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Eugene Stefanovich
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Joined: 24 Mar 2005
Posts: 519

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

Arnold Neumaier wrote:
 Quote: Eugene Stefanovich wrote: Arnold Neumaier wrote: All renormalization until today follows the same pattern, including yours. One does certain formal computations at finite cutoff and at some point where it no longer harms moves the cutoff to infinity, being left with approximate formulas at some (fixed or variable) loop order which no longer contain a cutoff and have finite values. In Tomonaga-Schwinger-Feynman theory the cutoff-dependent counterterms are added to the Hamiltonian. When the cutoff momentum is finite, everything is finite: the Hamiltonian, the counterterms, the S-matrix. However, when the cutoff goes to infinity the conterterms do not have a finite limit. That's what I call "the Hamiltonian is infinite". The S-matrix does have a definite limit, but the Hamiltonian does not. This is only because you ignore wave function renormalization, which is of course necessary to get finite matrix elements for the Hamiltonian. The limiting Hamiltonian is perturbatively well-defined in the physical Hilbert space obtained as limit of renormalized Hilbert spaces at finite cutoff, as the cutoff goes to infinity. For this, it is best to work in the functional Schroedinger picture. Maybe I should be even more precise: The Hamiltonian in T-S-F theory is a polynomial function of creation and annihilation operators of (bare) particles. In the limit of infinite cutoff the coefficients of the terms in this polynom do not tend to finite values. But this is irrelevant. Nobody uses these coefficients, and they have no physical meaning. What _is_ relevant is that the renormalized matrix elements phi|H|psi> = lim_{Lambda to inf}

There are two important features in my approach that distinguish it
from other approaches. First is the
requirement of full dressing which means that the interaction part

lim_{Lambda to inf} V(Lambda)

of the full Hamiltonian H_0 + V(Lambda)
should have zero matrix elements between states

lim_{Lambda to inf} <vacuum| V(Lambda) |anything> = 0
lim_{Lambda to inf} <one-particle| V(Lambda) |anything> = 0

If these conditions are not satisfied, then
(in disagreement with experiment) vacuum is not a no-particle
eigenstate of the Hamiltonian and one particle constantly
emits and reabsorbs virtual particles during its time evolution.

Second is the use of the instant form relativistic dynamics.
I already presented arguments why this form should be preferable.
I'll summarize it here for the record. The kinematical character
of space translations and rotations is abundantly clear from
everyday observations of interacting macroscopic systems.
The changes due to translations and rotations of observers
are always described as purely geometrical. The dynamical
(interaction-dependent) character of time translations is also
obvious. Then, the group properties of the Poincare group leave no
other choice but to admit that boosts are dynamical as well,
i.e., interactions have the instant form.

There are two caveats. First, the macroscopic observations mentioned
above refer only to long-range interactions, such as EM and gravity.
They do not say anything about short-range weak and strong forces.
However, in chapter 13 of the book I demonstrate that only instant form
is compatible with observed slowing-down of decays of fast moving
particles. This makes it very likely that weak interactions have
instant form as well.

Second, as long as you are interested only in the S-matrix and
energies of bound states (99.9% of high energy physics is about those
properties), any form of dynamics can be used equally well.
Front form or point form may give you some benefits in calculations.
However, for the description of wave functions (they are very difficult
to observe experimentally!) and their transformations
wrt various inertial transformations (including time translations,
boosts, rotations, etc.) only the instant form will do.

 Quote: This is discussed at length in my recent article hep-th/0503076 where I use a simple model of QFT as an example. I would appreciate any comments/opinions/ideas about this paper. I'd have appreciated if you had acknowledged the discussions with me that made you aware of the work by Glazek and Wilson.

I apologize for this omission. I have made corrections in
the text of hep-th/0503076.

Eugene Stefanovich
Franz Heymann
science forum Guru Wannabe

Joined: 03 Feb 2005
Posts: 282

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message
news:422EB74C.5020806@univie.ac.at...
 Quote: Franz Heymann wrote: "Arnold Neumaier" wrote in message news:422AC49F.3030502@univie.ac.at... Franz Heymann wrote: Unstable particles are modelled exactly like stable particles, namely as external lines in a Feynman diagram. Virtual particles in Feynman diagrams are exactly those which are not given by external lines. Hence what is real and what is virtual is not affected by a diagram rotation - this only affects what is input oand what is output. I am not happy with that. Let me illustrate my unhappiness with a specific example. Take the resonant elastic scattering case: pi + p --> Z --> pi + p I hope we both agree that the Z is produced as a real but unstable particle in the s-channel. No. Z is a virtual particle since it is an intermediate line. You are doing nothing to alleviate my unhappiness. If the Z were a virtual prticle, it would not be on its mass shell and a mass measurement would not be possible by studying the reaction. It is not on its (complex) mass shell when it is in an internal line as in your diagram. When measured, one has instead a diagram where the Z is detected, and hence on an external line of Feynman diagrams associated to the coresponding cross section! In ordinary scattering experiemnts the existence of the Z shows in resonances of the cross section. This is evidence for the presence of the Z in the action, and can be used to calculate properties of Z, but it is Not a measurement of Z as a particle. My undertanding is that the line corresponding to the Z is the trajectory of a real Z travelling from its birth to its decay. This interpretation of Feynman diagrams is popular but completely unfounded, and gives misleading conclusions if carried beyond the most superficial view. But it is observed, and its (complex) mass is actually measured. It is _not_ observed in an experiment where the Feynman diagram you drew is part of the calculation. It _can_ be observed, but only in a process analyzed with Feynman diagrams where the Z is an external line.

