# Are massless antineutrinos in the Standard Model right-handed or right-chiral?

For massless fermions, their chirality (determining under which representation of the Lorentz group they transform) and their helicity (projection of spin onto three-momentum) eigenvalues are the same. For massless antifermions, it's exactly opposite (see e.g. Itzykson & Zuber, Eq. (2-103) and the text below).

When we talk about which particles couple to $$W$$ bosons in the Standard Model, many sources say "left-handed fermions and right-handed antifermions". Knowing that many references use "..-handed" in terms of "..-chiral", this is not a problem for fermions (since chirality and helicity is the same in the massless limit), but for antifermions, this distinction is important.

Question: Is an antineutrino that participates in the interaction with a $$W$$ boson right-handed or right-chiral?

My thoughts are as follows: The fermions in the SM are implemented via left-chiral doublets and right-chiral singlets. The $$W$$ bosons only interact with the doublets. This interaction can be written like (e.g. Cottingham & Greenwood's "Introduction to the SM of particle physics", Eq.(12.15)): $$\mathcal L = -\frac{g_2}{\sqrt 2} \nu_{eL}^\dagger \tilde\sigma^\mu e_L W_\mu^+ + ...$$ where $$\psi$$ is the left-chiral doublet. Since a Hermitian conjugation swaps the (0,1/2) and (1/2,0) reps of the Lorentz group, this means that e.g. a left-chiral $$e_L$$ and a right-chiral $$\nu_L^\dagger$$ come together. Conclusion: the $$W$$ bosons interact with left-chiral fermions and right-chiral antifermions. This would imply that (massless) neutrinos and antineutrinos in the SM are both left-handed, since a right-chiral antineutrino is left-handed.

For example, in QED, a left chiral Weyl spinor field $$\psi$$ with charge $$1$$ annihilates a left helicity particle with charge $$1$$ and creates a right helicity particle with charge $$-1$$. And the right chiral Weyl spinor field $$\psi' \equiv \psi^\dagger$$ has charge $$-1$$ and annihilates a right helicity particle with charge $$-1$$, and creates a left helicity particle with charge $$1$$.
So if you see a left helicity particle with charge $$1$$, it is meaningless to ask which chirality it is. It can be annihilated by a left chiral field, and it can be created by a right chiral field. Which field you write the Lagrangian with is purely a matter of convention; sometimes people even use both, depending on which is more convenient at the moment.