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?

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Apr 16, 2021 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
    Commented Apr 16, 2021 at 11:05
  • 1
    $\begingroup$ 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
    Commented Apr 16, 2021 at 11:08
  • $\begingroup$ Did you work this through? :) $\endgroup$
    – BjornW
    Commented Jan 18, 2023 at 9:03

1 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 $$ (\gamma^{\mu})^{\dagger}~=~ \gamma^0\gamma^{\mu}\gamma^0,\qquad (\gamma^0)^2~=~{\bf 1}.\tag{B} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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