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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?

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

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  • $\begingroup$ 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). $\endgroup$ – R. Rankin Dec 14 '18 at 11:18
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Not exactly, one can choose gamma-matrices that satisfy the anti-commutation conditions, but do not satisfy the hermiticity conditions.

EDIT: (11/19/2016) However, one can argue that the hermiticity condition for gamma-matrices can be derived from hermiticity of the Dirac Hamiltonian (https://arxiv.org/pdf/physics/0703214.pdf). Nevertheless, one can live with gamma-matrices that do not satisfy the condition, but are still related to the gamma-matrices of standard representations via a similarity condition.

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