# Where is the gauge symmetry in an ideal Bose gas?

It seems in the literature that there is a certain notion of a “macroscopic wavefunction” associated with a Bose-Einstein system (see this PSE answer) which exhibits a global $$U(1)$$-phase symmetry. After the BEC phase transition, this symmetry is supposedly broken corresponding to the sudden vanishing of the chemical potential (which is possibly an order parameter for that symmetry). There is another PSE answer which suggested I should probably read the second chapter of Quantum Liquids by A. J. Leggett to understand this gauge-symmetric aspect of Bose Einstein condensation, which I tried but find very different from the intuition I have been given.

For a background on how I have understood Bose Einstein condensation, please read the section “Background” below. I do not find any place in the language I am familiar with to describe anything like gauge symmetry and that the chemical potential tracks that symmetry.

Can somebody explain to me how gauge-theoretic perspectives about the BEC can be developed from the background I have?

Thank you!

### Background:

The Bose distribution is defined by average occupation numbers, $$n_p \equiv (e^{\beta(\varepsilon_p - \mu)}-1)^{-1}$$, of single-particle states $$|p\rangle$$ with energies $$\varepsilon_p = p^2/2m$$, such that $$\sum_p n_p = N$$ and $$\mu< \min_p \varepsilon_p = 0$$ (in order to ensure that $$n_p \ge 0$$).

The partition function is given by

$$\mathcal Z = \sum_{N=0}^\infty \ \sum_{\{n_p\ |\sum n_p=N\}} \exp\left(-\beta\sum_p(\varepsilon_p-\mu)n_p \right) \ = \prod_p\sum_{n_p=0}^\infty e^{-\beta(\varepsilon_p-\mu)n_p} = \prod_p \frac{1}{1-e^{-\beta(\varepsilon_p-\mu)}}\,.$$

The grand potential is $$\Phi = -\beta^{-1}\log\mathcal Z$$, and the average particle number is

$$N \equiv -\left( \frac{\partial \Phi}{\partial \mu}\right) = \sum_pn_p \approx \frac{V}{(2\pi\hbar)^3}\int d^3p\ n_p = \frac{V}{\lambda^3}Li_{3/2}\left(e^{\beta\mu}\right) \equiv N_{\text{excited}}\,,$$

where the sum is approximated by an integral assuming the large momentum states are occupied more significantly. $$V$$ is the volume occupied by $$N$$ particles, thus defining a specific volume $$v = V/N$$. Here, $$\lambda = \hbar\sqrt{2\pi\beta/m}$$ is the so-called thermal wavelength. In terms of the specific volume $$v$$, we can find the dependence of the chemical potential on the temperature with constant $$v$$.

$$\frac{\mu(T)}{kT_c} = \begin{cases} \sim 0 &\text{for } TT_c\,. \end{cases}$$

As you see below, at $$T=T_c(v)=2\pi\hbar^2m^{-1}(\zeta(3/2)v)^{-2/3}$$, a phase transition occurs.

$$\hspace{4cm}$$

If $$N_0$$ is the number of bosons in the ground state, and $$N = N_0 + N_{\text{excited}}$$, then in the thermodynamic limit,

$$\lim_{\substack{N\to\infty \\ v \text{ fixed} }} \frac{N_0}{N} = \begin{cases} 1-(T/T_c)^{3/2} &\text{for } T < T_c\,,\\ 0 &\text{for } T > T_c\,. \end{cases}$$

This is Bose Einstein Condensation.

• Are you familiar with the conection between phase transitions and symmetry breaking more generally? (Landau theory, order parameters, renormalization, etc.) – By Symmetry Jan 24 at 15:26
• @BySymmetry Yes. Are you going to write an answer? Thanks. – Nanashi No Gombe Jan 28 at 10:17