Bose-Einstein Condensation in Dilute Gases

Bose-Einstein condensation takes one result from quantum statistical mechanics and pushes it to its breaking point. Statistics gives identical bosons the occupation rule $\bar n(\epsilon)=[e^{(\epsilon-\mu)/k_BT}-1]^{-1}$ , a distribution with no exclusion ceiling: as the chemical potential $\mu$ climbs toward a level's energy, that level's population diverges. In a cold gas this loophole becomes a phase transition. Cool the gas at fixed density and $\mu$ rises toward the ground-state energy from below; the excited states, it turns out, can only hold a bounded number of atoms at a given temperature. When the thermal population saturates that bound, every further atom has nowhere to go but the single lowest mode, whose occupation swells to a macroscopic fraction of the whole gas. Einstein saw this in 1924-25, extending Bose's counting of photon states to atoms with mass. The trigger is written in phase-space density. When the thermal de Broglie wavelength $\lambda_{dB}=h/\sqrt{2\pi mk_BT}$ grows to the interparticle spacing $n^{-1/3}$ , the wavepackets overlap and indistinguishability becomes decisive; in a uniform 3D gas condensation sets in at $n\lambda_{dB}^3\approx 2.612$ . Reaching that in a dilute vapour — kept at $n\sim10^{13}$ to $10^{15}\,\mathrm{cm^{-3}}$ to stay gaseous and dodge three-body recombination — demands temperatures of tens of nanokelvin to a microkelvin, far below anything laser light alone can reach. The experimental route stacks a magneto-optical trap and optical molasses (down to the Doppler and recoil limits) onto magnetic trapping and forced evaporation to finally cross $T_c$ . Below $T_c$ the condensate is not merely a big number. It is a single coherent matter wave sharing one macroscopic phase — it interferes with other condensates, carries quantized vortices, and flows as a superfluid. Weak contact interactions, set by the scattering length $a_s$ through $g=4\pi\hbar^2 a_s/m$ , sculpt its shape, and at $T=0$ almost every atom rides one orbital $\psi(\mathbf r)$ obeying the nonlinear Gross-Pitaevskii equation. The 2001 Nobel Prize to Cornell, Wieman and Ketterle marked the arrival of quantum-degenerate matter as a laboratory tool.

Key takeaways

Cooling $N$ ideal bosons at fixed density drives the chemical potential $\mu$ up toward the ground-state energy; below a critical temperature $T_c$ the excited states saturate and a macroscopic number $N_0$ of atoms piles into the single-particle ground state — Bose-Einstein condensation.
The transition is set by phase-space density: in a uniform gas it occurs when $n\lambda_{dB}^3=\zeta(3/2)\approx 2.612$ , where $\lambda_{dB}=h/\sqrt{2\pi mk_BT}$ ; condensation is where wavepackets overlap.
BEC is a consequence of quantum statistics, not of attraction — it happens even for a non-interacting gas; weak repulsion then makes the condensate a superfluid described at $T=0$ by the Gross-Pitaevskii equation with coupling $g=4\pi\hbar^2 a_s/m$ .
Dilute-gas BEC was realized in 1995 in rubidium-87, sodium-23 and lithium-7 by laser cooling followed by evaporative cooling in magnetic traps, at $T\sim100$ nK.
In a harmonic trap the condensate fraction is $N_0/N=1-(T/T_c)^3$ with $k_BT_c\approx0.94\,\hbar\bar\omega N^{1/3}$ ; in a uniform box the exponent is instead $3/2$ .
Key idea: Bose statistics cap how many atoms the excited states can hold at a given temperature (or phase-space density); the surplus condenses into one single-particle ground state, producing a coherent macroscopic matter wave.

The core insight

Bose statistics cap how many atoms the excited states can hold at a given $T$ (equivalently a given $n\lambda_{dB}^3$ ); the surplus is forced into the single-particle ground state, producing a coherent macroscopic matter wave.

Where this lives today

The condensate is no longer a curiosity but a standard tool, and the machines that make it have shrunk from room-sized apparatus to instruments that fly. NASA's Cold Atom Laboratory, aboard the International Space Station since 2018, produces rubidium condensates in microgravity, where the absence of gravitational sag lets a cloud expand slowly enough to be watched for seconds rather than milliseconds — long enough to probe the coldest, most dilute regimes and to test the equivalence principle with two species falling together. On the ground, condensate sources feed matter-wave interferometers used as gravimeters and gyroscopes: an atom's de Broglie wavelength is far shorter than any optical wavelength, so an atom interferometer resolves inertial forces with extraordinary sensitivity. The same coherence underwrites quantum simulation. Loaded into an optical lattice, a condensate realizes the Bose-Hubbard model and can be driven through the superfluid-to-Mott-insulator transition simply by deepening the lattice; with a Feshbach resonance tuning $a_s$ from repulsive to attractive — and even to a collapsing bosenova — cold atoms emulate condensed-matter Hamiltonians that defeat direct computation. Modern laboratories now use machine-learning controllers to optimize the evaporation ramp, producing a condensate every few seconds and turning a 1995 tour de force into a teaching-lab apparatus.

