# Does Haag's theorem say covariant transformation of interacting field is not possible?

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?

• 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. – Slereah May 9 '18 at 15:27

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.