When lasers are used to cool atoms or ions, etc where does the heat go?

According to the first law of thermodynamics, sourced from wikipedia "In any process in an isolated system, the total energy remains the same."

So when lasers are used for cooling in traps, similar to the description here: http://optics.colorado.edu/~kelvin/classes/opticslab/LaserCooling3.doc.pdf where is the heat transferred?

From what I gather of a cursory reading on traps, whether laser, magnetic, etc the general idea is to isolate the target, then transfer heat from it, thereby cooling it.

I don't understand how sending photons at a(n) atom(s) can cause that structure to shed energy, and this mechanism seems to be the key to these systems.

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There are two thermodynamic aspects to laser cooling that are worth mentioning. The first, as others have noted, has to do with the frequency of the light that is absorbed and emitted. In Doppler cooling, the laser is tuned slightly below the frequency that the atom wants to absorb. An atom moving toward the laser sees that light shifted slightly up in frequency, and is thus more likely to absorb it than an atom at rest or moving away from the laser is. When it absorbs the light, it also picks up the momentum associated with that photon, which is directed opposite the momentum of the atom (since the atom is headed in the opposite direction from the light), and thus slows down.

When the atom spontaneously emits a photon a short time later, dropping back to the ground state, the emitted photon has exactly the resonant frequency of the atomic transition (in the atom frame). That means that its frequency is a little higher (in the lab frame) than the frequency of the absorbed light. Higher frequency means higher energy, so the atom has absorbed a low-energy photon and emitted a high-energy photon. The difference in energy between the two has to come from somewhere, and it comes out of the kinetic energy of the moving atom.

(You might reasonably ask what happens with the momentum of the emitted photon; the emission process also gives the atom a kick in the direction opposite the emitted photon's direction. If this happens to be exactly the same as the direction of the laser, the resulting kick sends the atom back to the same initial velocity; however, the direction of spontaneous emission is random, and as a result, it tends to average out over many repeated absorption-emission cycles. On average, the atom loses one photon's worth of momentum with each cycle, and a corresponding amount of kinetic energy.)

The other aspect, which you didn't ask about, but is worth mentioning, is the entropy. On the surface, it might seem that even though energy is conserved, there would be a thermodynamics problem here because the system moves from a higher entropy (lots of fast-moving atoms) to a lower entropy (lots of slow-moving atoms). The difference is made up in the entropy of the light field. Initially, you have a single monochromatic beam of photons, all with the same frequency, all headed in the same direction, which is about as low-entropy as it gets. After the cooling, those photons have been scattered out into all different directions, and with a range of frequencies, so the entropy of the light is much greater, and more than makes up for the decrease in the entropy of the atoms.

That's probably way more detail than you wanted/needed, but there you go.

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I'm accepting this answer as it goes into slightly more detail than Dr Motl's answer. I wish I could accept both. – Tommy Hinrichs Feb 26 '11 at 18:53
Can you explain further why the change in light doesn't violate the 2nd law? It seems to me like you're taking thermal energy, which cannot be used to do work, and converting it into high-energy photons, which can. – endolith Jun 9 '11 at 20:42

The most widespread type of laser cooling is "Doppler cooling". The laser light, coming from the sides, is adjusted to a frequency that is slightly below the natural frequency of the atom.

So the atom will only absorb the light if it is moving against the laser beam. Consequently, the velocity will drop. When it absorbs the photon, the energy goes to its excitation. Later, the atom will emit a photon again but in a random direction. So the average magnitude of the atom's velocity after these two steps drops.

Both steps in the process - emission and absorption - are obviously energy-conserving, much like all other phenomena in the lab. The energy conservation is essential to calculate what happens.

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Finally I've got it :-) Thanks :-) – BarsMonster Feb 25 '11 at 10:09
Is that really sufficient? Before the interaction, the photon has one energy and the atom has another. After the interaction the photon has the same energy but the atom has (much) less. Where did the difference go? – Floris May 24 '14 at 2:15
Floris, as I said, the energy is perfectly conserved in every step and it's needed to do calculations. You need to consider the energy carried by the photon; the kinetic energy of the atom as a whole; and the excitation energy of the atom internally. – Luboš Motl May 30 '14 at 4:48

Just to expand on Dr. Motl's answer:

On average - when laser cooling is working properly and cooling the atoms - the photons emitted by the atoms are slightly higher energy than the photons absorbed. That's where the energy goes.

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The photon's energy goes into putting the atom into an excited state. Its momentum goes into slowing down the atom. Both are conserved.

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