Bose-Einstein Condensation in the Ideal Gas

Cool a gas of ordinary atoms far enough and something happens that has no classical precedent: a finite, macroscopic fraction of them abruptly crowd into one and the same quantum state — the single lowest-energy level of the box that holds them. They do this with no attraction pulling them together, no chemistry, no phase boundary in real space. The only ingredient is that the atoms are identical bosons, and identical bosons like to share. This is Bose–Einstein condensation, and it is the cleanest phase transition in physics: a transition driven by statistics alone. The mechanism is a counting accident with dramatic consequences. The excited states of the gas — every level above the ground state — can hold only so many bosons at a given temperature. As you cool at fixed density, that ceiling drops. The moment the number of atoms you actually have exceeds what the excited states can carry, the surplus has nowhere to go but the ground state, whose occupation can grow without bound. A single quantum level, normally holding a negligible share, suddenly swells to hold a sizeable fraction of the whole gas. This article isolates that mechanism. We take the Bose–Einstein occupation numbers as given — the mean occupation $\langle n_\varepsilon\rangle = 1/(e^{\beta(\varepsilon-\mu)}-1)$ and the ceiling $\mu < \varepsilon_0$ are derived in aq-sm-bose-einstein-distribution — and ask the one question that makes condensation inevitable: how many particles can the excited states hold? We will find that answer is finite in three dimensions, that its finiteness is a razor-thin property of the geometry of space, and that it fails entirely in two dimensions or one. The critical temperature and the condensate fraction that follow are the subject of aq-sm-bec-critical-temperature-condensate-fraction.

Key takeaways

Bose–Einstein condensation is a phase transition of an ideal gas driven by quantum statistics alone — no interactions, no forces.
The chemical potential of a Bose gas cannot exceed the ground-state energy: $\mu\le\epsilon_0$ . Choosing the ground-state energy as $\epsilon_0=0$ gives $\mu\le0$ and therefore $0<z=e^{\beta\mu}\le 1$ . For $\mu>\epsilon_0$ , the Bose occupation of the ground state would cease to be physical.
At fixed $T$ and $V$ , the excited states can hold at most $N_{\text{ex}}^{\max} = \frac{V}{\lambda_T^3}\,\zeta\!\left(\tfrac32\right)$ , because $\mu \le \varepsilon_0$ and hence $z \le 1$ . For $N < N_{\text{ex}}^{\max}$ , no macroscopic condensate is required; when $N > N_{\text{ex}}^{\max}$ , the excess particles must occupy the ground state macroscopically.
Condensation is a razor's-edge property of three dimensions: the excited-state integral converges in $d = 3$ but diverges for the uniform gas in $d \le 2$ , so there is no free-space BEC in two dimensions or one.
The ground state is not represented by the continuum density of states: $g(\epsilon)\propto\sqrt{\epsilon-\epsilon_0}$ describes the continuum of excited states and misses the discrete ground-state level. The ground state must therefore be treated separately, with $N_0=\frac{1}{e^{\beta(\epsilon_0-\mu)}-1} =\frac{z}{1-z}$ when $\epsilon_0=0$ .
Key idea: the excited states saturate at $N_{\text{ex}}^{\max} = \zeta(3/2)\,V/\lambda_T^3$ when $\mu \to 0^-$ ; any excess is forced to occupy the single ground state macroscopically — that overflow is the condensate.

The core insight

The excited states of an ideal Bose gas can hold at most $N_{\text{ex}}^{\max} = \zeta(3/2)\,V/\lambda_T^3$ particles as $\mu \to 0^-$ ; any excess must macroscopically occupy the single ground state, and that forced overflow is Bose–Einstein condensation.

