I am reading the introduction of The mathematics of the Bose gas and its condensation by E. Lieb, R. Seiringer, J. Solovej and J. Yngavason. The authors explain how, in the free case, the density in the thermodynamic limit becomes: $$\lim_{L\to \infty}\frac{\langle N\rangle}{L^{3}} \equiv \rho =\int_{\mathbb{R}^{3}}\frac{1}{e^{\beta(\frac{|p|^{2}}{2m}-\mu)}-1}dp \tag{1}\label{1}$$ which is a monotonic increasing function of $\mu$ which attains its maximum at $\mu \to 0^{-}$. In this limit, the density becomes critical $\rho_{c}(\beta)$.

It is clear to me that the boundedness of the density is an absurd and the problem lies in taking the limit $L\to \infty$ and then taking $\mu\to 0^{-}$. However, there is an observation in the paper which I don't quite follow. It goes like this:

This phenomenon was discovered by Einstein, and the resolution is that the particles exceeding the critical number all go into the lowest energy state. In mathematical terms, this means that we have to take the limit $\mu \to 0^{-}$ simultaneously with $L \to \infty$ to fix the density at some number $> \rho_{c}$.

I understand that the whole point of the Bose-Einstein condensation is that, at low temperatures, a huge number of particles fall into the same quantum state. However, I do not understand how this follows from the formulae presented before. Why particles falling into the lowest energy state solves the problem that (\ref{1}) is bounded? In other words, can we see the necessity of this condition right from (\ref{1})? And why is this equivalent to saying that both limits $\mu \to 0$ and $L \to \infty$ have to be taken simultaneously?


1 Answer 1


The occupancy $\frac{\langle N\rangle}{L^{3}}$ needs to always be positive to be physical. That means that your $e^{\beta(\frac{|p|^{2}}{2m}-\mu)}$ needs to always be $\geq 1$. Hence, for a fixed temperature and energy range ($\propto p^2$), the occupancy is fixed and bounded.

Only one state, the one where the exponent is such that the denominator is 0 ($p^2/2m = \mu$) has infinite occupancy. This is the ground state.

So, if you particle number at temperature $T$ exceeds the fixed occupancy, it can only be "fit" in the energy level with infinite occupancy, the ground state.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.