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...


1 Answer 1


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 $$ U(t) = \operatorname{diag}\bigl( \mathrm e^{\mathrm i \alpha t}, \mathrm e^{\mathrm i \beta t}, \mathrm e^{\mathrm i \gamma t} \bigr) . $$

  • 1
    $\begingroup$ After posting the answer, I realized that this is an old question. Unsure why it showed up under active questions, but hopefully the answer will help somebody at some point. $\endgroup$
    – Noiralef
    Commented Aug 6, 2021 at 13:25

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