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

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up vote 7 down vote accepted

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. :)

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@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 :( –  Boy Simone Feb 11 '11 at 1:19
    
>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. –  wsc Feb 11 '11 at 1:24
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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. –  genneth Feb 11 '11 at 9:41
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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.

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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.

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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.

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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." –  David Z Feb 11 '11 at 0:27
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