I was trying to prove the identity $$\overline{\displaystyle{\not}{a}\displaystyle{\not}{b}\dots \displaystyle{\not}{p}} = \displaystyle{\not}{p}\dots \displaystyle{\not}{b}\displaystyle{\not}{a}$$. On simplifying the LHS I end up with $$\overline{ \displaystyle{\not}{p}^{}} \dots \overline{ \displaystyle{\not}{b}^{}} \ \overline{ \displaystyle{\not}{a}^{}}$$.

I'm wondering if $$\overline{ \displaystyle{\not} p} = \gamma^0 \displaystyle{\not} p^{\dagger} \gamma^0 = \displaystyle{\not} p$$ as this would let me complete the final step.

• Does the overbar mean complex conjugate? For the mostly plus metric we do have $\gamma_a^\dagger = \gamma_0 \gamma_a \gamma_0$. – mike stone Oct 19 '19 at 19:11
• Hi! The overbar is defined as $\overline{\gamma} = \gamma^0 \gamma^{\dagger} \gamma^0$ – A quarky name Oct 19 '19 at 19:46

Usually, gamma-matrices are chosen in such a way that they satisfy the hermiticity condition $$\gamma^{\mu\dagger}=\gamma^0\gamma^\mu\gamma^0$$, and that is what you need, but this choice is not mandatory.
• Ah, thanks! Do the elements of the 4 vector $p_{\mu}$ also satisfy $\gamma^0 p_{\mu}^{\dagger} \gamma^0 = p_{\mu}$ – A quarky name Oct 19 '19 at 19:39
• @Aquarkyname : Yes, as $p^\dagger_\mu=p_\mu$ and typically $(\gamma^0)^2=1$. – akhmeteli Oct 19 '19 at 19:59
• Why does $p_{\mu}^{\dagger} = p_{\mu}$. Does the Hermitian adjoint not act on the elements of the complex 4 vector? – A quarky name Oct 19 '19 at 20:05
• Ah, I should have used a different letter. $p$ is just an arbitrary 4-vector, not the momentum operator. To show the identity above, I would also need to show $a_{\mu}^{\dagger} = a_{\mu}$ for instance – A quarky name Oct 20 '19 at 10:56