I maintain that it is an external line which has its far end
terminated in the decay of the resonance.

In atomic physics, a photon may be absorbed by an atom in its ground
state. The excited state very quickly decays back to the ground state
by the emission of a photon.

Are you now saying that the excited state is not a real, but a virtual
state?

 Quote: One can also see it by computing the momentum balance and check that the momentum does not lie on the Z's (complex) mass shell, as would be needed for a real Z. Oh? Sorry, but the Z is no less real than the pi0. If the pi0 can decay into two photons, then it can also be created bytwo photons. For the pi0 the situation is the same. If measured, the cross section for pi0 production or decay comes from an analysis with external pi0 lines, while in other processes it is a virtual particle without objective meaning.

I fear me I will never understand you, and you will never understand
to me, if indeed you wish to make one.

--
Franz
"A first-rate laboratory is one in which mediocre scientists can
produce outstanding work"
P.M.S. Blackett
Arnold Neumaier
science forum Guru

Joined: 24 Mar 2005
Posts: 379

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

Franz Heymann wrote:
 Quote: "Arnold Neumaier" wrote in message news:422AC49F.3030502@univie.ac.at... Franz Heymann wrote: Unstable particles are modelled exactly like stable particles, namely as external lines in a Feynman diagram. Virtual particles in Feynman diagrams are exactly those which are not given by external lines. Hence what is real and what is virtual is not affected by a diagram rotation - this only affects what is input oand what is output. I am not happy with that. Let me illustrate my unhappiness with a specific example. Take the resonant elastic scattering case: pi + p --> Z --> pi + p I hope we both agree that the Z is produced as a real but unstable particle in the s-channel. No. Z is a virtual particle since it is an intermediate line. You are doing nothing to alleviate my unhappiness. If the Z were a virtual prticle, it would not be on its mass shell and a mass measurement would not be possible by studying the reaction.

It is not on its (complex) mass shell when it is in an internal line as

When measured, one has instead a diagram where the Z is detected,
and hence on an external line of Feynman diagrams associated to
the coresponding cross section!

In ordinary scattering experiemnts the existence of the Z shows in
resonances of the cross section. This is evidence for the presence
of the Z in the action, and can be used to calculate properties of Z,
but it is Not a measurement of Z as a particle.

 Quote: My undertanding is that the line corresponding to the Z is the trajectory of a real Z travelling from its birth to its decay.

This interpretation of Feynman diagrams is popular but completely
unfounded, and gives misleading conclusions if carried beyond the
most superficial view.

 Quote: But it is observed, and its (complex) mass is actually measured.

It is _not_ observed in an experiment where the Feynman diagram
you drew is part of the calculation.

It _can_ be observed, but only in a process analyzed with
Feynman diagrams where the Z is an external line.

 Quote: One can also see it by computing the momentum balance and check that the momentum does not lie on the Z's (complex) mass shell, as would be needed for a real Z. Oh? Sorry, but the Z is no less real than the pi0. If the pi0 can decay into two photons, then it can also be created bytwo photons.

For the pi0 the situation is the same. If measured, the cross
section for pi0 production or decay comes from an analysis with
external pi0 lines, while in other processes it is a virtual
particle without objective meaning.

Arnold Neumaier
Arnold Neumaier
science forum Guru

Joined: 24 Mar 2005
Posts: 379

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

Eugene Stefanovich wrote:
 Quote: Arnold Neumaier wrote: Eugene Stefanovich wrote: Arnold Neumaier wrote: I wonder how _you_ predict a relativistic quantum Boltzmann equation for an plasma in an electromagnetic field, which is surely part of QED. No problem, RQD approach has a well-defined finite Hamiltonian in which particles interact via direct instantaneous potentials, just as in ordinary non-relativistic quantum mechanics. This Hamiltonian can be directly plugged into the time evolution operator, or diagonalized to find the stationary states. With this Hamiltonian, you can define the partition function exp(-\beta H) and all kinds of thermodynamical and statistical quantities just as in ordinary quantum statistics. The Boltzmann equation is not the Boltzmann distribution but a kinetic equation for the 1-particle phase space density. If you would be able to reproduce the qantum Boltzmann equation of Klimontovich with your formalism, I'd be very surprised. He is doing real dynamical work with QED, and gets manifestly covariant equations that are useful in practice. Repeat this feat before making again your exaggerated claims! Are you saying then that in order to get non-zero-temperature dynamics one needs to have something else beside the Hamiltonian?

