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Fermions at Work: The Degenerate Fermi Gas and Condensation

The Pauli principle forces fermions to stack into a 'Fermi sea', explaining the stiffness of metals and the survival of white dwarfs — while bosons do the opposite, collapsing into a single quantum state as they condense.

The Fermi sea at absolute zero

Because no two fermions can share a state, at T = 0 they cannot all fall to the ground level. Instead they fill every state from the bottom up, one per state, until the particles run out. The energy of the highest filled state is the Fermi energy \varepsilon_F; the occupied states form the degenerate Fermi gas and their boundary in momentum space is the Fermi surface.

The Fermi sea: every state up to the Fermi energy is filled (one fermion each), every state above is empty. At T = 0 the boundary is a razor-sharp step; warming smears it over a shell of width about k_B T around ε_F.

\varepsilon_F = \frac{\hbar^{2}}{2m}\left(3\pi^{2} n\right)^{2/3}, \qquad p_F = \hbar\left(3\pi^{2} n\right)^{1/3}

The Fermi energy and Fermi momentum of a 3D gas of spin-½ fermions at density n. Denser gases have deeper, higher-energy Fermi seas.

Worked example: the Fermi energy of a metal

Treat the conduction electrons of copper as a free Fermi gas and estimate how 'quantum' they are.

  1. Get the density. Copper contributes about one conduction electron per atom, giving n \approx 8.5\times10^{28}\ \text{m}^{-3}.
  2. Plug into \varepsilon_F. With the electron mass m, the formula gives \varepsilon_F \approx 1.1\times10^{-18}\ \text{J} \approx 7.0\ \text{eV}.
  3. Convert to a Fermi temperature. T_F = \varepsilon_F/k_B \approx 8\times10^{4}\ \text{K} — hundreds of times room temperature. Room temperature is 'ice cold' to these electrons, so the gas is deeply degenerate.

Degeneracy pressure and heat capacity

Even at T = 0 the stacked fermions carry enormous kinetic energy — they cannot stop, because the low states are full. This produces a purely quantum degeneracy pressure that owes nothing to temperature.

P = \frac{2}{5}\,n\,\varepsilon_F = \frac{\hbar^{2}}{5m}\left(3\pi^{2}\right)^{2/3} n^{5/3}

The degeneracy pressure of a non-relativistic Fermi gas. It is this pressure, from electrons, that holds up a white dwarf against gravity; from neutrons, a neutron star.

Warming the gas only excites the thin surface shell, so the heat capacity is suppressed by the factor T/T_F. A careful (Sommerfeld) expansion gives a heat capacity linear in T, not the classical constant.

C_V = \frac{\pi^{2}}{2} N k_B \,\frac{T}{T_F}, \qquad T_F = \varepsilon_F / k_B

The electronic heat capacity: linear in T and tiny (suppressed by T/T_F ≪ 1). Its linear signature is a fingerprint of degenerate fermions, seen in every metal at low temperature.

Bose-Einstein condensation: the opposite instinct

Bosons do the reverse. Cool a gas of massive bosons (unlike photons, their number is fixed) and the chemical potential rises toward the ground-state energy. Below a critical temperature the excited states simply cannot hold all the particles, and a macroscopic fraction avalanches into the single lowest state — Bose-Einstein condensation.

k_B T_c = \frac{2\pi\hbar^{2}}{m}\left(\frac{n}{\zeta(3/2)}\right)^{2/3}, \qquad \zeta(3/2) \approx 2.612

The BEC transition temperature. Equivalently, condensation sets in exactly when the degeneracy parameter reaches nλ³ = ζ(3/2) ≈ 2.612 — the quantum clouds have finally overlapped.

\frac{N_0}{N} = 1 - \left(\frac{T}{T_c}\right)^{3/2}, \qquad T < T_c

The condensate fraction: below T_c a growing share of the atoms occupies the single ground state, reaching 100% at T = 0. This N₀/N is the order parameter of the transition.