QED Lagrangian - real or complex?

Question

I'm confused about the use of complex numbers in the QED Lagrangian: $$\mathcal{L}=\bar{\psi}(i\gamma^{\mu}\partial_{\mu} - m)\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e\bar{\psi}\gamma^{\mu}A_\mu\psi.$$

Clearly, the Dirac field spinor has complex components. The $\gamma^\mu$ matrices involve imaginary numbers.

Is there some algebraic magic that means that $\mathcal{L}$ always comes out real, or is it complex? And if it's complex then how does the action $\int{\mathcal{L}d^4x}$ come out real? What about $A^\mu$ - are its values real, and if so, how does the RHS of the EOM $\partial_{\nu}F^{\nu\mu}=e\bar{\psi}\gamma^\mu\psi$ come out real given that the $\gamma^\mu$ matrices involve imaginary components?

Answer 1

OP's Lagrangian density is up to a total divergence term equal to $${\cal L}~=~\bar{\psi}(\frac{i}{2}\gamma^{\mu}\underbrace{\stackrel{\leftrightarrow}{\partial}_{\!\mu}}_{=\stackrel{\rightarrow}{\partial}_{\mu}-\stackrel{\leftarrow}{\partial}_{\mu}} - m)\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e\bar{\psi}\gamma^{\mu}A_\mu\psi,\tag{A}$$ which in turn is real. Here we use the conventions $$ (\gamma^{\mu})^{\dagger}~=~ \gamma^0\gamma^{\mu}\gamma^0,\qquad (\gamma^0)^2~=~{\bf 1}.\tag{B} $$