Evaporative Cooling

Evaporative cooling increases the phase-space density of a trapped gas by selectively removing its most energetic particles and allowing elastic collisions to rethermalize those that remain. The basic cycle is

$$\text{truncate the high-energy tail} \longrightarrow \text{escape} \longrightarrow \text{elastic rethermalization} \longrightarrow \text{repeat at a lower threshold}.$$

Particle loss is the resource that exports energy and entropy. Cooling is efficient only when an escaping particle carries substantially more than the average energy per trapped particle.

Three statements that sound similar are physically different:

1. The temperature fell. This may result from adiabatic decompression and need not increase phase-space density.
2. Energetic particles were spilled. A rapid depth ramp can discard particles before collisions establish a colder equilibrium.
3. Evaporation increased phase-space density. Selective loss and rethermalization jointly produced a favorable gain relative to atom loss.

The central experimental competition is between good collisions, which redistribute energy elastically, and bad processes, which remove or heat particles without sufficient energetic selectivity.

Required background. Optical dipole traps distinguishes escape depth from curvature; scattering length sets the low-energy elastic cross section; the Maxwell–Boltzmann limit supplies the classical density and energy distributions used below.

Truncating the Energy Distribution

Measure single-particle energy $\varepsilon$ from the trap minimum. Let the lowest escape threshold be $\varepsilon_t$. The truncation parameter is

$$\eta = \frac{ \varepsilon_t }{ k_{\mathrm B}T }.$$

A large $\eta$ makes loss energetically selective but suppresses the available tail exponentially. A small $\eta$ speeds loss but throws away particles that carry too little excess energy. There is no universal optimum because elastic rate, background lifetime, inelastic loss, trap geometry, and quantum statistics all matter.

Truncated Boltzmann model

A standard classical ansatz is

$$f(\mathbf r,\mathbf p) = \mathcal A \exp \left[ - \frac{ p^2/(2m)+V(\mathbf r) }{ k_{\mathrm B}T } \right] \Theta \left[ \varepsilon_t - \frac{ p^2 }{ 2m } - V(\mathbf r) \right].$$

This truncated Boltzmann ansatz keeps a thermal exponential below the knife and vanishes above it; it is not an exact equilibrium of an open escaping gas. The model assumes sufficient ergodicity: collisions and trap dynamics mix phase space well enough that the distribution is approximately a function of total energy.

For a three-dimensional harmonic trap in the semiclassical regime $k_{\rm B}T\gg\hbar\omega_i$,

$$g(\varepsilon) = \frac{ \varepsilon^2 }{ 2(\hbar\bar\omega)^3 }, \qquad \bar\omega = \left( \omega_x\omega_y\omega_z \right)^{1/3}.$$

Before truncation, the normalized classical energy distribution is

$$p(\varepsilon) = \frac{ \varepsilon^2 }{ 2(k_{\mathrm B}T)^3 } e^{-\varepsilon/(k_{\mathrm B}T)}.$$

The equilibrium population above the threshold is

$$P \left( \varepsilon>\varepsilon_t \right) = e^{-\eta} \left( 1+\eta+\frac{\eta^2}{2} \right).$$

This tail fraction is not the evaporative loss rate. Evaporation is a collision-flux problem: collisions must place an outgoing particle above the threshold in a geometry from which it can escape.

(a) The untruncated harmonic distribution $p(x)=x^2e^{-x}/2$ with $x=\varepsilon/(k_{\rm B}T)$; the shaded tail is removed by a knife at $\eta=6$. Its area is a probability, not a rate. (b) With $\eta'=\eta+\kappa=8.75$ and constant cross section, the normalized elastic rate grows as $(N/N_i)^{(6-\eta')/3}$ at fixed confinement but falls as $(N/N_i)^{\eta'/3}$ in the ideal power-lowered trap. Both curves start at one; evaporation proceeds toward smaller $N/N_i$. The vertical axis is logarithmic.

Physical Implementations of the Knife

The escape threshold can be controlled in several ways.

