# Rotating Frame Transformation for three-level system

I am trying to solve a task from an exercise sheet:

We consider a three-level system ($$\Lambda$$-configuration) with three eigenstates $$|1\rangle$$, $$|2\rangle$$ and $$|3\rangle$$. The levels $$|1\rangle$$ and $$|3\rangle$$ are coupled by a probe beam with Rabi frequency $$\Omega_d$$ and angular frequency $$\omega_d$$. Additionally, Level $$|3\rangle$$ is coupled to $$|2\rangle$$ by a strong drive field with Rabi frequency $$\Omega_p$$ and angular frequency $$\omega_d$$. We denote the decay rates for spontaneous emission between the levels by $$\gamma_{31}$$ and $$\gamma_{32}$$.

a) Construct the Hamiltonian of the system and show that in an appropriate rotating frame it can be written as $$H = \left( \begin{matrix} 0 & 0 & \Omega_p \\ 0 & \Delta_p-\Delta_d & \Omega_d \\ \Omega_p & \Omega_d & \Delta_p \\ \end{matrix} \right)$$ where $$\Delta_p$$ = $$\omega_3-\omega_1-\omega_p$$ and $$\Delta_d = \omega_3-\omega_2-\omega_d$$. It is useful to introduce the operators $$\sigma_{13} = |1\rangle\langle3|$$ and $$\sigma_{23} = |2\rangle\langle3|$$.

I started by setting up the free and the interaction Hamiltonian as: $$H = H_0+H_{int}$$ and

$$H_0 = \hbar(\omega_1 \sigma_{11} +\omega_2 \sigma_{22} +\omega_3 \sigma_{33})$$

With the rotating wave approximation, I can ommit the quickly oscillating terms in the interaction Hamiltonian:

$$H_{int} = \hbar\left(\Omega_p\left( \sigma_{13}e^{i\omega_p t} + \sigma_{31}e^{-i\omega_p t} \right)+ \Omega_d\left( \sigma_{23}e^{i\omega_d t} + \sigma_{32}e^{-i\omega_d t} \right) \right)$$

Now putting this into a Matrix and setting $$\omega_1 = 0$$ I get:

$$H = \left( \begin{matrix} 0 & 0 & \Omega_pe^{i\omega_p t} \\ 0 & \omega_2 & \Omega_de^{i\omega_d t} \\ \Omega_pe^{-i\omega_p t} & \Omega_de^{-i\omega_d t} & \omega_3 \\ \end{matrix} \right)$$

But is this what I am supposed to do here? I feel like I am missing something because first I am not using the spontaneous emission rates in the Hamiltonian at all and second I am left with those exponential functions in the off-diagonal terms and the terms on the diagonal are not in the required form.

My guess would be to find some rotation Matrix such that $$H' = RHR^\dagger$$ but I have no idea how this matrix would look like.

I would really appreciate some help on what I'm missing here, google and my script didn't really help me out this time...

Transforming to a rotating frame means that you use a time-dependent unitary transformation $$U(t)$$. The transformed Hamiltonian is $$H'(t) = U(t) H(t) U(t)^\dagger + \mathrm i \hbar\, \dot U(t) U(t)^\dagger$$ (see for example this answer for an explanation of the second term).
In your concrete problem, since a) it is a homework exercise and therefore hopefully not too complicated and b) we really only want to get rid of some phases in the Hamiltonian, a good ansatz for the transformation would be $$U(t) = \operatorname{diag}\bigl( \mathrm e^{\mathrm i \alpha t}, \mathrm e^{\mathrm i \beta t}, \mathrm e^{\mathrm i \gamma t} \bigr) .$$