Evaporative Cooling

Evaporative cooling begins exactly where every optical method is forced to stop, and it aims at a single number handed to it by statistical mechanics. Laser cooling can prepare a dense microkelvin cloud, but it is bounded below by the recoil limit inherited from the quantum of photon momentum, and at the densities where it might reach further, photon reabsorption and light-assisted collisions heat and eject atoms. The target it cannot hit is Bose-Einstein condensation, where the phase-space density $\rho = n\lambda_{\mathrm{dB}}^3$ must climb to $\zeta(3/2)\approx 2.612$ — the point where the thermal de Broglie wavelength grows to the interparticle spacing and quantum statistics takes over. Evaporation is the purely thermodynamic, light-free technique that bridges the six-order-of-magnitude gap between them. The mechanism is almost embarrassingly simple. A conservative trap of depth $U_0$ holds the gas; the most energetic atoms in the Boltzmann tail escape over the rim, and because each carries away far more than the mean energy, the remainder is left colder. Elastic collisions refill the depleted tail and re-establish a thermal distribution at lower temperature, and lowering $U_0$ keeps it running. The whole art is in the timescales: rethermalizing elastic collisions must overwhelmingly outnumber the inelastic and background collisions that eject atoms without cooling. When the elastic rate rises as the gas is compressed, cooling self-accelerates — the runaway regime that reaches degeneracy in finite time and made 1995 the year the first atomic condensates appeared.

Key takeaways

Evaporative cooling removes the highest-energy atoms from a trapped gas and lets elastic collisions rethermalize the remainder to a lower temperature, driving phase-space density toward degeneracy.
It is the essential stage after laser cooling, which stalls near the recoil limit; evaporation covers a factor $\sim 10^6$ in $\rho$ up to the BEC threshold $\rho=\zeta(3/2)\approx 2.612$ .
The trap depth $U_0=\eta\,k_\mathrm{B}T$ is lowered progressively, often with a radiofrequency knife on a magnetic trap; atoms above the cut escape carrying $(\eta+\kappa)\,k_\mathrm{B}T$ , far more than the mean.
Cooling follows $T\propto N^{\alpha}$ with $\alpha=(\eta+\kappa)/3-1$ ; because $\rho\propto N^{\,1-3\alpha}$ , a $\sim 70$ atom-number loss buys a factor $\sim 10^5$ in phase-space density.
Runaway evaporation needs $\alpha>1$ and a good-to-bad ratio $R=\gamma_\mathrm{el}/\gamma_\mathrm{loss}\gtrsim 150$ ; spin-polarized fermions cannot rethermalize (s-wave forbidden) and are cooled sympathetically.
Key idea: cutting away the energetic tail of a trapped thermal gas and letting elastic collisions rethermalize the remainder lowers the temperature far faster than it lowers the density, so phase-space density climbs toward quantum degeneracy — provided elastic collisions overwhelmingly dominate over loss.

The core insight

Cutting away the energetic tail of a trapped thermal gas and letting elastic collisions rethermalize the remainder lowers the temperature far faster than it lowers the density, so phase-space density climbs toward quantum degeneracy provided elastic collisions dominate over loss processes.

Where this lives today

Every quantum-gas apparatus built since 1995 ends its cooling sequence with evaporation. In magnetic traps the RF knife of equation (6) is swept from tens of megahertz down to the trap bottom over $10$ – $60$ s; the same idea drives microwave evaporation on atom chips, whose steep traps condense in under a second. All-optical machines drop magnetic fields entirely: a crossed optical dipole trap sets its depth by laser intensity, so ramping the power down forces evaporation, and because it works at any field it can sit on a Feshbach resonance and tune the scattering length $a$ — hence $\sigma$ and the elastic rate — for optimal cooling. That capability opened the frontier beyond simple condensates: sympathetic cooling of a fermionic isotope by a bosonic coolant produced the first degenerate Fermi gases and reaches the BEC–BCS crossover, and evaporation near a Feshbach resonance is the standard route to ultracold molecules, molecular BECs, and the atoms behind lattice quantum simulators and clocks. Whenever a headline reports a record-low temperature or a novel quantum phase, an evaporation ramp — tuned by exactly the exponents above — is doing the final, quiet work.

The physical picture

Picture a cup of coffee cooling by evaporation: the fastest water molecules escape as vapor, carrying more than their share of energy, and the liquid left behind is cooler. A trapped atomic gas does the same, with one decisive advantage — the rim of the cup is a trap depth you can dial down at will, forcing evaporation to continue even as the gas cools and its tail would fall below the barrier. The indispensable ingredient is rethermalization. Skimming the tail alone leaves a truncated, non-thermal distribution; it is elastic collisions that redistribute energy among the survivors, refilling the tail and re-establishing a Boltzmann distribution at a genuinely lower $T$ . Only a few ( $\sim 3$ ) collisions per atom are needed, so cooling proceeds on the elastic-collision timescale $\gamma_\mathrm{el}^{-1}=(n\sigma\bar v)^{-1}$ . The first figure shows the truncated distribution and an escaping atom's disproportionate energy.

Figure

A truncated thermal gas in a trap of depth U₀ = η k_BT with η = 6. Atoms in the Boltzmann tail above the cut escape carrying ≈ 7 k_BT each — far more than the mean 3 k_BT — and elastic collisions refill the tail to rethermalize the remainder at lower T

Because an escaping atom carries $\bar\epsilon_\mathrm{ev}=(\eta+\kappa)\,k_\mathrm{B}T$ with $\eta\sim 5$ – $10$ , while a trapped atom's mean energy is only $3\,k_\mathrm{B}T$ , each departure removes roughly twice its fair share — which is why one can throw away $99\%$ of the atoms and still gain a factor $\sim 10^6$ in phase-space density. In a harmonic trap the cloud also shrinks as it cools, so density need not fall as fast as number; if it holds up well enough that $\gamma_\mathrm{el}=n\sigma\bar v$ actually rises while the gas cools, evaporation runs away. The second figure traces that runaway, $\rho$ against $N$ .

Figure

Runaway trajectory in a harmonic trap. With cooling exponent α = 4/3 the phase-space density scales as ρ ∝ N⁻³, so losing 99% of the atoms (N falls ×73) lifts ρ from 6.6×10⁻⁶ after the MOT to the BEC threshold ρ = 2.612, while T drops from 100 μK to ≈ 330 nK

The core equations

The quantity every step is trying to raise is the phase-space density: the number of atoms inside a cube one thermal de Broglie wavelength on a side — the single dimensionless number that decides whether the gas is classical or quantum, and the sole target of the whole ramp.

\rho = n\,\lambda_{\mathrm{dB}}^3, \qquad \lambda_{\mathrm{dB}} = \sqrt{\frac{2\pi\hbar^2}{m\,k_\mathrm{B} T}} \qquad (1)

where: $n$ — peak number density; $\lambda_{\mathrm{dB}}$ — thermal de Broglie wavelength; $m$ — atomic mass; $T$ — temperature. A uniform Bose gas condenses at $\rho=\zeta(3/2)\approx 2.612$ ; a MOT delivers $\rho\sim 10^{-6}$ , so evaporation must supply a factor of a million. Rethermalization runs on elastic collisions, so their rate is the clock of the whole process: a density times a cross section times a mean relative speed. For indistinguishable bosons the cross section carries a symmetry factor that fermions cannot share.

