2
$\begingroup$

I'm trying to prove that the QED Lagrangian $$\mathscr{L}=\bar{\psi}(i\!\!\not{\!\partial}-m)\psi - \frac{1}{4}F^{\mu\nu}F_{\mu\nu} - J^\mu A_\mu$$ Is invariant under P and C. The two fields transform as $$A^\mu(x) \xrightarrow{P} - \eta^*\mathscr{P}^\mu{}_\nu A^\nu(\mathscr{P}x), \qquad A^\mu(x) \xrightarrow{C} \xi^* A^\mu(x)$$ $$\psi(x) \xrightarrow{P} \eta^*\gamma^0 \psi(\mathscr{P}x), \qquad \psi(x) \xrightarrow{C} i\xi^*\gamma^2 \psi^*(x)$$

I start with Parity, I have proved that $$\bar{\psi}(x) \gamma^\mu\psi(x) \xrightarrow{P} \left(\eta \bar{\psi}(\mathscr{P}x)\gamma^0\right)\gamma^\mu\left(\eta^* \gamma^0\psi(\mathscr{P}x)\right) =\mathscr{P}^\mu{}_\nu\bar{\psi}(\mathscr{P}x)\gamma^\nu\psi(\mathscr{P}x)$$ But then, if the derivative transform as $\frac{\partial}{\partial x^\mu} \xrightarrow{P}\mathscr{P}_\mu{}^\nu \frac{\partial}{\partial x^\nu} $ Then $$\bar{\psi}(x) \!\!\not{\!\partial}\psi(x) \xrightarrow{P} \bar{\psi}(\mathscr{P}x)\mathscr{P}^\mu{}_\nu \gamma^\nu\mathscr{P}_\mu{}^\lambda \partial_\lambda\psi(\mathscr{P}x) =\bar{\psi}(\mathscr{P}x) \gamma^\mu \frac{\partial}{\partial x^\mu}\psi(\mathscr{P}x)$$ Which is not really Parity invariant isn't it? I have the same problem with the term photon kinetic term, so I suspect the solution will be the same in both cases. On the other hand, for the term $A^\mu J_\mu$ I find it invariant only under the assumption that the photon has intrinsic parity $\eta_\gamma=-1$, which I know it's the case, but how can I prove it?

On the other hand, for C I find $$\bar{\psi} \gamma^\mu\psi \xrightarrow{C} -\bar{\psi}\gamma^\mu \psi$$ But I don't think that $\partial_\mu$ should transform under C, so it seems that the term $\bar{\psi}\!\!\not{\!\partial}\psi$ changes sign, so it's not invariant. Also for the term $A_\mu J^\mu$ I find it only invariant if $\xi_\gamma=-1$ which again I know it's the case, but I don't know how to prove it.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.