Where this lives today

Bose–Einstein condensates are now a standard laboratory platform rather than an exotic endpoint. Ultracold atoms in optical lattices—periodic potentials created by interfering laser beams—provide clean, tunable quantum simulators. By controlling lattice depth, interactions, and filling, experiments can realize Hubbard-type models and observe quantum phase transitions that are difficult to access directly in conventional materials. Atom interferometers built on condensates measure gravity and rotation with extraordinary precision, and matter-wave "atom lasers" extract coherent beams of atoms the way an optical laser extracts light. Each of these rests on the counting argument of this article: cool below the point where the excited states saturate, and a macroscopic number of atoms occupy one quantum state, coherent across the whole sample. The mechanism also explains why some transitions happen and others cannot. Its sharp dependence on dimensionality is equally important: the uniform ideal gas has a finite-temperature condensation transition in three dimensions, but not in one or two dimensions in the thermodynamic limit. Two-dimensional interacting Bose systems can nevertheless exhibit Berezinskii–Kosterlitz–Thouless physics, in which superfluidity emerges through a topological transition rather than through conventional Bose–Einstein condensation. The presence or absence of a finite excited-state ceiling, decided by a single convergence exponent, controls what quantum matter is possible in each dimension.

The physical picture

Think of the excited states as a bucket. At fixed volume, their maximum population is finite at a given temperature, and that capacity decreases as the gas cools. Each atom is smeared out over a thermal de Broglie wavelength $\lambda_T = h/\sqrt{2\pi m k_B T}$ , derived and owned by aq-sm-quantum-concentration-classicality-criterion; as $T$ falls, $\lambda_T$ grows, wavepackets fatten, and the number of distinct thermal cells that can each hold a boson goes down. At fixed atom number $N$ , you eventually reach the point where the bucket simply cannot hold everyone. The overflow does not spill onto the floor — it pours into the one place that has no capacity limit at all: the single ground state. Why can one state absorb an unlimited number of bosons? Because bosons carry no exclusion. The ground-state occupation is $N_0=\frac{1}{e^{\beta(\epsilon_0-\mu)}-1}$ . Choosing the ground-state energy as $\epsilon_0=0$ , this becomes $N_0=z/(1-z)$ . As $\mu\to\epsilon_0$ from below, $z\to1^{-}$ , so the ground-state occupation can become macroscopic. That would be impossible for fermions (Pauli caps each level at one), and it is the whole reason condensation is a boson phenomenon. The quantum permission for the pile-up is owned by aq-sm-bose-einstein-distribution; here we exploit its consequence. The control knob is the phase-space density $n\lambda_T^3$ — the number of atoms per thermal cell. When $n\lambda_T^3 \ll 1$ , the gas is nondegenerate and Maxwell–Boltzmann statistics is a good approximation. As $n\lambda_T^3$ approaches unity, quantum statistical effects become important; in three dimensions, Bose–Einstein condensation occurs when the excited states can no longer accommodate all particles. Everything about BEC is a statement about that one dimensionless number crossing order unity.

Figure

The excited-state bucket overflows as the gas cools — surplus atoms pour into the single ground state

The figure holds the whole story. Read left to right as a cooling movie: the blue thermal cloud (excited states) shrinks, the ceiling drops from generous, to exactly full at $T_c$ , to sparse — and the red block of condensate appears in the ground state only once the ceiling has fallen below $N$ . Nothing forces atoms into the ground state above $T_c$ ; below it, they have no other home. Now the punchline figure — the quantity that is the transition. In the thermodynamic limit, the condensate fraction is zero for $T\ge T_c$ and becomes nonzero for $T<T_c$ , increasing continuously as the temperature is lowered:

Figure

The order parameter: condensate fraction switches on at T_c and climbs to one at absolute zero

The kink at $T_c$ —the condensate fraction emerging continuously from zero with a finite slope—is the signature of a continuous transition. There is no latent heat, no discontinuity in $N_0/N$ itself; the order parameter grows smoothly out of zero. (The exact form and the slope belong to aq-sm-bec-critical-temperature-condensate-fraction; here it is the payoff that motivates the counting.)

