---
title: Bose-Einstein Condensation Derived
module: Bosonic Systems
moduleNumber: 8
lessonNumber: 5
order: 805
summary: >
  For a gas of conserved bosons the excited states can hold only a finite number
  of particles at fixed temperature, set by the Bose function at unit fugacity.
  When the total exceeds that ceiling the surplus collapses into the single
  ground state, which the continuum density-of-states integral misses and which
  must be restored by hand. This fixes the critical temperature, the condensate
  fraction, and the fact that a uniform gas condenses only in three or more
  dimensions.
topics: [Bosonic Systems]
sources:
  - book: Schroeder
    ref: "Ch. 7 — Quantum Statistics; §7.6 Bose-Einstein Condensation"
  - book: Pathria & Beale
    ref: "Ch. 7 — Ideal Bose Systems; §7.1"
  - book: Kardar (Statistical Physics of Particles)
    ref: "Ch. 7 — Quantum Statistical Mechanics; §7.4"
  - book: Reif
    ref: "Ch. 9 — Quantum Statistics of Ideal Gases; §9.11–9.12"
draft: false
---

Photons and phonons are non-conserved bosons whose chemical potential is pinned
at zero. A gas of atoms is different: the particle number is conserved, so the
chemical potential is a free parameter fixed by the requirement that the
occupations sum to $N$. For bosons that constraint cannot always be met by the
excited states alone. Below a critical temperature the excited states saturate,
and the leftover particles have nowhere to go but the ground state, which they
occupy in macroscopic number. This is Bose-Einstein condensation, derived here
from the ideal-gas [grand-canonical
framework](/statistical-mechanics/quantum-statistics/ideal-quantum-gases-general-framework)
rather than surveyed.

## Fixing the fugacity

The mean occupation of a single-particle state of energy $\varepsilon$ is the
Bose-Einstein form,

$$
\langle n(\varepsilon)\rangle = \frac{1}{z^{-1}e^{\beta\varepsilon} - 1},
\qquad z = e^{\beta\mu},
$$

with $z$ the fugacity. Every occupation must be non-negative, so for the ground
state at $\varepsilon = 0$ the fugacity is bounded, $0 \le z < 1$, equivalently
$\mu \le 0$. The ground-state occupation itself is

$$
N_0 = \frac{z}{1 - z},
$$

which stays finite for any $z < 1$ but diverges as $z \to 1$. Everything about
condensation lives in how $z$ approaches $1$.

For a free gas in a box the single-particle density of states is

$$
g(\varepsilon) = \frac{2\pi V (2m)^{3/2}}{h^3}\,\varepsilon^{1/2},
$$

and the number in excited states is the integral of $g(\varepsilon)\langle
n\rangle$. Substituting $x = \beta\varepsilon$ turns it into a standard Bose
function,

$$
N_{\mathrm{exc}} = \int_0^\infty \frac{g(\varepsilon)\,\d\varepsilon}
{z^{-1}e^{\beta\varepsilon} - 1}
= \frac{V}{\lambda^3}\,g_{3/2}(z),
\qquad
g_{3/2}(z) = \sum_{\ell=1}^\infty \frac{z^\ell}{\ell^{3/2}},
$$

where $\lambda = h/\sqrt{2\pi m k_B T}$ is the thermal de Broglie wavelength.[^bec-int]
The Bose function $g_{3/2}(z)$ increases monotonically with $z$ on $[0,1]$ and
reaches a finite maximum at $z = 1$,[^zeta]

$$
g_{3/2}(1) = \zeta\!\left(\tfrac{3}{2}\right) = 2.612.
$$

## The saturation ceiling

The number the excited states can hold is bounded, because $g_{3/2}(z) \le
2.612$. The largest excited population at temperature $T$ is

$$
N_{\mathrm{exc}}^{\max} = \frac{V}{\lambda^3}\,\zeta\!\left(\tfrac{3}{2}\right)
= 2.612\,\frac{V}{\lambda^3}.
$$

