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This was inspired by this question. According to Wikipedia, a Majorana neutrino must be its own antiparticle, while a Dirac neutrino cannot be its own antiparticle. Why is this true?

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3 Answers 3

here is how I understood the answer. Imagine that we look at the electron in our reference frame, and find that it travels in z direction and has spin projection +1/2. This one we call right-handed electron. Since it is massive, there exist reference frames in which observers see it as left-handed electron. For example those observers that travel faster than the electron in the z-direction. Its charge is Lorentz invariant quantity, so everyone agrees that this is electron and not positron. Massive electron is then described by four degrees of freedom (4 basic spinors), left and right handed electron and left and right-handed positron; it is Dirac field. Imagine now that we have left handed neutrino in our reference frame. Since we have lot of evidence that neutrino is massive, we can see that there are reference frames in which right handed neutrino is seen. The right handed neutrino is not observed in any experiment so far, and their mass could be very high. We only observed left handed neutrino and right handed antineutrino. So we can consistently demand that the observer in the other reference frame actually sees right handed antineutrino, without introducing right handed neutrino and left handed antineutrino. For charged particle this would not work. So we can describe neutrino by only two degrees of freedom; it is then Majorana neutrino. But the neutrinos are charged under lepton number symmetry which is global symmetry of the Standard Model. Being global means that it does not create dynamics, and there is no fundamental reason why it cannot be broken. If (local) gauge symmetry is explicitly broken then it leads to inconsistency of the theory, meaning that non-physical degrees of freedom could arise. So Majorana particles are their own antiparticles only if we do not take into account lepton number symmetry.

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In crude terms I think it amounts to the following:

Consider, for instance, a (Dirac) fermion creation operator: $c_j^\dagger$. A Majorana fermion is somehow the "real" Part of a Dirac fermion:

$m_j = c_j + c_j^\dagger$

(conventions on normalization differs). Hence a Majorana fermion transforms into itself under charge conjugation.

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This is not good as an answer either, because you can then take a pair of Majorana spinors and make a Dirac spinor out of them, and the charge conjugation eigenstates are combinations of the two. There are many different charge conjugations possible in noninteracting theories, the interesting restrictions come when when you make particles charged (hence the name "charge conjugation"). I have a hard time answering this question, because I find the idea of Dirac neutrinos laughably absurd, it is so obvious that neutrinos are Majorana. Also in (3+1)d, Majorana equals Weyl. –  Ron Maimon Dec 9 '11 at 5:37
    
@Ron yes but as you said you need two Majorana fermions to make a Dirac one –  lcv Dec 12 '11 at 20:05

Because the spinor of the Majorana neutrino is an eigenstate of the charge conjugation operator. This is different from the case of a Dirac spinor that will change under the effect of the same operator.

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3  
But why can't a Dirac spinor be an eigenstate of the charge conjugation operator? This is only a partial answer. –  Peter Shor Dec 2 '11 at 15:56
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This depends on the way the Dirac equation is formulated. In order to produce a Majorana spinor, you need to perform an unitary transformation on the $\gamma$s matrices of the Dirac equation. This will change the behavior of your solution under charge conjugation, that is given by the $\gamma_2$ matrix. –  Jon Dec 2 '11 at 16:00

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