Kohn Sham, calculation of electronic density

[Derivator]

hi,

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?

 

[daveyrocket]

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.

 

[alxm]

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 V_s. If you have V_s, then your single determinant description is fine, as long as the true ground state is "non-interacting pure-state-V_s 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.

 

[Derivator]

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?

 

[alxm]

No it is not. Try reading http://www.diss.fu-berlin.de/diss/servlets/MCRFileNodeServlet/FUDISS_derivate_000000002262/02_kap2.pdf"

 

[Derivator]

hi alxm,

you said:

If you have V_s, then your single determinant description is fine, as long as the true ground state is "non-interacting pure-state-V_s representable".

In Kohn-Sham-Scheme one assumes "non-interacting pure-state-V_s 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?

 

[Derivator]

i repost my question:

hi alxm,
In Kohn-Sham-Scheme one assumes "non-interacting pure-state-V_s 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?