\section{Appendices}
\input{labframesolution}

\subsection{{Linear Albegra}
\label{ss:linearalgebra}}
\def\op{{\omega^\prime}}
Determine the eigenvalues and eigenvectors of a Hermitian matrix
$$H=\begin{pmatrix}A & B\cr B^* & -A\end{pmatrix}$$
The eigenvalues
\begin{eqnarray*}
0&=&(A-\lambda)(-A-\lambda)-|B|^2\\
\rightarrow \lambda_\pm &=&\pm\sqrt{ A^2 +|B|^2}
\end{eqnarray*}
but for conevience define $\lambda = +\sqrt{A^2+|B|^2}$.
The eigenvector equation 
\begin{eqnarray*}
\begin{pmatrix}A & B\cr B^* & -A\end{pmatrix}\begin{pmatrix}a\cr b\end{pmatrix}&=&\lambda \begin{pmatrix}a\cr b\end{pmatrix}\\
Aa+Bb &=&\lambda a,\ \ \ B^*a-Ab=\lambda b \\
b(Aa+Bb)&=& a(B^* a-Ab)\\
0&=&Bb^2+2 a b A -a^2 B^*\\
\rightarrow a&=&\frac{2Ab \pm\sqrt{(2b A)^2+4|B|^2b^2}}{-2B^*}\\
 a&=&\frac{b}{B^*}\left(A \pm\sqrt{(A)^2+|B|^2}\right) = \frac{b}{B^*}(A\pm\lambda)\\
\end{eqnarray*}
Normalize
\begin{eqnarray*}
|b|^2+|a|^2 &=& 1 = |b|^2\left(\frac{(A+\lambda)^2}{|B|^2} +1\right) \\
\rightarrow |b| &=&b= e^{i\theta}\left(\frac{|B|^2}{(A+\lambda)^2+|B^2|}\right)^{1/2}\\
 a &=& \frac{e^{i\theta}|B|}{B^*}\frac{ (A+\lambda)}{\sqrt{(A+\lambda)^2+|B^2|}}\\
 a &=& \frac{ (A+\lambda)}{\sqrt{(A+\lambda)^2+|B^2|}}\
\end{eqnarray*}
That last step follows if we choose $a$ to be real which requires that $B^*=|B|e^{i\theta}$. 

Now let's check to be sure
\begin{eqnarray*}
\frac{1}{N}\begin{pmatrix} A & B\cr B^* & -A\end{pmatrix}\begin{pmatrix} (A+\lambda)\cr e^{i\theta}|B|\end{pmatrix}&=&\lambda \frac{1}{N}\begin{pmatrix} (A+\lambda)\cr e^{i\theta}|B|\end{pmatrix}\\
\begin{pmatrix} A(A+\lambda) +B|B|e^{i\theta} \cr (A+\lambda)B^*  -A|B|e^{i\theta}\end{pmatrix}&=&\lambda\begin{pmatrix} (A+\lambda)  \cr e^{i\theta}|B|\end{pmatrix}
\end{eqnarray*} 
where $N=((A+\lambda)^2+|B^2|)^{1/2}=(2\lambda(\lambda+A))^{1/2}$.
The orthogonal eigenvectors with eigenvalues $\pm\lambda$ are
\begin{equation}
v_+ = \frac{1}{N}\begin{pmatrix} A+\lambda\cr B^*\end{pmatrix},\ \ \ v_- =\frac{1}{N}\begin{pmatrix}B\cr -(A+\lambda)\end{pmatrix}
\end{equation}

The matrix $H$ is diagonalized by the similarity transformation 
$$S^{-1}HS = \tilde H = \begin{pmatrix} \lambda_+ & 0 \cr 0 & \lambda_-\end{pmatrix}$$
where the matrix $S$ is constructed from the eigenvectors 
$$S = \frac{1}{N}\begin{pmatrix} A+\lambda  & B \cr B^* & -(A+\lambda)\end{pmatrix}$$
The general solution to the Schrodinger equation
$$\tilde H\psi = i\partder{}{t}\psi$$
are
\begin{eqnarray*}
\tilde\psi&=& \tilde a_0\begin{pmatrix} 1 \cr 0\end{pmatrix}e^{-i\lambda t}+\tilde b_0\begin{pmatrix} 0 \cr 1\end{pmatrix}e^{i\lambda t}
\end{eqnarray*}
Transform to the lab basis
\begin{eqnarray*}
\begin{pmatrix} a(t)\cr b(t)\end{pmatrix}&=& S\begin{pmatrix} \tilde a_0 e^{-i\lambda t} \cr \tilde b_0 e^{i\lambda t}\end{pmatrix}\equiv {\cal S}(t)\begin{pmatrix} \tilde a_0\cr \tilde b_0\end{pmatrix}\\
