Fermi and Bose statistics

  • Thread starter jostpuur
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  • #1
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Main Question or Discussion Point

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

[tex]
\psi(x_1,x_2,x_3,x_4)
[/tex]

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

[tex]
\approx \psi'(x_{12}, x_{34})
[/tex]

where [itex]x_{12}[/itex] and [itex]x_{34}[/itex] are some kind of approximate coordinates for the particles B.

How can these ideas made more rigor? We have

[tex]
\psi(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}, x_{\sigma(4)}) = \varepsilon(\sigma) \psi(x_1,x_2,x_3,x_4),
[/tex]

and we want to prove

[tex]
\psi'(x_{12}, x_{34}) = \psi'(x_{34}, x_{12}).
[/tex]
 

Answers and Replies

  • #2
2,111
18
One could try to prove that just by

[tex]
\psi(x_1,x_2,x_3,x_4) = \psi(x_3,x_4,x_1,x_2).
[/tex]

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

[tex]
\psi(x_1,x_2,x_3,x_4) = -\psi(x_2,x_1,x_3,x_4)
[/tex]

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

[tex]
\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})
[/tex]

and the confusion continues.
 
Last edited:
  • #3
blechman
Science Advisor
779
8
One could try to prove that just by

[tex]
\psi(x_1,x_2,x_3,x_4) = \psi(x_3,x_4,x_1,x_2).
[/tex]

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.
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

[tex]
\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)
[/tex]

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".


What would

[tex]
\psi(x_1,x_2,x_3,x_4) = -\psi(x_2,x_1,x_3,x_4)
[/tex]

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

[tex]
\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})
[/tex]

and the confusion continues.
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
2,111
18
I am unfamiliar with this term "approximation wavefunction" - can you tell me where you saw it?
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
blechman
Science Advisor
779
8
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"

https://www.physicsforums.com/showthread.php?t=190325
 
  • #6
I have a naive question about Fermi Dirac distribution. In solid state electronics we learnt that doping an intrinsic semiconductor with n type material shifts the fermi level close to conduction band increasing probability of finding an electron in conduction band. But at the same time from Fermi Dirac distribution, the probability of finding electron in valence band is also increasing. How can both of them possible at the same time, given sum of probability finding electron in conduction band and valence band is 1.
 

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