 Solution

For a real vector relativistic quantum field, $V^{\mu }=V^{\mu †}$. Its transformed field must also be a Hermitian operator, so the intrinsic charge-conjugation phase satisfies ${\eta }_C={\eta }_C^{\ast }$; combined with $|{\eta }_C|=1$, this givesThe quantum electrodynamics interaction is $-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. Since the Dirac electromagnetic current is odd under charge conjugation, charge conjugation invariance of this interaction requires the photon field to be odd too: