Quantum Mechanics – Why Rubidium-87 is Used for Bose-Einstein Condensates

atomic-physicsbose-einstein-condensatequantum mechanics

Rubidium-87 was the first and the most popular atom for making Bose–Einstein condensates in dilute atomic gases. Even though rubidium-85 is more abundant, rubidium-87 has a positive scattering length, which means it is mutually repulsive, at low temperatures. This prevents a collapse of all but the smallest condensates. It is also easy to evaporatively cool, with a consistent strong mutual scattering.

I am confused by the above paragraph, from Wikipedia's article "Isotopes of rubidium".

It says that Rubidium-87 atoms are mutually repulsive, which prevents a collapse……

But isn't a 'collapse' exactly what you want from a BEC? You DON'T want want them to be repulsive, but to come together in one single quantum state, unlike a fermion?

P.S.: Does Rb-87, with an odd number of both protons and electrons (albeit an even number of neutrons) only 'become' a boson when paired up with another Rb-87 atom? Or does the odd number of electrons and odd number of protons within a single atom like Rb-87 'cancel out' or add up' to an even number, thereby making that individual atom a boson?

I'm still confused about which isotopes are considered bosons….

Can ANY isotope be or become a boson?

Best Answer

For a BEC, you want atoms to be in the same quantum state, not necessarily at the same position.

For a BEC, the temperature is low enough so that the de Broglie wavelength $\lambda_{\mathrm{dB}} \propto 1/\sqrt{T}$ is larger than the interatomic spacing $\propto n^{-1/3}$, $n$ being the density. This means that the wave nature of the atoms is large enough for it to be felt by other atoms, in other words atoms "see" each other even without exactly sitting on top of each other. This is just to further justify the claim that you don't need atoms at the same position. Actually, if you had a perfect box potential of side $L$, and you reached BEC, then the atoms will macroscopically occupy the ground state $ |\Psi|^2 \propto \sin^2(x/L)$ which is very much extended. If you let $L\rightarrow \infty$, the atomic distribution becomes flat. So, again, very much atoms not at the same positions.

Ok, so now interactions and collapse.

First of all, BEC is a non-interacting effect. It is not driven by a competition of interaction terms, but solely by Bose-Einstein statistics. It is experimentally interesting that BEC seems to exist also in interacting systems, though there is no general theoretical proof. By BEC in an interacting system I mean macroscopic occupation of the ground state + Off-Diagonal Long-Range Order (ODLRO) — so not all superfluids are BECs. Let me also point out that you need interactions to reach a BEC as you need to reach thermal equilibrium.

The interaction strength among weakly interacting Bose-condensed bosons is quantified by a $g n$ term in the Hamiltonian, where $g$ is $4\pi\hbar^2 a/m$ (Gross-Pitaevski equation). You can make this interaction attractive with $a<0$ and repulsive with $a>0$, where $a$ is the scattering length and it is given by $a(B) = a_0 f(B)$, where $a_0$ is the background scattering length in the presence on no external magnetic field $B$ ($f$ is some function).

The pressure of a weakly interacting Bose-condensed gas is (at $T=0$): $$ P = -\frac{\partial E}{\partial V} = \frac{1}{2}gn^2.$$

Because $n^2$ is always positive, the condition for stability (i.e. not to collapse) is $P>0$ and hence $g>0 \Rightarrow a>0$ i.e. a repulsive system. With a positive pressure, the gas expands until it hits a wall (e.g. the confining potential). But if $P<0$ then the system is intrinsically unstable and collapses.

Rb-87 is "easy" because its background scattering length is positive and therefore trivially allows for a stable BEC. K-39, on the other hand, has a negative background scattering length so its "BEC" would collapse (and eventually explode). But its scattering length can be made repulsive by the use of a Feshbach resonance (applying a field $B$ to change $a$) so that it can undergo BEC.

Related Question