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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?

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    $\begingroup$ As @Toffomat mentioned, try to evaluate what $\mathcal{L}^*$ comes out to be. Clearly some terms are real by construction, such as the terms involving only $A_\mu$s. The other can be found very easily. $\endgroup$
    – Quiver
    Apr 16 '21 at 10:57
  • $\begingroup$ OK I take it from these comments that the answer is "algebraic magic makes it real", which I will work my way through. Thanks. $\endgroup$
    – HenryH
    Apr 16 '21 at 11:05
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    $\begingroup$ Note that when calculating the complex conjugate, there are serveral places where there exist different conventions (sign of the metric, Hermitean/anti-Hermitean $\gamma$s etc.), so be careful $\endgroup$
    – Toffomat
    Apr 16 '21 at 11:08
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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 $$ (\gamma^{\mu})^{\dagger}~=~ \gamma^0\gamma^{\mu}\gamma^0,\qquad (\gamma^0)^2~=~{\bf 1}.\tag{B} $$

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