# Derivation of the adjoint of Dirac equation

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).

• In step 1 and 3a you should still have a minus sign in front of m unless it and step 3 are typos. – Triatticus Nov 21 '18 at 12:42
• Your actual error is in step 3b, you factor a $\gamma^0$ out but this requires that both terms inside the parenthesis have a $\gamma^0$ to factor, you can fix this by changing $m$ to $\gamma^0 m \gamma^0$ – Triatticus Nov 21 '18 at 12:50
• @Triatticus Oh, now I get it. – Wang Yun Nov 21 '18 at 12:56

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