Bose-Einstein condensation — concept

WHY does BEC happen at all?

WHAT makes bosons special? They obey Bose–Einstein statistics: any number of bosons may occupy the same single-particle state. The mean occupation of a state with energy $\varepsilon$ is

$\bar n(\varepsilon) = \frac{1}{e^{(\varepsilon-\mu)/k_BT}-1}.$

HOW do we constrain $\mu$? Occupations must be non-negative, so $e^{(\varepsilon-\mu)/k_BT} > 1$ for all states. The lowest energy is $\varepsilon_0 = 0$, so we need

$\mu \le 0 \quad\text{(for bosons, with ground state at }0).$

The ground-state occupation is $N_0 = \frac{1}{e^{-\mu/k_BT}-1}.$ As $\mu \to 0^-$, $N_0 \to \infty$. This is the escape valve: when the excited states saturate, $\mu$ pins itself just below $0$ and the ground state quietly soaks up the rest.

Deriving the saturation number (from scratch)

Step 1 — Count states. Sum over excited states → integral with density of states. Why this step? For a large box the levels are dense; replacing the sum by an integral is the continuum approximation. The catch (remember it!): the integral misses the $\varepsilon=0$ state because the density of states $g(\varepsilon)\propto\sqrt\varepsilon$ vanishes there — so we must track $N_0$ separately.

The density of states (derived from counting $k$-shells, $N(k)=\frac{V}{(2\pi)^3}\cdot\frac{4}{3}\pi k^3 \cdot g$): $g(\varepsilon)\,d\varepsilon = \frac{gV}{4\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\sqrt{\varepsilon}\,d\varepsilon.$

Step 2 — Maximum excited population occurs at $\mu = 0$: $N_{\text{exc}}^{\max}(T)=\int_0^\infty \frac{g(\varepsilon)\,d\varepsilon}{e^{\varepsilon/k_BT}-1}.$

Step 3 — Substitute $x=\varepsilon/k_BT$. Why? To pull all $T$-dependence out front and expose a pure number. $N_{\text{exc}}^{\max}=\frac{gV}{4\pi^2}\left(\frac{2mk_BT}{\hbar^2}\right)^{3/2}\underbrace{\int_0^\infty\frac{\sqrt{x}\,dx}{e^x-1}}_{\Gamma(3/2)\,\zeta(3/2)}.$

Using $\Gamma(3/2)=\tfrac{\sqrt\pi}{2}$ and the thermal de Broglie wavelength $\lambda_T=\dfrac{h}{\sqrt{2\pi m k_BT}}$, this collapses beautifully to

WHAT it means: the excited states can hold at most $\sim g\,\zeta(3/2)$ particles per thermal volume $\lambda_T^3$. Push the real density higher (or cool $T$ so $\lambda_T$ grows) and the rest condense.

The transition temperature

Set $n_{\text{exc}}^{\max}(T_c)=n$ (the actual density) — the point where excited states are just full: $n=g\frac{\zeta(3/2)}{\lambda_{T_c}^3}\;\Rightarrow\; \lambda_{T_c}^3=\frac{g\,\zeta(3/2)}{n}.$

Solving for $T$ (using $\lambda_T\propto T^{-1/2}$):

Condensate fraction below $T_c$

For $T<T_c$, $\mu\approx 0$, so $N_{\text{exc}}(T)\propto T^{3/2}$. Compare to its value at $T_c$ (where $N_{\text{exc}}=N$): $\frac{N_{\text{exc}}(T)}{N}=\left(\frac{T}{T_c}\right)^{3/2}.$ Everything else is in the ground state:

Worked examples

Common mistakes (Steel-manned)

Active recall

Connections

• Bose-Einstein statistics — the occupation law that makes condensation possible.
• Thermal de Broglie wavelength — $\lambda_T=h/\sqrt{2\pi mk_BT}$ sets the overlap criterion.
• Density of states — its $\sqrt\varepsilon$ form causes the ground-state subtlety.
• Chemical potential — pinning at $0^-$ is the mechanism of condensation.
• Riemann zeta function — $\zeta(3/2)$ appears from the saturation integral.
• Phase transitions — BEC as a continuous transition with no latent heat from interactions.
• Superfluidity — BEC underlies superfluid $^4$He and atomic condensates.
• Fermi-Dirac statistics — contrast: Pauli exclusion forbids condensation for fermions.

Concept Map

Hinglish (regional understanding)