFT is time-symmetric, and it
>>conserves entropy
>
> Entropy is a statistical law and has no meaning for individual
> particles

This goes poor. In previous responses I have tried to avoid to
discuss thermal and statistical mechanics question, maintaining this
on-topic, but your insistence on such nonsensical statements deserves
some reply.

First, entropy is a physical quantity, not a "statistical law".

Second, there exists a well-known theorem that the fundamental
equation of motion in QFT conserves entropy.

Third, causality in QFT is not based in entropy or the second law
of thermodynamics.

(...)

>>This is the reason which we define the S-matrix as
>>
>>S = U(+oo, -oo)
>>
>>and not otherwise.
>
> even that is just a human convention

Nope. It is an outcome of *all* our observation and experiments: time
flows from past to future and we prepare particles in initial states
then collide them and study the final states of resulting particles.

Probably, this is my last post to you or Fred in this thread because
it has very little research content and because it seems that both of
you already decided that rest of people here (including moderators)
are wrong.

Fred has just opened a thread in SPF for noticing rejection here [A]:

"Your posting is inappropriate for sci.physics.research since it
contains elementary misunderstandings concerning the quantization of
the interacting electromagnetic field.

With kindest regards,
Hendrik van Hees.
Co-moderator for sci.physics.research"

And you have replied [B]:

"It seems to me patently clear that what you say is correct, and that
any elementary misunderstandings are on the part of the moderator.

I have noted too that Igor's debating style includes snipping the
part of a post which refutes his position and responding as though
the refutation was not given."

[A] http://groups.google.com/group/sci.physics.foundations/browse_thread/thread/89d3425dc0305d13#

[B] http://groups.google.com/group/sci.physics.foundations/msg/0995e1e8d3461880

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