Feynman diagrams with long fermion lines

Question

how do I calculate the interference of these two diagrams, where the fermions are electrons?

The matrix element of the first diagram is calculated as $\frac{1}{(p_3+p_4)^2}\bar{u}(p_2)\gamma_\mu u(p_1) \bar{u}(p_3) \gamma^\mu v (p_4)$, the amplitude neglecting the electron mass is calculated as $\frac{1}{(p_3+p_4)^4} (\gamma p_2) \gamma_\mu (\gamma p_1) \gamma_\nu (\gamma p_3) \gamma^\mu (\gamma p_4) \gamma^\nu $, the second diagram is just calculated by exchanging $p_2$ and $p_3$.

When I attempt to calculate the interference, I can't just write a matrix element for it and multiply it with the complex conjugate since now there are not two distinct fermion lines but one large one. Therefore, none of the spinors can be commuted and I cannot properly combine spinors into momenta. My only idea would be to directly write the amplitude as $\frac{1}{(p_3+p_4)^2 (p_2+p_4)^2}(\gamma p_1)\gamma_\nu (\gamma p_2) \gamma^\mu (\gamma p_4) \gamma^\nu (\gamma p_3) \gamma_\mu$.

Is this correct and if not how do I calculate diagrams with longer fermion lines like this.

Answer 1

Let $q_1$ and $q_2$ be the exchanged photon's momentum of each diagram, and for sake of brevity I will abbreviate $u(p_a)=u_a$ so as to limit the expression length.

The first Diagram has a matrix element:

$$ \mathcal{M}_1 = \frac{i}{q_1^2}(\bar{u}_3\gamma_{\mu}v_4)(\bar{u}_2\gamma^{\mu}u_1) $$

The second:

$$ \mathcal{M}_2 = \frac{i}{q_2^2}(\bar{u}_2\gamma_{\mu}v_4)(\bar{u}_3\gamma^{\mu}u_1) $$ Then: $$ \frac14\sum_{\text{spins}}|\mathcal{M}_1\bar{\mathcal{M}}_2|= \frac{1}{4q_2^2q_1^2}\sum_{\text{spins}}(\bar{u}_3\gamma_{\mu}v_4)(\bar{u}_2\gamma^{\mu}u_1)(\bar{u}_1\gamma^{\nu}u_3)(\bar{v}_4\gamma_{\nu}u_2) $$

This is, I assume where you get stuck, so we do as normal and rearrange the terms to put pairs of spinors together:

$$ \frac{1}{4q_2^2q_1^2}\sum_{\text{spins}}(\bar{u}_3\gamma_{\mu}v_4)(\bar{u}_2\gamma^{\mu}u_1)(\bar{v}_4\gamma_{\mu}u_2)(\bar{u}_1\gamma^{\mu}u_3) \to \frac{1}{4q_2^2q_1^2}\sum_{\text{spins}}(\bar{u}_3\gamma_{\mu}v_4)(\bar{v}_4\gamma_{\nu}u_2)(\bar{u}_2\gamma^{\mu}u_1)(\bar{u}_1\gamma^{\nu}u_3) $$

Now just form all the easier spin sums you can to get:

$$ \frac{1}{4q_2^2q_1^2}\sum_{\text{spins}}(\bar{u}_3\underbrace{\gamma_{\mu}\not{p_4}\gamma_{\nu}\not{p_2}\gamma^{\mu}\not{p_1}\gamma^{\nu}}_{Q_{ij}}u_3) \to \frac{1}{4q_2^2q_1^2}\sum_{ij}\bar{u}_{3i} Q_{ij} u_{3j} = \frac{1}{4q_2^2q_1^2}\sum_{ij}Q_{ij}(u_3\bar{u}_3)_{ji} $$

This just leaves you with:

$$ \frac{1}{4q_2^2q_1^2}\mathrm{Tr}(Q\not{p_3}) = \frac{1}{4q_2^2q_1^2}\mathrm{Tr}(\gamma_{\mu}\not{p_4}\gamma_{\nu}\not{p_2}\gamma^{\mu}\not{p_1}\gamma^{\nu}\not{p_3}) $$