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

Kohn Sham, calculation of electronic density

  1. Jun 29, 2011 #1

    in the Kohn-Sham-Scheme one calculates the electronic density as follows:

    [itex]\rho =2 \sum_{i=1}^{N/2} \phi^*_i \phi_i[/itex]

    where the [itex]\phi[/itex] are the Kohn-Sham orbitals.

    This looks like the density of a closed shell slater determinant. But am I correct, that this way of calculating the denisty is nothing more than an ansatz in the Kohn-Sham scheme? Or is there a justification, why the density [itex]\rho[/itex] of the interacting system should be given by the above sum?
  2. jcsd
  3. Jun 29, 2011 #2
    The Kohn-Sham ansatz is that it is possible to construct a non-interacting system with the same density as the given interacting system. That way of calculating the density is correct for a non-interacting system, which is the same as the density of the interacting system by assumption.
  4. Jun 30, 2011 #3


    User Avatar
    Science Advisor

    Any DFT book (Parr-Yang for instance) goes into great detail on the assumptions and derivations behind the Kohn-Sham scheme.

    But in short, the ansatz is really that you can describe the true ground-state density in terms of a non-interacting "reference system" of electrons, under the influence of an effective potential Vs. If you have Vs, then your single determinant description is fine, as long as the true ground state is "non-interacting pure-state-Vs representable".

    A few some non-degenerate ground states exist where this is known not to be the case, but there's not a lot of knowledge on when this becomes important in general. As per the quote I provided in your other thread on KS-DFT, there's a bit of a tendency to attribute errors to the single-determinant description that are actually due to the inaccuracy of the functional.

    In general, for a non-degenerate ground state there does exist a set of orbitals ('natural orbitals', by definition) which diagonalize the single-particle density matrix. So that much isn't an approximation.
  5. Jun 30, 2011 #4
    thank you for your answers.

    just to be absolutely shure that i got it right.

    the way of calculating the density by the formula
    [itex]\rho =2 \sum_{i=1}^{N/2} \phi^*_i \phi_i[/itex]
    is a direct consequence of the assumption of non-interacting V-representabillty.

    Could you please confirm or deny this?
  6. Jun 30, 2011 #5


    User Avatar
    Science Advisor

    No it is not. Try reading http://www.diss.fu-berlin.de/diss/servlets/MCRFileNodeServlet/FUDISS_derivate_000000002262/02_kap2.pdf" [Broken]
    Last edited by a moderator: May 5, 2017
  7. Jul 1, 2011 #6
    hi alxm,

    you said:

    In Kohn-Sham-Scheme one assumes "non-interacting pure-state-Vs representabillity", thus the single determinant description is fine, thus the way of calculating the density as [itex]\rho =2 \sum_{i=1}^{N/2} \phi^*_i \phi_i[/itex] is fine. (In this sense my last post (no. 4) should be understand)

    Can you confirm this?
  8. Jul 13, 2011 #7
    i repost my question:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook