\subsection{Perturbative Solution for a Time Dependent Hamiltonian}
\label{ss:revisit}
Our goal is to solve~\ref{eq:pitchingse} to second order in $\beta_z^0$.
(It is convenient to adjust the phase of the vertical oscillation so that when $\omega_v\rightarrow 0$, the
Hamiltonian takes the form of the system with fixed pitch, and to that end include a phase shift $\phi=\pi/2$.)  
\begin{eqnarray*}
\Omega(t)+i\Pi(t) &=& -\Omega_0\sin(\omega_v t+\pi/2) +i\Pi_0\cos(\omega_v t+\pi/2)\\ 
 &=& -\Omega_0\cos(\omega_v t) -i\Pi_0\sin(\omega_v t)\\ 
&=&\frac{1}{2}\left(-\Omega_0(e^{i\omega_v t}+e^{-i\omega_v t}) -\Pi_0(e^{i\omega_v t}-e^{-i\omega_v t})\right)\\
&=&\frac{-1}{2}\left((\Omega_0+\Pi_0)e^{i\omega_v t}+(\Omega_0 -\Pi_0)e^{-i\omega_v t})\right)\\
&=&\left(X_0e^{i\omega_v t}+Y_0e^{-i\omega_v t})\right)
\end{eqnarray*}
where
\begin{eqnarray*}
\Omega_0&=&-z_0\frac{eB}{mc}\left(a_\mu\frac{\gamma}{\gamma+1}\frac{\omega_v}{c}\right)\beta_{xy}(t)\\
\Pi_0&=&-z_0\frac{eB}{mc}\left(a_\mu+\frac{1}{\gamma+1}\right)\frac{\omega_v^2}{c\omega}\beta_{xy}(t)
\end{eqnarray*}
At the magic momentum, and with the substitution $\beta_z^0=-z_0\omega_v/c$

\begin{eqnarray}
X_0 &=& -\frac{\Omega_0+\Pi_0}{2}= -\frac{1}{2}\beta_z^0\omega_a(\frac{\gamma}{\gamma+1} +\gamma\frac{\omega_v}{\omega})\beta_{xy}\label{eq:x0}\\
Y_0&=&-\frac{\Omega_0-\Pi_0}{2} =-\frac{1}{2}\beta_z^0\omega_a(\frac{\gamma}{\gamma+1} - \gamma\frac{\omega_v}{\omega})\beta_{xy}\label{eq:y0}\\
\end{eqnarray}
Also
\begin{eqnarray*}
\omega_0(t) &=& \omega_0-\pi_0\sin^2(\omega_v t+\pi/2) = \omega_0 -\onehalf\pi_0(1+\cos 2\omega_v t)
=(\omega_0-\onehalf\pi_0)-\frac{1}{2}\pi_0\cos 2\omega_v t\\
&\equiv& \omega_0^\prime - \pi^\prime_0\cos 2\omega_v t\\
\end{eqnarray*}
with $$\omega_0^\prime=\omega_0-\frac{\pi_0}{2}=\frac{eB}{mc}\left(a_\mu+\frac{1}{\gamma}-a_\mu\frac{\gamma}{\gamma+1}\frac{(\beta_z^0)^2}{2}\right)
\ \ {\rm and}\ \ \pi_0^\prime=\frac{eB}{mc}\left(a_\mu\frac{\gamma}{\gamma+1}\frac{(\beta_z^0)^2}{2}\right)$$
\def\op{{\eta}}
and $$\op = (\omega_0^\prime-\omega) = \omega_a(1-\frac{\gamma}{\gamma+1}\frac{(\beta_z^0)^2}{2})$$
Then we can write, 
\begin{eqnarray*}
{\cal H}&=& {\cal H}_0 + {\cal H}^\prime
\end{eqnarray*}
where in the rotating frame
\begin{eqnarray*}
{\cal H}_0 &=& -\frac{\hbar}{2}\begin{pmatrix} \omega_0^\prime-\omega & 0 \cr 0 & -\omega_0^\prime +\omega\end{pmatrix}
=\hbar\begin{pmatrix}\eta/2 & 0 \cr 0 & -\eta/2\end{pmatrix}\\\\
{\cal H}^\prime &=&\begin{pmatrix}H_{aa}^\prime & H_{ab}^\prime \cr H_{ba}^\prime & H_{bb}^\prime\end{pmatrix}
= -\frac{\hbar}{2}\begin{pmatrix}\pi_0^\prime\cos 2\omega_v t & X_0 e^{i\omega_v t}-Y_0 e^{-i\omega_v t}   \cr
X^*_0 e^{-i\omega_v t}-Y^*_0 e^{i\omega_v t} & -\pi_0^\prime  \cos 2 \omega_v t\end{pmatrix}   
\end{eqnarray*}
The solutions to the unperturbed hamiltonian, (${\cal H}_0$) are
\begin{eqnarray*}
a(t) &=& a_0 e^{-i\op t/2}\\
b(t) &=& b_0 e^{i\op t/2}
\end{eqnarray*}
The general solution to the Schrodinger equation for the full Hamiltonian is written
\begin{equation}
\psi(t)=c_a(t)\psi_a e^{-i\op t/2}+c_b(t)\psi_b e^{i\op t/2}
\end{equation}
$\psi_a$ and $\psi_b$ are a pair of eigenkets of ${\cal H}_0$.
Substitution into the Schrodinger equation and some rearrangement gives us
\begin{eqnarray}
\dot c_a &=& \frac{-i}{\hbar}[c_a H^\prime_{aa}+c_b H^\prime_{ab}e^{i\op t}]\nonumber\\
\dot c_b &=& \frac{-i}{\hbar}[c_b H^\prime_{bb}+c_a H^\prime_{ba}e^{i\op t}]\label{eq:de}
\end{eqnarray}
Note that all of the terms in the perturbative part of the Hamiltonian are proportional to pitch angle, $\beta_z^0$
or ${\beta_z^0}^2$. We cannot solve the differential equations exactly, 
but for small pitch angle, the perturbation is small, and
we solve by successive approximations. 
At $t=0$, the muon is in the state $c_a(0),c_b(0)$
with polarization 
\begin{eqnarray*}
\vev{\bfs}&=& \begin{pmatrix}c_a & c_b\end{pmatrix}\bm\sigma\begin{pmatrix} c_a\cr c_b\end{pmatrix}\\
&=&  \begin{pmatrix}c_a & c_b\end{pmatrix}\sigma_x\begin{pmatrix} c_a\cr c_b\end{pmatrix}\bfhx
+  \begin{pmatrix}c_a & c_b\end{pmatrix}\sigma_y\begin{pmatrix} c_a\cr c_b\end{pmatrix}\bfhy
+  \begin{pmatrix}c_a & c_b\end{pmatrix}\sigma_z\begin{pmatrix} c_a\cr c_b\end{pmatrix}\bfhz\\
&=&  \begin{pmatrix}c_a & c_b\end{pmatrix}\begin{pmatrix}0 & 1 \cr 1 & 0\end{pmatrix}\begin{pmatrix} c_a\cr c_b\end{pmatrix}\bfhx
+  \begin{pmatrix}c_a & c_b\end{pmatrix}\begin{pmatrix}0 & -i \cr i & 0\end{pmatrix}\begin{pmatrix} c_a\cr c_b\end{pmatrix}\bfhy
+  \begin{pmatrix}c_a & c_b\end{pmatrix}\begin{pmatrix}1 & 0 \cr 0 & -1\end{pmatrix}\begin{pmatrix} c_a\cr c_b\end{pmatrix}\bfhz\\
\end{eqnarray*}
If we choose $c_a(0)=c_b(0)=\onroot2$ then the polarization $\bfs(0) = s_x$ is initially in the x-direction and
the zeroth order solution is $c_a^{(0)}=c_b^{(0)}= \onroot2$.
To compute the first order approximation substitute the zeroth order solution into ~\ref{eq:de} and integrate
\begin{eqnarray}
 c^{(1)}_a &=& \frac{-i}{\hbar}\int_0^t[c^{(0)}_a H^\prime_{aa}+c^{(0)}_b H^\prime_{ab}e^{i\op \tp}]d\tp\nonumber\\
 c^{(1)}_b &=& \frac{-i}{\hbar}\int_0^t[c^{(0)}_b H^\prime_{bb}+c^{(0)}_a H^\prime_{ba}e^{-i\op \tp}]d\tp\nonumber
\end{eqnarray}
We know that in the case of a fixed pitch angle ($\beta_z$ constant), the precession frequency depends on $\beta_z^2$,
so we will need to go at least to second order. But now that is straightforward.
\begin{eqnarray}
 c^{(2)}_a &=& \frac{-i}{\hbar}\int_0^t[c^{(1)}_a H^\prime_{aa}+c^{(1)}_b H^\prime_{ab}e^{i\op \tp}]d\tp\nonumber\\
 c^{(2)}_b &=& \frac{-i}{\hbar}\int_0^t[c^{(1)}_b H^\prime_{bb}+c^{(1)}_a H^\prime_{ba}e^{-i\op \tp}]d\tp\nonumber
\end{eqnarray}
Then we will have
\begin{eqnarray*}
c_a(t) &=& c_a^{(0)}+ c_a^{(1)}+ c_a^{(2)} + \ldots\\
c_b(t) &=& c_b^{(0)}+ c_b^{(1)}+ c_b^{(2)} + \ldots
\end{eqnarray*}
In order to simplify the calculation it is helpful to take advantage of the known symmetries of ${\cal H}^\prime$, in particular
$H_{bb}^\prime = -H_{aa}^\prime$ and $H_{ba}^\prime = {H_{ab}^\prime}^*$, and introduce a bit of notation.
Define
\begin{eqnarray*}
A_1&=& \frac{1}{\hbar}\int_0^t[ H^\prime_{aa}]d\tp\\
B_1&=&\frac{1}{\hbar}\int_0^t[ H^\prime_{ab}e^{i\op \tp}]d\tp
\end{eqnarray*}
Then
\begin{eqnarray*}
c_a^{(1)}&=& -i(c_a^{(0)}A_1 +c_b^{(0)}B_1 \rightarrow \frac{-i}{\sqrt{2}}(A_1+B_1)\\
c_b^{(1)}&=& i(c_a^{(0)} A_1- c_v^{(0)}B_1^*) \rightarrow \frac{i}{\sqrt{2}}(A_1-B_1)
\end{eqnarray*}
where we set $c_a^{(0)}=c_b^{(0)}=\onroot2$.
Next since
\begin{eqnarray*}
 c^{(2)}_a &=& \frac{-i}{\hbar}\onroot2\int_0^t[-i(A_1+B_1) H^\prime_{aa} -i(-A_1+B_1^*) H^\prime_{ab}e^{i\op \tp}]d\tp\\
 c^{(2)}_b &=& \frac{-i}{\hbar}\onroot2\int_0^t[-i(-A_1+B_1^*)(-H^\prime_{aa})-i(A_1+B_1)(H^\prime_{ab}e^{i\op \tp})^*]d\tp
\end{eqnarray*}
it is convenient to also define
\begin{eqnarray*}
 C_2 &=& \frac{1}{\hbar}\int_0^t[A_1 H^\prime_{aa}]d\tp\\
 D_2 &=& \frac{1}{\hbar}\int_0^t[B_1 H^\prime_{aa}]d\tp \\
 E_2 &=& \frac{1}{\hbar}\int_0^t[(-A_1) H^\prime_{ab}e^{i\op \tp}]d\tp \\
 F_2 &=& \frac{1}{\hbar}\int_0^t[B_1^* H^\prime_{ab}e^{i\op \tp}]d\tp
\end{eqnarray*}
so that
\begin{eqnarray*}
c_a^{(2)}&=& -\onroot2(C_2+D_2+E_2+F_2)\\
c_b^{(2)}&=& -\onroot2(C_2-D^*_2-E^*_2+F^*_2)
\end{eqnarray*}
In that last step we use the fact that $H^\prime_{aa}$ is real, and with the initial condition that $c_a(0)=c_b(0)=\onroot2$, $A_1$ is also real.
