2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.