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.
 A: So here is my first take, which does not answer all of your questions.
The coherent interactions with the lasers, including stimulated redistribution of photons from one beam to the other, lead to a gradient of the dressed states' energies: the dipole force. This force is conservative and provides the spatially oscillating optical potential depending on the internal m-level of the atom, as shown by the blue and red lines of your picture. A moving atom in some internal m state can run "up-hill" losing kinetic energy and "down-hill" gaining kinetic energy. Because the dipole force is conservative it cannot provide cooling by its own.
The key point in this cooling scheme is optical pumping between different m-levels, which is a dissipative process and represented by the curved arrow going up and down in the picture. Optical pumping means that a laser photon is absorbed and then spontaneously emitted to free space, possibly changing the internal m-level. An important thing to note is that optical pumping is "slow" in a sense that it does not follow the motion of the atom adiabatically but with some delay. The rates for optical pumping depend on the natural linewidth and detuning of the excited state, intensity and polarisation of the electric field (at the atom's position) and the Clebsch-Gordan coefficients. Sisyphus cooling works because optical pumping from m=1/2 to m=-1/2 is more likely to occur when the atom is at the top of the m=1/2 potential hill, compared to the reverse optical pumping process which contributes to heating. An analogous argument is true starting from the m=-1/2 level.
In this more probable optical pumping process, the energy of the absorbed photon is less than the energy of the spontaneously emitted photon, this is where the energy is carried away. The difference in energies, equal to the difference of the optical potentials for the two m-levels, is compensated by a loss of kinetic energy.
To analyze the temperature limit of a cooling scheme one has to compare the friction forces to the competing diffusion (heating). The recoil limit is a lower limit for Sisyphus cooling because it corresponds to the diffusion of one spontaneously emitted photon initially at rest. This spontaneous emission is not between two ground states, there is an intermediate excited state involved which is only hinted at in your picture. I'm not sure whether it is trivial or not that Sisyphus cooling can reach the recoil limit, maybe someone else can fill in. I think recalling some early analysis where the result is on the same order but higher than the recoil limit.
In fact, the recoil limit is fundamental for cooling methods in free space which solely rely on (randomly directed) spontaneous emission. Methods that can go beyond the recoil limit are, for example, sideband cooling where a trap absorbs the photon momentum, velocity selection methods (e.g. velocity-selective coherent population trapping), cavity cooling or evaporative cooling.
