I am trying to understand better the rotating wave approximation (RWA).

Consider an atom modeled as a two level system, interacting with a Laser. I have the dipole momentum operator $$\vec{D} = d \left( \vec{\epsilon_d} \sigma_{-} + \vec{\epsilon_d}^{*} \sigma_{+}\right) \, .$$

The electric field is $$\vec{E} = E_0 \left( \vec{\epsilon} e^{j(\omega_L t + \phi_L)} + \vec{\epsilon}^{*} e^{-j(\omega_L t + \phi_L)} \right) \, .$$

We have the Hamiltonian of our system $$H=\frac{\hbar \omega_q}{2} \sigma_z - \vec{D} \cdot \vec{E}$$ where $\hbar \omega_q$ is the bare energy of our two level system.

After few calculations, we can write it as \begin{align} H &= \frac{\hbar \omega_L}{2}\sigma_z + \frac{\hbar (\omega_q - \omega_L)}{2}\sigma_z \\ &- dE_0 (\vec{\epsilon_d} \cdot \vec{\epsilon} e^{j(\omega_L t + \phi_L)} \sigma_{-} + \vec{\epsilon_d} \cdot \vec{\epsilon}^{*}e^{-j(\omega_L t + \phi_L)} \sigma_{-} \\ &+ \vec{\epsilon_d}^{*} \cdot \vec{\epsilon}e^{j(\omega_L t + \phi_L)} \sigma_{+} + \vec{\epsilon_d}^{*} \cdot \vec{\epsilon}^{*}e^{-j(\omega_L t + \phi_L)} \sigma_{+}) \, . \end{align}

The usual trick to show the RWA approximation is to go in interaction picture taking $\frac{\hbar \omega_L}{2}\sigma_z$ as the non interacting part. Doing that, we end up with \begin{align} H^I &= \frac{\hbar (\omega_q - \omega_L)}{2}\sigma_z \\ &-dE_0 (\vec{\epsilon_d} \cdot \vec{\epsilon}e^{j(\phi_L)} \sigma_{-} + \vec{\epsilon_d} \cdot \vec{\epsilon}^{*}e^{-j(2*\omega_L t + \phi_L)} \sigma_{-}\\ &+ \vec{\epsilon_d}^{*} \cdot \vec{\epsilon}e^{j(2*\omega_L t + \phi_L)} \sigma_{+} + \vec{\epsilon_d}^{*} \cdot \vec{\epsilon}^{*}e^{-j(\phi_L)} \sigma_{+}) \end{align}

And at this point we say that we will neglect the fast oscillating term.

Here, we had to go to interaction picture to see what we could neglect. Isn't it possible to directly see it on the first Hamiltonian $H$?

The R.W.A approximation is equivalent to directly neglect the terms $$\vec{\epsilon_d} \cdot \vec{\epsilon}^{*}e^{-j(\omega_L t + \phi_L)} \sigma_{-} $$ and $$\vec{\epsilon_d}^{*} \cdot \vec{\epsilon}e^{j(\omega_L t + \phi_L)} \sigma_{+}$$ in the first Hamiltonian $H$.

However, they are exactly of the same order as the two others that wont be neglected. So I am a little confused.

Is there a direct argument before using the interaction picture trick to see why we neglect those terms?

  • 1
    $\begingroup$ It's better to use the inline $\exp$ when you have a lot of stuff in the exponential... easier to read. $\endgroup$
    – DanielSank
    Feb 8, 2019 at 23:54
  • $\begingroup$ I'm happy to see this question because the rotating wave approximation does not seem to be well-motivated in any book or paper that I know of. Also, the issue of when the RWA fails to work is very close to my own heart. $\endgroup$
    – DanielSank
    Feb 8, 2019 at 23:57
  • $\begingroup$ I think the point of RWA is to remove the explicit time dependency of the Hamiltonian. By directly removing those 2 terms you still get time dependent H $\endgroup$
    – donnydm
    Nov 1, 2019 at 6:49

1 Answer 1


Choose a basis such that the non interacting Hamiltonian is diagonal with eigenenergies $E_i$. Write your interaction Hamiltonian in matrix form. Then take an term of $H_{ij}$ (i - row, j - column) which has a time dependence $H_{ij} \sim e^{i\omega t}$. The interaction picture time dependence will then be $e^{i (\omega + E_i - E_j)t}$. So you keep this term if $\omega + E_i - E_j \approx 0$. In particular diagonal terms survive if $\omega \approx 0$.