\end{eqnarray*}
where $${\cal S}(t) = \begin{pmatrix} S_{11}e^{-i\lambda t} & S_{12}e^{i\lambda t}\cr S_{21}e^{-i\lambda t}& S_{22}e^{i\lambda t}\end{pmatrix}$$
Then
\begin{eqnarray*}
\begin{pmatrix} a(t)\cr b(t)\end{pmatrix}&=& {\cal S}(t)\begin{pmatrix} \tilde a_0\cr \tilde b_0\end{pmatrix}= {\cal S}(t) S^{-1}\begin{pmatrix}a \cr b\end{pmatrix}\\
\end{eqnarray*}
Since
\begin{eqnarray*}
\begin{pmatrix} a_0\cr b_0\end{pmatrix}&=&S\begin{pmatrix} \tilde a_0  \cr \tilde b_0 \end{pmatrix}\\
\end{eqnarray*}
in the lab basis
\begin{eqnarray*}
\begin{pmatrix} a(t)\cr b(t)\end{pmatrix}&=& {\cal S}(t)\begin{pmatrix} \tilde a_0\cr \tilde b_0\end{pmatrix}= {\cal S}(t) S^{-1}\begin{pmatrix} a_0 \cr b_0\end{pmatrix}
=\frac{1}{N}\begin{pmatrix} (A+\lambda)e^{-i\lambda t}  & Be^{i\lambda t} \cr B^*e^{-i\lambda t} & -(A+\lambda)e^{i\lambda t}\end{pmatrix}
 \frac{1}{N}\begin{pmatrix} A+\lambda  & B \cr B^* & -(A+\lambda)\end{pmatrix}\\
\end{eqnarray*}
\begin{eqnarray*}
\begin{pmatrix} a \cr b\end{pmatrix}&=& \frac{1}{N^2}\begin{pmatrix} (A+\lambda)^2 e^{-i\lambda t} + |B|^2 e^{i\lambda t} & (A+\lambda)Be^{-i\lambda t}-B(A+\lambda)e^{i\lambda t}\cr\cr
 (A+\lambda)B^*e^{-i\lambda t}-B^*(A+\lambda)e^{i\lambda t} &  |B|^2 e^{-i\lambda t} + (A+\lambda)^2 e^{i\lambda t}                                   \end{pmatrix}
\begin{pmatrix} a_0\cr b_0\end{pmatrix}\\ \\
&=& \frac{1}{N^2}\begin{pmatrix} ((A+\lambda)^2+|B|^2)\cos(\lambda t) + i(|B|^2-(A+\lambda)^2)\sin(\lambda t) & -2i(A+\lambda)B\sin(\lambda t)\cr\cr
 -2i(A+\lambda)B^*\sin(\lambda t) &((A+\lambda)^2+|B|^2)\cos(\lambda t) - i(|B|^2-(A+\lambda)^2)\sin(\lambda t)                                 \end{pmatrix}
\begin{pmatrix} a_0\cr\cr b_0\end{pmatrix}\\ \\
&=& \frac{1}{N^2}\begin{pmatrix} ((A+\lambda)^2+|B|^2)\cos(\lambda t) + i(-2A(A+\lambda))\sin(\lambda t) & -2i(A+\lambda)B\sin(\lambda t)\cr\cr
 -2i(A+\lambda)B^*\sin(\lambda t) &((A+\lambda)^2+|B|^2)\cos(\lambda t) + i(2A(A+\lambda))\sin(\lambda t)                                 \end{pmatrix}
\begin{pmatrix} a_0\cr\cr b_0\end{pmatrix}\\
&=& \begin{pmatrix} \cos(\lambda t) + i\frac{(-2A(A+\lambda))}{N^2}\sin(\lambda t) & -2i\frac{(A+\lambda)B}{N^2}\sin(\lambda t)\cr\cr
 -2i\frac{(A+\lambda)B^*\sin(\lambda t)}{N^2} &\cos(\lambda t) + 2iA\frac{(A+\lambda))}{N^2}\sin(\lambda t)                                 \end{pmatrix}
\begin{pmatrix} a_0\cr\cr b_0\end{pmatrix}\\ \\
\end{eqnarray*}
Then 
\begin{eqnarray*}
\begin{pmatrix} a\cr\cr b\end{pmatrix}
&=& \begin{pmatrix} \cos(\lambda t) + i\frac{-2A(A+\lambda)}{2\lambda(A+\lambda)}\sin(\lambda t) & -2i\frac{(A+\lambda)B}{2\lambda(A+\lambda)}\sin(\lambda t)\cr
