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Bound Neutron States?

  1. May 20, 2009 #1
    Hey all,

    My year 13 physics students stumped me with this one: Why don't Neutron-Neutron (or P-P for that matter) states exist?

    Thanks in anticipation...

    Mr T
  2. jcsd
  3. May 20, 2009 #2
    It would be really nice to have a first principles calculation showing this, but I'm fairly sure such a thing does not exist.

    The textbook argument is this: the nucleon-nucleon force is spin dependent; specifically, parallel spins are favoured. If you had a proton-proton or neutron-neutron system, then due to them being fermions, the spins would have to be anti-parallel. Deuterium can be spin-parallel, and the nucleon-nucleon force is only just enough to hold it together (for instance, no excited bound state of it exists).
  4. May 21, 2009 #3


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    Gold Member

    For NN you could just look at the potentials holding it to gether vs pushing it apart.
    I think you can do some first order approximations for QCD (say, a heavy and light meson exchange yukawa potential) + gravity (very very weak) + spin+etc.

    I think you can show that there should be no bound states. The reason I mention gravity is that there IS a bound state for a neutron particle, its called a neutron star. Gravity is finally large enough at that level to pull in the same amount that the repulsive forces (degeneracy pressure+temp) are pushing out.
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