I've read that for a Bose-Einstein gas in 1D there's no condensation. Why this happenes? How can I prove that?


4 Answers 4


The claim is often that there is no condensation in $d<3$. The other answers are correct, but let's be clear, there are actually two assumptions present in the claim:

  1. Assume you have $N$ noninteracting bosons in $d$-dimensions in a hypervolume $L^d$

  2. Assume that these bosons have an energy-momentum relationship of $E(p) = Ap^s$.

Now, the way we calculate the critical temperature ($1/\beta_c$) for BEC requires satisfying the equation $$\int_0^\infty \frac{\rho(E)dE}{e^{\beta_c E}-1}=N$$

where $\rho(E)$ is the density of states. Whether this integral is convergent or not depends on the values of both $s$ and $d$. The details of the proof are up to you though. :)

  • $\begingroup$ @mbq and @wsc Thank's for your relevant and targeted answers. So, if I have understand, condensate exist ONLY in three dimension? Not in 1 or 2, or (if we imagine it's possible, but there is a mathematical speculation) in >3D ;) PS i vote +1 at your answers, but I can't select mor than 1 accepted answer :( $\endgroup$
    – Boy S
    Feb 11, 2011 at 1:19
  • $\begingroup$ >3D is just fine, and in fact if you work it out carefully, you'll see that this common proof fails to deny the existence of a condensate in 2D if the bosons have a linear dispersion, E(p)~p. But this is just math. Zoran Hadzibabic does some truly beautiful experiments with quasi-2D BECs. $\endgroup$
    – wsc
    Feb 11, 2011 at 1:24
  • 2
    $\begingroup$ There is a little more to this; for the not so degenerate case of purely non-interacting bosons, a BEC is a kind of quantum condensation where a continuous ($U(1)$) symmetry is broken; in general, such a symmetry breaking will yield arbitrarily low energy (Goldstone) bosons. In below 3D, these bosons will, at any finite temperature, be infinite in number, signalling a failure of the theory, i.e. it is not actually symmetry broken. In 1D this is absolute; in 2D the divergence is only logarithmic, so for a small sample it is indistinguishable from a broken symmetry state. $\endgroup$
    – genneth
    Feb 11, 2011 at 9:41

This is obviously pure geometry; precisely because the density of states of zero energy is not approaching 0 in $d<3$ (it behaves like $E^{\frac{d-2}{2}}$); (probably) the simplest proof can be done by showing that this explodes the number of particles.


It is necessary to clarify that a uniform, non interacting Bose gas (considered to be confined in a periodic box) in thermal equilibrium does not have a macroscopic occupation of the zero momentum mode if $d<3$. This is not quite accurate for $d=2$ as macroscopic occupation is achieved at T=0, or rather the critical temperature tends to zero in the limit of $N \to \infty$, $V \to \infty$, $N V = {\rm const}$.

This is however not the case if one has external potentials and makes no continuum approximation in the thermodynamics. Additionally attractive condensates $(a_s < 0)$ can form stable, self localised states (solitons) even without confinement in $d=1$. Such states satisfy the conditions for off diagonal long range order required for BEC.


Look at the derivation of the critical temperature of a Bose gas. In there, you should get nonsensical results for one dimension. This is because the density of states even of non-interacting particles depends on the dimension.

  • 5
    $\begingroup$ Although technically correct, I don't think this answer is very helpful. It doesn't really seem to answer the question, it just says, essentially, "look it up." $\endgroup$
    – David Z
    Feb 11, 2011 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.