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Pauli exclusion regarding nucleons 
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#1
Mar3114, 02:20 AM

P: 8

I think this is more or less a quick question.
So deuteron (pn) is an isosinglet in the state [itex]00> =\frac{1}{\sqrt{2}}(pnnp)[/itex] since it cannot be part of the isotriplet that includes pp and nn, since these violate pauli exclusion. That's fine. So how is it that we can have atoms like [itex] He^3=ppn [/itex]? How does this not violate Pauli exclusion with two protons bound in the nucleus (both with isospin state [itex] \frac{1}{2} \frac{1}{2}> [/itex]). It makes some sense if I think of this as a bound state of a proton and deuteron, with the deuteron being a sort of "nucleon" in its own right, where we combine [itex] p(pn)=p(d)=\frac{1}{2} \frac{1}{2}> 00> [/itex], but I can't seem to reconcile that with the fact that there are still two identical nucleons (the protons) in a bound state, which sounds like it should violate Pauli exclusion for the same reason as before. What have I misunderstood? 


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