# How this spinor identity is shown?

In one QFT problem it is asked to prove the following identity:

$$\overline{u}_\sigma(p)\gamma^\mu u_{\sigma'}(p)=2\delta_{\sigma\sigma'}p^\mu.$$

Considering $u_\sigma$ the basis solutions to the Dirac equation usually written as

$$u_\sigma(p)=\begin{pmatrix}\sqrt{p\cdot \sigma}\xi_\sigma \\ \sqrt{p\cdot \overline{\sigma}}\xi_{\sigma}\end{pmatrix}.$$

Now my idea was to simply expand everything. So I have

$$\overline{u}_\sigma(p)\gamma^\mu u_{\sigma'}(p)=u_\sigma(p)^\dagger \gamma^0 \gamma^\mu u_{\sigma'}(p)$$

but now

$$\gamma^0\gamma^\mu=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & \sigma^\mu \\ \overline{\sigma}^\mu & 0\end{pmatrix}=\begin{pmatrix}\overline{\sigma}^\mu & 0 \\ 0 & \sigma^\mu\end{pmatrix}$$

thus we need to compute

$$\overline{u}_\sigma(p)\gamma^\mu u_{\sigma'}(p)=\begin{pmatrix}\sqrt{p\cdot \sigma}\xi_\sigma \\ \sqrt{p\cdot \overline{\sigma}}\xi_{\sigma}\end{pmatrix}^\dagger\begin{pmatrix}\overline{\sigma}^\mu & 0 \\ 0 & \sigma^\mu\end{pmatrix}\begin{pmatrix}\sqrt{p\cdot \sigma}\xi_{\sigma'} \\ \sqrt{p\cdot \overline{\sigma}}\xi_{\sigma'}\end{pmatrix}$$

expanding this I got

$$\overline{u}_\sigma(p)\gamma^\mu u_{\sigma'}(p)=\xi_\sigma^\dagger \sqrt{p\cdot \sigma}\overline{\sigma}^\mu \sqrt{p\cdot \sigma}\xi_{\sigma'}+\xi_\sigma^\dagger\sqrt{p\cdot\overline{\sigma}}\sigma^\mu \sqrt{p\cdot \overline{\sigma}}\xi_{\sigma'}^\dagger.$$

When $\mu = 0$ since $\sigma^0 = 1$ it is simple to recover the expected result with $p^0 = E$.

Now, I feel the only way to continue is to insert each value of $\mu$ there. But wait a minute, this seems like a terrible approach. Is there a better way to solve this, to get the result for general $\mu$ once and for all?

Another idea is: the spinors are solutions to the Dirac Equation in momentum space so

$$\gamma^\mu p_\mu u_{\sigma}(p)=m u_{\sigma}(p).$$

Hence if we pick $\overline{u}_\sigma(p) \gamma^\mu u_{\sigma'}(p)$ and contract with $p_\mu$ we have

$$\overline{u}_\sigma(p) \gamma^\mu p_\mu u_{\sigma'}(p)=m \overline{u}_\sigma(p)u_{\sigma'}(p)=2m^2\delta_{\sigma \sigma'}$$

but now $m^2=p_\mu p^\mu$ hence

$$(\overline{u}_\sigma(p) \gamma^\mu u_{\sigma'}(p)-2\delta_{\sigma\sigma'}p^\mu)p_\mu=0$$

But then I'm also stuck. How this identity is shown?