Rotating Wave Approximation - Derivation

Derivation

Given the above definitions the interaction Hamiltonian is

\begin{align} H_1 &= -\vec{d}\cdot\vec{E} \\ &=-\left(\vec{d}_\text{eg}|\text{e}\rangle\langle\text{g}|+\vec{d}_\text{eg}^*|\text{g}\rangle\langle\text{e}|\right) \cdot\left(\vec{E}_0e^{-i\omega_Lt}+\vec{E}_0^*e^{i\omega_Lt}\right) \\ &=-\left(\vec{d}_\text{eg}\cdot\vec{E}_0e^{-i\omega_Lt} +\vec{d}_\text{eg}\cdot\vec{E}_0^*e^{i\omega_Lt}\right)|\text{e}\rangle\langle\text{g}| -\left(\vec{d}_\text{eg}^*\cdot\vec{E}_0e^{-i\omega_Lt} +\vec{d}_\text{eg}^*\cdot\vec{E}_0^*e^{i\omega_Lt}\right)|\text{g}\rangle\langle\text{e}| \\ &=-\hbar\left(\Omega e^{-i\omega_Lt}+\tilde{\Omega}e^{i\omega_Lt}\right)|\text{e}\rangle\langle\text{g}| -\hbar\left(\tilde{\Omega}^*e^{-i\omega_Lt}+\Omega^*e^{i\omega_Lt}\right)|\text{g}\rangle\langle\text{e}|, \end{align}

as stated. The next step is to find the Hamiltonian in the interaction picture, . The required unitary transformation is

,

where the last step can be seen to follow e.g. from a Taylor series expansion, and due to the orthogonality of the states and we have

\begin{align} H_{1,I} &\equiv U H_1 U^\dagger \\ &=-\hbar\left(\Omega e^{-i\omega_Lt}+\tilde{\Omega}e^{i\omega_Lt}\right)e^{i\omega_0t}|\text{e}\rangle\langle\text{g}| -\hbar\left(\tilde{\Omega}^*e^{-i\omega_Lt}+\Omega^*e^{i\omega_Lt}\right)|\text{g}\rangle\langle\text{e}|e^{-i\omega_0t} \\ &=-\hbar\left(\Omega e^{-i\Delta t}+\tilde{\Omega}e^{i(\omega_L+\omega_0)t}\right)|\text{e}\rangle\langle\text{g}| -\hbar\left(\tilde{\Omega}^*e^{-i(\omega_L+\omega_0)t}+\Omega^*e^{i\Delta t}\right)|\text{g}\rangle\langle\text{e}|\ . \end{align}

Now we apply the RWA by eliminating the counter-rotating terms as explained in the previous section, and finally transform the approximate Hamiltonian back to the Schrödinger picture:

\begin{align} H_1^{\text{RWA}}&=U^\dagger H_{1,I}^{\text{RWA}} U \\ &=-\hbar\Omega e^{-i\Delta t}e^{-i\omega_0t}|\text{e}\rangle\langle\text{g}| -\hbar\Omega^*e^{i\Delta t}|\text{g}\rangle\langle\text{e}|e^{i\omega_0t} \\ &=-\hbar\Omega e^{-i\omega_Lt}|\text{e}\rangle\langle\text{g}| -\hbar\Omega^*e^{i\omega_Lt}|\text{g}\rangle\langle\text{e}|. \end{align}

The atomic Hamiltonian was unaffected by the approximation, so the total Hamiltonian in the Schrödinger picture under the rotating wave approximation is

H^\text{RWA}=H_0+H_1^{\text{RWA}} = \hbar\omega_0|\text{e}\rangle\langle\text{e}| -\hbar\Omega e^{-i\omega_Lt}|\text{e}\rangle\langle\text{g}| -\hbar\Omega^*e^{i\omega_Lt}|\text{g}\rangle\langle\text{e}|.

Read more about this topic:  Rotating Wave Approximation