As $T$ falls, $\lambda \propto T^{-1/2}$ grows, and this ceiling drops. Above the
critical temperature $N_{\mathrm{exc}}^{\max} > N$, so the constraint $N =
N_{\mathrm{exc}}(z)$ is met at some $z < 1$ with a negligible ground-state
occupation. When $T$ drops to the point where the ceiling equals $N$, the
excited states are exactly full at $z = 1$, and any further cooling leaves a
surplus $N - N_{\mathrm{exc}}^{\max}$ that must occupy the ground state.

$$
% caption: The maximum number of particles the excited states can hold falls as temperature drops. Above the critical temperature it exceeds N; below, it undershoots N and the surplus condenses into the ground state.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.4,0) node[right] {T};
\draw[->, black] (0,0) -- (0,3.4) node[above] {population};
% N level
\draw[black, dashed] (0,2.4) -- (6.0,2.4);
\node[black, anchor=east] at (-0.05,2.4) {N};
% ceiling curve ~ T^{3/2}
\draw[acc, thick, smooth] plot coordinates {(0,0)(0.8,0.35)(1.6,0.98)(2.4,1.8)(3.0,2.4)(3.8,3.2)};
\node[text=acc, anchor=west] at (0.7,0.9) {excited-state ceiling};
% Tc marker where curve crosses N
\draw[black, dashed] (3.0,0) -- (3.0,2.42);
\node[black, anchor=north] at (3.0,-0.05) {$T_c$};
% shading condensed region
\fill[acc!10] (0,0) rectangle (3.0,3.3);
\node[text=black, anchor=south] at (1.4,2.6) {condensed};
\end{tikzpicture}
$$

## Why the ground state must be separated

The continuum integral is blind to the condensate. Because $g(\varepsilon)
\propto \varepsilon^{1/2}$ vanishes at $\varepsilon = 0$, the ground state
carries zero weight in $N_{\mathrm{exc}}$, and no value of $z$ makes the integral
exceed $2.612\,V/\lambda^3$. Yet the true ground-state occupation $N_0 = z/(1-z)$
diverges as $z \to 1$. Replacing the sum over states by an integral silently
drops a term that becomes macroscopic. The correct bookkeeping splits the ground
state off explicitly,

$$
N = \underbrace{\frac{z}{1 - z}}_{N_0} + \underbrace{\frac{V}{\lambda^3}\,g_{3/2}(z)}_{N_{\mathrm{exc}}}.
$$

Above $T_c$ the second term carries essentially all of $N$ and $z$ sits below
$1$. Below $T_c$ the second term is stuck at its ceiling $2.612\,V/\lambda^3$ with
$z$ pinned within $O(1/N)$ of $1$, and the first term holds the rest. The fugacity
does not pass through $1$; it approaches $1$ and stops, and $N_0$ jumps from
microscopic to macroscopic across $T_c$.

$$
% caption: The ground-state occupation from the exact formula diverges as the fugacity approaches unity, while the continuum integral for the excited states saturates at a finite ceiling; the gap between them is the condensate the integral omits.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.0,0) node[right] {fugacity z};
\draw[->, black] (0,0) -- (0,3.4) node[above] {occupation};
\draw[black, dashed] (5.2,0) -- (5.2,3.3);
\node[black, anchor=north] at (5.2,-0.05) {z to 1};
% ground state diverging
\draw[black, thick, densely dashed, smooth] plot coordinates {(0,0.02)(2,0.28)(3.4,0.62)(4.2,1.05)(4.7,1.65)(5.0,2.55)(5.12,3.2)};
\node[text=black, anchor=east] at (4.6,2.3) {ground state};
% excited saturating
\draw[acc, thick, smooth] plot coordinates {(0,0)(1,0.75)(2,1.4)(3,1.9)(4,2.25)(5.2,2.5)};
\node[text=acc, anchor=north west] at (2.2,1.55) {excited ceiling};
\end{tikzpicture}
$$

