\section{Zero magnetic field}
\label{sec:straight}
It is instructive to study the case of an electrostatic focusing channel in the absence of a magnetic field. We expect that at the magic momentum,
the projection of polarization on velocity $\bm\beta\cdot\bfs$ will be constant. Let's see how that happens. 

Now $\omega_0=\omega=0,$ and  $\beta_{xy}\rightarrow \beta_x$). 
\begin{eqnarray*}
{\cal H}&=&\begin{pmatrix} 0 &  i\Pi(t)\cr  -i\Pi(t) & 0\end{pmatrix}\\
&=& R\begin{pmatrix} \Pi(t)&0 \cr 0&  -\Pi(t)\end{pmatrix} R^{-1}
\end{eqnarray*}
where $$R=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & i\cr i & 1\end{pmatrix}$$ 
Then
\begin{eqnarray*}
\partder{}{t}\begin{pmatrix}a \cr b\end{pmatrix}&=& \frac{i}{2}R\begin{pmatrix} \Pi(t)&0 \cr 0&  -\Pi(t)\end{pmatrix} R^{-1}\begin{pmatrix}a \cr b\end{pmatrix}\\
\partder{}{t}R^{-1}\begin{pmatrix}a \cr b\end{pmatrix}&=& \frac{i}{2}R\begin{pmatrix} \Pi(t)&0 \cr 0&  -\Pi(t)\end{pmatrix} R^{-1}\begin{pmatrix}a \cr b\end{pmatrix}\\
\partder{}{t}\begin{pmatrix}\bar a \cr \bar b\end{pmatrix}&=& \frac{i}{2}\begin{pmatrix} \Pi(t)&0 \cr 0&  -\Pi(t)\end{pmatrix} \begin{pmatrix}\bar a \cr \bar b\end{pmatrix}\\
\end{eqnarray*}
The now decoupled Schrodinger equations are straightforward to solve.
\begin{eqnarray*}
\int\frac{\dot {\bar a}}{\bar a} dt&=& \frac{i}{2}\Pi(t)\ dt=\frac{i}{2} \frac{e}{ mc} \left( a_\mu + \frac{1}{\gamma+1}\right)E_z\beta_x(t)\ dt\\
\ln {\bar a}&=& i\kappa \frac{ek z_0}{2mc}\int \cos\omega_v t \beta_x^0 \cos(\beta_z^0\sin\omega_v t) \ dt \\
&\sim& i\kappa \frac{ekz_0}{2mc}\int\cos\omega_v t \beta_x^0 (1-\onehalf(\beta_z^0\sin\omega_v t)^2) \ dt \\
&\sim& i\kappa \frac{ek z_0}{2mc\omega_v} \beta_x^0\left(\sin\omega_v t -\frac{1}{3!}{\beta_z^0}^2\sin^3\omega_v t)\right) \\
&\sim& \frac{i}{2}\kappa \gamma_0 \beta_z^0 \beta_x^0\left(\sin\omega_v t\right) 
\end{eqnarray*}
where $\kappa= \left(a_\mu + \frac{1}{\gamma+1}\right)$ and we keep to order $\beta_z^2$ or equivalently $z_0^2$.
Then
\begin{eqnarray*}
\bar a&\sim& \bar a_0 \exp({\frac{i}{2} \kappa \gamma_0\beta_z^0 \beta_x^0\sin\omega_v t})\\
\bar b&\sim& \bar b_0 \exp({-\frac{i}{2} \kappa\gamma_0 \beta_z^0 \beta_x^0\sin\omega_v t})
\end{eqnarray*}
Recall
\begin{eqnarray*}
\begin{pmatrix}a \cr b\end{pmatrix}&=&R\begin{pmatrix}\bar a& \bar b\end{pmatrix} 
\end{eqnarray*}
and 
\begin{eqnarray*}
\vev{s_x} &=& \begin{pmatrix}a^* & b^*\end{pmatrix}\sigma_x \begin{pmatrix} a \cr b\end{pmatrix}\\
&=& \begin{pmatrix}\bar a^* &\bar b^*\end{pmatrix}R^\dagger \sigma_x R\begin{pmatrix} \bar a \cr \bar b\end{pmatrix}
=\begin{pmatrix}\bar a^* &\bar b^*\end{pmatrix} \sigma_x \begin{pmatrix} \bar a \cr \bar b\end{pmatrix}\\
&=& \bar a^* \bar b + \bar b^*\bar a\\
\vev{s_y}&=& \begin{pmatrix}\bar a^* &\bar b^*\end{pmatrix}R^\dagger \sigma_y R\begin{pmatrix} \bar a \cr \bar b\end{pmatrix} = -|\bar a|^2+|\bar b|^2\\
\vev{s_z}&=& \begin{pmatrix}\bar a^* &\bar b^*\end{pmatrix}R^\dagger \sigma_z R\begin{pmatrix} \bar a \cr \bar b\end{pmatrix} = |\bar a|^2-|\bar b|^2 - i(\bar a^* b-\bar b^* a)
\end{eqnarray*}
If we assume $\bar a_0=\bar b_0$, so that the polarization is initially in the $x$-direction,
\begin{eqnarray}
\vev{s_x}&=& \cos(\kappa\gamma_0\beta_z^0\beta_x^0\sin\omega_v t)\nonumber\\
\vev{s_y}&=& 0\label{eq:spinpol}\\
\vev{s_z} &=&\sin(\kappa\gamma_0\beta_z^0\beta_x^0\sin\omega_v t)\nonumber
\end{eqnarray}
Using ~\ref{eq:betaz2o},~\ref{eq:betax2o}, and~\ref{eq:spinpol} we find that
\begin{eqnarray*}
\bfs\cdot\bmbetahat &=& \frac{\beta_x^0\cos(\kappa\gamma_0\beta_z^0\beta_x^0\sin\omega_v t)\cos(\beta_z^0\sin\omega_v t) + \sin(\kappa\gamma_0\beta_z^0\beta_x^0\sin\omega_v t)\sin(\beta_z^0\sin\omega_v t)}
{[{\beta_x^0}^2\cos^2(\beta_z^0\sin\omega_v t) +\sin^2(\beta_z^0\sin\omega_v t)]^{1/2}}
%&=&\cos\left((\kappa\beta_z^0\beta_x^0-\beta_z^0)\sin\omega_v t\right) \sim 1-\onehalf(\kappa\beta_z^0\beta_x^0-\beta_z^0\sin\omega_v t)^2 +\ldots
\end{eqnarray*}
The rate of change of $\bfs\cdot\bmbetahat$ will include even harmonics of $\omega_v$.
At the magic momentum, $\bfs\cdot\bmbetahat$ is independent of time ($\kappa\gamma_0\beta_x^0\sim 1$). (See ~\ref{ss:magic} for details.)
