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Majorana fermions are their own antiparticles, and Weyl fermions are just Majorana fermions without mass. However, I haven't been able to find any source that says whether a Weyl fermion is its own antiparticle.

My suspicion is that the question is meaningless. My impression is that "X is its own antiparticle" means "the mass eigenstates are mapped to themselves under charge conjugation". In the absence of a mass, we can choose any basis we want, so the question doesn't have a well-defined answer.

Then again, we can unambiguously pick out particles and antiparticles for a massless complex scalar field using the $U(1)$ charge. But I'm not sure if such a charge exists for Weyl spinors.

Is a Weyl fermion its own antiparticle? Generally, what does 'being your own antiparticle' mean?

knzhou
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  • Absence of mass, or even charge, does not make the notion of antiparticle meaningless, and a Weyl fermion is not its own antiparticle: the antiparticle is distinguished by opposite helicity. See neutrino vs. antineutrino, for instance http://www.physics.upenn.edu/~pgl/neutrino/now2000/node4.html – udrv Apr 20 '16 at 15:28
  • On a 2nd thought, you might want to look up "Majorana neutrino", which indeed is its own antiparticle. These slides give a quick overview of the neutrino problem and the difference between Weyl and Majorana neutrinos: http://www.desy.de/~troms/teaching/SoSe12/slides/neutrinoII_b.pdf – udrv Apr 20 '16 at 15:39
  • @udrv I agree with what you said, but on the level of the Lagrangian, isn't a Weyl fermion just a Majorana neutrino with zero mass? How can the property of 'being its own antiparticle' change discontinuously? – knzhou Apr 20 '16 at 16:39
  • @urdv I think there are two slightly different notions of 'antiparticle' being used interchangably; this is related to my other question. It's fine for Majorana, but for Weyl the two notions give different answers. – knzhou Apr 20 '16 at 18:36
  • I think you are onto a subtle point. There is a fundamental difference between a massless Weyl fermion and a massive Majorana one, and it has to do with the distinction between chirality and helicity. In essence, a massless fermion has no rest frame, its helicity cannot be flipped by a boost that reverses its direction of motion, and in this case chirality = helicity. Otoh, a massive fermion always has a rest frame and chirality $\neq$ helicity. A chirally left-handed Majorana fermion can have right-handed helicity, but remains chirally left-handed under boosts. – udrv Apr 21 '16 at 03:50
  • This is because helicity refers to spin orientation vs. direction of motion (physical property), while chirality concerns the type of Lorentz group representation (symmetry). Unfortunately particle physics is not exactly my strong suit and I feel kinda rusty on the topic, but I think you can find some good pointers in https://fliptomato.wordpress.com/2008/01/04/spinors-chirality-and-majorana-mass/ – udrv Apr 21 '16 at 03:51

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Consider a single Weyl fermion, which has equation of motion $$\sigma^\mu \partial_\mu \psi = 0.$$ Just like the Dirac equation, the Weyl equation is linear in momenta and hence has negative energy solutions. We then perform the usual procedure to regard these as positive energy solutions (e.g., historically, we might imagine them as holes in the Dirac sea). These two classes of solutions, $\psi$ and $\psi^c$, are related by charge conjugation.

When we say "X is its own antiparticle", we mean that solutions to the equation of motion for X are mapped to themselves under charge conjugation. However, this is a basis-dependent statement. Majorana and Weyl spinors pick out different important bases.

  • A Majorana spinor has a mass term, which couples $\psi$ and $\psi^c$ together. Then it's natural to consider the mass eigenstates, which are $\psi \pm \psi^c$. These states are mapped to themselves by charge conjugation, which is why Majorana fermions are their own antiparticles.
  • A Weyl spinor may be coupled to a gauge field in the usual way. Then $\psi$ and $\psi^c$ have conjugate gauge charge (for example, opposite $U(1)$ charges). Then it's natural to consider eigenstates that have definite charge, in which case our basis is $\psi$ and $\psi^c$. These are mapped to each other by charge conjugation, so Weyl spinors are not their own antiparticles.
knzhou
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