Bose-Einstein Condensation

5.1 Ideal Bose Gas

For bosons, the average occupation of a single-particle state of energy $\varepsilon$ is

$\langle n_\varepsilon \rangle = \frac{1}{e^{\beta(\varepsilon - \mu)} - 1}$

The chemical potential must satisfy $\mu \leq \varepsilon_0$ (the lowest single-particle energy) to prevent negative occupation numbers.

5.2 Density of States and Critical Temperature

For a 3D free Bose gas with $\varepsilon = \hbar^2 k^2 / (2m)$The density of states is $g(\varepsilon) = (V/4\pi^2)(2m/\hbar^2)^{3/2}\sqrt{\varepsilon}$. The number of particles in excited states ($\varepsilon > 0$) is

$N_{\mathrm{ex} = \int_0^\infty \frac{g(\varepsilon)\, d\varepsilon}{e^{\beta \varepsilon} - 1} = V\left(\frac{mk_BT}{2\pi\hbar^2}\right)^{3/2}\,\zeta\!\left(\frac{3}{2}\right)}$

Where $\zeta(3/2) \approx 2.612$ is the Riemann zeta function.

Theorem 5.1 (BEC critical temperature). The maximum number of particles that can be accommodated in excited states is achieved at $\mu = 0$. When $N$ exceeds this maximum, the excess condenses into the ground state. The critical temperature is

$T_c = \frac{2\pi\hbar^2}{mk_B}\left(\frac{n}{\zeta(3/2)}\right)^{2/3}$

Where $n = N/V$.

Proof. Setting $N = N_{\mathrm{ex}^{\max}}$ at $\mu = 0$ and solving for $T$:

$n = \left(\frac{mk_B T_c}{2\pi\hbar^2}\right)^{3/2}\,\zeta\!\left(\frac{3}{2}\right)$

$T_c = \frac{2\pi\hbar^2}{mk_B}\left(\frac{n}{\zeta(3/2)}\right)^{2/3} \qquad \blacksquare$

5.3 Condensate Fraction

Below $T_c$, $\mu \approx 0$ and the condensate fraction is

$\frac{N_0}{N} = 1 - \left(\frac{T}{T_c}\right)^{3/2}$

This follows from $N_0 = N - N_{\mathrm{ex}}$ with $\mu = 0$:

$N_{\mathrm{ex} = N\left(\frac{T}{T_c}\right)^{3/2}}$

5.4 Thermodynamic Properties below $T_c$

The energy below $T_c$:

$U = \int_0^\infty \frac{\varepsilon\, g(\varepsilon)\, d\varepsilon}{e^{\beta\varepsilon} - 1} = V\left(\frac{mk_BT}{2\pi\hbar^2}\right)^{3/2}\,(k_BT)\,\frac{3}{2}\,\zeta\!\left(\frac{5}{2}\right) \cdot \Gamma\!\left(\frac{5}{2}\right)$

$= \frac{3}{2}\,Nk_BT_c\,\zeta\!\left(\frac{5}{2}\right)\Big/\zeta\!\left(\frac{3}{2}\right)\,\left(\frac{T}{T_c}\right)^{5/2}$

The heat capacity:

$C_V = \frac{15}{4}\,Nk_B\,\zeta\!\left(\frac{5}{2}\right)\Big/\zeta\!\left(\frac{3}{2}\right)\,\left(\frac{T}{T_c}\right)^{3/2} \propto T^{3/2}$

This contrasts with the constant $C_V = \frac{3}{2}Nk_B$ above $T_c$ (equipartition). There is a cusp (discontinuity in the derivative) at $T_c$Characteristic of a phase transition.

5.5 Worked Example: BEC in Rubidium-87

Problem. Estimate $T_c$ for a gas of $N = 10^4$ rubidium-87 atoms confined in a harmonic trap with frequency $\omega_{\mathrm{ho} = 2\pi \times 100}$ Hz.