# Simplified Hartree-Fock Computations on Second-Row Atoms

Simplified Hartree-Fock Computations on Second-Row Atoms

Modern computational quantum chemistry has developed largely from applications of the Hartree–Fock method to atoms and molecules [1–3]. A simple representation of a many-electron atom is given by a Slater determinant constructed from occupied spin-orbitals:

N

Ψ(1⋯N)=

1

N!

| ||||

ϕ N |

| ||||

ϕ N |

| ||||

⋯ |

| ||||

ϕ N |

where spin-orbitals are products of the form

ϕ(x)=ψ(r)

α

{

β

A Slater determinant automatically satisfies the Pauli exclusion principle, requiring that spin-orbitals be, at most, singly occupied. The Hamiltonian for an -electron atom, in hartree atomic units, is given by:

N

H=--+

N

Σ

i=1

1

2

2

∇

Z

r

i

N

Σ

i>j=1

1

r

ij

The corresponding approximation to the total energy is then given by

E=+-

N

Σ

i=1

H

i

N

Σ

i>j=1

J

ij

K

ij

where , and are, respectively, the core, coulomb and exchange integrals.

H

J

K

This Demonstration carries out a simplified Hartree–Fock computation on the ground states of the atoms He to Ne, to 10. Assume a single Slater determinant, thus possibly shortchanging open-shell configurations. The computation involves only , and orbitals, approximated by modified Slater-type orbitals:

Z=2

1s

2s

2p

ψ

1s

3/2

α

π

-αr

e

ψ

2s

3

5

β

π-αβ+

2

α

2

β

α+β

3

-βr

e

ψ

2p{x,y,z}

5/2

γ

π

-γr

e

These are normalized and mutually orthogonal, which simplifies the computation. The orbital parameters are chosen so as to minimize the total energy, in accordance with the variational principle. You can vary these using the sliders. Actually, since the possible values of vary over inconveniently large ranges, the sliders vary internal parameters that determine them. The values of are shown in the graphic.

α,β,γ

α,β,γ

α,β,γ

The blue bar shows the exact energy of the atomic ground state (actually, the exact nonrelativistic value) in hartrees. The red bar shows the calculated energy, given the displayed values of . You should try to adjust the three parameters to get the lowest possible energy. You can never reach the exact energy, however, since a single Hartree–Fock determinant fails to account for correlation energy, which involves instantaneous electron-electron interactions.

α,β,γ