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  Anomaly and renormalization procedure choice

+ 3 like - 0 dislike

I've heard an argument, which provides independence of quantum breaking of naive classical symmetry - so-called anomaly - on regularization choice: If we get non-local (here this means non-polynomial) symmetry breaking term in one regularization, it will appear in other ones, since it can't be removed by adding local counterterm. I don't like this argument. It is insufficiently strict, or it is not finished. Could you please comment this argument and provide alternative (or finish it)?

asked Oct 17, 2016 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
recategorized Oct 18, 2016 by Dilaton

 it will appear in other ones

Do you mean some chirality breaking regularization will happen in other cases, or some non-local chirality breaking regularization will happen in other cases? Although in either case I don't think there can be a general yet rigourous argument for it, without specifying enough other conditions.

@JiaYiyang : I've meaned that the chiral anomaly provides the presence of non-local term in each possible regularization. Then if we observe it in one of regularizations, we can conclude that independently on the regularization type it will appear always.

This isn't quite the case already in lattice regularization: a naively discretized Dirac operator has both exact gauge invariance and chiral invariance, the anomaly manifests itself as the fermion doubling problem.

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