Radio-frequency evaporation

In a magnetic trap, an RF field drives atoms from a trapped Zeeman state to an untrapped or anti-trapped state on a resonant magnetic-field surface. Sweeping the RF frequency lowers the energy at which atoms are removed while leaving the trap’s central curvature approximately independent.

Microwave evaporation

Microwave transitions can provide hyperfine-state-selective removal and can distinguish species or internal states more strongly than an RF knife.

Lowering optical power

In a red-detuned optical trap, reducing laser power lowers the escape depth. It also usually lowers the trap frequencies:

$$\omega_i \propto \sqrt{U_0}$$

for fixed beam geometry. Depth and confinement therefore change together. This reduces density and may slow the elastic collision rate.

Tilting or opening the trap

A magnetic-field gradient, gravity, an auxiliary optical field, or a controlled saddle can lower one escape barrier while preserving much of the local curvature. This separates the energy knife from confinement and can accelerate evaporation.

Species-selective removal

In a mixture, the knife may remove only one component. Other components cool sympathetically through interspecies elastic collisions.

For mechanical escape, use the lowest accessible saddle of the complete potential, including gravity and field gradients. An RF or microwave knife can remove atoms at a resonant spin-flip surface below that mechanical barrier. Its threshold is the energy needed to reach the removal surface. Use the lowest operative channel, together with its transition probability and accessibility; a static trap saddle alone does not define an RF knife.

Rethermalization

Immediately after hot particles leave, the remaining distribution is not an equilibrium distribution at a lower temperature. Elastic collisions redistribute energy and refill the high-energy tail.

Define the density-averaged elastic collision rate

$$\Gamma_{\mathrm{el}} = \bar n \left\langle \sigma(v_{\mathrm{rel}}) v_{\mathrm{rel}} \right\rangle.$$

For a dilute, nondegenerate gas of equal masses in a deep harmonic trap,

$$n_0 = N \left( \frac{ m\bar\omega^2 }{ 2\pi k_{\mathrm B}T } \right)^{3/2},$$

and

$$\bar n = \frac{1}{N} \int n^2(\mathbf r)\,d^3r = \frac{ n_0 }{ 2^{3/2} }.$$

The Maxwellian mean relative speed is

$$\left\langle v_{\mathrm{rel}} \right\rangle = 4 \sqrt{ \frac{ k_{\mathrm B}T }{ \pi m } }.$$

When the cross section is effectively constant,

$$\Gamma_{\mathrm{el}} \propto \frac{ N\bar\omega^3\sigma }{ T }.$$

This is the collision rate per particle; the total number of binary collision events per time is $N\Gamma_{\rm el}/2$. It is distinct from the rate at which those events remove particles.

Rethermalization time

Write

$$\tau_{\mathrm{th}} \simeq \frac{ \xi }{ \Gamma_{\mathrm{el}} },$$

where $\xi$ is the number of collisions per particle required for the measured relaxation mode. It is of order a few for equal-mass, isotropic cross-dimensional relaxation but can be much larger for unequal masses, anisotropic scattering, poor overlap, or weak ergodicity.

A ramp should be slow enough that the gas remains near a truncated equilibrium:

$$\tau_{\mathrm{ramp}} \gg \tau_{\mathrm{th}}$$

locally. It should also be fast compared with the time over which background loss, three-body loss, and technical heating dominate. These requirements can conflict.

Elastic Collision Requirement

For purely elastic $s$-wave scattering, real scattering length $a$, and negligible effective-range corrections, the distinguishable-particle cross section is approximately

$$\sigma(k) = \frac{ 4\pi a^2 }{ 1+k^2a^2 }.$$

Here $k=\mu v_{\rm rel}/\hbar$ is the relative wave number and $\mu$ the reduced mass. For identical bosons in one internal state, exchange doubles this result:

$$\sigma_{\mathrm B}(k) = \frac{ 8\pi a^2 }{ 1+k^2a^2 }.$$

Identical spin-polarized fermions cannot scatter in the $s$ wave. Their leading allowed partial wave is suppressed at low energy. Evaporative cooling of fermions therefore commonly uses two spin states or sympathetic cooling with another component.

Increasing $|a|$ does not improve evaporation without limit:

• the cross section is bounded by unitarity;
• three-body recombination can grow strongly;
• mean-field effects alter the trap and density;
• hydrodynamic opacity can prevent escaping atoms from leaving cleanly;
• near a resonance, the cross section becomes energy dependent.

The relevant figure of merit is not $\Gamma_{\mathrm{el}}$ alone but its ratio to all bad rates:

$$\mathcal R = \frac{ \Gamma_{\mathrm{el}} }{ \Gamma_1 + \Gamma_2 + \Gamma_3 + \Gamma_{\mathrm{heat}} }.$$

The denominator is schematic because heating is not always expressible as a particle-loss rate. A useful experiment measures each contribution rather than assigning every deviation to “imperfect evaporation.”

Energy Removed per Evaporated Particle

In a deep classical three-dimensional harmonic trap, neglecting the small truncation correction,

$$E = 3Nk_{\mathrm B}T.$$

All energies here are measured from the instantaneous trap minimum. For the exactly truncated ansatz the mean instead is

$$\frac EN=3k_{\rm B}T\frac{P(4,\eta)}{P(3,\eta)},\qquad P(s,\eta)=\frac1{\Gamma(s)}\int_0^\eta x^{s-1}e^{-x}\,dx.$$

At $\eta=8$ it is $2.913k_{\rm B}T$, rather than $3k_{\rm B}T$. The simple scaling laws below consistently use the high-$\eta$ limit; they should not be combined with exact finite-cutoff moments as if exact. Kinetic theory gives the energy carried away per evaporated particle as

$$\varepsilon_{\mathrm{ev}} = \left( \eta+\kappa \right) k_{\mathrm B}T.$$

The quantity $\kappa k_{\mathrm B}T$ is the mean excess above the threshold. For an energy-independent cross section in the high-$\eta$ harmonic regime ($\eta$ sufficiently above $4$),

$$\kappa \simeq \frac{ \eta-5 }{ \eta-4 }.$$

Thus $\kappa$ approaches $1$ for large $\eta$. Exact finite-depth expressions contain incomplete gamma functions and depend on trap shape and cross-section energy dependence.

Evaporation cools because

$$\varepsilon_{\mathrm{ev}} > \frac{E}{N}.$$

In the harmonic classical model, this requires

$$\eta+\kappa > 3.$$

Increasing phase-space density requires the stronger condition derived below.

Ideal Energy and Number Scaling

First consider fixed trap frequencies, constant $\eta$, negligible bad loss, and no external heating. If $dN<0$ particles evaporate, energy conservation gives

$$dE = \left( \eta+\kappa \right) k_{\mathrm B}T dN.$$

Because $E=3Nk_{\mathrm B}T$,

$$\frac{ d\ln T }{ d\ln N } = \frac{ \eta+\kappa-3 }{ 3 }.$$

As $N$ decreases, $T$ decreases when $\eta+\kappa>3$.

Phase-space density

For one classical component in a harmonic trap, define

$$\mathcal D = N \left( \frac{ \hbar\bar\omega }{ k_{\mathrm B}T } \right)^3.$$

This equals the peak phase-space density in the Maxwell–Boltzmann limit. For a spin mixture, use the population of the component whose degeneracy is being assessed.

At fixed $\bar\omega$,

$$\frac{ d\ln\mathcal D }{ d\ln N } = 4 - \eta - \kappa.$$

The evaporation efficiency is conventionally

$$\gamma_{\mathrm{evap}} = - \frac{ d\ln\mathcal D }{ d\ln N }.$$

Therefore, in the ideal harmonic model,

$$\gamma_{\mathrm{evap}} = \eta+\kappa-4.$$

Phase-space density increases when

$$\eta+\kappa > 4.$$

A quoted efficiency should specify the interval over which the logarithmic slope was fit. A single endpoint ratio can hide stagnation or a change of regime.