 -2i\frac{(A+\lambda)B^*\sin(\lambda t)}{2\lambda(A+\lambda)} &\cos(\lambda t) + 2iA\frac{(A+\lambda))}{2\lambda(A+\lambda)}\sin(\lambda t)                                 \end{pmatrix}
\begin{pmatrix} a_0\cr\cr b_0\end{pmatrix}\\ \\
&=& \begin{pmatrix} \cos(\lambda t) - i\frac{A}{\lambda}\sin(\lambda t) & -i\frac{B}{\lambda}\sin(\lambda t)\cr
 -i\frac{B^*\sin(\lambda t)}{\lambda} &\cos(\lambda t) + i\frac{A}{\lambda}\sin(\lambda t)                                 \end{pmatrix}
\begin{pmatrix} a_0\cr\cr b_0\end{pmatrix}\\ \\
\end{eqnarray*}
If $A=-(\omega_0-\omega)/2$, $B=-\tilde\Omega/2$ and $\omega^\prime = 2\lambda=\sqrt{(\omega_0-\omega)^2+|\tilde\Omega|^2}$
\begin{eqnarray*}
\begin{pmatrix} a\cr\cr b\end{pmatrix}&=& \begin{pmatrix} \cos(\opt2) - i\frac{\omega-\omega_0}{\op}\sin(\opt2) & i\frac{\tilde\Omega}{\op}\sin(\opt2)\cr\cr
 i\frac{\tilde\Omega^*\sin(\opt2)}{\op} &\cos(\opt2) +i\frac{(\omega-\omega_0)}{\op}\sin(\opt2)                                 \end{pmatrix}
\begin{pmatrix} a_0\cr\cr b_0\end{pmatrix}\\ \\
\end{eqnarray*}
The above is the solution in the rotating frame. In the lab frame
\begin{eqnarray*}
\begin{pmatrix}a \cr b\end{pmatrix}_{lab}&=&R^{-1}\begin{pmatrix} a \cr b\end{pmatrix}= \begin{pmatrix} e^{i\omega t/2} & 0 \cr 0 & e^{-i\omega t/2}\end{pmatrix}\begin{pmatrix} a \cr b\end{pmatrix}\\
&=& \begin{pmatrix} (\cos(\opt2) - i\frac{\omega-\omega_0}{\op}\sin(\opt2))e^{i\omega t/2} & -i\frac{\tilde\Omega}{\op}\sin(\opt2)e^{i\omega t/2}\cr\cr
 -i\frac{\tilde\Omega^*\sin(\opt2)}{\op}e^{-i\omega t/2} &(\cos(\opt2) +i\frac{(\omega-\omega_0)}{\op}\sin(\opt2))e^{-i\omega t/2}                                 \end{pmatrix}
\begin{pmatrix} a_0\cr\cr b_0\end{pmatrix}
\end{eqnarray*}

Recall that $\vev{\bm\sigma}=\vev{\sigma_x}\bfhx+\vev{\sigma_x}\bfhy+\vev{\sigma_z}\bfhz$
\begin{equation}
\vev{\sigma_x} = \begin{pmatrix}a^* & b^*\end{pmatrix}\begin{pmatrix}0 & 1 \cr 1 & 0\end{pmatrix}\begin{pmatrix} a \cr b\end{pmatrix} = a^*b + b^*a
\end{equation}
If at $t=0, \sigma_x = 1$, then $a_0=b_0=\frac{1}{\sqrt{2}}$ then
\begin{eqnarray*}
a^* &=& \frac{1}{\sqrt{2}}\left((\cos(\opt2) + i\frac{\omega-\omega_0}{\op}\sin(\opt2))e^{-i\omega t/2}  +i\frac{\tilde\Omega^*}{\op}\sin(\opt2)e^{-i\omega t/2}\right)\\
b&=&  \frac{1}{\sqrt{2}}\left(-i\frac{\tilde\Omega^*\sin(\opt2)}{\op}e^{-i\omega t/2} +(\cos(\opt2) +i\frac{(\omega-\omega_0)}{\op}\sin(\opt2))e^{-i\omega t/2}  \right)\\
a^* b&=& \frac{1}{2}\left((\cos^2(\opt2) -\frac{(\omega_0-\omega)^2}{\op^2}\sin^2(\opt2))e^{-i\omega t} + \frac{({\tilde\Omega^*})^2}{\op^2}\sin^2(\opt2)e^{-i\omega t}
 +2i\frac{\omega-\omega_0}{\op}\cos(\opt2)\sin(\opt2)e^{-i\omega t}              \right)\\
\end{eqnarray*}
The polarization, ($\tilde\Omega = \Omega + i\Pi$), 
\begin{eqnarray}
\vev{s_x}&=& \left(\cos^2(\opt2) -\frac{(\omega_0-\omega)^2-\Omega^2+\Pi^2}{\op^2}\sin^2(\opt2)\right)\cos\omega t 
