# Fermion+Fermion = Boson ?

1. Jun 2, 2005

### Nicky

In thinking about Bose-Einstein condensation, an apparent contradiction came to mind ...

Take the usual "particle in a box" potential and start filling it with identical spin-1/2 particles, two at a time. Also impose the rule that each pair of particles added must have opposite spin, i.e. net spin of zero for the pair. Since the individual particles are fermions, they will occupy higher and higher energy levels, due to the exclusion principle. However, we can also view each inserted pair as a single boson. Since the occupation number of a boson state can rise as high as you want, it would seem that all the pairs fall into the ground state (at zero temperature), and their energies are all the same.

So which picture of the system's energy is correct, Fermi, Bose, or some combination of the two? Quantum Mechanics must have an answer to this question, but I can't figure it out. Can anyone out there explain how to reconcile these two pictures of the system?

2. Jun 2, 2005

### Norman

It all depends on if the individual fermions form a physical bound state and what is the force that is binding them...
If they physically pair up, then they are bosons. If they do not the are still just 2 fermions and obey fermi-dirac statistics. If you are thinking of BCS theory- phonons serve as the "glue" that holds the electrons in the bound state and allows them to condense. Without some force or "glue" holding the electrons in the bound state, they would still obey fermi-dirac statistics. The force that holds the fermions in a bound state is the key point here.
Hope that helps.

3. Jun 2, 2005

### mathman

Simple example: Protons and electrons are fermions. Hydrogen atom is boson.

4. Jun 2, 2005

### Nicky

Hmm .. that makes sense in the two extreme regimes:
1) Bose-like limit: (fermion-fermion interaction energy) >> (fermi level in well)
2) Fermi-like limit: (fermion-fermion interaction energy) << (fermi level in well)

But what happens in the grey area between these two extremes?:
3) (fermion-fermion interaction energy) ~ (fermi level in well)

It seems like each particle pair is partly two fermions and partly a single boson, with the relative magnitude of each depending on the strength of the interaction. I'm trying to imagine how it can be decomposed as such through a change of basis, but can't figure out what a basis looks like that has both fermionic and bosonic characteristics.

5. Jun 2, 2005

### Norman

Hmm... what does the fermi level mean here?

6. Jun 2, 2005

### Nicky

Fermi level is clearly defined in the weak-interaction limit, but I guess it is not very well defined if there are significant particle-particle interactions. This is part of my confusion about such systems -- does the particle-particle interaction need to be exactly zero in order to treat them as fermions, or can the exclusion principle apply when the interaction is small, but finite? If the latter is true, then there must be a smooth transition from mostly fermionic to mostly bosonic behavior, and I have no idea how to model that with creation/annihilation operators. For example, which commutation rules apply .. is it some combination of the two?

7. Jun 2, 2005

### Norman

This is a great question! I will have to think on it. As for the smooth transition issue- smooth in what sense is the question. What sort of transition is this? I believe that is a pertinent question. I think part of the problem here is talking about a somewhat fictitious situation. For a transition from fermi to bose, I would think this has been studied already. Or maybe it hasn't- I am always amazed at what has not been looked into or better yet what has no satisfactory answer. I am too tired to start nosing around now, but I will look into it in the morning.

Maybe some of the greater gurus around here can shed a little light.

Cheers,
Norm

8. Jun 11, 2005

### CarlB

This is a cool question. I am responding not with the intention of shedding any light on the subject, but instead to bring it back up to the front of the QM question list.

Carl

9. Jun 11, 2005

### Blackforest

Posted by Mathman:"Simple example: Protons and electrons are fermions. Hydrogen atom is boson." If it is a pertinent example, then the transition described by Nicky must include some polarisation phenomenon, since H <---> (H+= p+) + (e-). This is also meaning that the intern coherence of the hydrogen atom could be related to a description of this grey area... Do I formulate a good picture of your interrogation?

10. Jun 13, 2005

### mikeyork

The simple answer to this question is that it depends on whether or not the fermions combine into a bosonic eigenstate of composite spin. If they do then the composite system will be a boson. Obviously, the internal fermions must be in a valid multi-fermion eigenstate.

11. Jun 17, 2005

### Nicky

After thinking about this question a lot, I still don't have a definitive answer, but here is another piece of the puzzle, perhaps:

Since we are talking about 'particles in a box', let's use the one-particle eigenstates as a basis, with wavefunctions of the form sin(x*Pi*n/L). In this basis, the one-particle hamiltonian can be written as a matrix with the energy eigenvalue of each state on the diagonal (proportional to (n/L)^2, I think). If there are no particle-particle interactions, then the particles just stack up in the available states as they are added to the box. If we look at the first four states, the hamiltonian matrix looks something like this (sorry, I don't know how to do it in LaTeX on this site):

1 0 0 0
0 4 0 0 = H
0 0 9 0
0 0 0 16

Now add interaction between particles in the box. The true QM representation of the system is a superposition of many-particle states, but that is very complicated to calculate. It is easier to approximate using the effective potential that each particle sees, due to the presence of all the other particles, rather like the Hartree -Fock method. In this approximation, off-diagonal elements appear in the hamiltonian matrix. For simplicity's sake, we can let all these off-diagonal elements be the same:

1 v v v
v 4 v v = H
v v 9 v
v v v 16

When this matrix is diagonalized, you get a different spectrum of energy eigenvalues than the non-interacting system. Here are the calculated eigenvalues for a very weak interaction and a very strong interaction:

v=-0.001
E = { .9999994730, 4.000000048, 9.000000180, 16.00000030 }

v=-100
E = { -292.5801520, 102.2679079, 106.7365227, 113.5757214 }

So as the interaction gets stronger, the states are closer together relative to their total energy. The first calculation is the "fermi-like" limit, and the second is the "bose-like" limit.

12. Jun 18, 2005

### CarlB

Nicky, that is a nice calculation. Here's the limiting case (which also shows how to write matrices in LaTex):

$$\left(\begin{array}{cccc} 0&1&1&1 \\ 1&0&1&1 \\ 1&1&0&1 \\ 1&1&1&0 \end{array}\right)$$

The above matrix clearly has a maximum eigenvalue of 3 with eigenvector of (1,1,1,1). I think that the other eigenvectors are (1,i,-1,-i), (1,-i,-1,i) and (1,-1,1,-1) and that these have eigenvalues of magnitude 1.

So in the limit, you have three eigenvectors with the same eigenvalue, and one eigenvector with triple the value (which is abut the behavior you're begining to see with your example).

Carl