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?

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    $\begingroup$ Are integrals necessary to make calculations in celestial mechanics or do they just make the calculations easier? In both cases you don’t really need any specific definition / construct, you can always get away with not defining it at the cost of making the explanation more involved. It’s only a useful construct $\endgroup$ May 23 at 11:12
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    $\begingroup$ You can do perturbative QFT in interaction picture purely in terms of operator matrix elements, without ever referring to a virtual particle. The terms are just conveniently labeled by virtual particles $\endgroup$ May 23 at 11:13
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    $\begingroup$ > "PBS Spacetime is typically a reliable source" Why do you think so? $\endgroup$ May 23 at 13:07
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    $\begingroup$ As for the statement about nonperturbative Yang-Mills: The idea is that we can indeed do these calculations in e.g a lattice QCD setting and we do have algorithms to do so, however placing the theory on a lattice leads to restrictions and problems on its own (and is perturbative in some other sense as well) . As pointed out above, these virtual particles play an important role in perturbative calculations. I guess the better idea to understand this would be that the idea of a particle often does not make sense at all in non-perturbative calculations, not only virtual particles $\endgroup$ May 23 at 19:51
  • $\begingroup$ In the first papers on perturbative QED, the differences between physical and virtual particles were already obvious. This was many years before Wilson proposed lattice gauge theory. $\endgroup$ May 23 at 21:42

1 Answer 1


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.

  • $\begingroup$ And this brings me once more to the chronic stickler question: without "rigorous" QFT or, more appropriately, without a rigorous formulation of the Standard Model, to what extent are claims that it can account for "all known physics" justified versus similar claims for other, rigorously-formulated theories when confined to their claimed domains of applicability - particularly "in the wild" physics outside of the very specific situation of a particle accelerator? $\endgroup$ Oct 19 at 2:42

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