The core equations

Everything follows from the Bose occupation formula together with one bookkeeping rule: separate the discrete ground state from the continuum of excited states. We begin with the mean occupation of a single-particle level:

\langle n_\varepsilon\rangle=\frac{1}{e^{\beta(\varepsilon-\mu)}-1},\qquad \mu\le \varepsilon_0=0 \qquad (1)

where: $\langle n_\epsilon\rangle$ is the mean occupation of a single-particle state of energy $\epsilon$ ; $\beta=1/(k_BT)$ ; and $\mu$ is the chemical potential, bounded by the ground-state energy $\epsilon_0$ . Choosing the zero of energy so that $\epsilon_0=0$ , we have $\mu\le0$ , and therefore the fugacity $z=e^{\beta\mu}$ satisfies $0<z\le1$ . The bound is required for the Bose occupation to remain physical. That single inequality is the seed of the whole transition. The total particle number is a sum over every level. The temptation is to turn it into an integral immediately, but that discards the one state we care about. We keep it explicit:

N=\underbrace{\frac{z}{1-z}}_{N_0\ \text{(ground)}}+\underbrace{\frac{V}{\lambda_T^3}\,g_{3/2}(z)}_{N_{\text{ex}}\ \text{(excited)}},\qquad z=e^{\beta\mu} \qquad (2)

where: $N_0 = z/(1-z)$ is the ground-state occupation when $\varepsilon_0 = 0$ ; $\lambda_T = h/\sqrt{2\pi m k_B T}$ is the thermal de Broglie wavelength; and $g_{3/2}(z)$ is the Bose–Einstein function describing the excited-state population. In the thermodynamic limit, $z \to 1^-$ at the transition, while below $T_c$ the excited-state population remains finite and the excess particles occupy the ground state macroscopically. Read the split physically: the first term is a single state that can hold anything, the second is the entire continuum of excited states, and the transition is the moment the second term hits its ceiling. The Bose–Einstein function is the workhorse. It is defined by an integral that turns into a tidy series:

g_\nu(z)=\frac{1}{\Gamma(\nu)}\int_0^\infty \frac{x^{\nu-1}\,\mathrm{d}x}{z^{-1}e^{x}-1}=\sum_{k=1}^\infty \frac{z^k}{k^\nu} \qquad (3)

where: $\nu$ — the order, fixed by dimension and by which quantity you compute ( $3/2$ for particle number in 3D, $5/2$ for energy and pressure); $\Gamma(\nu)$ — the gamma function. The single fact that governs condensation is what happens at $z = 1$ : the series becomes $\sum 1/k^\nu = \zeta(\nu)$ , which is finite when $\nu > 1$ and divergent when $\nu \le 1$ . In three dimensions $\nu = 3/2 > 1$ , so the excited-state count has a ceiling; that ceiling is the entire story. The three zeta values that appear across the ideal-Bose-gas problem are worth memorizing:

\zeta\!\left(\tfrac32\right)\approx 2.612,\qquad \zeta\!\left(\tfrac52\right)\approx 1.341,\qquad \zeta(3)\approx 1.202 \qquad (4)

where: $\zeta(3/2)$ sets the excited-state ceiling (and hence the critical density) in a 3D box; $\zeta(5/2)$ governs energy, pressure and heat capacity; $\zeta(3)$ sets $T_c$ for a harmonic trap, the geometry of real atomic-gas experiments. All three are pure numbers, computed once and reused everywhere.

The physical picture, made quantitative

Derivation AThe excited states have a finite ceiling — and it lives or dies on dimensionality

The claim that drives condensation is that the excited states can hold only a bounded number of particles. This is not obvious: a priori, raising $\mu$ toward the ground energy could let the excited states accommodate ever more particles. We now show that, for a uniform ideal Bose gas, the excited-state population is bounded above in three dimensions, evaluate that bound, and—most importantly—trace its finiteness to the power of energy in the density of states. This also shows why the same mechanism fails in one and two dimensions in the thermodynamic limit. The strategy is to write the excited count as a Bose integral, push $\mu$ to its maximum $\mu \to 0^-$ (which maximizes every occupation), and ask whether the resulting integral converges.

Step 1. Write the excited count as a Bose integral. The excited states form a continuum with the free-particle density of states $g(\varepsilon) = (V/4\pi^2)(2m/\hbar^2)^{3/2}\,\varepsilon^{1/2}$ , owned by aq-sm-density-of-states-free-particles. The excited population is

N_{\text{ex}}=\int_0^\infty \frac{g(\varepsilon)\,\mathrm{d}\varepsilon}{z^{-1}e^{\beta\varepsilon}-1}.

