How to prove this spinor identity? 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?
 A: In fact, the calculation is even simpler than the one you started making above.
The starting point is the observation that $$\overline{u}_\sigma(p)\gamma^\mu u_{\sigma'}(p)$$
is an expression that transforms as a fourvector under a Lorentz transformation. This can easily be seen, as the kinetic term of the Lorentz invariant Dirac lagrangian (in the x-domain) is
$$\overline{\psi}_\sigma(x)\gamma^\mu \partial_\mu\psi_{\sigma'}(x)$$
Without the partial derivate, we get a fourvector.
So, if we can equate the first expression to a fourvector in one specific frame, the equation is valid in any frame.
Let us take the simplest frame, i.e. the rest frame of the fermion. In that frame, your result
$$\overline{u}_\sigma(p)\gamma^\mu u_{\sigma'}(p)=\xi_\sigma^\dagger \sqrt{p\cdot \sigma}\overline{\sigma}^\mu \sqrt{p\cdot \sigma}\xi_{\sigma'}+\xi_\sigma^\dagger\sqrt{p\cdot\overline{\sigma}}\sigma^\mu \sqrt{p\cdot \overline{\sigma}}\xi_{\sigma'}^\dagger.$$ reduces to
$$\overline{u}_\sigma(p)\gamma^\mu u_{\sigma'}(p)=m\ \xi_\sigma^\dagger \overline{\sigma}^\mu \xi_{\sigma'}+m \ \xi_\sigma^\dagger\sigma^\mu \xi_{\sigma'}=m\ \xi_\sigma^\dagger (\overline{\sigma}^\mu+ \sigma^\mu)\xi_{\sigma'}$$
The right hand side of this equation is obviously $0$ in case $\mu =1,2$ or $3$ and equal to $2m\delta_{\sigma\sigma'}$ in case $\mu=0$. Now, these are exactly the components of the fourvector $p^\mu$ in the rest frame (up to a factor $2\delta_{\sigma\sigma'}$) . So, the expression$$\overline{u}_\sigma(p)\gamma^\mu u_{\sigma'}(p)=2\delta_{\sigma\sigma'}p^\mu.$$ which holds in the rest frame, should hold in any frame.
A: Use $(\gamma^\mu k_\mu-m)u_\alpha=0$ and $\bar u_\beta(\gamma^\mu k_\mu-m)=0$ to see that
$$
m\bar u_\alpha \gamma^\nu u_\beta = k_\mu \bar u_\alpha \gamma^\nu\gamma^\mu u_\beta
\\
= \frac 12 k_\mu \bar u_\alpha(\{\gamma^\nu,\gamma^\mu\}+ [\gamma^\nu, \gamma^\mu])u_\beta\\
=k_\mu \bar u_\alpha\left(g^{\mu\nu} + \frac 12 (m-m) \gamma^\nu\right) u_\beta\\
= k^\nu  \bar u_\alpha u_\beta\\
=2m k^\nu\delta_{\alpha\beta}
 $$
