4
$\begingroup$

The neutron is thought to consist of three tightly bound quarks, each with spin 1/2. Simple addition of angular momentum would tell us that the resulting system (neutron) could have either spin 1/2 or 3/2. So why does the neutron have spin 1/2 and not 3/2 (or a combination of both for that matter)?

I have knowledge of QM, group theory and classical field theory, so if possible I would like an answer in terms of these. If QFT is a necessity in the explanation though, then that's fine - I'll come back to this post after I learn it.

$\endgroup$
  • 1
    $\begingroup$ It's just a name. As you may see from your baryon spectroscopy, the spin 1/2 isodoublet member ddu state is called neutron, while the spin 3/2 isoquartet member state is called the $\Delta^0$, instead. These particles are spin and isospin eigenstates, but linear combinations of them feature in amplitudes and intermediate rates. Just read up on the quark model. $\endgroup$ – Cosmas Zachos Mar 1 '18 at 1:52
  • 1
    $\begingroup$ @SigmaAlpha The two different eigenstates have very different energy (mass), which means that only one (the low energy state which we call a neutron) is stable. $\endgroup$ – dmckee --- ex-moderator kitten Mar 1 '18 at 2:10
  • 1
    $\begingroup$ @SigmaAlpha In principle, yes, QCD holds the answers. In practice, the question of "where do baryons get their spin?" is an active research question, because QCD gets enormously complicated in the low-energy limit. $\endgroup$ – probably_someone Mar 1 '18 at 4:34
  • 1
    $\begingroup$ @Jon. *::facepalm:: Well, in most nuclear contexts. Or by comparison to a delta. Yeah. That's what I meant. Yeah. $\endgroup$ – dmckee --- ex-moderator kitten Mar 1 '18 at 15:47
  • 1
    $\begingroup$ @Jon I believe that dmckee meant that the neutron is stable against decays mediated by the strong force, while the delta isn't, so the lifetimes are many orders of magnitude different. Similarly, if proton decay is ever observed, I don't think that chemists will start referring to the proton as a radionuclide. (And yes, both of you gentlemen already know these things, but the distinction is interesting enough to make clear.) $\endgroup$ – rob Mar 1 '18 at 20:15

Your Answer

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

Browse other questions tagged or ask your own question.