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.

  • $\begingroup$ 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. $\endgroup$ Oct 1, 2019 at 19:16

1 Answer 1


First, neutrinos are now known to have mass, meaning "sterile" right-handed neutrinos with zero electroweak interaction likely exist.

Second, chirality is only conserved when it coincides with helicity. What's really conserved is angular momentum.

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.

  • $\begingroup$ The fact that neutrinos have mass does not imply that right-handed neutrinos exist. This conclusion would only follow if the mass is a Dirac mass, but it could also be a Majorana mass. $\endgroup$ Oct 1, 2019 at 8:42
  • $\begingroup$ @flippiefanus Edited. But why not just edit the otherwise helpful answer to a confused question, rather than angrily downvoting? $\endgroup$ Oct 1, 2019 at 15:25
  • $\begingroup$ The downvoter is not me $\endgroup$ Oct 2, 2019 at 4:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.