= Sisyphus cooling
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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.<Laser cooling below the Doppler limit by polarization gradients: simple theoretical models by Jean Dalibard and Claude Cohen-Tannoudji>[§3 A, pp. 2027–2028, Figs. 3–4]{ref}
To make the geometry and signs explicit, take an original choice of phase origin for equal opposing waves:
$$
\boldsymbol{\mathcal E}(z)=E_0\left(\hat{\mathbf x}e^{ikz}+i\hat{\mathbf y}e^{-ikz}\right),\quad
k=\frac{2\pi}{\lambda},\quad
\boldsymbol\epsilon_\pm=\frac{\hat{\mathbf x}\pm i\hat{\mathbf y}}{\sqrt2},\quad
w_\pm(z)=\frac{1\pm\cos(2kz)}2.
$$
The real <electric field> is $\operatorname{Re}[\boldsymbol{\mathcal E}(z)e^{-i\omega_Lt}]$; $\lambda$ 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 $\boldsymbol\epsilon_+,\boldsymbol\epsilon_-,\boldsymbol\epsilon_+$; 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 gives
$$
U_+(z)=-U_0-\frac{U_0}{2}\cos(2kz),\qquad
U_-(z)=-U_0+\frac{U_0}{2}\cos(2kz),\qquad U_0>0.
$$
Both <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.
\Image[/sisyphus-polarization-gradient-energy-cycle.png]
{title=Polarization geometry, shifted ground states and a Sisyphus energy cycle}
{description=Original schematic: pumping arrows join occupied hilltops to opposite minima. Ideal completed cycles neglect recoil; actual transfers are stochastic.}
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, $\hbar(\omega_{\rm out}-\omega_L)=U_0$. 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> $E_R=(\hbar k)^2/(2m)$ sets a different scale from $k_BT_D=\hbar\Gamma/2$. 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.<Nobel Lecture: Manipulating atoms with photons by Claude Cohen-Tannoudji>[p. 712, §§II B and III A]{ref}
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