= Solution
Write $\gamma=ge/(2m)$, retaining the printed <magnetic moment> convention. The displayed two-dimensional <matrices> are <Pauli matrices>, and the <spin> <Hamiltonian operator> is
$$
\boxed{H=-\frac{\gamma\hbar}{2}
\begin{pmatrix}B_z&B_x-iB_y\\B_x+iB_y&-B_z\end{pmatrix}}.
$$
For an x-directed field let $\omega=\gamma B_x/2$. The <Schrödinger equation> gives $i\dot c_+=-\omega c_-$ and $i\dot c_-=-\omega c_+$. With $c_+(0)=1,c_-(0)=0$, these equations yield
$$
\boxed{|\Psi(t)\rangle=\cos(\omega t)|+z\rangle+i\sin(\omega t)|-z\rangle,\qquad
\omega=\frac{geB_x}{4m}}.
$$
Equivalently, $U(t)=e^{i\omega t\sigma_x}=I\cos\omega t+i\sigma_x\sin\omega t$. If $e$ denotes a positive <elementary charge> magnitude, a free <electron> is conventionally assigned a minus sign in its <magnetic moment>; that alternative reverses the rotation sense. The expressions here consistently use the moment specified in the paper.
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