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.


There are lots of questions on this site about chirality versus helicity, but you can clear them all up by just remembering that helicity is a property of particles (since it's defined in terms of a particle's momentum and spin), and chirality is a property of fields (since it's defined in terms of a field's Lorentz transformation properties). Fields are mathematical constructs that create and annihilate particles. It is meaningless to say a particle has the "same" chirality and helicity, because particles don't have chirality, fields do.

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.

In the Standard Model with massless neutrinos, all neutrinos have left helicity and all antineutrinos have right helicity. Under the standard conventions, the neutrino field is left chiral, but you often also work with its conjugate which is right chiral. As you can see from the above paragraphs, which chirality is "really" in play is actually a meaningless question!

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    $\begingroup$ Thank you very much, this is a great answer. Let me try to summarize the situation for massless neutrinos: the field content of the SM Lagrangian concerning interactions with the W bosons are left-chiral neutrino fields and right-chiral antineutrino fields. The particles that can scatter off the W bosons are left-handed neutrinos and right-handed antineutrinos. Is this correct? $\endgroup$ – pmpyt Jun 9 at 5:43

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