# Bound Neutron States?

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

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).

Hepth
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.