---
title: Bose-Einstein Condensation and the Fermion Gas
module: Bosonic Systems
moduleNumber: 8
lessonNumber: 1
order: 801
summary: >
  Below a critical temperature a boson gas drops a macroscopic fraction of its
  particles into the single ground state — Bose-Einstein condensation, the
  mechanism behind superfluid helium and the dilute-atom condensates cooled to
  nanokelvin. The same statistics applied to a photon gas reproduces Planck's
  blackbody spectrum. Fermions do the opposite: forbidden from sharing states,
  they fill every level up to the Fermi energy, and that filled sea governs the
  electrons in metals and the pressure that holds up a white dwarf.
topics: [Bosonic Systems]
sources:
  - book: Tipler & Llewellyn
    ref: "Ch. 8 Statistical Physics; §8-3 The Bose-Einstein Condensation, §8-4 The Photon Gas, §8-5 Properties of a Fermion Gas"
  - book: Tipler & Llewellyn
    ref: "Ch. 3 Quantization of Charge, Light, and Energy; §3-2 Blackbody Radiation"
draft: false
---

The [quantum distributions](/statistical-mechanics/quantum-statistics/quantum-statistics-bose-einstein-and-fermi-dirac)
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

$$
N = \int_0^\infty \frac{g(E)}{e^{\alpha}e^{E/kT} - 1}\,\d E
  = \frac{2\pi V}{h^3}(2mkT)^{3/2}\int_0^\infty \frac{x^{1/2}\,\d x}{e^{\alpha}e^{x} - 1}
$$

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.[^tl-tc]

> **Definition (Bose-Einstein critical temperature).** The temperature at which
> the excited states can no longer hold all $N$ bosons is
> $$
> T_c = \frac{h^2}{2\pi m k}\left[\frac{N}{2.612\,V}\right]^{2/3}.
> $$
> Below $T_c$ a macroscopic number of particles occupies the single ground
> state — **Bose-Einstein condensation**.

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,

$$
N = N_0 + \frac{2\pi V}{h^3}(2mkT)^{3/2}\int_0^\infty \frac{x^{1/2}\,\d x}{e^{x} - 1},
$$

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

> **Theorem (Condensate fraction).** For $T < T_c$ the fraction of bosons in the
> ground state is
> $$
> \frac{N_0}{N} \approx 1 - \left(\frac{T}{T_c}\right)^{3/2}.
> $$

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

$$
% caption: Fraction of a boson gas in the ground state against temperature. It
% is zero above the critical temperature and rises to unity at absolute zero as
% one minus the three-halves power of the reduced temperature.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.2,0) node[right] {T};
\draw[->, black] (0,0) -- (0,3.2) node[above] {$\frac{N_0}{N}$};
\draw[black, dashed] (0,3.0) -- (5.2,3.0);
\node[font=\scriptsize, anchor=east] at (-0.05,3.0) {1};
\draw[acc, thick, smooth] plot coordinates
  {(0,3.0)(0.6,2.85)(1.3,2.63)(2,2.32)(2.6,1.97)(3.2,1.5)(3.7,1.05)(4.1,0.6)(4.4,0.28)(4.6,0)};
\draw[acc, thick] (4.6,0) -- (5.9,0);
\draw[black, dashed] (4.6,0) -- (4.6,3.05);
\node[font=\scriptsize, anchor=north] at (4.6,-0.05) {$T_c$};
\node[text=acc, font=\scriptsize, anchor=west] at (0.9,2.55) {condensate fraction};
\end{tikzpicture}
$$

## 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,

$$
\rho = \rho_s + \rho_n,
$$

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.[^tl-super]

$$
% caption: Specific heat of liquid helium against temperature. The sharp peak at
% 2.17 kelvin resembles the Greek letter lambda and marks the transition from
% normal fluid to superfluid.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.4,0) node[right] {T};
\draw[->, black] (0,0) -- (0,3.2) node[above] {C};
\draw[acc, thick, smooth] plot coordinates
  {(0.3,0.7)(1.0,0.85)(1.8,1.1)(2.6,1.45)(3.2,1.95)(3.55,2.85)};
\draw[acc, thick, smooth] plot coordinates
  {(3.7,2.85)(3.9,1.6)(4.3,1.35)(5.0,1.2)(5.9,1.12)};
\draw[black, dashed] (3.62,0) -- (3.62,2.9);
\node[font=\scriptsize, anchor=north] at (3.62,-0.05) {2.17 K};
\node[text=acc, font=\scriptsize, anchor=west] at (3.75,2.7) {lambda point};
\end{tikzpicture}
$$

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](/quantum-mechanics/matter-waves/the-uncertainty-principle) 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](/condensed-matter/superconductivity/superconductivity-phenomenology).

