# In David Griffiths Elementary Particles Question 7.27, why do the spinors and Bjorken and Drell gamma tensor commute?

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.

Hi and welcome to Physics stackexchange. You are right in saying that the polarization vector for any given momentum commutes with (a) gamma matrices (b) spinors (representing both particles and anti-particles) The reason is because each and every component of four-vectors is simply a scalar quantity. On the other hand, each component of the gamma matrices is a matrix (i.e. 4x4 traceless matrix), which in principle does not commute with other gamma matrices. The spinors are columns and rows that are to be multiplied with gamma matrices in a way s.t. the result is a scalar, hence this is why the polarization vector also commutes with that. A typical scalar quantity can be given for example as $$\bar{u}(\vec{p})\gamma^{\mu}u(\vec{q})\epsilon_{\mu}^*(\vec{k})$$ I can move the polarization vector (or any kind of vector) wherever I want in this small example. I can not, however, move matrices and columns/rows. If there are any more questions, please do not hesitate to ask.