Why we may use the smooth $g(\varepsilon)$ here: it correctly counts every state except the one at $\varepsilon = 0$ , which we have already pulled out as $N_0$ in equation (2).

Step 2. Maximize the excited population by pushing $\mu$ to its upper bound. Every Bose occupation increases monotonically with $\mu$ , so $N_{\text{ex}}$ is largest as $\mu \to \varepsilon_0^-$ . Choosing $\varepsilon_0 = 0$ , this means $z \to 1^-$ . Setting $z = 1$ in the limiting integral, substituting $x = \beta\varepsilon$ and collecting constants gives

N_{\text{ex}}^{\max}=\frac{V}{4\pi^2}\left(\frac{2mk_BT}{\hbar^2}\right)^{3/2}\int_0^\infty \frac{x^{1/2}\,\mathrm{d}x}{e^{x}-1}.

Why $\mu = 0$ is the right place to look: below $T_c$ the fugacity is pinned there, so this is the actual working value, not a mathematical extreme.

Step 3. Evaluate the integral — and note it converges. The integral is $\int_0^\infty x^{1/2}/(e^x-1)\,\mathrm{d}x = \Gamma(3/2)\,\zeta(3/2) = (\sqrt\pi/2)\,\zeta(3/2)$ , a finite number. This finiteness is the crux. Near $x = 0$ the integrand behaves as $x^{1/2}/x = x^{-1/2}$ , and $\int_0 x^{-1/2}\,\mathrm{d}x$ converges. Assembling the prefactor with $\lambda_T = h/\sqrt{2\pi m k_B T}$ gives

N_{\text{ex}}^{\max}=\frac{V}{\lambda_T^3}\,\zeta\!\left(\tfrac32\right).

Why this is the ceiling: $g_{3/2}(z)$ rises monotonically on $(0,1]$ and reaches its largest value $\zeta(3/2)$ exactly at $z = 1$ — there is nowhere higher to go.

Step 4. Read off the dimensional watershed. In $d$ dimensions the density of states scales as $g(\varepsilon)\propto\varepsilon^{d/2-1}$ , so the number integrand behaves near $\varepsilon = 0$ as $\varepsilon^{d/2-1}/\varepsilon = \varepsilon^{d/2-2}$ . The integral $\int_0 \varepsilon^{d/2-2}\,\mathrm{d}\varepsilon$ converges only when $d/2 - 2 > -1$ , i.e. $d > 2$ . So the excited-state ceiling is finite in 3D ( $d/2-2 = -\tfrac12$ , converges), marginally divergent in 2D ( $=-1$ , log-divergent), and badly divergent in 1D ( $=-\tfrac32$ ). Why this settles the existence of BEC: only a finite ceiling can be exceeded — where the ceiling is infinite the excited states swallow any $N$ and nothing is ever forced into the ground state. *The key results — $\int x^{1/2}/(e^x-1)\,\mathrm{d}x = \Gamma(3/2)\zeta(3/2)$ , the value $\zeta(3/2)=2.612$ , the prefactor collapse to $V\zeta(3/2)/\lambda_T^3$ , and the convergence exponents for $d=1,2,3$ — can each be confirmed symbolically and numerically.*

Go deeper · Derivation B · 2 minWhy the ground state must be reinstated by hand

Derivation BWhy the ground state must be reinstated by hand

There is a subtle trap in the standard sum-to-integral conversion, and getting it wrong makes condensation vanish from the equations entirely. The goal here is to show that the smooth density of states $g(\varepsilon)\propto\varepsilon^{1/2}$ assigns exactly zero weight to the ground state, so that the very state that condenses is silently discarded — and that restoring it as a separate term $N_0$ is not a cosmetic fix but the mathematically mandatory step that makes equation (2) correct. The route is to compare the discrete ground term against what the integral does at $\varepsilon = 0$ .

