Timeline for Why doesn't the deuterium nucleus have spin $0$?
Current License: CC BY-SA 3.0
4 events
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| May 26, 2019 at 19:10 | comment | added | user4552 | If I'm understanding correctly, then this seems to be making an argument about existence of spin singlet and triplet states. Both exist, it's just that the singlet state is lower in energy. Now it may happen, as it does in deuterium, that one of these states is not bound, but that's a different issue. | |
| Jul 13, 2017 at 22:30 | comment | added | rob♦ | Nucleons, which are color singlets, don't have a color degree of freedom. That makes me doubt that the $-1$ needed for exchange antisymmetry could be hiding in the color part of the wavefunction. Where you write "flavor," I wrote "isospin"; the heavy flavors don't contribute. If you'd like me to elaborate further, please convert this to a new question and I'll answer it. | |
| Jul 11, 2017 at 19:03 | review | Late answers | |||
| Jul 11, 2017 at 19:47 | |||||
| Jul 11, 2017 at 18:43 | history | answered | Miguel Garcia | CC BY-SA 3.0 |