## 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.[^tl-bec]

- **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](/condensed-matter/molecular-spectra/lasers-and-masers).

$$
% caption: Velocity distribution of a trapped atom cloud through the transition.
% A broad thermal hill develops a central condensate spike, which grows until
% only condensed atoms remain at the lowest temperature.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% panel 1: thermal
\draw[->, black] (0,0) -- (2.6,0);
\draw[->, black] (0,0) -- (0,2.4);
\draw[acc, thick, smooth] plot coordinates {(0.2,0.35)(0.7,0.75)(1.2,1.0)(1.7,0.72)(2.2,0.35)};
\node[font=\scriptsize, anchor=north] at (1.2,-0.05) {above $T_c$};
% panel 2: emerging spike
\begin{scope}[xshift=3.4cm]
\draw[->, black] (0,0) -- (2.6,0);
\draw[->, black] (0,0) -- (0,2.4);
\draw[acc, thick, smooth] plot coordinates {(0.2,0.3)(0.6,0.55)(1.0,0.72)(1.2,1.85)(1.4,0.72)(1.8,0.5)(2.2,0.28)};
\node[font=\scriptsize, anchor=north] at (1.2,-0.05) {near $T_c$};
\end{scope}
% panel 3: pure condensate
\begin{scope}[xshift=6.8cm]
\draw[->, black] (0,0) -- (2.6,0);
\draw[->, black] (0,0) -- (0,2.4);
\draw[acc, thick, smooth] plot coordinates {(0.9,0.1)(1.15,1.1)(1.2,2.15)(1.25,1.1)(1.5,0.1)};
\node[font=\scriptsize, anchor=north] at (1.2,-0.05) {below $T_c$};
\end{scope}
\end{tikzpicture}
$$

## 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

$$
f_{ph}(E) = \frac{1}{e^{E/kT} - 1}.
$$

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

> **Theorem (Planck radiation law).** The energy density of cavity radiation
> between frequencies $f$ and $f + \d f$ is
> $$
> u(f)\,\d f = \frac{8\pi f^2}{c^3}\,\frac{hf}{e^{hf/kT} - 1}\,\d f.
> $$

This is identical to the empirical [blackbody
formula](/quantum-mechanics/old-quantum-theory/blackbody-radiation-and-the-planck-quantum). 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.

$$
% caption: The cavity radiation spectrum. The classical mode density rises as
% frequency squared and would diverge, but the Bose-Einstein energy per mode
% falls off exponentially at high frequency, giving the peaked Planck curve.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.6,0) node[right] {frequency};
\draw[->, black] (0,0) -- (0,3.3) node[above] {u(f)};
% Rayleigh-Jeans divergence (f^2)
\draw[black, thick, dashed, smooth] plot coordinates {(0,0)(0.8,0.19)(1.6,0.77)(2.4,1.73)(3.0,2.7)(3.35,3.2)};
\node[text=black, font=\scriptsize, anchor=west] at (2.7,3.0) {classical (diverges)};
% Planck curve
\draw[acc, thick, smooth] plot coordinates {(0,0)(0.6,0.55)(1.2,1.4)(1.8,2.05)(2.3,2.25)(2.9,2.05)(3.6,1.55)(4.3,1.05)(5.0,0.65)(5.7,0.38)(6.3,0.22)};
\node[text=acc, font=\scriptsize, anchor=west] at (3.4,1.75) {Planck};
\end{tikzpicture}
$$

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.

> **Definition (Fermi energy).** The Fermi-Dirac distribution is
> $$
> f_{FD}(E) = \frac{1}{e^{(E - E_F)/kT} + 1},
> $$
> where $E_F$ is the **Fermi energy**. At $E = E_F$ the occupation is exactly
> $\tfrac{1}{2}$ at every nonzero temperature.[^tl-fermi]

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.

$$
% caption: The Fermi-Dirac occupation at three temperatures. At absolute zero it
% is a step falling from one to zero at the Fermi energy; warming rounds the
% step over a width of about kT while leaving deep states full.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.4,0) node[right] {E};
\draw[->, black] (0,0) -- (0,3.0) node[above] {f(E)};
\draw[black, dashed] (0,2.6) -- (6.0,2.6);
\node[font=\scriptsize, anchor=east] at (-0.05,2.6) {1};
\draw[black, dashed] (0,1.3) -- (3.2,1.3);
\node[font=\scriptsize, anchor=east] at (-0.05,1.3) {$\frac{1}{2}$};
% T=0 step
\draw[acc, very thick] (0,2.6) -- (3.2,2.6) -- (3.2,0);
% T>0 rounded
\draw[black, thick, densely dashed, smooth] plot coordinates
  {(0,2.58)(1.0,2.55)(1.8,2.45)(2.4,2.2)(2.8,1.75)(3.2,1.3)(3.6,0.85)(4.0,0.42)(4.5,0.16)(5.0,0.06)(5.6,0.02)};
\draw[black, dashed] (3.2,0) -- (3.2,2.62);
\node[font=\scriptsize, anchor=north] at (3.2,-0.05) {$E_F$};
\node[text=acc, font=\scriptsize, anchor=south] at (1.3,2.62) {at absolute zero};
\node[text=black, font=\scriptsize, anchor=west] at (3.75,0.75) {warmed};
\end{tikzpicture}
$$

