# Why does an external laser drive only couples certain levels?

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.

• Are you just asking why only a specific photon energy (wavelength) equals a specific delta energy between two given electron levels? – Carl Witthoft Oct 16 '14 at 14:17
• What is a "standard lambda system"? – garyp Oct 16 '14 at 17:26
• @garyp I think the OP refers to an atomic level scheme similar to the first figure on this page. – Mark Mitchison Oct 16 '14 at 20:41

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.