So, I have been trying to teach myself out of the David Griffiths Elementary Particles textbook, and I am trying my best to understand the conventions. I am looking at question 7.27 (found on page 270 of the Second, Revised Edition), which says the following:
Derive the amplitudes (Equation 7.133 and 7.134) for pair annihilation, $e^++e^-\rightarrow\gamma+\gamma$.
If I can understand how to obtain the first amplitude, Equation 7.133, then I think I am golden with understanding it enough to obtain 7.134, so I am just going to focus on Equation 7.133. Equation 7.133 is the following:
$\mathcal{M}_1=\frac{g_e^2}{(p_1-p_3)^2-m^2c^2}\bar{v}(2)\not\!\epsilon_3(\not\!p_1-\not\!p_3+mc)\not\!\epsilon_3u(1)$
First, I defined the momenta in the system. I said that the incoming electron had a momenta of $p_1$, the positron of $p_2$, the fermion that mediates the two as $q$, then the photon connects at the vertex with the electron as $p_3$ and the final photon as having a momentum of $p_4$. This seems to be the 'standard' practice for labelling momenta Griffiths has for starting these problems with a second processes, so I have the same labels as the answer keys I found.
I then got the following integral by following the rules of Feynman calculus:
$\int[\epsilon_\mu^*(4)[\bar{v}(2)(ig_e\gamma^\mu)(\frac{i(\gamma^\mu q_\mu + mc)}{(p_1-p_3)^2-m^2c^2})(ig_e\gamma^\nu)u]\epsilon^*_\nu(3)] \times (2\pi)^4\delta^4(p_1-q-p_3)\delta^4(p_2+q-p_4)d^4q$
What I would personally get after this is the following:
$-ig_e^2[\epsilon_\mu^*(4)\bar{v}(2)\gamma^\mu(\frac{\gamma^\mu p_\mu(1)-\gamma^\mu p_\mu(3) + mc}{(p_1-p_3)^2-m^2c^2})\gamma^\nu u\epsilon^*_\nu(3)] \times (2\pi)^4\delta^4(p_2+p_1-p_3-p_4)d^4q$
What I don't understand is how every answer key I could find instead got the following:
$-ig_e^2[\bar{v}\not\!\epsilon^*(4)(\frac{\not p_\mu(1)- \not p_\mu(3) + mc}{(p_1-p_3)^2-m^2c^2})\not\!\epsilon^*(3)u] \times (2\pi)^4\delta^4(p_2+p_1-p_3-p_4)d^4q$
From how I understand it, $\not\! a \equiv a^\mu\gamma_\mu$; after all, that is the definition that is provided on page 249 of the textbook.
What this seems to be saying, however, is that $\bar{v}\not\!\epsilon^*(4)=\bar{v}(\epsilon^*)^\mu(4)\gamma_\mu$. But, given my solution, this means that $\bar{v}(\epsilon^*)^\mu(4)\gamma_\mu=\epsilon_\mu^*(4)\bar{v}\gamma^\mu$.
Additionally, by the same logic, $(\epsilon^*)^\nu(3)\gamma_\nu u=\gamma_\nu u \epsilon^*_\nu(3)$.
Is this right? Do $(\epsilon^*)^\mu$, $\gamma_\mu$, $u$, and $\bar{v}$ commute like this? Or am I missing something else? Any and all help would be tremendous! I felt like I understood Chapter 6 when I set up Feynman calculus with the 'toy model' Griffiths creates, but I'm feeling a bit lost for actual QED.
