Pauli exclusion regarding nucleons

[dimwatt]

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?

 

[Bill_K]

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.