\subsection{Muon at rest in a static field}
For orientation, and to get the signs right, 
let's be sure we understand how it looks in a static field.
In our experiment the magnetic field is in the positive z-direction. 
Therefore, the spin $+\onehalf$ muon has energy $-\bm\mu\cdot\bfB= -\hbar\omega_0/2 = -\frac{ge}{2mc}\frac{\hbar}{2} B_z$ is less than zero, since
the electric charge and $B_z$ are both positive. That is, $\omega_0>0$. Then $$H=-\frac{\hbar}{2}\begin{pmatrix} \omega_0 & 0 \cr 0 & -\omega_0\end{pmatrix}.$$
The solutions to Schrodinger's equation
\begin{equation}
i\hbar\partder{}{t}\begin{pmatrix}a\cr b\end{pmatrix} = H\begin{pmatrix}a\cr b\end{pmatrix}
\end{equation}
are $a(t)=a_0e^{i\omega_0 t/2}$ and $b(t) = b_0e^{-i\omega_0 t/2}$ where $\psi(t) = \begin{pmatrix}a\cr b\end{pmatrix}$. 
The polarization is the expectation value of the spin operator 
\begin{equation}
{\bf S}=\onehalf\vev{\bm\sigma} =\onehalf\begin{pmatrix}a^* & b^*\end{pmatrix}\bm\sigma\begin{pmatrix} a \cr b\end{pmatrix}
\end{equation}
If $a_0=b_0=\frac{1}{\sqrt{2}}$ then 
\begin{eqnarray*}
\vev{s_x}&=&\frac{1}{4}\begin{pmatrix}e^{-i\omega_0 t/2} & e^{i\omega_0 t/2}\end{pmatrix}\begin{pmatrix}0 & 1 \cr 1 & 0\end{pmatrix}
\begin{pmatrix}e^{i\omega_0 t/2}\cr e^{-i\omega_0 t/2}\end{pmatrix} = \onehalf \cos\omega t\\
\vev{s_y}&=&\frac{1}{4}\begin{pmatrix}e^{i\omega_0 t/2} & e^{-i\omega_0 t/2}\end{pmatrix}\begin{pmatrix}0 & -i \cr i & 0\end{pmatrix}
\begin{pmatrix}e^{i\omega_0 t/2}\cr e^{-i\omega_0 t/2}\end{pmatrix} = -\onehalf \sin\omega t\\
\vev{s_z}&=& |a|^2-|b|^2=0
\end{eqnarray*}
and the polarization vector rotates in a clockwise direction as anticipated, (the same direction as the muon velocity).
