# Dirac spinor parity

I'm not sure I understand the effect of a parity transform on a Dirac spinor $\left( \begin{array}{c} \psi_R\\ \psi_L\\ \end{array} \right)$. I've been given the definitions $P\psi=\gamma_0\psi$, which would mean swapping $\psi_L$ and $\psi_R$. I've also been told that $P\bar{\psi}=\bar{\psi}\gamma_0$. Based on the definition $\bar{\psi}=\psi^{\dagger}\gamma_0$, then $$P\bar{\psi}=\psi^{\dagger}\gamma_0^2 = \psi^{\dagger} = \left( \begin{array}{c} \psi_R^* && \psi_L^*\\ \end{array} \right)$$ But then the parity transform wouldn't have had any effect on $\bar{\psi}$, which doesn't seem right, so where did my reasoning go wrong?

I've looked at these two questions:

But I don't really understand the notation, or whether it addresses my problem.

## 1 Answer

What does $\bar \psi$ look like in the first place? Well, it is $\psi^\dagger \gamma^0 = (\psi_L^*\ \psi_R^*)$ according to your choice of $\psi$. It makes sense that parity swaps these slots.

$$(\psi_L^*\ \psi_R^*) \stackrel{P}\to (\psi_R^*\ \psi_L^*)$$