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:


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.


1 Answer 1


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.

  • 1
    $\begingroup$ Thank you so much, that makes a lot of sense! If I didn’t get a somewhat quick response, I figured my next step would be expanding it out by components and writing out the sums, but I knew that would be an absolute pain and a mess. $\endgroup$
    – Joshua G-F
    Commented Jun 27, 2022 at 7:14
  • $\begingroup$ Yeah I think there is no need for that... You can just keep in mind that the components of the Lorentz vectors commute with everything else inside a Feynman amplitude! Good job on your question ;) $\endgroup$
    – schris38
    Commented Jun 27, 2022 at 7:25

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