Instead of the Hamiltonian one can also have the action. This are
two equivalent starting point and the latter is much superior in the
relativistic case.

Try to compute the hydrodynamic limit of your theory and see
whether you can get it into the form of the relativistic perfect
fluid in an electromagnetic field. (There is no freedom here -
you must arrive exactly at these equations!) You'll see that you have
_much_ more difficulty in your formulation than if you stay always
on the action level model.

You say that your theory is equivalent. But it is _not_ equivalent in
terms of complexity of computations, where yours is much worse.

 Quote: Please don't get me wrong. I am not saying that I have done what Klimontovich has done. I am simply saying that I do not see any obstacle why kinetics, statistics, and thermodynamics cannot be modeled with RQD Hamiltonian.

And I say that one doesn't need RQD to do all that, although it is
a full dynamical description of QED is far off the mark!
But it may have a corner of usefulness, and if you'd moderate your
claims to that, and show by actual calculations where this
corner is, I'd have no longer any objection.

But with all the metaphysical baggage that you tenaciously defend,
what you present is misleading and hence subject to my attacks.

 Quote: I think it can, and probably even much easier, because in RQD one do not need to perform the renormalization step. The time evolution (or kinetics) predicted by RQD would be different from that predicted by Klimontovich, because we would be using different Hamiltonians.

Well, give a proof. Many people dream of the power of their ideas.
Few succeed in demonstrating it. Only the latter are recognized.
Hic Rhodos, hic salta!

 Quote: So, in principle, the two theories could be distinguished by their predictions of observable effects. However, I doubt that experimental observations of plasma's dynamics are sensitive to radiative corrections.

If there were a different prediction presented with some authority
(which in particular involves removing your superluminal assertions)
some experimentalists would probably be motivated to push for a
discrimination.

Independent of experiment, working out the alternative kinetics
would be a good exercise in teaching you humility.

Arnold Neumaier
Eugene Stefanovich
science forum Guru

Joined: 24 Mar 2005
Posts: 519

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

Arnold Neumaier wrote:
 Quote: Eugene Stefanovich wrote: Arnold Neumaier wrote: I wonder how _you_ predict a relativistic quantum Boltzmann equation for an plasma in an electromagnetic field, which is surely part of QED. No problem, RQD approach has a well-defined finite Hamiltonian in which particles interact via direct instantaneous potentials, just as in ordinary non-relativistic quantum mechanics. This Hamiltonian can be directly plugged into the time evolution operator, or diagonalized to find the stationary states. With this Hamiltonian, you can define the partition function exp(-\beta H) and all kinds of thermodynamical and statistical quantities just as in ordinary quantum statistics. The Boltzmann equation is not the Boltzmann distribution but a kinetic equation for the 1-particle phase space density. If you would be able to reproduce the qantum Boltzmann equation of Klimontovich with your formalism, I'd be very surprised. He is doing real dynamical work with QED, and gets manifestly covariant equations that are useful in practice. Repeat this feat before making again your exaggerated claims!

Are you saying then that in order to get non-zero-temperature dynamics
one needs to have something else beside the Hamiltonian? I always
thought that information about interactions contained in the Hamiltonian
is sufficient to make any kind of predictions about stationary and
dynamical properties of the system at any temperature.

Please don't get me wrong. I am not saying that I have done what
Klimontovich has done. I am simply saying that I do not see any
obstacle why kinetics, statistics, and thermodynamics cannot be modeled
with RQD Hamiltonian. I think it can, and probably even much easier,
because in RQD one do not need to perform the renormalization step.
The time evolution (or kinetics) predicted by RQD would be different
from that predicted by Klimontovich, because we would be using different
Hamiltonians. So, in principle, the two theories could be distinguished
by their predictions of observable effects. However, I doubt that
experimental observations of plasma's dynamics are sensitive to
radiative corrections, and that they can serve as a guide for theory at
this level.