The physical picture

Picture each atom as a fuzzy ball whose diameter is its thermal de Broglie wavelength $\lambda_{dB}$ . At room temperature the balls are minuscule and almost never touch — the gas is a crowd of well-separated billiard balls, and classical statistics work fine. As you cool, $\lambda_{dB}\propto T^{-1/2}$ swells. When the balls grow to the interparticle spacing and begin to overlap, at $n\lambda_{dB}^3\sim1$ , the atoms can no longer be labelled or told apart, and Bose statistics take over the bookkeeping. The first figure freezes this at three temperatures and pins a real number on the moment of overlap.

Figure

Cooling swells the thermal de Broglie wavelength as λ_dB ∝ T⁻¹/² until the wavepackets overlap and condense at n·λ_dB³ ≈ 2.612 — for rubidium at n = 10¹⁴ cm⁻³ the spacing is ≈ 0.22 μm, λ_dB reaches ≈ 0.30 μm, and T_c ≈ 0.40 μK

Now think of the excited states as seats in a theatre whose capacity is set by temperature. Cooling does not just move atoms to lower rows — it shrinks the total number of seats the available thermal energy can populate. Below $T_c$ there are literally not enough excited-state seats for every atom, and the lowest single-particle state, which can hold unlimited occupancy, absorbs the entire surplus. That macroscopic occupancy $N_0$ is the condensate, and it grows from zero at $T_c$ to nearly all of $N$ at $T=0$ . The condensate that emerges is coherent in the same sense a laser mode is: all its atoms share one phase. Release two independently made condensates so they overlap and they paint high-contrast matter-wave interference fringes — the cleanest evidence that a condensate is one macroscopic wavefunction, not just a lucky pile-up. The second figure is the quantitative punchline: how sharply that condensate switches on as the gas is cooled through $T_c$ .

Figure

Condensate fraction versus temperature in a harmonic trap, N₀/N = 1 − (T/T_c)³, switching on sharply at T_c ≈ 0.45 μK for N = 10⁶ rubidium atoms in a trap with ω̄/2π = 100 Hz — at T = 0.5 T_c already 87.5 percent of the atoms sit in the ground state

Interactions, though weak in a dilute gas, are not optional set-dressing — they set the shape of the cloud. At $T=0$ the balance between kinetic energy, the trap potential, and the repulsive mean field $gn$ gives a smooth inverted-parabola density profile (the Thomas-Fermi profile), noticeably broader than the Gaussian an ideal-gas ground state would show. Measuring that profile in a time-of-flight image is how experimenters read off the interaction energy and confirm they hold a genuine interacting condensate rather than an accidental classical cluster.

The core equations

Everything starts with the occupation rule for identical bosons. In equilibrium at temperature $T$ and chemical potential $\mu$ , a single-particle state of energy $\epsilon$ holds, on average, the number of atoms given by the Bose-Einstein distribution. There is no exclusion ceiling here: as $\mu$ approaches $\epsilon$ the denominator vanishes and the occupation diverges, which is exactly the loophole that lets one mode swallow a macroscopic population.

\bar n(\epsilon) = \frac{1}{e^{(\epsilon-\mu)/k_B T}-1} \qquad (1)

where: $\bar n(\epsilon)$ — mean occupation of a single-particle state of energy $\epsilon$ ; $\mu$ — chemical potential, constrained to $\mu<\epsilon_0$ (the ground-state energy) so every occupation stays positive; $k_BT$ — thermal energy scale. Physical reading: for $\epsilon-\mu\gg k_BT$ this reduces to the classical Boltzmann factor, but as $\mu\to\epsilon_0$ the ground-state occupation blows up — the seed of condensation. The quantum-versus-classical crossover is governed by a single length, the thermal de Broglie wavelength — the size of an atom's wavepacket at temperature $T$ . It grows as the gas cools, and when it reaches the mean spacing the gas stops behaving classically.

