As asked in How is a Bose-Einstein condensate produced from sodium atoms that do not have an integer spin? , Sodium 23 has been used experimentally to form a Bose Einstein condensate.

Sodium 23 is the only naturally occurring sodium isotope. Its components are:

11 protons

12 neutrons

11 electrons

Since fermion pairings are only possible for even number of fermions presumably of the same type, I don't understand how the odd 1 proton and odd 1 electron can be reduced, such that all fermions are paired, which is a necessary condition for the measurements taken on the lump to reflect Bose Einstein statistics?

I don't fully understand the answer given in the ref: "Sodium-23 has nuclear spin of 3/2, making it a fermion. There are 12 paired neutrons, 10 paired protons, and one leftover unpaired proton. The unpaired proton sits in a shell state which contributes the spin of 3/2. But the Bose-Einstein condensate is formed by atomic sodium, not nuclear sodium. The 11 electrons in a neutral sodium atom contribute an unpaired spin of 1/2 to the total. The full sodium atom is a composite boson, so it can form a Bose-Einstein condensate."

What has happened to the odd 1 proton and odd 1 electron which has enabled all present fermions to reduce to bosons in this lump of matter?

  • $\begingroup$ Well, the proton is composed of quarks yet we can treat it as a composite particle without regard to the quarks in many situations. The same holds true here - the particle overall acts as a boson provided you don't poke at it too deeply to uncover constituents. $\endgroup$
    – Jon Custer
    Nov 9, 2022 at 18:51
  • $\begingroup$ @JonCuster so the odd 1 proton and odd 1 electron have essentially "paired" into a boson? I am thinking also of lithium which is non reducible to a Bose Einstein condensate. Why can sodium be reduced to BEC, but not lithium? $\endgroup$
    – James
    Nov 9, 2022 at 18:52
  • 2
    $\begingroup$ The "pairing" isn't really necessary; that pairing-up just guarantees that the total spin ends up being relatively small. What matters is that the total spin of an even number of spin-1/2 particles must have integer spin. $\endgroup$
    – march
    Nov 9, 2022 at 18:57

1 Answer 1


The simples case is hydrogen, which consists of a single proton and a single electron. Both are fermions, but the hydrogen atom is a boson. The hydrogen atom has different properties then its components. It's a composite entity, and it makes sense to think of the hydrogen atom as a particle -- although it is not an elementary particle. Personally, I feel it is natural to talk about the hydrogen wave function in atomic physics, and I rarely have to stress that it is a composite particle. At "low energy" the hydrogen atom can be described as a single particle.

The spin-statistics theorem now tells us, that if we take two hydrogen atoms and interchange their position, that we and up with the same wave function. This leads to the "bunching" effect at low temperatures, which is called Bose-Einstein condensation.

  • $\begingroup$ Thank you, that clarifies the question. In the case of elements not reducible to a Bose Einstein condensate, is it then exactly the set of molecules which have odd numbers of neutron (because proton+electron will always be even), or is there a better condition for deciding which composite molecule's wavefunction will be symmetric and reduce to Bose Einstein condensate, and which other composite molecule's wavefunction will be anti-symmetric and lead to Fermi Dirac condensate at low temperature? $\endgroup$
    – James
    Nov 10, 2022 at 2:23
  • $\begingroup$ silly related question: if the hydrogen atoms bunch together into the same state, and swapping electrons between atoms inverts the wavefunction (because electrons are fermions) how does each electron remember which atom it's associated with? $\endgroup$
    – user253751
    Nov 10, 2022 at 3:09
  • $\begingroup$ @James: Yes, for an atom the number of neutrons determines whether it is a boson or a fermion. $\endgroup$
    – Semoi
    Nov 10, 2022 at 17:07
  • $\begingroup$ @user253751: The BEC assumes a non-interacting gas. In addition, we consider the atoms as the entity and therefore swapping the electrons between the atoms does not make sense in the context of a (standard BEC). If you are interested in the question you described, you should look into many-body theories. I reckon a good starting point could be the Hartree Fock method. $\endgroup$
    – Semoi
    Nov 10, 2022 at 17:14
  • $\begingroup$ user253751 as per comment by @Semoi, this would suggest that this fermion to boson process happens as "internal affairs" of each atom, and individual atoms never quite lose all their individual-ness in the process. Otherwise, atoms with odd neutron numbers can also freely form BEC, because the odd unpaired neutron will simply pair with another odd unpaired neutron from a neighboring atom. Thus, it would seem that fermion-pairing in BEC happens within each individual atom household, in contrast to Cooper pairing in superconductivity which is considered as a regional effect involving phonons. $\endgroup$
    – James
    Nov 10, 2022 at 17:28

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