Rotating wave approximation and classical Rabi oscillations: why don't the fast oscillating terms seem negligible in the initial frame?

Question

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.

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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?

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.

Answer 2

This is a common confusion. Since all frames are unitarily equivalent, you can make the approximation in any frame you like! But indeed, if you simply cross out highly oscillatory terms, then you get the wrong result in most frames. The answer to your questions is that rotating terms can only be neglected when the rest of the Hamiltonian does not scale with the oscillation frequency. For instance, on resonance, in the lab frame, the non-oscillating term grows linearly with the frequency, the RWA fails. For details see https://arxiv.org/abs/2111.08961