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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.

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    $\begingroup$ 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
    Apr 3, 2022 at 22:02
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    $\begingroup$ 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. $\endgroup$ Apr 3, 2022 at 22:10

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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…

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

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  • $\begingroup$ Indeed, not "only": the singly weak NC decay is helicity suppressed as well. $\endgroup$ Apr 1 at 21:28

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