Whenever I have encountered the rotating wave approximation, I have seen "the terms that we are neglecting correspond to rapid oscillations in the interaction Hamiltonian, so they will average to 0 in a reasonable time scale" as the justification for its use. However, it is not completely clear to me why does this justify that the Hamiltonian that we obtain is a good approximation of the original one, and I was wondering if there is a more rigorous version of the justification, whether it is for a particular system, or in a more general case.

As an example, something that would be a satisfying answer would be a result of the form "If you consider an arbitrary state of the system and any time t large enough, and evolve the system according to the RWA Hamiltonian, we obtain with high probability a state close to the one we would obtain under evolution of the original Hamiltonian". "t large enough", "close" and "high probability" would preferably have some good quantitative description.


The rotating wave approximation (RWA) is well justified in a regime of a small perturbation. In this limit you can neglect the so-called Bloch-Siegert and Stark shifts. You can find an explanation in this paper. But, in order to make this explanation self-contained, I will give an idea with the following model

$$H=\Delta\sigma_3+V_0\sin(\omega t)\sigma_1$$

being, as usual $\sigma_i$ the Pauli matrices. You can easily work out a small perturbation series for this Hamiltonian working in the interaction picture with

$$H_I=e^{-\frac{i}{\hbar}\sigma_3t}V_0\sin(\omega t)\sigma_1e^{\frac{i}{\hbar}\sigma_3t}$$

producing, with a Dyson series, the following next-to-leading order correction

$${\cal T}\exp\left[-\frac{i}{\hbar}\int_0^tH_I(t')dt'\right]=I-\frac{i}{\hbar}\int_0^t dt' V_0\sin(\omega t')\sigma_1e^{\frac{2i}{\hbar}\Delta\sigma_3t'}+\ldots.$$

Now, let us suppose that your system is in the eignstate $|0\rangle$ of the unperturbed Hamiltonian. You will get

$$|\psi(t)\rangle=|0\rangle-\frac{i}{\hbar}\int_0^t dt' V_0\sin(\omega t')e^{-\frac{2i}{\hbar}\Delta t'}\sigma_1|0\rangle+\ldots$$ $$=|0\rangle-\frac{1}{2\hbar}\int_0^t dt' V_0\left(e^{i\omega t'-\frac{2i}{\hbar}\Delta t'}-e^{-i\omega t'-\frac{2i}{\hbar}\Delta t'}\right)\sigma_1|0\rangle$$

Now, very near the resonance $\omega\approx2\Delta$, one term is overwhelming large with respect to the other and one can write down


but in the original Hamiltonian this boils down to

$$H_I=V_0\sigma_1\sin(\omega t)\left(\cos(2\Delta t)+i\sigma_3\sin(2\Delta t)\right)$$ $$=\frac{V_0}{2}\sigma_1\left(\sin((\omega-2\Delta)t)+\sin((\omega+2\Delta)t)\right)$$ $$+\frac{V_0}{2}\sigma_2\left(\cos((\omega-2\Delta)t)-\cos((\omega+2\Delta)t)\right)$$ $$\approx \frac{V_0}{2}\sigma_2$$

with all the counter-rotating terms properly neglected with the condition $\omega\approx 2\Delta$ applied. It is essential to emphasize that, as the applied field increases, this approximation becomes even less reliable and it is just the leading order of a perturbation series in a near-resonance regime.

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    $\begingroup$ Good ol' fashioned quantum optics experiments like the micromaser sat comfortably in the safe zone with the RWA. However, newer systems such as superconducting quantum circuits live in a much more strongly coupled domain. We have to be more and more careful about how we apply the RWA, and I find that I can only use it as a rule of thumb, or that an RWA provides rather stringent inequalities required to be satisfied before a system behaves in a useful way to me (on paper). $\endgroup$ – qubyte Nov 30 '11 at 16:09
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    $\begingroup$ Good answer overall, but it has to be said that $\omega \approx 2\Delta$ doesn't make sense --- how close do they have to be? My understanding is that it depends on how long you observe it for, i.e. you need $(\omega-2\Delta)t \ll 1$. In practice this time is actually something like the coherence time for the system, which is why @MarkS.Everitt mentioned the SC quantum circuits as a system where the coherence time is much longer. $\endgroup$ – genneth Nov 30 '11 at 21:53
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    $\begingroup$ @genneth: This is an operative matter. What you get are Rabi oscillations and normally a detuning $\delta=\omega-2\Delta$ is kept. Then, in a experimental setup, you vary the external frequency $\omega$ and observe when the detuning crosses the zero. So, $\omega\approx 2\Delta$ is a very concrete resonance condition. $\endgroup$ – Jon Nov 30 '11 at 22:00
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    $\begingroup$ @Jon: perhaps I'm not clear, but the reason I say it doesn't make sense is because they're dimensional quantities; you're going to have to multiply by some time scale to get something to compare to a "natural" number like one. $\endgroup$ – genneth Dec 2 '11 at 15:58
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    $\begingroup$ - Also, this is not really important, but it also seems to me that there is a $\Delta$ missing in the exponents in the first formula for $H_I$. $\endgroup$ – Abel Molina Dec 26 '11 at 15:59

I'm completely agree with you both, to invoke the RWA it is mandatory a driving close to resonance, and consider a weak driving strength (weak in comparison with the proper frequency of the system). However, how close? it's a good question. What we need is just consider a detuning which leads to a quasi time-independent behaviour for the propagating terms (for the counter-propagating ones... you already know), it means the detuning should be bounded as: $\delta < 2\Delta$.

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    $\begingroup$ As it is, this is not really an answer. At least define the terms in your inline equation please. $\endgroup$ – qubyte Feb 8 '12 at 7:30
  • $\begingroup$ I did not want to answer the original question, sorry for that, I should clicked on the comments bottom. I wanted to make a comment to the point 2, where the user genneth brought the point about the bound condition on the detuning. For me Jon's answer was OK, and the initial statement of genneth as well ($(\omega - 2\Delta)t < 1$), but the conclusion $\omega \approx 2\Delta$ is not correct at all. It is clear from Jon's answer the condition &\delta < 2\Delta& to invoke the RWA. $\endgroup$ – V. Leyton Feb 8 '12 at 9:34
  • $\begingroup$ Since this is not an answer, please consider placing the content into a comment as originally intended, and then remove this original. $\endgroup$ – qubyte Feb 9 '12 at 16:59

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