Subject: A question on DFT

From: "Armando Navarro" <[email-masked]>
Date: Mon, 31 Jan 2000 17:20:22 +0100

Dear members:
I have a doubt about the Kohn-Sham implementation of DFT theory.
Provide we would have the "true" functional, we can in principle =
determine the "true" electronic density. But, it could be always =
represented as a single determinant of Kohn-Sham molecular orbitals?
Thanks in Advance
Armando Navarro, Ph-D student
Facultade de quimica
Departamento de quimica Organica
Universidade de Santiago de Compostela

e-mail:[email-masked]

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From: Jan Labanowski <[email-masked]>
Date: Mon, 31 Jan 2000 16:26:48 -0500 (EST)

> Universidade de Santiago de Compostela

At the same time, you do not have to do restricted calculations. You
can do unrestricted UKS calculations even for singlets (why not?...).

The result may be spin contaminated (from autopsy, it usually is
when you have a multiconfigurational beast like a polyradical -- you have
to remember that you cannot start SCF with the same alpha and beta
charge densities, coefficients, etc., since there would be nothing
in the SCF method to kick you out from the local solution: F(alpha) = F(beta)).
The contaminated spin state is by definition a combination or pure parent and
higher spin states (all of them may be formally multideterminant).
Hence, the "pure" DFT function (state) derived from the unrestricted
calculations need not be a single determinant but can be a multideterminant.
While we do not necessarily know what is the form of these functions (states)
(though there is a way to find out, since formally the highest possible
multiplet is when all electrons are unpaired, which will be a single
determinant, and we can back substitute), we can guess that they are
multideterminant.

Hence, the pure spin parent function will be most likely also a combination
of determinants. And, while we are doing a single determinant DFT, and
do not count twice the correlation by introducing MCSCF, we actually
work on the multiconfigurational states.

And by the way, purists will tell you that KS orbitals are not MOs...
(I am joking here, I will not get involved in religious wars... I just
do not have the ammo... I am not a DFT guru, but it is a damn interesting
problem. I wish I knew this stuff better...).

Jan
[email-masked]

Jan K. Labanowski            |    phone: 614-292-9279,  FAX: 614-292-7168
Ohio Supercomputer Center    |    Internet: [email-masked] 
1224 Kinnear Rd,             |    http://www.ccl.net/chemistry.html
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From: [email-masked]
Date: Mon, 31 Jan 100 23:30:22 +0100 (MET)

Yes, any density can be represented as a single slater determinant
(this is actually very easy to be seen) -- however you need a
complete one-particle basis set to represent the orbitals.

The thing to keep in mind is that in general there is no single slater
determinant BUILT FROM ORBITALS EXPANDED IN A GIVEN BASIS SET, which
represents the density of, say, the full-CI wavefunction in the
same basis set.

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