\gamma_\mathrm{el} = n\,\sigma\,\bar v, \qquad \sigma = 8\pi a^2, \qquad \bar v = \sqrt{\frac{8\,k_\mathrm{B} T}{\pi m}} \qquad (2)

where: $\sigma$ — s-wave cross section for identical bosons; $a$ — s-wave scattering length; $\bar v$ — mean relative speed. For distinguishable particles $\sigma=4\pi a^2$ ; identical bosons interfere constructively to double it, while spin-polarized identical fermions have $\sigma\to 0$ and cannot rethermalize at all. The whole scheme is governed by how deep the trap is relative to the gas temperature. That ratio — the truncation parameter $\eta$ — fixes both how much energy an escaping atom removes and how many sit in the tail, the central trade-off.

\eta = \frac{U_0}{k_\mathrm{B} T}, \qquad \bar\epsilon_\mathrm{ev} = (\eta + \kappa)\,k_\mathrm{B} T \qquad (3)

where: $U_0$ — trap depth (or RF-set cut energy); $\eta$ — ratio of escape to thermal energy; $\bar\epsilon_\mathrm{ev}$ — energy carried off per escaping atom; $\kappa\sim 1$ — excess above the barrier. An atom must reach $U_0=\eta k_\mathrm{B}T$ to leave; large $\eta$ removes more energy per atom but leaves exponentially fewer ( $\propto e^{-\eta}$ ) in the tail. Removing atoms that each carry more than the mean energy must lower the average, and a one-line energy balance turns that into a power law linking temperature to atom number. The exponent $\alpha$ measures how efficiently evaporation trades atoms for coldness.

\frac{\mathrm{d}\ln T}{\mathrm{d}\ln N} = \frac{\bar\epsilon_\mathrm{ev}}{\bar\epsilon} - 1 \equiv \alpha \quad\Longrightarrow\quad T\propto N^{\alpha} \qquad (4)

where: $N$ — atom number; $\bar\epsilon$ — mean total energy per trapped atom ( $3k_\mathrm{B}T$ in a 3D harmonic trap); $\alpha$ — efficiency exponent. Since $\bar\epsilon_\mathrm{ev}>\bar\epsilon$ , $\alpha>0$ : temperature falls as atoms are lost, and typical $\alpha\sim 1$ – $2$ means modest loss produces large cooling. What we actually care about is not $T$ or $N$ alone but their combination $\rho$ . In a harmonic trap $n\propto N T^{-3/2}$ , so $\rho\propto N T^{-3}$ , and the cooling law then gives $\rho$ as a steep negative power of the atom number.

\rho \propto N\,T^{-3} \propto N^{\,1-3\alpha} \qquad (5)

where: $1-3\alpha$ is negative whenever $\alpha>1/3$ , so $\rho$ climbs as $N$ falls. For $\alpha=4/3$ it is $-3$ : a factor $70$ loss in $N$ multiplies $\rho$ by $70^3\approx 3\times 10^5$ — why evaporation works so well. In a magnetic trap the depth is not a physical wall but a resonance condition: a radiofrequency field flips spins to untrapped states, but only where the local Zeeman splitting matches the drive. Sweeping the frequency down lowers the effective rim — the "RF knife".

h\,\nu_\mathrm{RF} = \mu_B\,g_F\,m_F\,\lvert B(r_\mathrm{cut})\rvert \qquad (6)

where: $\nu_\mathrm{RF}$ — knife frequency; $\mu_B$ — Bohr magneton; $g_F$ — Landé g-factor; $m_F$ — magnetic sublevel; $B(r_\mathrm{cut})$ — field at the resonant surface. Only atoms energetic enough to climb to $r_\mathrm{cut}$ are spin-flipped away; the cut is a surface of constant potential, so it selects on energy. Finally, feasibility is a race: every good (elastic) collision helps, every bad (inelastic or background) collision ejects an atom without cooling. The ratio of the two rates decides whether runaway is reachable.

