# Charge conjugation operator and gamma matrices

The gamma matrices are defined by their anticommutation relations, which are symmetrical in permutations of $\gamma_1, \gamma_2, \gamma_3$. Given this symmetry, why is the change conjugation operator $\gamma_2$, rather than some symmetrical expression in $\gamma_1, \gamma_2, \gamma_3$?

• Why make it more complicated than it needs to be? – Demosthene Apr 22 '17 at 1:49
• @Demosthene No doubt $\gamma_2$ is a simple expression for the charge conjugation operator, but why does it lack the expected symmetry? – Sergei Patiakin Apr 22 '17 at 8:18

The charge conjugation operator $C$ cannot be expressed as a representation-invariant polynomial in $\gamma^0, \gamma^1, \gamma^2, \gamma^3$. Proof: Under a spinor basis change $U$, the gamma matrices transform as $\gamma^\mu \rightarrow U \gamma^\mu U^{-1}$, so any polynomial $P$ will transform likewise. But the charge conjugation operator transforms as $C \rightarrow U^* C U^{*-1}$, so cannot be expressed by any $P$.
In the Dirac representation, $C$ happens to be given by $\gamma^2$. This is a coincidence due to our choice of basis - in another basis it will not be true. As shown above, no polynomial expression can hold for every basis, no matter whether the expression is symmetrical in $\gamma^{1,2,3}$ or not.