Two Level System

Lets start from a two level system. We know a two level system has a Hamiltonian like

$H_0=\frac{\hslash {\omega }_0}{2}\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$

Now we want to perturb this Hamiltonian by an oscillating electric field. This has the form

$H_I=-\vec{d}\cdot \vec{E}$

We can write the electric field as $\vec{E}=E_0\cos (\omega t)\overrightarrow{ϵ}$ so inside we have

$H_I=-E_0\cos (\omega t)\vec{d}\cdot \overrightarrow{ϵ}$

Lets write this in matrix form

$⟨0|H_I(t)|0⟩=-E_0\cos (\omega t)⟨0|\mathbf{d}\cdot \mathbf{ϵ}|0⟩$

$⟨1|H_I(t)|1⟩=-E_0\cos (\omega t)⟨1|\mathbf{d}\cdot \mathbf{ϵ}|1⟩$

$⟨0|H_I(t)|1⟩=-E_0\cos (\omega t)⟨0|\mathbf{d}\cdot \mathbf{ϵ}|1⟩$

$⟨1|H_I(t)|0⟩=-E_0\cos (\omega t)⟨1|\mathbf{d}\cdot \mathbf{ϵ}|0⟩$

Now the dipole operator is defined as

$\vec{d}=e\vec{r}$

A definition is then the Rabbi Frequency

$\Omega =\frac{E_0⟨0|\mathbf{d}\cdot \mathbf{ϵ}|1⟩}{\hslash }$

Now we can assume the polarization is along a single axis, say the x axis so that

$\vec{d}\cdot \overrightarrow{ϵ}=ex$

Now we know $x$ is a odd function so we therefore a transition can only occur for opposite parity states, so the diagonal elements are 0. Thus we have

$H_I=\left(\begin{array}{cc}0& \hslash \Omega \cos (\omega t)\\ \hslash \Omega \cos (\omega t)& 0\end{array}\right)$

Assuming the Rabbi frequency is real. So in total we have that

$H=\left(\begin{array}{cc}\frac{\hslash {\omega }_0}{2}& \hslash \Omega \cos (\omega t)\\ \hslash \Omega \cos (\omega t)& -\frac{\hslash {\omega }_0}{2}\end{array}\right)$

Now we can write the cosine as

$\cos (\omega t)=\frac{e^{i\omega t}+e^{-i\omega t}}{2}$

Which is two vectors in the complex plane rotating in opposite directions, rewriting the matrix then we have

$H=\left(\begin{array}{cc}\frac{\hslash {\omega }_0}{2}& \frac{\hslash \Omega }{2}(e^{i\omega t}+e^{-i\omega t})\\ \frac{\hslash \Omega }{2}(e^{i\omega t}+e^{-i\omega t})& -\frac{\hslash {\omega }_0}{2}\end{array}\right)$

Now notice what’s happening, there is two components of the light rotating in opposite directions. Lets transform our Hamiltonian into one of these rotating frames, and it is important which one is picked. To do this we need to derive an equation that does this transformation. The Schrodinger equation reads as

$\frac{d}{dt}\psi =-\frac{i}{\hslash }H\psi$

Now apply a unitary transformation on both sides

$U\frac{d}{dt}\psi =-\frac{i}{\hslash }UH\psi$

And remembering that $U^{†}U=I$ we can insert this so

$U\frac{d}{dt}\psi =-\frac{i}{\hslash }(UH{U}^{†})U\psi$

The product rule states that

$\frac{d}{dt}(U\psi )=\dot{U}\psi +U\dot{\psi }$

Rearranging and plugging this in to the above equation reads

$\frac{d}{dt}(U\psi )-\dot{U}\psi =-\frac{i}{\hslash }(UH{U}^{†})U\psi$

$\frac{d}{dt}(U\psi )=\dot{U}\psi -\frac{i}{\hslash }(UH{U}^{†})U\psi$

And inserting the identity again we get

$\frac{d}{dt}(U\psi )=\dot{U}{U}^{†}U\psi -\frac{i}{\hslash }(UH{U}^{†})U\psi$

$\frac{d}{dt}(U\psi )=[\dot{U}{U}^{†}-\frac{i}{\hslash }(UH{U}^{†})]U\psi$

But now we can define ${\psi }_T=U\psi$ and $H_T=i\hslash \dot{U}{U}^{†}+UH{U}^{†}$ and we now have

$\frac{d}{dt}{\psi }_T=-\frac{i}{\hslash }{H}_T{\psi }_T$

Okay so lets pick a direction of the electric field. We should pick the positive omega otherwise we will be driving at twice the resonance frequency if you work though the math

$U=e^{i\omega t}$

So plugging this into the transformed Hamiltonian formula we get

$H=\left(\begin{array}{cc}-\frac{1}{2}(\omega -{\omega }_0)\hslash & e^{i\omega t}\Omega \hslash \cos (\omega t)\\ e^{-i\omega t}\Omega \hslash \cos (\omega t)& \frac{1}{2}(\omega -{\omega }_0)\hslash \end{array}\right)$

$H=\left(\begin{array}{cc}-\frac{1}{2}\hslash (\omega -{\omega }_0)& \frac{\hslash \Omega }{2}\left(1+e^{i2\omega t}\right)\\ \frac{\hslash \Omega }{2}\left(1+e^{-i2\omega t}\right)& \frac{1}{2}\hslash (\omega -{\omega }_0)\end{array}\right)$

Now to finish the rotating wave approximation we drop the fast oscillating terms

$H=\left(\begin{array}{cc}-\frac{1}{2}\hslash (\omega -{\omega }_0)& \frac{\hslash \Omega }{2}\\ \frac{\hslash \Omega }{2}& \frac{1}{2}\hslash (\omega -{\omega }_0)\end{array}\right)$

Now just writing this is a nicer form we have

$H=\left(\begin{array}{cc}\frac{\hslash \delta }{2}& \frac{\hslash \Omega }{2}\\ \frac{\hslash \Omega }{2}& -\frac{\hslash \delta }{2}\end{array}\right)$

Where $\delta ={\omega }_0-\omega$. This is the new Hamiltonian once the perturbation from the Electric field has been added. So lets find its eigen values

$\det (H-\lambda I)={\lambda }^2-\frac{{\hslash }^2}{4}({\delta }^2+{\Omega }^2)$

${\lambda }_{\pm }=\pm \frac{\hslash }{2}\sqrt{{\delta }^2+{\Omega }^2}$

Far Detuned Limit

Now suppose $\delta >>\Omega$ then we can write

${\lambda }_{\pm }=\pm \frac{\hslash }{2}\sqrt{{\delta }^2+{\Omega }^2}=\pm \frac{\hslash \delta }{2}\sqrt{1+\frac{{\Omega }^2}{{\delta }^2}}=\pm \frac{\hslash \delta }{2}(1+\frac{{\Omega }^2}{2{\delta }^2})$

${\lambda }_{\pm }=\pm (\frac{\hslash \delta }{2}+\frac{\hslash {\Omega }^2}{4\delta })$

Which is important for Optical Dipole Trap’s.

References

[1] Atomic Physics. Foot. 2008