Notice that these transformations do not alter the chirality of particles. A left-handed neutrino would be taken by charge conjugation into a left-handed antineutrino, which does not interact in the Standard Model. --http://en.wikipedia.org/wiki/C-symmetry
The excerpt above seems to unambiguously answer this question. But, then:
You can easily convince yourself (exercise II.1.9) that the charge conjugate of a left handed field is right handed and vice versa. --Quantum Field Theory in a Nutshell, A. Zee
These statements appear to be contradictory. What's going on here?
Also, it does seem easy to convince myself of Zee's comment (following Zee's convention that $\psi \to \psi_c = \gamma^2 \psi^\ast$):
Suppose $\psi$ is left-handed (i.e. $P_L \psi = \psi$ and $P_R \psi = 0$), then
$P_L \psi_c = P_L \gamma^2 \psi^\ast = \gamma^2 P_R \psi^\ast = \gamma^2 (P_R \psi)^\ast = 0$
$P_R \psi_c = P_R \gamma^2 \psi^\ast = \gamma^2 P_L \psi^\ast = \gamma^2 (P_L \psi)^\ast = \psi_c$ .
Therefore, it appears that Zee's comment is correct. Can anyone help me understand why the two quotes above are or are not in contradiction?