A very naive question about Weyl spinor

Question

Suppose we consider a left-handed Weyl spinor. Is its anti-fermion right-handed or still left-handed?

Answer 1

It depends on how you define left/right-handedness for an anti-fermion:

1. For a left-handed fermion $\gamma_5\psi_L = -\psi_L$. Its anti-fermion (Dirac conjugate) $\bar{\psi}_L$ has the chirality: $$ \bar{\psi}_L\gamma_5= {\psi}^\dagger_L\gamma_0\gamma_5 = -{\psi}^\dagger_L\gamma_5\gamma_0 = - (\gamma_5{\psi}_L)^\dagger\gamma_0 = -(-{\psi}_L)^\dagger\gamma_0= \bar{\psi}_L. $$ We may regard it as right-handed given that $\bar{\psi}_L\gamma_5 = \bar{\psi}_L$. And this is the underlying reason we can NOT have a mass term like $m\bar{\psi}_L\psi_L$ (unless for a Majorana fermion), because of the different chirality.
2. In terms of gauge interaction, the anti-fermion $\bar{\psi}_L$ still behaves as a left-handed fermion since it interacts with the left-handed-fermion-related gauge fields (e.g. weak $SU(2)$).

So the answer depends on which one of the above attributes is of your chief concern.