Suppose we consider a left-handed Weyl spinor. Is its anti-fermion right-handed or still left-handed?
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1$\begingroup$ You are probably only getting confused because you talk vaguely about spinors and handedness, while neglecting to mention if you care about particles or fields, or chirality or helicity. If you just completely ban all use of the word "handed", then the answer will be obvious! $\endgroup$– knzhouCommented Dec 3, 2019 at 19:50
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$\begingroup$ You may find the comments of @knzhou useful physics.stackexchange.com/questions/407788/… $\endgroup$– SRSCommented Jan 3, 2020 at 7:06
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1 Answer
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It depends on how you define left/right-handedness for an anti-fermion:
- 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.
- 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.
