Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pauli exclusion regarding nucleons

  1. Mar 31, 2014 #1
    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?
  2. jcsd
  3. Mar 31, 2014 #2


    User Avatar
    Science Advisor

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Pauli exclusion regarding nucleons