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Antisymmetric wavefunction

  1. Mar 19, 2009 #1

    I have always read the texts in which they have mentioned that for the electrons which are fermions the wave function should be antisymmetric, but I have not yet found a good proof for that.

    In some books they have mentioned the pauli's exclusion principle and some relations, but still the concept is not clear to me. I hope you would kindly help me understand this matter.

  2. jcsd
  3. Mar 19, 2009 #2


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    It's quite simple. Assume you have a two-particle wave function [tex]\psi(r_1,r_2)[/tex], where rx are the coordinates, including spin for the respective particles. What happens, or rather, what can happen to that wave function if the particles are interchanged? I.e. how does [tex]\Psi(r_1,r_2)[/tex] relate to [tex]\Psi(r_2,r_1)[/tex]? Well, if the two particles are indisinguishable, the two wave functions should be physically equivalent; identical with respect to all measurable properties. Which means the only difference is at most a difference in the phase of the wavefunction. We can write:
    where 'x' is a phase factor (e^i*w). Particle interchange is the equivalent of multiplying by 'x'.

    If you change the particles back again, you multiply by x again. But obviously this is the same as our original wavefunction so:
    [tex]\Psi(r_1,r_2)=x^2\Psi(r_1,r_2)[/tex] and [tex]x^2 = 1[/tex].

    So x = ±1. If x = 1, the wave function is symmetric with respect to particle interchange ([tex]\psi(r_1,r_2)=\psi(r_2,r_1)[/tex]) then that's what's defined as a boson. Correspondingly if x = -1, the wave function is antisymmetric ([tex]\psi(r_1,r_2)=-\psi(r_2,r_1)[/tex]) and that's what's called a fermion.

    Now, if you form a multi-particle wave function in terms of single-particle wave functions, then it's a simple product in the case of bosons:
    [tex]\Psi(r_1,r_2) = \psi_1(r_1)\psi_2(r_2)[/tex]
    The two particles have no problems being in the same state, since obviously if [tex]\psi_1=\psi_2[/tex] the condition [tex]\Psi(r_1,r_2)=\Psi(r_2,r_1)[/tex] is still obeyed.

    But this simple product wavefunction does not work for fermions since it doesn't follow the anti-symmetry condition. The simplest two-particle wave function that does, in terms of single-particle wave functions is:
    [tex]\Psi(r_1,r_2)=\frac{1}{\sqrt{2}}(\psi_1(r_1)\psi_2(r_2) - \psi_1(r_2)\psi_1(r_1))[/tex]
    (where the 1/sqrt(2) is for normalization)
    More generally, for any number of particles, this takes the form of a Slater determinant.

    This obeys the anti-symmetry condition, since swapping r1 and r2 leads to a sign change:
    [tex]\psi_1(r_1)\psi_2(r_2) - \psi_1(r_2)\psi_2(r_1) = -(\psi_1(r_2)\psi_2(r_1) - \psi_1(r_1)\psi_2(r_2))[/tex]

    But can the particles be in the same state now? No. If [tex]\psi_1=\psi_2[/tex] then the two products are identical and the wave function becomes zero. This is the Pauli exclusion principle: Bosons can occupy the same states while having the same coordinates, Fermions can not.

    (Now that I've spent my time writing an explanation that's already given in hundreds of text books, maybe it should be added to the PF "library"?)
  4. Mar 19, 2009 #3


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    you can add the post in PF library if you want :-)

    Also, please refer to some reference books and/or free additional web-references :-)
  5. Mar 19, 2009 #4
    The sentences you have written about the pauli exclusion really made me feel and understand the antisymmetric nature of the wavefunctions for the fermions. Thanks so much for putting time on my answer.
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