<aside> 💡 Under the limit of dipole approximation ($\lambda_L \gg a_0$) and RWA approximation ($|\Delta| \ll \omega_0, \omega_L$), the atom-photon interacting Hamiltonian of the two-level system could be written as,

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

Using Pauli operator, RWA Hamiltonian is written as

$$ \hat{H}{\rm RWA}=\frac{\hbar}{2} \begin{bmatrix} 0 && \Omega \\ \Omega^* && -2\Delta \end{bmatrix} = \frac{\hbar}{2}\left(\Omega\Re\hat{\sigma}^x+\Omega_\Im\hat{\sigma}^y -2\Delta \hat{n}\right) $$

</aside>

Atom-photon interacting two-level system

The time-dependent Schrodinger equation (TDSE) is written as,

$$ i\hbar \partial_t\left|\psi \right> = \hat{H}\left|\psi \right> $$

The case of atom-photon interaction, the whole Hamiltonian $\hat{H}$ is decomposed into two terms $\hat{H}=\hat{H}{\rm atom} +\hat{H}{\rm atom-photon}$, (1) atomic Hamiltonian $\hat{H}{\rm atom}$ and (2) atom-photon interaction term $\hat{H}{\rm atom-photon}$. The atomic two-level $\{\left|\tilde{g}\right>, \left|\tilde{e}\right>\}$ is energetically separated as $\hbar\omega_0$.

$$ \hat{H}_{\rm atom} = \hbar \begin{bmatrix}0 \\ && \omega_0\end{bmatrix} $$

The incident field is plane AC-field, $\vec{E}(t)=\vec{E}_0(t) \cos(\omega_L t-\vec{k}_L\cdot \vec{x})$.

Under the dipole approximation limit $\lambda_L \gg a_0$, the $\vec{k}_L \cdot \vec{x}$ term is ignorable, where the wavelength is $\lambda_L=2\pi/|\vec{k}_L|$ and $a_0$ is a Bohr radius.

The atom-photon interaction term becomes,

$$ \hat{H}_{\rm atom-photon} = \vec{\mu}\cdot\vec{E}(t)=\hbar\begin{bmatrix} && \Omega\cos(\omega_L t) \\ \Omega^*\cos (\omega_L t)&& \end{bmatrix} $$

where $\Omega=\vec{\mu}\cdot\vec{ E}_0/ \hbar$ is Rabi frequency. Finally, the total Hamiltonian is written as

$$ \hat{H} = \hbar\begin{bmatrix} 0 && \Omega \cos(\omega_L t) \\\Omega^* \cos(\omega_L t) && \omega_0\end{bmatrix} $$

Rotating wave approximation (RWA) Hamiltonian

By unitary transform $\hat{U}$, the quantum state evolves as $\left|\psi\right> \rightarrow \left|\psi'\right>=\hat{U}\left|\psi\right>$.

Therefore, TDSE becomes

$$ i\hbar \partial_t (\hat{U}^{\dagger}\left|\psi'\right>) = \hat{H}(\hat{U}^{\dagger}\left|\psi'\right>) $$

$$ i\hbar \partial_t \left|\psi'\right> = [\hat{U}\hat{H}\hat{U}^{\dagger}-i\hbar \hat{U}(\partial_t \hat{U}^{\dagger})]\left|\psi'\right> $$