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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.

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  • $\begingroup$ After working through some examples, I'm pretty sure $P_L$ and $P_R$ actually project helicity, not chirality. Except I just saw three textbooks say the opposite. $\endgroup$ – knzhou Apr 21 '16 at 4:54
  • $\begingroup$ No $P_L$ and $P_R$ definitely project chirality not helicity. $\endgroup$ – JakobH Apr 21 '16 at 6:09
  • $\begingroup$ @JakobH Okay. Now, charge conjugation flips $P_L$ and $P_R$ eigenstates, but it does not change the Lorentz group representation. So which of these two notions is the 'real' chirality? $\endgroup$ – knzhou Apr 21 '16 at 6:11
  • $\begingroup$ Why shouldn't charge conjugation flip Lorentz representations? A spinor in the $(1/2,0)$ representation becomes a spinor in the $(0,1/2)$ representation under charge conjugation. $\endgroup$ – JakobH Apr 21 '16 at 6:25
  • $\begingroup$ Majorana fermions are not chiral, and they do not correspond to the (1/2,0) representation - that's what Weyl spinors are. $\endgroup$ – ACuriousMind Apr 21 '16 at 11:36
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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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  • $\begingroup$ Thanks for the answer! I guess my core confusion is that a Majorana fermion, written in two-component notation, is either left or right chiral but not both. But upon embedding it in a Dirac spinor, which is just a change of notation, there are both left-chiral and right-chiral components. What is the interpretation of the right-chiral component? $\endgroup$ – knzhou Apr 21 '16 at 6:27
  • $\begingroup$ @knzhou Yes, a Dirac spinor is simply a convenient notation. The point is that chirality is Lorentz invariant, but not conserved in time. Therefore in order to describe nature, we always need to take into accound that every particle can appear as left- and right-chiral. To keep track of the left- and right-chiral parts we write them into one object. You can work with Weyl spinors, but then it's harder to keep track of this chirality change in time $\endgroup$ – JakobH Apr 21 '16 at 6:34
  • $\begingroup$ Sorry, I'm now really confused. Are you saying a Majorana particle's chirality changes? I thought we used those to represent left-handed neutrinos. $\endgroup$ – knzhou Apr 21 '16 at 6:36
  • $\begingroup$ @knzhou Yes! As a MIT student I suppose you have access to SpringerLink, i.e. free Springer ebooks? If yes, have a look at chapter 8.8 in link.springer.com/book/10.1007/978-3-319-19201-7 . There this is shown explicitly $\endgroup$ – JakobH Apr 21 '16 at 6:40
  • $\begingroup$ Thanks for the reference! It made a lot of sense, and I'll definitely read it more closely later. $\endgroup$ – knzhou Apr 21 '16 at 6:52

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