# What happens to Boson and Fermi gases at very low temperature?

At low temperatures Fermi gases pile up to the state with the Fermi-energy having one particle in each state and Boson gases form Bose-Einstein condensates. However, the only derivations I have seen of this (I have looked for others) work in the continuum limit. In other words they say that: $$\bar N=\sum_i n_i =\int g(\varepsilon)n(\varepsilon)d\varepsilon$$ To do this, however, analogous to the case of an ideal gas we need to be in the limit: $$T\gt \gt \frac{\hbar^2}{mk_B V^{1/3}}$$ or something very similar. This puts a limit on how low the temperature an actually be before these standard derivations break down. My question is whether there is any change if we move below this limit in temperature? (if there is anything wrong with my reasoning please let me know).

• Where do you get this temperature criterion from ? To work in the continuum limit you just need to have a great number of states availiable, which is the case in e.g. a metal, even at very low temperatures. – Dimitri Apr 7 '16 at 16:34
• The point is that one takes the thermodynamic limit $V \to \infty$. The rationale behind this seemingly formal manipulation is that one wants to study a system that is so large that its precise volume is unimportant, i.e. there are no finite-size or boundary effects. The continuum limit is always valid so long as $V$ is large enough. For example, in a homogeneous and isotropic system you need $2\pi\hbar V^{-1/3}$ to be smaller than any experimentally resolvable momentum scale. – Mark Mitchison Apr 7 '16 at 16:40

The famous Fermi-Dirac and Bose-Einstein average occupations, $$\overline{n_i} = \frac{1}{e^{(\epsilon_i-\mu)/kT} \pm 1},$$ are only exact in the grand canonical ensemble (GCE) where the total particle number is a flexible (fluctuating) quantity. That flexibility, and the assumption of noninteracting particles, is what allows us to treat each single-particle state as statistically independent and derive the above result. The thermodynamic limit is not required.
In contrast if you fix total particle number $N$ to some exact, finite value, you are in canonical ensemble, in which case the above formula only holds approximately. The problem is that each state's $\overline{n_i}$ is not statistically independent of the others. BUT, if you take thermodynamic limit then the result converges to the GCE.
It turns out for metals the Fermi-Dirac formula is very valid in practice, since metals are almost always connected to some electrode which fixes the $\mu$, rather than having a fixed $N$. And so, nanotechnologists are comfortable to use Fermi-Dirac statistics even in tiny single-electron systems, like quantum dots, without trouble.
In contrast, Bose condensates in practice are not attached to a particle bath, so they cannot freely choose their particle number. In the GCE Bose gas, the relative variance in $N$ becomes enormous at low temperatures, in striking contrast to what actually happens if you reduce the temperature of a bag of bosons. (In fact, many argue that such a thing as "condensation" is infeasible in the GCE, rather one obtains an instability due to the large $N$ variance and extreme sensitivity to $\mu$.)