Why do bosons prefer to occupy the same ground state?

[Shaybay92]

I am doing some research on Bose-Einstein condensates and was hoping someone could give me a non-mathematical reason as to why bosons 'want' to occupy the same ground state. I think its details come from Bose-statistics, but is there a simplified way of explaining it? Thanks

 

[Matterwave]

Basically, if the particles are indistinguishable, and they can occupy the same state, then there are simply more microstates where the particles are bunched up than where they are split apart. And then, the leap of statistical mechanics is that all microstates are equally possible.

As an example, think of 2 bosons which can occupy only 2 possible states. There is 1 state where both are in the lower state, 1 state where both are in the top state, and 1 state where they are split apart. If any of these 3 states are equally likely, then there is a 2/3 chance you have the bosons bunched together and a 1/3 chance of them being split apart. This is different than if the particles were distinguishable, since if they were distinguishable, there would be TWO states where the particles are split apart (particle 1 up, particle 2 down, or vice versa). And the chances would be equally likely that the particles be bunched together or split apart.

 

[Shaybay92]

Oh so its just like the basic statistical physics explanation... am I right to say however, that they all want to occupy the GROUND state? rather than just be together in any state?

 

[Matterwave]

As there are simply 2 states, there is no "ground" state or otherwise. It's been quite a while since I've done this, and I'm quite rusty...perhaps someone more knowledgeable can answer your question better. =]

 

[Drakkith]

The ground state is the minimum energy state correct? To get a bose-einstein condensate you have to get them all to about this energy level. I don't know what you mean when you ask why do they WANT to get to the ground state. They don't want anything, we made them go into the ground state, right? At a low enough energy level the bosons simply start to act "together" and quantum effects get seen at a macroscopic level due to that.

Edit: Just read a bit on it, and I think its simply that because bosons are allowed to occupy the same states, reducing the energy of them causes them to fall into the lowest state possible all at the same time.

 

[Shaybay92]

Oh, so the fact that we have lowered them to around ground state means most of them will, statistically, be in the ground state. So this tendency is purely a statistical one?

 

[Matterwave]

In the end, all of these statistics is just counting the number of microstates. I don't really remember how to do this though! Haha.

 

[Drakkith]

Oh, so the fact that we have lowered them to around ground state means most of them will, statistically, be in the ground state. So this tendency is purely a statistical one?

I don't know what you mean by that. They actually fall into a specific state, right?

 

[Shaybay92]

If we take most of the energy out of the system so there's practically no energy, then there may be a few bosons in higher energy states but most of them will be in the lowest possible state, that is what I got from it.

Are you saying there is some other reason why they go to the ground state?

 

[Mike Pemulis]

There are a couple of different issues here.

Consider a classical ideal gas. If you take a lot of energy out of the system, you would expect to find most of the particles in the ground state. Big surprise. What OP is asking is, does quantum statistics give the bosons any "extra" stickiness -- all things being equal, are they more likely to clump together than classical particles?

The answers is yes, and this is what Matterwave was getting at in the first post. The key is that in classical mechanics, it is always possible in principle to follow each particle. So if State 1 is the same as State 2, except that two identical particles switch places, classical mechanics counts those as different states, but quantum mechanics does not. In the counting of microstates, State 1 and State 2 get lumped together, and are only counted once. Because Stat Mech is all about how microstates are counted, this difference has a profound effect on the physical behavior.

For a simple example, consider a system with three states (A, B, C) occupied by three identical structureless particles. The states are all of equal energy and equally likely. We can write down all the microstates for fermions, classical particles, and bosons.

Notation for fermions and bosons: Because the particles are not labeled, the microstates consist of counting the number of particles in each state, without distinguishing among the particles. A typical microstate might have the form: A2, B1, C0, which means "2 particles in State A, 1 particle in State B, and 0 in State C."

Notation for classical particles: Because the particles are labeled, call the particles x, y, anx z. A typical microstate has the form: Ay, Cxz.

Fermions:

A1, B1, C1

Total: 1

Classical particles:

Axyz
Bxyz
Cxyz
Ax, Byz
Ax, Cyz
Ay, Bxz
Ay, Cxz
Az, Bxy
Az, Cxy
Bx, Ayz
Bx, Cyz
By, Axz
By, Cxz
Bz, Axy
Bx, Cxy
Cx, Ayz
Cx, Byz
Cy, Axz
Cy, Bxz
Cz, Axy
Cz, Bxy
Ax, By, Cz
Ax, Bz, Cy
Ay, Bx, Cz
Ay, Bz, Cx
Az, Bx, Cy
Az, By, Cx

Total: 27

Bosons:

A1, B1, C1
A2, B1, C0
A2, B0, C1
A1, B2, C0
A0, B2, C1
A1, B0, C2
A0, B1, C2
A3, B0, C0
A0, B3, C0
A0, B0, C3

Total: 10

So then, we have:

Probability that all particles are in a different state:

P(fermions) = 1
P(classical) = 2/9
P(bosons) = 1/10

Probability that exactly two particles are in the same state:

P(fermions) = 0
P(classical) = 2/3
P(bosons) = 3/5

Probability that all particles are in the same state:

P(fermions) = 0
P(classical) = 1/9
P(bosons) = 3/10

So it's apparent that, because of pure statistics, bosons clump together more readily than classical particles. Something similar would happen for more complicated systems, as you could probably work out.

 

[Shaybay92]

Thanks for the detailed response Mike. I can see how just statistically working it out, it shows that bosons are more likely to be together... However, I came across this article which describes a sort of 'avalanche' process:

http://eve.physics.ox.ac.uk/Personal/ruprecht/BEC/pw2/pw2.html

"For the second class of particles, the bosons, there is no restriction as to the number of particles that can occupy a given quantum state, and indeed the probability of a particle falling into a given state increases in proportion to the number of particles already occupying it. This can lead to BEC , an avalanche process which occurs when the particles are close enough for their wavefunctions to overlap..."

Is it just bad wording, or is it actually true that bosons in a particular state force others to fall into that state? I suppose this is what I was getting at in my original post.