Then
\begin{equation}
\psi(t) = \begin{pmatrix}a(t)\cr b(t)\end{pmatrix}=\begin{pmatrix}c_a(t)  e^{-i\op t/2} \cr c_b(t) e^{i\op t/2}\end{pmatrix}
\end{equation}
The observable is the projection of the polarization onto the velocity.
\begin{eqnarray*}
{\bm\sigma\cdot\bm\beta} = \begin{pmatrix}\beta_z & \beta_x -i\beta_y \cr \beta_x+i\beta_y & -\beta_z\end{pmatrix}
=\begin{pmatrix}-\beta_z^0\sin\omega_v t & -\beta_{xy}^0\cos(\beta_z^0\sin\omega_v t)e^{-i\omega t}\cr -\beta_{xy}\cos(\beta_z^0\sin\omega_v t)e^{i\omega t} & \beta_z^0\sin\omega_v t\end{pmatrix}
\end{eqnarray*}
with expectation value
\begin{eqnarray*}
&=&-(|a|^2-|b|^2)\beta_z^0\sin(\omega_v t) -(a^* b) \beta_{xy}^0\cos(\beta_z^0\sin\omega_v t)e^{-i\omega t}
-(b^*a)\beta_{xy}\cos(\beta_z^0\sin\omega_v t)e^{i\omega t}\\
&=&-(|a|^2-|b|^2)\beta_z^0\sin(\omega_v t) -2\Re\left((a^* b) \beta_{xy}^0\cos(\beta_z^0\sin\omega_v t)e^{-i\omega t}\right)
\end{eqnarray*}
Let's write out $a(t)$ and $b(t)$
\begin{eqnarray*}
a(t)&=&\onroot2\left(1 -i(A_1+B_1)-(C_2+D_2+E_2+F_2)\right)e^{-i(\op+\omega) t/2}\\
a^*(t)&=&\onroot2\left(1 +i(A_1+B^*_1)-(C_2+D^*_2+E^*_2+F^*_2)\right)e^{i(\op+\omega) t/2}\\
b(t)&=&\onroot2\left(1 +i(A_1-B^*_1)-(C_2-D^*_2-E^*_2+F^*_2)\right)e^{i(\op+\omega) t/2}\\
b^*(t)&=&\onroot2\left(1 -i(A_1-B_1)-(C_2-D_2-E_2+F_2)\right)e^{-i(\op+\omega) t/2}
\end{eqnarray*}
where that factor $e^{i\omega t/2}$ transforms from rotating back to laboratory frame.
The observables are linear combinations of $|a|^2-|b|^2|$ and $a^* b$  To second order in the small parameter ${\beta_z^0}$
\begin{eqnarray}
|a|^2-|b|^2&\sim&\left(1 -i(B_1-B_1^*) -(2C_2+D_2+D_2^*+(E_2+E_2^*)+(F_2-F_2^*)+|B_1|^2\right)\nonumber\\
&-&\left(1 +i(B_1-B_1^*) -(2C_2 -(D_2+D_2^*)-(E_2+E_2^*)+(F_2-F_2^*)+|B_1|^2\right)\nonumber\\
&\sim& -2i\left( (B_1-B_1^*) -2(D_2+D_2^*) -2(E_2+E_2^*)\right)\nonumber\\
&\sim& -2i(B_1-B_1^*)\label{eq:a2b2}\\
a^* b&\sim& \onehalf\left(1+2iA_1 + (B_1^*)^2 -2(C_2+F_2^*)\right)e^{i(\op+\omega)t}\nonumber\\
&\sim& \onehalf\left(1+2iA_1 + (B_1^*)^2 -2F_2^*\right)e^{i(\op+\omega)t}\label{eq:asb}
\end{eqnarray}
where in the end we dropped all terms that are order higer than ${\beta_z^0}^2$. (Note that $A_1\sim \beta_z^2$, $B_1\sim \beta_z$, $F_2 \sim \beta_z^2$ and $C_2,D_2,E_2\sim \beta_z^3$).
We need only to evaluate $A_1,B_1$ and $F_2$.
