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


*

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

*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. :)
A: 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.  
A: 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.