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.

$$
% caption: Occupied fermion energies, the density of states times the Fermi
% factor, at three temperatures for a material with Fermi energy near five
% electron volts. Warming moves a thin shell of electrons from just below the
% Fermi energy to just above it.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.6,0) node[right] {E};
\draw[->, black] (0,0) -- (0,3.0) node[above] {n(E)};
% E^{1/2} envelope
\draw[black, dashed, smooth] plot coordinates {(0,0)(1,1.15)(2,1.63)(3,2.0)(4,2.3)(4.2,2.36)};
% T=0: sqrt up to EF then cliff
\draw[acc, very thick, smooth] plot coordinates {(0,0)(1,1.15)(2,1.63)(3,2.0)(4.0,2.3)(4.2,2.36)};
\draw[acc, very thick] (4.2,2.36) -- (4.2,0);
% warmed curve rounded at EF
\draw[black, thick, densely dashed, smooth] plot coordinates
  {(0,0)(1,1.13)(2,1.6)(3,1.95)(3.6,2.13)(4.0,2.1)(4.2,1.75)(4.5,1.0)(4.9,0.42)(5.4,0.15)(5.9,0.05)};
\draw[black, dashed] (4.2,0) -- (4.2,2.4);
\node[font=\scriptsize, anchor=north] at (4.2,-0.05) {$E_F$};
\node[text=acc, font=\scriptsize, anchor=south] at (2.2,2.55) {absolute zero};
\node[text=black, font=\scriptsize, anchor=west] at (4.6,1.0) {warmed};
\end{tikzpicture}
$$

## 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](/statistical-mechanics/foundations/classical-statistics-and-equipartition): the conduction
electrons should add $\tfrac{3}{2}R$ 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](/condensed-matter/free-electron-fermi-gas/free-electron-gas-and-conduction),
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.[^tl-degen]

| | Bose-Einstein condensate | Degenerate Fermi gas |
| --- | --- | --- |
| Spin | integer | half-integer |
| Ground-state occupation | macroscopic ($N_0$ near $N$) | one particle per state |
| Onset | sharp transition at $T_c$ | gradual below $\sim T_F$ |
| Energy at $T = 0$ | all in ground state | filled sea up to $E_F$ |
| Physical example | superfluid $^4$He, cold-atom BEC | electrons in metals, neutron star |

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](/astrophysics-cosmology/orientation/stellar-death-final-states). 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.

[^tl-tc]: **Tipler & Llewellyn**, _Modern Physics_, §8-3 (The Bose-Einstein Condensation), Eqs. 8-47 to 8-52 — the normalization integral, its maximum value at $\alpha = 0$, the critical temperature, and the $(T/T_c)^{3/2}$ condensate fraction.
[^tl-super]: **Tipler & Llewellyn**, _Modern Physics_, §8-3 (Liquid Helium, Experimental Characteristics of Superfluid $^4$He), Eq. 8-45 — the lambda point, London's two-fluid model, and the superfluid as the condensate.
[^tl-bec]: **Tipler & Llewellyn**, _Modern Physics_, §8-3 (The Bose-Einstein Condensate) — the Wieman-Cornell rubidium condensate: laser cooling, magnetic trapping, spin polarization, and evaporative cooling to below $100\ \mathrm{nK}$.
[^tl-photon]: **Tipler & Llewellyn**, _Modern Physics_, §8-4 (The Photon Gas), Eqs. 8-53 to 8-57 — $\alpha = 0$ for photons, the photon density of states, and the recovery of the Planck spectrum from Bose-Einstein statistics.
[^tl-fermi]: **Tipler & Llewellyn**, _Modern Physics_, §8-5 (Properties of a Fermion Gas), Eqs. 8-67 to 8-69 — the Fermi energy, the step distribution at $T = 0$, and the $kT$-wide softening of the step at finite temperature.
[^tl-degen]: **Tipler & Llewellyn**, _Modern Physics_, §8-5 (Quantum Degenerate Fermion Gas) — the Jin-DeMarco degenerate $^{40}$K gas and the finite zero-temperature energy of the filled Fermi sea.