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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).

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    $\begingroup$ In step 1 and 3a you should still have a minus sign in front of m unless it and step 3 are typos. $\endgroup$
    – Triatticus
    Nov 21, 2018 at 12:42
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    $\begingroup$ 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 $\gamma^0 m \gamma^0$ $\endgroup$
    – Triatticus
    Nov 21, 2018 at 12:50
  • $\begingroup$ @Triatticus Oh, now I get it. $\endgroup$
    – Wang Yun
    Nov 21, 2018 at 12:56

2 Answers 2

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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.

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  • $\begingroup$ Thanks for your answer.....I made a mistake in my derivation $\endgroup$
    – Wang Yun
    Nov 21, 2018 at 12:57
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When you pull $\gamma^0$ to the left, you have forgotten the mass term which should have been multiplied before by $1=(\gamma^0)^2$ so you'd get $-\psi^{\dagger}(i\partial_{\mu} (\gamma^0 \gamma^{\mu} \gamma^0)+(\gamma^0)^2m)=0$

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  • $\begingroup$ Then, you multiply by $\gamma^0$ from the right. $\endgroup$
    – Gomez Bock
    Apr 21 at 14:46

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