research34 min studyLesson 484 in Physics

Evaporative cooling and phase-space density

Track entropy-selective atom loss, rethermalization, and trap geometry to distinguish a colder gas from a more quantum-degenerate gas.

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What you’ll understand

  • Compute phase-space density using an explicit thermal-wavelength convention.
  • Explain why evaporation requires elastic rethermalization and preferential energy loss.
  • Separate harmonic-trap phase-space gain from changes caused by decompression.

A conservative trap does not cool a cloud merely by holding it. Evaporative cooling selectively removes particles with above-average energy and allows the remainder to rethermalize through elastic collisions. Both atom number and temperature decrease, and their competition determines whether quantum degeneracy becomes more accessible. The central quantity is often phase-space density rather than temperature alone. Lowering temperature while diluting the gas too strongly can leave that density unchanged or even reduce it.

Degeneracy compares wave extent with spacing

Define the thermal de Broglie wavelength using Planck's constant divided by the square root of two pi times mass and thermal energy. Its cube multiplied by density is a dimensionless phase-space measure. The numerical convention matters when comparing a quoted threshold. In a uniform ideal three-dimensional Bose gas, condensation begins at a phase-space density of about 2.612, rather than exactly one. The rough statement that wavelengths overlap describes a scale, not the exact threshold.

A harmonically trapped gas has an inhomogeneous density and a different global critical-temperature relation. In the classical thermal regime, its central phase-space density can be written using atom number and the geometric mean angular trap frequency. This formula is useful before degeneracy, but should not be extrapolated through condensation as a complete description of the density profile. Fermionic degeneracy instead compares temperature with the Fermi temperature for a declared component density and spin multiplicity.

λT=h2πmkBT,D=nλT3,D0=N(ℏωˉkBT)3\lambda_T=\frac h{\sqrt{2\pi mk_BT}},\qquad \mathcal D=n\lambda_T^3,\qquad \mathcal D_0=N\left(\frac{\hbar\bar\omega}{k_BT}\right)^3
λT is in metres, n in m⁻³, and D is dimensionless. D₀ is the central classical phase-space density in a three-dimensional harmonic trap with geometric mean angular frequency ω̄. The uniform ideal Bose threshold is ζ(3/2) ≈ 2.612.↔ Scroll wide equations horizontally.

Select energy and then wait for collisions

Atoms escaping over a finite barrier or through an energy-selective radio-frequency knife remove energy above a chosen cutoff. Elastic collisions repopulate the high-energy tail and produce a new near-thermal distribution. The cutoff relative to thermal energy controls both how many atoms leave and how much energy each removes. Very aggressive truncation can leave a nonequilibrium cloud, while an extremely high cutoff removes too few atoms before background loss becomes important.

Efficient evaporation needs a favorable ratio of elastic rethermalization to inelastic or background loss. At low energy, identical bosons in one internal state have an s-wave cross section twice the distinguishable-particle result. Identical spin-polarized fermions cannot use s-wave collisions, so a second spin component or another species is often needed for efficient cooling. Close to a scattering resonance, finite collision energy limits the cross section, and simply extrapolating its zero-energy square-law growth is invalid.

η=UcutkBT,Γel∼nσvˉrel,σboson≃8πa2(∣ka∣≪1)\eta=\frac{U_{\mathrm{cut}}}{k_BT},\qquad \Gamma_{\mathrm{el}}\sim n\sigma\bar v_{\mathrm{rel}},\qquad \sigma_{\mathrm{boson}}\simeq8\pi a^2\quad(|ka|\ll1)
η is a dimensionless cutoff parameter, Γel in s⁻¹, a the s-wave scattering length in metres, and k the relative collision wavenumber. The cross section is for identical bosons in one internal state in the zero-energy s-wave limit; distinguishable particles give 4πa².↔ Scroll wide equations horizontally.

