# Conservation of energy (or lack thereof) in Doppler cooling [duplicate]

I did not find my question answered elsewhere, so here it is.

I'm doing a project in my first optics course, and we are reading a bit about Doppler cooling. I understand that a laser is tuned to a frequency where the photon energy $E_p = h \nu$ corresponds to an energy slightly below the energy $E_{abs}$ required to excite the gas atoms, and that an atom moving towards the laser at a high speed (relative to rest of the gas) will experience a blueshift that allows the atom to absorb the photon, thus slowing down the atom.

I also understand that the photon is re-emitted in a random direction, which of course means that the atom's velocity in the direction of the laser will be decreased. What I'm not sure I understand is how the re-emission of the photon doesn't just bring the atom back to the same kinetic energy (but with a different direction of motion).

I have not been able to find a source explaining this, but here's my own attempt at an explanation (which I am not sure is correct, which is why I am asking here): The absorbed photon has an energy (in the frame of the laboratory) of $E_1 = h \nu_1$ which is slightly less than the absorption energy $E_{abs}$. But after the absorption the atom's speed is reduced, thus the photon will be emitted with a different frequency $\nu_2$ (in the frame of the laboratory). Since the speed of the atom (relative to the frame of the laboratory) has been reduced by the absorption, the energy $E_2$ of the emitted photon will be closer to $E_{abs}$, thus $|E_{abs} - E_2| > |E_{abs} - E_1|$ (the last step of the explanation only works if the atom's speed perpendicular to the laser is not greater than the speed parallel to the laser for most atoms, but since the laser is tuned to only hit the fast moving atoms, this seems improbable enough to not invalidate the explanation).

Is this explanation correct or have I made mistake? I have taken a special relativity course but no quantum mechanics yet, so please keep the quantum mechanics to a minimum if possible.

## marked as duplicate by Neuneck, DavePhD, Danu, Brandon Enright, Kyle KanosMay 24 '14 at 13:51

• After writing my answer, I did some more googling and came across physics.stackexchange.com/q/5851 . It looks like a duplicate to me. – Floris May 23 '14 at 23:12
• I'm not sure the questions are necessarily the same, or at least they are stated differently (although I don't know how similar is too similar). In any case, the answers on that other question definitely covered my question too. – Johnny Hansen May 23 '14 at 23:31
• The questions are quite different but the answer applies. So even if this is not closed as a duplicate I think that the link between the two is useful to record. – Floris May 24 '14 at 1:07

Here are some intuitive arguments; I believe they are valid but it's been a very long time since I thought about these things - so I am open to comments / improvements.

The simple description of Doppler cooling:

Because of the detuning, the probability of a photon absorption is greatest for an atom moving towards the light source - and in the process of absorbing the photon, the atom loses momentum in the direction it was going.

When it re-emits the photon, it will be in a random direction; therefore there will be no net impact on the momentum of the atoms. This is sufficient to see that the atoms "slow down" in the beam direction, but not enough to see that they lose energy in total.

To see that the energy is lower, imagine that the atom had an initial momentum $2p$ where $p$ = momentum of the photon. After the absorption, the atom has momentum $p$. Now if we re-emit the photon at 90 degrees to the original direction, the net momentum is $\sqrt{2}p$ and thus the kinetic energy of the atom, which scales with momentum squared, is lower than it was. Only if the photon is re-emitted in the original direction do you see no net change in energy.

This does beg the question: what happened to the energy? It appears that we did no work on the system - the same amount of energy came in (one photon) as went out (same photon, barring tiny relativistic effects that are much smaller than the drop in kinetic energy of the atom). I don't think that the shift in the wavelength explains this.

I suspect this is where entropy comes in: since this process actually decreases the entropy of the system, we must have done work on the system: $\Delta U = -\Delta S\cdot T$. So energy needed to reduce the entropy was taken from the kinetic energy of the atom.

Here's a stab at it:

I don't believe you need to consider the change in frequency of the scattered light to understand how Doppler Cooling works - indeed, a conservation of momentum argument is usually more insightful.

In a nutshell, you've got the core part of it down. The laser field that the atom interacts with is red-detuned in frequency to an excited state transition such that atoms with any velocity in the direction of the beam path sees the frequency of the laser Doppler shifted by some amount proportional to its speed closer to its transition frequency. This Doppler shift makes the photons appear to have a higher energy in the atomic frame. It's not as if the photon really has two separate energies, but rather this is really just a sort-of addition between the energy of the photon and a portion of the kinetic energy of the moving atom. The atom and the photon are meeting half-way, as it were.

In the momentum picture, the atom experiences a "collision" with the photon going in the opposite direction, leading to a reduction in the velocity of the atom along that direction. When the photon is remitted, it is emitted in a random direction. While the atom now moves with a new velocity in a direction opposite to the emitted photon, the photon is emitted with energy matching the transition energy - slightly higher than the frequency of the absorbed light.

While no net energy is lost in our system when the photon is absorbed (the collision is elastic), the way the energy is spread between the atom and the photon is not the same as when the photon is emitted. In this subsequent "inelastic" anticollision (emission), the photon leaves with more than it started. The number of scattering events is high enough that through the course of many emissions the net contribution to motion of the atom not along the beam path and atom trajectory averages to 0.

Basically, dumping a unit of momentum consistently from one specific direction into a bunch of random directions that cancel out. You're still conserving energy in the system, but a small bit of energy is being removed in the form of emitting photons in other directions.

You may also be interested in looking at the Doppler Cooling limit (when absorbing and remitting a photon does not lead to a further loss of kinetic energy due to the absorption and emission process both behaving elastically), or also how magnetic fields are used in conjunction to prepare atomic traps and optical molasses.

• I don't believe that the difference in energy between the photon arriving and leaving explains the difference in kinetic energy of the atom from before the absorption to after the emission. – Floris May 24 '14 at 1:22
• I'm not sure how compelling I find it either, but if energy is conserved only in the same frame, the photon being absorbed at a Doppler shifted frequency will certainly be emitted at a different frequency relative to the atom. The atom is traveling at a different velocity when it emits compared to when it absorbs. If the frequency of both events is that of the transition in both cases with shifting taken into account, the photon leaving must have a different energy than that being absorbed. Entropy alone tells us state-wise what must be changing, but not the mechanism or change involved. – user28754 May 24 '14 at 21:04
• Aaaand I see this was marked as duplicate. Will check out that other answer ... – user28754 May 24 '14 at 21:21