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Pauli exclusion regarding nucleons

by dimwatt
Tags: exclusion, nucleons, pauli
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dimwatt
#1
Mar31-14, 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}}(pn-np)[/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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Bill_K
#2
Mar31-14, 05:04 AM
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Quote Quote by dimwatt View Post
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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