R = \frac{\gamma_\mathrm{el}}{\gamma_\mathrm{loss}} = \frac{n\sigma\bar v}{\gamma_\mathrm{bg} + \gamma_\mathrm{inel}} \qquad (7)

where: $\gamma_\mathrm{bg}$ — background-gas loss rate (set by vacuum quality); $\gamma_\mathrm{inel}$ — inelastic loss (spin-exchange, three-body recombination $\propto n^2$ ); $R$ — good-to-bad ratio. Runaway needs $R\gtrsim 150$ ; the $n^2$ three-body loss caps the density and sets the endpoint.

Derivation AEnergy per evaporated atom and the cooling exponent

The goal is to turn the claim "hot atoms leave, so the gas cools" into the cooling law $T\propto N^\alpha$ , pinning the exponent $\alpha$ to the one knob the experimenter controls, the truncation parameter $\eta$ . We model the cloud as a Boltzmann distribution truncated at $U_0=\eta k_\mathrm{B}T$ and balance the energy the escaping tail removes against the mean energy of the survivors.

Step 1. In a 3D harmonic trap, equipartition over three kinetic and three potential quadratic degrees of freedom gives $\bar\epsilon = 3\,k_\mathrm{B}T$ . Why: this is the baseline — an escaping atom cools the gas only insofar as it carries more than $3k_\mathrm{B}T$ .

Step 2. An atom leaves only if its energy reaches the barrier $U_0=\eta k_\mathrm{B}T$ ; averaging over the escaping slice of the tail adds a small $\kappa\,k_\mathrm{B}T$ ( $\kappa\sim 1$ ), so $\bar\epsilon_\mathrm{ev}=(\eta+\kappa)\,k_\mathrm{B}T$ . Why: the trap depth enters here as a physical energy — a deeper cut means each departing atom takes more.

Step 3. Impose energy balance. Removing $\mathrm{d}N$ atoms ( $\mathrm{d}N<0$ ) each carrying $\bar\epsilon_\mathrm{ev}$ gives $\mathrm{d}E=\bar\epsilon_\mathrm{ev}\,\mathrm{d}N$ ; expanding $\mathrm{d}(N\bar\epsilon)=\bar\epsilon\,\mathrm{d}N+N\,\mathrm{d}\bar\epsilon$ with $\bar\epsilon\propto T$ gives $\mathrm{d}\ln T/\mathrm{d}\ln N=\bar\epsilon_\mathrm{ev}/\bar\epsilon-1$ . Why: a bookkeeping identity becomes a relation between fractional temperature drop and fractional atom loss.

Step 4. Substitute the two energies to read off $\alpha=(\eta+\kappa)/3-1$ . For $\eta=6,\ \kappa\approx 1$ this is $\alpha\approx 7/3-1=4/3\approx 1.33$ , comfortably above $1$ . Why: raising $\eta$ raises $\alpha$ and efficiency, but (next card) also thins the tail and slows the process — so $\eta$ is optimized, not maximized.

Go deeper · Derivation B · 1 minThe runaway condition — when cooling accelerates itself

Derivation BThe runaway condition — when cooling accelerates itself

The central question is not merely whether the gas cools, but whether it cools faster and faster as atoms are lost, so degeneracy is reached in finite time. The goal is to find how the elastic collision rate $\gamma_\mathrm{el}$ — the clock of the process — evolves as the atom number falls, and the condition under which it increases. Combining the density scaling with the cooling law gives a sharp threshold on $\alpha$ .

Step 1. For a Boltzmann gas in a 3D harmonic potential the peak density is $n=N/(2\pi k_\mathrm{B}T/m\bar\omega^2)^{3/2}\propto N\,T^{-3/2}$ , with $\bar\omega$ the geometric-mean trap frequency. Why: as it cools the gas contracts, so density does not simply track $N$ — the $T^{-3/2}$ factor lets $\gamma_\mathrm{el}$ hold up or grow.

Step 2. Assemble the rate: $\gamma_\mathrm{el}=n\sigma\bar v\propto (N\,T^{-3/2})(T^{1/2})=N\,T^{-1}$ , using $\bar v\propto T^{1/2}$ and $\sigma$ roughly constant. Why: falling $N$ pushes the rate down, falling $T$ pushes it up — the outcome hinges on how fast $T$ drops with $N$ .

