# Deriving adjoint Dirac equation from Lagrangian

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

$$\mathcal{L}=i\bar{\psi}\gamma^\mu\partial_\mu\psi-m\bar{\psi}\psi$$

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

$$\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)}\right)-\frac{\partial\mathcal{L}}{\partial\psi}=0$$

Which yields:

$$\partial_\mu(i\bar{\psi}\gamma^\mu)-m\bar{\psi}=0$$

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 $$\bar{\psi}(i\gamma^\mu\partial_\mu+m)=0.$$

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.
• 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 $i$? Jan 21 at 20:01
• @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. Jan 21 at 20:03