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I was reading the following book/series of lectures located here http://isites.harvard.edu/fs/docs/icb.topic521209.files/QFT-Schwartz.pdf; it's an introduction to quantum field theory. I'm at the part about vacuum polarization in QED, regularization, and renormalization.

On page 179 the author gives the one-loop corrections to the effective electron's electric charge as a function of the momentum (they also state it as the effective fine structure constant/effective momentum dependent potential, etc). Here's my question; a little bit later on the author says that the loop corrections to the effective charge/running fine structure constant/etc are known to all orders. This would imply that the photon propagator is known to all orders in perturbation theory (this is also implied by the diagram that they've drawn on the same page). So the question is; is this accurate? Is the EXACT photon propagator in quantum electrodynamics?

EDIT: I'd just like to clarify what exactly I am asking. I am not asking if QED is a complete physical theory. It most certainly is not! A Landau pole exists at arbitrarily high energies (absurdly high energies; before posting this question I searched for similar questions and I remember reading something about the Landau pole in QED being in excess of 100 orders of magnitude larger than the plank energy), and more to the point at energies far lower than the Landau pole you ought to use electroweak theory instead. What I am asking is whether the propagator for the photon is known to ALL orders in perturbation theory (one loop corrections, two loop corrections, three loop corrections, four loop corrections, etc). I am not asking whether or not such a propagator is part of a physically complete QFT.

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At the bottom of p.178 of the reference you cite, there is a bold, framed statement:

QED has a Landau Pole: perturbation theory breaks down at short distances

which, if I understand it correctly, would imply a negative answer to your question. The basic idea seems to be that the effective (renormalized) coupling constant obtained from perturbative calculations diverges at a finite (albeit very large) energy. But perturbation theory assumes a small coupling, so it clearly cannot apply up to there: it predicts its own breakdown. In particular, this would mean no perturbative QED result can be strictly said to be "exact": its domain of validity will always be limited to low enough energies.

This does not mean that QED itself is necessarily doomed, it could be an artifact of the perturbative expansion (and the wikipedia page linked above mentions that non-perturbative numerical computations indeed appear to suggest so).

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  • $\begingroup$ Hey. I'm really sorry Luzanne. I didn't mean to edit your comment, I meant to edit mine! Not sure how I managed that. If anyones looking at this to approve the edit please don't. My apologies Luzanne! $\endgroup$
    – chuckstables
    May 18, 2017 at 1:22
  • $\begingroup$ Ok, thanks for the clarification, I was a bit confused what to do with it. Knowing what happened, I have rejected it, so no worry that someone else would accept it. $\endgroup$
    – Luzanne
    May 18, 2017 at 1:24
  • $\begingroup$ Thanks :). I didn't even realize that I could edit others comments. Apparently that's something I can do. $\endgroup$
    – chuckstables
    May 18, 2017 at 1:26
  • $\begingroup$ @chuxley No worries :) Regarding your point, I would make a distinction btw a theory being physically complete (no theory is, really), a theory being mathematically consistent (QED might be, although it has not been proven yet), and the question whether a certain calculation gives the exact result in a given (hopefully mathematically consistent) theory. I understand Landau pole as meaning that no perturbative calculation can give an exact result in QED. $\endgroup$
    – Luzanne
    May 18, 2017 at 1:38
  • $\begingroup$ @chuxley Then there is also the question of resuming: even if you have the terms order by order, it doesn't mean that there sum converges, in fact it is generally accepted that it won't. But I remember vaguely from my QFT course that there are clever way of "resuming" to go around this issue. Not sure about the status of this in QED though... $\endgroup$
    – Luzanne
    May 18, 2017 at 1:44
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I think yes for your question. As far as I understand Quantum electrodynamics is the most complete and perfect part of Standard Model. Here you don't need extra branches or in other words the electrodynamics does not require a phenomena to be seen through some other particles than photons. On the other hand in QCD a process can be seen through many different paths and particle combinations, so we need to discuss branching ratios. This leads us to find more and more perturbative solutions. So photonic propagator is exact.

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  • $\begingroup$ As an aside; isn't QED an incomplete theory? I think the actual theory that involves electromagnetism in the standard model is the electroweak theory, and it uses a completely different lagrangian density, though at lower energies it behaves like QED in many ways. (I think it unifies the weak and electromagnetic forces). $\endgroup$
    – chuckstables
    May 17, 2017 at 6:41
  • $\begingroup$ Then....What is the answer of question posted above? $\endgroup$
    – Sami Khan
    May 17, 2017 at 12:16
  • $\begingroup$ @chuxley Hmm, by this standard, no theory could ever be deemed complete right? Because we don't have a full theory of quantum gravity (yet...) $\endgroup$
    – Luzanne
    May 17, 2017 at 16:19
  • $\begingroup$ That's a good point! What I should've said was that I don't think that QED is TECHNICALLY part of the standard model, although I completely agree with Sami that the predictions made by QED are probably the most precise and well verified that have ever been made by a theory (magnetic moment of the electron is a great example). $\endgroup$
    – chuckstables
    May 18, 2017 at 1:25
  • $\begingroup$ The people who are spending a lot of effort on the construction of gamma-gamma colliders will disagree that "other particles are not needed". If all you care about are photons, then you are done after Maxwell. Since photons can't interact with each other or anything else in the simplified theory, there really is no physics here. $\endgroup$
    – FlatterMann
    May 31 at 13:58

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