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Pauli exclusion, symmetry, and electric repulsion

  1. Jun 30, 2014 #1
    I have a few questions about the Pauli exclusion principle:

    1. Why do physicists believe that the symmetry in the wavefunction we assign to particles (indistinguishability) is due to an actual restriction in the physical state space that the particles can occupy (the attributes following from assuming "indistinguishability" is something "fundamental") versus the inability of our measurements to distinguish between two particles?

    It seems to me that (as in classical physics), if all present measurements fail to distinguish between two particles (electron A and electron B in a well), then there is nothing lost (or gained) (relative to that set of distinguishing measurements) between representing the state with or without the symmetry. Introduce a measurement that treats electron A and electron B differently (we suddenly discover a new distinguishing property or something), and now you can no longer adequately represent the state symmetrically.

    (hypothesis A: The two particles have some required symmetry to the actual physical state that nature uses to do it's thing, hence we cannot detect any difference between Phi(x1,x2) and Phi(x2,x1)

    hypothesis B: We have no measurements that can distinguish between two exchanged particles, so physics can be represented by a symmetric (or antisymmetric) wavefunction, which may be a reduced projection of the actual state space nature uses to do it's thing

    We may have no reasons to favor the more complicated hypothesis B, but do we have any reasons to reject it?)

    2. The Pauli exclusion principle is invoked to explain why electrons cannot occupy the same state. The antisymmetry of their wavefunction is imposed to enforce this. But if electrons were bosons, the electrostatic repulsion between them would *still* require that Phi(x,x) = 0 for all states of finite energy. What is the difference between the behavior of a "helium atom" with standard electrons versus ones that have "boson electrons" (which are nonetheless prevented from occupying the same state due to electric repulsion). Is there any difference? The square of an antisymmetric function and a symmetric function where the diagonals are forced to be zero seems like it would be drawn from the same set. If no distinction were made between fermions and bosons, would the same behavior arise from the presence or absence of interparticle forces that go to infinity as particles are forced into identical states?
     
    Last edited: Jun 30, 2014
  2. jcsd
  3. Jun 30, 2014 #2

    Bill_K

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    It's called the Spin-Statistics Theorem, and it's an absolutely fundamental result in QFT. Half-integer spin particles must be fermions, and integer spin particles must be bosons.

    QM is totally different in this respect from classical physics. The difference makes itself evident in thermodynamics, for example, where classical statistics leads to the Gibbs Paradox.

    Saying that the two-particle wavefunction ψ(x1, x2) vanishes when x1 = x2 does not prevent the two particles from being in the same state.
     
  4. Jun 30, 2014 #3
    I haven't regarded the Gibbs paradox as paradoxical in a while. Isn't the real point of it that it forces you to recognize the fundamental subjectivity of any given entropy measure? Why wouldn't something similar apply to quantum physics?
     
  5. Jun 30, 2014 #4
    "Saying that the two-particle wavefunction ψ(x1, x2) vanishes when x1 = x2 does not prevent the two particles from being in the same state. "

    Well, it certainly means there is a zero amplitude for finding them in the same position state, right? This restriction would also apply to prevent everything from sitting in the seperable single-particle-ground-level energy state, wouldn't it? The more such restricted particles you add, the higher the joint energy, just as with fermion electrons.
     
    Last edited: Jun 30, 2014
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