Geometry and unwanted losses change the gain

At fixed harmonic frequencies, central phase-space density scales as number divided by temperature cubed. A modest temperature reduction can therefore compensate substantial number loss. If lowering optical power also weakens confinement, the trap frequencies fall and reduce this gain. Adiabatic decompression of an ideal classical harmonic cloud lowers its temperature in proportion to trap frequency and preserves phase-space density at fixed number. It is cooling in a thermometer sense without the same entropy reduction per particle as selective evaporation.

Three-body recombination and light-assisted losses often remove atoms from the densest part of the cloud and can heat the survivors. A high collision rate is therefore not sufficient by itself: increasing density may improve rethermalization while worsening inelastic loss. Report evaporation efficiency over a stated trajectory of number, temperature, and confinement, together with the thermometry and density calibration. Reaching a fitted temperature below a target does not alone establish condensation or superfluidity.

γev=−dln⁡D0dln⁡N,kBTc(0)=ℏωˉ[Nζ(3)]1/3\gamma_{\mathrm{ev}}=-\frac{d\ln\mathcal D_0}{d\ln N},\qquad k_BT_c^{(0)}=\hbar\bar\omega\left[\frac N{\zeta(3)}\right]^{1/3}
γev is an efficiency defined along the measured cooling trajectory. The second relation is the ideal three-dimensional harmonic-trap Bose transition estimate, with ζ(3) ≈ 1.202; finite-size and interaction shifts are omitted. The classical D₀ relation alone is not valid throughout the condensed regime.↔ Scroll wide equations horizontally.

Lose atoms while gaining degeneracy

During an illustrative classical evaporation segment at fixed harmonic frequencies, number falls from N to N/2 while temperature falls from T to T/2. Find the central phase-space-density gain and the logarithmic evaporation efficiency for that segment.

  1. The harmonic expression gives D0∝N/T3\mathcal D_0\propto N/T^3 when the trap frequencies stay fixed. Number loss contributes a factor 1/2.

  2. The temperature reduction contributes (T/(T/2))3=8(T/(T/2))^3=8. Combining the factors gives a phase-space-density ratio of 4.

  3. The finite-segment logarithmic efficiency is −ln⁡4/ln⁡(1/2)=2-\ln4/\ln(1/2)=2. This comparison assumes both endpoints remain in the classical thermal regime used for the formula.

The result

Central phase-space density increases fourfold despite losing half the atoms. The segment has logarithmic evaporation efficiency 2.

Explore the model

Interactive lab

Interactive lab · probability

Removing a tail changes an ensemble

Reset and set BOTH Success probability p to 0.5 and Trials n to 10; predict mean 5. Then Reset and set BOTH p to 0.5 and n to 20; predict mean 10 and variance 5. This exact binomial distribution illustrates ensemble moments and tails only. It does not remove outcomes, simulate evaporation, sample random atoms, or compute thermal phase-space density.

Removing a tail changes an ensembleThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA cooling claim needs the distribution and its moments; an analogy based on a tail is not an evaporation solver.

Check your understanding

An ideal classical harmonic cloud is adiabatically decompressed at fixed number so T and every trap frequency both halve. What happens to D₀?

Try it yourself

In a dilute low-energy identical-boson gas, double scattering length at fixed density and relative-speed distribution. Estimate the change in elastic collision rate. State two reasons this extrapolation can fail near a resonance.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the s-wave zero-energy cross section, then examine its collision-energy assumption and competing loss channels.

Compare with the solution

Within ∣ka∣≪1|ka|\ll1, cross section scales as a2a^2, so doubling a multiplies σ and the estimated elastic rate by four. Near resonance, |ka| may no longer be small and unitarity limits the cross section. Inelastic losses such as three-body recombination can also become important and alter density and temperature. Therefore a fourfold zero-energy rate estimate does not guarantee faster successful evaporation in a resonant gas.

Take it with you

  • Phase-space density depends on density, temperature, and a declared thermal-wavelength convention.
  • Evaporation needs selective energy removal followed by elastic rethermalization.
  • Changing trap geometry or enhancing inelastic loss can offset an apparent temperature improvement.

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