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For context, I watched PBS Spacetime's video on virtual particles (link goes to relevant timestamp) where they say that virtual particles aren't mathematically necessary, because the lattice version of nonperturbative QFT doesn't use them, and yet still makes all the same predictions as perturbative QFT. I was satisfied with that, until I had a brief exchange with someone in the comments of this answer where he says that, in most cases, it's impossible to actually do computations in nonperturbative QFT, and, when I asked if this was just due to not having sufficiently efficient algorithms, he said

Note that in particular that even establishing the existence of a non-perturbative Yang-Mills QFT (which is what the standard model / QCD / QFD are) is a millennium problem.

which implies that we don't currently even have a nonperturbative version of the standard model and that it's unclear whether one exists. However, PBS Spacetime is, in my experience, typically a reliable source for high-level explanations, so I wouldn't have expected them to mention nonperturbative QFT as the reason not to think virtual particles are physical if that theory wasn't actually useful for nontrivial calculation. Is it that most physicists think there probably is a nonperturbative version of the standard model and it just hasn't been discovered/created yet?

Qmechanic
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Mikayla Eckel Cifrese
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1 Answers1

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The current non-existence of a solution to the Yang-Mills millenium problem doesn't really have anything to do with perturbation theory. Instead it has to do with mathematical rigor.

At the physics level of rigor, we have both perturbative and non-perturbative formulations of Yang-Mills theories. For instance, lattice models are non-perturbative and frequently used to investigate e.g. QCD at strong coupling, and a crucial feature of instantons is also that they are invisible to pure perturbation theory (see e.g. this question and its answers).

At the mathematical level of rigor, both perturbative and non-perturbative approaches require establishing the formal existence of quantum fields as operator-valued distributions obeying the Wightman axioms or something mathematically equivalent to this. For instance, the usual rigorous formulation of perturbation theory is via Epstein-Glaser renormalization, which also requires establishing the Wightman axioms (in particular their causality condition) first.

Such a formal and rigorous construction is unknown in four dimensions for Yang-Mills theories like the Standard Model, and that is what the millennium problem is about, so both perturbative and non-perturbative QFT are similarly nonrigorous. The claim that the Yang-Mills millennium problem somehow means that perturbative QFT is more well-defined than non-perturbative QFT is just wrong. The claim that in many cases perturbative approaches are the only ones that are computationally tractable is correct, but this is not directly related to the problems with rigor.

"Perturbative" and "non-perturbative" are orthogonal issues to "rigorous" or "non-rigorous". We lack rigorous QFT, not non-perturbative QFT.

ACuriousMind
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