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I was always wondering how is it that all the quantum optics levels schemes are depicted as if the laser couples only two certain levels with some frequency. For exmaple the standard lambda system which has three levels is driven by two external lasers. and each of the lasers only couples a single transition?

Why is it that each of the laser doesn't couple all of the possible transitions?

thank you.

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  • $\begingroup$ Are you just asking why only a specific photon energy (wavelength) equals a specific delta energy between two given electron levels? $\endgroup$ – Carl Witthoft Oct 16 '14 at 14:17
  • $\begingroup$ What is a "standard lambda system"? $\endgroup$ – garyp Oct 16 '14 at 17:26
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    $\begingroup$ @garyp I think the OP refers to an atomic level scheme similar to the first figure on this page. $\endgroup$ – Mark Mitchison Oct 16 '14 at 20:41
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This is typically because of optical selection rules which forbid certain types of transitions. The most usual case is where the states are, in order of energy, $S$, $D$ and $P$ states, driven by a reasonably-intense laser. In this case, the coupling to the EM field is usually a dipole coupling, which means that the atomic operator that does the transitions is the position operator.

As it happens, it is impossible to couple an $S$ and a $D$ states using a dipole: $$⟨D|\mathbf r|S⟩\equiv0.$$ There are a number of ways to see this, all of which have the Wignert-Eckart theorem at their core. More simply, though, a wavefunction like $\mathbf r|S⟩$ has a $|P⟩$ character, and must therefore be orthogonal to anything with a $|D⟩$ character.

Of course, this is only ever an approximation. For the particular case of $S$ and $D$ states, there will be quadrupole terms which can indeed couple the two; however, these tend to require much higher intensities and have much narrower linewidths, so they can be safely ignored unless you are making a concerted effort to address those transitions.

This is fairly specific to the atomic case, and lambda schemes are too widely spread over the map to give an answer that will cover them all. However, there are plenty of systems with relevant selection rules which make the couplings very close to zero. Most importantly, when we treat lambda schemes, it is in the understanding that that zero is a (very good) approximation. Moreover, if you're trying to do a treatment that's detailed enough to include the kind of higher-order terms that make that coupling nonzero, other things will break first: there will be a multitude of other effects and states to deal with before things like quadrupole couplings become important.

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  • $\begingroup$ I see, but if I have as in a lambda system, two "ground levels" with no coupling between them but both of them are coupled to some mutual excited state. Then driving the system with a single frequency should couple both of the ground states to the excited level (of course not with the same strength since usually the levels are non-degenerate) ? This is different than what is usually seen in the books, where each laser couples only one transition (has nothing to do with the possible couplings..) $\endgroup$ – Koby Yavilberg Oct 16 '14 at 13:57
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    $\begingroup$ That is very unclear to me. Can you rephrase? What exactly is the question? $\endgroup$ – Emilio Pisanty Oct 16 '14 at 14:12
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    $\begingroup$ @KobyYavilberg Often the two ground states will have non-zero spin projections in different directions. You can selectively couple to just one transition by using polarised light, which changes the angular momentum projection by a definite amount according to the direction of polarisation and the selection rules Emilio has described. So even if the ground sub-levels are completely degenerate, and the two lasers are at the same, resonant frequency, each laser can be made to couple to only one transition by choosing different polarisation states. $\endgroup$ – Mark Mitchison Oct 16 '14 at 20:39

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