The deuteron wavefunction must be antisymmetric if the isospin representation is used (since a proton and a neutron are not identical particles, the wavefunction
need not be antisymmetric in general). Apart from their isospin, the two nucleons also have spin and spatial distributions of their wavefunction. The latter is symmetric if the deuteron is symmetric under parity (i.e. have an "even" or "positive" parity), and antisymmetric if the deuteron is antisymmetric under parity (i.e. have an "odd" or "negative" parity). The parity is fully determined by the total orbital angular momentum of the two nucleons: if it is even then the parity is even (positive), and if it is odd then the parity is odd (negative).
The deuteron, being an isospin singlet, is antisymmetric under nucleons exchange due to isospin, and therefore must be symmetric under the double exchange of their spin and location. Therefore, it can be in either of the following two different states:
*Symmetric spin and symmetric under parity. In this case, the exchange of the two nucleons will multiply the deuterium wavefunction by (−1) from isospin exchange, (+1) from spin exchange and (+1) from parity (location exchange), for a total of (−1) as needed for antisymmetry.
*Antisymmetric spin and antisymmetric under parity. In this case, the exchange of the two nucleons will multiply the deuterium wavefunction by (−1) from isospin exchange, (−1) from spin exchange and (−1) from parity (location exchange), again for a total of (−1) as needed for antisymmetry.
In the first case the deuteron is a spin triplet, so that its total spin s is 1. It also has an even parity and therefore even orbital angular momentum l ; The lower its orbital angular momentum, the lower its energy. Therefore, the lowest possible energy state has , .
In the second case the deuteron is a spin singlet, so that its total spin s is 0. It also has an odd parity and therefore odd orbital angular momentum l. Therefore, the lowest possible energy state has , .
Since  gives a stronger nuclear attraction, the deuterium ground state is in the ,  state.
The same considerations lead to the possible states of an isospin triplet having ,  or , . Thus the state of lowest energy has , , higher than that of the isospin singlet.
The analysis just given is in fact only approximate, both because isospin is not an exact symmetry, and more importantly because the strong nuclear interaction between the two nucleons is related to angular momentum in spin–orbit interaction that mixes different s and l states. That is, s and l are not constant in time (they do not commute with the Hamiltonian), and over time a state such as ,  may become a state of , . Parity is still constant in time so these do not mix with odd l states (such as , ). Therefore, the quantum state of the deuterium is a superposition (a linear combination) of the ,  state and the ,  state, even though the first component is much bigger. Since the total angular momentum j is also a good quantum number (it is a constant in time), both components must have the same j, and therefore . This is the total spin of the deuterium nucleus.
To summarize, the deuterium nucleus is antisymmetric in terms of isospin, and has spin 1 and even (+1) parity. The relative angular momentum of its nucleons l is not well defined, and the deuteron is a superposition of mostly  with some .