Step 3. Insert $T\propto N^\alpha$ to get $\gamma_\mathrm{el}\propto N^{1-\alpha}$ , i.e. $\mathrm{d}\ln\gamma_\mathrm{el}/\mathrm{d}\ln N = 1-\alpha$ . Why: the competition collapses into one exponent, so "does the rate rise?" becomes a one-line inequality.

Step 4. Read off the threshold: $\gamma_\mathrm{el}$ rises as $N$ falls exactly when $\alpha>1$ . With $\alpha=(\eta+\kappa)/3-1$ this needs $\eta+\kappa>6$ , so $\eta\gtrsim 5$ – $6$ . Why: this is why practical ramps hold $\eta$ near $6$ and demand $R\gtrsim 150$ — only then does rethermalization outrun atom loss and the finite trap lifetime.

Go deeper · Derivation C · 1 minPhase-space density gain — the exponent that makes it worthwhile

Derivation CPhase-space density gain — the exponent that makes it worthwhile

Efficiency is ultimately measured not in temperature but in phase-space density, the one number that must reach $2.612$ . The goal is to express $\rho$ directly as a power of atom number, so that "how many atoms must I sacrifice?" gets a concrete answer. Combining the definition of $\rho$ , the density scaling, and the cooling law yields a surprisingly large negative exponent — what makes an apparently wasteful process the workhorse of cold-atom physics.

Step 1. With $\lambda_{\mathrm{dB}}^3\propto T^{-3/2}$ and $n\propto N T^{-3/2}$ , the definition $\rho=n\lambda_{\mathrm{dB}}^3$ gives $\rho\propto N\,T^{-3}$ . Why: $\rho$ is far more sensitive to temperature (cube) than to atom number (first power) — cooling is worth far more than atoms.

Step 2. Insert $T\propto N^\alpha$ to obtain $\rho\propto N^{\,1-3\alpha}$ . Why: the sign and size of $1-3\alpha$ decide whether losing atoms is a net gain.

Step 3. For $\alpha=4/3$ the exponent is $1-4=-3$ , so $\rho\propto N^{-3}$ : cutting $N$ by $70$ multiplies $\rho$ by $70^3\approx 3.4\times 10^5$ . Why: the whole payoff — a few hundred thousand in $\rho$ for a factor of seventy in atom number, the MOT-to-condensate gap.

Limiting cases

The character of evaporation is set by the balance of collision rates and the quantum statistics of the atoms:

Case	Result	Meaning
Runaway (good/bad ratio large)	$\gamma_\mathrm{el}$ rises as $T\downarrow$	$\alpha>1$ : elastic rate grows during cooling, so rethermalization accelerates and $\rho$ climbs to degeneracy in finite time. Needs $R\gtrsim 150$ .
Marginal / decaying	$\gamma_\mathrm{el}$ falls as $T\downarrow$	$\alpha<1$ : density drop outpaces temperature gain; evaporation slows and stalls before degeneracy. Common in weak or shallow traps.
Truncation cut $\eta=U_0/k_\mathrm{B}T$	$\eta\gtrsim 6$ optimal	Large $\eta$ : efficient but slow (tail $\propto e^{-\eta}$ ). Small $\eta$ : fast but wasteful. Ramps lower $U_0$ to hold $\eta$ roughly constant.
Identical fermions	$\sigma_{s\text{-wave}}\to 0$	Antisymmetry forbids s-wave collisions between identical fermions ( $\sigma_{p\text{-wave}}\propto T^2$ ). No rethermalization — use two spin states or a boson.
Three-body dominated	$\gamma_\mathrm{inel}\propto n^2$	At high density, recombination overtakes elastic gain and heats the cloud; sets the endpoint, favouring a Feshbach zero-crossing.

