EVERY QFT text I've ever examined states that if there is an external vector potential, $A_\mu$, then one writes the Dirac eq.(or Klein-Gordon eq.) using a covariant derivative to include this U(1) gauge field, $A_\mu$. No problem. Wikipedia states that one must also include the contribution due to the electron self-field, $A_\mu'$. Again, makes perfect sense. Thus if there is no external 4-vector potential, the Dirac equation should exhibit the internal one! However, all texts state the "free" Dirac equation, without an accompanying vector potential.

If I now solve the free Dirac equation, the spinorial solutions can be used to construct the Dirac charged current. Using this current as the source, I can then solve the classical D'Alembertian eq. for the corresponding internal self-field $A_\mu'$.

If I insert $A_\mu'$ back into the Dirac equation, the solution must be different than those obtained from the free Dirac equation. I am simply wondering what the interpretation of the solutions will be now?


The problem is that this is not the right way to solve the Dirac equation interacting with an electromagnetic field. The method you are using assumes there is a classical field around the single-particle Dirac electron, and uses this field to find the motion of the electron. This is a meaningless approximation. The field produced by an electron is entangled with the electron, so if you have an electron bouncing around in an external potential, and it emits a photon, the photon tells you where the electron is, and collapses the wavefunction of the electron partially from the information in the photon. This collapse is a sign of entanglement.

In order to describe the self-field properly, you don't use a classical field, you need to quantize the electromagnetic field. In this case, you find the usual Feynman expansion for quantum electrodynamics, and the self-field calculation that you are doing corresponds to an electron emitting and absorbing the same photon.

The effect of this diagram is to alter the charge and mass of the electron from the bare values, and this is usually absorbed into the perturbation series by adding counterterms. The result is an electron field whose travelling waves obey the uncorrected Dirac equation.

It might be possible to interpret the self-energy in terms of a Dirac equation moving in its own generated electromagnetic field, I don't know, but it wouldn't be physically right--- the right thing is the Feynman diagram, which is simple anyway.


I agree with the core of Ron Maimon's answer. Indeed, the electron's self-field is properly taken into account in quantum electrodynamics and leads (among other things) to renormalization of charge and mass. As a result, the Dirac equation with renormalized mass and charge, but without electron self-field, turns out to be a very good approximation.

However, that does not necessarily mean that it would be wrong "to interpret the self-energy in terms of a Dirac equation moving in its own generated electromagnetic field". This approach was taken by Barut in his "self-field electrodynamics" and published in dozens of articles, including several articles in Phys. Rev. Unfortunately, I don't have time to find references to his journal articles, but a review and all references can be found in A.O. Barut, "Foundations of Self-Field Quantumelectrodynamics", in: "New Frontiers in Quantum Electrodynamics and Quantum Optics", Ed. by A.O. Barut, NATO ASI Series V.232, 1991, p. 358 (or around this page). Barut claims that his results are very close to those of quantum electrodynamics, so his approach may have some sound basis. However, his work remained an unfinished business after his death. I am not making any judgement here on the validity of his approach, just wanted to emphasize that there may be some open questions there.

  • $\begingroup$ I mostly agree (although I didn't read Barut, and it is easy to fudge these types of calculations and get the right answer by wrong reasoning). I really am not sure how much of QED you can reproduce from this type of thing in this case, Dirac field plus self-field, which is why I didn't say it was wrong. Feynman in his "quantum electrodynamics" treats all the electromagnetic processes are semiclassical, and if you do the right tricks, you can get the full Feynman expansion. But I never liked these half-way measures, especially in cases where you can get the full right answer in simple ways. $\endgroup$ – Ron Maimon Aug 21 '12 at 7:29
  • $\begingroup$ @Ron Maimon: Well, you did say that "this is not the right way to solve the Dirac equation interacting with an electromagnetic field" and that "It might be possible to interpret the self-energy in terms of a Dirac equation moving in its own generated electromagnetic field, I don't know, but it wouldn't be physically right":-) But again, this is an open question, and you may be right thinking that this is a dead end. As for getting "the full right answer in simple ways", QED is not "simple ways" - it took humanity at least 20 years and innumerable man-hours to develop it:-). $\endgroup$ – akhmeteli Aug 21 '12 at 10:27
  • $\begingroup$ @Ron Maimon: Let me just explain why there might be a possibility that QED is overcomplicated. As nightlight noticed, there is an off-the-shelf mathematical trick (an extension of the Carleman linearization) that embeds a system of partial differential equations into a quantum field theory (see, e.g., my article in Int'l Journ. of Quantum Information ( akhmeteli.org/akh-prepr-ws-ijqi2.pdf ), the end of Section 3. There is also a substantially updated version at arxiv.org/abs/1111.4630, where the case of the Dirac-Maxwell electrodynamics we discuss now is treated much better. $\endgroup$ – akhmeteli Aug 21 '12 at 10:50
  • $\begingroup$ I know QED can be simplified into semiclassical stuff for the photon, but that's because the photon field is free. Feynman did this by integrating out the photon field in the path integral, and then he could interpret all the photon vertices as semiclassical emission and absorption in the appropriate single-quantum limit. I don't care for these tricks, as it is clear that the physically correct thing is QED, and so much so that I won't bother looking at alternatives, as time is limited. $\endgroup$ – Ron Maimon Aug 21 '12 at 15:58
  • $\begingroup$ Thanx RM & AK. However, my gut feeling is that 2nd quantization is Not mandatory to answer my question. In particular, lets focus on the Klein-Gordon eq. It should be possible to view it as a scalar field eq., in which a charged meson say, is propagating in a region Devoid of Other external fields. We neglect virtual pair creation/annihilation & any other QFT concerns. Now my original question remains. If I insert the obtained self-field Au' back into the KGE & solve it, the solution Must now be different from that obtained w/the `Free KGE'. $\endgroup$ – user14071 Oct 15 '12 at 14:10

Perhaps read work by Asim O Barut to understand the problem more fully. You certainly can describe a single electron-positron-photon system with a self-coupled "classical" field.

The reason why this is not done is rather simple.

It calls into question the entire foundation of conventional Quantum Field Theory since the given treatment is not consistent with the assumption of point-like particles.

However, the given treatment is completely consistent with the interpretation of wavefunctions as a material matter wave as first proposed by Erwin Schroedinger in his last paper of 1926.

The fact that most physicists know nothing of the above is a tribute to the Education System.

The fact that Barut and co-workers got the right answers to order alpha, but that nobody paid attention, suggests (to me) that Academia is not the place to go for new knowledge.

The fact that the given system of equations lies within the nonlinear quantum mechanics of Steven Weinberg, and that he did not pay attention, suggests (to me) that some folks may not actually be as smart as they first appear. Who knows? Life is like that.


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