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I am reading Fujikawa's paper for axial anomaly: https://doi.org/10.1103/PhysRevD.21.2848

In equation (2.15), the anomalous part of axial transformation is regularized by $$\begin{align*} \mathcal{A}(x) = \lim\limits_{M\rightarrow \infty} \Big(\sum_n \phi_n^\dagger(x)~ \gamma^5~e^{-(\frac{\lambda_n}{M})^2} \phi_n(x) \Big) \end{align*}\tag{2.15}$$ where $M$ is taken to $\infty$. In realistic condensed matter system, however, all regularization is of finite energy. If I naively take $M$ to be finite in the following calculation, from the fourth to the fifth line of equation (2.15), the Taylor expansion would include some higher order term, such as \begin{align} & \frac{1}{M^4}Tr\Big[ \gamma^5 [\gamma^\mu, \gamma^\nu] [\gamma^\rho, \gamma^\sigma] [\gamma^\alpha, \gamma^\beta] [\gamma^\gamma, \gamma^\delta]\Big] F_{\mu\nu} F_{\rho \sigma} F_{\alpha \beta} F_{\gamma \delta}\\ \propto& \frac{1}{M^4}(E\cdot B) \cdot (E^2 + B^2) \end{align} which in principle can contribute to the anomalous $\partial_\mu j_5^\mu$, but is not of topological origin as far as I can tell. Does it mean that the finite regularization can actually modify the anomaly, or I understood something wrong?

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After taking the trace, the result will be independent of $M$. Individual terms such that thatyou exhibit may be non zero, but the sum of them must cancel.

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  • $\begingroup$ I checked that the coefficient for this term really cancelled out, kind of amazing. Thanks! $\endgroup$
    – hikotatsu
    Jan 13 at 22:41
  • $\begingroup$ Another related question: when one integrate out a massive Dirac fermion in (2+1)-D, there would be a k=1/2 Chern-Simons term. However, if the regularization scale is set to finite, naively speaking the k=1/2 would be modified. Is that there is also a way to see (in some direct calculation) that the coefficient should be kept invariant even when the regularization scale is finite? $\endgroup$
    – hikotatsu
    Jan 14 at 0:47

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