Charge conjugation of a Dirac field (${\psi }^c=C{\bar{\psi }}^T$)

The unitary charge-conjugation operator acts on field operators and commutes with numerical chiral projectors, so $\hat{C}(P_L\psi ){\hat{C}}^{-1}=P_L{\psi }^c$. This component is left-chiral. The algebraic conjugate of an already projected spinor instead satisfies $(P_L\psi {)}^c=P_R{\psi }^c$, because its adjoint carries the opposite projector. Keeping the two operations separate avoids a false chirality conclusion.

If $C{\gamma }^{\mu T}{C}^{-1}=-{\gamma }^{\mu }$, transposition shows that $C^T{C}^{-1}$ commutes with every gamma matrix. Irreducibility of the complex Dirac representation and the Schur lemma make it scalar. Transposing $C^T=zC$ again yields $z^2=1$, so $C^T=\pm C$.

For ${\psi }^C={\eta }_{C}C{\bar{\psi }}^T$ and ${\gamma }^{0T}={\gamma }^0=-C{\gamma }^0{C}^{-1}$, the Dirac adjoint transforms as ${\bar{\psi }}^C=-{\eta }_C^{\ast }{\psi }^T{C}^{†}$. Anticommuting the components of the fermion bilinear gives $(\bar{\psi }\psi {)}^C=\bar{\psi }(C^{†}C{)}^T\psi$. Invariance of the scalar bilinear therefore fixes $C^{†}C=I$.