Proving Even Parity for this Expression

[Markus Kahn]

Homework Statement

Prove that
$$\bar { \psi } ( x ) \gamma ^ { \mu } \partial _ { \mu } \psi ( x )$$
has even parity for $\gamma^\mu$ a set of matrices satisfying the Clifford algebra and $\psi$ a bispinor (an element that transforms under the $(\frac{1}{2},0)\oplus (0,\frac{1}{2} )$ representation of the Lorentz group) and $\bar\psi=\psi^\dagger\gamma^0$ the adjoint spinor.

Relevant Equations

Not entirely sure but I think if we write $\psi$ in terms of left/right-handed Wely spinors the parity operation is given by
$\psi(x)\longmapsto \psi'(x')=\gamma^0\psi(x).$

My idea was straight forward calculation:
$$\begin{align*}\bar { \psi }' ( x' ) \gamma ^ { \mu } \partial _ { \mu }' \psi ( x' ) &= \psi^\dagger\gamma^{0\dagger}\gamma^0\gamma^\mu \partial_\mu'\gamma^0\psi = \bar\psi\underbrace{\gamma^0\gamma^\mu\gamma^0}_{=\gamma^{\mu\dagger}=-\gamma^\mu} \partial_\mu'\psi = - \bar\psi\gamma^\mu\partial_\mu'\psi\\
&\overset{!}{=}\bar { \psi } ( x ) \gamma ^ { \mu } \partial _ { \mu } \psi ( x ) .\end{align*}$$
For this to be true the parity should therefore transform $\partial_\mu\mapsto -\partial_\mu$ which doesn't really make sense to me (I can't see why $\partial_0$ should pick up a sign under parity).. I found in a post here on physics formus (I sadly can't find the link right now) that $\partial_\mu\mapsto\partial^\mu$ under parity, which honestly confused me completely...

I assume that I'm doing something completely wrong, but I just don't really see what...

 

[Nate810]

The correct transformation is indeed $\partial_\mu\mapsto\partial^\mu$. This is the natural result when you consider that the components of the four-vector $\partial^\mu$ transform as the components of a contravariant vector, while the components of the four-vector $\partial_\mu$ transform as the components of a covariant vector. Under a parity transformation, the contravariant components of a four-vector are simply exchanged with the covariant components, so we have $\partial^\mu\mapsto\partial_\mu$ and $\partial_\mu\mapsto\partial^\mu$.