My goal is to deduce the adjoint of Dirac equation:
$$
\overline \psi (i\gamma^\mu \partial_\mu+m)=0 \tag{1}
$$
My process: I started with Dirac equation $(i\gamma^\mu \partial_\mu-m)\psi=0$. Taking the Hermitian adjoint of Dirac equation, I got
$$
\psi^\dagger(-i(\gamma^\mu)^\dagger\partial_\mu-m)=0 \tag{2}
$$
As we all know, the hermitian adjoint of $\gamma^\mu$ is that $(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$. Substituating $(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$ into equation (2), I got
$$
\psi^\dagger(-i\gamma^0\gamma^\mu\gamma^0\partial_\mu-m)=0 \\
(\psi^\dagger\gamma^0)(-i\gamma^\mu\gamma^0\partial_\mu-m)=0 \tag{3}
$$
By $\overline \psi= \psi^\dagger \gamma^0$, we have
$$
\overline \psi (i\gamma^\mu\gamma^0 \partial_\mu+m)=0 \tag{4}
$$
We can see that equation (4) is different with equation (1). (we all know that equation (1) is the right form) I have tried multiplying from the right of eq. (4) by $\gamma^0$(using $(\gamma^0)^2=1$), the adjoint equation
$$
\overline \psi (i\gamma^\mu\partial_\mu+m\gamma^0)=0 \tag{5}
$$
Equation (5) is still different with equation (1).
I am frustrated. I hope that someone could help me to build the process from eq.(5) to eq.(1).
Best Answer
Start with $$ i\gamma^\mu\partial_\mu\psi-m\psi=0. $$ Take its h.c.: $$ -i\psi^\dagger\gamma^0\gamma^\mu\gamma^0\partial_\mu-\psi^\dagger m=0. $$ Multiply by $\gamma^0$ from the right, and use $\bar{\psi}=\psi^\dagger\gamma^0$, $$ i\bar{\psi}\gamma^\mu\partial_\mu+\bar{\psi}m=0. $$
The mistake is in your step (3) as pointed out by @Triatticus.