I mean, do these emitted photons go out of the system without any other happening. Because they are in higher frequency and so more energy, can they be used as a source of energy?
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2 Answers
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In many cases, you can see laser cooled atoms in a MOT with the naked eye. They look like a little fuzzy glowing dot at about the same color as the trapping laser. So there are indeed lots of photons leaving the system. This spontaneously emitted light is what ultimately carries away the entropy and make cooling possible.
You could try to capture this energy, just as you can try to gather any emitted light from any object. But I'm not sure why one would want to. It is relatively dim compared to room light, and since it is being emitted in all directions it would be difficult to collect very much of it.
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1$\begingroup$ I want to add that firstly photon energy gain is small, the density of trapped atoms is small amd also wall plug efficiency of laser is small. Chances of converting this process for energy production is negligence. $\endgroup$– hsinghalJun 12, 2016 at 17:44
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$\begingroup$ @hsinghal Yes, these are all good additional reasons why this wouldn't be practical (btw, if you will forgive a minor English correction the last word should be "negligible"). $\endgroup$– RococoJun 12, 2016 at 18:04
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2$\begingroup$ yes you are right it was my smartphone which is smarter than me, and I could not edit the comment now. $\endgroup$– hsinghalJun 12, 2016 at 18:06
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Laser cooling works by irradiating a e.g. an ensemble of gas atoms with light that has slightly less energy than a excitable transition in the atom. If the frequency is chosen right, the transition is only addressable by a 2nd order process which will consume kinetic energy from the atom.
If I understood your question right, you are asking what is happening after an atom absorbed a photon by this process. The atom will relax at some point of time and emit a photon with the energy of the transition. This photon might now leave the cooling volume or be reabsorbed. The presence of alternative, non-radiative, relaxation channels will endanger our goal to cool the gas. Non-radiative decay would mean that the energy of the photon would (at least in parts) end up raising the temperature of the gas.
If there is no non-radiative channel, reabsorption does not hurt. There may be other interactions which one could think of but for low density gasses they are not very likely.
To make optical cooling work you have to find a transition which should only have optical recombination channels. In addition you will have to surpass all other mechanisms which deposit energy in your system.
