Does Only the Divergent Part of the Integral Contribute to the Chiral Anomaly?

[André H Gomes]

I am following Pierre Ramond's book "Field Theory - A Modern Primer", chapter 8, section 8.9, "Anomalies". My question starts in what follows from p.304.

Homework Statement

He draws the triangular diagram with one axial vector at one of the vertices and, using the standard Feynman rules, writes down the (superficially divergent) integral for the scattering amplitude of this process, eq. (8.9.61). He uses dimensional regularization, with dimension d=2ω.

With some work on the integral and trace, we find that the integral has some pieces which, separately, diverge as ω→2 and other pieces which are immediately finite. But, gathering the divergent pieces together, we end up with another finite result.

Now Pierre Ramond goes on to take the divergence of this result and basically says that only the divergent pieces (which are finite when taken as a whole) will contribute to the divergence (giving a finite nonvanishing result, the anomaly).

My question is: is there some argument that justifies this statement, that the (superficially) finite part of the integral just doesn't contribute to the divergence (and therefore doesn't contribute to the anomaly)?

Homework Equations

The Attempt at a Solution

I tried some "brute force" calculation: I gathered all (superficially) finite pieces and took the divergence of it. I did the calculation by myself, not following Ramond's one, and found that the divergence of the whole didn't vanish, so I probably made some mistake during the evaluation --- I shall repeat then following Ramond's results.

But, afterall, what I'm really looking for is not just a "computational answer"... I'm more interested in a argument for it. For example, following the derivation of other books, we discover that the divergence of this diagram would vanish at first glance, after a naive shift of the momenta integration variable, but this turns out to be wrong because the integral is (superficially) divergent so we are not allowed to do this naive shifts: they actually generate relevant (nonvanishing) surface terms, which end up giving rise to the anomaly. Following this approach, I'm thinking if we could somehow justify on these grounds whether or not only the (superficially) divergent part of the diagram contributes to the anomaly.

 

[mark726]

There are a few different arguments that can be made to justify the statement that only the divergent pieces of the integral contribute to the anomaly.

Firstly, it is important to note that the anomaly arises from the integration over the full range of momenta, including both the ultraviolet and infrared regions. In dimensional regularization, the ultraviolet divergences are regulated by the introduction of the parameter d=2ω, while the infrared divergences are regulated by the introduction of a mass scale μ. When we take the divergence of the integral, we are essentially looking at the behavior of the integral as the momentum cutoffs go to infinity (for the ultraviolet) and zero (for the infrared). Since the (superficially) finite pieces of the integral do not have any divergences in these limits, they do not contribute to the overall divergence of the integral.

Secondly, the (superficially) finite pieces of the integral are typically associated with terms that are already known to be renormalized in the theory. This means that these terms are already taken into account in the renormalization procedure and do not contribute to the anomaly. It is only the new, divergent pieces that arise from the integration over the ultraviolet and infrared regions that contribute to the anomaly.

Finally, there are also arguments based on the symmetries of the theory. In order for an anomaly to arise, there must be a breaking of a symmetry in the quantum theory that is preserved in the classical theory. The divergent pieces of the integral typically arise from terms that break this symmetry, while the (superficially) finite pieces do not. Therefore, it makes sense that only the divergent pieces would contribute to the anomaly.

Overall, while it is important to check the calculation to ensure that there are no mistakes, there are also good theoretical justifications for the statement that only the divergent pieces of the integral contribute to the anomaly.