<aside> 💡 By the finite light field $\Omega \neq 0$, the eigenstates of the RWA Hamiltonian become

$$ \left|\lambda_+\right> = \cos\frac{\theta}{2} \left|g\right>+\sin\frac{\theta}{2} \left|e\right> \\ \left|\lambda_-\right> = -\sin\frac{\theta}{2}\left|g\right>+\cos\frac{\theta}{2}\left|e\right> $$

where the eigenenergies are

$$ \lambda_{\pm} = \frac{-\hbar\Delta \pm \hbar\sqrt{\Delta^2 + |\Omega|^2}}{2} $$

and the mixing angle is

$$ \tan \theta = \frac{\Omega}{\Delta} $$

• AC-Stark shift on $\left|g\right>$ by far-off detuned light ($|\Delta| \ll \Omega$): Red detuned case ($\Delta < 0$): $-\hbar \Omega^2/4|\Delta|$ Blue detuned case ($\Delta > 0$): $+\hbar \Omega^2/4|\Delta|$ </aside>

Without the driving field, i.e. $\Omega = 0$, the RWA Hamiltonian is diagonal,

$$ \hat{H}_{\rm RWA}(\Omega=0)=\frac{\hbar}{2} \begin{bmatrix} 0 && 0 \\ 0 && -2\Delta \end{bmatrix} $$

However, if there is a finite field, i.e. $\Omega \neq 0$, the RWA Hamiltonian is not diagonal,

$$ \hat{H}_{\rm RWA}=\frac{\hbar}{2} \begin{bmatrix} 0 && \Omega \\ \Omega^* && -2\Delta \end{bmatrix} $$

To find the eigenvalue of the RWA Hamiltonian matrix, let’s solve the characteristic polynomial

$$ \begin{bmatrix}-\lambda && \hbar\Omega/2 \\ \hbar\Omega^*/2 && -\hbar\Delta-\lambda\end{bmatrix} \rightarrow \lambda(\hbar\Delta+\lambda)-\hbar^2|\Omega|^2/4 = 0 $$

$$ \lambda_{\pm} = \frac{-\hbar\Delta \pm \hbar\sqrt{\Delta^2 + |\Omega|^2}}{2} $$

By changing the basis $\left|\psi\right> \rightarrow \left|\lambda\right>=\hat{U}\left|\psi\right>$, where $\left|\lambda\right>$ is eigenstate, the TDSE becomes,

$$ i\hbar \partial_t \left|\lambda\right> = [\hat{U}\hat{H}_{\rm RWA}\hat{U}^{\dagger}-i\hbar \hat{U}(\partial_t \hat{U}^{\dagger})]\left|\lambda\right> $$

Because $\left|\lambda\right>$ is eigenstate so that $\hat{U}\hat{H}_{\rm RWA}\hat{U}^{\dagger}$ is a diagonal matrix.

$$ \hat{U}\hat{H}{\rm RWA}\hat{U}^{\dagger}=\begin{bmatrix} \lambda+ && \\ && \lambda_- \end{bmatrix} $$

The inertial term $-i\hbar \hat{U} (\partial_t \hat{U}^{\dagger})$ makes an off-diagonal element.

The unitary operator $\hat{U}$ can be written as,

$$ \hat{U}=\begin{bmatrix} \cos\frac{\theta}{2} && \sin\frac{\theta}{2}\\ -\sin\frac{\theta}{2} && \cos\frac{\theta}{2} \end{bmatrix} $$

where $\left|g\right> \rightarrow \left|\lambda_+\right> = \cos\frac{\theta}{2} \left|g\right>+\sin\frac{\theta}{2} \left|e\right>$, and $\left|e\right> \rightarrow \left|\lambda_-\right> = -\sin\frac{\theta}{2}\left|g\right>+\cos\frac{\theta}{2}\left|e\right>$.

$$ \hat{U}\hat{H}_{\rm RWA}\hat{U}^{\dagger} = \frac{\hbar}{2} \begin{bmatrix} \cos\frac{\theta}{2} && \sin\frac{\theta}{2}\\ -\sin\frac{\theta}{2} && \cos\frac{\theta}{2} \end{bmatrix} \begin{bmatrix} 0 && \Omega \\ \Omega^* && -2\Delta \end{bmatrix} \begin{bmatrix} \cos\frac{\theta}{2} && -\sin\frac{\theta}{2}\\ \sin\frac{\theta}{2} && \cos\frac{\theta}{2} \end{bmatrix}
$$

$$ = \frac{\hbar}{2}\begin{bmatrix} \Omega_\Re\sin\theta-\Delta(1-\cos\theta) & \Omega_\Re\cos\theta-\Delta\sin\theta+i\Omega_\Im \\ \Omega_\Re\cos\theta-\Delta\sin\theta-i\Omega_\Im & -\Omega_{\Re}\sin\theta-\Delta(1+\cos\theta) \end{bmatrix} =\begin{bmatrix} \lambda_+ && \\ && \lambda_- \end{bmatrix} $$

For simplicity, let’s assume that the imaginary part of the Rabi frequency is zero $\Omega_{\Im}=0$,