Eugene Stefanovich.
Franz Heymann
science forum Guru Wannabe

Joined: 03 Feb 2005
Posts: 282

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message
news:422AC49F.3030502@univie.ac.at...
 Quote: Franz Heymann wrote: "Arnold Neumaier" wrote in message news:4225B962.6070905@univie.ac.at... Franz Heymann wrote: "Arnold Neumaier" wrote in message news:421F2395.9070900@univie.ac.at... PD wrote: If I see a peak in the invariant mass of a combination of two electrons seen in a detector, then I take it that I've seen evidence of a particle decaying into those two electrons, especially if other parameters of the final state points to a fixed set of quantum numbers (such as spin=0 or spin=1). If that peak has some width to it, that's evidence of the finite lifetime of the decaying particle, and indeed any decay that is "off-peak" in this invariant mass can be said to point to the decay of a virtual particle. Neutral pions, J/psi's, Z bosons, all fit this description. This doesn't seem to me to be relegated to a "mathematical trick". Virtual particles are as "real" as real particles -- the boundary is a soft, fuzzy one. You mix up unstable particles and virtual particles. These are distinct concepts. What you see as a peak in a spectrum is a resonance, the signature of an unstable particle. And if you turn the Feynman diagram which describes the production of this unstable particle through 90 deg, why does this real particle magically becomes a virtual particle It doesn't! Unstable particles are modelled exactly like stable particles, namely as external lines in a Feynman diagram. Virtual particles in Feynman diagrams are exactly those which are not given by external lines. Hence what is real and what is virtual is not affected by a diagram rotation - this only affects what is input oand what is output. I am not happy with that. Let me illustrate my unhappiness with a specific example. Take the resonant elastic scattering case: pi + p --> Z --> pi + p I hope we both agree that the Z is produced as a real but unstable particle in the s-channel. No. Z is a virtual particle since it is an intermediate line.

You are doing nothing to alleviate my unhappiness.
If the Z were a virtual prticle, it would not be on its mass shell and
a mass measurement would not be possible by studying the reaction.

My undertanding is that the line corresponding to the Z is the
trajectory of a real Z travelling from its birth to its decay.
It just happens that its life time is exceedingly short.
My understanding is that the s-channel Feynman diagram which produces
the excited Z is simply a conflation of two separate Feynman diagrams,
one in which p+pi interact to produce the Z, followed by one in which
the Z decays.

(It occurs to me to warn readers that the Z of which we speak is not
the Z0 boson, but a nucleon excited state.)

 Quote: The line appears only in the standard perturbative formalism for the computation of the in/out cross section not involving Z, and would not be there if that cross section was computed by another method. That's why it is virtual only.

But it is observed, and its (complex) mass is actually measured.

 Quote: One can also see it by computing the momentum balance and check that the momentum does not lie on the Z's (complex) mass shell, as would be needed for a real Z.

Oh? Sorry, but the Z is no less real than the pi0.
If the pi0 can decay into two photons, then it can also be created
bytwo photons.
Feel free , therefore to draw a Feynman diagram showing a pi0 produced
in the s-channel by an interaction of two photons, and decaying.

They both happen to have imaginary parts to their masses.

--
Franz
"A first-rate laboratory is one in which mediocre scientists can
produce outstanding work"
P.M.S. Blackett
Arnold Neumaier
science forum Guru

Joined: 24 Mar 2005
Posts: 379

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

Eugene Stefanovich wrote:
 Quote: Arnold Neumaier wrote: Eugene Stefanovich wrote: Igor Khavkine wrote: Let me summarize my understanding of renormalization. I hope this time I'll be more clear. I see five historical stages in the development of this subject: AFTER 1949 (Tomonaga-Schwinger-Feynman): It was decided to fix the above problem by adding infinite counterterms to the Hamiltonian. ("Infinite" means that matrix elements of the Hamiltonian on the states formed by creation operators of particles are infinite. No. Infinite only means that the limit where the cutoff goes to infinity does not exist. That's exactly what I meant. At any finite value of the cutoff, both the Hamiltonian and the counterterms are finite. If it were not so, one couldn't do renormalization and get something finite. The problem solved by Tomonaga, Schwinger and Feynman was that they discovered how to produce a well-defined limiting theory for cutoff to infinity which allows to extract finite values for quantities that can be compared with experiment. Agreed. They found how to make the S-matrix finite and accurate in the limit of infinite cutoff. All renormalization until today follows the same pattern, including yours. One does certain formal computations at finite cutoff and at some point where it no longer harms moves the cutoff to infinity, being left with approximate formulas at some (fixed or variable) loop order which no longer contain a cutoff and have finite values. In Tomonaga-Schwinger-Feynman theory the cutoff-dependent counterterms are added to the Hamiltonian. When the cutoff momentum is finite, everything is finite: the Hamiltonian, the counterterms, the S-matrix. However, when the cutoff goes to infinity the conterterms do not have a finite limit. That's what I call "the Hamiltonian is infinite". The S-matrix does have a definite limit, but the Hamiltonian does not.

This is only because you ignore wave function renormalization,
which is of course necessary to get finite matrix elements
for the Hamiltonian.

The limiting Hamiltonian is perturbatively well-defined in the
physical Hilbert space obtained as limit of renormalized Hilbert
spaces at finite cutoff, as the cutoff goes to infinity.
For this, it is best to work in the functional Schroedinger picture.

 Quote: In the Glazek-Wilson theory and in RQD both the Hamiltonian and the S-matrix have finite limits at infinite cutoff. I probably haven't stressed enough that I am always talking about theories in the limit of infinite cutoff.

It is less prone to misunderstanding if you avoid the sloppy use
of 'infinite' and talk about the limiting process.

