I have been studying the axial anomaly and everywhere I see the calculation of the triangle loop using dimensional regularisation (see for example pages 661-664 of section 19.2 of Peskin). In the ‘t Hooft prescription for the $\gamma^5$ they divide the Lorentz space into the usual 4 dimensional one and the rest of dimensions (inside the integration), so the loop momentum can be written as (eq. 19.53) \begin{equation} l=l_\parallel+l_\perp \end{equation}

“ Where the first term has nonzero components in dimensions 0,1, 2, 3 and the second term has nonzero components in the other d—4 ($-2\epsilon$) dimensions.”

Then, we arrive to this integral

\begin{equation} \int \frac{d^dl}{(2\pi)^d}\frac{l_\perp^2}{(l^2-\Delta)^3} \end{equation}

and what I do not understand is the following replacement for it (eq 19.57) \begin{equation} l_\perp^2\to\frac{d-4}{d}l^2 \end{equation} “under the symmetrical integration”.

I do understand when we do similar things for an even integral for which any odd term is zero so

\begin{equation} l_\mu l_\nu\to\frac{1}{d}l^2g_{\mu\nu} \end{equation}

But in the case with the $l_\perp$ I do not see a proper derivation, could someone help me with that? Thanks in advance.

  • $\begingroup$ Just a side note, “the symmetrical integration” sort of sleight of hands only works when the Feynman integral is either convergent or logarithmically divergent. $\endgroup$ – MadMax Aug 5 '19 at 14:19
  • $\begingroup$ As manifested in OP's example, whenever the pseudo scalar $\gamma_5$ is in the picture, dimensional regularization feels as awkward as Uncle Joe talking about internet. $\endgroup$ – MadMax Aug 5 '19 at 14:28

As the integral has rotational invariance in $d$ dimension, each $\ell$-component should yield the same value. There are $d-4$ non-zero components in $\ell_\perp$ and $d$ non-zero components in $\ell$, we should thus have \begin{equation} \frac{1}{d-4}\int \frac{d^4 \ell}{(2\pi)^4} \frac{{\ell}_\perp^2}{(\ell^2-\Delta)^3} = \frac{1}{d}\int \frac{d^4 \ell}{(2\pi)^4} \frac{{\ell}^2}{(\ell^2-\Delta)^3} \end{equation} Therefore \begin{equation} \mathcal{I} = \frac{d-4}{d} \int \frac{d^4 \ell}{(2\pi)^4} \frac{{\ell}^2}{(\ell^2-\Delta)^3} \end{equation}

  • $\begingroup$ If the answer is correct, can you please accept it. Thanks $\endgroup$ – Oбжорoв Aug 8 '19 at 8:14

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