 +2\left(-\frac{\Omega\Pi}{\op^2}\sin^2(\opt2)-\frac{\omega_0-\omega}{\op}\cos(\opt2)\sin(\opt2)\right)\sin\omega t\nonumber              \\
\vev{s_y}&=& -\left(\cos^2(\opt2) -\frac{(\omega_0-\omega)^2-\Omega^2+\Pi^2}{\op^2}\sin^2(\opt2)\right)\sin\omega t 
 +2\left(-\frac{\Omega\Pi}{\op^2}\sin^2(\opt2)-\frac{\omega_0-\omega}{\op}\cos(\opt2)\sin(\opt2)\right)\cos\omega t  \nonumber        \\
\vev{s_z}&=& \onehalf(|a|^2-|b|^2)= -\frac{(\omega_0-\omega)\Omega}{\op^2}\sin^2(\opt2) + \frac{\Pi}{\op}\cos(\opt2)\sin(\opt2)\label{eq:polarizationefande}
\end{eqnarray}
\subsection{Electrostatic vertical focusing in a uniform magnetic field}
\label{ss:checkeom}
The equations of motion are
\begin{eqnarray}
\frac{d\beta_z}{dt}&=&\frac{e}{\gamma m c}E_z(1 - \beta_z^2)\label{eq:eomezapp}\\
\frac{d\beta_x}{dt}&=&\frac{e}{\gamma m c}(\beta_y B_z - \beta_x\beta_z E_z)\label{eq:eomexapp}\\
\frac{d\beta_y}{dt}&=&-\frac{e}{\gamma m c}(\beta_x B_z + \beta_y\beta_z E_z)\label{eq:eomeyapp}
\end{eqnarray}
The magnetic field $\bfB= B_z\bfhz$, and the electric field $\bfE = E_z\bfhz = -kz\bfhz$. 
Let's substitute the proposed solution (to second order in the small parameter ($\beta_z^0$)) into the above equations of motion to confirm. 
\begin{equation}
\bm\beta = 
 -\beta^0_{xy}\cos(\beta_z^0\sin\omega_v t)(\cos(\omega t)\bfhx -\sin(\omega t)\bfhy) -\sin(\beta_z^0\sin\omega_v t) \bfhz\label{eq:betathree}
\end{equation}
Dropping terms beyond second order in $\beta_z^0$ the first of the three equatons of motion becomes
\begin{eqnarray*}
\frac{d\beta_z}{dt}&=&\frac{e}{\gamma m c}E_z(1 - \beta_z^2)=-\frac{e k}{\gamma_0 mc}z(1-\beta_z^2)^{3/2} \\
&=&-\frac{\omega_v^2}{c}z_0\cos\omega_v t(1 - \beta_z^2)^{3/2}=-\omega_v\beta_z^0\cos\omega_v t(1 - \beta_z^2)^{3/2}\sim -\omega_v\beta_z^0\cos\omega_v t
\end{eqnarray*}
The time derivative of $\beta_z$ from~\ref{eq:betathree} is
\begin{eqnarray*}
\der{\beta_z}{t}&=&  -\beta_z^0\omega_v\cos\omega_v t\cos(\beta_z^0\sin\omega_v t)\sim -\beta_z^0\omega_v\cos\omega_v t\\
\end{eqnarray*}
in agreeement with the preceding.
The second of the three equations of motion is
\begin{eqnarray*}
\frac{d\beta_x}{dt}&=&\frac{e}{\gamma m c}(\beta_y B_z - \beta_x\beta_z E_z)= \omega \beta_y - \beta_x\beta_z\omega_v \beta_z^0
\end{eqnarray*}
The time derivative of $\beta_x$ given in~\ref{eq:betathree} is
\begin{eqnarray*}
&=&\beta_{xy}^0\left(\omega\cos(\beta_z^0\sin\omega_v t)\sin\omega t+\beta_z^0\omega_v\cos\omega_v t\sin(\beta_z^0\sin\omega_v t)\cos\omega t \right)\\
&\sim& -\omega\beta_y - E_z \beta_z \beta_x)\\
\end{eqnarray*}
as required.
%\subsection{Lorentz Boost}
%\label{ss:lorentzboost}
\input{lorentz}
\input{magicmomentumelectrostatic}
\input{TimeDepPert}
\input{pitch_omegav_dependence_rewritten}
