# Question about the trace in simplifying $e\!-\!e$ cross section [duplicate] In computing a cross section for the mutual scattering of two electrons (Zee, Sec II.6), I have gotten the expression factorized so that it contains a term

\begin{align*} \tau^{\mu\nu}&=\frac{1}{2}\sum_s\sum_S\,\bar u(p,s) \gamma^\mu u(P,S) \bar u(P,S)\gamma^\nu u(p,s) ~~. \end{align*}

I want to simplify with

$$\begin{equation*} \sum_s u_\alpha(p,s)\bar u_\beta(p,s)=\left(\frac{\gamma^\mu p_\mu+m}{2m}\right)_{\!\alpha\beta}~~. \end{equation*}$$

My first step is two rewrite $$\tau^{\mu\nu}$$ with matrix multiplication indices so that I can group the two outer spinors as required for the identity. This yields

\begin{align*} \tau^{\mu\nu}&=\frac{1}{2}\sum_s\sum_S\,\bar u_\alpha(p,s) \gamma^\mu_{\alpha\beta} u_\beta(P,S) \bar u_\sigma(P,S)\gamma^\nu_{\sigma\rho} u_\rho(p,s) ~~. \end{align*}

By moving the $$u_\rho$$ on the right over to the far left, I obtain

\begin{align*} \tau^{\mu\nu}&= \frac{1}{2\big(2m\big)^2}\,\big( \gamma^\lambda p_\lambda+m \big)_{\!\rho\alpha} \gamma^\mu_{\alpha\beta} \big( \gamma^\kappa P_\kappa+m \big)_{\beta\sigma}\gamma^\nu_{\sigma\rho} ~~. \end{align*}

Zee says this is equal to

\begin{align*} \tau^{\mu\nu}&= \frac{1}{2\big(2m\big)^2}\,\text{tr}\big( \gamma^\lambda p_\lambda+m \big) \gamma^\mu \big( \gamma^\kappa P_\kappa+m \big) \gamma^\nu ~~, \end{align*}

but I am not seeing how this trace epxression follows. Any pointers?

• Traces come from contracting all the indices, and if you look at the indices you'll see they are all contracted together. Dec 7 '20 at 23:17
• I thought the trace comes contracting a single index.? Could you say a little more? In my thinking I could contract $\alpha$ and $\sigma$ to write $\text{tr}[( /\!p +m \big)_{\!\rho} \gamma^\mu_{\beta}]\text{tr}[( /\! P +m \big)_{\beta}\gamma^\nu_{\rho}]$ but this makes little sense to have a single matrix index on these objects. Do you have a formula for a trace over many indices? Dec 7 '20 at 23:26
• As in that whole string of gamma matrices can just be written as $M_{\rho \rho}$ and that's the definition of a trace. Dec 8 '20 at 0:23
• In index notation $(AB)_{ij}=A_{ik}B_{kj}$ so $\text{tr}(AB)=(AB)_{ii}=A_{ik}B_{ki}$. This extends to any number of matrices. When adjacent indices are contracted, and the last is contracted back to the first in circular fashion, you get the trace of the product. Dec 8 '20 at 1:46
• It was sufficient for me to take the trace of a matrix as the trace of a product of two and then four matrices by adding more summed indices. Dec 8 '20 at 11:29