Arnold Neumaier
Igor Khavkine
science forum Guru

Joined: 01 May 2005
Posts: 607

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

On Tue, 08 Mar 2005 06:49:15 +0000, Eugene Stefanovich wrote:
 Quote: Igor Khavkine wrote: I do not see how you can keep making the claim that there "are no differences between bare and dressed particles." By all conventional definitions, they are different. Perhaps this case is once again, a question of interpreting terminology. However, the conventional meaning has been pointed out to you several times in the past. That's exactly why I emphasised the historical aspect. You are right that we must discuss calculations within a defined framework. Within Tomonaga-Schwinger-Feynman theory and within Glazek-Wilson approach there is a difference between bare and physical particles. Bare particle operators are present in the Hamiltonian, however single bare particle states are not eigenstates of the Hamiltonian. Physical particles are complex linear combinations of multi-bare-particle states. They are eigenstates of the full interacting Hamiltonian. They have experimentally measured masses and charges. These, I believe, are "conventional definitions" valid for the two mentioned theories.

Physical particle states are independent of the formulation you use to
calculate them. Bare particles are usually defined to be eigenstates of
individual non-interacting theories that are used to build up your
interacting Hamiltonian. Again, this is independent of the formulation you
are using. Before you object that bare particles are unphysical, yes, they
are unphysical. No-one ever claimed otherwise. However, they are necessary
in the formulation of the theory and at intermediate calculational steps.
Moreover, the results of QFT calculations make no reference to bare
particles, only physical ones. Bare particles are no more harmful than a
choice of coordinates in Euclidean space when doing a calculation whose
result will in the end be coordinate independent.

 Quote: In RQD, there is no distinction between bare and dressed particles. The particles whose creation and annihilation operators are present in the Hamiltonian are eigenstates of the same Hamiltonian. There is no need for dressing in RQD, because we are working with fully dressed particles already.

See, right in the above paragraph is where your terminology departs from
the conventional one. The particle operators that are present in the
Hamiltonian in no way shape or form defines what a "bare" particle is. The
Hamiltonian can be rewritten in a myriad of different ways with different
particle operators.

Second, your claim that there is no need for dressing in your theory. Once
again, what you mean is different from what your words say. Did you write
down your "RQD" Hamiltonian from thin air? No, then you must have started
with a formulation of QED that pieces together two non-interacting
theories into one interacting one. Therefore you have used bare particles
at least as an initial step to calculate the coefficients in your
expansion of the QED Hamiltonian. Even if you have completely eliminated
references to bare particles from your formulation, that does not give you
the right to claim that (1) you have not used bare particles nor (2) that
bare particles are the same as physical ones. Both of these claims are
false simply by definition.

 Quote: And instead of agreeing with me on this correction, as I've said before, I would rather you change your terminology. Moreover, getting rid of virtual particles is not a virtue of a theory, simply because virtual particles are a name for squiggly lines drawn by theorists on paper while doing QFT calculations. There is no absolute need to draw them in the first place. If you claim to have eliminated virtual particles, then the only thing I can congratulate you on is saving dead trees. You then agree with me that virtual particles are just artefacts of doing quantum field theory a'la Tomonaga-Schwinger-Feynman?

If by "a'la Tomonaga-Schwinger-Feynman" you mean the technique involving
writing diagrams on paper and calling their internal lines "particles",
then yes.

 Quote: You then agree with me that there are no degrees of freedom associated with these artefacts?

This statement is devoid of meaning, hence I can neither agree nor
disagree with it.

 Quote: You then agree with me that during interaction the momentum is transferred between charges instantaneously without any involvement of (virtual) carriers?

No. On the one hand, there are "virtual particles" which are figments of
interpretation. On the other hand, there is a physical effect: particle A
starts wiggling at space-time event (x,t) and particle B starts feeling
the effect of the wiggling at space-time event (x',t'). The space-time
separation between (x,t) and (x',t') is both calculable and measurable.
Obviously, what is one hand cannot have anything to do with what's on the
other.

Moreover, (both classical and quantum) theory predicts the space-time
separation between (x,t) and (x',t') to be time-like. To my knowledge,
violations of effect have never been observed experimentally. No, I do not
agree with you that "superluminal" propagation of light pulses in some
materials is evidence to the contrary.

