Why does causal perturbation theory in the sense of Epstein Glaser fall under algebraic QFT rather than heuristic QFT in renormalization?
Asked
Active
Viewed 741 times
3
-
Related: http://physics.stackexchange.com/q/6530/2451 and links therein. – Qmechanic May 12 '15 at 09:52
-
I'm not fully certain what your question here is. What do you mean with "fall into the favor of"? – ACuriousMind May 12 '15 at 17:01
-
Belong to the category of I should say – user41508 May 12 '15 at 18:16
1 Answers
3
Because its statements and methods are fully rigorous, hence constitute mathematical physics. No unproved or unprovable claims are made. See https://www.physicsforums.com/insights/causal-perturbation-theory/ for an exposition of the principles and results.
There is other work on renormalization in relativistic QFT that is fully rigorous and hence qualifies as algebraic QFT, for example Salmhofer's work, which treats renormalization in a rigorous formal power series framework. See Salmhofer's book ''Renormalization: an introduction'', Springer 2013.
Arnold Neumaier
- 44,880