# Why can't a Weyl Fermion have mass?

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?

• 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? – doublefelix Oct 6 '19 at 13:39