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The angular momentum commutation relations implyThus $S_{\pm }|s,\sigma ⟩$ has the same total-spin eigenvalue and magnetic quantum number $\sigma \pm 1$. Its squared norm follows fromChoosing the conventional positive phases gives the spin ladder operator formula

For $s=1$, in the ordered basis $|1,1⟩,|1,0⟩,|1,-1⟩$,Since $S_x=(S_{+}+S_{-})/2$,

The incident state is the spin coherent state obtained by rotating $|1,1⟩$ through $\theta$ about the $y$ axis. Its amplitudes in the $S_z$ basis areup to an irrelevant phase convention. The Born rule therefore gives

The spin raising operator and spin lowering operator areThe angular momentum commutation relations implyso $S_{\pm }|s,\sigma ⟩$ is proportional to $|s,\sigma \pm 1⟩$. Sinceits squared norm isChoosing the conventional positive phase gives

Using the spin ladder operators,and thereforeIn the ordered product basis of the tensor product of quantum systemsthe matrix representation isThe two parallel-spin states are already eigenvectors. Diagonalizing the central block gives the four energy eigenvaluesThis is the spectrum of the Two-spin Heisenberg Hamiltonian in an opposing longitudinal field.

The asserted ground-state ordering requires the antiferromagnetic case $\alpha >0$. As $B\to 0$,The corresponding state at $E_{-}$ is the unique spin-one-half singlet statewhereas the three states at the first excited energy form the spin-one-half triplet state.

Indeed, for the total angular momentum operatorone has at $B=0$The Hamiltonian then has rotational symmetry, so energy depends only on the total-spin sector. The singlet has total spin $s=0$ and multiplicity one; the triplet has $s=1$ and multiplicity $2s+1=3$. This rotational symmetry explains why the triplet energy eigenspace has dimension three.