# Fermi and Bose statistics

1. Oct 10, 2007

### jostpuur

Suppose we have particles of kind B, that consist of two fermions of kind F. Now the particles B satisfy the Bose statistics. But what precisely does this mean? If we have four F particles, the system is described by a wave function

$$\psi(x_1,x_2,x_3,x_4)$$

Suppose the particles 1 and 2 are bounded and form one particle B, and 3 and 4 are bounded too. Then it should be possible to approximate this system as a two particle system

$$\approx \psi'(x_{12}, x_{34})$$

where $x_{12}$ and $x_{34}$ are some kind of approximate coordinates for the particles B.

How can these ideas made more rigor? We have

$$\psi(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}, x_{\sigma(4)}) = \varepsilon(\sigma) \psi(x_1,x_2,x_3,x_4),$$

and we want to prove

$$\psi'(x_{12}, x_{34}) = \psi'(x_{34}, x_{12}).$$

2. Oct 10, 2007

### jostpuur

One could try to prove that just by

$$\psi(x_1,x_2,x_3,x_4) = \psi(x_3,x_4,x_1,x_2).$$

but there's still something I don't like this. What precisely is that "approximation wave function"? I haven't seen any definitions for such things anywhere.

What would

$$\psi(x_1,x_2,x_3,x_4) = -\psi(x_2,x_1,x_3,x_4)$$

mean? The particle $x_{12}$ doesn't move anywhere, but the wave function changes sign. So we could also continue

$$\psi(x_1,x_2,x_3,x_4) = \psi(x_3,x_4,x_1,x_2) = -\psi(x_3,x_4,x_2,x_1) \quad\underset{\textrm{maybe}}{\implies}\quad \psi'(x_{12},x_{34}) = -\psi'(x_{34},x_{12})$$

and the confusion continues.

Last edited: Oct 10, 2007
3. Oct 24, 2007

### blechman

I am unfamiliar with this term "approximation wavefunction" - can you tell me where you saw it? In the limit that you have four fermions, both pair of them forming bound states that do not interact, it is exactly true that the 4-fermion wavefunction factorizes into a product of 2 wavefunctions

$$\Psi_F(x_1,x_2,x_3,x_4) = \psi_F(x_1,x_2)\psi_F(x_3,x_4)-\psi_F(x_1,x_3)\psi_F(x_2,x_4)+\psi_F(x_1,x_4)\psi_F(x_2,x_3)$$

Note at this stage, everything's still fermionic, and you need all of these terms with the signs to be sure to maintain the antisymmetry with all particles. As I say, as long as there are no 3-or-more-body interactions between the fermions, this is an exact equation. If you cannot entirely ignore such interactions, then there are corrections to this result. Maybe that's what you mean by "approximate".

I don't think it means anything! You have to be very careful about what it is you are exchanging. For example, if you are exchanging two of the fermions (in or out of the bound state) then you must use the fermionic wavefunction. However, if you are exchanging two bosons (PAIRS of fermions) you can either use the bosonic wavefunction, OR you can still use the fermionic wavefunction - both give you the same answer. But if you exchange two fermions inside a boson, then you are exchanging two fermions and must use the fermion wavefunction. It is incorrect to talk about bosons in that case.

I always get a headache keeping track of this stuff! ;-)

4. Oct 25, 2007

### jostpuur

I haven't seen it anywhere. But surely there must be some kind of approximation wave function. Everybody talks that pairs of fermions can form bosonic particles, and it wouldn't make sense unless there was some wave function for the new bosonic particles.

5. Oct 25, 2007

### blechman

The only thing I can think of is that you must ignore the substructure of composite bosons made of fermions if you want to write down a bosonic wavefunction, and that is an approximation. I see this point was mentioned in the thread "bosonic atoms"