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

A Symmetrisation of wave function for fermions

  1. Apr 27, 2017 #1
    The wave function for fermions has to be anti-symmetric with respect to exchange of positions of electrons, but what if it depends on wave vector as well. Does they have to be exchanged as well, in other words, for two-electron system what is correct

    Ψ(r1,k1,r2,k2) = - Ψ(r2,k1,r1,k2)


    Ψ(r1,k1,r2,k2) = - Ψ(r2,k2,r1,k1)
  2. jcsd
  3. Apr 27, 2017 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    The wave vector and the position are just different representations of the same Hilbert space. You can use either to represent your state but not both at the same time.

    That being said, it is the overall state that should be antisymmetric under the exchange of the electrons. If you have an additional degree of freedom that the state depends on (spin comes to mind) then you need to make the full state antisymmetric. If the state in the additional degree of freedom is already antisymmetric your spatial wavefunction will be symmetric. Positronium in the spin-0 state comes to mind.
  4. Apr 27, 2017 #3
    Thanx for quick replay,
    That somewhat clarifies my problem.
    However I have particles in continuum, in the presence of some potential, that are described by Coulomb (Coulomb-like to be more precise) wave function that does depend on both, wave vector and position at the same time.
    The system is in singlet state, so spin related part can be taken out.
    Last edited: Apr 27, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted