For deuterium, both the proton and neutron preferentially occupy $1s_{1/2}$ orbitals which have orbital angular momentum $l=0$ and spin $s=1/2$. By coupling their orbital and spin angular momenta for a new conserved quantum value $j$, one gets a spin of $1/2$ for each particle. j-j coupling then leads to either a $I=1$ or $I=0$ total nuclear spin corresponding to symmetric and anti-symmetric spin states respectively. Since the the spatial wavefunctions for the s orbitals are each of even parity, their coupled wavefunctions would be expected to be of even parity as well. So, exchanging each of the nucleons would spatially symmetric. It is theorized that deuterium represents an anti-isymmetric isospin state. Thus the total wavefunction would be anti-symmetric in the spin one state which apparently is the only bound state for deuterium. The total wavefunction being symmetric in the spin 0 state would violate spin-statistics theorem.
It is also known that even-even nuclei have protons that pair with protons and neutrons that pair with neutrons. These pairings involve identical particles and coupling their identical spatial wavefunctions would lead to even parity and hence be spatially symmetric under exchange. Isospin being symmetric, their combined spins would need to be anti-symmetric to avoid violating spin statistics. Which is why these nuclei have total spin $I=0$.
If one were to set aside isospin considerations, the proton and neutron in the deuterium are distinct particles (their masses are not the same), so their angular momentum can be coupled to a spin one or a spin zero state without violating spin-statistics. It appears that the spin one (parallel spins) state is energetically more favorable and in fact is the only bound state, with singlet state unbounded and dissociated. However, with even-even nuclei, the pairings involve identical particles so the energetically more favorable spin one state is blocked by the Pauli exclusion principle.
So my question is there a way to explain without invoking the concept of isospin, why neutron-neutron and proton-proton pairings in even-even nuclei have anti-symmetric spins whereas the deuterium nucleons pair up with symmetric spins.