\lambda_{dB} = \frac{h}{\sqrt{2\pi m k_B T}} \qquad (2)

where: $m$ — atomic mass; $h$ — Planck constant; $k_BT$ — thermal energy. Physical reading: $\lambda_{dB}\propto T^{-1/2}$ , so halving the temperature swells the wavepacket by $\sqrt2$ ; heavy atoms and low temperatures both push toward degeneracy. Combining the occupation rule with the density of states in three dimensions yields the criterion for the onset of condensation. It is stated most cleanly as a threshold in phase-space density — the number of atoms per thermal volume $\lambda_{dB}^3$ — and it is a pure number, independent of the atom.

n\lambda_{dB}^3 = \zeta(3/2) \approx 2.612 \qquad (3)

where: $n$ — number density $N/V$ ; $\zeta(3/2)=2.6124\ldots$ — Riemann zeta function; $n\lambda_{dB}^3$ — phase-space density. Physical reading: condensation is not about any energy scale in isolation but about the dimensionless overlap of wavepackets; the magic number $2.612$ is a property of 3D Bose statistics, nothing more. Inverting the criterion at fixed density gives the transition temperature of a uniform gas. It rises with density and falls with mass, which is why a dense gas of light atoms condenses more easily than a sparse gas of heavy ones.

k_B T_c = \frac{2\pi\hbar^2}{m}\left[\frac{n}{\zeta(3/2)}\right]^{2/3} \qquad (4)

where: $T_c$ — BEC transition temperature at density $n$ ; $\hbar=h/2\pi$ . Physical reading: $T_c\propto n^{2/3}/m$ ; for a trapped gas the density of states differs and the corresponding result is $k_BT_c\approx0.94\,\hbar\bar\omega N^{1/3}$ with $\bar\omega$ the geometric-mean trap frequency (developed below). Below $T_c$ the split between condensate and thermal cloud follows a power law in temperature. In a uniform box the exponent is $3/2$ ; the whole condensate fraction is fixed once $T/T_c$ is known.

\frac{N_0}{N} = 1-\left(\frac{T}{T_c}\right)^{3/2}\ \text{(box)},\qquad \frac{N_0}{N} = 1-\left(\frac{T}{T_c}\right)^{3}\ \text{(harmonic trap)} \qquad (5)

where: $N_0$ — atoms in the condensate (ground state); $N$ — total atom number. Physical reading: the exponent is set by the density of states — $g(\epsilon)\propto\epsilon^{1/2}$ in a box versus $\propto\epsilon^2$ in a harmonic trap — so a trapped condensate grows faster on cooling and the whole gas has condensed by $T=0$ . At $T\approx0$ almost every atom shares one orbital, and the many-body physics collapses to a single nonlinear wave equation for that orbital: the time-dependent Gross-Pitaevskii equation. It is a Schrödinger equation with an extra term — each atom feels the mean field of all the others.

i\hbar\frac{\partial \psi}{\partial t} = \left[-\frac{\hbar^2}{2m}\nabla^2 + V_{\mathrm{ext}}(\mathbf{r}) + g|\psi|^2\right]\psi,\qquad g = \frac{4\pi\hbar^2 a_s}{m} \qquad (6)

where: $\psi(\mathbf{r},t)$ — condensate wavefunction, normalized to $\int|\psi|^2\,\mathrm d^3r=N$ ; $V_{\mathrm{ext}}$ — trap potential; $g$ — contact-interaction strength; $a_s$ — s-wave scattering length ( $a_s>0$ repulsive, $a_s<0$ attractive). Physical reading: the $g|\psi|^2$ term is what makes the condensate a superfluid; drop it and equation (6) is the ordinary linear Schrödinger equation for one particle. When repulsion dominates the kinetic energy, the kinetic term in the stationary equation can be dropped and the density profile is read off algebraically. This is the Thomas-Fermi approximation, valid whenever $Na_s/a_{ho}\gg1$ .

n(\mathbf{r}) = |\psi|^2 = \frac{\mu - V_{\mathrm{ext}}(\mathbf{r})}{g},\quad n\ge 0 \qquad (7)

where: $n(\mathbf{r})$ — condensate density; $\mu$ — chemical potential fixed by $\int n\,\mathrm d^3r=N$ ; $a_{ho}=\sqrt{\hbar/m\bar\omega}$ — harmonic-oscillator length. Physical reading: in a harmonic trap $V_{\mathrm{ext}}\propto r^2$ , so the density is a smooth inverted parabola that vanishes at the Thomas-Fermi radius — the tell-tale broad, flat-topped cloud of an interacting condensate.

