Sunday, October 11, 2015

operators - 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?



Answer



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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