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I'm having some trouble with what I think should be an easy calculation. On qutip.org/docs/2.2.0/examples/me/ex-25.html they calculate the steady state for a master equation, more specifically they take an example from nanomechanics: they c a low-frequency nanomechanical resonator (with frequency $\omega_m$ lower than temperature) coupled to a high-frequency (optical) resonator (with frequency $\omega_r$ higher than temperature).

By applying a driving field to the high-frequency resonator with a frequency matching the frequency difference between the resonator ($\omega_d = \omega_r - \omega_m$), the two oscilators can effectively be brought into resonance in a rotating frame, allowing for excitation transfer from the low-frequency mechanical resonator to the high-frequency resonator (i.e., sideband cooling of the mechanical resonator).

They then go on to write that the Hamiltonian of this system is given by

$H = \omega_r a^\dagger a + \omega_m b^\dagger b + g a^\dagger a (b + b^\dagger) + A \sin(\omega_d t) (a + a^\dagger)$

which I fully accept. However, they then write that using the rotating wave approximation this can be rewritten as

$H = (\omega_r-\omega_d) a^\dagger a + \omega_m b^\dagger b + g a^\dagger a (b + b^\dagger) + \frac{1}{2} A (a + a^\dagger)$

This I do not see at all. The rotating wave approximation I know I have seen in the Jaynes Cummings model, where there are some phase factors that you recombine in a clever way such that some of them are much faster than the others, and you subsequently drop them. However, I see no such phase factors in the original question here, and I am honestly not sure what kind of steps are being taken in the above. Any help is much appreciated!

Some specific confusion comes from the fact that we get a term $\omega_r - \omega_d$, but no $\omega_m - \omega_d$.

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The rotating-wave approximation requires a transformation to a rotating frame. In particular, use the transformation $U(t) = {\rm e}^{{\rm i} \omega_d t a^\dagger a}$, i.e. transform your states as $\lvert \psi(t)\rangle \to \lvert \tilde{\psi}(t)\rangle = U(t) \lvert \psi(t)\rangle $. Now, in the original frame, the evolution is given by the Schroedinger equation ($\hbar = 1$) $$ {\rm i}\lvert \dot{\psi}(t)\rangle = H\lvert \psi(t)\rangle. $$ As an easy exercise, show that in the rotating frame the Schroedinger equation reads as $${\rm i}\lvert \dot{\tilde{\psi}}(t)\rangle = \tilde{H}(t)\lvert \tilde{\psi}(t)\rangle, $$ where $$\tilde{H}(t) = U(t) H U^\dagger(t) - \omega_d a^\dagger a. $$ You should now be able to prove the desired result using the rotating-wave approximation, which consists of neglecting "counter-rotating" terms oscillating at frequency $2\omega_d$.

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  • $\begingroup$ That is what I had in mind indeed. But what that does not tell me is why $w_r$ is special, compared to $\omega_m$, in the sense that we get the $\omega_r-\omega_d$ term but not the $\omega_m - \omega_d$ term. Is there some implicit assumption regarding the low/high frequency part of the story? $\endgroup$ – user129412 May 17 '16 at 17:19
  • $\begingroup$ There is nothing special about any of the frequencies. This transformation is just one possible choice out of (infinitely) many. It was apparently chosen to remove the tricky $\sin(\omega_d t)$ term, leading to an approximately time-independent Hamiltonian. This driving term only couples to the $a$ oscillator, and not the $b$ one. It is easiest to see why it comes out this way if you just work through it yourself. If you want a $\omega_m - \omega_d$ term, use a transformation like ${\rm e}^{{\rm i}\omega_d b^\dagger b}$. But this will make the radiation pressure term time-dependent. $\endgroup$ – Mark Mitchison May 17 '16 at 17:24
  • $\begingroup$ Well, that is incredibly stupid, somehow I missed the fact that the other one only had b terms. I guess that clarifies the situation. $\endgroup$ – user129412 May 17 '16 at 17:27
  • $\begingroup$ A short follow up. Say I also add a drive to the b part, so that $H = \omega_r a^\dagger a + \omega_m b^\dagger b + g a^\dagger a (b + b^\dagger) + A \sin(\omega_d t) (a + a^\dagger) + + A \sin(\omega_d t) (b + b^\dagger)$. How would this change my transformation? Do I simply use $e^{i \omega_d (a^\dagger a + b^\dagger b)}$? $\endgroup$ – user129412 May 18 '16 at 13:43
  • $\begingroup$ @user129412 you should try it and see. But as I said, the radiation pressure term (coupling between a and b) will make your life difficult since it will not transform nicely. $\endgroup$ – Mark Mitchison May 18 '16 at 16:11

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