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I was wondering why a dilute gas (e.g. rubidium) forms a BEC at low temperatures rather than a crystal. My (naive) reasoning goes as follows: The dominant interaction between two atoms at low temperatures is governed by the Van-der-Waals force which is well described by the Lennard-Jones-Potential $$V(r)=\frac{a}{r^{12}}-\frac{b}{r^6},$$ and usually a,b are chosen such that there is a minimum < 0. Now this seems to support bound states and I would expect all the atoms to form a crystal then. Below its melting point, Rb is cristalline indeed. However if we have a dilute gas of Rb and cool it sufficiently, a BEC is formed. This is a surprise to me because the cooling of the gas relies on interatomic interactions, so why do these interactions not produce a crystal? (Why do the atoms in the dilute gas not bunch together and form a little crystal?)

Which point am I missing?

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You are missing nothing. The thermodynamical ground-state of an interacting Bose-Einstein condensate is a crystal and not the BEC state -- see e.g. Ketterle et al: "Making, probing and understanding Bose-Einstein condensates". However, the gas is cooled very rapidly such that it is not in thermal equilibrium. Hence, if we keep a BEC in its trap and wait for some time, we observe that the atoms "are lost" due to three-body interactions. These three-body interactions are the first step towards the crystalline state.

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  • $\begingroup$ "very rapidly" makes it sound like it's quenched. it's not. It's just a metastable state so it's very slow to reach the true (solid) ground state. $\endgroup$ – SuperCiocia Jun 25 '20 at 19:02
  • $\begingroup$ A term such as "very rapidly" is only meaningful, if we compare it to a second time scale, which acts as reference. I hoped it would be clear from the context that this reference time scale is the 3-body relaxation time. $\endgroup$ – Semoi Jun 25 '20 at 19:13
  • $\begingroup$ Thank you very much for the nice answer! $\endgroup$ – Simon Jun 25 '20 at 19:49
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Just to rephrase what @Semoi said.

The true ground state is indeed a solid, a crystalline solid. In order to get there though, the atoms need to form bonds with each other. The ability to form bonds is enhanced if the atoms are closer together, i.e. if the gas is denser. Indeed, three-body interactions (needed to form a molecule, because two-body ones would not conserve momentum and energy) scale as $a^4 n^2$, where $n$ is the density and $a$ the scattering length (quantifying the interaction strength).

To get (and keep) a BEC, you want to delay the onset of molecular bonds and hence three-body interactions. So you want to operate in a regime of low $n$ and low $a$. The BEC is a metastable state, not the true equilibrium state.

Hence, you need weakly interacting (low $a$) and dilute (low $n$) gases.

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  • $\begingroup$ thank you very much! $\endgroup$ – Simon Jun 26 '20 at 8:27

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