Why are both C and CP violation necessary for baryogenesis?

Question

Sakharov condition states both C and CP violation is necessary for baryogenesis. Now consider, for example, a theory with B-number violating interaction and C-violation. Therefore, if $p\to e^+\gamma$ is an allowed B-violating process, C-violation would imply that the C-conjugated process $\bar{p}\to e^-\gamma$ would occur at a different rate. Shouldn't therefore, C-violation be sufficient for baryogenesis? Why do we also need CP-violation?

Answer 1

Assume You have only $C$-violation. Then it implies that the rate $\Gamma$ of hypothetical process $p^{-} \to e^{-}\gamma$ won't be equal to the rate of hypothetical process $p^{+} \to e^{+}\gamma$, but only for the given helicities $L/R$. Say, $$ \tag 1 \Gamma\big(p_{L}^{+} \to e^{+}_{R}\gamma_{L}\big) \neq \Gamma(p_{L}^{-} \to e^{-}_{R}\gamma_{L}), $$ and $$ \tag 2 \Gamma\big(p_{R}^{+} \to e^{+}_{L}\gamma_{R}\big) \neq \Gamma(p_{R}^{-} \to e^{-}_{L}\gamma_{R}) $$ (note that the "photon" $\gamma$ has helicities $\pm 1$ while the "electron" $e$ and the "proton" $p$ have helicites $\pm \frac{1}{2}$). (Added) This is because the $C$-transformation changes the particle on corresponding antiparticle without changing the helicity.

But let's assume that such processes respect $CP$-symmetry, (added) under which the left/right particle is changed on right/left antipatrticle. Then there must be $$ \tag 3 \Gamma (p^{-}_{L} \to e^{-}_{R}\gamma_{L}) = \Gamma (p^{+}_{R} \to e_{L}^{+}\gamma_{R}) $$ Let's add separately the left and the right hand-sides of $(1)$ and $(2)$ and use $(3)$. We obtain $$ \Gamma\big(p_{R}^{+} \to e^{+}_{L}\gamma_{R}\big) + \Gamma\big(p_{L}^{+} \to e^{+}_{R}\gamma_{L}\big) = \Gamma(p_{L}^{-} \to e^{-}_{R}\gamma_{L})+ \Gamma(p_{R}^{-} \to e^{-}_{L}\gamma_{R}), $$ and no total baryon-asymmetry will be generated.

Therefore we require the CP-asymmetry.