Two constraints of $\bar\psi$ from equations of motion for Free Dirac Field Lagrangian

$$\mathcal{L}=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi,$$ taking Euler-Lagrange equation on $$\bar\psi$$ gives the more familiar Dirac equation $$(i\gamma^\mu\partial_\mu-m)\psi=0$$ and its adjoint version $$\bar\psi(i\gamma^\mu\stackrel{\leftarrow}{\partial_\mu}+m)=0 .$$

Taking E-L equation on $$\psi$$ however gives $$\bar\psi(i\gamma^\mu\partial_\mu-m)=\partial_\mu(\bar\psi i\gamma^\mu)$$ which clearly constrains $$\bar\psi$$ differently. What is going on here?

• The r.h.s. of your third equation should not be there. I would advice you to check your derivation. Mar 14, 2023 at 6:51
• For some reason you kept the $\partial_\mu$-term on the l.h.s. of the last equation, which is supposed to be $\frac{\partial\mathcal{L}}{\partial\psi}$. If you drop it, you're fine. Mar 14, 2023 at 17:56

In quantum field theory fields $$(\phi)$$are operators but positions $$x$$ are not. So $$\partial_\mu$$ is also not an operator. That means conjugate of $${\partial_\mu\psi}$$ is just $${\partial_\mu\bar\psi}$$
So the adjoint of EOM with respect to $$\psi$$ is actually $$(i\gamma^\mu\partial_\mu+m) \bar\psi =0$$
Now use product rule on EL equation on $$\psi$$ to see that they are the same.