research34 min studyLesson 489 in Physics

Unitary Fermi gases and universal scales

Define density and Fermi scales for a balanced mixture and identify the simultaneous resonance, range, temperature, and state assumptions behind universal thermodynamics.

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What you’ll understand

  • Use total and per-spin density conventions consistently in kF.
  • Distinguish unitarity from weak BCS pairing and the molecular BEC limit.
  • Apply zero-temperature universal thermodynamics only within its range assumptions.

A balanced gas of two fermionic internal states can interact strongly in the s-wave channel even though identical same-state fermions cannot. Near a broad scattering resonance, its scattering length can exceed the interparticle spacing while the interaction range remains much shorter. This produces the unitary regime, where many equilibrium properties are set by density, mass, temperature, and dimensionless many-body functions. Universality removes short-range detail under stated limits; it does not make the strongly correlated many-body problem noninteracting or exactly solved.

Declare which density defines the Fermi surface

For two equally populated states, define total density as the sum of both components. The Fermi wavenumber of either component can then be written using three pi squared times total density, or equivalently six pi squared times one component's density. Mixing these conventions changes the inferred energy and temperature scales. The Fermi energy is a density reference scale even for an interacting gas; it is not automatically equal to the actual chemical potential.

Opposite internal states permit low-energy s-wave scattering. To reach the zero-range unitary regime, require inverse scattering length small compared with kF and effective range small compared with inverse kF. Merely tuning to a pole does not remove a large effective range near a narrow resonance. The gas must also be described by the intended dimensionality, spin balance, and near-equilibrium state. Trap averages require a spatial model rather than direct substitution of a single global density.

n=n↑+n↓,n↑=n↓=n2,kF=(3π2n)1/3,EF=ℏ2kF22m,TF=EFkBn=n_\uparrow+n_\downarrow,\quad n_\uparrow=n_\downarrow=\frac n2,\quad k_F=(3\pi^2n)^{1/3},\quad E_F=\frac{\hbar^2k_F^2}{2m},\quad T_F=\frac{E_F}{k_B}
n is total density in m⁻³ and kF is in m⁻¹. EF is in joules and TF in kelvins. For a separately specified per-spin density nσ, use kF = (6π²nσ)^{1/3}. These definitions assume a balanced homogeneous three-dimensional two-component mixture.↔ Scroll wide equations horizontally.

Pairing crosses from many-body to molecular physics

On the weak negative-a side, attractive fermions can form extended Cooper pairs through a many-body instability even though there is no universal free-space shallow dimer. At positive a, tightly bound dimers can form and eventually behave as composite bosons when their size is small relative to spacing. The balanced zero-temperature crossover connects these limits through strong pairing. A pairing signal and a superfluid phase are related but distinct: a finite-temperature system can have pairing correlations without long-range phase coherence.

At resonance, the two-body cross section is limited at finite relative momentum instead of diverging with a squared. Thermodynamic properties depend on universal functions of reduced temperature and any additional declared dimensionless parameters such as polarization. These functions must be measured or calculated using controlled many-body methods. Inserting an infinite scattering length into a weak-coupling GP interaction energy is not a valid way to predict a unitary Fermi gas.

1∣kFa∣≪1,kF∣re∣≪1,σ↑↓(k)⟶4πk2(zero-range unitarity)\frac1{|k_Fa|}\ll1,\qquad k_F|r_e|\ll1,\qquad \sigma_{\uparrow\downarrow}(k)\longrightarrow\frac{4\pi}{k^2}\quad\text{(zero-range unitarity)}
a and re are in metres. The cross section is for distinguishable opposite-spin particles in the relative s-wave channel, with k the relative collision wavenumber rather than automatically kF. Many-body transport and thermodynamics additionally depend on quantum statistics and state preparation.↔ Scroll wide equations horizontally.

