# Compton Scattering Feynman diagram integral expression

I'm trying to write down the integral expression according to the feynman-rules for this Diagram of an electron with compton scattering and a one-loop correction:

Feynman Diagram http://s8.postimg.org/qd1n6zt9v/DSCN9963.jpg ![Compton Scattering][1]

$$(-ie)^4\int \frac{d^4k}{2\pi} \gamma_{\mu}\frac{i}{p-k-m+i\epsilon}\gamma_{\nu}\frac{-ig^{\mu \nu}}{k^2 + i \epsilon}\gamma_{\sigma}\frac{-ig^{\sigma\rho}}{k_1^2+i\epsilon}\frac{i}{p+k_1-m+i\epsilon}\\\gamma^{\alpha}\frac{-ig^{\alpha \beta}}{k_2^2+i\epsilon}\frac{i}{p+k_1+k_2-m+i\epsilon}$$

Is that somehow correct? I have the impression that the indices are not balanced. Is it correct that i only integrate over the first part with the loop?

From right to left:

• outgoing electron $e^-$ spinor: $\bar{u}(p_2)$
• QED vertex: $ie\gamma^\nu$
• outgoing photon: $e^*(k_2)$
• electron propagator: $\frac{i}{\not{q}-m+i0}$
• incoming photon: $e(k_1)$
• QED vertex: $ie\gamma^\mu$
• incoming electron $e^-$ spinor: $u(p_1)$

There is no photon propagator in this process, and also, only one electron propagator. You should rewrite your expression since it's wrong.

• So for the diagram without the photon loop, I get: $(ie)^2 u(p+k_1+k_2) \gamma^{\nu}e^{*}(k_2) \frac{i}{\gamma^{\nu}(p+k_1)-m+i\epsilon} e(k_1)\gamma^{\mu}u(p)$ ... is that correct? Jun 20, 2015 at 10:47
• Yes. It's correct. And there is no integral here. But when there exist a loop correction in Compton scattering, you should insert a loop integral in this amplitude. That integral will be divergent. To deal with this divergence, you should use one of regularization methods, and then renormalization. Jun 20, 2015 at 12:45
• @LêDũng I am under impression that OP is trying to calculate loop corrections, not the tree-level diagram. Jun 20, 2015 at 12:55
• Can you load your diagram again? I cannot see your diagram Mechanix Jun 20, 2015 at 12:59
• i was trying to add a picture from here: s8.postimg.org/qd1n6zt9v/DSCN9963.jpg That's my diagram. The only difference is that i have a loop correction in the incoming electron. Jun 20, 2015 at 13:02