For parallel spins, our instructor told us that it is because of pauli exclusion principle. Here is my doubt : if total spin is 1. That means spin part of wavefunction is symmetric, but if the spatial part of the wave function is anti-symmetric , the overall state will be anti-symmetric, which is good. How does then pauli exclusion principle rule out this possibility? Please, clarify.
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A neutron-neutron is comprised of inner quarks which are not stable in such a configuration. – twisted manifold Jan 17 '20 at 18:02
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So, the spatial part of the wavefunction can only be symmetric? (In that case, it is clear that the state wont exist) – Farman Ullah Jan 17 '20 at 18:08
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1@FarmanUllah: You're oversimplifying the question. The exclusion principle rules out the existence of some states but not others. Your question is a duplicate, so please take some time to study the answers to that one. – Jan 17 '20 at 18:26
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Space symmetric s-states are normally the lowest energy ones. – Cosmas Zachos Jan 17 '20 at 19:35
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@CosmasZachos: I think you're misapplying something that doesn't apply here. Presumably you mean something like the bonds in an H2 molecule? This has to do with specifics of the strong force. – Jan 17 '20 at 21:07
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@Ben Crowel. To be sure. It sort of holds in quark spectroscopy. It would be nice to know how often this is violated in nuclear physics. – Cosmas Zachos Jan 17 '20 at 22:46