Igor
Eugene Stefanovich
science forum Guru

Joined: 24 Mar 2005
Posts: 519

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

Igor Khavkine wrote:
 Quote: On Sun, 06 Mar 2005 07:51:28 +0000, Eugene Stefanovich wrote: Igor Khavkine wrote: Let me summarize my understanding of renormalization. I hope this time I'll be more clear. I see five historical stages in the development of this subject: [...history sinpped...] History is useful. But it is not relevant when discussing calculations within a defined framework. AFTER 2001 (RQD): Greenberg-Schweber and Glazek-Wilson ideas are joined together. "Similarity-dressing" unitary transformation is applied to the QED Hamiltonian with counterterms. This is equivalent to full reformulation of the theory (in each perturbation order) through observable dressed particles. The transformed Hamiltonian has finite matrix elements on the dressed particle states. There are no differences between bare and dressed particles. There is no non-trivial vacuum. There are no clouds of virtual particles. There is no longer need for renormalization. The S-matrix is exactly the same (accurate) as in T-S-F and G-W theories. I do not see how you can keep making the claim that there "are no differences between bare and dressed particles." By all conventional definitions, they are different. Perhaps this case is once again, a question of interpreting terminology. However, the conventional meaning has been pointed out to you several times in the past.

That's exactly why I emphasised the historical aspect. You are right
that we must discuss calculations within a defined framework.
Within
Tomonaga-Schwinger-Feynman theory and within Glazek-Wilson approach
there is a difference between bare and physical particles. Bare particle
operators are present in the Hamiltonian, however single bare particle
states are not eigenstates of the Hamiltonian. Physical particles
are complex linear combinations of multi-bare-particle states. They are
eigenstates of the full interacting Hamiltonian. They have
experimentally measured masses and charges. These, I believe, are
"conventional definitions" valid for the two mentioned theories.

In RQD, there is no distinction between bare and dressed particles.
The particles whose creation and annihilation operators are present
in the Hamiltonian are eigenstates of the same Hamiltonian. There is no
need for dressing in RQD, because we are working with fully dressed

 Quote: And instead of agreeing with me on this correction, as I've said before, I would rather you change your terminology. Moreover, getting rid of virtual particles is not a virtue of a theory, simply because virtual particles are a name for squiggly lines drawn by theorists on paper while doing QFT calculations. There is no absolute need to draw them in the first place. If you claim to have eliminated virtual particles, then the only thing I can congratulate you on is saving dead trees.

You then agree with me that virtual particles are just artefacts of
doing quantum field theory a'la Tomonaga-Schwinger-Feynman?
You then agree with me that there are no degrees of freedom associated
with these artefacts? You then agree with me that during interaction
the momentum is transferred between charges instantaneously without
any involvement of (virtual) carriers?

Eugene Stefanovich.
Eugene Stefanovich
science forum Guru

Joined: 24 Mar 2005
Posts: 519

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

Arnold Neumaier wrote:
 Quote: Eugene Stefanovich wrote: Igor Khavkine wrote: Let me summarize my understanding of renormalization. I hope this time I'll be more clear. I see five historical stages in the development of this subject: AFTER 1949 (Tomonaga-Schwinger-Feynman): It was decided to fix the above problem by adding infinite counterterms to the Hamiltonian. ("Infinite" means that matrix elements of the Hamiltonian on the states formed by creation operators of particles are infinite. No. Infinite only means that the limit where the cutoff goes to infinity does not exist.

That's exactly what I meant.

 Quote: At any finite value of the cutoff, both the Hamiltonian and the counterterms are finite. If it were not so, one couldn't do renormalization and get something finite. The problem solved by Tomonaga, Schwinger and Feynman was that they discovered how to produce a well-defined limiting theory for cutoff to infinity which allows to extract finite values for quantities that can be compared with experiment.

Agreed. They found how to make the S-matrix finite and accurate in the
limit of infinite cutoff.

 Quote: All renormalization until today follows the same pattern, including yours. One does certain formal computations at finite cutoff and at some point where it no longer harms moves the cutoff to infinity, being left with approximate formulas at some (fixed or variable) loop order which no longer contain a cutoff and have finite values.

In Tomonaga-Schwinger-Feynman theory the cutoff-dependent
counterterms are
added to the Hamiltonian. When the cutoff momentum is finite, everything
is finite: the Hamiltonian, the counterterms, the S-matrix. However,
when the cutoff goes to infinity
the conterterms do not have a finite limit. That's what I call
"the Hamiltonian is infinite". The S-matrix does have a definite limit,
but the Hamiltonian does not.

In the Glazek-Wilson theory and in RQD both the Hamiltonian and
the S-matrix have finite limits at infinite cutoff.

I probably haven't stressed enough that I am always talking about
theories in the limit of infinite cutoff. That's the only limit in
which one can hope to obtain accurate results.

Eugene Stefanovich.
Arnold Neumaier
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Joined: 24 Mar 2005
Posts: 379

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

Eugene Stefanovich wrote:

 Quote: Arnold Neumaier wrote: I wonder how _you_ predict a relativistic quantum Boltzmann equation for an plasma in an electromagnetic field, which is surely part of QED. No problem, RQD approach has a well-defined finite Hamiltonian in which particles interact via direct instantaneous potentials, just as in ordinary non-relativistic quantum mechanics. This Hamiltonian can be directly plugged into the time evolution operator, or diagonalized to find the stationary states. With this Hamiltonian, you can define the partition function exp(-\beta H) and all kinds of thermodynamical and statistical quantities just as in ordinary quantum statistics.