## Critical temperature and condensate fraction

At $T_c$ the excited states just hold all $N$ particles with $z = 1$:

$$
N = 2.612\,\frac{V}{\lambda_c^3}
\quad\Longrightarrow\quad
\left(\frac{N}{V}\right)\lambda_c^3 = \zeta\!\left(\tfrac{3}{2}\right).
$$

Writing $\lambda_c = h/\sqrt{2\pi m k_B T_c}$ and solving for the temperature,

> **Theorem (BEC critical temperature).** A uniform ideal Bose gas of number
> density $n = N/V$ condenses below
> $$
> k_B T_c = \frac{2\pi\hbar^2}{m}\left(\frac{n}{\zeta(3/2)}\right)^{2/3}
> = \frac{h^2}{2\pi m}\left(\frac{n}{2.612}\right)^{2/3}.
> $$
> The condition $n\lambda_c^3 = 2.612$ says the transition occurs when the thermal
> wavelength grows to the interparticle spacing, so neighboring wave packets
> overlap.

Below $T_c$ the excited population is $N_{\mathrm{exc}} = 2.612\,V/\lambda^3$ with
$z = 1$, and dividing by the same expression at $T_c$ gives $N_{\mathrm{exc}}/N =
(\lambda_c/\lambda)^3 = (T/T_c)^{3/2}$. The rest is condensate.

> **Theorem (Condensate fraction).** For $T < T_c$,
> $$
> \frac{N_0}{N} = 1 - \left(\frac{T}{T_c}\right)^{3/2}.
> $$
> It rises continuously from zero at $T_c$ to unity at $T = 0$, where every
> particle sits in the ground state.

The critical temperature drawn against density is a phase boundary: the condition
$n\lambda^3 = 2.612$ is a curve $T_c \propto n^{2/3}$ separating the normal gas
from the condensed phase.

$$
% caption: The condensation boundary in the density-temperature plane. The line n lambda cubed equals the Bose constant separates the normal gas below and to the right from the condensed phase above and to the left.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.0,0) node[right] {density n};
\draw[->, black] (0,0) -- (0,3.6) node[above] {T};
% boundary Tc ~ n^{2/3}
\fill[acc!10] (0,0) -- plot[domain=0.1:3.8, samples=60] (\x,{1.35*pow(\x,0.667)}) -- (0,3.4) -- cycle;
\draw[acc, very thick, smooth] plot[domain=0.1:3.8, samples=60] (\x,{1.35*pow(\x,0.667)});
\node[text=acc, anchor=west] at (3.95,3.0) {$T_c$ boundary};
\node[text=black, anchor=east] at (2.0,2.5) {condensed};
\node[black, anchor=north west] at (3.4,1.1) {normal gas};
\end{tikzpicture}
$$

$$
% caption: The condensate fraction against reduced temperature. It is zero above the critical temperature and rises as one minus the three-halves power below it, reaching unity at absolute zero.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.0,0) node[right] {$T/T_c$};
\draw[->, black] (0,0) -- (0,3.4) node[above] {$N_0/N$};
\draw[black, dashed] (0,2.9) -- (5.4,2.9);
\node[black, anchor=east] at (-0.05,2.9) {1};
\draw[acc, very thick, smooth] plot coordinates
  {(0,2.9)(0.7,2.62)(1.4,2.28)(2.1,1.87)(2.7,1.45)(3.3,0.98)(3.9,0.5)(4.4,0.16)(4.6,0)};
\draw[acc, very thick] (4.6,0) -- (5.6,0);
\draw[black, dashed] (4.6,0) -- (4.6,2.95);
\node[black, anchor=north] at (4.6,-0.05) {1};
\node[text=acc, anchor=west] at (0.6,2.35) {condensate fraction};
\end{tikzpicture}
$$

