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If a particle with non-degenerate spectrum starts in some eigenstate, and the frequency of the external EM field matches some transition frequency, then this would lead the particle to do periodic absorption and emission of the photons with Rabi frequency. But in real measurements we usually just see some more or less intense absorption.

What I don't quite understand is how do we relate the parameters of Rabi cycles like Rabi frequency to the intensity of absorption? Why do we in fact see any absorption at all if it's more or less compensated by emission? Shouldn't we shine the light for exactly $n$ Rabi periods plus a $\frac14$ of the period to get perfect absorption?

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What I don't quite understand is how do we relate the parameters of Rabi cycles like Rabi frequency to the intensity of absorption? Why do we in fact see any absorption at all if it's more or less compensated by emission?

Rabi oscillation is a concept based on an approximate description of a single atom. This description does not include any irreversible element so it cannot by itself explain systematic absorption in medium.

If the model is to manifest behaviour consistent with systematic absorption of light, it has to include some form of friction - mechanism which allows the charge oscillations of the atoms to be not in phase with the external electric field.

One way to do this is to include mutual interaction of atoms in the medium into the description. The friction (and absorption) is then macroscopically visible result of the reversible interaction between charged particles.

Simpler but less accurate would be to just add some damping term into the Schroedinger (or equivalent) equation to make the oscillations of the atoms' electric moment lag in phase behind the external electric field.

Often all this is neglected and the absorption coefficient is obtained by a formal trick - assuming the atoms jump between discrete states irreversibly with probability given by the Fermi golden rule formula (see time-dependent perturbation theory). Then the absorption coefficient is calculated to be proportional to square of the transition dipole moment of the pair of eigenfunctions whose frequency of oscillation matches the external frequency. The absorption coefficient does not depend on the amplitude of the external electric field though, so it is not simply a function of the Rabi frequency.

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  • $\begingroup$ So, if I just include two atoms into the description, would I then get the necessary friction? E.g. take a hydrogen molecule instead of a single atom? This seems unbelievable: what's the difference whether there're two particles or four? In fact, the hydrogen molecule is very similar to lithium atom, but per your description the hydrogen molecule would systematically absorb while lithium atom wouldn't, because the former is two atoms and the latter is a single one. $\endgroup$
    – Ruslan
    Commented Dec 14, 2014 at 6:44
  • $\begingroup$ Actually I think I start to understand what you mean. You suggest to take an atom in its eigenstate, then add another atom to the system, putting it at some large distance moving towards the first one. Then the state of the system would be approximately eigenstate, and the collision of the atoms would perturb the process of transition in such a way that the oscillation are not in phase with external field. Then the energy would dissipate into other modes, and the light would actually be irreversibly absorbed. Do I understand your point correctly? $\endgroup$
    – Ruslan
    Commented Dec 14, 2014 at 18:47
  • $\begingroup$ What I meant were two things. The first is that if we connect one harmonic oscillator to a huge number of harmonic oscillators, the first oscillator will move in a way that exhibits tendency to lose energy to the large system, although absolute rest won't be reached, some fluctuating motion will persist. The other way which works even for two oscillators is to model their EM interaction by retarded EM fields. Because the force is not in phase with the motion then, the motion will be damped and the system energy will be lost to EM energy propagating out of the system. $\endgroup$
    – Ján Lalinský
    Commented Dec 15, 2014 at 18:52
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Spontaneous emission rate ($\Gamma$) plays an important role. If $\Gamma=0$ than indeed one would excite atoms and de-excite back in ground state by stimulated emission with emission of photon in same direction. As a result as you pointed out one would not observe absorption. In real life spontaneous emission will lead to emission of photons in random direction and we see it as an absorption in experiment.

If you have that Rabi frequency > $\Gamma$ than one can excite atom by shinning the pulse of light during the 1/2 of the period(also known as $\pi$ pulse). Also one can put atom in superposition of ground and excited state when pulse area is $\pi/2$. This is an important part of many single atom experiments e.g. Ramsey interferometer.

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  • $\begingroup$ I actually was under the impression that spontaneous emission would be somewhat suppressed in presence of stimulated one. Am I wrong? $\endgroup$
    – Ruslan
    Commented Dec 14, 2014 at 18:33
  • $\begingroup$ To elaborate: my impression is based on what I know of semiconductor laser operation: under the lasing threshold all its emission is due to spontaneous mechanism. But once the current is higher than lasing threshold, the broad part of emission spectrum becomes smaller, and the stimulated emission spectrum shaped by cavity appears. $\endgroup$
    – Ruslan
    Commented Dec 14, 2014 at 18:53
  • $\begingroup$ There are ways to suppress or enhance spontaneous emission(like Purcell effect), but in most simple case where you just shine light on the atoms (like absorption spectroscopy) this should not happen. I would say lasers are quite different story since photons that are emitted by one atom will de-excite another atom, and you can get the effect you talked about. Here I think also $\Gamma$ is not suppressed(see rate eq.) but I would rather say it is dominated by stimulated. You might want to take a look in some QO textbook(e.g. Scully) on deriving absorption profile in semiclassical picture... $\endgroup$
    – sa101
    Commented Dec 14, 2014 at 22:43

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