The Boltzmann equation is not the Boltzmann distribution but a kinetic
equation for the 1-particle phase space density. If you would be able
to reproduce the qantum Boltzmann equation of Klimontovich with your
formalism, I'd be very surprised. He is doing real dynamical work with
QED, and gets manifestly covariant equations that are useful in practice.

Repeat this feat before making again your exaggerated claims!

Arnold Neumaier
Eugene Stefanovich
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Joined: 24 Mar 2005
Posts: 519

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

Arnold Neumaier wrote:

 Quote: I wonder how _you_ predict a relativistic quantum Boltzmann equation for an plasma in an electromagnetic field, which is surely part of QED.

No problem, RQD approach has a well-defined finite Hamiltonian in which
particles interact via direct instantaneous potentials, just as in
ordinary non-relativistic quantum mechanics. This Hamiltonian can be
directly plugged into the time evolution operator, or diagonalized to
find the stationary states. With this Hamiltonian, you can define the
partition function exp(-\beta H) and all kinds of thermodynamical and
statistical quantities just as in ordinary quantum statistics.

However, I do not know if this is really needed to understand the
behavior of plasma in electromagnetic fields. The radiative corrections
provided by QED are normally small. They are not so prominent even
in pure simple systems at zero temperature, such as hydrogen atom
or two-particle scattering. I am wondering if there is any
experimental observations of radiative corrections in plasma behavior
at high temperature? Isn't it sufficient for practical calculations
to use just simple Coulomb
1/r approximation for interparticle potentials?

If there are prominent radiative corrections in the dynamical behavior
of plasma, then there should be a chance to check which dynamical theory
is better: RQD or "closed path formalism".

Eugene Stefanovich.
Frank Hellmann
science forum beginner

Joined: 16 May 2005
Posts: 26

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: Negative energy particles, Mass Gap & 3rd Law

 Quote: Lower energy states have MORE states, rather than less! This is the characteristic property of a system whose temperature is negative.

Precisely. Thus physics is perfectly symmetrical under exchanging all
positive energies with negative energies, however, if you have both in
one and the same system, the most likely state for them to be in is
+inf and -inf respectively.

And as long as we have time reflection invariance we have a setup
symmetric between positive and negative energy states, this suggests
our energy range (The spectrum of the Hamiltonian) must be bounded on
one side.

Frank
Eugene Stefanovich
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Joined: 24 Mar 2005
Posts: 519

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

Igor Khavkine wrote:
 Quote: On Tue, 08 Mar 2005 06:49:15 +0000, Eugene Stefanovich wrote: Igor Khavkine wrote: In RQD, there is no distinction between bare and dressed particles. The particles whose creation and annihilation operators are present in the Hamiltonian are eigenstates of the same Hamiltonian. There is no need for dressing in RQD, because we are working with fully dressed particles already. See, right in the above paragraph is where your terminology departs from the conventional one. The particle operators that are present in the Hamiltonian in no way shape or form defines what a "bare" particle is. The Hamiltonian can be rewritten in a myriad of different ways with different particle operators. Second, your claim that there is no need for dressing in your theory. Once again, what you mean is different from what your words say. Did you write down your "RQD" Hamiltonian from thin air? No, then you must have started with a formulation of QED that pieces together two non-interacting theories into one interacting one. Therefore you have used bare particles at least as an initial step to calculate the coefficients in your expansion of the QED Hamiltonian. Even if you have completely eliminated references to bare particles from your formulation, that does not give you the right to claim that (1) you have not used bare particles nor (2) that bare particles are the same as physical ones. Both of these claims are false simply by definition.

I agree, I do not have any idea how to write the RQD Hamiltonians in
terms of dressed particles from first principles. In order to get to
this Hamiltonian one should go through a painful set of steps:
canonical gauge quantization -> renormalization -> dressing.
On this way you'll meet a bunch of objects which, in my opinion,
have no any relevance to physics. These are bare and virtual particles,
quantum fields, gauges, etc. They play useful mathematical roles at
some stages of the above derivation, but they are just parts of a
3-step mathematical trick. This mathematical trick
is designed to guess the correct form of the Hamiltonian, and it
reaches this goal, however the physical significance of this procedure
remains obscure to me. IMHO physics starts AFTER we wrote the correct
(dressed particle) form of the Hamiltonian. Everything we did BEFORE
that (Lagrangians, action, quantum fields, bare particles, gauges, etc.)
is no longer relevant after the Hamiltonian is written. With this
Hamiltonian we can answer any question having relevance to observations.

You are right, I do need bare particles BEFORE the correct Hamiltonian
is written. I need them to go through the 3 steps mentioned above.
I hate to do that, but I do not have other choice at the moment.
My point was that I do not need bare or virtual particles AFTER
the RQD Hamiltonian is written.

