# Hartree product and the Slater determinant

I am very new to theoretical chemistry so my question may sound very basic to most. It is known that the Hartree product does not satisfy the antisymmetry principle for electrons while the Slater determinant is able to do so. Thus, the electronic wavefunction is often written as a Slater determinant, rather than a Hartree product. However, I am just wondering if the Hartree product can still be used to approximate the electronic wavefunction or if it is still being use to do so in some theoretical methods? And if the Hartree product is used, what other measures are implemented to ensure that the antisymmetry principle is not violated? Also, in Density Functional Theory, which is used to approximate the wavefunction, the Hartree product or the Slater determinant?

• The Hartree product will always produce a physically unsound wave function guess; it cannot obey the antisymmetry principle. The Slater determinant is the easiest formalism to guarantee it. It is therefore used on most applications. – Martin - マーチン Nov 4 '18 at 15:06
• @Martin-マーチン But are there any cases where it is used instead? Because I was just reading this introductory DFT text, which made no mention of Slater determinants under the DFT section and only talked about Slater determinants under the section on HF... Anyway, is Slater determinant even relevant to DFT since the wavefunction isn' really computed? – Tan Yong Boon Nov 4 '18 at 16:16
• @TanYongBoon strictly speaking, you don't need Slater determinants for DFT because it solves for the energy using the density and so you can do orbital free/pure DFT that doesn't use Slater determinants or orbitals at all. In practice, we often formulate DFT using the Kohn-Sham equations, which computes orbitals/Slater determinants for a fictitious 1 particle potential. – Tyberius Nov 4 '18 at 17:04
• @TanYongBoon basically, the Hartree product can't be used unless it is antisymmetrized and the most intuitive way to do that is to form a Slater determinant from the same set of orbitals. – Tyberius Nov 4 '18 at 17:06
• @Tyberius If I may enquire further, how is the Hartree product different from the Slater determinant in terms of approximating the wavefunction? Yes, we know that only the latter satisfies antisymmetry principle and that it is also more complicated to compute for large N. But are there any other advantages of using a Slater determinant over Hartree product as an approximation of the wavefunction? – Tan Yong Boon Nov 4 '18 at 23:51

## 1 Answer

The Hartree product can definitely be used for approximate calculations; as you might know Hartree developed his method for atoms in 1927 (for reference, the Schrödinger equation was discovered in 1926) and he implicitely used a Hartree product for the wave function, although the connection to the exact solution of the many-body Schrödinger equation via the variational principle wasn't know. It was Slater in 1930 (and, independently, Fock) to introduce the determinant form which bears his name. When using the Hartree product one has to account for the Pauli principle "by hand", by allowing only two electrons to occupy the same spatial orbital. By starting with a Hartree product, enforcing the Pauli principle and applying the variational principle one obtains the Hartree equations, which are similar to the Hartree-Fock ones but are missing the exchange term, usually indicated by $$K$$; this term is, however, smaller than the electrostatic term $$J$$ so Hartree calculations are meaningful and give useful results, although of course very approximate ones.

Kohn-Sham DFT calculations use in principle the Slater product form, but in practice, depending on the exchange-correlation functional used, they lead to equations which may be more similar to the Hartree ones (i.e., without the exchange term) than to the Hartree-Fock ones. In the original LDA (local density approximation) functional the exchange term is dropped and replaced by a functional of the electron density which is much simpler to compute. As a result, DFT-LDA is similar in complexity to the Hartree method, not to the Hartree-Fock (a rather significant simplification). However, most modern functionals for molecules (as opposed to solid state calculations), e.g. B3LYP, re-introduce the exchange term (with a scaling coefficient).

The Hartree form of the wave function can be also used to compute expectation values (e.g., to compute dipole moments); you can check by looking a the Slater-Condon rules that the results you obtain with Hartree functions are, for one-body operators, the same you'd get using the proper Slater-determinant form.