Step 1. Isolate the ground term in the discrete sum. The exact particle number is $N = \sum_i \langle n_i\rangle$ . Split off the lowest level:

N=\underbrace{\frac{1}{z^{-1}-1}}_{N_0=z/(1-z)}+\sum_{i\ne 0}\langle n_i\rangle.

Why keep it separate: the $\varepsilon = 0$ term is the one that misbehaves as $z \to 1$ — it is the only term that can become macroscopic.

Step 2. Watch the ground term become macroscopic while the excited-state integral stays finite. As $z\to 1^{-}$ , the formal ground-state occupation $N_0=\frac{z}{1-z}$ diverges in the thermodynamic description, reflecting the fact that a single bosonic state can accommodate arbitrarily many particles. By contrast, the excited-state population remains finite and is capped at $N_{\text{ex}}^{\max}=\zeta\!\left(\tfrac32\right)\frac{V}{\lambda_T^3}$ . For a finite system, of course, $N_0 \le N$ ; the divergence signals that $N_0$ can become $O(N)$ , i.e. macroscopic. Why this is the whole point: two things that both count "the number of particles" disagree — the discrete ground state can blow up, the continuum cannot see it.

Step 3. See why the continuum integral misses the ground state. For the three-dimensional free-particle gas, $g(\epsilon)\propto\epsilon^{1/2}$ , so the continuum density of states vanishes at $\epsilon=0$ . The discrete ground state is therefore not represented correctly by the continuum integral. So converting $\sum_{i\ne 0}\to\int g\langle n\rangle\,\mathrm{d}\varepsilon$ is fine for the excited states but assigns zero to the ground state. Reinstating $N_0 = z/(1-z)$ as a standalone term — as equation (2) does — is therefore mandatory, not optional. The reason is that the continuum density of states describes states distributed over energy intervals, whereas the ground state is a single discrete level at $\epsilon=0$ . A single level has zero measure in the continuum energy integral and must therefore be added separately. *Both results — $N_0 = z/(1-z)\to\infty$ as $z\to 1^-$ and the finiteness of the excited integral — can be confirmed by direct symbolic computation.*

Limiting cases

Case	Result	Meaning
3D free gas	$N_{\text{ex}}^{\max}=\zeta(3/2)\,V/\lambda_T^3$ , finite	the ceiling exists; exceed it and the surplus condenses — the textbook BEC
2D free gas	$g_1(z)=-\ln(1-z)\to\infty$	excited count diverges as $z\to 1$ ; any density is absorbed, so no condensation at $T>0$
1D free gas	integrand $\sim\varepsilon^{-3/2}$ at $0$	worse divergence than 2D; no finite-temperature BEC
3D harmonic trap	ceiling set by $\varepsilon^2$ density of states	trapping steepens $g(\varepsilon)$ , restoring (indeed strengthening) the transition — the experimental geometry
$T\to 0$	$N_0/N\to 1$ , $\mu\to 0^-$	essentially the entire gas in the ground state; the thermal cloud vanishes

Common trapThe continuum integral already contains the condensate

The single most common error in ideal-Bose-gas calculations is to compute $N$ from the excited-state integral alone and then conclude, below $T_c$ , that the gas cannot accommodate all $N$ particles. The missing particles are the condensate. For the three-dimensional uniform ideal gas, the continuum density of states $g(\epsilon)\propto\epsilon^{1/2}$ describes the excited states but does not represent the discrete ground state at $\epsilon=0$ . Always write $N=N_0+N_{\rm ex}$ , $N_0=\frac{z}{1-z}$ , keeping the ground-state occupation separate. The integral describes the thermal cloud; $N_0$ describes the condensate. In the thermodynamic limit, $N_0$ becomes macroscopic below $T_c$ .

Worked examples

1 · When does a rubidium gas cross into the quantum regime? (confidence builder)

A gas of $^{87}$ Rb atoms ( $m = 87\times 1.66\times 10^{-27}\ \mathrm{kg} = 1.44\times 10^{-25}\ \mathrm{kg}$ ) has number density $n = 1.0\times 10^{19}\ \mathrm{m^{-3}}$ (that is $10^{13}\ \mathrm{cm^{-3}}$ ). At what temperature does the phase-space density reach the condensation threshold $n\lambda_T^3 = \zeta(3/2)$ ?

