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?

  • $\begingroup$ Also, how can an atom with odd numbers be a boson? Like rubidium-87? $\endgroup$
    – Kurt Hikes
    Sep 22, 2020 at 7:16
  • $\begingroup$ For your last question, an odd number of nucleons + one valence electron makes a boson with even total spin. So, for example, 39 K and 7 Li are bosons, while 40 K and 6 Li are fermions. $\endgroup$
    – Rococo
    Mar 21, 2021 at 19:05
  • $\begingroup$ @Rococo Why is it only number of nucleons plus the one valence electron? Why isn't number of nucleons plus total number of electrons? $\endgroup$
    – user196574
    Jun 15, 2021 at 20:27
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    $\begingroup$ @user196574 Sorry, it was slightly sloppy wording. All the electrons do formally contribute in the same way to the bosonic/fermionic nature, but there are always an even number in filled shells. So if you only remember the number of valence electrons (which is often all you bother to remember, since they determine the chemical properties), that is enough along with the nucleons to know if it is a boson or fermion. $\endgroup$
    – Rococo
    Jun 16, 2021 at 14:44

2 Answers 2


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.


I'll add a couple notes to the nice answer by @SuperCiocia.


Regarding attractive vs. repulsive interactions. Your original intuition that you would want attractive interactions for a BEC is understandable. You want the atoms to be very cold and densely formed so that they bose condense. Surely attractive interactions would bring the atoms closer? This chain of reasoning is incorrect. As SuperCiocia points out, the BEC transition is a non-interacting effect. If you had a non-interacting gas of bosons which was cold and dense enough it would form a BEC. Of course, again as SuperCiocia points out, in practice interactions are required for thermalization but that is a detail from the perspective of the phase transition.

So that said, we should understand that BEC doesn't happen because "the atoms are all attracted to eachother in a clump"*. Once we rid ourselves of this misconception we can then ask how does the introduction of interactions into the problem change things? The answer is short.

Strong attractive interactions cause the atoms to violently fly towards eachother resulting in inelastic collisions in which atoms can gain so much energy that they are ejected from the trap which is holding the atoms. The dynamics in light of attractive interactions is that you will just see rapid atom loss and no condensation. This is the collapse of the condensate.

What about repulsive interactions? A BEC can survive in spite of repulsive interactions. The atoms will be a little further apart than they would be without interactions but much of the major physics is unchanged.

So you should think like this: 1) theoretically our starting point is always a non-interacting BEC. Then we add in interactions. 2) If the interactions are attractive we get collapse. 3) If the interactions are repulsive things are changed/renormalized slightly but much of the essential physics is unchanged.

Why Rb-87?

Rb-87 was more attractive for initial BEC than Rb-85 because Rb-87 supports an accessible cycling transition which could be used with early magneto optical trap and optical molasses technologies for laser cooling. Rb-85 does not support such a transition so more sophisticated laser cooling stages would have been necessary for the initial stages of cooling towards BEC. That is to say, Rb-87 probably wasn't specially chosen for its natural abundance

Rb-85 has attractive interactions and Rb-87 has repulsive interactions. For the reasons described above, this makes Rb-87 more favorable for realizing a BEC than Rb-85. Why does Rb-85 have attractive interactions and Rb-87 has repulsive? As far as I can tell the sign of the scattering length depends on the exact position of the continuum relative to the last atom-atom bound state, and whether the scattering length is positive or negative is basically 50/50 random chance for any given isotope. That is, it just happens to be the case that Rb-87 is more favorable with its repulsive interactions. The choice had nothing to do with natural abundance.

Why Rb among all other atoms? I can't speak much to this since I've mostly worked with Rb so far but I can point out that BECs of Na (for which the nobel prize was shared) and Li were formed shortly after the Rb BEC so I don't think we should infer something especially unique about Rb compared to other elements that it was the first to be condensed.

All of that said, I think your main question was really a confusion about interactions and BEC collapse which I think has been answered by now. It turns out that Rb has the right sign for interactions so that was helpful for it historic condensation.

*The question of why BEC does happen I will leave for you to research on your own or ask another question. The short story is that it is a thermodynamic transition that depends essentially on the fact that you have bosonic statistics and the density of states for the system.

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    $\begingroup$ Interestingly, there is a phase of matter well characterised by "the atoms are all attracted to eachother in a clump" at that temperature - it's called a solid. (In other words, most of the work in creating a BEC is to keep the cloud at a low enough density to stop it from condensing; repulsive interactions are a large part of that.) $\endgroup$ Aug 2, 2020 at 20:35
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    $\begingroup$ I can't confirm this, but I've heard that the reason that rubidium was the first element to be be condensed is that it happens to have a transition at a frequency very close to the frequency of the lasers used in CD-ROM drives, so there were already lots of cheap and reasonably high-quality lasers available near the right frequency for rubidium. $\endgroup$
    – tparker
    Mar 20, 2021 at 3:33
  • $\begingroup$ @tparker I've heard that also. But I'm curious if there is a longer list of more substantial reasons why Rb has persisted to have a reputation as "easy to work with" even as laser technology has improved. I'd be curious for a more thorough answer to this question which compares all of the Alkali atoms in terms of level structure, mass etc. $\endgroup$
    – Jagerber48
    Mar 20, 2021 at 3:58
  • $\begingroup$ In addition, 87 Rb is one of the most effective atoms for sub-Doppler cooling, which helps with getting a high phase-space density starting point for evaporation. This comes back to its electronic structure; for example the small excited-state hyperfine splitting. $\endgroup$
    – Rococo
    Mar 21, 2021 at 19:17
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    $\begingroup$ @flevinBombastus now that I look into that part of the answer is just plain wrong. 85Rb has a cycling transition on F=3 -> F’=4. I’ll correct the answer when I get the chance. $\endgroup$
    – Jagerber48
    Sep 11, 2022 at 23:48

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