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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,

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

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

1 Answer 1

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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 $$P_R C \gamma^0 P_L \sim (1+\gamma^5)\gamma^2 (1-\gamma^5) = \gamma^2 (1-\gamma^5) =\gamma_2(\gamma_0)^2(1-\gamma_5)\sim C\gamma_0P_L.$$

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  • $\begingroup$ Why could i take $C$ as you defined? I didn't get it. $\endgroup$
    – LSS
    Commented Apr 24, 2023 at 4:59
  • $\begingroup$ We can't determine a phase factor of $C$; it is arbitrary. Different textbooks use a different phase choice. This is a famous story. $\endgroup$
    – Siam
    Commented Apr 25, 2023 at 0:20
  • $\begingroup$ Didn't know it. Why is it arbitrary? $\endgroup$
    – LSS
    Commented Apr 25, 2023 at 18:34
  • $\begingroup$ $C$ 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. $\endgroup$
    – Siam
    Commented Apr 25, 2023 at 22:45

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