# Charge+Parity operator lead left-handed to right-handed

So i need to show that the, if $$\psi$$ is left-handed,

$$C\gamma^0\psi^*$$

Is right-handed.

So, we know that, for any $$\psi$$, $$P_L \psi$$ is left handed.

Also, for any $$\omega$$, is right-handed, $$P_R \omega$$ is right-handed

And so, here, i need to show that

$$P_R C \gamma^0\psi^* = P_R C \gamma^0(P_L \psi)^* = P_R C \gamma^0 P_L^{*} \psi^{*} = C \gamma^0 P_L^{*} \psi^{*}$$

Where i have used the property that, if $$\epsilon$$ is right-handed, $$P_R \epsilon = \epsilon$$

So, what we need in another words, is to show that

$$P_R C \gamma^0 P_L^{*} = C \gamma^0 P_L^{*}$$

But, instead, when i evaluate the product, i get that

$$P_R C \gamma^0 P_L^{*} = 0$$

Indeed,

$$P_R C \gamma^0 P_L^{*} \propto (1+\gamma^5)\gamma^2 \gamma^0 (1-\gamma^5) = \gamma^2 \gamma^0 - \gamma^5 \gamma^2 \gamma^0 \gamma^5 = \gamma^2\gamma^0 - \gamma^2 \gamma^0 = 0$$

• I deleted my answer, since the interpretation of $C\gamma^0\psi^*$ is not clear to me. Commented Apr 21, 2023 at 20:29

If you can take a charge conjugate matrix as $$C=i\gamma_2\gamma_0$$, then $$C\gamma_0\sim\gamma_2$$ and your final equations become $$P_R C \gamma^0 P_L \sim (1+\gamma^5)\gamma^2 (1-\gamma^5) = \gamma^2 (1-\gamma^5) =\gamma_2(\gamma_0)^2(1-\gamma_5)\sim C\gamma_0P_L.$$

• Why could i take $C$ as you defined? I didn't get it.
– LSS
Commented Apr 24, 2023 at 4:59
• We can't determine a phase factor of $C$; it is arbitrary. Different textbooks use a different phase choice. This is a famous story.
– Siam
Commented Apr 25, 2023 at 0:20
• Didn't know it. Why is it arbitrary?
– LSS
Commented Apr 25, 2023 at 18:34
• $C$ is defined by $C(\gamma_\mu)^T C^{-1}=-\gamma_\mu.$ Firstly, the transpose of gamma matrices depends on their representation, so $C$ depends on the choice of the representation of $\gamma$s. Also, we can’t determine the phase factor of $C$ from this equation because the phase factor of $C^{-1}$ cancels that of $C$. Additional information is briefly summarized in Wikipedia.
– Siam
Commented Apr 25, 2023 at 22:45