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In the ordered product basis $(|\uparrow \uparrow ⟩,|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\downarrow \downarrow ⟩)$, the Hamiltonian operator has matrixThe middle block has characteristic polynomial $(E+\alpha /4{)}^2-{\beta }^2-{\alpha }^2/4$. Hence the exact energy eigenvalues areThis is the Two-spin Heisenberg Hamiltonian in an opposing longitudinal field, with its interaction and field parameters rescaled to the paper's units. For positive $\beta$ and $|\alpha |\ll \beta$, the Taylor expansion giveswhich agrees with the first-order result of part (c). For positive $\alpha$ and $|\beta |\ll \alpha$,These agree with the second-order triplet and singlet results. If $\alpha <0$, their association with the signs $\pm$ reverses; if $\beta <0$, the two antiparallel product-state labels reverse. The unordered exact spectrum and the perturbative expressions with their signed denominators remain valid.

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.