Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

As we know, fermions are subject to exchange interactions that limit the densities they can achieve. However bosons (simple or composite) are not constrained by this, which implies physical phenomena like the Bose-Einstein condensate, where the main component is Helium-4

Question: Are there physical limits to the density that can be achieved with Bose-Einstein condensates made up from composite bosons?

my suspicion is that there is a critical density above which the fermion character of the fermion components will play up, but i couldn't find any mention about this anywhere

share|improve this question
2  
Notice that even bosons like cold atoms (or He-4) have a massive potential barrier to interpenetration. And notice that fermions like electrons can be packed to give arbitrarily large fermi levels. The question is not quite correct... –  genneth Feb 3 '11 at 17:36
add comment

2 Answers

up vote 1 down vote accepted

It's a good question. The answer is that the bound on the density is given by the requirement that the interactions between the bosons have to remain weak for the Bose-Einstein condensate to exist. In practice, the helium-4 atoms have to be further away from each other than their radius.

Why it is so? Well, if you're talking about the bosons occupying the "same state", it really means that you are constructing a multi-particle state in the multi-particle theory. If you want the energy of this state to be simply given by the sum of the energies of the individual bosons - i.e. $N$ times the energy of the one-particle state - you must guarantee that you have the right Hamiltonian which is essentially the Hamiltonian for the bosons in an external potential, without any significant interaction term in between the bosons.

A sufficiently strongly interacting Hamiltonian for the bosons couldn't be solved that easily.

If you try to push the composite bosons really close to each other, i.e. by lowering the temperature extremely close to the absolute zero, the interactions between them will start to matter which will prevent you from approximating the Hamiltonian by a sum of many one-particle terms. Consequently, the right description is in terms of the component particles - which are often fermions.

It's believed by many condensed matter physicists that the ultimate state of any bound matter very near the absolute zero is a superconductor or a Fermi liquid - and I don't know. Consider helium-3 as an example. I am actually not sure what one gets at superextremely low temperatures.

share|improve this answer
1  
Excellent answer Lubos. There could be other factors other than the hardcore repulsion that could contribute to the critical density at which the condensate is no longer the lowest ground state. –  user346 Feb 3 '11 at 18:28
    
I would add that there is a general theorem which states that Fermi liquids are not stable at low temperatures. Usually a superconducting state takes over. –  genneth Feb 3 '11 at 18:32
1  
Oh, I see, so I may have completely confused the two phases. ;-) –  Luboš Motl Feb 3 '11 at 18:55
add comment

There's a repulsive van der Waals force between the atoms. This shows up in the second quantized formalism as $$H = \int d^3x \left[ \frac{\hbar^2}{2m} \nabla \Psi^\dagger \cdot \nabla \Psi - \mu\Psi^\dagger \Psi \right] + \int d^3x\, d^3y\, \frac{1}{2}V\left( \vec{y} - \vec{x} \right) \Psi^\dagger\left( \vec{x} \right) \Psi^\dagger\left( \vec{y} \right) \Psi\left( \vec{x} \right) \Psi\left(\vec{y}\right)$$ The density is given by the expectation value $\left\langle \Psi^\dagger\Psi \right\rangle$, but you can see the repulsive potential acts as a brake requiring far greater and greater pressure to achieve higher and higher densities.

Beyond a certain density, the van der Waals approximation treating two atoms close to each other as independent subsystems breaks down.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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