$U(1)$ abelian/axial/chiral anomaly in 4D

Question

I am reading $U(1)$ abelian/axial/chiral anomaly in 3+1 dimensions using the path integral method (Fujikawa). Am I wrong in assuming that the anomaly can be cancelled by introducing a counter term in the Lagrangian that exactly cancels the anomalous divergence of the $U(1)$ axial current? Literature sources beg to differ but I cannot understand why it is so. Any reference would be very much useful.

Answer 1

This Bardeen counterterm is an elusive beast, I must say. Yet I will share what I have found and understand:

Define a $\mathrm{U}(1)$ gauge theory by writing its action in left- and right-handed chiral spinors as

$$ S_{\mathrm{chiral}}[A] = \int \bar \psi_L (\mathrm{i} {\not \hspace{-4px} \partial} - \not \hspace{-5px} A)\psi_L + \bar \psi_R (\mathrm{i} {\not \hspace{-4px} \partial} - \not \hspace{-5px} A)\psi_R $$

and observe that, with $j = j_R+j_L$ and $j^5 = j^5_R - j^5_L$, we have

$$ \partial_\mu j^\mu = 0 \; \text{and} \; \partial_\mu j^{5\mu} = \frac{1}{24\pi^2}F \wedge F = \frac{1}{12\pi^2}\mathrm{d}A\wedge\mathrm{d}A$$

Now, this seems to give us the anomaly directly. However, we can also look at introducing an auxiliary field $B_\mu$ coupled to the axial current as

$$ S_{\mathrm{aux}}[A,B] = \int \bar \psi_L (\mathrm{i} {\not \hspace{-4px} \partial} - \not \hspace{-5px} A + \not B)\psi_L + \bar \psi_R (\mathrm{i} {\not \hspace{-4px} \partial} - \not \hspace{-5px} A - \not B)\psi_R$$

$B_\mu j^5_\mu$ is a gauge invariant operator, so it should not ruin our theory. Yet it does, as one finds for the Ward identities

$$ \partial_\mu j^\mu = \frac{1}{6\pi}\mathrm{d}A\wedge\mathrm{d}B \; \text{and} \; \partial_\mu j^{5\mu} = \frac{1}{12\pi^2}(\mathrm{d}A\wedge\mathrm{d}A + \mathrm{d}B\wedge\mathrm{d}B)$$ This inspires us to add the Bardeen counterterm

$$S_{\mathrm{Bardeen}}[A,B] = \frac{1}{6\pi^2}\int A \wedge B \wedge \mathrm{d}A$$

to the auxiliary action. Now, the currents fulfill

$$ \partial_\mu j^\mu = 0 \; \text{and} \; \partial_\mu j^{5\mu} = \frac{1}{4\pi^2}(\mathrm{d}A\wedge\mathrm{d}A + \frac{1}{3}\mathrm{d}B\wedge\mathrm{d}B)$$

and we have indeed that the gauge invariant perturbation by $B$ does not destroy gauge invariance anymore. Thus, what we have got rid of through the counterterm is the gauge anomaly, not the axial anomaly. Note that this is indeed a renormalization in the usual sense, since $A \wedge B \wedge \mathrm{d}A$ produces some additional coupling/Feynman diagrams.

What most likely confused you is that many sources state that there is no local counterterm for the axial anomaly. This is entirely true, since $\int F \wedge F$ is a topological term, the second Chern class, and hence not local. You could add that to the action to try and kill the axial anomaly, but this would not be a local term, and thus not a good thing to do.