4
$\begingroup$

If we have a set of linear symmetry currents $J^{\mu}_{\alpha}$ and attempt to find if they are anomalous, we find that if we change the regularization procedure, the anomaly will get mixed around the currents. In particular, for ordinary massless QED, the canonical form of the anomaly is purely in the chiral symmetry. But it is possible to do the integral such that the gauge symmetry is also anomalous.

Why is this an acceptable fact? In the case of QED, there is the regularization independent fact that there is an anomaly, but where the anomaly resides is regularization dependent. So what does this mean? A different way of calculating the anomaly produces a different theory?

My understanding is the fundamental idea of the renormalization group is that observable quantities cannot depend on the method of regularization. Anomalies are observable, otherwise it could not have been used to explain the neutral pion decaying to two photons amplitude.

$\endgroup$
4
  • $\begingroup$ If you make sure that you use a gauge invariant regularization is there an ambiguity? $\endgroup$
    – octonion
    Commented Aug 28, 2019 at 1:56
  • $\begingroup$ I believe so, the calculation is done in all generality in Weinberg's QFT book, volume 2. He simply shifts the loop momentum variables. I don't think that breaks any symmetry like Pauli Villars does. $\endgroup$
    – fewfew4
    Commented Aug 28, 2019 at 2:00
  • $\begingroup$ Well, the whole anomaly appears as the ambiguity that may be fixed by demanding the conservation of some currents at expense of others. E.g. the usual triangle anomaly contains an ambiguity with one parameter that contributes to both vector and axial anomaly and requiring the vector current conservation (that also may happen because the regularization you use enforces that symmetry) fixes that parameter. Is it what you are talking about? $\endgroup$
    – OON
    Commented Aug 28, 2019 at 7:38
  • $\begingroup$ @OON Yes this is what I am talking about. It appears that specifying the action is not enough to uniquely identify a quantum field theory. You must further choose what symmetries to hold to get a unique theory. Thus if there are $N$ global symmetries of the action, there are at most $2^N$ different theories the action specifies. $\endgroup$
    – fewfew4
    Commented Aug 29, 2019 at 0:58

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.