Step 1. At the ideal-gas condensation point, $n\lambda_T^3=\zeta\!\left(\tfrac32\right)\simeq 2.612$ . Therefore $\lambda_{T_c}=\left[\frac{\zeta(3/2)}{n}\right]^{1/3}=\left(\frac{2.612}{1.0\times10^{19}}\right)^{1/3}\simeq 6.39\times10^{-7}\,\mathrm{m}$ .

Step 2. The corresponding thermal wavelength is $\lambda_{T_c}\simeq 6.39\times10^{-7}\,\mathrm{m}=639\,\mathrm{nm}$ . This is hundreds of nanometres — far larger than the atomic size — showing that the matter waves have become strongly overlapping.

Step 3. Invert $\lambda_T = h/\sqrt{2\pi m k_B T}$ for temperature: $T = h^2/(2\pi m k_B \lambda_T^2)$ . Plugging $h = 6.626\times 10^{-34}$ , $m = 1.44\times 10^{-25}$ , $k_B = 1.381\times 10^{-23}$ , $\lambda_T = 6.39\times 10^{-7}$ :

T = \frac{(6.626\times 10^{-34})^2}{2\pi(1.44\times 10^{-25})(1.381\times 10^{-23})(6.39\times 10^{-7})^2}.

Step 4. Evaluating $T_c=\frac{h^2}{2\pi m k_B\lambda_{T_c}^2}$ gives $T_c\simeq8.6\times10^{-8}\,\mathrm{K}=86\,\mathrm{nK}$ . Thus, at this dilute density, ideal-gas Bose condensation occurs at temperatures of order $10^{-7}\,\mathrm{K}$ . The numerical value is specific to the chosen density and ideal-gas approximation; trapping, interactions, and finite-size effects modify the experimental transition.

2 · How many atoms can the excited states hold, and how big is the overflow? (exam level)

Take the same rubidium gas in a box of volume $V = 1.0\times 10^{-15}\ \mathrm{m^3}$ (a $0.1\ \mathrm{mm}$ cube), so $N = nV = 1.0\times 10^4$ atoms. At $T = 43\ \mathrm{nK}$ (half the threshold temperature of Example 1), find the excited-state ceiling $N_{\text{ex}}^{\max}$ and the condensate number $N_0$ .

Step 1. At $T = 43\ \mathrm{nK}$ the thermal wavelength grows as $\lambda_T \propto T^{-1/2}$ , so halving $T$ multiplies $\lambda_T$ by $\sqrt 2$ : $\lambda_T = 639\ \mathrm{nm}\times\sqrt 2 = 904\ \mathrm{nm}$ .

Step 2. The excited-state ceiling is $N_{\text{ex}}^{\max} = \zeta(3/2)\,V/\lambda_T^3$ . Compute $\lambda_T^3 = (9.04\times 10^{-7})^3 = 7.39\times 10^{-19}\ \mathrm{m^3}$ , so

N_{\text{ex}}^{\max} = 2.612\times\frac{1.0\times 10^{-15}}{7.39\times 10^{-19}} = 2.612\times 1354 = \mathbf{3536\ atoms}.

Step 3. The gas has $N = 10^4$ atoms but the excited states can hold only $3536$ . The surplus is forced into the ground state:

N_0 = N - N_{\text{ex}}^{\max} = 10^4 - 3536 = \mathbf{6464\ atoms}.

Step 4. Sanity check against the fraction law: at $T/T_c = 0.5$ , $N_0/N = 1 - 0.5^{3/2} = 1 - 0.354 = 0.646$ , so $N_0 = 6464$ atoms — the two routes agree exactly. Nearly two-thirds of the gas now occupies a single quantum state, a macroscopic matter wave. The overflow computed by hand is the condensate; the counting is the physics.

