Bose-Einstein condensation is a phenomenon that theoretically happens for free Bose gases because Bose-Einstein distribution holds for free Bose gases. But in superconductors and superfluids, there are interactions between the bosons.

  1. Are superconductors and superfluids really Bose condensates?

  2. In reality, is it possible to see the Bose condensation in a system of free Bosons?

  • $\begingroup$ YouTube : helium at 0K $\endgroup$
    – Ismasou
    Aug 13, 2017 at 17:16
  • 1
    $\begingroup$ The non-interacting Bose gas is an unrealistic limit, in that some infinitesimal interactions are necessary to reach thermal equilibrium. Nonetheless, a cloud of trapped alkali atoms is very much in the weakly interacting limit compared to superfluids, and BEC in this system was recognized with the 2001 Nobel Prize: nobelprize.org/nobel_prizes/physics/laureates/2001/press.html . $\endgroup$
    – Rococo
    Aug 13, 2017 at 19:38

1 Answer 1


Yes, Einstein considered a purely statistical effect, where condensation occurs although interactions are absent. However, for most people the accumulation of particles in the ground state is the key point of a Bose-Einstein condensate. Therefore, most people agree that atomic, molecular, exciton-polariton and photonic Bose-Einstein condensates have been realized. They adopt the modern definition of a BEC (link).

So, to answer your first question, most BEC experts consider superfluid helium as a BEC. In order to convince yourself, you could read the introduction of the Nobel prize speeches.

The second question is harder to answer, because it depends on your definition of "free bosons". E.g. if you work with an atomic BEC, you can use a so called Feshbach resonance to switch off 2-body interactions. However, since you always have some fluctuations in your B-fields, this won't be perfect. Furthermore, the BEC needs to be confined in space. Hence, even if you switch off all 2-body and higher order interactions, the BEC still won't be truly free.


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