# Pauli exclusion regarding nucleons

1. Mar 31, 2014

### 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?

2. Mar 31, 2014

### Bill_K

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.