### How we can improve QED?

Let us now count arguments for and against the currently accepted form of the renormalized QED. We start with positive things:

- QED is relativistically invariant
- QED is cluster separable
- The S-matrix of QED perfectly agrees with experiment.

Now the negatives:

- The Hamiltonian of QED contains infinite counterterms
- QED interaction leads to unphysical processes of creation and absorption of extra particles from vacuum and 1-particle states

Is it possible to reformulate the theory so as to keep the positive stuff and get rid of the negative? Yes. The answer is here. The Hamiltonian of QED is apparently wrong, but according to Ekstein, we can modify the Hamiltonian without changing the S-matrix (i.e., without changing the agreement with experiment). This can be done by a unitary transformation. Moreover, if we are careful enough, we can choose this transformation such that both the relativistic invariance and the cluster separability are not affected. Now, the question is, whether one can simultaneously eliminate bad properties 1. and 2? And the answer is, again, 'yes'! One can find a unitary transformation of the Hamiltonian such that all interaction terms become finite, and all terms that act non-trivially on the vacuum and 1-particle states are eliminated. In one shot we kill two rabbits: with the new Hamiltonian the renormalization is not needed anymore and we have a clean physically satisfactory definitions of the vacuum and 1-particle states. This is called the "unitary dressing transformation". More about it in another post.

**Update**So, the conclusion is that the QED Hamiltonian is simply wrong and must be substituted by another Hamiltonian. Why did we use a wrong Hamiltonian for almost 80 years? One reason is in three "good" properties I listed above. This is a respectable reason. Another reason is that the Hamiltonian of QED was derived from correspondence with classical Maxwell's theory of electromagnetism. This is, actually, not a good reason. As we discussed here , Maxwellian description of radiation by a continuous wave is not accurate. There are numerous other flaws in Maxwell's theory, one of them being inadequate description of the "radiation reaction" force. So, derivation of quantum electrodynamics by "quantizing" classical theory is, at best, a heuristic trick.

## 12 Comments:

I agree with you and that is why I propose another Hamiltonian in my "Reformulation instead of Renormalizations" (RiR). The standard QED difficulties are due to self-action, in my opinion. This kind of "interaction" term never worked flawlessly. It always needed some "counter-terms" to get rid of bad physical and mathematical features intrinsic to it.

In my "Atom as a "Dressed" Nucleus" ("Atom" for short) and in RiR I proposed a physical and mathematical solution for a dressed electron based on "interaction" insight.

It would be nice to learn your opinion about my proposal, privately or via your blog, whatever.

Best regards,

Vladimir Kalitvianski

Hi Vladimir,

We have discussed your approach quite extensively in the past. Let me summarize my opinion here. (I must admit that I wasn't able to follow exactly your logic. There could be two explanations for that. First is my ignorance and laziness. Second, perhaps you wasn't able to explain things clearly enough. So, please forgive me if I misunderstood your ideas.)

It seems that the major claim of your work is that the "Novel QED" must be based on the Hamiltonian (60) from arXiv:0811.4416. Two major questions come to mind: 1) How this Hamiltonian was derived? 2) Can this Hamiltonian describe known experimental facts?

In answering the first question, I understand that you think that the physical electron must be considered as a kind of bound state of one electron and attached electromagnetic field (photons). So, it forms some kind of complex system that you call "electronium". Then you compare the electronium with two classical particles connected by a spring or with an atom, and try to demonstrate the idea of renormalization on these examples. Personally, I don't understand the meaning of these analogies. It looks like a lot of handwaving to me. I am also not convinced how these analogies justify the choice of your Hamiltonian ansatz (60). Moreover, your Hamiltonian is written in terms of quantities that you never defined: pi_c, eta_c, u_c, j^0, etc.

Ok, let us assume that you just guessed your Hamiltonian. To show that this guess has some value, you must demonstrate that it can reproduce at least some well-known results. To get some confidence I would like to see a few simple low-order calculations with your Hamiltonian: scattering cross-sections, energy levels of the hydrogen atom, electron's g-factor, etc. If your guessed Hamiltonian can reproduce those, then the lack of rigorous derivation can be forgiven. Unfortunately, you do not provide any such calculations.

So, both questions 1) and 2) remain unanswered for me. Perhaps you have a good idea, but in my opinion you need to do extra work in order to convince me that your approach is a better alternative to QED.

This comment has been removed by the author.

