My question is about zero-temperature ground state of a Bose system. Suppose that the system stabilizes a BEC order parameter, say $\langle b^+ \rangle$, and fixes its phase. Is this a superfluid? And vice versa. In 2D Bose-Hubbard model one stabilizes regions of Mott insulators and everywhere else the system is in superfluid state. Is this state any different from BEC?


1 Answer 1


Short answer: the terminology is not always used consistently.

A superfluid and a BEC are certainly logically different. Superfluidity refers to the ability of some liquid to flow without any viscosity, and some associated properties, such as the ability to "expel" the angular momentum of its container, which is known as the Hess–Fairbank effect*. In other words, superfluidity is a list of macroscopic properties. BEC, on the other hand, is the occupation by O(N) atoms of a single state of a Bose system.

In practice, since superfluidity is always (?) caused by the creation of some BEC in a very broad sense (including condensation of Cooper pairs, for example), people sometimes use both almost interchangeably. However, as the linked answer mentions, there are edge cases in which you have one and not the other. One of these cases is a non-interacting Bose gas. As a result, sometimes one will also see the following distinction made: a system in which the particles have no or negligible interactions is called a 'BEC', while a strongly interacting system that condenses is called a 'superfluid'. So, for example, a trapped dilute alkali gas is often called a BEC, but He-4 is called a superfluid. As far as I know there is no hard cutoff between the two terms.

For example, in the context of the 3D Bose-Hubbard model with interactions, the state with a macroscopic wavefunction is usually called a superfluid. However, it does still have macroscopic occupation of some state (which might be a complicated state due to the interactions), and arguably deserves to be called a BEC for that reason as well. In practice, if you do so you might annoy people who think of a BEC as exclusively a non-interacting Bose gas below Tc.

In 2D (or 1D) the situation is slightly more complicated, because a true BEC will not form in the absence of a confining potential. This is another "edge case," in which the low dimensionality weakens the tendency to Bose condense. So in 2D it is also possible in general to have either a non-BEC or a BEC superfluid.

*I use the term "expel" to draw an analogy with the corresponding property in superconductors, the Meissner effect.

Edit: response to comments:

Okay, I think a few things are being confused here regarding different definitions of BEC:

  1. Notice that the definition of BEC that I gave actually doesn't say anything about long-range order. Some people do define it that way instead though.

  2. Specifically, a BEC is said (by some!) to have "off-diagonal LRO." This is a particular type of order that is different from, for example, magnetic order. In fact, the "normal" order of magnets and crystals is diagonal long-range order.

For the sake of concreteness, a formal definition is that ODLRO is when $\rho_N(x,x')\neq 0$ as $|x-x'|\rightarrow \infty$, where $\rho_N$ is the N-particle (usually one or two-particle) density matrix.

  1. Non-zero ODLRO corresponds with the definition given above of BEC in many, but not all cases. A simple example is that non-zero ODLRO is only formally possible in an infinite system, so under this criterion a finite system is never a "true" BEC. So all these similar competing definitions of BEC are another complication on top of the distinctions between BEC and superfluidity. Hey, no one said it was going to be simple.
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    $\begingroup$ +1 Nicely fleshed out the comparison, should resolve OP's confusion. $\endgroup$
    – Ellie
    Commented Feb 25, 2016 at 2:39
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    $\begingroup$ Nice answer. I would claim that "people who think of a BEC as exclusively a non-interacting Bose gas below Tc" deserve to be annoyed, since they have a totally useless definition of BEC! $\endgroup$ Commented Feb 25, 2016 at 10:52
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    $\begingroup$ Thank you, Rococo, for the answer. It is indeed helpful. So basically if we leave aside the noninteracting gas, and use the broad definition of BEC, i.e. a state that breaks rotational symmetry (states that support LRO, namely condensates, superconductors, ordered magnets...), then all superfluids will fall under this definition of BEC. However there is one more question: is it true that all states supporting "some" LRO (i.e. breaking of the rotational symmetry) are superfluids (have no viscosity and exhibit Hess–Fairbanks or Meissner effects)? Thanks! $\endgroup$
    – danport
    Commented Feb 26, 2016 at 1:38
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    $\begingroup$ Note that you can also have a superfluid without a BEC (long range order), as for exemple in the case of (2D) helium films or cold atoms confined in a 2D trap. $\endgroup$
    – Adam
    Commented Apr 8, 2016 at 9:46
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    $\begingroup$ Hi, Kurt: O(N) means 'of order N,' where N is the total number of atoms. In other words: in a BEC the number of atoms occupying the ground state, $N_0$, may be 100% or 10%, or 1% of the total number of atoms, but the key point is that it is some fixed fraction, such that $N_0/N$ approaches a constant between 0 and 1 as $N$ goes to infinity. This is an example of 'big O notation' , used to express the leading order dependence of some term without worrying about the exact form: en.wikipedia.org/wiki/Big_O_notation $\endgroup$
    – Rococo
    Commented Dec 27, 2020 at 18:55

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