# Chirality in Feynman diagrams

In quantum mechanics, chirality is preserved for massless particles.

But in Standard Model, is there any reason for chirality to be preserved for massless particles in annihilation diagrams? (unless it would not be allowed like right-handed neutrinos)

For example, if I have a left-handed neutrino and antineutrino colliding, propagating as z-boson...

Is there any reason for the outgoing massless particles to be only left? Or can they be right also? Outgoing particles could be massless electron and positron.

I don't think Z-boson conserves any chirality, but I'm not 100% sure.

• You may have garbled your question. The Z may flip chirality for charged leptons, but not for neutrinos (since these are chargeless), so a left neutrino needs to merge with a right antineutrino to form a Z. Run to your PDG booklet review! Then write the relevant couplings in your question. Words are more ambiguous than formulas. – Cosmas Zachos Oct 1 '19 at 19:16

A left-handed neutrino + a left-handed anti-neutrino cannot annihilate under the standard model. $$\nu_L + \bar\nu_L \to Z^0$$ is impossible, as $$\bar\nu_L$$ is sterile. In fact $$\psi_L + \bar\psi_L\to (?)$$ is impossible for any $$\psi$$ under the SM. But any annihilations of hypothetical massless Weyl fermions would conserve angular momentum.