# Bose-Einstein Condensation in 2d?

1. Jan 11, 2008

### T-7

I'm wondering -- would you get the phenom. of Bose-Einstein condensation in a 2D system. My intuition is "no", but I'm having a hard time saying why. I suppose I could try and work it out mathematically (no time, right now) -- perhaps someone has? -- but a physical explanation would be nice too.

2. Jan 11, 2008

### Gokul43201

Staff Emeritus
Look up the Mermin Wagner theorem. Then, if you feel like going further, look up the Kosterlitz-Thouless (1973) paper.

3. Apr 6, 2008

### Minich

One can get BEC in 2D if energy of bozon particle is proportional to momentum:

E=const*p

See for example pairion condensation in 2D in the book:

Quantum Statistical Theory of Superconductivity
by Shigeji Fujita and Salvador Godoy
http://www.ebooks.kluweronline.com

4. Apr 7, 2008

### Mute

It depends on the kind of potential the particles are in. I don't have a good physical reason handy for why you shouldn't normally get BEC in 2d, but here's a mathematical reason for a given case. For free particles,

$$E = \frac{\hbar^2}{2m}\left(\frac{\pi}{L}\right)^2n^2$$,

where $$n^2 = n_x^2 + n_y^2$$
and [itex]k_x = \pi n_x/L[/tex].

If you compute the density of states, you find $$g(E) = \frac{L^2m}{2\pi\hbar^2}$$.

The number of particles in the system is then approximately equal to (this doesn't really count particles in the ground state)

$$N \approx \frac{L^2mk_BT}{2\pi\hbar^2}\int_0^{\infty}\frac{dx}{e^x-1}$$

If you suppose we fix N, then the temperature is given by

$$T \approx \left[\frac{L^2mk_B}{2\pi N\hbar^2}\int_0^{\infty}\frac{dx}{e^x-1}\right]^{-1}$$

But the integral there diverges, and so the temperature tends to zero - i.e., you a temperature of zero for BEC to occur in this 2d system.

Really, the crux of the matter is the density of states. You need a density of states such that the Bose integral will not diverge. The constant g(E) in this example was the first sign of danger. The density of states needs to be a higher power of E in order for that integral to converge. This determines what kind of potentials might results in a BEC in 2d.

5. Mar 28, 2010

### akrejci

The cause of BE condensation is a filling of all excited states. Once these states are filled, the ground state is the only place for particles to go, so all additional particles will condense in the ground state.

The equation for N in the previous post is actually the number of particles in an excited state. When this integral is taken, you find N excited, which ends up having no limit.

Nex = AmkbT/(2*pi*hbar2)*(-ln(1-z)*kbT

Because there is no limit to Nex, you can add more and more particles, and they will go into an excited or ground state, rather than only a ground state.

6. Mar 29, 2010

### Cthugha

This is only true for infinite 2D systems. For some finite 2D systems you can see several "smoking guns" of a BEC like long-range off-diagonal order, spontaneous build-up of coherence, quantized vortices and half-vortices and a linearized Bogoliubov excitation spectrum. Although this looks a bit like cheating - a true BEC is considered to be an infinite system - truely infinte systems are of course never realized experimentally.

See for example the famous paper by Kasprzak et al. for an example of a 2D BEC: http://www.nature.com/nature/journal/v443/n7110/abs/nature05131.html (J. Kasprzak et al., Boseâ€“Einstein condensation of exciton polaritons, Nature 443, 409-414 (2006)).

If you want the hard theoretical analysis, Gokul43201 already pointed you to the important papers.

7. Apr 8, 2010

### genneth

I consider the definitive textbook on these things to be Quantum Liquids by Leggett. He deals with the realities of experiment very carefully from the theoretical side.

As has been alluded to, there is some argument as to what BEC is in realistic systems of finite sizes, trapping potentials and finite lifetimes.

The simple answers have been given above:

1. Mermin-Wagner says that for free particles you can't get a proper BEC.
2. You can if you assert an appropriate trap, e.g. harmonic.
3. Condensates in 2D go through a Kosterlitz-Thouless transition. This is because in 2D it is entropically favourable to create dipole pairs of vortices, which mess up long range order and give an algebraic decay ("almost" short range). At high T, these unbind and exponentially decaying order is restored.
4. Leggett prefers to use a different definition of condensation; using this, free 2D systems have no difficulty condensing, simply displays different phenomenology.

8. May 10, 2010

### puzzly

For a physical explanation, you could consider long wavelength phonons destroying the coherence in an infinite system. However, in a real system, finite size effects give a lower bound on the phonon energy and so it requires more energy for a phonon to show up and start tearing the condensate apart...

9. May 11, 2010

### stephenhky

Mermin-Wagner theorem tells you that if there is a long-range-order (existence of BEC) in two-dimensional systems, the fluctuations will drive the system to somewhere else, destroying the condensate. Yes, as puzzly said, it is the long-wavelength phonon that destroys the coherence of the infinite system.
Another way to view it is that the atoms have fewer freedom to move in two-dimensional space than three-dimensional space. The effectively stronger interaction between atoms make the long-range correlation impossible.

For Kosterlitz-Thouless transition, it is about the existence of another phase at low-temperature with quasi-long-range order, despite the absence of true long-range-order. In 2D Bose gas, it is the superfluid phase.

But since bosonic systems are all prepared in a trap, the finite-size effects play a big role. I suggest reading the experimental report PRL 102, 170401 (2009)