I know that the Dirac equation is $$i\gamma^{\mu}\partial_{\mu}\psi=m\psi$$.

How do I use this to show that $$(\partial_{\mu}\bar{\psi})\gamma^{\mu}=im\bar{\psi}?$$


closed as off-topic by Mike, John Rennie, Jon Custer, Yashas, unsym Mar 24 '17 at 3:53

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By definition, $\bar{\psi}\equiv \psi^\dagger\gamma^0$. Knowing that $\gamma^0\gamma^0 = 1$ (4x4 identity matrix), that $\psi$ is a 4 component vector and that $\gamma^0\gamma^{\mu\dagger}\gamma^0=\gamma^\mu$. Then:

$$(i\gamma^\mu\partial_\mu\psi)^\dagger\gamma^0 = (m\psi)^\dagger\gamma^0$$ Now, for any product of matrices we have $(AB)^\dagger = B^\dagger A^\dagger$. Thus,

$$-i\partial_\mu\psi^\dagger\gamma^{\mu\dagger}\gamma^0 = m\psi^\dagger\gamma^0$$

Inserting the identity matrix $1=\gamma^0\gamma^0$ between $\partial_\mu\psi^\dagger$ and $\gamma^{\mu\dagger}$ we have:

$$\partial_\mu\psi^\dagger\gamma^0\gamma^0\gamma^{\mu\dagger}\gamma^0 = im\bar{\psi}$$ $$\partial_\mu\bar{\psi}\gamma^0\gamma^{\mu\dagger}\gamma^0 = im\bar{\psi}$$

Plugging the last identity we finally have: $$\partial_\mu\bar{\psi}\gamma^{\mu} = im\bar{\psi}$$

Which is what you wanted!


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