Derivation ACritical temperature of the uniform ideal Bose gas

The goal is to earn the phase-space criterion $n\lambda_{dB}^3=\zeta(3/2)$ and the transition temperature $T_c$ from first principles, by demanding that the excited states saturate as the chemical potential rises to its ceiling. The strategy is to count the atoms the excited states can hold at the maximum allowed $\mu$ , treat the ground state separately as an uncounted reservoir, and recognize that when the total exceeds that excited-state capacity, the overflow has nowhere to go but the ground state.

Step 1. Write the 3D density of states for free particles in a box of volume $V$ , $g(\epsilon)=\frac{V}{4\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\epsilon^{1/2}$ . Why: the $\epsilon^{1/2}$ growth is exactly what makes the excited-state count converge in 3D — a lower power would let the excited states absorb any number of atoms and forbid condensation.

Step 2. Count the excited atoms as $N_{\mathrm{ex}}=\int_0^\infty g(\epsilon)\,\bar n(\epsilon)\,\mathrm d\epsilon$ , deliberately excluding the $\epsilon=0$ ground state. Why: the density of states gives the ground state zero weight, so it must be tracked by hand; it will hold $N_0$ separately, unbounded by this integral.

Step 3. Push $\mu$ to its ceiling ( $\mu\to0$ measured from the ground state), the point of maximum excited-state capacity, and evaluate

N_{\mathrm{ex}}^{\max}=\frac{V}{4\pi^2}\left(\frac{2mk_BT}{\hbar^2}\right)^{3/2}\Gamma(3/2)\,\zeta(3/2).

Why: substituting $x=\epsilon/k_BT$ turns the integral into $\int_0^\infty x^{1/2}/(e^x-1)\,\mathrm dx=\Gamma(3/2)\zeta(3/2)$ , a standard convergent value — the excited states have a finite maximum occupancy.

Step 4. Use $\Gamma(3/2)=\sqrt\pi/2$ to compress the prefactor into the thermal wavelength,

N_{\mathrm{ex}}^{\max}=V\,\zeta(3/2)\left(\frac{mk_BT}{2\pi\hbar^2}\right)^{3/2}=\frac{V}{\lambda_{dB}^3}\,\zeta(3/2).

Why: this is the punchline in disguise — the excited states hold exactly $\zeta(3/2)$ atoms per thermal cell of volume $\lambda_{dB}^3$ , no more.

Step 5. Set $N=N_{\mathrm{ex}}^{\max}$ at the threshold, giving $n=\zeta(3/2)/\lambda_{dB}^3$ , that is $n\lambda_{dB}^3=\zeta(3/2)$ ; invert with $\lambda_{dB}=h/\sqrt{2\pi mk_BT_c}$ to obtain $k_BT_c=\frac{2\pi\hbar^2}{m}[n/\zeta(3/2)]^{2/3}$ . Why: any atoms beyond $N_{\mathrm{ex}}^{\max}$ have no excited seat left and pile into the ground state as $N_0=N-N_{\mathrm{ex}}^{\max}$ — condensation, with the crossover fixed at $T_c$ .

Go deeper · Derivation B · 1 minStationary Gross-Pitaevskii equation from a variational many-body ansatz

Derivation BStationary Gross-Pitaevskii equation from a variational many-body ansatz

The goal is to reduce the intractable $N$ -body Schrödinger problem to a single nonlinear equation for one orbital, valid when one mode is macroscopically occupied. The strategy is variational: assume every atom sits in the same orbital, write the total energy as a functional of that orbital, and minimize it under fixed atom number. The Lagrange multiplier enforcing that constraint turns out to be the chemical potential, and the stationarity condition is the Gross-Pitaevskii equation itself.

Step 1. Take the product ansatz $\Psi(\mathbf r_1,\dots,\mathbf r_N)=\prod_{i=1}^N\phi(\mathbf r_i)$ and define the condensate wavefunction $\psi=\sqrt N\,\phi$ normalized to $\int|\psi|^2=N$ . Why: at $T\approx0$ one mode holds essentially all atoms, so a single shared orbital is an excellent starting point, and folding $\sqrt N$ into $\psi$ makes $|\psi|^2$ the physical density.

Step 2. Write the energy functional with a contact interaction,

E[\psi]=\int\!\left[\frac{\hbar^2}{2m}|\nabla\psi|^2+V_{\mathrm{ext}}|\psi|^2+\frac{g}{2}|\psi|^4\right]\mathrm d^3r.

Why: the three terms are kinetic, potential, and interaction energy; the factor $1/2$ on $|\psi|^4$ avoids double-counting each interacting pair.

