Bose-Einstein Condensation and the Fermion Gas

The quantum distributions differ from the classical one only by a $\mp 1$ in the denominator, and for an ordinary gas that difference is imperceptible: the states are so numerous and so sparsely occupied that a boson gas is barely distinguishable from a classical one. The difference becomes everything at low temperature. Bosons pour into the ground state; fermions stack up to a sharp energy ceiling. This lesson works out both limits and the two experimental systems that display them cleanly: superfluid helium and the laser-cooled atomic condensate, then the photon gas as a boson application, and finally the degenerate Fermi gas.

The critical temperature

Replacing the discrete spectrum of a boson gas by a continuous density of states $g(E)\propto E^{1/2}$ loses the ground state, because $g(0)=0$. For fermions that loss is harmless: at most two particles sit in any state, and discarding two out of $10^{22}$ changes nothing. For bosons it is fatal, because any number can occupy one state. The normalization

with $x=E/kT$ cannot be satisfied below a certain temperature. The constant $\alpha$ cannot be negative — a negative $\alpha$ would make $f_{BE}$ negative at small $E$, which is meaningless — so the integral is largest when $\alpha =0$, where it equals $2.315$. That caps the density the continuous formula can account for, and setting it equal to the actual $N/V$ defines the critical temperature.¹

Inserting the density of liquid helium gives $T_c\approx 3.1\mathrm{K}$, close to the observed transition at $2.17\mathrm{K}$, a fair result given the ideal-gas assumption for a liquid. Below $T_c$ the ground-state occupation $N_0$ must be counted separately,

and the fraction of particles left in excited states scales as $(T/T_c{)}^{3/2}$, so the condensate fraction is

The condensate grows continuously from zero at $T_c$ to the whole population at absolute zero.

Superfluid helium

When liquid ${}^4$He is cooled through $2.17\mathrm{K}$ — the lambda point, named for the shape of its specific-heat anomaly — it changes from an ordinary fluid (helium I) into a superfluid (helium II) that flows with viscosity near zero. London's two-fluid model treats helium II as a mixture of a normal fluid with the properties of helium I and a superfluid of zero viscosity,

with the superfluid identified as the condensate: the atoms that have dropped into the ground state and can carry no viscous drag. As $T$ falls from the lambda point the superfluid fraction rises from zero to one, exactly the condensate curve above.²

Helium II shows the superfluid directly. It conducts heat better than any metal, so local hot spots cannot form and the vigorous boiling of helium I stops abruptly at the lambda point even as evaporation continues. It flows through microscopic channels in packed powder that block any normal liquid, and it creeps as a thin film up and over the wall of its container until levels equalize. Only the two helium isotopes superfluid near absolute zero, because every other boson solidifies well above its condensation temperature; helium stays liquid because its atoms are light and their zero-point motion is large enough to melt the solid at low pressure. The rarer isotope ${}^3$He is a fermion and cannot condense as a single atom, yet it too superfluids, at about $2\mathrm{mK}$, once its atoms pair into effective bosons of integer spin — the same pairing idea that underlies superconductivity.

The dilute-atom condensate

Most atoms have integer ground-state spin and are bosons, but the level spacing in a macroscopic box is around $10^{-20}\mathrm{eV}$, so at any reachable temperature the atoms spread thinly over an enormous number of levels and no one state is macroscopically occupied. Cooling a gas to condense it the ordinary way fails: the atoms crowd together, interact through their outer electrons, and begin to act like fermions before they reach the ground state. Wieman and Cornell solved this in 1995 by forming the condensate directly from a supersaturated vapor, never letting it reach the solid equilibrium state.³

• Laser cooling. Six diode-laser beams tuned below resonance slow atoms in the low-speed tail of the Maxwell distribution; about $10^7$ rubidium atoms collect at roughly $1\mathrm{mK}$ in the beam intersection.
• Magnetic trapping and spin polarization. A shaped magnetic field squeezes the spin-polarized cloud, which reaches equilibrium as a vapor long before the solid can form.
• Evaporative cooling. Letting the warmest atoms escape a controlled leak carries off kinetic energy and cools the remaining few thousand atoms to below $100\mathrm{nK}$, where they fall into the ground state of the trap.

