Bose-Einstein Condensation

Cool particles enough and they all collapse into the same quantum state — they stop being individuals and become one giant quantum object. This is the mechanism behind superfluidity, and the reason $\gamma = 2$.

When Atoms March in Lockstep

Imagine a parade. At high temperature, everyone is walking in random directions at random speeds — a chaotic crowd. Cool them down and they slow down. Cool them enough and something remarkable happens: everyone suddenly snaps into perfect lockstep. Same direction, same speed, same phase. They are no longer individuals — they are one collective entity.

This is Bose-Einstein condensation (BEC). Below a critical temperature, a macroscopic number of particles all drop into the exact same quantum state — the lowest-energy state available. They become a single quantum object that you can see with the naked eye (well, with a camera — the first BEC was about 10 micrometers across).

Why does this happen?

It is a consequence of quantum mechanics plus one rule: bosons are allowed to share a quantum state. (Fermions — electrons, protons, neutrons — are not. That is the Pauli exclusion principle.)

At high temperature, there are so many available states that sharing is rare — each particle occupies its own state. But as you cool the system, the number of thermally accessible states shrinks. Eventually, the particles have no choice but to pile into the ground state. This pile-up is BEC.

First predicted by Einstein in 1924 (building on Bose's work on photon statistics). First created in a lab in 1995 (Cornell, Wieman, and Ketterle — Nobel Prize 2001). Now routinely produced in hundreds of labs worldwide.

Bosons vs Fermions

All particles in the universe fall into two categories:

BEC is a boson phenomenon. Fermions can form BEC-like states only if they pair up first (Cooper pairs in superconductors, paired helium-3 atoms in superfluid $^3$He), because a pair of fermions is a composite boson.

Thermal de Broglie Wavelength

Every particle has a quantum wavelength that grows as it cools:

$$\lambda_{\text{dB}} = \frac{h}{\sqrt{2\pi m k_B T}}$$

At high $T$, $\lambda_{\text{dB}}$ is tiny — particles are well-separated point-like objects. As $T$ drops, $\lambda_{\text{dB}}$ grows. When it becomes comparable to the inter-particle spacing $n^{-1/3}$, the quantum waves of neighboring particles overlap. At that point, the particles can no longer be treated as distinguishable — quantum statistics takes over, and BEC occurs.

Critical Temperature

BEC sets in when $n \lambda_{\text{dB}}^3 \sim 1$, giving:

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

where $\zeta(3/2) \approx 2.612$ is a value of the Riemann zeta function. Below $T_c$, a macroscopic fraction of particles occupies the ground state.

Condensate Fraction

The fraction of particles in the ground state is:

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

At $T = 0$, all particles are condensed. At $T = T_c$, the condensate vanishes.

Contact Interactions and $\gamma = 2$

Real atoms interact. In a dilute BEC, the dominant interaction is short-range (contact) — characterized by a single number, the s-wave scattering length $a_s$. This gives a mean-field interaction energy per unit volume:

$$\mathcal{E}_{\text{int}} = \frac{g}{2}n^2, \quad g = \frac{4\pi\hbar^2 a_s}{m}$$

The pressure is then:

$$P = n\frac{\partial \mathcal{E}_{\text{int}}}{\partial n} - \mathcal{E}_{\text{int}} = \frac{g}{2}n^2 \propto \rho^2$$

This is a polytrope with $\gamma = 2$. Not fitted, not assumed — derived from the quantum mechanics of contact interactions.

Ideal Bose Gas: Partition Function

Grand canonical partition function

For non-interacting bosons, the mean occupation of single-particle state $\mathbf{k}$ with energy $\varepsilon_k$ is:

$$\langle n_\mathbf{k} \rangle = \frac{1}{e^{(\varepsilon_k - \mu)/k_BT} - 1}$$
This is the Bose-Einstein distribution. $\mu$ is the chemical potential (must be $\leq 0$ for bosons, since occupation numbers must be non-negative).
Total particle number

Sum over all states. Split the ground state ($\varepsilon_0 = 0$) from the excited states:

$$N = N_0 + \sum_{\mathbf{k} \neq 0} \frac{1}{e^{(\varepsilon_k - \mu)/k_BT} - 1}$$

