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  How do we know for sure a theory is non-renormalizable?

+ 6 like - 0 dislike

In quantum field theory, we are looking for a Lagrangian that is, amongst other, renormalizable. But how do we determine whether or not a theory is renormalizable? Is this purely done by power counting due to Weinberg? This question is already answered in the a previous question.

My question result from the fact that the Yang-Mills Lagrangian was considered to be non-renormalizable, and thus non-physical, for a decade until Veltman and 't Hooft found a method to regularize the theory. Keeping this in mind, is it possible that there are theories that we today consider to be non-renormalizable, and thus non-physical, which actually are renormalizable but we haven't (yet) discovered a way to do this?

I apologize in advance if my question is vague and if I'm using the wrong terminology, but I'm very new to the idea of renormalizability.

This post imported from StackExchange Physics at 2014-04-05 17:25 (UCT), posted by SE-user Hunter
asked Feb 24, 2014 in Theoretical Physics by Hunter (520 points) [ no revision ]
retagged Apr 5, 2014
related post physics.stackexchange.com/questions/88884/… ; P.S. non-renormalizable Lagrangians are accetable as effective theories

This post imported from StackExchange Physics at 2014-04-05 17:25 (UCT), posted by SE-user user26143
@user26143 thanks for that! I will now update my question (I don't think it answers my second question).

This post imported from StackExchange Physics at 2014-04-05 17:25 (UCT), posted by SE-user Hunter

If a theory is power counting non-renormalizable, then for sure it is not renormalizable.(EDIT: I realize this is too bold an assertion, power counting is not enough to make us sure of nonrenomalizablity)

This post imported from StackExchange Physics at 2014-04-05 17:25 (UCT), posted by SE-user Jia Yiyang

1 Answer

+ 3 like - 0 dislike

Renormalizability is _defined_ by that finitely many counter terms suffice to have a well-defined renormalized perturbation theory. This can be decided by a combinatorial analysis, which for some theories simply hadn't been completed in early times. The news in Veltman and 't Hooft's arguments was that symmetry-breaking introduces a special situation where the couplings are related in a specific way such that finitely many counterterms suffice as if the broken symmetry were absent.

Nonrenormalizable does not mean nonphysical. It only means that infinitely many counterterms are needed to fully specify the theory perturbatively. Most of these terms are, however, suppressed by huge factors, so that only a few are needed to extract the physics at a given scale. This leads to effective theories.

Some nonrenormalizable theories become renormalizable after a nonlinear field transformation, at least in 1+1 dimensions.

answered Apr 13, 2014 by Arnold Neumaier (15,787 points) [ no revision ]

Ahh ok, so Veltman and 't Hooft discovered that if a theory is renormalizable before symmetry breaking, then it must also be renormalizable after symmetry breaking?

And so if I (or someone else) were to write down a new Lagrangian, then it is always possible to tell if the theory is renormalizable or not (by using the combinatorial analysis you mentioned)?

Thanks for your reply!

Yes to the first question.

To the second one: It is not easy to do such an analysis: The combinatorial analysis of Veltman and 't Hooft took many pages. Even the renormalizability analysis of scalar field theories to all orders took a number of years to be complete without missed cases of overlapping divergences. But I believe that all cases of interest (with Poincare symmetry) have now been covered, so that the answer should be yes. (But I might have misinterpreted the available evidence - did not check the truth myself.)

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