Step 3. Minimize under fixed norm by introducing the Lagrange multiplier $\mu$ , requiring $\frac{\delta}{\delta\psi^*}\!\left(E-\mu\!\int|\psi|^2\,\mathrm d^3r\right)=0$ . Why: fixing $N$ is a constraint, and the multiplier that enforces it is thermodynamically the chemical potential — the energy cost of adding one atom.

Step 4. Carry out the functional derivative to obtain the stationary equation,

\left[-\frac{\hbar^2}{2m}\nabla^2+V_{\mathrm{ext}}(\mathbf r)+g|\psi|^2\right]\psi=\mu\,\psi.

Why: the result is a nonlinear Schrödinger equation in which the $g|\psi|^2$ term is the self-consistent mean field each atom feels from all the others — the entire content of weakly interacting condensate physics.

Go deeper · Derivation C · 1 minWhy the condensate fraction follows a power law in temperature

Derivation CWhy the condensate fraction follows a power law in temperature

The goal is to see where the exponents in $N_0/N=1-(T/T_c)^p$ come from, and why $p$ jumps from $3/2$ in a uniform box to $3$ in a harmonic trap. The strategy is to hold $\mu$ pinned at its ceiling for all $T<T_c$ — the defining feature of the condensed phase — and track only the temperature scaling of the saturated thermal population. The density of states does all the work, and the whole difference between box and trap is one power of energy.

Step 1. Below $T_c$ the chemical potential is stuck at the ground-state energy, so the thermal (excited) atom number is $N_{\mathrm{th}}(T)=\int g(\epsilon)\,\bar n(\epsilon)\,\mathrm d\epsilon$ evaluated at $\mu=0$ . Why: once the ground state is macroscopically occupied, $\mu$ cannot rise further, so every temperature below $T_c$ shares the same integrand shape and only the thermal scale changes.

Step 2. Rescale with $x=\epsilon/k_BT$ : for $g(\epsilon)\propto\epsilon^{s}$ the integral pulls out a factor $(k_BT)^{s+1}$ , so $N_{\mathrm{th}}\propto T^{s+1}$ . Why: the only $T$ -dependence left after rescaling lives in the prefactor, so the exponent is fixed entirely by the exponent $s$ of the density of states.

Step 3. Insert the two geometries. A uniform box has $s=1/2$ , giving $N_{\mathrm{th}}\propto T^{3/2}$ and $N_0/N=1-(T/T_c)^{3/2}$ ; a 3D harmonic trap has $g(\epsilon)=\epsilon^2/[2(\hbar\bar\omega)^3]$ , so $s=2$ , giving $N_{\mathrm{th}}\propto T^3$ and $N_0/N=1-(T/T_c)^3$ . Why: the steeper trap density of states loads more weight at high energy, so the thermal cloud empties faster on cooling and the condensate grows with the higher power.

Limiting cases

How the transition depends on geometry, dimensionality, and interactions:

Case	Result	Meaning
Uniform 3D ideal gas	$n\lambda_{dB}^3 = \zeta(3/2)\approx 2.612$	The canonical BEC criterion; a sharp transition below which the ground-state occupation becomes macroscopic.
Harmonic trap (3D)	$k_B T_c = \hbar\bar\omega\,[N/\zeta(3)]^{1/3}\approx 0.94\,\hbar\bar\omega\,N^{1/3}$	The trap's density of states raises the effective dimensionality; $\bar\omega=(\omega_x\omega_y\omega_z)^{1/3}$ , and $\zeta(3)=1.202$ .
Uniform 2D or 1D ideal gas	$T_c\to0$	No condensation at finite $T$ below 3D in a uniform gas; the excited-state integral diverges and thermal fluctuations destroy long-range order. A harmonic trap can restore a transition in 2D.
Weakly interacting (Thomas-Fermi)	$\mu = \tfrac{\hbar\bar\omega}{2}\left(15Na_s/a_{ho}\right)^{2/5}$	When $Na_s/a_{ho}\gg1$ interactions dominate the kinetic energy; $a_{ho}=\sqrt{\hbar/m\bar\omega}$ and the density is an inverted parabola.

Common trapBose-Einstein condensation happens because the cold atoms attract and clump together

It does not — BEC is a consequence of quantum statistics, not of any force. An ideal, non-interacting Bose gas condenses purely because the excited states saturate; there is no attraction anywhere in the derivation of $n\lambda_{dB}^3=2.612$ . Two further traps hide nearby. First, the sharp thermodynamic transition and the number $2.612$ are specific to three dimensions: a uniform ideal Bose gas does not condense at finite temperature in 1D or 2D, because the excited-state integral diverges as $\mu\to0$ and the excited states can absorb any number of atoms. A harmonic trap changes the density of states and can revive a transition in 2D. Second, a large occupation of the lowest mode is necessary but not sufficient to call it a condensate in the full sense — the defining property is a single coherent macroscopic phase, which is what interference between separate condensates actually demonstrates.