There is more about renormalization and dressing in my recent paper
hep-th/0503076

 Quote: And instead of agreeing with me on this correction, as I've said before, I would rather you change your terminology. Moreover, getting rid of virtual particles is not a virtue of a theory, simply because virtual particles are a name for squiggly lines drawn by theorists on paper while doing QFT calculations. There is no absolute need to draw them in the first place. If you claim to have eliminated virtual particles, then the only thing I can congratulate you on is saving dead trees. [...] You then agree with me that there are no degrees of freedom associated with these artefacts? This statement is devoid of meaning, hence I can neither agree nor disagree with it. You then agree with me that during interaction the momentum is transferred between charges instantaneously without any involvement of (virtual) carriers? No. On the one hand, there are "virtual particles" which are figments of interpretation. On the other hand, there is a physical effect: particle A starts wiggling at space-time event (x,t) and particle B starts feeling the effect of the wiggling at space-time event (x',t'). The space-time separation between (x,t) and (x',t') is both calculable and measurable. Obviously, what is one hand cannot have anything to do with what's on the other.

If you say that interaction is retarded, then, in order to comply with
conservation laws (for momentum and angular momentum, in particular)
you should admit that there is some time interval during which the
momentum transferring between two interacting particles is en route
between the particles. There should be some material carrier which
takes this momentum from one particles, keeps it for some time,
and then gives it to the other particle. There should be some "degrees
of freedom" associated with this carrier. You can call this carrier
"virtual particles" or "fields", I do not care. The important thing is
that there should be some material stuff which can possess momentum,
angular momentum, energy, etc, but which so far has escaped direct
observations. The whole point of the RQD approach is to reformulate
the theory in such a way that this mysterious stuff is not present
in the theory just as it is not seen in experiments.

 Quote: Moreover, (both classical and quantum) theory predicts the space-time separation between (x,t) and (x',t') to be time-like.

I disagree that QED predicts anything about the speed of propagation of
interaction. There are statements about the commutativity of fields
at spacelike intervals in QED, but I have no idea how this is related
to the
speed of propagation of interaction. Probably there is a gap in
my education, but I haven't seen a solution of a dynamical
(time-dependent) problem relevant to the propagation of interaction
between charges based on the QED Hamiltonian. Arnold Neumaier
claims that such solutions exist within the "closed path" formalism, but
I haven't found anything satisfactory yet. In my view QED is good
only for predictions of the S-matrix which disregards any time
dependence by definition and cannot say anything about the speed of
interactions.

 Quote: To my knowledge, violations of effect have never been observed experimentally.

To my knowledge, the retarded character of electromagnetic interactions
has not been confirmed by experiment either.
(I am not talking here about the propagation of EM waves, of course)

Eugene Stefanovich.
Eugene Stefanovich
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Joined: 24 Mar 2005
Posts: 519

Posted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles?

Arnold Neumaier wrote:

 Quote: All renormalization until today follows the same pattern, including yours. One does certain formal computations at finite cutoff and at some point where it no longer harms moves the cutoff to infinity, being left with approximate formulas at some (fixed or variable) loop order which no longer contain a cutoff and have finite values. In Tomonaga-Schwinger-Feynman theory the cutoff-dependent counterterms are added to the Hamiltonian. When the cutoff momentum is finite, everything is finite: the Hamiltonian, the counterterms, the S-matrix. However, when the cutoff goes to infinity the conterterms do not have a finite limit. That's what I call "the Hamiltonian is infinite". The S-matrix does have a definite limit, but the Hamiltonian does not. This is only because you ignore wave function renormalization, which is of course necessary to get finite matrix elements for the Hamiltonian. The limiting Hamiltonian is perturbatively well-defined in the physical Hilbert space obtained as limit of renormalized Hilbert spaces at finite cutoff, as the cutoff goes to infinity. For this, it is best to work in the functional Schroedinger picture.

Maybe I should be even more precise: The Hamiltonian in T-S-F theory
is a polynomial function of creation and annihilation operators of
(bare) particles. In the limit of infinite cutoff the coefficients
of the terms in this polynom do not tend to finite values.
(I simply say that they are infinite). RQD approach suggests to
apply a unitary transformation to this Hamiltonian (you can also
consider this as a unitary redefinition of particles, i.e., the
transition from bare to dressed particles) which makes the coefficients
in the Hamiltonian finite (in the limit of infinite cutoff).

This is discussed at length in my recent article hep-th/0503076
where I use a simple model of QFT as an example. I would appreciate

 Quote: In the Glazek-Wilson theory and in RQD both the Hamiltonian and the S-matrix have finite limits at infinite cutoff. I probably haven't stressed enough that I am always talking about theories in the limit of infinite cutoff. It is less prone to misunderstanding if you avoid the sloppy use of 'infinite' and talk about the limiting process. I'll try to be less sloppy in the future. You can change

the word 'infinite' in my previous posts to the
'does not have a finite value in the limit of infinite cutoff'.

Eugene Stefanovich.

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