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