In the thermodynamic limit, convert the sum to an integral over energy using the density of states $g(\varepsilon) = \frac{4\pi V}{h^3}(2m)^{3/2}\varepsilon^{1/2}$:

$$N_{\text{exc}} = \frac{V}{\lambda_{\text{dB}}^3} \, g_{3/2}(z)$$
where $z = e^{\mu/k_BT}$ is the fugacity and $g_{3/2}(z) = \sum_{l=1}^\infty z^l / l^{3/2}$ is the polylogarithm.
Critical temperature

BEC occurs when $\mu \to 0$ ($z \to 1$) and the excited states can no longer accommodate all particles. At $T_c$:

$$n = \frac{N}{V} = \frac{g_{3/2}(1)}{\lambda_{\text{dB}}^3(T_c)} = \frac{\zeta(3/2)}{\lambda_{\text{dB}}^3(T_c)}$$

Solving for $T_c$:

$$T_c = \frac{2\pi\hbar^2}{mk_B}\left(\frac{n}{\zeta(3/2)}\right)^{2/3}$$
$\zeta(3/2) = g_{3/2}(1) \approx 2.612$. For rubidium-87 at $n \sim 10^{14}$ cm$^{-3}$: $T_c \sim 100$ nK.

Mean-Field Interaction and Equation of State

Scattering length and coupling

Two bosons interacting via a short-range potential can be characterized by a single parameter — the s-wave scattering length $a_s$. The effective coupling constant is:

$$g = \frac{4\pi\hbar^2 a_s}{m}$$
This is exact in the Born approximation for $k a_s \ll 1$ (dilute limit). $a_s > 0$ = repulsive; $a_s < 0$ = attractive.
Mean-field energy

Each particle interacts with the local density $n$ of all other particles. The interaction energy density is:

$$\mathcal{E}_{\text{int}} = \frac{g}{2}n^2 = \frac{2\pi\hbar^2 a_s}{m}n^2$$
The factor $1/2$ avoids double-counting pairs.
Pressure

Using $P = n(\partial \mathcal{E}/\partial n) - \mathcal{E}$ (or equivalently $P = -\partial F / \partial V$ at fixed $N$):

$$P = \frac{g}{2}n^2 = \frac{2\pi\hbar^2 a_s}{m}n^2$$
Polytropic form

Since $n = \rho/m$:

$$P = \frac{2\pi\hbar^2 a_s}{m^3}\rho^2 = K\rho^2$$

This is a polytrope with $\gamma = 2$.

Key Results

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

Coupling constant: $g = 4\pi\hbar^2 a_s / m$

Equation of state: $P = \frac{g}{2m^2}\rho^2 \implies \gamma = 2$

Expert Notes

Feshbach Resonances

The scattering length $a_s$ is not a fixed property of a particle species — it can be tuned using an external magnetic field near a Feshbach resonance. At the resonance, $a_s \to \pm\infty$, giving access to the strongly interacting regime. This allows experimentalists to explore the full range from weak ($na_s^3 \ll 1$) to strong ($na_s^3 \sim 1$) interactions in a single apparatus.

At a Feshbach resonance, the mean-field (GP) description breaks down and beyond-mean-field effects become dominant.

Beyond Mean-Field: Lee-Huang-Yang Correction

The mean-field energy is the leading term in an expansion in the gas parameter $\sqrt{na_s^3}$. The next correction (Lee-Huang-Yang, 1957) gives:

$$\mathcal{E} = \frac{g n^2}{2}\left(1 + \frac{128}{15\sqrt{\pi}}\sqrt{na_s^3} + \cdots\right)$$

This modifies the equation of state: $\gamma$ is no longer exactly 2, but acquires corrections of order $\sqrt{na_s^3}$. For dilute condensates ($na_s^3 \ll 1$), these corrections are negligible.

Cosmological BEC vs Laboratory BEC

The UFC superfluid is a BEC, but it differs from lab condensates in fundamental ways:

The relativistic treatment replaces $g = 4\pi\hbar^2 a_s/m$ with a quartic self-coupling $\lambda$ in the Lagrangian $\mathcal{L} \supset -\lambda|\phi|^4/4$. The equation of state in the ultrarelativistic limit gives $c_s = c/\sqrt{3}$ — the same as radiation, but from a completely different mechanism.