Origins · 1 minBose-Einstein Condensate and how this was discovered

Bose-Einstein Condensate

predicted 1924–25, realized 1995 · from Dhaka and Berlin to Boulder

The concept has two fathers separated by a translation. In 1924 Satyendra Nath Bose, working in Dhaka, re-derived Planck's radiation law by treating photons as indistinguishable and counting states in a new way; his manuscript was rejected by the Philosophical Magazine. He sent it to Einstein, who recognized its depth, translated it into German himself, and had it published. Einstein then did what Bose had not: he applied the same statistics to a gas of massive atoms with conserved number and found, in two papers of 1924 and 1925, that below a critical temperature a macroscopic fraction must occupy the lowest state. For seventy years this was theory. Only in 1995, using laser and evaporative cooling to reach nanokelvin temperatures, did Cornell and Wieman in Boulder (with $^{87}$ Rb) and Ketterle at MIT (with $^{23}$ Na) produce the condensate directly — the 2001 Nobel Prize. The predicted overflow, seen at last as a spike of atoms sharing one wavefunction.

History

1924

— Bose counts photon states by indistinguishability and re-derives Planck's law; his paper is first rejected, then championed and translated by Einstein.

1924

— Einstein extends the counting to a gas of massive atoms with conserved particle number.

1925

— Einstein predicts that below a critical temperature a macroscopic fraction of such a gas condenses into the ground state.

1938

— London proposes Bose–Einstein condensation as the mechanism behind the superfluid transition of liquid helium-4.

1995

— Cornell and Wieman (rubidium-87) and Ketterle (sodium-23) achieve the first atomic Bose–Einstein condensates.

2001

— Cornell, Wieman, and Ketterle share the Nobel Prize in Physics for the achievement.

Going further

Texts. Pathria & Beale, Statistical Mechanics, 3rd ed., Ch. 7 (Ideal Bose Systems) · Kardar, Statistical Physics of Particles, Ch. 7 · Landau & Lifshitz, Statistical Physics Part 1, §62 · Reif, Fundamentals of Statistical and Thermal Physics, Ch. 9 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 7 · Pethick & Smith, Bose–Einstein Condensation in Dilute Gases, Ch. 2. All formulas are standard ideal-Bose-gas results. Key checkpoints: the Bose integral $\int_0^\infty x^{1/2}/(e^x-1)\,\mathrm{d}x = \Gamma(3/2)\zeta(3/2) = (\sqrt\pi/2)(2.612)$ and the zeta values $\zeta(3/2)=2.612$ , $\zeta(5/2)=1.341$ , $\zeta(3)=1.202$ ; the excited-state ceiling $N_{\text{ex}}^{\max} = \zeta(3/2)V/\lambda_T^3$ ; the convergence criterion $d > 2$ from the near-zero exponent $\varepsilon^{d/2-2}$ ; the 2D divergence $g_1(z) = -\ln(1-z)$ ; and the worked numbers ( $T\approx 86\ \mathrm{nK}$ at $n = 10^{13}\ \mathrm{cm^{-3}}$ ; $N_{\text{ex}}^{\max}=3536$ and $N_0 = 6464$ for the box example).

Prerequisites. Quantum Statistics of Ideal Gases: Occupation Numbers · Density of States for Free Particles. Leads to. Critical Temperature and Condensate Fraction · Thermodynamics of the BEC Transition · Photon Gas and Planck Radiation Law.

Why does the chemical potential have to stay below the ground energy, and what happens right at the boundary?

Each occupation is $\langle n_\varepsilon\rangle = 1/(e^{\beta(\varepsilon-\mu)}-1)$ . For the ground state $\varepsilon = 0$ , this is positive only if $e^{-\beta\mu} > 1$ , i.e. $\mu < 0$ . If $\mu$ crept above $0$ , the ground occupation would go negative — physically impossible. As $\mu \to 0^-$ the ground occupation $N_0 = z/(1-z)$ diverges, and that divergence is not a breakdown but the equation correctly reporting that a single state is about to hold a macroscopic number of atoms. The boundary $\mu = 0$ is the onset of condensation itself.

Is BEC really a phase transition if there are no interactions? Phase transitions are supposed to come from forces.