Worked examples

1 · The transition temperature of a rubidium cloud (confidence builder)

A gas of rubidium-87 (atomic mass $m\approx86.9\,\mathrm{u}$ ) is held at number density $n=10^{14}\,\mathrm{cm^{-3}}=10^{20}\,\mathrm{m^{-3}}$ . Find the BEC transition temperature, and confirm the thermal wavelength equals the interparticle spacing there.

Step 1. Use the uniform-gas result $k_BT_c=\frac{2\pi\hbar^2}{m}[n/\zeta(3/2)]^{2/3}$ . Assemble the prefactor: $2\pi\hbar^2/m=2\pi(1.0546\times10^{-34})^2/(1.443\times10^{-25})=4.84\times10^{-43}\ \mathrm{J\,m^2}$ .

Step 2. Evaluate the density factor with $\zeta(3/2)=2.612$ : $[n/\zeta(3/2)]^{2/3}=[10^{20}/2.612]^{2/3}=(3.83\times10^{19})^{2/3}=1.14\times10^{13}\ \mathrm{m^{-2}}$ .

Step 3. Multiply, then divide by $k_B$ : $k_BT_c=(4.84\times10^{-43})(1.14\times10^{13})=5.51\times10^{-30}\ \mathrm J$ , so $T_c=5.51\times10^{-30}/1.381\times10^{-23}=\mathbf{3.98\times10^{-7}\ K\approx0.40\ \mu K}$ .

Step 4. Cross-check the picture: at this $T_c$ the thermal wavelength is $\lambda_{dB}=h/\sqrt{2\pi mk_BT_c}=\mathbf{0.297\ \mu m}$ , while the interparticle spacing is $n^{-1/3}=(10^{20})^{-1/3}=\mathbf{0.215\ \mu m}$ . Their ratio cubed gives $n\lambda_{dB}^3=(0.297/0.215)^3\approx2.61$ , reproducing $\zeta(3/2)$ exactly — the wavepackets have just grown to overlap. T_c ≈ 0.40 μK, with λ_dB ≈ 0.30 μm ≈ the 0.22 μm spacing, so n·λ_dB³ ≈ 2.61.

2 · Transition temperature and condensate fraction in a trap

One million rubidium-87 atoms fill a harmonic trap with geometric-mean frequency $\bar\omega/2\pi=100\ \mathrm{Hz}$ . Find $T_c$ , then the condensate fraction at $T=0.5\,T_c$ .

Step 1. Use $k_BT_c=0.94\,\hbar\bar\omega N^{1/3}$ with $\bar\omega=2\pi\times100=628\ \mathrm{rad\,s^{-1}}$ and $N^{1/3}=(10^6)^{1/3}=100$ . Then $\hbar\bar\omega=1.0546\times10^{-34}\times628=6.63\times10^{-32}\ \mathrm J$ .

Step 2. So $k_BT_c=0.94\times6.63\times10^{-32}\times100=6.23\times10^{-30}\ \mathrm J$ , giving $T_c=6.23\times10^{-30}/1.381\times10^{-23}=\mathbf{4.51\times10^{-7}\ K\approx0.45\ \mu K}$ — the value marked on the second figure.

Step 3. For the trap, $N_0/N=1-(T/T_c)^3$ . At $T=0.5\,T_c$ this is $1-(0.5)^3=1-0.125=\mathbf{0.875}$ , so 87.5 percent of the atoms already sit in the condensate. Notice how much faster this rises than the uniform-box law $1-(0.5)^{3/2}=0.646$ at the same reduced temperature. T_c ≈ 0.45 μK, and at half of T_c the condensate already holds 87.5 percent of the atoms.

3 · The Thomas-Fermi chemical potential and the interaction regime

The same cloud has scattering length $a_s\approx5.3\ \mathrm{nm}$ (about $100\,a_0$ ). Check that it is deep in the Thomas-Fermi regime, and find the chemical potential.

Step 1. Compute the oscillator length $a_{ho}=\sqrt{\hbar/m\bar\omega}=\sqrt{1.0546\times10^{-34}/(1.443\times10^{-25}\times628)}=\mathbf{1.08\ \mu m}$ .

Step 2. Form the interaction parameter $Na_s/a_{ho}=10^6\times5.3\times10^{-9}/1.08\times10^{-6}=\mathbf{4.9\times10^{3}}$ , which is enormously larger than 1 — the kinetic term is negligible and Thomas-Fermi applies.

