0
$\begingroup$

In $\beta^+$ decay, a proton (consisting of 2 up and 1 down quarks) decays into a neutron (1 up and 2 down quarks), a positron and an electron neutrino.

enter image description here

(Image source: https://commons.wikimedia.org/wiki/File:Feynman-beta-plus-decay.svg)

If I'm not mistaken, all quarks, the positron and the neutrino each have a spin of 1/2. So in the overall process, there is a net gain of spin 1. How is this possible? Where does the extra spin come from?

(If spin conservation didn't matter on the other hand, I wouldn't see any argument for the production of the neutrino.)

$\endgroup$
  • $\begingroup$ Spin and angular momentum are vector quantities. You can have two spin-1 objects that form a system with a net angular momentum of zero. $\endgroup$ – dukwon Jul 18 '17 at 10:16
1
$\begingroup$

The addition of angular momenta in quantum mechanics is rather more complicated than it is in classical mechanics. In short, it is possible for two spin-1/2 particles to couple in such a way (with "opposite" angular momenta) that the system as a whole behaves as a spin-0 system. They can also couple in a way that will give total system spin 1, but that option is ruled out in this case because of the constraint you note.

$\endgroup$
  • $\begingroup$ Thanks. So if you can't use the spin, why does the neutrino have to be produced? $\endgroup$ – jmb Jul 18 '17 at 10:25
  • $\begingroup$ From a modern perspective, or as a historical development? $\endgroup$ – Emilio Pisanty Jul 18 '17 at 10:31
  • $\begingroup$ From a modern perspective. I know it was originally postulated because of the continuous energy distribution of the electron/positron. $\endgroup$ – jmb Jul 18 '17 at 10:32
  • $\begingroup$ From a modern perspective, it is an integral part of the Standard Model, which wouldn't survive without the neutrinos, so the question makes very little sense. In this particular reaction the neutrino is required because otherwise you'd have a different reaction, which can range over a wide range of possibilities with multiple constraints, so again the question makes very little sense. You can argue that the neutrino is there to preserve lepton number (or more specifically electron number) or that the only available $e$, $W^+$ vertices involve a neutrino, though. $\endgroup$ – Emilio Pisanty Jul 18 '17 at 11:58
  • 2
    $\begingroup$ From a modern perspective the neutrino is present because lepton number must be conserved at the $W$-$e^+$ vertex. Though historical development of lepton-number as quantity we expect to be conserved was driven in part by the observation of neutrinos in beta decay processes. $\endgroup$ – dmckee Jul 18 '17 at 15:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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