# How do I get the amplitude for the one-loop photon self-energy?

I am studying Maggiore's book on QFT and I am stuck in the amplitudes of one-loop corrections in QED. Could someone clearly explain me how do I get the following amplitude from the respective diagram?

$$i\Pi_{\mu\nu}(q) = \int\frac{d^4 k}{(2\pi)^4}Tr\left[(-ie\gamma_\mu)\frac{i(\require{cancel}\cancel k+m)}{k^2-m^2+i\epsilon}(-ie\gamma_\nu)\frac{i(\cancel p-\cancel k+m)}{(p-k)^2-m^2+i\epsilon}\right].$$

In particular, I don't understand why does not appear the polarization $$\varepsilon_\mu(p)$$ (and its complex conjugate), since there are photons external legs, and how the trace comes up. I have a similar problem with the fermion self-energy because does not appear any spin wave function either and still there are external fermion legs. So maybe by understanding the photon case I am able to reproduce the fermion's result.

• There are no polarization vectors because it is irrelevant whether or not the photons are external. We want to be able to plug this correction into a generic diagram. The trace emerges because of the loop. The gamma matrix from the left vertex should connect to both of the fermion propagators. That is, the leftmost gamma matrix should also multiply the rightmost propagator (in the amplitude). Try writing out the spin indices explicitly and you will see the trace. Commented Nov 12, 2015 at 17:28
• But one of the Feynman rules tells us that we have to put a polarization vector in the amplitude for each incoming photon and its conjugate for each outcoming photon. What am I missing?
– user74106
Commented Nov 12, 2015 at 18:20
• @IberêKuntz The rule is that if you have an external on shell photon then it gets a polarization factor. The diagram here is off-shell. It is a sub diagram that is to be inserted into a larger Feynman diagram. Try to read up the Feynman rules for off shell amplitudes. Commented Nov 12, 2015 at 18:29
• @Prahar, so $i\Pi_{\mu\nu}$ is part of the total amplitude $iM_{fi}$ and this does have the polarization vectors?
– user74106
Commented Nov 12, 2015 at 18:48
• @sawankt the reasons are laid out in any QFT book, say, Peskin and Schroeder Ch. 5 Commented Jun 24, 2021 at 8:54

The photon polarization vectors have been factored out. The full expression is $$\Pi=\epsilon_\mu\,\epsilon_\nu^*\,\Pi^{\mu\nu}$$ where $$\Pi^{\mu\nu}$$ is the tensor described in the OP. We don't bother writing the vectors $$\epsilon_\mu,\epsilon_\nu^*$$ because they are irrelevant for the present discussion. But they are there.
Now let us follow the fermionic lines counterclockwise. You can begin wherever you want, say the left vertex. This vertex leads to $$\gamma^\nu_{\alpha\beta}$$. Next we see the lower fermion propagator, that leads to $$(\not k-\not q+m)^{-1}_{\beta\delta}$$. Next we see the right vertex, with rule $$\gamma^\mu_{\delta\gamma}$$. Finally we see the upper fermion propagator, with value $$(\not k+m)^{-1}_{\gamma\alpha}$$. Putting everything together we get $$\gamma^\nu_{\alpha\beta}(\not k-\not q+m)^{-1}_{\beta\delta}\gamma^\mu_{\delta\gamma}(\not k+m)^{-1}_{\gamma\alpha}$$
Finally, note that by definition of matrix product, $$A_{\alpha\beta}B_{\beta\gamma}=(AB)_{\alpha\gamma}$$, and therefore this becomes $$=(\gamma^\nu(\not k-\not q+m)^{-1}\gamma^\mu(\not k+m)^{-1})_{\alpha\alpha}$$ which is nothing but the trace of the matrix inside, $$=\operatorname{tr}(\gamma^\nu(\not k-\not q+m)^{-1}\gamma^\mu(\not k+m)^{-1})$$ which agrees with the expression in the OP.