# Does charge conjugation affect parity?

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

$P_L \psi_c = P_L \gamma^2 \psi^\ast = \gamma^2 P_R \psi^\ast = \gamma^2 (P_R \psi)^\ast = 0$

and

$P_R \psi_c = P_R \gamma^2 \psi^\ast = \gamma^2 P_L \psi^\ast = \gamma^2 (P_L \psi)^\ast = \psi_c$ .

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?

I have messed around with some matrices and I think I have the answer. Essentially, I think there is a disagreement about the meaning of left- and right-handedness.

By applying the charge conjugation operation $\psi_{c}=C=i\gamma_{2}\psi^*$ in the Dirac basis, a spin-up positive energy electron $e^-(E,\mathbf{p},up)=u_{1}$ is converted into a negative energy electron $e^-(-E,-\mathbf{p},up)=u_{4}$. Now this state has apparently got negative helicity since the momentum and spin are opposite now (initially we assumed them parallel).

However, in terms of the physical particles, this state is just a positron of positive energy and opposite momentum $e^-(-E,-\mathbf{p},up)=e^+(E,\mathbf{p},up)=v_{4}$ and so we are back to calling it positive helicity.

Now I think Zee is using the convention that an L-projection operator kills a right-handed state and an R-projection kills a left-handed state. In this case, Zee is talking about positive and negative energy particles.

The wikipedia article is talking about the handedness of the spins of the physical particles seen in experiments, in which case, the left-projection operator for a particle is L and an antiparticle is R (and vice versa).

All very confusing.

Wikipedia page used Srednicki convention of $\mathcal{C}=i\gamma^0\gamma^2$ operator, while Zee definition is different by a $\gamma^0$ factor. If you repeat the calculation you just did with Srednicki definition of $\mathcal{C}$, you'll see the correctness of wiki comments.

• So, the only difference is that Srednicki uses $\mathcal{C} = i \gamma^0 \gamma^2$ and Zee uses $\mathcal{C} = \gamma^2$. It seems that by including the $\gamma^0$, then srednicki is also doing a partity transformation (since $\gamma^0$ gives a parity transformation). Thus, it seems srednicki is doing a $CP$ transformation, whereas Zee is doing a $C$ transformation. What do you think? – adhanlon Mar 7 '16 at 21:00