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I'm studying Rabi oscillations in a two-level system (TLS) with Hamiltonian \begin{align}H(t)=\omega_0 \sigma_z + \Omega \cos(\omega_1 t) \sigma_x.\end{align}

The model is usually treated within the rotating-wave approximation (RWA). I am now trying to reconcile the RWA with the adiabatic theorem.

Rotating-wave approximation

The model can be solved exactly within the rotating wave approximation (RWA) if the driving is weak ($\Omega \ll 1$) and near-resonant ($|\omega_0 - \omega_1| \ll \omega_0$). The system then shows the well known Rabi-oscillations, i.e. starting from the $\sigma_z |0\rangle=|0\rangle$, the probability of finding the state in either $|0\rangle, |1\rangle$ is (at resonance) \begin{align} p_1(t)=\cos^2(\Omega t/2) \\ p_2(t)=\sin^2(\Omega t/2) \\ \end{align}

Adiabatic theorem

We can also try to apply the adiabatic theorem to the model. The adiabatic approximation is justified, if $\dot{H}(t)$ is "small". More, precisely the adiabatic approximation neglects terms like \begin{align} \mu_{m,n}=\frac{\langle m(t)|\dot{H}(t)| n(t)\rangle}{E_m (t) - E_n(t)} \end{align}for $m\neq n$ and $| n(t)\rangle,E_n(t)$ the instantaneous eigenstates resp. eigenvalues of $H(t)$. In this model we have \begin{align} \mu\equiv \max_{m\neq n} |\mu_{m,n}|=\left |\frac{\Omega \omega_0 \omega_1 \cos (t \omega_1)}{\left(4 \Omega ^2 \sin ^2(t \omega_1)+\omega_0^2\right)^{3/2}} \right | \end{align} which is maximal for $t=n \cdot 2\pi$ with $\mu_\max=\Omega \omega_1/\omega_0^2$. The adiabatic theorem then states that any eigenstate stays an eigenstate and just acquires a dynamical and geometric phase. While the exact form for the instantaneous eigenstates of $H(t)$ are cumbersome to write down, it should be clear, that since the perturbation is weak, the instantaneous eigenstates of $H(t)$ are always very close to $|0\rangle, |1\rangle$. As a consequence there can be no Rabi oscillations while treating the system adiabatically. Numerical calculations confirm this.

Question

How can those two seemingly contradictive results be reconciled? It seems like both are valid in the same parameter regime, i.e. for weak driving $\Omega \ll 1$ and close to resonance. Since the Rabi-oscillations within the RWA are a very established result, I assume that I missed something when applying the adiabatic theorem but I can't find any mistake.

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  • $\begingroup$ You apply adiabatic approximation before or after RWA? $\endgroup$
    – Roger V.
    Commented Jul 11, 2022 at 14:52
  • $\begingroup$ When treating the problem with the adiabatic theorem I don't apply the RWA at all because it is not necessary. The point is that doing so I find that the adiabatic approximation should be valid in the same regime as the RWA, i.e. at weak driving and and close to resonance. $\endgroup$ Commented Jul 12, 2022 at 8:13

1 Answer 1

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There is an article that answers exactly this question:

https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.102.220401

The main argument of the paper is that one cannot rely solely in the condition to adiabaticity when the driving is resonant with the system. The example you have shown is in fact the paradigmatic model for such a breakdown in the adiabatic theorem, and your calculations are correct. In order to evolve the Rabi model adiabatically, the driving frequency should be made small instead of the intensity.

For more information, I refer you to the original reference.

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  • $\begingroup$ Welcome to Physics! In general, it's preferable to provide a summary of the article, since URLs may change in in the future. Link-only answers are frowned upon and may end up being deleted. Can you edit your answer to provide a brief summary of the answer in the article? $\endgroup$ Commented Jul 8, 2022 at 16:08
  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Jul 8, 2022 at 16:08
  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – Miyase
    Commented Jul 8, 2022 at 16:11
  • $\begingroup$ Thanks guys, I'll edit then $\endgroup$ Commented Jul 8, 2022 at 16:40

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