4
$\begingroup$

This question already has an answer here:

I have recently encountered Haag's theorem and according to Wikipedia:

Rudolf Haag postulated [1] that the interaction picture does not exist in an interacting, relativistic quantum field theory (QFT), something now commonly known as Haag's Theorem. Haag's original proof was subsequently generalized by a number of authors, notably Hall and Wightman,[2] who reached the conclusion that a single, universal Hilbert space representation does not suffice for describing both free and interacting fields. In 1975, Reed and Simon proved [3] that a Haag-like theorem also applies to free neutral scalar fields of different masses, which implies that the interaction picture cannot exist even under the absence of interactions.

Studying some particle physics, this kinda bothers me. Also the article says:

While some physicists and philosophers of physics have repeatedly emphasized how seriously Haag's theorem is shaking the foundations of QFT, the majority of QFT practitioners simply dismiss the issue.

And not to worry about this is exactly what my particle professor recommended. I understand that evidently physicists are doing pretty well in calculating their scattering amplitudes with the interaction picture, but this as an explanation is just saying that the end justifies the means. It doesn't really comfort me when there is still a mathematical proof saying that the interaction picture just cannot exist. Now, is it okay to worry? (Why not? - Besides the notion that it just works, which sounds a bit like magic to me)

$\endgroup$

marked as duplicate by Qmechanic Jul 3 '13 at 14:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ Related question: physics.stackexchange.com/q/3983 $\endgroup$ – twistor59 Jun 26 '13 at 18:06
  • $\begingroup$ When I first learned of Haag's theorem, I found this dissertation quite illuminating: d-scholarship.pitt.edu/8260 $\endgroup$ – Alfred Centauri Jun 26 '13 at 18:16
  • 4
    $\begingroup$ Haag's theorem is just saying that the passage from free fields to interacting ones is singular, much more so than in quantum mechanics. This is quite consistent with the calculations physicists perform. Naive perturbation theory contains lots of infinities. Renormalized perturbation theory removes the infinities, and in doing so abandons the interaction picture. (And lattice QFT avoids Haag's theorem by not even trying to start with non-interacting fields.) $\endgroup$ – user1504 Jun 26 '13 at 18:17
  • 1
    $\begingroup$ Would be great if @user1504 can turn his/her comment into an answer that can be upvoted $\endgroup$ – Slaviks Jun 26 '13 at 20:26
  • 1
    $\begingroup$ @Slaviks: I've put the answer in the related question twistor59 linked to. $\endgroup$ – user1504 Jun 26 '13 at 20:28