Hermitian properties of the gamma matrices

The gamma matrices $\gamma^{\mu}$ are defined by

$$\{\gamma^{\mu},\gamma^{\nu}\}=2g^{\mu\nu}.$$

There exist representations of the gamma matrices such as the Dirac basis and the Weyl basis.

Is it possible to prove the relation

$$(\gamma^{\mu})^{\dagger}\gamma^{0}=\gamma^{0}\gamma^{\mu}$$

without alluding to a specific representation?

• How do you define ${}^\dagger$ without a specific representation? – ACuriousMind Nov 19 '16 at 16:12

The answer is negative. If you have a representation of the generators of the algebra made of matrices $\gamma^\mu$ and $S$ is an invertible matrix, Dirac's commutation relations $$\{\gamma^{\mu},\gamma^{\nu}\}=2g^{\mu\nu}I$$ are valid also replacing $\gamma^\mu$ for $\gamma'^\mu := S\gamma^\mu S^{-1}$.
However, if $S$ is not unitary your second commutation relation generally fails as it is generally false that $$(S\gamma^\mu S^{-1})^\dagger = S(\gamma^\mu)^\dagger S^{-1}$$ for $S$ generic. This proves that the commutation relations $$(\gamma^{\mu})^{\dagger}\gamma^{0}=\gamma^{0}\gamma^{\mu}$$ are not consequence of the Dirac's ones, thus they are not valid in general, but they depend on the choice of the representation.
• My understanding is that you lose Lorentz invariance when you impose hermiticity conditions (since the Lorentz group isn't compact. (which is another way of saying the $S$ in the above isn't unitary in general). – R. Rankin Dec 14 '18 at 11:18