Rigorous justification for rotating wave approximation

Question

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.

Jon · Accepted Answer · 2021-11-04T07:30:28.613

23

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

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

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

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

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

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

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.

edited Nov 04 '21 at 07:30

answered Nov 30 '11 at 12:01

Jon

3,818

2

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). – qubyte Nov 30 '11 at 16:09
3

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. – genneth Nov 30 '11 at 21:53
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@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. – Jon Nov 30 '11 at 22:00
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@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. – genneth Dec 02 '11 at 15:58
@genneth: I think you should really be looking at something like $\omega/2\Delta \approx 1$, which is dimensionless. – qubyte Dec 04 '11 at 09:41
(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: – Abel Molina Dec 26 '11 at 15:56
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.

Abel Molina

Dec 26 '11 at 15:57

When it says "in the original Hamiltonian this boils down to", I don't understand the details, but I suspect that what it means is that if we had this Hamiltonian, that would produce the $|\phi \rangle$ that appears in the previous line. Am I right?

– Abel Molina Dec 26 '11 at 15:58

score 3 · Answer 2 · answered Sep 25 '21 at 00:40

The Magnus expansion is a simple, quantitative way to see why the rotating wave approximation (RWA) works, and what the leading order correction are. Given a time-dependent Hamiltonian $H(t)$, the dynamics over a time interval $T$ can be approximated by the series expansion

$$\bar{H} = \frac{1}{T}\int_0^T H(t) dt - \frac{i}{2T}\int_0^T dt\int_0^t dt' [H(t),H(t')]+...$$

valid if $\int_0^T \text{Tr}[H^2] \ll \pi$. If the Hamiltonian is periodic, as is typically the case in RWA problems, we can set $T$ to be the period. The first term is simply the RWA Hamiltonian, because oscillating terms average to zero over a single period. The second term scales proportional to $T$, so this leading order-correction goes to zero in the high-frequency limit $T\rightarrow 0$.

An advantage of the Magnus expansion over a Dyson series is that you can approximate the dynamics of periodic Hamiltonians with the time evolution operator $\bar{U} = e^{-iHt}$. Because $\bar{H}$ is Hermitian, $\bar{U}$ is exactly unitary, and thus you can use $\bar{U}$ to compute the corrected dynamics out to long times.

It's clear that this gives a time independent Hamiltonian that reproduces the exact dynamics whenever we're looking at the system at an integer multiple of T. But is it obvious that this Hamiltonian is useful for predicting the dynamics between 0 and T? — Jahan Claes, Oct 07 '21 at 13:32
As long as T is small, this expansion is valid, so it is very useful for getting the dynamics at intermediate timescales. Note that we can get the dynamics for any time t by breaking things up like t=nT + deltat, where n is an integer and deltat is also small — user34722, Oct 08 '21 at 17:35

score 1 · Answer 3 · answered Feb 07 '12 at 17:10

1

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: $\delta < 2\Delta$.

answered Feb 07 '12 at 17:10

3

As it is, this is not really an answer. At least define the terms in your inline equation please. – qubyte Feb 08 '12 at 07:30
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. – Feb 08 '12 at 09:34
Since this is not an answer, please consider placing the content into a comment as originally intended, and then remove this original. – qubyte Feb 09 '12 at 16:59

user45524 · Answer 4 · 2021-11-20T23:04:41.317

1

This has bugged me for some years, but finally I know a rigorous bound for the finite dimensional case. See Eq. (3.12) in https://arxiv.org/abs/2111.08961

The bound is on the difference between the full unitary evolution and the approximated one, in spectral norm. It scales inverse proportionally to the frequency, so that in the limit of infinite frequency the approximation becomes perfect.

You will see that the bound actually becomes worse roughly proportionally over the total evolution time, rather than better.

edited Nov 20 '21 at 23:04

answered Nov 19 '21 at 05:37

user45524

11

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Nov 19 '21 at 06:35
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – John Rennie Nov 19 '21 at 06:35

Rigorous justification for rotating wave approximation

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