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.

  • 2
    $\begingroup$ 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

1 Answer 1


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 $-\gamma_5$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.