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A pion is a spin-zero composite particle, so $S = S^z = 0$. A $\pi^-$ pion can decay into an antineutrino $\bar{\nu}$ and a negatively charged lepton $l$, each with spin-$1/2$. Let the direction of antineutrino motion be the positive $z$-axis. Since all antineutrinos are right-handed (neglecting neutrino masses), the neutrino must have $S^z = +1/2$. Conservation of $S^z_\text{tot} = 0$ then requires that the lepton have $S^z = -1/2$. But then the spin degrees of freedom $$| \uparrow \rangle_{\bar{\nu}} | \downarrow \rangle_l = \frac{1}{\sqrt{2}} \left[ \frac{1}{\sqrt{2}} \left( | \uparrow \rangle_{\bar{\nu}} | \downarrow \rangle_l - | \downarrow \rangle_{\bar{\nu}} | \uparrow \rangle_l \right) \right] + \frac{1}{\sqrt{2}} \left[ \frac{1}{\sqrt{2}} \left( | \uparrow \rangle_{\bar{\nu}} | \downarrow \rangle_l + | \downarrow \rangle_{\bar{\nu}} | \uparrow \rangle_l \right) \right]$$ seem to have components in both the $S_\text{tot} = 0$ (singlet) and $S_\text{tot} = 1$ (triplet) sectors. How is this compatible with the conservation of $S_\text{tot} = 0$?

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  • $\begingroup$ The left handed neutrino has helicity -1/2, so $S^z=-1/2$ in the direction of the neutrino's momentum. From conservation of angular momentum, the antilepton must then have spin +1/2, so must also be left-handed, in contrast to the weak coupling which dictates positive chirality for it. This can only happen by dint of the anti lepton's mass, allowing a chirality-helicity mismatch--the reason it decays to a μ instead an e. I can't see why you expect to violate angular momentum. $\endgroup$ – Cosmas Zachos Aug 27 '17 at 23:53
  • $\begingroup$ @CosmasZachos Oops, I mixed up left- and right-handed helicity. Edited my question to fix. But my confusion remains: the antineutrino's and lepton's spin degrees of freedom $| \uparrow \rangle_{\bar{\nu}} | \downarrow \rangle_l$ are not in the singlet configuration, so they do not have $S_\text{tot} = 0$, as conservation of angular momentum seems to require. $\endgroup$ – tparker Aug 28 '17 at 0:05
  • $\begingroup$ @CosmasZachos As I explain in my comment to Johnathan Gross's answer, the problem is that projecting the spin state down to the singlet configuration restores the full rotational invariance of the antineutrino spin (which is maximally mixed in the singlet state), so you'd measure an equal number of left-handed and right-handed antineutrinos, in violation of the experimental results. $\endgroup$ – tparker Aug 28 '17 at 1:08
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    $\begingroup$ I can't be too helpful, here, beyond my sense that the amp you wrote is somehow malformed. The pion would be stable for massless ν s and leptons, and your states would all reduce to zero. Spin is not a good little group for luxons, of course, but the minuscule ν mass should fix that, in principle: the decay is roughly dominated only by the square of the lepton's mass. If you converted the schematic Lorentz-invariant QFT amp to helicity pieces, maybe you'd parse out the source of the malformation. $\endgroup$ – Cosmas Zachos Aug 28 '17 at 14:47
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    $\begingroup$ And perhaps this; recall no split between spin and space.... $\endgroup$ – Cosmas Zachos Aug 29 '17 at 14:44
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Only the component of the spin wavefunction that has a nonzero projection onto $S=0$ will be present in the decay. That factor of $\frac{1}{\sqrt 2}$ gets factored in and reduces the decay probability by half (which is already taken into account).

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  • $\begingroup$ That would imply that the antineutrino's and lepton's spins are are entangled into the singlet state $\frac{1}{\sqrt{2}} \left( | \uparrow \rangle_{\bar{\nu}} | \downarrow \rangle_l - | \downarrow \rangle_{\bar{\nu}} | \uparrow \rangle_l \right)$ (don't worry about the normalization constant, that's not what's bothering me). But the problem is that the singlet state is fully rotationally symmetric, so if you measured $S^z$ for the antineutrino, then 50% of time you'd measure $S^z = +1/2$ (right-handed antineutrino) and 50% of the time you'd measure $S^z = -1/2$ (left-handed antineutrino) ... $\endgroup$ – tparker Aug 28 '17 at 1:04
  • $\begingroup$ ... contradicting the experimental result that all antineutrinos appear to be right-handed. $\endgroup$ – tparker Aug 28 '17 at 1:05
  • $\begingroup$ The singlet is antisymmetric, not symmetric. What matters is what you're measuring. What you will always measures is the lepton spin being opposite the neutrino spin. If you measure their combined spin, you will see a spin of 0. If you measure their individual spins, you will see the neutrino spin in the direction of its motion. $\endgroup$ – Johnathan Gross Aug 28 '17 at 1:14
  • $\begingroup$ The spin wavefunction is intimately linked with the spatial wavefunction. If the spin up and spin down don't make sense. Right and left hand motion only matters. "Spin down" corresponds to neutrinos traveling in the -z direction. $\endgroup$ – Johnathan Gross Aug 28 '17 at 1:16
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    $\begingroup$ It's neither. Antineutrinos are right handed. You cannot talk about the spin wavefunction without the spatial wavefunction. You cannot decouple the two. $\endgroup$ – Johnathan Gross Aug 28 '17 at 1:57

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