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We know, $2\otimes 2=3\oplus 1$. Thus we have a spin triplet of states and a spin singlet. Can we regard these states as the spin part of wavefunction for the excited states and the ground state of the deuteron nucleus? What is the difference between the members of a triplet?

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The spin part of the quantum state of any system consisting of two spin-$1/2$ particles (including a Deuterium nucleus) can be described as a general linear combination of the singlet and triplet states.

The the symbolic manipulation $2\otimes 2 = 3\oplus 1$ is telling you that the Hilbert space of the system of two spin-$1/2$ particles, which is simply the tensor product of the spin-$1/2$ Hilbert space with itself, admits an orthonormal basis for which $3$ of the states in the basis, the so-called triplet states, have total spin quantum number $s=1$, while one of the states in the basis, the so-called singlet state, has total spin quantum number $s=0$.

More explicitly, if we denote $|s, m\rangle$ as a state with \begin{align} S^2|s,m\rangle = \hbar^2 s(s+1) |s,m\rangle, \qquad S_z|s,m\rangle = \hbar m|s,m\rangle \end{align} where $S^2$ is the total spin squared operator, and $S_z$ is the $z$-component of the total spin operator, then the triplet states are \begin{align} |1,1\rangle, \qquad |1,0\rangle, \qquad |1, -1\rangle, \end{align} and the singlet state is \begin{align} |0,0\rangle, \end{align} and together they form a basis for the spin Hilbert space of the two-spin-$1/2$ system.

As can be seen explicitly in the notation, the triplet states are distinguished by the value of their "magnetic" quantum number $m$.

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