\subsection{Solution of Schrodinger Equation in lab frame}
\label{ss:labframesolution}
We aim to solve the coupled differential equations exactly.
$$ \left( \begin{array}{c} \dot{a}(t) \\ \dot{b}(t) \end{array} \right)
=  {i \over 2 }\left( \begin{array}{c} 
    a(t)\omega_0 + b(t)\Omega e^{i \omega t} \\
    a(t) \Omega e^{-i \omega t}  -b(t)\omega_0 \end{array} \right). $$

To solve these coupled differential equations first differentiate the $\dot{b}(t)$ equation,
$$\ddot{b} = {i\over 2}\left[(\dot{a}-i\omega a)\Omega e^{-i\omega t} - \dot{b}\omega_0\right]$$
 then substitute for $\dot{a}(t)$.  We have 
$$\ddot{b} = {i\over 2}\left[({i\over 2}(a\omega_0 + b\Omega e^{i \omega t})-i\omega a)\Omega e^{-i\omega t} - \dot{b}\omega_0\right]$$
and
$$\ddot{b} = {i\over 2}\left[(i a\Omega e^{-i\omega t}({\omega_0\over 2}-\omega)+i{b\over 2}\Omega^2 - \dot{b}\omega_0\right]$$
Then substitute the expression for $a(t)$ from the original equation
$$ a(t)  \Omega e^{-i \omega t} = -2 i \dot{b}(t)+ b(t) \omega_0. $$ and we get
$$\ddot{b} = {i\over 2}\left[(i\left( (-2i\dot{b}+ b \omega_0) e^{i \omega t}/\Omega  
\right)\Omega e^{-i\omega t}({\omega_0\over 2}-\omega)+i{b\over 2}\Omega^2 - \dot{b}\omega_0\right]$$
$$\ddot{b} = {i\over 2}\left[(i \left(-2 i \dot{b}+ b \omega_0 \
\right)({\omega_0\over 2}-\omega)+i{b\over 2}\Omega^2 - \dot{b}\omega_0\right]$$
$$\ddot{b} = {i\over 2}\left[(-2\dot{b}\omega+i{b\over 2}\left( \omega_0^2-2\omega_0\omega+\Omega^2\right)\right]$$
 $$\ddot{b} = -i\dot{b}\omega-{b\over 4}\left( \omega_0^2-2\omega_0\omega+\Omega^2\right)$$
 $$\ddot{b} = -i\dot{b}\omega-{b\over 4}\left({\omega^\prime}^2-\omega^2\right)$$
where 
\begin{eqnarray}
\omega_0^2-2\omega_0\omega + \Omega^2 &=& {\omega^\prime}^2-\omega^2\nonumber\\
\rightarrow {\omega^\prime}^2 &=& (\omega_0-\omega)^2+\Omega^2\label{eq:omegaplabframe}
\end{eqnarray}
 
Define $\gamma=i\omega$, and $\alpha = {b\over 4}({\omega^\prime}^2-\omega^2)$. Now our coupled second order differential equation looks like the equation of motion for a damped
harmonic oscillator.
 $$\ddot{b} = -\gamma\dot{b}-\alpha b$$
The general solution has the form
$$b= Ae^{i\theta t}$$
Substitution into the differential equation gives
$$-\theta^2 = -i\theta\gamma -\alpha$$
and $$\theta = {1\over 2}\left(i\gamma \pm\sqrt{-\gamma^2 + 4\alpha}\right)
={1\over 2}\left(-\omega\pm\sqrt{\omega^2 + ({\omega^\prime}^2-\omega^2)}\right)=-{\omega\over 2}\pm{{\omega^\prime}^2\over 2}$$
so
$$b= \left[Ce^{i{\omega^\prime\over 2}t }+ De^{-i{\omega^\prime \over 2}t}\right]e^{-i{\omega\over 2}t}$$
or even better
$$b= \left[A\cos({\omega^\prime t/2})+ B\sin({\omega^\prime t /2})\right]e^{-i{\omega\over 2}t}$$
The boundary conditions give us
$$b(0) = A = b_0$$ and
\begin{eqnarray*}
\dot{b}(0) &=& (-i\omega/2)A + (\omega^\prime/2)B = {i\over 2}(\Omega a_0 - \omega_0 b_0)\\
&\rightarrow& \omega b_0+i\omega^\prime B =  (\omega_0 b_0-\Omega a_0) \\
&\rightarrow& B = {i\over \omega^\prime}\left[(\omega-\omega_0)b_0+\Omega a_0\right]
\end{eqnarray*}
Then
