A partial answer, is that supposing the gamma matrices, block-diagonal , as $\begin{pmatrix}A&\\&\epsilon A\end{pmatrix}, \begin{pmatrix}&A\\\epsilon A&\end{pmatrix}$, where $A$ is hermitian or anti-hermitian, and $\epsilon =\pm1$, give constraints on $A$ and $\epsilon$ due to $(\gamma^0)^2= \mathbb Id_4, (\gamma^i)^2= - \mathbb Id_4$.

For instance, if $\gamma_0 = \begin{pmatrix}&A\\ \epsilon A&\end{pmatrix}$, then $(\gamma_0)^2 = \begin{pmatrix} \epsilon A^2&\\ &\epsilon A^2\end{pmatrix}$.

So, if $A$ is hermitian, we may choose $A$ such $A^2 = AA^\dagger = A^\dagger A = \mathbb Id_2$, and $\epsilon = 1$

If $A$ is anti-hermitian, we may choose $A$ such $A^2 = - AA^\dagger = - A^\dagger A= -\mathbb Id_2$, and $\epsilon=-1$

In the two cases, it is easy to see that $\gamma^0$ is hermitian.

So, with the above hypothesis about the gamma matrices, it is easy to see that $\gamma^0$ is hermitian and the $\gamma^i$ are anti-hermitian.

Now with the anti-commutation relations $\gamma^0 \gamma^i + \gamma^i \gamma^0 =0$, you have $\gamma^i= - \gamma^0 \gamma^i \gamma^0$ (remembering that $(\gamma^0)^2= \mathbb Id_4$), so you have $(\gamma^i)^\dagger= - \gamma^i = \gamma^0 \gamma^i \gamma^0$, and you have obviously $(\gamma^0)^\dagger= \gamma^0 = \gamma^0 \gamma^0 \gamma^0$

This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user Trimok