Why large eta is not automatically optimal

In a common high-$\eta$ harmonic approximation,

$$\Gamma_{\mathrm{ev}} \sim \left( \eta-4 \right) e^{-\eta} \Gamma_{\mathrm{el}},$$

up to a factor set by the collision-rate convention. Increasing $\eta$ improves energy removed per lost particle but suppresses evaporation exponentially. If evaporation becomes slower than background loss or heating, the measured efficiency falls.

Lowering an Optical Trap

When optical power is reduced, the changing trap does work on the gas. For a harmonic optical trap changed slowly compared with motion, with $\bar\omega\propto\sqrt{U_0}$ and negligible extra nonadiabatic excitation,

$$\dot E_{\mathrm{work}} = \frac{ E }{ 2 } \frac{ \dot U_0 }{ U_0 }.$$

At constant $\eta$ with threshold proportional to depth, define

$$\eta' = \eta+\kappa.$$

The ideal scaling law is

$$\frac{ N }{ N_0 } = \left( \frac{ U_0 }{ U_{0,\mathrm i} } \right)^{ \frac{ 3 }{ 2(\eta'-3) } }.$$

The phase-space-density gain remains

$$\frac{ \mathcal D }{ \mathcal D_0 } = \left( \frac{ N_0 }{ N } \right)^{\eta'-4}.$$

Thus the same ideal efficiency $\eta'-4$ emerges after the adiabatic work is included. What changes dramatically is the collision-rate trajectory.

Adiabatic decompression is not evaporation

If a harmonic trap is weakened adiabatically with no particle loss,

$$T \propto \bar\omega.$$

Then

$$\mathcal D = N \left( \frac{ \hbar\bar\omega }{ k_{\mathrm B}T } \right)^3$$

stays constant. The gas becomes colder in kelvin and larger in space, but it does not become more quantum degenerate.

Runaway Evaporation

Runaway evaporation means that the elastic collision rate increases as evaporation proceeds. This continually accelerates rethermalization and can permit a faster knife sweep.

Fixed confinement

For constant cross section and fixed $\bar\omega$,

$$\Gamma_{\mathrm{el}} \propto \frac{ N }{ T }.$$

Using the ideal scaling,

$$\frac{ \Gamma_{\mathrm{el}} }{ \Gamma_{\mathrm{el},0} } = \left( \frac{ N }{ N_0 } \right)^{ \frac{ 6-\eta' }{ 3 } }.$$

Because $N/N_0$ decreases, the collision rate rises when

$$\eta' > 6.$$

RF evaporation in a magnetic trap can approach this fixed-confinement regime.

Power-lowered optical confinement

For fixed beam geometry,

$$\bar\omega \propto \sqrt{U_0}, \qquad T \propto U_0$$

at constant $\eta$. With an energy-independent cross section,

$$\frac{ \Gamma_{\mathrm{el}} }{ \Gamma_{\mathrm{el},0} } = \left( \frac{ U_0 }{ U_{0,\mathrm i} } \right)^{ \frac{ \eta' }{ 2(\eta'-3) } }.$$

The exponent is positive, so $\Gamma_{\mathrm{el}}$ decreases as the trap is lowered. Ordinary power lowering is therefore not runaway in this ideal constant-cross-section model.

For a dilute classical gas near elastic $s$-wave unitarity, thermal averaging gives $\langle\sigma v_{\rm rel}\rangle\propto T^{-1/2}$, corresponding to an effective $\sigma\propto1/T$. The exponent becomes

$$\frac{ 6-\eta' }{ 2(\eta'-3) },$$

and runaway behavior can reappear for $\eta'>6$. This is a special energy-dependent-cross-section regime, not a general property of optical traps. The loss-weighted energy and thus $\kappa$ must also be recalculated for this energy-dependent cross section; importing $(\eta-5)/(\eta-4)$ unchanged is not controlled.

Preserving curvature

Tilt evaporation, auxiliary knives, dynamically shaped traps, and dimple geometries can lower the escape threshold without proportionally lowering all trap frequencies. The design objective is to preserve density and elastic rate while avoiding excessive three-body loss and opacity.