\begin{eqnarray}
A_1&=&\frac{1}{\hbar}\int_0^t H^\prime_{aa}d\tp = -\int_0^t \frac{\pi_0^\prime}{2}\cos 2\omega_v \tp d\tp = -\frac{\pi_0^\prime}{2}\frac{\sin 2\omega_v t}{2\omega_v}\label{eq:a1}\\
&&\lim_{\omega_v\rightarrow 0} A_1(\omega_v) = -\frac{\pi_0^\prime}{2}t \label{eq:a1limit}\\
B_1&=&\frac{1}{\hbar}\int_0^t H^\prime_{ab}e^{i\op \tp}d\tp = \int_0^t\onehalf(X_0 e^{i\omega_v \tp}+Y_0 e^{-i\omega_v \tp})e^{i\op \tp}d\tp 
=\onehalf\left(X_0\frac{e^{i(\omega_v+\op)t}-1}{i(\omega_v+\op)}+Y_0\frac{e^{-i(\omega_v-\op)t}-1}{-i(\omega_v-\op)}\right)\label{eq:b1}\\
%\lim_{\omega_v\rightarrow 0}B_1(\omega_v)&=&e^{i\op t/2}\frac{\sin(\op t/2)}{\op}(X_0+Y_0)=2X_0 e^{i\op t/2}\frac{\sin(\op t/2)}{\op}=-\Omega_0 e^{i\op t/2}\frac{\sin(\op t/2)}{\op} \\
B_1-B_1^*&=& -i\left(X_0\frac{\cos(\omega_v +\op)t-1}{\omega_v+\op}-Y_0\frac{\cos(\omega_v-\op)t-1}{\omega_v-\op}\right)\label{eq:b1b1}\\
%\lim_{\omega_v\rightarrow 0} B_1-B_1^*&=& -i\frac{\cos\op t-1}{\op}(X_0+Y_0) = -i\Omega_0\frac{\cos\op t-1}{\op}\label{eq:b1b1limit}\\
(B_1^*)^2&=&\frac{1}{4}\left(4(X_0^*)^2\frac{e^{-i(\omega_v+\op)t}\sin^2(\omega_v+\op)t/2}{(\omega_v+\op)^2} +2X_0^*Y_0^*\frac{e^{-2i\op t}- 2e^{-i\op t}\cos(\omega_v t)+1}{(\omega_v^2-\op^2)}\right.\nonumber\\
&+&\left. 4(Y_0^*)^2\frac{e^{i(\omega_v-\op) t}\sin^2(\omega_v-\op)t/2}{(\omega_v-\op)^2}\right)\nonumber\\
&=&e^{-i\op t}\left(X_0^2\frac{e^{-i(\omega_v)t}\sin^2(\omega_v+\op)t/2}{(\omega_v+\op)^2} +\frac{X_0Y_0}{2}\frac{e^{-i\op t}- 2\cos(\omega_v t)+e^{i\op t}}{(\omega_v^2-\op^2)}+Y_0^2\frac{e^{i(\omega_v) t}\sin^2(\omega_v-\op)t/2}{(\omega_v-\op)^2}\right)\nonumber\\
&=&e^{-i\op t}\left(X_0^2\frac{e^{-i(\omega_v)t}\sin^2(\omega_v+\op)t/2}{(\omega_v+\op)^2}+X_0Y_0\frac{(\cos\op t- \cos \omega_v t)}{(\omega_v^2-\op^2)}+Y_0^2\frac{e^{i(\omega_v) t}\sin^2(\omega_v-\op)t/2}{(\omega_v-\op)^2}\right)\nonumber\\
%\lim_{\omega_v\rightarrow 0}(B_1^*)^2&=&X_0^2e^{-i\op t}\left(\frac{\sin^2(\op t/2)}{\op^2}+\frac{\cos\op t - 1}{-\op^2} +\frac{\sin^2\op t/2}{\op^2}\right)=\Omega_0^2 e^{-i\op t}\frac{\sin^2\op t/2}{\op^2}\\
\end{eqnarray}
\begin{eqnarray}
F_2&=& \frac{1}{\hbar}\int_0^t B_1^*H_{ab}^\prime e^{i\op \tp} d\tp= 
\onehalf\int_0^t \left(X_0\frac{e^{-i(\omega_v+\op)\tp}-1}{-i(\omega_v+\op)}+Y_0\frac{e^{i(\omega_v-\op)\tp}-1}{i(\omega_v-\op)}\right)\onehalf(X_0e^{i\omega_v \tp}+Y_0 e^{-i\omega_v \tp})e^{i\op\tp} d\tp\nonumber\\
&=&\frac{1}{4}\left(\frac{1}{-i(\omega_v+\op)}\left(X_0^2(t -\frac{e^{i(\omega_v+\op)t}-1}{i(\omega_v+\op)}) +X_0Y_0(\frac{e^{-2i\omega_v t}-1}{-2i\omega_v}-\frac{e ^{-i(\omega_v-\op)t}-1}{-i(\omega_v-\op)})\right)\right.\nonumber\\
&+&\left. \frac{1}{i(\omega_v-\op)}\left(Y_0 X_0(\frac{e^{2i\omega_v t}-1}{2i\omega_v}-\frac{e^{i(\omega_v+\op)t}-1}{i(\omega_v+\op)})+Y_0^2 (t-\frac{e^{-i(\omega_v-\op)t}-1}{-i(\omega_v-\op)})\right)  \right)\label{eq:f2a}\\
%&=&\frac{X_0^2}{4}\left(\frac{1}{-i(\omega_v+\op)}\left((t -\frac{e^{i(\omega_v+\op)t}-1}{i(\omega_v+\op)}) +(\frac{e^{-2i\omega_v t}-1}{-2i\omega_v}-\frac{e ^{-i(\omega_v-\op)t}-1}{-i(\omega_v-\op)})\right)\right.\nonumber\\
%&+&\left. \frac{1}{i(\omega_v-\op)}\left((\frac{e^{2i\omega_v t}-1}{2i\omega_v}-\frac{e^{i(\omega_v+\op)t}-1}{i(\omega_v+\op)})+ (t-\frac{e^{-i(\omega_v-\op)t}-1}{-i(\omega_v-\op)})\right)  \right)\label{eq:f2b}\\
&=& \frac{1}{4}\left(\frac{it[(\omega_v-\op)X_0^2 - (\omega_v+\op)Y_0^2]}{\omega_v^2-\op^2}\right.\nonumber\\
&+&\left.X_0Y_0\left[\frac{e^{-i(\omega_v-\op)t}-1 + e^{i(\omega_v+\op)t}-1}{\omega_v^2-\op^2}