The signature is a sharp spike in the velocity distribution: a narrow peak of condensed atoms rising out of the broad thermal background, sharpening to a pure condensate as the last thermal atoms are removed. The condensate is a macroscopic quantum wave function — coherent matter, the atomic analog of the coherent light in a laser.

The photon gas and Planck's law

Photons have spin $1$ and are bosons, so the radiation in a cavity is a boson gas and its spectrum follows from Bose-Einstein statistics. The number of photons is not conserved — the cavity walls emit and absorb them freely — so the normalization that fixes $\alpha$ for a material gas does not apply. The count adjusts itself, forcing $\alpha =0$ and

The photon density of states, counting two polarizations, is $g_{ph}(E)\mathbf{d}E=(8\pi V{E}^2/c^3{h}^3)\mathbf{d}E$. The energy density in the interval $\mathbf{d}E$ is $u(E)\mathbf{d}E=E{g}_{ph}(E)f_{ph}(E)\mathbf{d}E/V$, and converting to frequency with $E=hf$ gives the spectrum.⁴

This is identical to the empirical blackbody formula. The factor $8\pi f^2/c^3$ is the classical mode density that produced the ultraviolet catastrophe; the Bose-Einstein factor $hf/(e^{hf/kT}-1)$ is the average energy per mode, which cuts off the high-frequency divergence. That the same spectrum arises whether the radiation is treated as distinguishable standing waves (Planck's route) or indistinguishable photons (the boson gas) is a statement of wave-particle duality.

The same photon gas describes the cosmic background radiation. Integrating the number density over the spectrum at the present $2.7\mathrm{K}$ gives about $4\times 10^8$ photons per cubic meter filling the universe.

The Fermi energy

Fermions run the argument in reverse. The exclusion principle forbids two from sharing a state, so there is no condensation into the ground state. Writing the normalization constant as $\alpha =-E_F/kT$ recasts the Fermi-Dirac distribution in terms of a single energy.

At absolute zero the distribution is a step. For $E<E_F$ the exponent is large and negative and $f_{FD}=1$; for $E>E_F$ it is large and positive and $f_{FD}=0$. Every state up to $E_F$ is filled, every state above it empty — a filled Fermi sea. This is the opposite of a boson gas, where all particles condense to the ground state; the exclusion principle forces fermions to stack upward, so the last one added sits at $E_F$ however cold the system.

At a temperature $T>0$ but with $kT\ll E_F$, only fermions within about $kT$ of $E_F$ can move: they alone find empty states within reach above them. A fermion deep in the sea cannot absorb $kT$ of energy, because the levels a step above it are already occupied. The step softens over a width $\sim kT$ and no more. Multiplying $f_{FD}$ by the density of states $g(E)\propto E^{1/2}$ gives the actual occupied-energy distribution: a filled band rising as $E^{1/2}$ up to $E_F$, its edge rounded by temperature.

The degenerate Fermi gas

Because only a shell of thickness $\sim kT$ near $E_F$ participates thermally, the number of electrons that can absorb heat is a small fraction $\sim kT/E_F$ of the total. This resolves the missing heat capacity of metals from the classical picture: the conduction electrons should add $\frac{3}{2}R$ by equipartition, but nearly all of them are locked deep in the Fermi sea and contribute almost nothing. The full electron contribution, worked out for metals, is linear in $T$ and small.

A gas of fermions cooled until the states fill smoothly from the ground state up to $E_F$ is a quantum degenerate Fermi gas — the fermion analog of the Bose-Einstein condensate, though the crossover is gradual rather than a sharp phase transition, and evaporative cooling works less well because the exclusion principle suppresses the collisions that rethermalize the gas. Jin and DeMarco reached this state in 1999 with ${}^{40}$K atoms, splitting them between two spin substates to keep collisions going. Its signature is thermodynamic: classically the total energy would fall to zero as $T\to 0$, but a degenerate Fermi gas retains the large zero-point energy of the filled sea and stays finite.⁶

The filled Fermi sea is not only the electron gas of a metal. The same degeneracy pressure — fermions resisting compression because the exclusion principle forbids them a lower state — supports a white dwarf against gravity and, for neutrons, a neutron star. The $\mp 1$ that distinguished the two quantum statistics ends, at the two extremes of temperature, in superfluids on one side and the stability of dead stars on the other.