= Solution
The <spin raising operator> and <spin lowering operator> are
$$
S_+=S_x+iS_y,
\qquad
S_-=S_x-iS_y.
$$
The <angular momentum commutation relations> imply
$$
[S_z,S_\pm]=\pm\hbar S_\pm,
\qquad
[S^2,S_\pm]=0,
$$
so $S_\pm|s,\sigma\rangle$ is proportional to $|s,\sigma\pm1\rangle$. Since
$$
S_\mp S_\pm=S^2-S_z^2\mp\hbar S_z,
$$
its squared <norm> is
$$
\begin{aligned}
\|S_\pm|s,\sigma\rangle\|^2
&=\langle s,\sigma|S_\mp S_\pm|s,\sigma\rangle\\
&=\hbar^2\{s(s+1)-\sigma(\sigma\pm1)\}.
\end{aligned}
$$
Choosing the conventional positive phase gives
$$
\boxed{
S_\pm|s,\sigma\rangle
=\hbar\sqrt{s(s+1)-\sigma(\sigma\pm1)}
\,|s,\sigma\pm1\rangle.}
$$
Using the <spin ladder operators>,
$$
\mathbf S^{(1)}\mathbin\cdot\mathbf S^{(2)}
=S_z^{(1)}S_z^{(2)}
+\frac12\left(S_+^{(1)}S_-^{(2)}
+S_-^{(1)}S_+^{(2)}\right),
$$
and therefore
$$
H=\alpha S_z^{(1)}S_z^{(2)}
+\frac\alpha2\left(S_+^{(1)}S_-^{(2)}
+S_-^{(1)}S_+^{(2)}\right)
+B\left(S_z^{(1)}-S_z^{(2)}\right).
$$
In the ordered product basis of the <tensor product of quantum systems>
$$
|\uparrow\uparrow\rangle,\quad
|\uparrow\downarrow\rangle,\quad
|\downarrow\uparrow\rangle,\quad
|\downarrow\downarrow\rangle,
$$
the <matrix representation> is
$$
H=
\begin{pmatrix}
\alpha\hbar^2/4&0&0&0\\
0&-\alpha\hbar^2/4+B\hbar&\alpha\hbar^2/2&0\\
0&\alpha\hbar^2/2&-\alpha\hbar^2/4-B\hbar&0\\
0&0&0&\alpha\hbar^2/4
\end{pmatrix}.
$$
The two parallel-spin states are already <eigenvectors>. Diagonalizing the central block gives the four <energy eigenvalues>
$$
\boxed{
E_{\uparrow\uparrow}=E_{\downarrow\downarrow}
=\frac{\alpha\hbar^2}{4},
\qquad
E_\pm=-\frac{\alpha\hbar^2}{4}
\pm\sqrt{B^2\hbar^2+\frac{\alpha^2\hbar^4}{4}}.}
$$
This 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 $B\to0$,
$$
E_-=-\frac{3\alpha\hbar^2}{4},
\qquad
E_+=E_{\uparrow\uparrow}=E_{\downarrow\downarrow}
=\frac{\alpha\hbar^2}{4}.
$$
The corresponding state at $E_-$ is the unique <spin-one-half singlet state>
$$
|0,0\rangle
=\frac{|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle}{\sqrt2},
$$
whereas the three states at the first excited energy form the <spin-one-half triplet state>.
Indeed, for the <total angular momentum operator>
$$
\mathbf S=\mathbf S^{(1)}+\mathbf S^{(2)},
$$
one has at $B=0$
$$
H=\frac\alpha2\left(
S^2-(S^{(1)})^2-(S^{(2)})^2
\right).
$$
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.
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