+\frac{\sin\omega_v t}{\omega_v}(\frac{e^{-i\omega_v t}}{-i(\omega_v+\op)} +\frac{e^{i\omega_v t}}{i(\omega_v-\op})\right]\right.\nonumber\\
&-&\left.X_0^2\frac{e^{i(\omega_v+\op)t}-1}{(\omega_v+\op)^2} -Y_0^2\frac{e^{-i(\omega_v-\op)t}-1}{(\omega_v-\op)^2}\right)\nonumber\\
&=& \frac{1}{4}\left(\frac{it[(\omega_v(X_0^2-Y_0^2)-\op(X_0^2+Y_0^2)]}{\omega_v^2-\op^2}
+ 2X_0Y_0\left[\frac{e^{i\op t}\cos\omega_v t-1}{\omega_v^2-\op^2}
+\frac{\sin\omega_v t}{\omega_v}\frac{(\omega_v\sin\omega_v t-i\op\cos\omega_v t)}{(\omega^2_v-\op^2)}\right]\right.\nonumber\\
&-&\left.X_0^2\frac{e^{i(\omega_v+\op)t}-1}{(\omega_v+\op)^2} -Y_0^2\frac{e^{-i(\omega_v-\op)t}-1}{(\omega_v-\op)^2}\right)\nonumber\\
&=& it\lambda/2 +F_{2-rem}
\end{eqnarray}
where we define
\begin{eqnarray*}
\lambda/2 &\equiv& \frac{1}{4}\left(\frac{[(\omega_v(X_0^2-Y_0^2)-\op(X_0^2+Y_0^2)]}{\omega_v^2-\op^2}\right)\\
F_{2-rem}&\equiv&\frac{1}{4}\left( 2X_0Y_0\left[\frac{e^{i\op t}\cos\omega_v t-1}{\omega_v^2-\op^2}
+\frac{\sin\omega_v t}{\omega_v}\frac{(\omega_v\sin\omega_v t-i\op\cos\omega_v t)}{(\omega^2_v-\op^2)}\right]
+X_0^2\frac{e^{i(\omega_v+\op)t}-1}{(\omega_v+\op)^2} -Y_0^2\frac{e^{-i(\omega_v-\op)t}-1}{(\omega_v-\op)^2}\right)\nonumber\\
\end{eqnarray*}
For the special case where the $\omega_v=0$, (zero electric field and fixed pitch angle)
\begin{eqnarray}
\lim_{\omega_v\rightarrow 0} F_2&=& \frac{1}{4}\left(\frac{-it(\op(X_0^2+Y_0^2)}{-\op^2}
+ 2X_0Y_0\left[\frac{e^{i\op t}-1}{-\op^2}
+\frac{i\op t}{\op^2}\right] - X_0^2\frac{e^{i\op t}-1}{\op^2} -Y_0^2\frac{e^{i \op t}-1}{\op^2}\right)\nonumber\\
&=& \frac{X_0^2}{2\op^2}\left(i \op t + \left[-(e^{i\op t}-1)+ i\op t\right] -(e^{i\op t}-1)\right)\nonumber\\
&=& \frac{X_0^2}{\op^2}\left(i \op t  -(e^{i\op t}-1)\right)=i\lambda^\prime/2 t  -\frac{X_0^2}{\op^2}(e^{i\op t}-1)\label{eq:lambdap}
 \label{eq:f4}
\end{eqnarray}
(Note that for $\omega_v=0$, $X_0=Y_0$.)

\label{eq:secondorderfshift}
%&=& \frac{X_0^2}{4}\left(\frac{-2i \op t}{\omega_v^2-\op^2}+\frac{2(e^{i \op t}\cos\omega_v t-1)}{\omega_v^2-\op^2}
%+\frac{\sin\omega_v t}{\omega_v}(\frac{2\omega_v\sin\omega_v t-2i\op\cos\omega_v t} {(\omega^2_v-\op^2)})\right.\\
%&-&\left.\frac{e^{i(\omega_v+\op)t}-1}{(\omega_v+\op)^2}-\frac{e^{-i(\omega_v-\op)t}-1}{(\omega_v-\op)^2}\right)\\
Then we can rewrite~\ref{eq:asb} as
\begin{eqnarray}
a^*b&\sim&\onehalf(1-it\lambda+2iA_1+(B_1^*)^2-2F_{2-rem}^*)e^{i(\eta+\omega)t}\nonumber\\
&\sim &\onehalf(1+2iA_1+(B_1^*)^2-2F_{2-rem}^*)e^{i(\eta+\omega-\lambda)t}\label{eq:asb1a}
\end{eqnarray}
and in the $\omega_v\rightarrow 0$ limit, when $A_1\rightarrow -\pi_0^\prime/2$ (see~\ref{eq:a1limit}), and $\lambda\rightarrow\lambda^\prime$ (see ~\ref{eq:lambdap}) 
\begin{eqnarray}
a^*b&\sim&\onehalf(1+(B_1^*)^2-2{F_{2-rem}^{\prime *} })e^{i(\eta+\omega+\lambda^\prime-\pi^\prime_0)t}\label{eq:asb1b}
\end{eqnarray}


(To get from ~\ref{eq:b1} to ~\ref{eq:b1b1} we use the fact that $X_0$ and $Y_0$ are real.) 

We learned in Appendix~\ref{ss:tdp} that the second order frequency shift appears in the observables as the coefficients of $it$ (imaginary time).
There is no such term in~\ref{eq:a2b2}. However, $F_2^*$ does contain such a term, $\lambda$ that contributes to a second order
frequency shift in $a^*b$. Recall that $a^*b$ corresponds to the polarization in the x-y plane. There is an additional
contribution to the frequency  shift, $\pi_0^\prime$ in the $\omega_v=0$ limit from the $A_1$ term.  

