Why $n$-$n$ and $p$-$p$ interaction do not exist just like $n$-$p$ interaction in a deutron?

Question

Why n-n and p-p interaction do not exist just like a bound n-p system i.e. a deutron?

Answer 1

Interactions of particles always exist, and the question is really asking why there are no bound states of nn and pp.

To get a bound state there has to be an effective potential where in order for the particles to be free energy must be supplied. In the np there is an effective attractive potential due to the spill over strong interaction, called a residual strong force, that attracts the two nucleons and forms a bound state.The energy of the system is such that the mass of the deuteron is smaller than the mass of the proton+neutron, due to special relativity algebra, and the neutron cannot decay, so a stable nucleus exists.

In the case of proton-proton the repulsion of the coulomb potential is such that the system is very unstable, that is what "negative binding energy" means:

Helium-2 is an extremely unstable isotope of helium. Its nucleus, a diproton, consists of two protons with no neutrons. According to theoretical calculations, it would have been much more stable (although still undergoing β+ decay to deuterium) if the strong force had been 2% greater. Its instability is due to spin–spin interactions in the nuclear force, and the Pauli exclusion principle, which forces the two protons to have anti-aligned spins and gives the diproton a negative binding energy.

In the case of nn , the neutrons can decay and their being neutral makes experiments difficult. It is assumed they cannot form bound states, only resonances in scattering or decay of larger nuclei.

The existence of a resonance is measured There is a paper when a dineutron with a binding energy of the dineutron to the nucleus it comes from, is measured

A bonded dineutron emission with the binding energy (Bdn) within limitations 1.3 MeV < Bdn < 2.8 MeV

There are Quantum Mechanical models that can show why there are bound and not bound dinucleon states not using the Pauli exclusion .

See this for example .

Show that the isospin structure of the interaction matrix element implies that the isosinglet deuteron is bound while the isotriplet, (pp, np, nn), is not.