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According to the answer on question 364576 this should be settled. But after looking for clear statements of the current situation on triviality of $\phi^4$ theory, I'm still not sure, because:

  1. In Scalar_field_theory we find the statement: "Michael Aizenman proved that the theory is indeed trivial, for space-time dimension $D \geq 5$. [6]".
    So that's not for $D=4$ (but this refers to the 1981 paper, so maybe the sentence just needs updating to the latest results?)
  2. In Quartic_interaction it is stated that "The triviality of both the $\phi^4$ model and the Ising model in $d\geq 4$ can be shown ... [2]", so then it actually is trivial for $D=4$.
  3. Reading arxiv abs 2305.05678 we still find the counterclaim that with an appropriate setup in 4 dimensions "one finds a non-trivial interacting theory".

So I'm wondering... is it settled now?

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3 Answers 3

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It sort of depends on what exactly you mean by "settled". Up to physics standards, the question is definitely settled: it has been understood, for many decades now, that $\phi^4$ in $d=4$ is trivial. The only open question is to prove this in a way that mathematicians will find convincing. This is very similar to the Yang-Mills mass gap problem: physicists have known that YM is gapped for many decades now, it only remains to prove it in a mathematically rigorous way. But the answer is definitely "known".

We actually know the answer in any dimension. In $4d$, the only non-trivial theory is non-abelian Yang-Mills with a handful of bosons and fermions. If you add too many matter fields, or make the gauge field abelian, you make the theory trivial. In lower dimensions, any gauge theory is non-trivial, but e.g. $3d$ $\phi^6$ is again trivial. In $2d$, any $V(\phi)$ is non-trivial.

This is the correct, known answer. It is not "settled" only in the sense that we don't know how to prove this rigorously, but only because we don't have a rigorous definition of QFT. As soon as we manage to define QFT in a rigorous way, one should be able to prove these claims rigorously; absolutely nobody expects a surprise here: if your rigorous definition of QFT does not imply triviality of $\phi^4$, or a YM gap, then your definition is wrong.

It is important to stress that "triviality" in this context does not mean that the theory is boring or useless. For example, $4d$ QED is trivial, but it is obviously a tremendously useful theory, it makes the most accurate physical prediction ever. In this context, "triviality" simply means "infrared triviality", i.e., the coupling constant becomes smaller and smaller the lower the energy of your experiment is. QED and $\phi^4$, and many other theories, are infrared trivial, but they are still non-trivial, interacting, useful low-energy effective theories.

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  • $\begingroup$ Thanks for the answer and even more for the straightening out the question! $\endgroup$ Mar 27 at 15:57
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    $\begingroup$ Triviality implies the whole theory is trivial, i.e. equivalent to the noninteracting theory. What we do in physics to get around that is: we assume the theory is physically wrong. A different theory is correct, which isn’t a quantum field theory at all. We don’t specify what that different theory is, because it doesn’t matter: it can be anything at high energies as long as it gives the perturbative expansion of the naive version of the original theory at low energies. $\endgroup$ Mar 27 at 17:48
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    $\begingroup$ @Prof.Legolasov You can of course use whatever language you want, but this is not how "trivial" is used in practice. For most people, triviality is just a synonym for infrared triviality. $\phi^4$, QED, etc., are all examples of this. Nobody would say that "whole" QED theory is trivial, non-interacting. This is simply not what people mean when they say a given theory is trivial. What they mean, instead, is that the theory becomes trivial when you remove the UV cutoff. But when we refer to $\phi^4$ and QED, we do not remove the cutoff, so these theories are not "whole trivial". $\endgroup$ Mar 27 at 20:28
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    $\begingroup$ @Prof.Legolasov Oh, absolutely, I do not subject my QFTs to something as silly as Wightman's axioms :-) QED makes the best prediction in the history of physics -- any axiomatic setting that does not include QED is incomplete in my book! $\endgroup$ Mar 27 at 20:40
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    $\begingroup$ @AccidentalFourierTransform I like my physics to be mathematically well defined to keep me honest about self-consistency, QED in my book is a set of highly successful perturbative techniques which are UV-incomplete :) Cheers! $\endgroup$ Mar 27 at 20:43
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It is settled now, $\phi^4$-theory is indeed quantum trivial in 4 spacetime dimensions. In 4 spacetime dimensions, a series of papers by Luscher and Weisz demonstrated early numerical evidence (by performing RG in a lattice discretisation) that the $\phi^4$ theory should be trivial. But an actual formal proof was not known until Aizenman, Duminil-Copin. As you point out the proof for quantum triviality in $\geq 5$ spacetime dimensions is easier and was carried out earlier.

The third paper you link is explicitly carries out calculations in the large-N limit of the $O(N)$ model (where you have an $N$-dimensional multiplet of scalar fields). The other papers you list are all talking about what people usually refer to as $\phi^4$, which is the case when $N = 1$.

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  • $\begingroup$ Thanks! And since you mention multiplets, can I still assume the result is the same with $\phi$ Real or Complex-valued? (With the latter just two uncoupled real fields that would seem obvious, but can there really not be any difference?) $\endgroup$ Mar 27 at 13:06
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Currently, the question in $D=4$ is settled by this paper of Michael Aizenman and Hugo Duminil-Copin.

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