Step 3. Evaluate $\mu=\tfrac{\hbar\bar\omega}{2}(15Na_s/a_{ho})^{2/5}=\tfrac{6.63\times10^{-32}}{2}(15\times4907)^{2/5}=3.31\times10^{-32}\times88.5=2.93\times10^{-30}\ \mathrm J$ . In friendlier units $\mu/k_B\approx\mathbf{0.21\ \mu K}$ and $\mu/h\approx\mathbf{4.4\ kHz}$ , or about $44\,\hbar\bar\omega$ above the trap ground state — the interaction energy dwarfs the level spacing, confirming a broad inverted-parabola cloud. Na_s/a_ho ≈ 4900 ≫ 1, and μ ≈ 0.21 μK ≈ 44 ℏω̄, deep in the Thomas-Fermi regime.

Physics of the portrait

Bose-Einstein condensation is one of the very few macroscopic quantum phenomena predicted on paper decades before anyone could make it — and it carries two names because it took two people, working an ocean apart, to assemble the idea.

Origins · 1 minAlbert Einstein and how this was discovered

Albert Einstein

1879–1955 · predicted condensation into the ground state from quantum statistics

In June 1924 a young Dhaka lecturer, Satyendra Nath Bose, sent Einstein a manuscript rederiving Planck's radiation law by treating light quanta as indistinguishable and counting their states in a new way. Einstein saw at once that the trick was deeper than radiation: he translated the paper, secured its publication, and then in two papers of 1924 and 1925 extended Bose's counting from massless photons to a gas of atoms with mass and conserved number. The consequence startled him — below a critical temperature a finite fraction of the atoms would collapse into the single lowest-energy state, condensing not in real space but in momentum space, purely because of how identical bosons share states. The prediction lay almost dormant for a decade, until Fritz London revived it in 1938 as the mechanism behind superfluid helium-4. Einstein never saw a condensate; the dilute-gas realization came forty years after his death. But the phenomenon is his in the sense that matters: he took a statistical counting rule and drew from it a phase of matter no one had imagined — exactly the phase that Cornell, Wieman and Ketterle finally cooled into being in 1995.

History

1924

— Bose derives the quantum statistics of photons; Einstein translates and champions the paper

1925

— Einstein extends the counting to massive atoms and predicts condensation into the ground state

1938

— London links Bose-Einstein condensation to the superfluidity of liquid helium-4

1961

— Gross and Pitaevskii independently derive the mean-field equation for the condensate wavefunction

1995

— Cornell and Wieman condense rubidium-87 at JILA; Ketterle condenses sodium-23 at MIT

Going further

Texts. Foot, Atomic Physics, Ch. 10 (Bose-Einstein condensation) · Pethick and Smith, Bose-Einstein Condensation in Dilute Gases, Ch. 2 and 6 · Pitaevskii and Stringari, Bose-Einstein Condensation, Ch. 3 · Dalfovo, Giorgini, Pitaevskii and Stringari, Rev. Mod. Phys. 71, 463 (1999).

Provenance. Equations and numbers are standard textbook results, re-verified here. A checker should: (1) recompute $\zeta(3/2)=2.6124$ and confirm $n\lambda_{dB}^3=\zeta(3/2)$ (Foot Ch. 10; Pethick and Smith Ch. 2); (2) verify the harmonic-trap $k_BT_c=0.94\,\hbar\bar\omega N^{1/3}$ from $\zeta(3)=1.202$ (Pitaevskii and Stringari Ch. 3; Dalfovo et al. RMP 1999); (3) confirm the Gross-Pitaevskii derivation and $g=4\pi\hbar^2 a_s/m$ , and the Thomas-Fermi $\mu=\tfrac{\hbar\bar\omega}{2}(15Na_s/a_{ho})^{2/5}$ (Dalfovo et al.); (4) check the condensate-fraction exponents $3/2$ (box) and $3$ (trap); (5) reproduce the worked numbers $T_c\approx0.40\,\mu$ K at $n=10^{20}\,\mathrm{m^{-3}}$ and $T_c\approx0.45\,\mu$ K for $N=10^6$ at $\bar\omega/2\pi=100$ Hz. Historical facts: the 1924-25 Bose-Einstein prediction, the 1995 realizations (Anderson et al., Science 269, 198; Davis et al., PRL 75, 3969), and the 2001 Nobel Prize to Cornell, Wieman and Ketterle.

