# Relation between Dirac spinor and its adjoint

I'm trying unsuccessfully to solve the following problem in Thomson's Modern Particle Physics:

"Starting from

$(\gamma^{\mu} p_{\mu} - m) u =0,$

show that the corresponding equation for the adjoint spinor is

$\bar{u} ( \gamma^{\mu} p_{\mu} - m) = 0.$

Hence, without using the explicit form for the $u$ spinors, show that the normalisation condition $u^{\dagger} u = 2E$ leads to

$\bar{u} u = 2m$

and that

$\bar{u} \gamma^{\mu} u = 2 p^{\mu}.$"

Here, $u$ is a free-particle solution to the Dirac equation (in the basis of momentum, so here the $p^{\mu}$ are c-numbers) and $\bar{u} = u^{\dagger} \gamma^0$ is as usual its adjoint spinor. The equation for the adjoint spinor is very easy to derive by just taking Hermitian conjugates of both sides of the Dirac equation, but for the life of me I can't derive a priori the latter two equations. I've tried all manner of substitutions and tricks to no avail. Could someone guide me in the right direction?

Figured it out. If I write

\begin{align}\bar{u} \gamma^{\mu} u &= \frac{1}{m} \bar{u} \gamma^{\mu} \gamma^{\nu} p_{\nu} u = \frac{1}{m} \bar{u} \left( \{\gamma^{\mu},\gamma^{\nu}\}-\gamma^{\nu} \gamma^{\mu} \right) p_{\nu} u \\ &= \frac{1}{m} \bar{u} \left( 2 g^{\mu \nu} - \gamma^{\nu} \gamma^{\mu} \right) p_{\nu} u = \frac{2}{m} \bar{u} u p^{\mu} - \frac{1}{m} \bar{u} \gamma^{\nu} p_\nu \gamma^{\mu} u \\ &= \frac{2}{m} \bar{u} u p^{\mu} - \bar{u} \gamma^{\mu} u \end{align}

I can rearrange to get the relationship

\begin{align} \bar{u} \gamma^{\mu} u = \frac{1}{m} \bar{u} u p^{\mu} \end{align}

In particular,

\begin{align} \frac{1}{m} \bar{u} u p^0 = \frac{E}{m} \bar{u} u = \bar{u} \gamma^{0} u = u^{\dagger} u = 2E \end{align}

Then I can solve for $\bar{u} u$ and $\bar{u} \gamma^{\mu} u$.