Worked examples

1 · Phase-space density of a freshly trapped cloud (confidence builder)

A magnetic trap holds $^{87}$ Rb at $T=100\ \mu$ K, $n=10^{18}\ \mathrm{m}^{-3}$ ( $=10^{12}\ \mathrm{cm}^{-3}$ ), $a=100\,a_0$ . Find the de Broglie wavelength, phase-space density, and elastic collision rate. How far is this from BEC, and how fast does it rethermalize?

Step 1. de Broglie wavelength. With $m=87\times 1.66\times 10^{-27}=1.445\times 10^{-25}$ kg and $\lambda_{\mathrm{dB}}=\sqrt{2\pi\hbar^2/(m k_\mathrm{B}T)}$ : here $2\pi\hbar^2=6.99\times 10^{-68}$ and $m k_\mathrm{B}T=2.00\times 10^{-52}$ , so $\lambda_{\mathrm{dB}}^2=3.50\times 10^{-16}\ \mathrm{m}^2$ and $\lambda_{\mathrm{dB}}=\mathbf{18.7\ nm}$ .

Step 2. Phase-space density. $\lambda_{\mathrm{dB}}^3=6.56\times 10^{-24}\ \mathrm{m}^3$ , so $\rho=n\lambda_{\mathrm{dB}}^3=\mathbf{6.6\times 10^{-6}}$ . Against the threshold $\rho=2.612$ , the cloud sits a factor $\approx\mathbf{4\times 10^{5}}$ below degeneracy — the gap evaporation must close.

Step 3. Elastic rate. With $\sigma=8\pi a^2=8\pi(5.29\times 10^{-9})^2=7.0\times 10^{-16}\ \mathrm{m}^2$ and $\bar v=\sqrt{8k_\mathrm{B}T/\pi m}=0.156\ \mathrm{m/s}$ : $\gamma_\mathrm{el}=n\sigma\bar v\approx\mathbf{110\ s^{-1}}$ . Since $\sim 3$ collisions rethermalize an atom, the cloud re-equilibrates in $3/\gamma_\mathrm{el}\approx\mathbf{27\ ms}$ — thousands of times faster than a good trap lifetime.

2 · How many atoms must we sacrifice to reach BEC?

Continue from Example 1 ( $\rho_i=6.6\times 10^{-6}$ , $T_i=100\ \mu$ K) at fixed $\eta=6$ , so $\alpha=4/3$ and $\rho\propto N^{-3}$ . What fraction of atoms survives to $\rho_f=2.612$ , and what is the final temperature?

Step 1. Required gain: $\rho_f/\rho_i=2.612/(6.6\times 10^{-6})=3.96\times 10^{5}$ .

Step 2. Atom loss. Since $\rho\propto N^{-3}$ , $\rho_f/\rho_i=(N_i/N_f)^3$ , so $N_i/N_f=(3.96\times 10^{5})^{1/3}=\mathbf{73.4}$ : you keep about $\mathbf{1.4\%}$ and discard $\mathbf{98.6\%}$ . From $N_i=10^{9}$ this leaves $N_f\approx\mathbf{1.4\times 10^{7}}$ condensed atoms.

Step 3. Final temperature. With $T\propto N^{\alpha}$ , $T_f/T_i=(1.36\times 10^{-2})^{4/3}$ ; taking logs, $\ln(1.36\times 10^{-2})\times 4/3=-5.73$ , so $T_f/T_i=e^{-5.73}=3.24\times 10^{-3}$ and $T_f\approx\mathbf{324\ nK}$ . Losing $99\%$ of the atoms cost a factor $\sim 300$ in temperature and bought $\sim 4\times 10^{5}$ in phase-space density — the trade that makes degeneracy reachable.

Physics of the portrait

Evaporative cooling connected two decades of atom-cooling ingenuity to the quantum-degenerate gases chased since the 1920s — turning a thermodynamic afterthought into precision engineering.