$$b= \left[b_0\cos({\omega^\prime t/2}) +{i\over \omega^\prime}\left[(\omega-\omega_0)b_0+\Omega a_0\right]\sin({\omega^\prime t /2})\right]e^{-i{\omega\over 2}t}$$
and
\begin{eqnarray*}
a&=&(-2i\dot b + b\omega_0)\frac{e^{i\omega t}}{\Omega}\\
&=& \left[\left(-2 i (-\frac{b_0\omega^\prime}{2}\sin(\omega^\prime t/2)+\frac{i}{2}[(\omega-\omega_0)b_0+\Omega a_0]\cos(\omega^\prime t/2)\right)e^{-i\omega t/2})\right. \\
&+& -2\frac{\omega}{2}\left[b_0\cos({\omega^\prime t/2}) +{i\over \omega^\prime}\left[(\omega-\omega_0)b_0+\Omega a_0\right]\sin({\omega^\prime t /2})\right]e^{-i{\omega\over 2}t} \\
&+& \left.\omega_0 \left[b_0\cos({\omega^\prime t/2}) +{i\over \omega^\prime}\left[(\omega-\omega_0)b_0+\Omega a_0\right]\sin({\omega^\prime t /2})\right]e^{-i{\omega\over 2}t}\right]\frac{e^{i\omega t}}{\Omega}\\
&=&\left(-2 i (-\frac{b_0\omega^\prime}{2}\sin(\omega^\prime t/2)+\frac{i}{2}[(\omega-\omega_0)b_0+\Omega a_0]\cos(\omega^\prime t/2))\right.\\
&+&\left.(\omega_0-\omega) \left[b_0\cos({\omega^\prime t/2}) +{i\over \omega^\prime}\left[(\omega-\omega_0)b_0+\Omega a_0\right]\sin({\omega^\prime t /2})\right]\right)\frac{e^{i{\omega\over 2}t}}{\Omega}\\
&=&\left( (ib_0\omega^\prime\sin(\omega^\prime t/2)+[(\omega-\omega_0)b_0+\Omega a_0]\cos(\omega^\prime t/2)\right.\\
&+&\left.(\omega_0-\omega) \left[b_0\cos({\omega^\prime t/2}) +{i\over \omega^\prime}\left[(\omega-\omega_0)b_0+\Omega a_0\right]\sin({\omega^\prime t /2})\right]\right)\frac{e^{i{\omega\over 2}t}}{\Omega}\\
&=&\left( (i b_0\omega^\prime\sin(\omega^\prime t/2)+[\Omega a_0]\cos(\omega^\prime t/2)\right.\\
&+&\left.(\omega_0-\omega) \left[{i\over \omega^\prime}\left[(\omega-\omega_0)b_0+\Omega a_0\right]\sin({\omega^\prime t /2})\right]\right)\frac{e^{i{\omega\over 2}t}}{\Omega}\\
&=&\left[(ib_0\omega^\prime -ib_0\frac{(\omega-\omega_0)^2}{\omega^\prime})\sin(\omega^\prime t/2))+ a_0\Omega(\cos(\omega^\prime t/2)+ \frac{i(\omega_0-\omega)}{\omega^\prime}\sin(\omega^\prime t/2))\right] \frac{e^{i\omega t/2}}{\Omega}\\
&=&\left[(-ib_0\frac{\Omega^2}{\omega^\prime})\sin(\omega^\prime t/2))+ a_0\Omega(\cos(\omega^\prime t/2)+ \frac{i(\omega_0-\omega)}{\omega^\prime}\sin(\omega^\prime t/2))\right] \frac{e^{i\omega t/2}}{\Omega}\\
&=&\left[(-ib_0\frac{\Omega}{\omega^\prime})\sin(\omega^\prime t/2))+ a_0(\cos(\omega^\prime t/2)+ \frac{i(\omega_0-\omega)}{\omega^\prime}\sin(\omega^\prime t/2))\right] e^{i\omega t/2}\\
&=&\left[a_0\cos(\omega^\prime t/2)+ \frac{i}{\omega^\prime}[(\omega_0-\omega)a_0-\Omega b_0]\sin(\omega^\prime t/2)\right] e^{i\omega t/2}
\end{eqnarray*}
or
\begin{equation}
\begin{pmatrix} a \cr b\end{pmatrix} 
= \begin{pmatrix} \left(\cos({\omega^\prime t/2}) -{i(\omega-\omega_0)\over \omega^\prime}\sin({\omega^\prime t /2})\right)e^{i{\omega\over 2}t} & -\frac{i \Omega}{\omega^\prime}\sin(\omega^\prime t /2)e^{i{\omega\over 2}t} \cr
   -\frac{i \Omega}{\omega^\prime}\sin(\omega^\prime t /2)e^{-i{\omega\over 2}t}& \left(\cos({\omega^\prime t/2}) +{i(\omega-\omega_0)\over \omega^\prime}\sin({\omega^\prime t /2})\right)e^{-i{\omega\over 2}t}
 \end{pmatrix}
\begin{pmatrix} a_0 \cr b_0\end{pmatrix}\label{eq:umatrixlabframe}
\end{equation}

