I'm trying to derive the adjoint Dirac equation from the Lagrangian:


To start, I plugged it into the Euler-Lagrange equation with my variation variable being $\psi$:


Which yields:


Next, I factored out the $i$, then used the anticommutation relation $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ to swap $\gamma^0$ (from the adjoint wavefunction) and $\gamma^\mu$. However, when I do this I don't get something resembling the Dirac adjoint; I have an extra $2\partial_\mu\psi^\dagger$ term, so I must have done something wrong (I end up getting $2\partial_\mu\psi^\dagger-\partial_\mu(\psi^\dagger\gamma^\mu\gamma^0)-m\bar{\psi}=0$). Could I just have a pointer in the right direction; I'm unsure how to proceed from what I have from the Euler-Lagrange equation to $\bar{\psi}(i\gamma^\mu\partial_\mu+m)=0.$


1 Answer 1


If you want to derive the equation of motion for $\bar \psi$, then you already have it. It's commonly written as

$$\bar \psi \left(i\gamma^\mu \overleftarrow{\partial}_\mu+m\right)=0$$

Where it's understood that we should act with the derivative on $\bar\psi$ as normal, i.e. $\partial_\mu \bar \psi$, but that the $\gamma^\mu$ must still multiply on the right.

  • $\begingroup$ I understand that's the equation, but how do I get from $\partial_\mu (i\bar{\psi}\gamma^\mu)-m\bar{\psi}=0$ to the equation you wrote. Does the derivative going to the right also conjugate $i$? $\endgroup$ Jan 21 at 20:01
  • $\begingroup$ @moboDawn_φ well your second term in the equation of motion $-\frac{\partial L}{\partial \psi}$ is off by a minus sign. Also your Lagrangian appears to be missing a factor of $i$ in the first term. $\endgroup$ Jan 21 at 20:03
  • $\begingroup$ Oh I see. Sorry, that was a stupid mistake on my part. $\endgroup$ Jan 21 at 20:04
  • 1
    $\begingroup$ @moboDawn_φ not stupid, just a small mistake. Happens all the time. :) $\endgroup$ Jan 21 at 20:28
  • $\begingroup$ Thanks for pointing it out to me! $\endgroup$ Jan 21 at 20:29

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