[ES]I understand that you think that the physical electron must be considered as a kind of bound state of one electron and attached electromagnetic field (photons). So, it forms some kind of complex system that you call "electronium".[/ES]

Yes, that's right.

[ES]Then you compare the electronium with two classical particles connected by a spring or with an atom, and try to demonstrate the idea of renormalization on these examples.[/ES]

No. First I consider classical particles and atoms in order to introduce the Novel QED's Hamiltonian (60). The latter comes last, not first.

[ES]Personally, I don't understand the meaning of these analogies. It looks like a lot of handwaving to me. I am also not convinced how these analogies justify the choice of your Hamiltonian ansatz (60).[/ES]

OK, let us forget for a moment the analogies. Did you find any error or lack of logic in the atomic and classical particle example expositions? It is extremely important to follow the physics and mathematics of the above examples to understand the Novel QED Hamiltonian derivation. I start with them on purpose, they cannot be skipped without missing the point.

[ES]Moreover, your Hamiltonian is written in terms of quantities that you never defined: pi_c,

eta_c, u_c, j^0, etc.[/ES]

They are the same as in Weinberg and other Coulomb gauge formulation sources. I just grouped the oscillator Hamiltonians together with the charge Hamiltonians and removed the self-action term [B]jA[/B].

I presented some non-relativistic estimations (scattering and the Lamb shift) and references to T. Welton's detailed calculations. He also obtained a good absolute value of the anomalous magnetic moment but of the opposite sign. It was because he did not take into account "the negative frequency" components in the mu estimations. So everything works fine in these estimations: no divergences and renormalizations, and good preliminary estimations.

I still hope that you will take time to read with a pencil my papers to understand my motivations and insights. It is not so difficult, believe me. At least, I tried hard to remove all unnecessary "too many particle" effects that shadow the essence of the problem.

Best regards,

Vladimir Kalitvianski.

[VK] OK, let us forget for a moment the analogies. Did you find any error or lack of logic in the atomic and classical particle example expositions? It is extremely important to follow the physics and mathematics of the above examples to understand the Novel QED Hamiltonian derivation. I start with them on purpose, they cannot be skipped without missing the point. [VK]

I admit that I haven't followed your classical example with pen and paper. The main reason is that I just don't believe that renormalization in QFT has any resemblance to properties of classical systems of particles. In my opinion the true origin of renormalization difficulties in QFT is a self-interaction caused by the presence of particle-number-changing terms in the Hamiltonian. In the classical two-particle model the number of particles does not change, so I just don't see the point of the analogy. Since I don't understand the justification of this classical model, I quickly lose interest in repeating your classical calculations.

On the other hand, I am willing to skip all your analogies and heuristic arguments, and fully accept your Hamiltonian if you demonstrate that you can consistently get good physical results with this Hamiltonian. Unfortunately, this part of your theory is not well-explained either.

[VK][[Ingredients in the Hamiltonian]] are the same as in Weinberg and other Coulomb gauge formulation sources. I just grouped the oscillator Hamiltonians together with the charge Hamiltonians and removed the self-action term [B]jA[/B]. {VK]

This is not clear from your text at all. Since this is the central issue of your approach, I would recommend you to spend more efforts in describing all components in your Hamiltonian in a detailed and explicit fashion. Even if these description would repeat some portions of the Weinberg's book...

[VK] I presented some non-relativistic estimations (scattering and the Lamb shift) and references to T. Welton's detailed calculations. He also obtained a good absolute value of the anomalous magnetic moment but of the opposite sign. It was because he did not take into account "the negative frequency" components in the mu estimations. So everything works fine in these estimations: no divergences and renormalizations, and good preliminary estimations. [VK]

Agreement with experiment is the most important selling point of any theory. Your brief statements and reference to Welton are not sufficient. It is very important to work out a few explicit examples in which well-known observable properties are calculated directly from your Hamiltonian (60).

You have done an impressive piece of work, but it would be helpful to improve your presentation style and fill in more important details to make your approach more convincing.

I completely agree with you that my works have not been finished yet. I strive to complete the relativistic calculations.

At the current stage I would like to discuss my concept, not the final numbers.

Yes, it is possible to reduce the problem of renormalizations to a classical mechanical problem of two particles. It is not connected with the particle number change in course of interactions but related to the wrong "self-action" ansatz for so called "welded" systems in the so called "mixed variable formulation". In other words, perturbative or exact corrections to the fundamental constants are legitimate in the mixed variable formulation if the system is assembled from known pieces and they are not legitimate for "welded" systems. It is interesting to see, believe me. You loose an opportunity to get the right insight with not reading my exposition. It is not so complicated mathematically and physically. In fact, it is rather elementary. What is highly unexpected is that we factually deal, since long-long ago, with "welded" compound systems but have not noticed it yet. My classical mechanical example is an easy introduction to this point of view.