Competing loss and anti-evaporation

Write positive particle-loss fluxes as $L_j>0$:

$$\dot N = - L_{\mathrm{ev}} - L_1 - L_2 - L_3.$$

The corresponding energy balance is

$$\dot E = - \varepsilon_{\mathrm{ev}}L_{\mathrm{ev}} - \varepsilon_1L_1 - \varepsilon_2L_2 - \varepsilon_3L_3 + \dot E_{\mathrm{heat}} + \dot E_{\mathrm{work}}.$$

One-body loss

Background collisions usually sample the cloud approximately uniformly. In a classical harmonic gas, a lost particle then carries the mean energy:

$$\varepsilon_1 \simeq 3k_{\mathrm B}T.$$

Ideal one-body loss changes $N$ but not $T$ directly, so it lowers phase-space density.

Density-dependent loss

A $q$-body process is weighted toward the dense trap center. In a classical harmonic gas with energy-independent local loss, the mean energy removed per lost particle is

$$\varepsilon_q = \frac32 \left( 1+\frac1q \right) k_{\mathrm B}T.$$

Hence

$$\varepsilon_2 = \frac94 k_{\mathrm B}T, \qquad \varepsilon_3 = 2k_{\mathrm B}T.$$

Both are below the ensemble mean $3k_{\mathrm B}T$. Rethermalization therefore raises the temperature: density-dependent loss is anti-evaporative. Recombination products can deposit additional energy, making the heating worse.

Technical heating

Intensity noise, pointing noise, photon scattering, and field fluctuations add energy without favorable particle removal. An observed plateau in $T$ can result from a balance between evaporative cooling and technical or recombination heating; it is not evidence that equilibrium has been reached.

Opacity and the Hydrodynamic Limit

Evaporated particles must leave. Let

$$\ell = \frac{ 1 }{ \bar n\sigma }$$

be a mean free path and $L$ a characteristic escape distance. When $\ell\ll L$, an energetic particle can recollide before exiting, return energy to the sample, or leave through a geometry-dependent channel.

For thermal particles, also compare $\Gamma_{\mathrm{el}}$ with trap frequencies. This is a related diagnostic, not an exact substitute for escape probability: hot particles can have different cross sections. A collision rate much larger than a motional rate signals hydrodynamic behavior along that axis. Runaway scaling must then be replaced by a transport model that includes opacity and anisotropic escape. Along a specified ballistic escape path, a more direct estimate is $P_{\rm no\ collision}=\exp[-\int n(\mathbf r)\sigma_{\rm eff}(E,\mathbf r)\,ds]$, where $\sigma_{\rm eff}=\langle\sigma v_{\rm rel}\rangle/v_{\rm hot}$ includes averaging over the target velocities.

The best elastic rate is therefore not infinite. Efficient evaporation requires enough collisions to rethermalize but enough mean free path for the hot products to escape.

Sympathetic Evaporative Cooling

Suppose component $A$ is selectively evaporated while component $B$ remains trapped. Interspecies elastic collisions transfer energy from $B$ to $A$, which exports it through evaporation.

Requirements include:

1. a large interspecies elastic-to-inelastic ratio;
2. spatial overlap throughout the ramp;
3. compatible trap depths and gravitational sag;
4. enough coolant heat capacity and particle number;
5. manageable differential light shifts and state changes.

For masses $m_A$ and $m_B$, the fractional energy-transfer efficiency per collision contains the kinematic factor

$$\chi_m = \frac{ 4m_Am_B }{ \left( m_A+m_B \right)^2 }.$$

Large mass imbalance slows thermalization. Differential sag can remove overlap precisely when the gas becomes coldest.

Sympathetic cooling is especially useful for:

• spin-polarized fermions with suppressed $s$-wave self-collisions;
• species that cannot be laser cooled efficiently;
• internal states that should not be exposed to the evaporation knife;
• mixtures used to assemble ultracold molecules.

The coolant can itself become too dilute or too degenerate to thermalize the target efficiently.

