Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pauli exclusion principle question (again)

  1. Sep 11, 2014 #1
    I've seen it stated in many places that the reason why atoms don't collapse is due to the pauli exclusion principle. The exclusion principle is given as a required anti-symmetry in the wavefunction of electrons.

    I don't understand how this principle was derived, or where it comes from. (I've seen the derivaiton in Bates, but all that proves is that wavefunctions that start either symmetric or asymmetric, that have a Hamiltonian that is symmetric with respect to particles (particles behave the same) will remain symmetric or antisymmetric.)

    It seems to me that the electrostatic repulsion of electrons suffices to explain the electronic structure of an atom. If you have a joint wave-function for 2 electrons, the electrostatic repulsion potential would enforce Phi(x,x) = 0 for states where x1=x2. If you had "boson electrons" where Phi(x2,x1) = +Phi(x1,x2), but Phi(x1,x1) = 0 due to the electrostatic repulsion, how you could distinguish between the behavior of an atom with "boson electrons" and one with "fermion electrons"? The conjugate product of the wavefunction with itself should come out the same in both cases.

    If you cannot distingish between these cases, then why is antisymmetry necessary to explain electronic structure, and not simply some sort of repulsion potential that blows up at 0 distance? (The repulsion potential ensures that the ground state wavefunction cannot be separable, and isn't equivalent to the independent ground state wavefunctions of single electrons in the same potential)

    What experiment could you build that could distinguish between particle symmetry vs. antisymmetry that isn't also confounded with fermions all having repulsion interactions in terms of other forces?
    Last edited: Sep 11, 2014
  2. jcsd
  3. Sep 11, 2014 #2


    User Avatar
    Gold Member

    I don't know about Pauli's line of thought, but the wave function for a two electron system, gives a finite probability for them being in the same place.
  4. Sep 11, 2014 #3


    User Avatar

    Staff: Mentor

    Why do you believe this to be true?
  5. Sep 11, 2014 #4


    User Avatar
    Gold Member

    This can help too.
  6. Sep 11, 2014 #5
    <quote>Why do you believe this to be true? </quote>
    For states of finite energy, Phi doesn't (and cant) diverge from zero wherever the hamiltonian blows up. I'll need to confirm this later (when I have more time), but it has been true in my investigations so far.
  7. Sep 11, 2014 #6


    User Avatar
    Science Advisor

    Pauli exclusion follows from the spin-statistics theorem of quantum field theory.
  8. Sep 11, 2014 #7
    An example: if electrons were bosons atomic orbitals would fill up differently. "Bosonic electrons" would all try to crowd into the lowest energy level of the atom instead of being forced into higher orbitals as the lower ones filled up. (They wouldn't entirely succeed, because of the electrostatic repulsion between electrons, but they would do a lot better than fermionic electrons).

    The antisymmetry requirement ##\phi(x_1, x_2) = -\phi(x_2, x_1)## is *much* stronger than the requirement ##\phi(x_1, x_2) = 0##. It's common to quote the Pauli exclusion principle as saying "no two electrons (with the same spin) can be in the same place at the same time," but this is a simplification intended for people who don't know about wave functions. The actual rule is the much stronger antisymmetry restriction.

    As Avodyne says, the fact that spin-1/2 particles like electrons have to be fermions can be proven in quantum field theory, but has to be taken on faith before you learn QFT.
  9. Sep 11, 2014 #8


    User Avatar
    Science Advisor

    The Hamiltonian does not blow up. The Coulomb singularity can be (and in practice is) compensated by a singularity in the kinetic energy operator. Or, re-formulated: There is a finite, non-zero chance that electrons of opposite spin sit right on top of each other, and they can do so just fine as long as the derivatives of the wave function are singular at the coalesence points. Search for Kato cusp conditions if you are interested in this topic.

    In the weak correlation limit (e.g., for core electrons), the positions of opposite spin electrons even become completely uncorrelated. That means, the chance of finding two opposite spin core electrons right on top of each other is not significantly lower than finding them anywhere else in individually equivalent positions.
  10. Sep 16, 2014 #9
    Thank you. I was wondering why "two electrons on top of each other" was permissible in any case
    (and was trying to pursue some wild goose chase where a magnetic force singularity cancelled the coulomb force singularity for opposite spins. I suppose this isn't necessary then.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook