# How is three-dimensional laser cooling possible?

I am a bit confused about laser cooling an atom in all three dimensions. I think I have understood the one-dimensional case: The atom absorbs doppler-shifted laser light and the momentum in this direction is reduced by the photon's momentum $$p_{\gamma}$$. When the excited state decays, the atom re-emits the photon in a random direction and the atom receives a momentum kick of $$p_{\gamma} \cos\theta$$ only.

However, in the 2D-case the atom also receives a momentum kick of $$p_{\gamma} \sin\theta$$ upon re-emission in the other coordinate as well, so the total energy did not change.

What am I missing?

• googling arxiv.org/abs/1409.2519 ,also this en.wikipedia.org/wiki/Laser_cooling#Doppler_cooling. It is not a costheta effect. Commented Oct 25, 2020 at 12:52
• @annav I can also ask differently: Where does the energy go when starting with a fixed amount of atoms with a certain energy? The amount of photons going in should be the same as the ones going out Commented Oct 25, 2020 at 14:23
• from the wiki "Since the initial momentum change was a pure loss (opposing the direction of motion), while the subsequent change was random (i.e., not pure gain), the overall result of the absorption and emission process is to reduce the momentum of the atom, therefore its speed". Think of the breitwigner shape of the photons of bound state energy transitions. The input photons are on the low side, but the random radoated after absorption photons will have the width, i.e. higher energies on the average. Commented Oct 25, 2020 at 14:41
• So in short, the spontaneously emitted photons have a larger energy than the absorbed ones because upon emission, the atom is slower? Commented Oct 25, 2020 at 14:46
• No , because there is a natural width/distribution to all quantum energy levels, and the way the laser is tuned towards the low side means the incoming photons are on the low side, whereas the ones randomly decaying will on the average be higher , taking away the extra energy that cools the sample. . think of a gaussian like shape. Commented Oct 25, 2020 at 16:15

In laser cooling, the laser light is red detuned, meaning it has lower energy than the atomic transition. An atom at rest cannot absorb it:

However, a moving atom now sees Doppler shifted laser beams. The light coming towards it from the right will be on resonance (for some velocity class) and will be absorbed:

The photon emitted by the laser has energy $$\hbar \omega < \hbar \omega_0$$. But when it is re emitted by the atom, however, it will have energy $$\hbar \omega_0 > \hbar \omega$$ ! So energy is conserved, but the emitted photon takes away some energy from the atomic cloud.

The atom receives a momentum kick upon absorbing the photon, and another one (essentially of equal magnitude) when spontaneously re-emitting it. BUT the absorbed photons always come from the same direction (the laser beams) whereas the spontaneously emitted photon is random. Over time, the random spontaneous emission averages to zero, only giving you a decrease in the momentum along each laser beam direction.

So for laser beams in $$6$$ orthogonal directions ($$\pm x, \pm y,$$ and $$\pm z$$) you get cooling in all directions.

## Limit of the above

This kind of "simple" laser cooling works until the Doppler temperature, set by the natural linewidth of the atom $$\Gamma$$: when the Doppler shifted frequency between right and left photons is less than $$\Gamma$$, the atom does not know which one to absorb because it cannot resolve it.

Eventually, the spontaneously emitted photon and the resulting momentum kick does limit the temperature you can reach, and that is called the recoil limit. Which is why, to get colder with light, you need to use conservative potentials and hence not rely on scattering.

## Applications to cold atoms

One of the main applications of laser cooling is to reach quantum degeneracy.

The degeneracy parameter $$D$$ goes as $$\exp(-S)$$ where $$S$$ is the entropy. To get quantum ($$D \sim 1$$), it is not enough to lose energy, you also need to lose entropy.

The incoming photon from the laser is coherent, hence has low entropy. The spontaneously emitted photon is random, hence has higher entropy. So you are also extracting entropy from the cold atomic gas.

• The breaking point for me was that the laser photon energy is smaller than the re-emitted photon. Thank you Commented Oct 26, 2020 at 6:18
• Okay, I thought about this again and another question appeared to me. Think of the case where the photon hits the atom head on and upon de-excitation, the photon is emitted in the very same direction as the laser photon. If I follow your argument, the laser photon has a lower energy than the emitted photon. Shouldn't the atom then also receive a larger momentum since it has emitted a photon with higher energy, effectively heating the atom? I know this must be wrong because both atom and photon have a higher energy in the end, but I don't see where... Commented Nov 16, 2020 at 12:52