Consider a Majorana fermion embedded in a Dirac spinor, $$\psi = \begin{pmatrix} \psi_L \\ i \sigma_2 \psi_L^* \end{pmatrix}.$$ The Majorana fermion $\psi_L$ is left-chiral, i.e. it transforms in the $(1/2, 0)$ representation of the Lorentz group.
Now, I've also been told that you can project out chirality components using $P_L = (1-\gamma_5)/2$ and $P_R = (1+\gamma_5)/2$. Then I would have expected that $$P_L \psi = \psi, \quad P_R \psi = 0$$ though this is clearly not the case.
The problem also appears when considering charge conjugation, $$C: \psi \to -i\gamma_2 \psi^*.$$ Charge conjugation does not affect a Majorana fermion, so it leaves the representation chirality alone. But on the other hand, if $P_L \psi = \psi$, then $$P_R (C\psi) = C\psi$$ so it flips the other kind of chirality.
What is the difference between these two notions of chirality? I think my problem is that I'm conflating properties of the field (the 'representation' chirality) and properties of individual quantum states (the $P_L/P_R$ chirality). But I haven't seen any textbook distinguish between the two.
Answer
I think your problem is mostly a problem of notation. If you write two Weyl spinors inside a Dirac spinor, you should use different symbols to avoud confusion, i.e.
$$\psi = \begin{pmatrix} \xi_L \\ i \sigma_2 \xi_L^* \end{pmatrix}.$$
Now, your object $\Psi$ has a left-chiral component $\xi_L$ and a right-chiral component $i \sigma_2 \xi_L^*$. (A Dirac spinor is an object that transforms according to the $(1/2,0) \oplus (0,1/2)$ representation.) Thus it should be no surprise that $P_R \Psi \neq 0$. The point of a Majorana fermion is that the left- and right-chiral components are not independent, i.e. the right-chiral component is simply the charge conjugate of the left-chiral component. A general Dirac spinor, in contrast reads
$$\psi = \begin{pmatrix} \xi_L \\ \eta_R \end{pmatrix},$$
with $i \sigma_2 \xi_L^* \neq \eta_R$. One way to think about Majorana spinors is as "real" Dirac spinors. See sidenote 12 here.
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