Pauli exclusion regarding nucleons

by dimwatt
Tags: exclusion, nucleons, pauli
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 P: 8 I think this is more or less a quick question. So deuteron (pn) is an isosinglet in the state $|00> =\frac{1}{\sqrt{2}}(pn-np)$ 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 $He^3=ppn$? How does this not violate Pauli exclusion with two protons bound in the nucleus (both with isospin state $|\frac{1}{2} \frac{1}{2}>$). 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 $p(pn)=p(d)=|\frac{1}{2} \frac{1}{2}> |00>$, 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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P: 4,160
 Quote by dimwatt What have I misunderstood?
Spin. The deuteron is a 3S1 state, which means the nucleons are both in the same spin state. In He3 on the other hand, the spins of the two protons are opposite.

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