The precession frequency in the rotating frame is $\omega^\prime$.
\begin{enumerate}
\item[]
if $\omega_v>0$, then 
\begin{equation}
\omega^\prime \sim \eta+\omega-\lambda=\omega_0^\prime-\frac{\pi_0}{2}+\lambda.\label{eq:exactgzero}
\end{equation}
\item[]
if $\omega_v=0$, then 
\begin{equation}
\omega^\prime \sim \eta+\omega-\lambda=\omega_0^\prime-\frac{\pi_0}{2}+\lambda^\prime -\pi_0^\prime=\omega_0^\prime-\pi_0+2\lambda.\label{eq:exactlzero}
\end{equation}
\end{enumerate}
There is a discontinuity in the precession frequency at $\omega_v=0$. If $\omega_v=0$ there is no oscillation, the pitch angle is fixed,
and $\vev{\beta_z^2}=(\beta_z^0)^2$. For $\omega_v>0$,$\vev{\beta_z^2}=\onehalf(\beta_z^0)^2$.  

At the magic momentum
\begin{eqnarray*}
X_0^2-Y_0^2 &=& (\beta_z^0)^2\omega_a^2\frac{\gamma^2}{\gamma+1}\frac{\omega_v}{\omega}\beta_{xy}^2\\
X_0^2+Y_0^2 &=& \onehalf(\beta_z^0)^2\omega_a^2\left(\frac{\gamma^2}{(\gamma+1)^2}+\gamma^2\frac{\omega_v^2}{\omega^2}\right)\beta_{xy}^2
\end{eqnarray*}
and 
\begin{eqnarray*}
\lambda &=&\frac{1}{4}(\beta_z^0)^2\omega_a^2\beta_{xy}^2\left(\frac{2\omega_v(\frac{\gamma^2}{\gamma+1}\frac{\omega_v}{\omega})
-\eta(\frac{\gamma^2}{(\gamma+1)^2}+\gamma^2\frac{\omega_v^2}{\omega^2})}
{\omega_v^2-\eta^2}\right)
\end{eqnarray*}
and since $\eta=\omega_a+{\cal{O}}(\beta_z^0)^2$ we might as well replace $\eta$ with $\omega_a$ on the right hand size
\begin{eqnarray}
\lambda &=&\frac{1}{4}(\beta_z^0)^2\omega_a^2\beta_{xy}^2\left(\frac{2\omega_v(\frac{\gamma^2}{\gamma+1}\frac{\omega_v}{\omega})
-\omega_a(\frac{\gamma^2}{(\gamma+1)^2}+\gamma^2\frac{\omega_v^2}{\omega^2})}
{\omega_v^2-\omega_a^2}\right)\label{eq:exactglambda}
\end{eqnarray}
If $\omega_v=0$, then 
\begin{eqnarray*}
\omega^\prime&=&\omega_a(1- (\beta_z^0)^2\left(\frac{\gamma}{\gamma+1} -\onehalf\frac{\gamma^2}{(\gamma+1)^2}\beta_{xy}^2\right))\\
&\sim&\omega_a(1- \onehalf(\beta_z^0)^2)
\end{eqnarray*}

If $\omega_v>0$
\begin{eqnarray}
\omega^\prime&=&\omega_a\left(1-\frac{(\beta_z^0)^2}{2}\left[\frac{\gamma}{\gamma+1} -
\frac{1}{2}\omega_a\beta_{xy}^2\left(\frac{2\omega_v(\frac{\gamma^2}{\gamma+1}\frac{\omega_v}{\omega})
-\omega_a(\frac{\gamma^2}{(\gamma+1)^2}+\gamma^2\frac{\omega_v^2}{\omega^2})}{\omega_v^2-\omega_a^2}\right)\right]\right)\label{eq:opomegav}
\end{eqnarray}
Using $$\frac{\omega_a}{\omega}=a_\mu\gamma$$ ~\ref{eq:opomegav} can be written (see ~\ref{ss:intermsofpitchangle} for details) as
\begin{eqnarray}
\omega^\prime&=&\omega_a\left(1-\frac{(\beta_z^0)^2}{4}\left[\frac{\gamma^2}{\gamma^2-1}
+\frac{ \omega_a^2\frac{1}{(\gamma^2-1)}}
{\omega_v^2-\omega_a^2}\right]\right)
\end{eqnarray}
and in terms of the pitch angle $\psi_0 \approx \frac{\beta_z^0}{\beta_{xy}}$
\begin{eqnarray}
\omega^\prime &=& \omega_a\left(1-\frac{\psi_0^2}{4}\left[1+\frac{\omega_a^2}{\gamma^2(\omega_v^2-\omega_a^2)}\right]\right)
\end{eqnarray}
as previously shown by Farley\cite{farley} and Kim\cite{kim}.

In the limit where $\omega_v\gg\omega_a$
\begin{eqnarray*}
\omega^\prime&=&\omega_a\left(1- \frac{(\beta_z^0)^2}{2}\left[\frac{\gamma}{\gamma+1}-\onehalf\frac{\omega_a}{\omega}\beta_{xy}^2\left(\frac{2\gamma^2}{\gamma+1}-\gamma^2\frac{\omega_a}{\omega}\right)\right]\right)\\
&=&\omega_a\left(1- \frac{(\beta_z^0)^2}{2}\left[\frac{\gamma}{\gamma+1}-\onehalf\gamma a_\mu(\gamma^2-1)\left(\frac{2}{\gamma+1}-\gamma a_\mu\right)\right]\right)\\
&=&\omega_a\left(1- \frac{(\beta_z^0)^2}{2}\left[\frac{\gamma}{\gamma+1}-\frac{\gamma}{2}\left(\frac{2}{\gamma+1}-\frac{\gamma}{\gamma^2-1}\right)\right]\right)\\
&=&\omega_a\left(1- \frac{(\beta_z^0)^2}{2}\left[\frac{\gamma}{\gamma+1}-\frac{\gamma}{2}\left(\frac{\gamma-2}{\gamma^2-1}\right)\right]\right)\\
&=&\omega_a\left(1- \frac{(\beta_z^0)^2}{4}\left[\frac{\gamma^2}{(\gamma^2-1)}\right]\right)\\
\end{eqnarray*}

The precession frequency $\omega^\prime$ is plotted as a function of $\omega_v$ in Figure~\ref{fig:opvsomegav} for muons at the magic momentum and with pitch angle $\beta_z^0=0.01$.