It is a genuine phase transition, and its interaction-free origin is precisely what makes it remarkable. The "force" is replaced by quantum statistics: indistinguishable bosons over-weight configurations where they share a state, and that statistical bias, sharpened in the thermodynamic limit, produces a sharp onset in the order parameter $N_0/N$ . For finite $N$ the ground occupation rises smoothly with no singularity; only as $N \to \infty$ does the kink at $T_c$ become sharp. Interactions shift the transition and enrich its physics (they are what make helium a true superfluid), but they are not needed to make it happen.

The article separated the ground state by hand. Doesn't that feel like cheating to get the answer you want?

It is the opposite of cheating — omitting it would be the error. The sum over states is exact; the approximation is replacing that sum by an integral, and integrals can only see quantities with finite energy-width. A single isolated level has zero width, so any smooth density of states $g(\varepsilon)$ misses it entirely. Below $T_c$ that one missed state holds a finite fraction of all the atoms, so ignoring it changes the answer at leading order, not as a small correction. Reinstating $N_0 = z/(1-z)$ restores what the integral approximation dropped; it is bookkeeping, not fudging.

Why does condensation depend so drastically on dimensionality? Three dimensions works, two doesn't — that seems arbitrary.

It comes from a single convergence exponent. The excited-state count near $\varepsilon = 0$ has an integrand scaling as $\varepsilon^{d/2-2}$ , and $\int_0 \varepsilon^{d/2-2}\,\mathrm{d}\varepsilon$ converges only when $d/2 - 2 > -1$ , i.e. $d > 2$ . In 3D the exponent is $-1/2$ (converges, finite ceiling, BEC exists); in 2D it is exactly $-1$ (log-divergent, no ceiling); in 1D it is $-3/2$ (worse). A finite ceiling is what allows the excited states to be "full" and force the overflow into the ground state. Change the density of states — for example by adding a harmonic trap, which steepens $g(\varepsilon)$ — and you can restore condensation even in low dimensions.

If a single quantum state can hold a macroscopic number of bosons, why can't a single state hold two fermions?

Because the two statistics come from opposite symmetries of the many-particle wavefunction, owned by aq-sm-bose-einstein-distribution and aq-sm-fermi-dirac-distribution. Bosons have symmetric states, which impose no limit on how many can share a single-particle level — the occupation runs $0, 1, 2, \ldots$ without bound, giving $N_0 = z/(1-z)$ that diverges as $z \to 1$ . Fermions have antisymmetric states, which vanish if two particles share the same state, capping each level at one. So condensation — a macroscopic pile-up in one state — is exclusively a boson phenomenon; the fermion analogue is instead the filled Fermi sea, where exclusion forces particles up to high energy.

Liquid helium becomes a superfluid near 2 K. Is that the same as the BEC described here?

It is closely related but not identical. In 1938 London proposed that the superfluid transition of helium-4 (the lambda transition at $2.17\ \mathrm{K}$ ) is essentially Bose–Einstein condensation, and plugging helium's density into the ideal-gas formula gives $T_c \approx 3.1\ \mathrm{K}$ — strikingly close. But helium is a dense, strongly interacting liquid, not a dilute gas: at absolute zero only about $8\%$ of its atoms sit in the condensate, not $100\%$ , and the interactions reshape the excitation spectrum into the phonon–roton dispersion that true superfluidity requires. So the ideal-gas BEC captures the qualitative essence (Bose statistics, macroscopic ground-state occupation) while interactions supply the quantitative details.

Do photons Bose-condense? They are bosons, and blackbody radiation is a Bose gas.

Ordinary blackbody photons do not condense, and the reason is instructive. Condensation requires a fixed number of particles with a chemical potential that can rise to the ground energy and saturate. Photons have no conserved number — walls absorb and emit them freely — so their chemical potential is pinned at $\mu = 0$ identically at every temperature (owned by aq-sm-photon-gas-planck-radiation). There is no fixed- $N$ constraint to overflow: as you cool, the photon number simply falls. This contrast is the sharpest way to see that number conservation, not just Bose statistics, is the essential ingredient of BEC. (Photons can be made to condense in engineered dye-filled microcavities where a cutoff makes their number effectively conserved, but that is a special construction.)