In QED, the fine structure constant $\alpha$ runs upwards in the UV, with a loop calculation (involving a geometric series of the vacuum polarisation diagram) indicating a divergence in $\alpha$ at $\sim 10^{286}\,\text{eV}$. It is often claimed (see, for instance, Schwartz, QFT and the Standard Model, section 21.2) that this means QED is an incomplete theory at high energies, or that it is not predictive at these energies, and that some UV completion is required.

However, QCD is another theory with a Landau pole (in the IR this time), at $\sim100\,\text{MeV}$. Neverthless, QCD is a theory valid down to arbitrarily low energies; it is merely non-perturbative in this regime. My understanding is that the Landau pole is an artefact of extrapolating a perturbative calculation of the coupling strength $\alpha_s$ into the non-perturbative regime. In fact, there is no divergence in $\alpha_s$, although explicitly calculating it is impossible (or perhaps not even meaningful) with current tools and understanding.

Therefore, whilst perturbation theory clearly breaks down in QED at very high energies, is it not possible that QED is a perfectly legitimate and consistent theory up to arbitrarily high energies, in much the same way that QCD is at low energies? Is the QED Landau pole really there?

Said another way, is there really any link between "the point at which perturbation theory breaks down" and "the point at which the theory stops being predictive"? Perhaps these are linked when we're working with an EFT with infinitely many terms whose coefficients are unconstrained, but if we postulate the QED Lagrangian as fundamental, is it not, at least in principle, predictive up to arbitrarily high energies?

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    $\begingroup$ Isn't it an ill-defined question? There could be multiple nonequivalent nonperturbative formulations of QED which agree with perturbative QED in the IR and behave differently in the UV. $\endgroup$ – Prof. Legolasov Sep 30 '17 at 14:17
  • $\begingroup$ The question here might be of some interest to you: physics.stackexchange.com/questions/486276/… $\endgroup$ – SRS Jan 2 '20 at 6:42

You are completely correct that the perturbative calculation of the Landau pole can't be trusted, as it will clearly become invalid long before the putative pole is reached. The only method that we know of that can give accurate predictions for the high-energy behavior of QED is numerical simulation. According to https://arxiv.org/abs/hep-th/9712244 and http://www.sciencedirect.com/science/article/pii/S092056329700875X, numerics suggests that QED is indeed quantum trivial (i.e. $e$ always renormalizes to zero for any choice of bare coupling), but not because of a Landau pole, which is the usual explanation. Instead, chiral symmetry breaking kicks in before the Landau pole is reached. So there is no Landau pole at high energies, but there is a different phase transition that causes QED to break down.

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    $\begingroup$ Thanks for the answer. Do you have any knowledge of why the presence of chiral symmetry breaking renders the theory quantum trivial? In QCD it doesn't, of course. $\endgroup$ – gj255 Sep 27 '17 at 17:28
  • $\begingroup$ @gj255 Nope, afraid not. $\endgroup$ – tparker Sep 27 '17 at 19:01

The IR Landau pole in QCD doesn't render the theory inconsistent, but it is a hint of serious trouble. It's a harbinger of confinement: It's telling you that, absent a miracle, the interactions between quarks and gluons at low energies are so intense that their correlation functions will cease to be well-defined when one tries to separate them by more than this scale. There are composite operators in QCD which make sense below the confinement scale, but they're necessarily complicated combinations of quarks and gluons, like glueballs and hadrons.

The same problem occurs in QED in reverse. Absent a mathematical miracle, the correlation functions between electrons will fail to be defined if you bring them too close together. You might get lucky and discover that there are analogues of the hadron operators that make sense to arbitrarily high energy. But this would be a mathematical curiousity: You'd have a QFT where the elementary fields emerged from their composites at low energies.. It's hard to imagine such a theory could be unitary.

  • $\begingroup$ Whatever device that prevents (zeno effect?) the string quark-antiquark breakdown as the quarks are separated will keep the energy scale up I assume? $\endgroup$ – Kevin Kostlan Sep 27 '17 at 21:09
  • $\begingroup$ @KevinKostlan I'm not sure I understand your question. $\endgroup$ – user1504 Sep 28 '17 at 0:31
  • $\begingroup$ Thanks for the answer. Are you suggesting that the fact QCD remains well-defined at low energies is somehow very surprising? $\endgroup$ – gj255 Sep 28 '17 at 2:11
  • $\begingroup$ I don't think that's surprising it's well-defined. But it's not obvious (to me, anyways) that it's non-trivial. Why should there be anything that survives below the confinement scale? That's the part that's interesting. $\endgroup$ – user1504 Sep 28 '17 at 3:00

This question posed in the title has only one possible answer. The theoretical Landau pole is not a problem in the theoretical Standard Model, nor in the experimentally achievable physics, because from a certain GeV energy range upwards the effects of electroweak interaction cannot be ignored.

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    $\begingroup$ This answer misses the point of the OP's question - they are asking whether QED is a logically self-consistent theory at high energies, not about how it gets modified by QCD and the weak force in the real world. $\endgroup$ – tparker Sep 27 '17 at 17:07
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    $\begingroup$ The landau pole still exists in the U(1)hypercharge of the electroweak theory, even after embedding QED as far as I’m concerned. This is usually interpreted as a harbinger for the need of a larger unified yang mills (hence GUT model building), among other properties. Grand unification sets a cutoff, preventing physical effects of the landau pole (so long as the GUT cutoff is below the pole). Most would agree when I say the standard model lacks UV completeness in large part due to the landau pole in U(1), so how you think this doesn’t affect the standard model is beyond me. $\endgroup$ – Alexander Nordal Jan 28 '20 at 0:31

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