# Rigorous justification for rotating wave approximation

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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.

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user Abel Molina

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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

$$|\psi\rangle\approx|0\rangle-\frac{V_0}{2\hbar}t\sigma_1|0\rangle+\ldots.$$

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.

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user Jon
answered Nov 30, 2011 by (345 points)
- To obtain that $H_I \approx \frac{V_0}{2} \sigma_2$, I see how the $\sin(\omega - 2\Delta)t$ term is approximated by $0$, and how the $\cos(\omega - 2\Delta)t$ term is approximated by one. However, I don't see how do the other two terms (the one for $\sin(\omega+2\Delta)t$, and the one for $\cos(\omega+2\Delta)t$) disappear.

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user Abel Molina
- 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 $H_I$.

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user Abel Molina
@AbelMolina: Sorry but I have not seen you comments before. Yes, the crucial point is that you are very near resonance. This would imply in the integrals that a term becomes dominant because it increases like t with respect to the other that is oscillating and bounded. This linear (in time) term is just proportional to your RWA Hamiltonian.

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user Jon
No problem. Which one is the "oscillating and bounder" term, then?

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user Abel Molina
@AbelMolina: Near the reasonance you will get a term increasing linearly with time: In $|\psi\rangle$, after performing integration near the resonance, this will go like $V_0\sigma_1t|0\rangle$ while for the other you will get $(e^{4i\Delta t}-1)/4\Delta$. Clearly, the term linear in time will become overwhelming large as time increases with respect to the other.

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user Jon
(Very belatedly) Thanks for your answer! Your paper looks interesting as well, it is encouraging to see that people have considered these issues before. There are still a few things that are unclear to me:

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user Abel Molina
- It is not clear to me how is "one term is overwhelming large with respect to the other". My best guess is that you mean that $\omega \approx 2\Delta/\hbar$, and also $(\omega - 2 \Delta/\hbar)t' \approx 0$, while this is not the case for $(\omega + 2 \Delta/\hbar) t'$. Then, if we split the integral for $|\phi(t) \rangle$ into one for the $e^{i\omega t' - \frac{2i}{\hbar}\Delta t'}$ term, and one for the $e^{i\omega t' + \frac{2i}{\hbar}\Delta t'}$ term, then the exponent in the first integral is approximately one in all the domain, and in the other one it rapidly averages to zero.

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user Abel Molina
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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: $\delta < 2\Delta$.

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user V. Leyton
answered Feb 7, 2012 by (10 points)
As it is, this is not really an answer. At least define the terms in your inline equation please.

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user qubyte
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.

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user V. Leyton
Since this is not an answer, please consider placing the content into a comment as originally intended, and then remove this original.

This post imported from StackExchange Physics at 2014-08-05 16:17 (UCT), posted by SE-user qubyte

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