Question

Why deuteron cannot exit in excited state?

Answer 1

Deuterium is light enough that isospin is a good symmetry: it's fair to neglect the electric charge and consider only the strong interaction between the proton and neutron. In that case we should expect essentially the same excitation structure in the diproton, the dineutron, and the neutron-proton system.

The proton and neutron are both fermions, and a state containing two of them must be antisymmetric under exchange. If they are bound without any orbital angular momentum, the only way to make the state antisymmetric is for the two particles to occupy a spin singlet. So the ground state of the diproton or dineutron must have spin zero. Since the diproton and dineutron are both unstable, isospin symmetry tells us that spin-zero deuteron should also be unstable.

In the stable deuteron the state is made antisymmetric by the isospin part of the wavefunction: the deuteron is a (symmetric) spin triplet but an (antisymmetric) isospin singlet. (It's true but irrelevant that the deuteron wavefunction is only mostly ss-wave; there's a small contribution of dd-wave with two units of orbital angular momentum, but that doesn't change the symmetry arguments or the total nucleon spin.)

If you prefer, you can turn this argument around. If it's the case that the strong interaction is more important than electrical repulsion in light nuclei, and ifyou found a stable two-nucleon state with zero angular momentum, you would expect to find that two-nucleon state for all allowed values of the charge: the dineutron, the deuteron, and the diproton. We don't find any evidence for stable dineutrons or diprotons, and we also don't find any spinless bound deuterons.

Answer 2

Answer:

It is possible the deuteron has excited state.

Explanation:

1.  T. Belgya, S.B. Borzakov, M. Jentschel, B. Maroti, Yu.N. Pokotilovski, L. Szentmiklosi, “Experimental search for the bound state singlet deuteron in the radiative n-p capture”, Phys. Rev. C99 044001, 2019.