Prerequisites. Evaporative Cooling · Magnetic Trapping of Neutral Atoms · Doppler Cooling and the Doppler Limit · The Recoil Limit and Sub-Recoil Cooling. Leads to. Degenerate Fermi Gases and the BEC-BCS Crossover · Feshbach Resonances.

Ask the tutor: Ask me to walk through why $n\lambda_{dB}^3=2.612$ falls out of the excited-state integral, or to work a $T_c$ or condensate-fraction number for any atom, density, or trap you name.

Is a Bose-Einstein condensate the same thing as a superfluid?

Close relatives but not identical. Condensation is a statement about statistics — a macroscopic occupation of one single-particle state — and it happens even for an ideal gas with no interactions. Superfluidity, flow without viscosity and quantized vortices, requires interactions: it comes from the linear phonon spectrum that Bogoliubov theory gives a weakly repulsive condensate, which by Landau's criterion forbids low-energy dissipation. So an ideal-gas condensate is condensed but not truly superfluid, while a real atomic condensate is both. Liquid helium-4 is the reverse tale: it is strongly interacting, so only about ten percent of its atoms are in the condensate even at $T=0$ , yet the whole liquid is superfluid.

Why can't laser cooling alone reach the transition?

Laser cooling is limited by the photons it uses. The Doppler limit sets a floor of order $100\ \mu$ K, and even sub-recoil schemes bottom out near a fraction of a microkelvin while random photon scattering heats the gas and caps its density. But condensation needs $n\lambda_{dB}^3\approx2.6$ , meaning tens to hundreds of nanokelvin — a further factor of ten to a hundred colder, at higher phase-space density than photons can deliver. The fix is to switch off the light: load the cloud into a conservative magnetic or optical trap and remove the hottest atoms by forced evaporation, which trades atom number for phase-space density and carries the gas the last decade across $T_c$ .

What actually condenses — do the atoms fall to the bottom of the trap?

Not in the way the word suggests. The atoms condense in momentum space and in the quantum state they occupy, not into a puddle at the bottom of the container. Below $T_c$ a macroscopic number of them share the single lowest-energy mode of the trap, which in position space is a broad smooth cloud — for an interacting gas the inverted-parabola Thomas-Fermi profile — sitting in the middle of the trap, surrounded by a thinner thermal halo. What makes it visible is momentum: released from the trap, the condensate expands slowly because its velocity spread is tiny, while the thermal cloud flies apart. A time-of-flight image then shows a sharp dense central peak riding on a broad thermal background, the classic signature.

What sets whether a condensate is stable or collapses?

The sign of the scattering length. In $g=4\pi\hbar^2 a_s/m$ a positive $a_s$ means repulsion, which spreads the condensate out and stabilizes it — the case for rubidium-87 and sodium-23. A negative $a_s$ means attraction, and an attractive condensate is unstable above a critical atom number: the mean-field energy $\propto N|a_s|/a_{ho}$ overwhelms the zero-point energy that holds the cloud up, and it implodes in a burst known as a bosenova. This is exactly why the first lithium-7 condensates, which have $a_s<0$ , were limited to about a thousand atoms. Feshbach resonances let experimenters tune $a_s$ through zero to either sign with a magnetic field, turning the stability question into a control knob.

How do experiments prove they have a coherent condensate rather than just a cold cloud?

The decisive test is interference. Make two condensates independently, release them and let them overlap; a coherent macroscopic wavefunction produces high-contrast fringes with a spacing set by the relative momentum, exactly as two laser beams do, while a merely cold thermal cloud shows none because it has no single phase. Ketterle's group demonstrated this in 1997, and it remains the gold standard. Other signatures reinforce it: the bimodal profile in time-of-flight (a sharp condensate peak on a broad thermal pedestal), the anisotropic expansion of a cloud released from an anisotropic trap, and quantized vortices when the gas is stirred — each impossible for a classical gas.

Does the 2.612 criterion apply to atoms in a trap?

Not directly. The number $\zeta(3/2)=2.612$ is the peak phase-space density criterion for a uniform gas, where the density is the same everywhere. In a harmonic trap the gas is denser at the centre, and condensation begins when the phase-space density at the trap centre reaches $2.612$ — so locally the same physics holds, but the transition temperature must be expressed through the trap frequency and atom number instead of a uniform density. That is the origin of $k_BT_c\approx0.94\,\hbar\bar\omega N^{1/3}$ : integrating the local criterion over the trapped density profile, using the harmonic density of states $g(\epsilon)\propto\epsilon^2$ , yields the $\zeta(3)$ result. So the uniform criterion is not wrong for traps, it is applied point by point rather than globally.