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.

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    $\begingroup$ Counterterms are a tool in renormalization, not a way to get rid of unwanted terms. I do not understand your question. Could you perhaps write down the "counterterm" you intend to introduce? $\endgroup$ – ACuriousMind Jul 26 '14 at 13:46
  • $\begingroup$ okay. I found out the specific name for it. Bardeen counter term. I am not sure why and how you can add the counter terms to the lagrangian. so, 1)What effects does the bardeen counter term produce? 2)does it cancel the anomalous divergence? 3)How does the anomaly cancel for the ABELIAN case? does it cancel at all? $\endgroup$ – SubhamDC Jul 27 '14 at 4:29

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.

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    $\begingroup$ I think the correct thing to say is that you cannot have counterterms which kill both anomalies (gauge anomaly and chiral anomaly). However, you may have counterterms which kills one of the anomalies. See for instance formulae ($(16),(17),(18)$ p. $25,26,27$) in this paper. $\endgroup$ – Trimok Jul 29 '14 at 10:52
  • $\begingroup$ @Trimok: I think you are right (and my last paragraphs probably wrong). Would you like to write an answer of your own, or would you rather that I rewrite mine? $\endgroup$ – ACuriousMind Jul 29 '14 at 17:02
  • $\begingroup$ I will not write an answer, so you may modify yours (however, I don't see any reaction of the OP to your answer, so...) $\endgroup$ – Trimok Jul 30 '14 at 8:21
  • $\begingroup$ @ACuriousMind Thanks for a concise answer. For people who are looking for a more elaborate answer, I would like to refer to the notes by Karl Landsteiner here (arxiv.org/pdf/1610.04413.pdf), because they give an excellent more in-depth explanation of the usage of Bardeen polynomials. These notes helped me a lot in understanding the subtleties. $\endgroup$ – Funzies Jun 22 at 16:01

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