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I am studying through the photon self-energy

$$ i\Pi_{\mu\nu}(q) = \int\frac{d^4 k}{(2\pi)^4}Tr\left[(-ie\gamma_\mu)\frac{i(\require{cancel}\cancel k+m)}{k^2-m^2+i\epsilon}(-ie\gamma_\nu)\frac{i(\cancel p-\cancel k+m)}{(p-k)^2-m^2+i\epsilon}\right]. $$ I don't know to expand this trace, some textbooks give direct results with out any expansion. It is helpful, if someone give a detailed expansion of this trace.

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    $\begingroup$ Can you share what you have tried? $\endgroup$ – Oбжорoв May 22 at 21:07
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Note that u can pull out of the trace the denominators and some constants.

$$\text{Tr}\left[(-ie\gamma_\mu)\frac{i(\require{cancel}\cancel k+m)}{k^2-m^2+i\epsilon}(-ie\gamma_\nu)\frac{i(\cancel p-\cancel k+m)}{(p-k)^2-m^2+i\epsilon}\right]=$$

$$=e^2\dfrac{1}{(k^2-m^2+i\epsilon)((p-k)^2-m^2+i\epsilon)}\text{Tr} \left[\gamma_\mu(\require{cancel}\cancel k+m)\gamma_\nu(\cancel p-\cancel k+m)\right].$$

Then note that the trace of an odd number of gamma matrices is zero so

$$\text{Tr} \left[\gamma_\mu(\require{cancel}\cancel k+m)\gamma_\nu(\cancel p-\cancel k+m)\right]=\text{Tr} \left[\gamma_\mu\require{cancel}\cancel k\gamma_\nu(\cancel p-\cancel k)\right]+m^2\text{Tr} \left[\gamma_\mu\gamma_\nu\right].$$

Finally use the well known results for the traces of gamma matrices and you should be good.

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