Bose-Einstein Condensation Derived

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 rather than surveyed.

Fixing the fugacity

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

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

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

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

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

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

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.

Why the ground state must be separated

The continuum integral is blind to the condensate. Because $g(\epsilon )\propto {\epsilon }^{1/2}$ vanishes at $\epsilon =0$, the ground state carries zero weight in $N_{\mathrm{exc}}$, and no value of $z$ makes the integral exceed $2.612V/{\lambda }^3$. 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,

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 $2.612V/{\lambda }^3$ 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 $T_c$.

Critical temperature and condensate fraction

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

Writing ${\lambda }_c=h/\sqrt{2\pi m{k}_B{T}_c}$ and solving for the temperature,

Below $T_c$ the excited population is $N_{\mathrm{exc}}=2.612V/{\lambda }^3$ 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.

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.

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 $\epsilon \propto p^2$, the density of states scales as $g(\epsilon )\propto {\epsilon }^{d/2-1}$, 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 $\epsilon =\hslash \omega (n_x+n_y+n_z)$ gives $g(\epsilon )\propto {\epsilon }^2$, and the excited count saturates at $\zeta (3)(k_{B}T/\hslash \omega {)}^3$. The critical temperature is

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.³

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 $2.612V/{\lambda }^3$ particles; below $T_c$ the surplus condenses into the ground state, which the ${\epsilon }^{1/2}$ continuum integral omits and which must be added by hand.
• The transition occurs at $n{\lambda }_c^3=\zeta (3/2)$, giving $k_B{T}_c=(2\pi {\hslash }^2/m)(n/\zeta (3/2){)}^{2/3}$ 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 $1-(T/T_c{)}^3$.