= Solution
In the ordered product <basis> $(|\uparrow\uparrow\rangle,|\uparrow\downarrow\rangle,|\downarrow\uparrow\rangle,|\downarrow\downarrow\rangle)$, the <Hamiltonian operator> has <matrix>
$$
H=\begin{pmatrix}
\alpha/4&0&0&0\\
0&-\alpha/4+\beta&\alpha/2&0\\
0&\alpha/2&-\alpha/4-\beta&0\\
0&0&0&\alpha/4
\end{pmatrix}.
$$
The middle block has <characteristic polynomial> $(E+\alpha/4)^2-\beta^2-\alpha^2/4$. Hence the exact <energy eigenvalues> are
$$
\boxed{\frac\alpha4\quad\text{(twice)},\qquad E_\pm=-\frac\alpha4\pm\sqrt{\beta^2+\frac{\alpha^2}{4}}.}
$$
This 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> gives
$$
E_\pm=-\frac\alpha4\pm\left(\beta+\frac{\alpha^2}{8\beta}+O(\alpha^4/\beta^3)\right),
$$
which agrees with the first-order result of part (c). For positive $\alpha$ and $|\beta|\ll\alpha$,
$$
E_+=\frac\alpha4+\frac{\beta^2}{\alpha}+O(\beta^4/\alpha^3),\qquad E_-=-\frac{3\alpha}{4}-\frac{\beta^2}{\alpha}+O(\beta^4/\alpha^3).
$$
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.
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