# How does Sisyphus cooling work in a photon picture?

Some years ago, during my masters degree, I took a short course on cold matter, which included a component on laser cooling and trapping taught by Ed Hinds. On the lecture on Sisyphus cooling, he makes the claim that

from the quantum point of view, this force is due to stimulated scattering of photons from one beam into the other.

It certainly sounds reasonable, so it just went into file in my head as-is, but I got called out on it in a recent comment, which states that

every cooling scheme needs spontaneous emission in one way or the other

and that definitely also sounds reasonable: for sure, any cooling scheme must involve some form of irreversible (or at least thermodynamically nontrivial) step at some point.

More to the point, the conflict mostly pointed out that I don't really understand how exactly this cooling scheme works. The usual understanding is that two counter-propagating light beams with opposite polarization will create a polarization grating, which will oscillate between linear and the two circular polarization, and this will introduce a position-dependent energy shift for the $m=±1/2$ ground state components via the dynamical Stark shift (i.e. light shift). The atom then rolls uphill, losing kinetic energy to potential energy in a reversible fashion, and then transitions down to the other curve, leaving it with yet another hill to climb just like Sisyphus was.

Here is, I guess, where I get lost: what is the precise nature of these transitions? Where exactly does the energy go, how much of it is there, and what fields intervene to do this? Saying that it's the original laser fields that are causing this transition seems disingenuous to me, as they are already in play in creating the optical lattice, but maybe there is a more rigorous way to account for both effects at the same time.

In addition to this, is the transition spontaneous or stimulated? If the latter, how does it square with the thermodynamics of cooling? In any case, where does the entropy in the centre-of-mass motion go? In the case of Doppler cooling this is relatively easy to see - the atom absorbs photons in an orderly fashion but it emits them spontaneously any which where - but here it's less clear where the energy is going and therefore it's also harder to keep track of that entropy.

Finally, how does the recoil limit arise for the scheme above? There are obviously some photon transfers between the beams to account for this, but the nature of the transition (between two ground states which can be arbitrarily close together, as the dynamical Stark splitting depends on the polarizability, which could be arbitrarily small) kind of obscures this - unless there were some form of scattering from one beam into the other one, which as above seems hard to pull out from the splitting.

• I don't think I'm comfortable with the "free-space" Sisyphus cooling, but when the system is put in an optical cavity, the dissipation occurs not by spontaneous emission but via leakage of the cavity field through the mirrors. I'm not sure what you have access to, but here's a review article where the first (non-Introduction) section is about cavity Sisyphus cooling. – march Mar 4 '16 at 17:44
• @march I made this comment about spontaneous emission, and I made it in a context comparing Doppler cooling to Sisyphus cooling. More precisely it should be "Every cooling scheme needs dissipation in some way or the other". You are absolutely right that cavity cooling methods can work without spontaneous emission, and that here losses through the mirror provide dissipation. – PiQuer Mar 6 '16 at 17:43