Then my Hamiltonian (60)is a straightforward generalization to many-particle (QED) case.

Vladimir.

Now, I think, we understand each other's positions well. Let me know when your relativistic calculations are complete.

Does that mean you will not read "RiR" to get a gist of it? Of course, you are free not to read it. But I am afraid that after my completing (successfully) the relativistic calculations, you will need the Hamiltonian (60) derivation and/or justification, and the problem returns to "RiR". Why not start from it right now?

Again, I do not insist. I am just trying to share my findings with peoples who are the most disposed to accept a "dressed" particle approach.

Vladimir.

Hi Vladimir,

there are few things that I would like to know before forming any opinion about your theory:

1) Can you derive your Hamiltonian from the well-established Hamiltonian of QED by a sequence of rigorous steps (not by analogies and handwavings)?

2) Can you prove that your theory is relativistically invariant?

3) Can you prove that your Hamiltonian leads to the same S-matrix as obtained in QED and confirmed by experiments?

So far, I don't see answers to these questions in your papers.

1) No, I can't since it is impossible. The standard QED Hamiltonian contains self-action which is only removed with the renormalizations in course of calculations. I do not follow this long way. I write directly the Novel QED Hamiltonian without self-action and without renormalizations.

2) Hamiltonian formulations in the Coulomb gauge is relativistically invariant, it is already proven. In particular, the four-fermion Coulomb interaction is itself relativistically invariant (can be written explicitly as such).

3) No, it leads to the zero-values S-matrix if one considers elastic or inelastic processes with finite number of final photons. It is the inclusive cross section that coincides with the standard QED inclusive picture (but in my approach it is obtained naturally, from the very beginning).

4) In the standard QED you have all you require. Are you happy with the standard QED?

Vladimir.

1) No, I can't since it is impossible. The standard QED Hamiltonian contains self-action which is only removed with the renormalizations in course of calculations. I do not follow this long way. I write directly the Novel QED Hamiltonian without self-action and without renormalizations.OK, so your Hamiltonian is just an heuristic guess. This is not bad, but it places an extra burden on you to clarify points 2) and 3)2) Hamiltonian formulations in the Coulomb gauge is relativistically invariant, it is already proven. In particular, the four-fermion Coulomb interaction is itself relativistically invariant (can be written explicitly as such).By relativistic invariance I mean Poincare commutation relations between interacting operators of energy and boost. In the case of traditional QED the proof is not trivial (see Appendix N in my book and Weinberg's papers cited there). You haven't even written your interacting boost operator, so the proof of the relativistic invariance is still on you.3) No, it leads to the zero-values S-matrix if one considers elastic or inelastic processes with finite number of final photons. It is the inclusive cross section that coincides with the standard QED inclusive picture (but in my approach it is obtained naturally, from the very beginning).Prove it. Take one simple example and perform all calculations from the beginning to the end. What about the hydrogen atom? Can you get the spectrum in your appraoch?4) In the standard QED you have all you require. Are you happy with the standard QED?As I wrote in the original posting, in QED I am not satisfiedwith infinite counterterms and with "unphys" interaction terms. These defects in the original QED Hamiltonian can be removed with step-by-step "unitary dressing" procedure. There is a proof of relativistic invariance of the resulting theory, and there are quite a few experimental predictions that can be extracted from the "dressed particle" QED and compared with experiment. See for example chapter 11 in the latest version (v11) of the book, where I make transition to classical electrodynamics.

>OK, so your Hamiltonian is just an heuristic guess.

No! It is not even a scientific guess!

It is a strict construction based on the energy-momentum conservation laws. It is unique, if you like. To simplify its understanding I reduced the problem to two interacting particles where no ambiguities are possible. You underestimate my efforts spent on introduction of the Hamiltonian (60) and the notions implemented in it.

>What about the hydrogen atom? Can you get the spectrum in your approach?

Of course, I can. It is outlined in "Atom..." as well as in T. Welton's paper (and in many textbooks).

I have all versions of your book as well as Weinberg's books.

As I said, I am going to fulfil the complete set of calculations.

Here I just wanted to learn your opinion about my "dressed" electron, but since you are still unaware of it, I quit.

Vladimir.

Post a Comment

<< Home