I How can laser detuning excite an atom?

From Wikipedia and other sources that I am reading, a laser detuning can excite an atom. However, for an atom to become excited its energy must be equal to the energy difference of two states. So how can laser detuning still excite an atom? The frequency from the laser would of course be near the frequency needed to excite an atom but this still goes against my seconds sentence.

Thanks in advance.
 

Mentz114

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From Wikipedia and other sources that I am reading, a laser detuning can excite an atom. However, for an atom to become excited its energy must be equal to the energy difference of two states. So how can laser detuning still excite an atom? The frequency from the laser would of course be near the frequency needed to excite an atom but this still goes against my seconds sentence.

Thanks in advance.
The Doppler effect. If the atom is moving relative to the light source then it may become resonant. See https://en.wikipedia.org/wiki/Optical_molasses

In a simple one-dimensional version, an optical molasses is made with two counterpropagating laser beams, the frequency of which is tuned slightly below an atomic absorption resonance. As a result, a motion of an atom (or ion) in the direction of one of the beams will lead to a Doppler shift so that the absorption rate for the counterpropagating beam is increased, whereas the absorption rate for the opposite laser beam is reduced. Effectively there is a dissipative light force which is always directed opposite to the motion and therefore serves to reduce that motion.
 
Thanks for the reply. My question comes from notes that discuss laser detuning in regards to rotating wave approximation. What relation does the Doppler effect have with RWA then? Or is the DE only an example?
 

Cthugha

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However, for an atom to become excited its energy must be equal to the energy difference of two states.
This is not really the case. As this topic is tagged with the term "Rabi oscillations", I guess this is what you are interested in. The photon energy must exactly equal the energy level separation in order to achieve deterministic excitation of the atom, which means Rabi oscillations that reach 100% probability of finding the atom in the excited state when performing a measurement. For detuned pumping you get a faster Rabi oscillation which, however, has a lower amplitude, so you do not get to 100% probability to find the atom in the excited state, but only lower numbers.

One should keep in mind that the eigenstates of interest in this case are NOT the bare energies of the atom and the light field, but the energies of the coupled system, the so called dressed states. In a nutshell, you take the matrix, where the atom and photon energies are on the diagonal and the interaction matrix elements coupling the states are on the off-diagonal and diagonalize this matrix. The resulting energies are the true eigenenergies of the coupled system and the corresponding modes are mixed light-matter modes. Microscopically, you can think of this in tems of the Autler-Townes or ac-Stark effect: The presence of a strong ac electromagnetic field (which means the laser light field) also changes the atom energy levels.
 

DrClaude

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While the answers above are good, I think they missed the main point, namely that most levels do not have a precise energy. All levels that can decay to lower energy states (so basically all of them except the ground state) have a width in energy due to that decay. A simple way to think about it is that the finite lifetime of an excited state corresponds to an uncertainty on the exact energy of the level (due to the time-energy uncertainty principle).
 

Cthugha

Science Advisor
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While the answers above are good, I think they missed the main point, namely that most levels do not have a precise energy.
Just for the record: I considered this reply as well, but as the OP explicitly used the tag "Rabi oscillations", I thought finite line widths are not the main point here. The OP may clarify this point, though.
 

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