# The Heisenberg Uncertainty in Bose Einstein condensates

What happens to the Heisenberg uncertainty principle, when a system reaches the Bose-Einstein condensed state?

In our statistical mechanics lecture, we derived the following formula for the fraction of ideal gas bosons existing in the condensed state: $$\eta = 1 - \frac{\zeta(\frac{3}{2})v}{\lambda_\beta^3}$$ with $$\zeta: \text{Zeta function}, v: \text{volume per particle}, \lambda_\beta: \text{thermal deBroglie Wavelength}$$.

So, if you consider the Heisenberg uncertainty $$\Delta x \Delta p \geq \hbar$$, and shrink the volume, the momentum uncertainty should grow. But because of the condensation, more and more particles populate the lowest momentum state, so $$\Delta p$$ should shrink.

I found a paper Observing properties of an interacting homogeneous Bose-Einstein condesate: Heisenberg-limited momentum spread, interaction energy and free-expansion dynamics; Gotlibovych, et. al.; ArXiv, where the team produces a completely coherent Bose-Einstein-condensate, so the problem seems to exist in practice. But what happens, if they tried to cool down the system even more, or shrink the volume?

• A real condensate will always be localized in an external potential. Therefore the ground state will have a finite momentum width (which is connected to the system volume). Note, that for the idealized case of a $p = 0$ ground state, there is no way of reducing the volume of the system (because the momentum eigenstates will span the whole space). – Sebastian Riese Jan 6 '16 at 9:09
• $\Delta p$ is not impulse or uncertainty in impulse, it's the uncertainty in momentum. – Johnathan Gross Dec 27 '17 at 5:34
• Sorry, that was a language issue, because in German 'momentum' is 'Impuls'. – Petroglyph Dec 27 '17 at 13:41