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I was wondering if, because for generic bounded operators, anti-distributivity applies, i.e. $$(AB)^{\dagger} = B^{\dagger}A^{\dagger},$$ the same is true of gamma matrices.

I was asked to prove $$\bar{\gamma_5} = -\gamma_5\\ (\gamma_5 \equiv i\gamma_0\gamma_1\gamma_2\gamma_3)$$ and $$\bar{M} = \gamma_0M^{\dagger}\gamma_0.$$

If anti-distributivity applies, however, we should end up with (keep in mind that $\gamma_{\mu}^{\dagger} = \gamma_0\gamma_{\mu}\gamma_0$ and $\gamma_0\gamma_0 = 1$) $$\bar{\gamma_5} = \gamma_0(-i\gamma^{\dagger}_3\gamma^{\dagger}_2\gamma^{\dagger}_1\gamma^{\dagger}_0)\gamma_0 = -i\gamma_3\gamma_2\gamma_1\gamma_0,$$ which is clearly not $-\gamma_5$. So surely the anti-distributive property cannot apply to gamma matrices, or am I missing something? Thanks in advance for answering.

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    $\begingroup$ Why do you say that your final answer is clearly not the same as $-\gamma_5$? Have you tried manipulating it further? $\endgroup$
    – knzhou
    Mar 10, 2020 at 5:54

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It looks like you are missing the defining condition for the Dirac gamma matrices

$$\{\gamma_a, \gamma_b\} = 2g_{ab} $$

All you have to do is anticommute the gamma matrices in your answer and you will get $-\gamma_5$

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