= Solution
The remaining cyclic <commutators> are $[S_y,S_z]=i\hbar S_x$ and $[S_z,S_x]=i\hbar S_y$. In a spin-$s$ irreducible representation,
$$
\boxed{S_z=\hbar m_s,\quad m_s=-s,-s+1,\ldots,s;\qquad S^2=\hbar^2s(s+1)}.
$$
For <spin> one-half, $S_i=\hbar\sigma_i/2$, appropriate to an <electron>'s <spin>. Each $S_i^2=\hbar^2I/4$, so this <Hamiltonian> is $H=A\hbar^2I/4$. The orthonormal vectors $(1,0)^T$ and $(0,1)^T$ are eigenstates with the same <energy> \b[$A\hbar^2/4$]; every normalized <superposition> is also an eigenstate. The $B$ term vanishes identically because <spin-one-half quadratic anisotropy is constant>.
For the supplied three-dimensional basis one consistent choice is
$$
S_y=i\hbar\begin{pmatrix}0&0&1\\0&0&0\\-1&0&0\end{pmatrix},\qquad
S_z=i\hbar\begin{pmatrix}0&1&0\\-1&0&0\\0&0&0\end{pmatrix}.
$$
These <Hermitian matrices> obey the cyclic <commutators> with the given $S_x$, and $S^2=2\hbar^2I$, representing <spin> one. Their squares give
$$
H=\hbar^2\operatorname{diag}(A-B,A+B,0).
$$
Thus the three normalized Cartesian basis vectors $e_1,e_2,e_3$ have energies \b[$\hbar^2(A-B)$, $\hbar^2(A+B)$ and $0$], respectively; degeneracies permit arbitrary orthonormal choices in the corresponding <eigenspaces>.
The parallel and antiparallel $S_x$ states are $|+x\rangle=(0,1,-i)^T/\sqrt2$ and $|-x\rangle=(0,1,i)^T/\sqrt2$. Evolution from the first gives
$$
|\psi(t)\rangle=\frac1{\sqrt2}\begin{pmatrix}0\\e^{-i(A+B)\hbar t}\\-i\end{pmatrix},\qquad
P_{-x}(t)=\sin^2\left(\frac{(A+B)\hbar t}{2}\right).
$$
This <spin-one reversal under quadratic anisotropy> is complete when
$$
\boxed{t=\frac{(2j+1)\pi}{|A+B|\hbar},\quad j=0,1,\ldots}
$$
provided $A+B\ne0$. If $A+B=0$, its initial state is stationary up to phase and complete reversal never occurs. Assume real $A,B$ for a Hermitian <Hamiltonian>.
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