Eq. 7.21 isE 0 ¼
X
nuclei AZA
ð
r 0 ðr 1 Þ
r 1 A
dr 1 #1
2
X^2 ni¼ 1cKS 1 ð 1 Þjr^21 jcKS 1 ð 1 Þþ1
2
ðð
r 0 ðr 1 Þr 0 ðr 2 Þ
r 12dr 1 dr 2 þEXC½r 0 Exc[r0] is a correction term to the electronic kinetic and potential energy; most
of this energy is (we hope!) treated classically by the other terms [1].
Reference
- Cramer CJ (2004) Essentials of computational chemistry. Wiley, Chichester, England, sec-
tions 8.3 and 8.4
Chapter 7, Harder Questions, Answers
Q5
Merrill et al. have said the “while solutions to the [HF equations] may be viewed as
exact solutions to an approximate description, the [KS equations] are approxima-
tions to an exact description!” Explain.
Solutions to the Hartree–Fock equations are exact solutions to an approximate
description because:
The HF equations are approximate mainly because they treat electron–electron
repulsion approximately (other approximations are mentioned in the answer sug-
gested for Chapter 5, Harder Question 1). This repulsion is approximated as
resulting from interaction between two charge clouds rather than correctly, as the
force between each pair of point-charge electrons. The equations become more
exact as one increases the number of determinants representing the wavefunctions
(as well as the size of the basis set), but this takes us into post-Hartree–Fock
equations. Solutions to the HF equations are exact because the mathematics of
the solution method is rigorous: successive iterations (the SCF method) approach an
exact solution (within the limits of the finite basis set) to the equations, i.e. an exact
value of the (approximate!) wavefunctionCHF.
The Kohn–Sham equations are approximations because the exact functional
needed to transform the electron density functionrinto the energy is unknown.
They are approximations to an exact description because the equations (as distinct
from methods of solving them) involve no approximations, with the ominous caveat
that the form of ther-to-EfunctionalExcis left unspecified.
Answers 641