I've read that for a Bose-Einstein gas in 1D there's no condensation. Why this happenes? How can I prove that?
Condensed Matter – Insights on Bose-Einstein Condensate in 1D Systems
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First and foremost, the BEC systems studied in detail today do not involve the formation of any bonds between atoms. Bose-Einstein Condensation is a quantum statistical phenomenon, and would happen even with noninteracting particles (though as a technical matter, that's impossible to arrange, but you can make a condensate and then manipulate the interactions so they are effectively non-interacting, and the particles remain a condensate).
The "high school physics" version of what happens at the BEC transition is this: particles with integer intrinsic spin angular momentum are "bosons," and many of them can occupy the same energy state. This is in contrast to particles with half-integer spin, such as electrons, termed "fermions," which are unable to be in exactly the same quantum state (this feature of electrons accounts for all of chemistry, so it's a Good Thing). When we talk about a confined gas of atoms, quantum mechanics tells us that we must describe it in terms of discrete energy states, spaced by a characteristic energy depending on the details of the confinement. Because of this, the two classes of particles have very different behaviors in large numbers.
The lowest-energy state for a gas of fermions is determined by the number of particles in the gas-- each additional particle fills up whatever energy state it ends up in, so the last particle added goes in at a much higher energy than the first particle added. For this reason, the electrons inside a piece of metal have energies comparable to the hot gas in the Sun, because there are so many of them that the last electron in ends up moving very rapidly indeed.
The lowest-energy state for a gas of bosons, on the other hand, is just the lowest-energy state available to them in whatever system is confining them. All of the bosons in the gas can happily pile into a single quantum state, leaving you with a very low energy.
It turns out that, as you cool a gas of bosons, you will eventually reach a point where the gas suddenly "condenses" into a state with nearly all of the particles occupying a single state, generally the lowest-energy available state. This happens with material particles because the wave-like character of the bosons becomes more and more pronounced as you lower the temperature. The wavelength associated with them, which at room temperature is many times smaller than the radius of the electron orbits eventually becomes comparable to the spacing between particles in the gas. When this happens, the waves associated with the different particles start to overlap, and at some point, the system "realizes" that the lowest-energy state would be for all the particles to occupy a single energy level, triggering the abrupt transition to a BEC.
This transition is a purely quantum effect, though, and has nothing to do with chemical bonding. In fact, strictly speaking, the dilute alkali metal vapors that are the workhorse system for most BEC experiments are actually a metastable state-- at the temperatures of these vapors, a denser gas would be a solid. They form a BEC, though, because the density of these gases is something like a million times less than the density of air. The atoms are too dilute to solidify, but dense enough to sense each others' presence and move into the same energy state.
The underlying physics is described in detail in most statistical mechanics texts, though it's often dealt with very briefly and in an abstract way. There are decent and readable descriptions of the underlying physics in The New Physics for the Twenty-first Century edited by Gordon Fraser, particularly the pieces by Bill Phillips and Chris Foot, and Subir Sachdev.
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.
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Best Answer
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. :)