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  Why isn't Quantum Yang-Mills Rigorous?

+ 6 like - 0 dislike
837 views

Obviously one of the major components of the Yang-Mills existence and mass gap problem of the Clay institute is the proof that 3+1d quantum yang-mills theory has rigorous foundations. This (I believe) involves demonstrating that such a theory in satisfies the wightman axioms, and, by the Osterwalder-Schrader theorem, also works in euclidean space. Is this true? Does the standard model fail to follow these axioms? If so, then what directly needs to be done to generate a rigorous a 3+1d Yang-Mills?

This post imported from StackExchange Physics at 2014-06-25 20:58 (UCT), posted by SE-user user47299
asked Jun 23, 2014 in Theoretical Physics by user47299 (50 points) [ no revision ]
both quantum Yang-Mills and clasical Yang-Mills can manifest various phenomena (e.g non-linear and others) which have not been studied fully (the mass gap is also another issue)

This post imported from StackExchange Physics at 2014-06-25 20:58 (UCT), posted by SE-user Nikos M.
@NikosM. what does this mean in the context of rigorous quantum gauge theories?

This post imported from StackExchange Physics at 2014-06-25 20:58 (UCT), posted by SE-user user47299

2 Answers

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You've got things slightly backwards. In constructive QFT, one almost always starts in Euclidean spacetime -- where it is 'easy' to define the path integral -- and then analytically continues to get correlation functions on Minkowski spacetime. (The Osterwalder-Schrader Theorem tells you when this analytic continuation 'works', meaning when the resulting Minkowski signature correlation functions satisfy the Wightman Axioms.) Some details here and here.

The Standard Model almost certainly does not satisfy the Wightman Axioms. The Wightman axioms are concerned with the continuum limit, and the continuum limit in the Standard Model probably doesn't exist, thanks to the Landau poles in the U(1) coupling and in the Higgs self-coupling.

But there's a good chance that the pure SU(3) Yang-Mills theory exists in the continuum limit. It's clear that the Euclidean theory is well-defined on any finite lattice. It's also clear that you can refine the lattice as fine as you want; Balaban, and Magnen, Seneor, & Rivasseau proved as much. The challenge is that not all observables remain well-defined when you make the volume of the lattice bigger and bigger. To construct the continuum theory, you have to figure out which observables remain well-defined in the infinite volume limit, and show that they obey some version of Osterwalder-Schrader. (Presumably, one proves the Euclidean version of the mass gap conjecture along the way; in constructive QFT, the existence of a mass gap is very helpful in providing a way to construct the theory on larger and larger volumes of space.) Then the OS theorem gives you the Minkowski theory and a Hilbert space. More detailed speculation here.

This post imported from StackExchange Physics at 2014-06-25 20:58 (UCT), posted by SE-user user1504
answered Jun 23, 2014 by user1504 (1,110 points) [ no revision ]
+ 2 like - 1 dislike

(add my comments as an answer)

Both quantum and classical Yang-Mills theory can manifest various phenomena (e.g non-linear and others) which have not been studied fully.

For example soliton solutions, asymptotic freedom, stability, confinement etc..

By the work of t'Hooft et al. Yang-Mills-type gauge theories are renormalizable

Yang-Mills (quantum and classical) is a generalisation of Maxwels' Electomagnetism, as such it is interesting, however due to non-linear form of the equations (plus quantization issues, non-abelianity) an (rigorous) analysis similar to EM is not easy to carry out (it is not even known if a Yang-Mills type theory can re-produce sth close to General Relativity).

This post imported from StackExchange Physics at 2014-06-25 20:58 (UCT), posted by SE-user Nikos M.
answered Jun 23, 2014 by Nikos M. (80 points) [ no revision ]

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