Fermi Degeneracy Pressure vs Bose Collapse Simulator

Trapped Quantum Gas

Bose-Einstein Condensate

Atom number N 400

Temperature T/T_c 0.00

Cloud radius (a_ho)

–

Condensed fraction

–

T_c (ħω)

–

Degeneracy energy

–

"Degeneracy energy" is the gas's total energy above N·E₀ — the excess kinetic energy Pauli exclusion forces onto fermions even at T=0. It is zero for a collapsed condensate.

How it works

N identical atoms sit in an isotropic 3D harmonic trap. Its single-particle energy shells are E_n = (n + 3/2)ħω with degeneracy g(n) = (n+1)(n+2)/2. How the atoms fill those shells depends entirely on their quantum statistics.

Bosons:  N0/N = 1 - (T/Tc)^3        Tc  ≈ (N/ζ(3))^(1/3) ħω/k
Fermions: f(E) = 1 / (exp[(E-μ)/kT] + 1)   TF ≈ (6N)^(1/3) ħω/k

Bosons, T < T_c — a macroscopic fraction condenses into the single n=0 ground state: the cloud shrinks toward a point-like core of radius ~a_ho, independent of N. At T=0 every atom is there.
Fermions, any T — the Pauli exclusion principle allows at most one fermion per quantum state. Even at T=0 the atoms must stack up through shell after shell to the Fermi energy E_F = ħω(6N)^(1/3), so the cloud stays a fixed-size "Fermi sea" ball and pushes outward with a real degeneracy pressure that has nothing to do with temperature.
Above T_c / T_F — quantum degeneracy washes out and both gases relax toward the same classical Boltzmann cloud, expanding smoothly as the temperature rises.

This is why a boson cloud can collapse to a point while a same-sized fermion cloud never can — the effect that keeps white dwarfs and neutron stars from collapsing under gravity is the same electron/neutron degeneracy pressure shown here in miniature for a trapped atomic gas.

Occupied quantum state Harmonic trap boundary (scale ref.)

⚠ Couldn't load the 3D engineThree.js failed to load from the CDN. Check your connection and reload.