## Dimensionality and the harmonic trap

Whether the excited states saturate depends on the density of states, hence on
dimension. For a free gas in $d$ dimensions with $\varepsilon \propto p^2$, the
density of states scales as $g(\varepsilon) \propto \varepsilon^{d/2 - 1}$, and
the excited population involves the Bose function $g_{d/2}(z)$. Its value at
$z = 1$ is $\zeta(d/2)$, which is finite only for $d/2 > 1$:

- **Three dimensions** ($d = 3$): $\zeta(3/2) = 2.612$ is finite, so the ceiling
  exists and the gas condenses.
- **Two dimensions** ($d = 2$): $g_1(z) = -\ln(1 - z)$ diverges as $z \to 1$, so
  the excited states absorb any $N$ at any $T > 0$. A uniform 2D Bose gas does
  not condense at finite temperature.

The trap geometry changes the effective density of states and can restore
condensation in lower dimensions. In an isotropic three-dimensional harmonic trap
of frequency $\omega$, the level spectrum $\varepsilon = \hbar\omega(n_x + n_y +
n_z)$ gives $g(\varepsilon) \propto \varepsilon^2$, and the excited count
saturates at $\zeta(3)(k_B T/\hbar\omega)^3$. The critical temperature is

$$
k_B T_c = \hbar\omega\left(\frac{N}{\zeta(3)}\right)^{1/3},
\qquad
\frac{N_0}{N} = 1 - \left(\frac{T}{T_c}\right)^3,
$$

with a cubic condensate fraction rather than the three-halves power of the box.
This is the geometry of the laser-cooled alkali condensates, where the trap
harmonic potential replaces the box walls and the cubic law is what the atom-cloud
images measure.[^bec-trap]

## Summary

- For conserved bosons the fugacity is fixed by $N = z/(1-z) + (V/\lambda^3)
  g_{3/2}(z)$; the ground-state term $z/(1-z)$ diverges as $z \to 1$ while the
  Bose function caps at $g_{3/2}(1) = \zeta(3/2) = 2.612$.
- The excited states hold at most $2.612\,V/\lambda^3$ particles; below $T_c$ the
  surplus condenses into the ground state, which the $\varepsilon^{1/2}$ continuum
  integral omits and which must be added by hand.
- The transition occurs at $n\lambda_c^3 = \zeta(3/2)$, giving
  $k_B T_c = (2\pi\hbar^2/m)(n/\zeta(3/2))^{2/3}$ and the condensate fraction
  $N_0/N = 1 - (T/T_c)^{3/2}$.
- A uniform gas condenses only for $d \ge 3$, since $\zeta(d/2)$ diverges at
  $d = 2$; a harmonic trap changes the density of states, restoring condensation
  with a cubic fraction $1 - (T/T_c)^3$.

[^bec-int]: Pathria & Beale, §7.1, reduces $N_{\mathrm{exc}}$ to $(V/\lambda^3)
g_{3/2}(z)$ and tabulates the Bose functions $g_\nu(z) = \sum_\ell z^\ell/\ell^\nu$;
$\zeta(3/2) = 2.6124$. Kardar, §7.4, gives the same construction.
[^bec-trap]: Pathria & Beale, §7.1, works the harmonic-trap case and the
$\varepsilon^2$ density of states; the cubic condensate fraction is what the 1995
alkali-atom experiments imaged. Schroeder, §7.6, treats the box gas.
[^zeta]: The **Riemann zeta function** is $\zeta(s)=\sum_{n=1}^{\infty} n^{-s}=1+2^{-s}+3^{-s}+\cdots$. It enters these gas integrals through the standard Bose result $\int_0^\infty \frac{x^{s-1}}{e^x-1}\,\d x=\Gamma(s)\,\zeta(s)$. Here $\zeta(3/2)\approx 2.612$ is a half-integer value characteristic of nonrelativistic Bose gases, while $\zeta(3)\approx 1.2021$ is Apéry's constant, with no closed form in terms of $\pi$.