 Solution

Use the PDF's convention in which $C$ multiplies ${\psi }^{\ast }$, rather than ${\bar{\psi }}^T$. Its unitarity converts the given gamma matrix identity intoApplying it twice givesFor real infinitesimal Lorentz transformation parameters,Therefore ${\psi }^c=C{\psi }^{\ast }$ transforms in exactly the same spinor representation as $\psi$. The coordinate pullback is real and transforms identically as well. This is the antilinear charge-conjugation intertwiner relation; exponentiating also gives $C{S}^{\ast }=SC$ for connected transformations.

For real mass, complex conjugation of the free Dirac equation gives $(-i{\gamma }^{\mu \ast }{\partial }_{\mu }-m){\psi }^{\ast }=0$. Multiply by $C$ and move it through the gamma matrices:ThusAs an explicit check, in the given Weyl representation of the gamma matrices one may take $C=i{\gamma }^2$. It is unitary and obeys the required conjugation identity. The free-equation conclusion does not assert that a charge-conjugated field retains the same charge in a fixed electromagnetic background.