In case you are interested in why it works:

For a perturbed Hamiltonian $H = H_0 + V(t)$ we define the interaction picture as $|\psi_I\rangle = e^{iH_0t} |\psi(t)\rangle$. This "basically" means that we are transforming into the "moving frame" of the unperturbed wavefunction. Especially if $V(t)$ is small then $|\psi_I\rangle$ will only change little in time.

Defining the interaction Hamiltonian as $H_I = e^{iH_0t}V(t)e^{-iH_0t}$ one can rewrite the Schrödinger equation as $i\partial_t |\psi_I\rangle = H_I|\psi_I\rangle$. This equation can be integrated over time to achieve:

$$|\psi_I(t)\rangle = |\psi_I(0)\rangle - i \int_0^t \mathrm{d}t' V_I(t')|\psi_I(t')\rangle$$ This equation cannot be solved in general. However we know two more things: 1. The wavefunction will still be dominated by the unperturbed part which is time independent. So we can treat the wavefunction as slowly varying over time. 2. The interaction potential can be split into a fast oscillating and a slow oscillating part (in your case the slow oscillating part is time independent)

Combining these two we see that for medium time scales $t \gg 1/\omega_{\text{fast}}$ the time integral $\int_0^t \mathrm{d}t' V_I(t')|\psi_I(t')\rangle$ will simply 'kill' all fast oscillating parts of $V_I$ leaving only the slow oscillating parts. So for medium times one can simply drop all fast oscillating parts of the interaction Hamiltonian. This is what is called the R.W.A.

  • $\begingroup$ Thanks for your answer. But I am asking why we don't see in the non interacting picture that things are negligible. If terms are negligible in the interacting picture, they also should be in the non interacting but for me it is not obvious at all. $\endgroup$
    – StarBucK
    Feb 9, 2019 at 8:43
  • $\begingroup$ Ordinary perturbation theory is justified because of some small expansion parameter (i.e. a coupling). This can then be seen directly from the Schrödinger Hamiltonian. The RWA does not rely on this. It is not so important the interaction is small. It is justified by the observation that that fast oscillating behavior cannot be observed (because in reality every measured value is a mean value over finite time range). However fast and slow oscillating must be taken relative to the systems most dominant oscillation frequency (which is the frequency of the non interacting system). $\endgroup$
    – toaster
    Feb 9, 2019 at 13:17
  • $\begingroup$ For me to be able to see "the terms oscillates fast" we would need to define a "speed of measurement". Nothing says that in my experiment I cannot be faster than $\omega_q + \omega_L$ for example. Also, why do you say that the dominant oscillation frequency is the one of the non interacting system ? $\endgroup$
    – StarBucK
    Feb 9, 2019 at 13:50
  • 1
    $\begingroup$ The second question is more complicated. The main important thing for RWA is to find a suitable $H_0$ (which my be part of the interaction part as in your case). This is highly non trivial. You need to guess what the dominant behavior of your system will be (otherwise it's not a approximation). If you guess a wrong $H_0$ then the RWA will give wrong results. I don't know if there is a nice recipe for this. In case of Rabi oscillations one uses the classical limit as reference. In the classical case a driven harmonic oscillator oscillates with the external frequency. $\endgroup$
    – toaster
    Feb 9, 2019 at 14:41
  • 2
    $\begingroup$ Sorry I am not sure what your non interacting and what your interacting Hamiltonian is. You write $H_0 = H_q + H_f$, but this is also the full Hamiltonian :D. Apart from that I think you got the point. It is non trivial to find a good choice for $H_0$. In this case we have a laser (which is strong), so we expect the system to follow the time dependence of the laser. However if our light source is weak (or just some scattered light hitting the atom) then would choose $H_0 = H_q = \hbar\omega_q/2\sigma_z$ (because it is now the dominant term). This will eventually result in Fermis golden rule. $\endgroup$
    – toaster
    Feb 13, 2019 at 18:19

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