1
$\begingroup$

I know neutral pions mostly decay electromagnetically to two photons, but I don't understand why the decay to two neutrino's is not possible.

Perhaps they violate parity, but someone at the link https://www.physicsforums.com/threads/is-neutrino-spin-parity-1-2-or-1-2.934771/ said it does not make sense to talk about parity for neutrino's since they are produced only via the weak interaction, which does not conserve parity.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ The answer is clear, thank you. I thought it had to do with the helicities of the neutrino's causing the weak suppression. $\endgroup$
    – Shean
    Commented Apr 3, 2022 at 22:02
  • 1
    $\begingroup$ Oh!, that too! You are right. In complete analogy to charged-pion decays, the weakly-interacting neutrino will be left-handed and the antineutrino right-handed, so, there will also be a helicity violation proportional to a power of the neutrino masses! The decay is triply doomed. $\endgroup$
    – Cosmas Zachos
    Commented Apr 3, 2022 at 22:10

2 Answers 2

Reset to default
5
$\begingroup$

Pion decay to two photons is electromagnetic (the original flavor chiral anomaly). But decay to two neutrinos could only go through a box diagram involving two Ws of opposite charge: the quarks or antiquarks hairpin of the pion wavefunction has to decay to a virtual $W^+ ~ W^-$ pair, which then connect to a similar hairpin involving a virtual charged lepton and a $\nu ~\bar \nu$ pair of decay product particles.

  • That, then, entails a doubly weak suppression, by, at the very least, a factor of $(m_\pi/M_W)^8$ ... twenty two orders of magnitude. Hopeless.

As for parity, indeed, the doubly weak interaction blasts parity to vapor, but further consider that a neutrino-anti neutrino pair has the parity of a quark-antiquark pair…

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Hopeless, but not forbidden, so it's "possible" but super rare. $\endgroup$
    – rfl
    Commented Apr 1, 2022 at 11:04
  • 1
    $\begingroup$ Wait until you start studying nucleon decay…. $\endgroup$
    – Cosmas Zachos
    Commented Apr 1, 2022 at 11:53
  • 1
    $\begingroup$ hahahaha, LOL :) $\endgroup$
    – rfl
    Commented Apr 1, 2022 at 15:16
1
$\begingroup$

I think the best answer is a chirality argument, much in the same way that $\Gamma( \pi^- \to \mu^- \overline{\nu}_\mu) \gg \Gamma (\pi^- \to e^- \overline{\nu}_e)$. See here.

A $\pi^0$ has a $J^P = 0^-$, so the final-state $\nu,\,\overline{\nu}$ would need to have back-to-back spins to conserve angular momentum. This would require either a right-handed chiral state neutrino or a left-handed chiral state antineutrino, forbidding this decay.

Also, Cosmas' answer is not quite correct -- decay to two neutrinos could proceed via the exchange of a virtual $Z$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Indeed, not "only": the singly weak NC decay is helicity suppressed as well. $\endgroup$
    – Cosmas Zachos
    Commented Apr 1 at 21:28

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.