I am interested to understand how positive entropy Bose Einstein condensation for cold atoms (say) behave. The way I think about it is as follows: We have an ideal pure state where every atom is in the same ground state which depends on a geometry of a certain trap and the atoms do not interact. Now, naively, I thought about the actual positive-entropy state looking as one of the two following alternatives:

(1) Every atom get excited with some small probability $p$.

A different picture is as follows:

(2) If the type of the intended ideal state puts every atom in the same ground state $A$ (where $A$, say, depends on the geometry of the trap), then the positive-entropy state is obtained by mixing this ideal state with a state which is Bose Einstein state w.r.t. a different ground state $B$ (or many such $B$ ;s).

I asked around a little and what I was told is that the $T>0$ theory is not part of the original theory discovered in the $1920 s$ but rather a more recent theory that was developed in the $1990 s$ . Moreover, the perturbation for the ideal pure state described by the $T>0$ state manifests pairwise interaction between the atoms, and that there are interesting singularities that occur immediately when $T>0$ no matter how small.

I will be very thankful for some explanations and references. Most helpful for me will be non-technical explanations.

Update: I am aware of one book on a related topic: The Poincare seminar 2003 on Bose Einstein Condensation - Entropy. (I did not get a hold of it yet but a few of the papers are here.) I will appreciate any information regarding the matter. Genneth recommend the book Quantum Liquids: Bose Condensation and Cooper Pairing in Condensed-Matter Systems (Oxford Graduate Texts) by Anthony Leggett.

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    $\begingroup$ Hi @Gil, are there any references you are working from currently? It sounds like you're asking what are the excitations above the ground state of a (free? interacting?) Bose gas. Is this right, or is there more to it than that? $\endgroup$
    – wsc
    Commented Mar 3, 2011 at 1:13
  • $\begingroup$ Dear wcs, This is essentially what I am asking about. And I am also interested what is the description which is relevant to experimental implementation of Bose-Einstein condensate. $\endgroup$
    – Gil Kalai
    Commented Mar 3, 2011 at 6:41
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    $\begingroup$ @Gil What do you mean by "positive entropy Bose-Einstein condenstate"? The term is neither known be me nor by Google. If you mean BEC at $T\neq 0$ (a widely investigated subject) then reformulate the question. $\endgroup$ Commented Mar 4, 2011 at 11:22
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    $\begingroup$ To my mind, the best theoretical description of all this is Quantum Liquids by Anthony Leggett: amazon.com/… $\endgroup$
    – genneth
    Commented Mar 5, 2011 at 8:44
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    $\begingroup$ @Gil once you have a BEC ground state and you want to look at excitations around that ground state, one usually looks at solutions of the Gross-Pitaevksii equation. Of course, this is also a mean-field (Hartree-Fock) based approach. $\endgroup$
    – user346
    Commented Mar 19, 2011 at 7:51

1 Answer 1


First, let me say I'm not sure what is meant by a BEC with $T\gt 0$. Condensation is a finite temperature phenomenon, which occurs due to the presence of pair-wise interactions (generally attractive, but pairing can happen even for repulsive potential) in a many-body system. For instance, in a superconductor below some critical temperature $T_c$, electrons with opposite momenta and spin (s-wave pairing) pair up to form a bound state called a Cooper pair.

The ground state of the unpaired electron gas for $T \gt T_c$ is characterized by the Fermi energy $E_F$. After condensation, the many-body system has a new ground state at energy $E_{bcs} = E_F - \Delta $, where $\Delta \sim k_B T_c$ is the binding energy of a Cooper pair. $\Delta$ is also known as the gap.

For $T\lt T_c$ all the electrons are not paired up due to thermal fluctuations. However, the number of unpaired electrons as a fraction of the total number of electrons (the condensate fraction) goes as $N_{free}/N_{all} = 1- (T/T_c)^\alpha$, where $\alpha\gt 0$. The number of free electrons drops rapidly as $T$ is decreased below $T_c$. In lab setups, BEC's generally undergo some form of evaporative cooling to get rid of particles with energies greater than $\Delta$. At this point the condensate can be treated as a gas of interacting (quasi)particles (cooper pairs) with an approximate hard-core repulsion.

So the gas, before and after condensate formation, is always at finite temperature! This is reflected, for instance, in the dependance of the condensate fraction on $T$ as mentioned above.

The mean-field solutions for low-energy excitations of the condensate are given by the Gross-Pitaevskii equation(GPE):

$$ \left( - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2} + V(r) + \frac{4\pi\hbar^2 a_s}{m} |\psi(r)|^2 \right) \psi(r) = \mu \psi(r) $$

where $a_s$ is the scattering length for the hard-core boson interaction, with $a \lt 0$ for an attractive interaction and $a \gt 0$ for a repulsive interaction.

Presumably one should be able to construct a canonical ensemble with solutions $\psi(k)$ ($k$ being a momentum label of the above equation), but this is by no means obvious because of the non-linearity represented by the $|\psi(r)|^2$ term. Here a "zero temperature" state would correspond to a perfect BEC with no inhomogeneities, i.e. the vacuum solution of the GPE. However, the entire system is at some finite temperature $T \lt T_c$ as noted above. The resulting thermal fluctuations will manifest in the form of inhomogeneities in the condensate, the exact form of which will be determined by the solutions of the GPE.

Of course, the GPE's regime of validity is that of dilute bose gases ($l_p \gg a_s$ - the average interparticle separation $l_p$ is much greater than the scattering length). For strong coupling I do not know of any similar analytical formalism. If I had to take a wild guess I'd say that the strong-coupling regime could be made analytically tractable by mapping it to a dual gravitational system, but that's another story altogether.

As $l_p$ approaches $a_s$ from above, the GPE breaks down and it will have singular solutions for any given $T$ and these are likely the singularities that you are referring to.

Reference: The single best reference I can suggest is Fetter and Walecka's book on many-body physics. I'm sure you can find more compact sources with a little effort. But generally the brief explanations leave one wanting for a comprehensive approach such as the one F&W provides.

This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.

  • $\begingroup$ Reasons for downvote? $\endgroup$
    – user346
    Commented Mar 19, 2011 at 19:15
  • $\begingroup$ @Approximist - no problem. One edit, coming right up. $\endgroup$
    – user346
    Commented Mar 19, 2011 at 19:40
  • $\begingroup$ @Deepak I am deleting my original comment (and this one subsequently) to leave this space open for mosre relevant discussion ;) $\endgroup$ Commented Mar 19, 2011 at 19:43
  • $\begingroup$ Dear Deepak, thank you for your answer. I was asking about the T<T_c domain. The way I thought about it (which may be wrong) was that there is some "ideal" pure state where all particles have the same ground state. (And with no interactions.) When T< T_c we are rather close to such "ideal" state. And my question was how these 0<T<T_c states differes from the "ideal" state. $\endgroup$
    – Gil Kalai
    Commented Mar 19, 2011 at 20:15
  • $\begingroup$ @Gil, its with a 'p', not a 'v'. And my question was how these 0<T<T_c states differes from the "ideal" state. - that is the question I attempted to answer, in my mind at least. The "ideal state" provides a background on which quasiparticles - excitations around the vacuum or ideal state - can propagate. In the limit that T->0, we approach this ideal ground state which will be characterized by increasing rigidity of the condensate fluid. Below some T'_c one might even obtain a crystalline lattice of bosons. I'd think of the ideal ground state as a droplet of condensate with no ... $\endgroup$
    – user346
    Commented Mar 19, 2011 at 20:26

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