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In https://www.physicsforums.com/threads/haags-theorem-perturbation-existence-and-qft.177865/#post-1384425 #2 post by meopemuk (Eugene) say that Haag's theorem says:

$$U(\Lambda)\Phi(x) U^{-1}(\Lambda) \neq \Phi(\Lambda x)$$ Or to quote directly:

The statement of Haag's theorem is that this interacting field cannot have a covariant transformation law with respect to the interacting representation of the Poincare group $U$.

Is this a correct understanding of Haag's theorem? I am asking this because this is usually not how Haag's theorem is presented. And if true, would this mean Poincare-invariant vacuum of interacting field does not exist?

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    $\begingroup$ You can't evolve field operators of the free theory at a time $t$ to interacting field operators, as is often done in perturbative QFT. But interacting theories still require a representation of the Poincaré group, it's just that it will act on operators unrelated to the free theory. $\endgroup$
    – Slereah
    May 9, 2018 at 15:27

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Haag tells you that if you assume $\Phi(x)$ and $\Phi(\Lambda x)$ satisfy the same commutation relations, then $U$ does not necessarily exist (in contrast with systems with a finite number of degrees of freedom). But Haag doesn't prove that $U$ doesn't exist; only that it need not.

In fact, the classical paper of Glimm&Jaffe on two-dimensional $\phi^4$ proves that $\phi$ does satisfy the covariant transformation laws in an interacting example, so it is factually false that $U$ doesn't exist. Sometime it does, sometimes it doesn't.

That being said, nowadays we don't usually build up QFT's from canonical commutation relations. You don't need to assume they hold, and so you may sidestep Haag. A lot of people working on rigorous QFT have abandoned the CCR as a fundamental ingredient anyway.

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  • $\begingroup$ Haag's theorem is not at all related to canonical commutation relations. It is related to distinct invariant states of an algebra of observables (w.r.t. the action of the Poincaré group), and their corresponding irreducible GNS representations (that are not, roughly speaking, unitarily equaivalent). $\endgroup$
    – yuggib
    Jun 26, 2018 at 16:47

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