Path to Quantum Degeneracy

The classical model is a guide, not a theory of the final state.

Bosons

For an ideal harmonically trapped Bose gas, condensation begins near

$$N \simeq \zeta(3) \left( \frac{ k_{\mathrm B}T_c }{ \hbar\bar\omega } \right)^3.$$

Equivalently, the global degeneracy parameter reaches

$$\mathcal D_c \simeq \zeta(3).$$

This $\mathcal D$ is a global trap parameter. Its equality with $n_0\lambda_{\rm dB}^3$ held only in the classical limit; the ideal Bose central critical value is $\zeta(3/2)$, not $\zeta(3)$.

Near and below the transition, Bose statistics, mean-field shifts, condensate growth, collective modes, and three-body loss modify the classical evaporation equations.

Fermions

For one spin component with $N_s$ particles,

$$k_{\mathrm B}T_F = \hbar\bar\omega \left( 6N_s \right)^{1/3}.$$

As $T/T_F$ decreases, Pauli blocking reduces available final states for collisions. A two-spin mixture can still evaporate effectively, but the thermalization law and heat capacity become quantum statistical.

Stopping criterion

The final trap should be chosen for the science measurement, not merely for the lowest temperature. Continuing evaporation can reduce particle number, increase relative fluctuations, worsen three-body loss, or lower interaction rates without improving the target observable.

Collision and evaporation times for rubidium

Consider a classical $^{87}\mathrm{Rb}$ gas with

$$N = 5.0\times10^5, \qquad T = 20\,\mu\mathrm K,$$

and geometric-mean frequency

$$\frac{ \bar\omega }{ 2\pi } = 500\,\mathrm{Hz}.$$

Take $m=1.443\times10^{-25}\,\mathrm{kg}$ and $a=100a_0$. First estimate the rate using the identical-boson threshold cross section, then check its energy dependence.

The peak density is

$$n_0 \simeq 1.18\times10^{19}\,\mathrm{m^{-3}},$$

and the collision-weighted density is

$$\bar n \simeq 4.16\times10^{18}\,\mathrm{m^{-3}}.$$

The threshold cross section and mean relative speed are

$$\sigma \simeq 7.04\times10^{-16}\,\mathrm{m^2}, \qquad \langle v_{\mathrm{rel}}\rangle \simeq 9.87\times10^{-2}\,\mathrm{m\,s^{-1}}.$$

Thus

$$\Gamma_{\mathrm{el}} \simeq 2.89\times10^2\,\mathrm{s^{-1}}.$$

In this threshold approximation, taking $\xi=2.7$ gives

$$\tau_{\mathrm{th}} \simeq 9.3\,\mathrm{ms}.$$

The threshold approximation is only a first estimate here. Set $q=ma^2k_{\rm B}T/\hbar^2\simeq0.1003$. Averaging the energy-dependent cross section over a Maxwell relative-speed distribution gives

$$\frac{\langle\sigma v_{\rm rel}\rangle} {8\pi a^2\langle v_{\rm rel}\rangle} =\int_0^\infty\frac{x e^{-x}}{1+qx}\,dx\simeq0.8433.$$

Thus $\Gamma_{\rm el}\simeq244\,\mathrm{s^{-1}}$, about $16\%$ lower. Using the same illustrative $\xi$ would give $11.1\,\mathrm{ms}$; a precise rethermalization rate also requires the appropriate energy-dependent collision kernel rather than keeping $\xi$ fixed.

The classical phase-space density is

$$\mathcal D = N \left( \frac{ \hbar\bar\omega }{ k_{\mathrm B}T } \right)^3 \simeq 8.6\times10^{-4}.$$

For $\eta=8$,

$$\kappa \simeq \frac{ 8-5 }{ 8-4 } = 0.75,$$

so the ideal efficiency is

$$\gamma_{\mathrm{evap}} \simeq 8+0.75-4 = 4.75.$$

For a rough timescale only, using the threshold rate in the conventional estimate

$$\Gamma_{\mathrm{ev}} \sim (\eta-4)e^{-\eta} \Gamma_{\mathrm{el}},$$

gives

$$\Gamma_{\mathrm{ev}} \sim 0.39\,\mathrm{s^{-1}}.$$

The gas rethermalizes in milliseconds, but the selected loss occurs on a seconds scale. Good vacuum and low heating remain essential.

If the ideal efficiency remained valid until $\mathcal D\simeq1$, the retained fraction would be

$$\frac{ N_f }{ N_i } \simeq \mathcal D_i^{1/\gamma_{\mathrm{evap}}} \simeq 0.23.$$

This is an optimistic classical extrapolation. Cross-section energy dependence, inelastic loss, trap weakening, and quantum statistics must be included in a real ramp.

Reading an evaporation trajectory

Plot $\ln\mathcal D$ against $\ln N$ over short ramp intervals, using measured trap frequencies, and compare the elastic rate with the measured rethermalization time. Track the actual knife threshold and heating at fixed knife. A temperature decrease alone can result from decompression, while a large fitted efficiency can hide an incorrect frequency calibration or changing spin populations. Near degeneracy, replace the classical density and collision model before interpreting the same plots.

Exercises

1. Population above an energy knife

For a classical gas in a three-dimensional harmonic trap, calculate the equilibrium fraction above a knife with $\eta=8$.

Solution

Use

$$P \left( \varepsilon>\varepsilon_t \right) = e^{-\eta} \left( 1+\eta+\frac{\eta^2}{2} \right).$$

At $\eta=8$,

$$P = e^{-8} \left( 1+8+32 \right) = 41e^{-8} \simeq 1.38\times10^{-2}.$$

About $1.4\%$ of an untruncated equilibrium distribution lies above the knife. This is not the rate at which collisions produce escaping atoms.

2. A sudden cut followed by rethermalization

Remove all atoms above $5k_{\rm B}T_i$ from an initially untruncated classical harmonic gas, then raise the knife so no further particles escape while the remaining atoms rethermalize at fixed confinement. Find $N_f/N_i$, $T_f/T_i$, and $\mathcal D_f/\mathcal D_i$.

Solution

The retained number fraction is $P(3,5)=0.87535$ and the retained energy fraction is $P(4,5)=0.73497$. After rethermalization,

$$\frac{T_f}{T_i}=\frac{P(4,5)}{P(3,5)}\simeq0.83964,\qquad \frac{\mathcal D_f}{\mathcal D_i} =\frac{P(3,5)}{[P(4,5)/P(3,5)]^3}\simeq1.4788.$$

This finite cut differs from an infinitesimal constant-$\eta$ trajectory. Rethermalization is essential to identify the final $T_f$; the immediate post-cut distribution has a cutoff and is not the final canonical gas.

3. Selectivity versus speed

Compare $\eta=6$ and $\eta=10$ using

$$\kappa = \frac{ \eta-5 }{ \eta-4 }$$

and the rate factor

$$f(\eta) = (\eta-4)e^{-\eta}.$$

Find the ideal efficiency and relative rate factor for each.

Solution

For $\eta=6$,

$$\kappa = \frac12, \qquad \gamma_{\mathrm{evap}} = 6+\frac12-4 = 2.5,$$

and

$$f(6) = 2e^{-6} \simeq 4.96\times10^{-3}.$$

For $\eta=10$,

$$\kappa = \frac56, \qquad \gamma_{\mathrm{evap}} \simeq 6.83,$$

and

$$f(10) = 6e^{-10} \simeq 2.72\times10^{-4}.$$

At the same elastic collision rate, the $\eta=10$ knife is about

$$\frac{ f(6) }{ f(10) } \simeq 18$$

times slower but ideally much more particle efficient. Background loss and heating decide which tradeoff is better.

4. Collision count versus evaporated particle count

A gas has $N=5\times10^5$ and per-particle elastic rate $\Gamma_{\rm el}=289\,\mathrm{s^{-1}}$. Its evaporation loss rate per atom is $\Gamma_{\rm ev}=0.39\,\mathrm{s^{-1}}$. Compare binary collision events with particles lost per second. What fraction of events remove one atom if each evaporative event removes exactly one?

Solution

Each binary collision is counted for two particles, so the event rate is $N\Gamma_{\rm el}/2=7.225\times10^7\,\mathrm{s^{-1}}$. The atom-loss flux is $N\Gamma_{\rm ev}=1.95\times10^5\,\mathrm{s^{-1}}$. Under the assumed one-atom escape per selected event, the fraction is $2\Gamma_{\rm ev}/\Gamma_{\rm el}\simeq0.00270$. The factor of two is a counting convention, not a change in collision physics.

5. Runaway at fixed confinement

In fixed harmonic confinement with constant cross section, show that $\Gamma_{\mathrm{el}}$ increases during evaporation when $\eta'=\eta+\kappa>6$.

Solution

At fixed confinement,

$$\Gamma_{\mathrm{el}} \propto \frac{ N }{ T }.$$

The ideal temperature scaling is

$$T \propto N^{(\eta'-3)/3}.$$

Hence

$$\Gamma_{\mathrm{el}} \propto N^{ 1-(\eta'-3)/3 } = N^{(6-\eta')/3}.$$

Because $N$ decreases, the collision rate increases only if the exponent is negative:

$$\eta' > 6.$$

If optical power lowering also reduces the trap frequencies, this fixed- confinement conclusion does not apply.

6. Adiabatic decompression

A classical gas in a harmonic trap is decompressed adiabatically so that every trap frequency is halved. No particles are lost. What happens to $T$ and $\mathcal D$?

Solution

Adiabatic invariance gives

$$T \propto \bar\omega.$$

Thus the temperature is halved. However,

$$\mathcal D = N \left( \frac{ \hbar\bar\omega }{ k_{\mathrm B}T } \right)^3$$

contains the ratio $\bar\omega/T$, which is unchanged. With constant $N$,

$$\mathcal D_f = \mathcal D_i.$$

The gas is colder but no more degenerate.

7. Anti-evaporation from three-body loss

Show that a local three-body process in a classical harmonic gas removes $2k_{\mathrm B}T$ per lost particle on average. Explain the resulting temperature change.

Solution

Three-body loss is weighted by $n^3(\mathbf r)$. Since

$$n(\mathbf r) \propto e^{-V(\mathbf r)/(k_{\mathrm B}T)},$$

the lost particles sample a spatial distribution proportional to

$$e^{-3V/(k_{\mathrm B}T)}.$$

For a three-dimensional harmonic potential, their mean potential energy is

$$\langle V\rangle_{\mathrm{loss}} = \frac{ 3 }{ 2 } \frac{ k_{\mathrm B}T }{ 3 } = \frac12k_{\mathrm B}T.$$

Their local kinetic distribution still contributes $3k_{\mathrm B}T/2$. Therefore

$$\varepsilon_3 = \frac32k_{\mathrm B}T + \frac12k_{\mathrm B}T = 2k_{\mathrm B}T.$$

This is below the ensemble mean $3k_{\mathrm B}T$. After rethermalization, the remaining energy is shared among fewer particles and the temperature rises.

8. Sympathetic cooling with unequal masses

Compute the mass-transfer factor

$$\chi_m = \frac{ 4m_Am_B }{ (m_A+m_B)^2 }$$

for mass ratio $m_B/m_A=10$. What other physical conditions determine efficient sympathetic cooling?

Solution

Set $m_B=10m_A$. Then

$$\chi_m = \frac{ 4(10) }{ (1+10)^2 } = \frac{ 40 }{ 121 } \simeq 0.331.$$

This mass factor is about one third of its equal-mass value. It is not the energy fraction exchanged in every individual collision: scattering angle, initial velocities, and the chosen relaxation observable also matter.

One must also measure the interspecies elastic and inelastic rates, spatial overlap and differential sag, both temperatures, coolant depletion, state-changing collisions, and whether either component becomes degenerate or hydrodynamic during the ramp.

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