 Sisyphus cooling

Finite optical pumping time lets an atom climb a light-shift hill before transfer into a lower magnetic sublevel. Spontaneous emission dissipates the gained potential energy; climbing alone is conservative.^[ref]

To make the geometry and signs explicit, take an original choice of phase origin for equal opposing waves:The real electric field is $\mathrm{Re}[\mathcal{E}(z)e^{-i{\omega }_{L}t}]$; $\lambda$ is wavelength. The fractions $w_{\pm }$ are squared projections onto the defined circular basis vectors, divided by total intensity. Their sum is one: interference changes electromagnetic polarization despite constant total intensity. At $z=0,\lambda /4,\lambda /2$ the field is respectively ${\mathbf{ϵ}}_{+},{\mathbf{ϵ}}_{-},{\mathbf{ϵ}}_{+}$; halfway between these positions it is linearly polarized. Specifying the vectors avoids an observer-dependent verbal handedness convention.

For an illustrative $J_g=1/2\to J_e=3/2$ angular momentum transition, use the squared Clebsch-Gordan coefficients $1$ and $1/3$ for the strong and weak circular couplings from each lower magnetic sublevel. Interchanging circular components interchanges these strengths. The weighted sums are $w_{+}+w_{-}/3=(2+\cos (2kz))/3$ and $w_{-}+w_{+}/3=(2-\cos (2kz))/3$. Rescaling their common negative shift givesBoth potential energies are negative relative to the unilluminated level. Each hill has height $U_0$, and opposite hills are displaced by $\lambda /4$. An atom starting at the $U_{+}$ minimum at $z=0$ loses $U_0$ of kinetic energy if it reaches $z=\lambda /4$ on that branch. A pump transfer there lowers its potential energy by $U_0$, leaving it at a $U_{-}$ minimum. The next uphill segment ends at $z=\lambda /2$, where the favoured transfer is reversed. Actual transfers are stochastic and need not occur exactly at these extrema.

For one completed climb and pump, conservation of energy gives $\Delta K=-U_0$, while the ground-state potential energy returns to its starting value. Neglecting recoil and treating the scattered photon as the sole outgoing radiation change, $\hslash ({\omega }_{\mathrm{out}}-{\omega }_L)=U_0$. A potential drop is an internal-state reset, not an automatic equal upward kick in kinetic energy. Random emission nevertheless adds momentum fluctuations and recoil energy, so useful cooling requires the dissipated energy to compete with that heating.

The recoil energy $E_R=(\hslash k{)}^2/(2m)$ sets a different scale from $k_B{T}_D=\hslash \Gamma /2$. Conventional Sisyphus cooling reaches the recoil regime when reducing the hill height also weakens cooling; it is not a claim of arbitrarily small temperature at vanishing intensity. Subrecoil techniques need further physics.^[ref]