Does Kohn-Sham DFT use Slater determinants?

Question

In the Hartree method, it is known that the wavefunction of the system does not obey the antisymmetry principle of fermions - that is when you swap two particles, they don't up a negative sign. Therefore, electrons are filled by making sure that Pauli's exclusion principle is obeyed in every step. A Slater determinant is considered to be an improvement because it does satisfy the many-body antisymmetry nature of fermions, and often takes the form (for say, a 2 particle system):

\begin{aligned}\Psi (\mathbf {x} _{1},\mathbf {x} _{2})&={\frac {1}{\sqrt {2}}}\{\chi _{1}(\mathbf {x} _{1})\chi _{2}(\mathbf {x} _{2})-\chi _{1}(\mathbf {x} _{2})\chi _{2}(\mathbf {x} _{1})\}\\&={\frac {1}{\sqrt {2}}}{\begin{vmatrix}\chi _{1}(\mathbf {x} _{1})&\chi _{2}(\mathbf {x} _{1})\\\chi _{1}(\mathbf {x} _{2})&\chi _{2}(\mathbf {x} _{2})\end{vmatrix}}\end{aligned}

My question is, are Slater determinants found in practice in Kohn-Sham Density Functional theory?

Answer 1

As Nike has already mentioned, Slater determinants occur in KS-DFT in much the same way that they do in Hartree-Fock as a way to combine a set of one-electron orbitals into an antisymmetric many-electron wavefunction. However in DFT, as the name suggests, we are typically more interested in the density. The Hohenberg-Kohn theorems show that DFT can, in principle, produce the exact ground state density and the density is used to compute various molecular properties.

So why might we care about the wavefunction in DFT? In the limit of a complete basis set, it has been proven that the DFT density constructed from a single Slater determinant can reproduce exactly the density of a multideterminant method like Configuration Interaction. It is noteworthy that systems which seem to require a multideterminant description can in some sense be represented exactly with just a single Slater determinant.

A recent paper [1] by Frank Jensen explored how this manifests in practice by trying to fit the density of finite basis full CI calculations using a DFT density constructed from orthogonal orbitals. While the end goal is reproducing the density, the interesting thing about it is that it shows that even for a sufficiently large finite basis, a single Slater determinant should suffice as the "square root of the density".

1. Philip Jakobsen and Frank Jensen Representing Exact Electron Densities by a Single Slater Determinant in Finite Basis Sets J. Chem. Theory Comput. 2020, XXXX, XXX, XXX-XXX DOI: 10.1021/acs.jctc.0c01029

Answer 2

A couple of comments perhaps to clarify some points.

The first Hohenberg-Kohn theorem states that the one-electron density determines uniquely the energy, for a non-degenerate ground state. The second H-K theorem states that this density can be obtained by employing the variational principle, i.e. the density is obtained by minimizing the energy as a functional of a trial density. If the exact functional was known, this would give the exact density. BUT, the density MUST come from a proper antisymmetric wave function, you CANNOT use any freely varying density as your trial. This is known as the representability condition.

In the Kohn-Sham version of DFT, this is ensured by parameterizing the density as a sum of orbital contributions, and fulfilling the antisymmetry requirement by placing them in a Slater Determinant with the usual orthonormality conditions for the orbitals.

It can be shown that any valid density can be written in the form of a set of orbitals in a single Slater Determinant. This is a mathematical proof, and assumes a complete set of functions, i.e. a complete basis set. This is never fulfilled in practice, and in the above mentioned paper, we investigated how large the basis set should be in order to reduce the error to something like 'chemical accuracy'. The answer is a QZP quality basis set, but also shows that the DFT basis set should be the same as that used for generating the reference exact density. Exact here means exact within the basis set, i.e. full-CI.

The mathematical proof thus shows that it should not be necessary to have more than one Slater Determinant in KS-DFT, as any density that can be expressed as arising from a wave function written as a sum of Slater Determinants can be exactly reproduced by a set of KS-orbitals, as long as the basis set is sufficiently large.

Answer 3

From the Wikipedia article on Kohn-Sham equations:

"As the particles in the Kohn–Sham system are non-interacting fermions, the Kohn–Sham wavefunction is a single Slater determinant constructed from a set of orbitals."

So the answer to your question:

"Are slater determinants found in practice in Kohn-Sham Density Functional theory?"

is yes.
Other users that have a deeper knowledge of DFT than me might tell you more!