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In the book An Introduction to Quantum Field Theory by Peskin and Schroeder there is a derivation of the anomalous magnetic moment of the electron. The Feynman diagram to be solved is this one: enter image description here

and the integral is the following:

\begin{equation} \int \frac{d^4 k}{(2\pi)^4} \frac{-i g_{\nu\rho}}{(k-p)^2 + i\epsilon} \overline{u}(p')(ie\gamma^\nu) \frac{i( \not{k} ' + m)}{k'^2 - m^2 + i\epsilon} \gamma^\mu \frac{i( \not{k} + m)}{k^2 - m^2 + i\epsilon} (ie\gamma^\rho)u(p) \end{equation}

This can be rewritten to: \begin{equation} 2ie^2 \int \frac{d^4 k}{(2\pi)^4} \frac{\overline{u}(p') \left[ \not{k} \gamma^\mu \not{k}' + m^2 \gamma^\mu - 2m(k + k')^\mu \right] u(p)}{\left((k-p)^2 + i\epsilon\right)\left(k'^2 - m^2 + i\epsilon \right)\left(k^2 - m^2 + i\epsilon \right)} \end{equation}

I'm stuck trying to get the $\not{k} \gamma^\mu \not{k}'$ part in the brackets. I have obtained the other ones using the identity $\gamma^\mu \gamma^\nu \gamma_\mu = - 2 \gamma^\nu$ as described in the book:

From the $m^2$ term in the product $( \not{k} ' + m) ( \not{k} + m)$ you get the $m^2 \gamma^\mu$. From the products involving one $m$ and one $k$ you get the $- 2m(k + k')^\mu$ and it just looks like $\gamma^\nu \not k ' \gamma^\mu \not k \gamma_\nu$ gives $\not k \gamma^\mu \not k'$ but I haven't found how. I've tried swapping $k$ and $k'$ using $\not k \gamma^\mu + \gamma^\mu \not k = 2k^\mu$ but I get a lot of terms that doesn't seem to cancel out.

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You have the following identity

$$\gamma^{\nu}\gamma^{\mu}\gamma^{\rho}\gamma^{\sigma}\gamma_{\nu}=-2\gamma^{\sigma}\gamma^{\rho}\gamma^{\mu}$$

This gives you that $$\gamma^{\nu}\not{k^{'}}\gamma^{\rho}\not{k}\gamma_{\nu}=-2\not{k}\gamma^{\rho}\not{k^{'}}$$

Which is exactly what you need to get the above expression.

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  • $\begingroup$ Note: I don't know how to insert proper slashes in Mathjax since the Slashed package is not supported, if someone more knowledgeable could edit the above that would be great! $\endgroup$ – ranques Jun 21 '16 at 16:30
  • $\begingroup$ That's it, I knew it was something easy. The webpage with all the identities may be useful later. $\endgroup$ – Nister Jun 21 '16 at 17:47

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