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In 2001, Andrew G. Truscott, Randall G. Hulet, and collaborators observed degeneracy pressure in trapped lithium. A Zeeman slower fed a dual magneto-optical trap; optical pumping prepared atoms for a cloverleaf apparatus for magnetic trapping of neutral atoms. Microwave evaporation removed energetic lithium-7 bosons; elastic collisions cooled lithium-6 fermions by sympathetic cooling.^[ref]

Sequential absorption imaging recorded both trapped clouds on a camera. Fits to axial profiles used the boson cloud as a thermometer. At common temperature, the nearly equal magnetic trapping potentials provided a comparison: the clouds had similar sizes near $T/T_F=1$, but the fermion cloud remained broader as cooling reached $T/T_F=0.25$. Figure 3 compared its radius with ideal Fermi gas and classical predictions.^[ref]

Varying the wait before imaging tested thermal equilibrium. Finite imaging resolution restricted the analysis to the axial direction; fits neglected interaction-induced density shifts. Cooling eventually suffered from the dwindling boson heat capacity. The result supports quantum occupation effects, without directly resolving individual occupied states or reaching zero temperature.^[ref]

These were neutral alkali atoms in a spatially varying trap. The Fermi energy of a model electron gas instead uses uniform Electrons, two equally populated spin states, and the Electron mass. The measured trapped-atom radii are not measurements of that electron model's $p_F$ or $E_F$; comparing their numerical Fermi energies requires the appropriate particle mass, internal-state populations, and confining potential.

For the uniform noninteracting Electron model with two spin states and kinetic energy of nonrelativistic particles $E=p^2/(2m_e)$, count all available one-particle states up to an energy $E\ge 0$:This extensive density of states has units of inverse energy; $g(E)/V$ is the density of states per unit volume. Availability and occupation are different: $g(E)$ is positive above $E_F$, but at zero temperature those states are empty. The occupied energy distribution is $g(E)f(E)$, with the Fermi-Dirac distribution reduced to $f(E)=1$ below $E_F$ and $f(E)=0$ above it. The convention at the exact boundary does not affect a continuum integral.

With the dimensionless quantity $x=E/E_F$, the occupied count per $dx$, divided by the total particle number, is $3\sqrt{x}/2$ for $0<x<1$. Its area is one. Weighting this by $E$ gives the internal energy and its mean per particle:Thus the highest occupied energy and the mean energy differ. For the Fermi energy of a model electron gas, $U/N\simeq 1.0155\mathrm{eV}$. At fixed $N$, $E_F\propto V^{-2/3}$, so the derivative $P=-(\partial U/\partial V{)}_N$ gives the electron degeneracy pressure $P=2n{E}_F/5$. The same state-counting geometry therefore predicts a nonzero pressure at zero temperature.