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My understanding was that a particle may have mass if there is a quadratic term in the fields without derivatives. For a single left-handed Weyl fermion, the following expression is lorentz invariant, hermitian, and quadratic in the fields:

$$\psi_L^T \epsilon \psi_L-\psi_L^\dagger \epsilon \psi_L^*$$ where $\epsilon = -i\sigma _y$, and $^T$ is the transpose. The whole thing works with $R \leftrightarrow L$ as well.

Why can't we throw an $m^2$ prefactor over the whole thing and call it a mass term? And if we can, then is there a reason I am missing why the sources I have read suggest that Weyl particles are massless?

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Yes, it is a mass term. Its the famous Majorana mass. You are coupling the left-handed particle to its right-handed antiparticle.

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  • $\begingroup$ I thought the Majorana particle was a superposition of two Weyl particles with a constraint $\psi_L = \epsilon \psi_R^*$. Of course this can be represented in 2 d.o.f. as well, but then we'd still have one kinetic term each for $L$ and $R$-particles. Writing the $\psi_R$ kinetic term with the Majorana constraint in terms of $\psi_L$ doesn't seem to reduce to just one Weyl kinetic term, though. Is there really no difference between a Majorana particle and a Weyl particle with Majorana mass term? $\endgroup$ – doublefelix Oct 6 '19 at 13:39
  • $\begingroup$ Exactly. In 3+1 dimensions a Majorana particle is Weyl particle with a Majorana mass term. This is what might be the case for neutrinos. If so, then no weak-interaction-sterile right-handed neutrino would be required. $\endgroup$ – mike stone Oct 6 '19 at 21:13
  • $\begingroup$ Great, and the last part of my question was why the kinetic terms don't match but I found an answer here (they do, if you try harder): physics.stackexchange.com/questions/429019/… $\endgroup$ – doublefelix Oct 6 '19 at 21:33

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