2
$\begingroup$

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 $\gamma_1, \gamma_2, \gamma_3$?

$\endgroup$
  • $\begingroup$ Why make it more complicated than it needs to be? $\endgroup$ – Demosthene Apr 22 '17 at 1:49
  • $\begingroup$ @Demosthene No doubt $\gamma_2$ is a simple expression for the charge conjugation operator, but why does it lack the expected symmetry? $\endgroup$ – Sergei Patiakin Apr 22 '17 at 8:18
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.