Is a Weyl fermion its own antiparticle? 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?
 A: 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.


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*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.

