 Solution

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.