I have read Rigorous justification for rotating wave approximation to have an idea of a rigorous proof of RWA approximation.
The main idea I have from this is that you can see it if you write a perturbative expansion of the solution to schrodingër equation : the high frequencies, when integrated will be negligible to the low frequencies.
I will take an analog example here. My point is to check if I understood correctly the justification of this approximation.
Why we can neglect high frequencies
Let's consider a Hamiltonian :
$$H(t)=V_0 \left( \widehat{A} e^{j \omega_1 t} + \widehat{B} e^{j \omega_2 t} + hc \right) $$
$hc$ means hermitic conjugate.
(To make a link with the other post, $H(t)$ plays the role of the hamiltonian in interaction picture here).
The solution to $i\hbar \frac{d}{dt}|\psi (t) \rangle = H(t) |\psi (t) \rangle$ is :
$$ |\psi(t)\rangle = Texp \left(-\frac{i}{\hbar} \int_0^t H(t')dt' \right) |\psi(0)\rangle$$
I will thus compute : $Texp \left(-\frac{i}{\hbar} \int_0^t H(t')dt' \right)$
$$Texp \left(-\frac{i}{\hbar} \int_0^t H(t')dt' \right)=1+\frac{V_0}{i \hbar} \left(A \int_0^t dt' e^{j \omega_1 t'} + B \int_0^t dt' e^{j \omega_2 t'} + hc \right)+...$$
The order one of this expansion will have terms that behave like : $\frac{V_0}{\hbar \omega_1}$ and $\frac{V_0}{\hbar \omega_2}$.
If I continue the calculation up to the second order of the expansion, I will have terms like $\frac{V_0^2}{\hbar \omega_1 \hbar \omega_2}$, $\left(\frac{V_0}{\hbar \omega_1} \right)^2$ and $\left(\frac{V_0}{\hbar \omega_2} \right)^2$.
Allright.
Now I can do the assumptions :
If I have $\omega_1 \ll \omega_2$, thus :
$$ \frac{V_0}{\hbar \omega_1} \gg \frac{V_0}{\hbar \omega_2} $$
Also :
$$\frac{V_0^2}{\hbar^2 \omega_1^2} \gg \frac{V_0^2}{\hbar \omega_1 \hbar \omega_2} \gg \frac{V_0^2}{\hbar^2 \omega_2^2}$$
Thus, at a fixed order of the expansion, the contribution coming from $\omega_2$ are negligible with the contribution coming from $\omega_1$.
However, for example, I don't know if :
$$\frac{V_0}{\hbar \omega_1} \gg \frac{V_0^2}{\hbar \omega_1 \hbar \omega_2}$$
Thus, even if at order $1$, the contribution from $\omega_2$ are negligible in comparison the the contribution from $\omega_1$. The contributions from $\omega_2$ that comes from the order $2$ are probably not negligible with respect to the $\omega_1$ contribution coming at order $1$.
Because of that I cannot say that I can remove the frequencies $\omega_2$ from my problem. If I do it, I will do a wrong approximation.
However, if I add the condition "The interaction is weak". Which means I can only study the first term of the developpment.
Which also means :
$$ \frac{V_0}{\hbar \omega_1} \ll 1$$ $$ \frac{V_0}{\hbar \omega_2} \ll 1$$
Then only the first term of the developpment will dominate. And at this order if $\omega_1 \ll \omega_2$ then the contribution of frequency $\omega_2$ is negligible.
Is my explanation correct ?
Did I understood correctly why we both need "the interaction is weak" and "one of the frequency is higher with respect to the other one".