Scale invariance constrains thermodynamics

In a homogeneous balanced zero-temperature unitary gas, dimensional analysis makes the ground-state energy proportional to the noninteracting Fermi energy per particle with one dimensionless coefficient. Differentiating that energy at fixed particle number gives pressure, and differentiating at fixed volume gives chemical potential. The same coefficient appears in all three relations. Its numerical value is a many-body result, not determined by dimensional analysis alone.

Finite temperature replaces that one ground-state coefficient with thermodynamic functions and introduces a superfluid transition at a reduced temperature determined by the interacting system. Precision pressure and compressibility measurements provide stronger evidence for thermodynamics than an arbitrary fit of a cloud width. Finite effective range, spin imbalance, nonequilibrium expansion, and losses can all complicate comparison. Pauli suppression can make a resonant two-component Fermi gas long lived, but it does not imply that every inelastic channel is absent.

EN=35ξBEF,μ=ξBEF,P=25ξBnEF(T=0)\frac EN=\frac35\xi_B E_F,\qquad \mu=\xi_BE_F,\qquad P=\frac25\xi_BnE_F\quad(T=0)
ξB is the dimensionless Bertsch parameter, distinct from a Bose healing length. E is total ground-state energy, μ energy per added particle, and P pressure in pascals. These relations assume a balanced homogeneous three-dimensional zero-range unitary gas in its ground state.↔ Scroll wide equations horizontally.

Use one illustrative universal coefficient consistently

A homogeneous balanced unitary gas has EF/kB = 1.0 μK. Use an illustrative ξB = 0.40, supplied for arithmetic rather than as a current best estimate. Find μ/kB, (E/N)/kB, and P/(nkB) at zero temperature.

  1. The chemical-potential temperature scale is μ/kB=ξBEF/kB=0.40 μK\mu/k_B=\xi_BE_F/k_B=0.40\,\mu\mathrm K. It differs from the defining TF = 1.0 μK.

  2. The energy per particle is (E/N)/kB=(3/5)(0.40)(1.0)=0.24 μK(E/N)/k_B=(3/5)(0.40)(1.0)=0.24\,\mu\mathrm K.

  3. The pressure-to-density temperature scale is P/(nkB)=(2/5)(0.40)(1.0)=0.16 μKP/(nk_B)=(2/5)(0.40)(1.0)=0.16\,\mu\mathrm K. These obey P = 2E/(3V), the same zero-range scale relation.

The result

The three scales are 0.40 μK for μ/kB, 0.24 μK for energy per particle divided by kB, and 0.16 μK for P/(nkB).

Explore the model

Interactive lab

Interactive lab · function

Fermi energy is quadratic in momentum

Reset and set BOTH Coefficient a to 0.5 and Position x to 1; predict y = 0.5. Then Reset and set BOTH a to 0.5 and x to 2; predict y = 2. Treat x and y as normalized momentum and free kinetic-energy scales. The exact quadratic does not solve a Fermi distribution, pairing gap, universal equation of state, or interacting-gas expansion.

Fermi energy is quadratic in momentumThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 reference kinetic-energy scale can be simple even when the many-body coefficient multiplying it is strongly correlated.

Check your understanding

For a balanced two-component three-dimensional gas, which convention uses the total density n correctly?

Try it yourself

At fixed a and effective range, increase the total density of a balanced uniform gas by eight. Find changes in kF and EF. Initially |kFa| = 20 and kF|re| = 0.01; evaluate the new two dimensionless parameters.

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

Give me a hint

Fermi wavenumber scales as density to the one-third power, and energy scales as its square.

Compare with the solution

The new kF is twice its initial value and EF is four times larger. Consequently |kFa| becomes 40, while kF|re| becomes 0.02. The scattering-length condition improves but the finite-range condition becomes less strongly separated, though it is still small in this example. Raising density does not automatically improve every universality condition; temperature must also be compared with the changed TF.

Take it with you

  • Total and per-spin density conventions give different-looking but equivalent Fermi formulas.
  • Unitarity requires large scattering length and small effective range on the many-body scale.
  • Dimensional analysis constrains universal relations while their many-body coefficients need independent determination.

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