At $\omega_v\ll\omega^\prime$, $\omega^\prime =\omega_a(1-\frac{1}{4}(\beta_z^0)^2)$, whereas if $\omega_v\gg\omega^\prime$, $\omega^\prime=(\omega_a(1-\frac{1}{4}(\beta_z^0)^2\frac{\gamma^2}{\gamma^2-1})$.
In the g-2 experiment $\omega_v\sim 10\omega^\prime$.
\begin{figure}
\centering
\includegraphics{figures/pitch_numerical_analytic.pdf}
\caption{The simulation (points) integrates the equations of motion and the Thomas-BMT equation to determine $\bmbetahat\cdot\bfs(t)$ and the frequency
is extracted by a fit. $\beta_z^0 = 0.01$. The fitted frequency represents an average precession frequency over 2000 turns of the muon around the ring.
The solid curve is Equation~\ref{eq:opomegav} assembled from Equations~\ref{eq:exactgzero} with~\ref{eq:exactglambda}. 
When the betatron frequency $f_v$ is very low, there are an insufficient number of betatron periods to determine an average precession frequency, as evidenced
by the discrepancy at low frequency in the plot. The nominal precession frequency, $\omega_a=\frac{eB}{mc}a_\mu = 2\pi (0.23{\rm MHz}$).
}
\label{fig:opvsomegav}
\end{figure}

\begin{comment}
In evaluating the $\omega_v\rightarrow 0$ limit we note (see ~\ref{eq:x0} and \ref{eq:y0})
in the the $\omega_v\rightarrow 0$ limit, $ X_0=Y_0=-\Omega_0/2$, $X_0Y^* = |X_0|^2=|Y_0|^2$).
\subsection*{$\omega_v=0$ limit}
As long as we are evaluating in the $\omega_v=0$ limit, let's complete that calculation. Then we can compare to the exact solution for the fixed pitch to see if
we are on the right track.
So, in the $\omega_v=0$ limit
\begin{eqnarray*}
|a|^2-|b|^2&\sim& -2i(B_1-B_1^*)= 2\frac{\Omega_0}{\op}(\cos\op t-1)\\
a^*b&\sim& \onehalf(1-i\pi_0^\prime t +\onehalf\frac{\Omega_0^2}{\op^2} e^{-i\op t}(1-\cos\op t)-\onehalf\frac{\Omega_0^2}{\op^2}(-i\op t - (e^{-i\op t}-1)) e^{i(\op+\omega)t}\\
a^*b&\sim& \onehalf(1-i\pi_0^\prime t +\onehalf\frac{\Omega_0^2}{\op^2} \left(e^{-i\op t}(2\sin^2\opt2)+ (e^{-i\op t}-1) +i\op t \right) e^{i(\op+\omega)t}\\
&\sim& \onehalf(1-i\pi_0^\prime t +\onehalf\frac{\Omega_0^2}{\op^2} \left(e^{-i\op t}(2\sin^2\opt2)- e^{-i\op t/2}(2i\sin\op t/2) +i\op t \right) e^{i(\op+\omega)t}\\
&\sim& \onehalf(1-i\pi_0^\prime t +\onehalf\frac{\Omega_0^2}{\op^2} \left(e^{-i\op t}(2\sin^2\opt2- 2ie^{i\op t/2}\sin\op t/2) +i\op t \right) e^{i(\op+\omega)t}\\
&\sim& \onehalf\left(1 +\onehalf\frac{\Omega_0^2}{\op^2}( e^{-i\op t}(2-\cos\op t)-1)\right)e^{i(\pi_0^\prime +\frac{\Omega_0^2}{2\op})t} e^{i(\op+\omega)t}\\
\end{eqnarray*}
Finally
\begin{eqnarray*}
\lim_{\omega_v\rightarrow 0} \vev{\bm\sigma\cdot\bm\beta}&\sim& -(|a|^2-|b|^2)\beta_z^0 - 2\Re(a^* b)\beta_{xy}e^{-i\omega t}\\
&=&2\frac{\Omega_0}{\op}(\cos\op t-1)\beta_z^0 - (\cos \lambda t) \left(1+ \frac{\Omega_0^2}{\op^2}\left(\cos(\lambda-\op) t[1 -\onehalf\cos\op t]-\onehalf \right)\right)\beta_{xy}^0
\end{eqnarray*}
where, (and here $\op=\omega_a-\pi_0/2)$,
\begin{eqnarray*}
\lambda&=& \op-\pi_0^\prime+\frac{\Omega_0^2}{\op} = \omega_a-\pi_0/2-\pi_0^\prime+\frac{\Omega_0^2}{2\op}
 = \omega_a-\pi_0+\frac{\Omega_0^2}{2\op}
\end{eqnarray*}
Compare to
\begin{eqnarray}
\bm\beta\cdot s
&=& \onehalf\left(1+\cos(\op t) -\frac{1}{\op^2}[(\omega-\omega_0)^2-\Omega^2](1-\cos\op t)\right)\beta_{xy} + \beta_z\frac{(\omega-\omega_0)\Omega}{\op^2}(1-\cos(\op t))\nonumber\\
&\sim& \onehalf\left(1 +\cos\op t - (1-2\frac{\Omega^2}{(\omega-\omega_0)^2})(1-\cos\op t)\right)\beta_{xy}+\beta_z\frac{(\omega-\omega_0)\Omega}{\op^2}(1-\cos\op t)\\
&\sim& \onehalf\left(2\cos\op t + 2\frac{\Omega^2}{(\omega-\omega_0)^2})(1-\cos\op t)\right)\beta_{xy}+\beta_z\frac{(\omega-\omega_0)\Omega}{\op^2}(1-\cos\op t)\\
&\sim& \left(\cos\op t + \frac{\Omega^2}{(\omega-\omega_0)^2})(1-\cos\op t)\right)\beta_{xy}+\beta_z\frac{(\omega-\omega_0)\Omega}{\op^2}(1-\cos\op t)
\end{eqnarray}
What about $\op$? (The notation needs to change as I have used $\op$ for two different things. Meanwhile using $\op$ as it was defined for the fixed pitch.
