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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.

Shean
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2 Answers2

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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…

Cosmas Zachos
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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 $Z$.

Josh
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