\section{Rotating frame}
\label{sec:rotatingframe}
It is convenient to transform to a 'rotating frame', in which the effective Hamiltonian is independent of time.
Then the Schrodinger equation (\ref{eq:schrodinger}) can be solved practically by inspection ({\it See Griffiths Quantum Mechanics, 1st edition, problem 9.7, p. 305}).
Define
$$R = \begin{pmatrix} e^{-i\omega t/2} & 0 \cr 0 & e^{i\omega t/2}\end{pmatrix}$$
Then
$$R{\cal H}R^{-1}=  R\begin{pmatrix} \omega_0 & \Omega e^{i\omega t} \cr \Omega e^{-i\omega t} & -\omega_0\end{pmatrix} R^{-1} =  \begin{pmatrix} \omega_0 & \Omega \cr \Omega & -\omega_0\end{pmatrix} $$ 
and the Schrodinger equation~\ref{eq:schrodinger} can be written as
\begin{eqnarray*}
i\der{}{t}\begin{pmatrix}a\cr b\end{pmatrix}  &=& -R^{-1}\frac{1}{2} \begin{pmatrix} \omega_0 & \Omega \cr \Omega  & -\omega_0\end{pmatrix}R\begin{pmatrix} a\cr b\end{pmatrix}\\
\rightarrow i R\der{}{t}\begin{pmatrix}a\cr b\end{pmatrix} &=& -\frac{1}{2} \begin{pmatrix} \omega_0 & \Omega \cr \Omega  & -\omega_0\end{pmatrix}R\begin{pmatrix} a\cr b\end{pmatrix}\\
  i\der{}{t}\left[R\begin{pmatrix}a\cr b\end{pmatrix}\right] -i\der{R}{t}\begin{pmatrix} a\cr b\end{pmatrix}  &=& -\frac{1}{2} \begin{pmatrix} \omega_0 & \Omega \cr \Omega  & -\omega_0\end{pmatrix}R\begin{pmatrix} a\cr b\end{pmatrix}\\
  i\der{}{t}\left[R\begin{pmatrix}a\cr b\end{pmatrix}\right] -\frac{\omega}{2}\sigma_z R\begin{pmatrix} a\cr b\end{pmatrix}  &=& -\frac{1}{2} \begin{pmatrix} \omega_0 & \Omega \cr \Omega  & -\omega_0\end{pmatrix}R\begin{pmatrix} a\cr b\end{pmatrix}\\
  i\hbar\der{}{t}\begin{pmatrix}\tilde a\cr \tilde b\end{pmatrix} &=&{\cal H^\prime}\begin{pmatrix} \tilde a\cr \tilde b\end{pmatrix}= -\frac{\hbar}{2} \begin{pmatrix} \omega_0-  \omega & \Omega \cr \Omega  & -\omega_0+\omega\end{pmatrix}\begin{pmatrix} \tilde a\cr \tilde b\end{pmatrix}\\
\end{eqnarray*}
where $$\begin{pmatrix} \tilde a\cr \tilde b\end{pmatrix} = R\begin{pmatrix} a \cr b\end{pmatrix}$$ 
In the rotating frame the Hamiltonian is independent of time and the characteristic frequencies are simply the eigenvalues.
$${\omega^\prime}^\pm = \pm\sqrt{(\omega_0-\omega)^2 + \Omega^2}$$ 
%same as ~\ref{eq:omegap}.

We next find the eigenvectors, take appropriate linear combinations and transform from the rotating frame back to the fixed frame. 
See Appendix~\ref{ss:linearalgebra} for details of the linear algebra.