Origins · 1 minWolfgang Ketterle and how this was discovered

Wolfgang Ketterle

b. 1957 · evaporative cooling and the road to Bose-Einstein condensation

Evaporative cooling was not born in the alkali labs that made it famous. Harald Hess proposed it in 1986 for magnetically trapped atomic hydrogen, and Masuhara and coworkers in the Greytak–Kleppner group at MIT demonstrated it in 1988, cooling spin-polarized hydrogen far below what any optical method could reach. The decisive transplant to alkali atoms came in the mid-1990s: with the RF "knife" for forced evaporation refined in his and others' hands, Wolfgang Ketterle drove sodium from a laser-cooled cloud to a Bose-Einstein condensate in 1995, weeks after Cornell and Wieman condensed rubidium in Boulder by the same route. His 1996 review with van Druten laid down the scaling theory used here — the cooling exponent, the runaway condition, and the good-to-bad ratio that every apparatus is now designed around. The 2001 Nobel Prize to Cornell, Wieman, and Ketterle recognized a chain of ideas in which evaporation was the last, indispensable link.

History

1986

— Hess proposes evaporative cooling of magnetically trapped atomic hydrogen

1988

— Masuhara and coworkers demonstrate evaporative cooling of spin-polarized hydrogen at MIT

1995

— Cornell, Wieman and Ketterle reach Bose-Einstein condensation using forced evaporation

1996

— Ketterle and van Druten publish the scaling theory of evaporative cooling

1999

— DeMarco and Jin reach a degenerate Fermi gas by two-spin-state evaporative cooling

Going further

Texts. Foot, Atomic Physics, §10 · Ketterle & van Druten, Adv. At. Mol. Opt. Phys. 37, 181 (1996), "Evaporative cooling of trapped atoms" — the scaling theory used here · Anderson et al., Science 269, 198 (1995), first BEC · Pethick & Smith, Bose–Einstein Condensation in Dilute Gases, Ch. 4 · Demtröder, Laser Spectroscopy.

Provenance. A verifier should confirm: the BEC threshold $\rho=\zeta(3/2)\approx 2.612$ (and $\zeta(3)\approx 1.202$ for the harmonic-trap central condition); the identical-boson cross section $\sigma=8\pi a^2$ versus $4\pi a^2$ ; the density scaling $n\propto N T^{-3/2}$ giving $\gamma_\mathrm{el}\propto N^{1-\alpha}$ and the runaway condition $\alpha>1$ ; the cooling exponent $\alpha=(\eta+\kappa)/3-1$ ; and $\rho\propto N^{1-3\alpha}$ . The worked numbers ( $^{87}$ Rb, $a=100\,a_0$ , $T=100\ \mu$ K, $n=10^{18}\ \mathrm{m}^{-3}$ : $\lambda_{\mathrm{dB}}=18.7$ nm, $\rho=6.6\times 10^{-6}$ , $\gamma_\mathrm{el}\approx 110\ \mathrm{s}^{-1}$ ; $\rho\propto N^{-3}$ giving $73\times$ atom loss and $T_f\approx 324$ nK) are recomputed in verify_aq-optics-evaporative-cooling.py.

Prerequisites. Magnetic Trapping of Neutral Atoms · The Recoil Limit and Sub-Recoil Cooling · Magneto-Optical Trap. Leads to. Bose-Einstein Condensation in Dilute Gases · Degenerate Fermi Gases and the BEC–BCS Crossover · Feshbach Resonances.

Ask the tutor: Ask me to walk an evaporation ramp step by step — plug in your species, trap frequencies, and starting density, and we will track $\eta$ , $\alpha$ , the atom loss, and the temperature all the way to the BEC threshold.

Why can't laser cooling alone reach Bose-Einstein condensation?