\begin{eqnarray*}
\op_f&=&\left((\tilde\omega_0-\omega)^2+\Omega^2\right)^{1/2}\sim \left((\omega_a-\pi_0)^2+\Omega^2\right)^{1/2}\sim\omega_a\left(\omega_a+\frac{-2\omega_a\pi_0+\pi_0^2+\Omega_0^2}{2\omega_a}\right)
\sim \omega_a(1-\pi_0 +\onehalf\frac{\Omega^2}{\omega_a})
\end{eqnarray*}
where $\tilde\omega_0 = \omega_a+\omega +\pi_0$

\subsection{Second order frequency shift}
Let's think about the second order frequency shift (the coefficient of $it$ in the first term in square brackets in ~\ref{eq:secondorderfshift})
\begin{eqnarray*}
\lambda_2&=& \frac{1}{4}\frac{[(\omega_v(X_0^2-Y_0^2)-\op(X_0^2+Y_0^2)]}{\omega_v^2-\op^2}
\end{eqnarray*}
Since
\begin{eqnarray*}
X_0^2-Y_0^2 &=& \frac{1}{4}(\beta_z^0\omega_a\beta_{xy})^2\frac{\gamma^2}{\gamma+1}\frac{\omega_v}{\omega}\\
X_0^2+Y_0^2 &=& \frac{1}{4}(\beta_z^0\omega_a\beta_{xy})^2\left(\left(\frac{\gamma}{\gamma+1}\right)^2+\left(\gamma\frac{\omega_v}{\omega}\right)^2\right)
\end{eqnarray*}
In the $\omega_v\rightarrow 0$ limit, $X_0^2-Y_0^2\rightarrow 0$ and
\begin{eqnarray*}
\lim_{\omega_v\rightarrow 0} \left(-\op\frac{X_0^2+Y_0^2 }{\omega_v^2-\op^2}\right)&\rightarrow& \frac{1}{4\op}(\beta_z^0\omega_a\beta_{xy})^2\left(\frac{\gamma}{\gamma+1}\right)^2
\end{eqnarray*}
Then we recover the frequency shift for fixed pitch.
In the $\omega_v\gg \op$ limit
\begin{eqnarray*}
\lim_{\omega\gg\op} \omega_v\frac{X_0^2-Y_0^2}{\omega_v^2-\op^2} &\rightarrow& \frac{1}{4\omega}(\beta_z^0\omega_a\beta_{xy})^2\frac{\gamma^2}{\gamma+1}\\
\lim_{\omega_v\gg\op} -\op\frac{X_0^2+Y_0^2}{\omega_v^2-\op^2}&\rightarrow& \frac{1}{4}(\beta_z^0\omega_a\beta_{xy})^2\gamma^2\frac{\op}{\omega^2}\\
\rightarrow \lambda_2\rightarrow \frac{1}{4\omega}(\beta_z^0\omega_a\beta_{xy})^2\left(\frac{\gamma^2}{\gamma+1}-\gamma^2\frac{\op}{\omega}\right)
\end{eqnarray*}
In both limits, the frequency shift is independent of $\omega_v$.
Recall $$\frac{\op}{\omega} = \frac{\omega_a(1-\frac{\gamma}{\gamma+1}\frac{(\beta_z^0)^2}{2})}{\frac{\omega_a}{a_\mu\gamma}}
=a_\mu\gamma(1-\frac{\gamma}{\gamma+1}\frac{(\beta_z^0)^2}{2})\sim a_\mu\gamma$$
and at the magic momentum
$$a_\mu\gamma = \frac{\gamma}{\gamma^2-1}.$$
Then
\begin{eqnarray*}
 \lambda_2&\rightarrow& \frac{1}{4\omega}(\beta_z^0\omega_a\beta_{xy})^2\left(\frac{\gamma^2}{\gamma+1}-\frac{\gamma^3}{\gamma^2-1}\right)\\
&=&\frac{1}{4\omega}(\beta_z^0\omega_a\beta_{xy})^2\left(\frac{\gamma^2(\gamma-1)-\gamma^3}{\gamma^2-1}\right)\\
&=&\frac{1}{4\omega}(\beta_z^0\omega_a\beta_{xy})^2\left(\frac{-\gamma^2}{\gamma^2-1}\right)\\
&\sim&\frac{a_\mu\gamma}{4\op}(\beta_z^0\omega_a\beta_{xy})^2\left(\frac{-\gamma^2}{\gamma^2-1}\right)\\
&\sim&\frac{1}{4\op}\frac{\gamma}{\gamma^2-1}(\beta_z^0\omega_a\beta_{xy})^2\left(\frac{-\gamma^2}{\gamma^2-1}\right)\\
\end{eqnarray*}
\end{comment}
