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Fermion-Boson question

  1. Nov 10, 2011 #1
    For two identical particles, we have
    thus [itex]\psi(x_1,x_2,t)=e^{i\phi}\psi(x_2,x_1,t)[/itex]
    also [itex]\psi(x_2,x_1,t)=e^{i\phi}\psi(x_1,x_2,t)[/itex]
    that means [itex]e^{2i\phi}=1[/itex] or [itex]e^{i\phi}=\pm 1[/itex]

    For fermions
    Why does it happen to be different when we just switch the position of these particles?
  2. jcsd
  3. Nov 11, 2011 #2
    It actually turns out that if you instantaneously switch the positions of two particles, there is no physical reason for the wavefunction to remain the same. When people talk about statistics they have in the back of their mind a slow (adiabatic) process which switches the two particles.

    Let me give you a simple illustration.. you have two electrons sitting in a 2D plane, localized at two points, and we have total incoherence (wavefunctions do not interfere). Now imagine that you slowly rotate both electrons about the centre of the line joining them by 180 degrees clockwise, so that the two particles have their positions swapped. As you observe, if the particles are truly identical then the physical state of the system has not changed. However since the system has undergone an adiabatic process, it can actually acquire a phase factor. Now imagine that you rotate the electrons again by another 180 degrees clockwise, so that the process has completed a full loop. Then the system has acquired double the original phase factor.

    The phase that a system picks up when it undergoes adiabatic evolution around a full loop is called the Berry phase, and there is a famous theorem that says that this phase is determined entirely by the number of topological singularities enclosed by the loop. When you adiabatically vary the system, you drive the parameters (in this case the positions) around a "parameter space" (in this case, two copies of the 2D plane, one for each electron), and if there is a topological singularity in this space, what people sometimes call the monopole charge, then the Berry phase that the system picks up can be different to zero (or an integer multiple of 2pi).

    If there are no topological singularities, that is, there are no monopole charges penetrating the 2D plane, then the system picks up a phase of 1 when it goes around the full loop. That means the phase it picks up around hte half-loop where you simply exchange the particles is either 1 or -1, corresponding to bosons and fermions.

    If you have two electrons with Coulomb repulsion, then there is indeed a topologically singular point: when the two electrons are sitting at the same point, their Coulomb repulsion becomes infinite. You can imagine that you have one electron fixed at one point and you move another electron around it in a circle. Each time the second electron goes around the first electron, the first electron acts as a "magnetic monopole" if you like and the electron picks up a phase (the geometric phase). So in 2D, it turns out that we do not have fermions or bosons at all, we have anyons, and the phase factor that the state acquires when you adiabatically exchange the positions of two particles (what is called the "statistical phase") can be arbitrary.

    In 3D, you can always deform the loop so that it avoids the monopole, so the Berry phase you pick up when you move one particle around the other is always equal to 1. The Berry phase you pick up when you simply swap two particles is always equal to either 1 or -1, so we haev fermions and bosons.

    In 1d, it is impossible to swap two particles without bringing them close together, if there is an interaction between the two particles then we do not have statistics at all.
  4. Nov 12, 2011 #3
    Wow! Thanks for taking the effort to write this.

    I have looked into this a little bit, and it is possible to get the same sort of result in 3D, though not with particles. You have to have an object with extent in the third dimension is great enough that it is not possible for a loop to pass over or under it and must go around. I always think of pieces of dry spaghetti, but the cooked stuff will work just as well.

    Such things exist in the cores of neutron stars. The core is superfluid and so rotates in columns like a BEC. The columns extend through the core, can't get out of the core, and cannot cross one another in any simple way, so the criteria are more or less met.
  5. Nov 12, 2011 #4


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    tommyli, You've given a nice description of the Berry phase, but unfortunately it has nothing to do with the original question.
    Completely incorrect. The particle interchange that's involved in quantum statistics IS instantaneous.
  6. Nov 12, 2011 #5
    Hi Bill, I was trying to address question about statistical phase. It doesn't matter whether the particle interchange is instantaneous or not, the result is that interchange of particles twice gives an overall phase of +1, and interchange of particles once therefore an overall phase of +1 or -1, and this is particle statistics as it is usually introduced when you learn quantum mechanics. This makes sense only when you assume that interchanging particles twice does not give a nontrivial overall phase.

    I do not doubt that for identical particles instantaneous exchange can only change the phase. This is the criterion for identical particles. However the assumption that exchanging particles twice cannot contribute a phase is completely incorrect. The entire concept of instantaneous exchange is a mathematical abstraction that is completely violated when topology is nontrivial. This is central to the fractional quantum Hall effect- you can directly turn a fermionic system into a bosonic system via statistical phase coupled to the gauge interaction.

    Do you agree that for interacting particles in 1D, particle statistics is poorly defined?
    Last edited: Nov 12, 2011
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