Two walls stop it. First, optical cooling is bounded below by the recoil limit — the single-photon recoil energy floors the temperature of any method that scatters photons — so laser cooling parks the gas at microkelvin, far above degeneracy. Second, at the densities needed to raise phase-space density, reabsorbed photons and light-assisted collisions heat the cloud and eject atoms. The net result is $\rho$ stuck near $10^{-6}$ , six orders of magnitude short of $2.612$ . Evaporation sidesteps both walls because it uses no light at all — a thermodynamic process driven by elastic collisions.

What sets the timescale of evaporative cooling?

The elastic collision rate, $\gamma_\mathrm{el}=n\sigma\bar v$ . Rethermalization — the step that actually lowers the temperature — needs only about three collisions per atom, so the gas re-equilibrates in $\sim 3/\gamma_\mathrm{el}$ , roughly $27$ ms for the Rb cloud in the worked example. The full ramp is far longer, tens of seconds in a magnetic trap, because the depth must be lowered slowly enough to keep the gas near equilibrium and the escaping tail fraction ( $\propto e^{-\eta}$ ) is small. Keeping $\gamma_\mathrm{el}$ high relative to the loss rate is the whole design goal — which is why steep atom chips can evaporate in under a second.

Why does a deeper cut (larger η) cool more per atom but take longer?

The two effects pull opposite ways. A deeper cut forces an escaping atom to climb higher, so it removes more energy, $\bar\epsilon_\mathrm{ev}=(\eta+\kappa)k_\mathrm{B}T$ ; this raises the cooling exponent $\alpha=(\eta+\kappa)/3-1$ and makes each lost atom worth more coldness. But the tail fraction above the cut falls as $e^{-\eta}$ , so with a deep cut very few atoms are eligible to evaporate and the process crawls. In practice the depth is lowered continuously to hold $\eta$ near $6$ — deep enough for runaway cooling, shallow enough that the tail keeps refilling.

What is the "RF knife" and how does it choose which atoms to remove?

In a magnetic trap the potential is the Zeeman energy $\mu_B g_F m_F B(r)$ , so more energetic atoms swing out to larger $\lvert B\rvert$ . A radiofrequency field spin-flips atoms to untrapped $m_F$ states, but only resonantly — where $h\nu_\mathrm{RF}$ matches the local Zeeman splitting, on a single surface of constant field $B(r_\mathrm{cut})$ . Since that surface is also one of constant potential energy, the knife removes exactly the atoms energetic enough to reach it. Sweeping $\nu_\mathrm{RF}$ downward shrinks $r_\mathrm{cut}$ , lowering the effective depth continuously — far more controllable than physically lowering a barrier.

Why is evaporation useless for a single-spin-state Fermi gas?

Rethermalization needs elastic collisions, and at microkelvin temperatures only s-wave collisions survive (higher partial waves freeze out, the p-wave cross section vanishing as $T^2$ ). But two identical fermions in the same internal state have an antisymmetric spatial wavefunction, which forbids s-wave scattering outright — the cross section is exactly zero, so evaporation just spills atoms without cooling. The fix is a distinguishable partner: prepare two spin states so unlike fermions can collide, or evaporate a bosonic coolant and let thermal contact drag the fermions down — sympathetic cooling, the route to the first degenerate Fermi gases.

What ultimately limits how cold and dense evaporation can go?

Loss processes that grow with density. Three-body recombination — two atoms form a molecule while the third carries off the binding energy — has a rate $\propto n^2$ , so as the gas is compressed this inelastic loss eventually overtakes the elastic gain, heating and depleting the cloud. Background-gas collisions set a competing, density-independent lifetime that demands excellent vacuum. The ratio $R=\gamma_\mathrm{el}/\gamma_\mathrm{loss}$ captures the balance, and runaway needs $R\gtrsim 150$ . In practice one condenses just as $\rho$ crosses $2.612$ — sometimes at reduced density near a Feshbach zero-crossing — then decompresses to grow a larger, longer-lived condensate.