Page 45 - 71 the abc of dft_opt

90 CHAPTER 11. SIMPLE EXACT CONDITIONS
Exercise 40 Fermi-Amaldi correction
An approximation for the exchange energy is E X [n] = −U[n]/N. This is exact for one or
two (spin-unpolarized) electrons. Show that it is not size-consistent.
This problem becomes acute when there is more than one center in the external potential.
Chapter 11 +
Consider, e.g. H 2 . As the nuclear separation grows from zero, the charge density spreads out,
so that a local approximation produces a smaller and smaller fraction of the correct result.
Simple exact conditions +
Exercise 41 Stretched H 2
+
Calculate the LSD error in the exchange energy of H 2 in the limit of infinite separation.
1
In this chapter, we introduce various simple exact conditions, and see how well LDA 11.3 Lieb-Oxford bound
meets these conditions.
Another exact inequality for the potential energy contribution to the exchange-correlation
11.1 Size consistency functional is the Lieb-Oxford bound:
U XC [n] ≥ C LO E LDA [n] (11.4)
An important exact property of any electronic structure method is size consistency. Imagine X
you have a system which consists of two extremely separated pieces of matter, called A and where C LO ≤ 2.273. It takes a lot of work to prove this result, but it provides a strong bound
B. Their densities are given, n A (r) and n B (r), and the overlap is negligible. Then a size- on how negative the XC energy can become.
consistent treatment should yield the same total energy, whether the two pieces are treated
separately or as a whole, i.e., Exercise 42 Lieb-Oxford bound
What does the Lieb-Oxford bound tell you about & unif (r s )? Does LDA respect the Lieb-Oxford
C
E[n A + n B ] = E[n A ] + E[n B ]. (11.1) bound?
Configuration interaction calculations with a finite order of excitations are not size-consistent,
but coupled-cluster calculations are. Most density functionals are size-consistent, but some 11.4 Bond breaking
with explicit dependence on N are not. Any local or semilocal functional is size-consistent.
Another infamous difficulty of DFT goes all the way back to the first attempts to understand
H 2 using quantum mechanics. In the absence of a magnetic field, the ground state of two
11.2 One and two electrons
electrons is always a singlet. At the chemical bond length, a Hartree-Fock single Slater deter-
For any one electron system minant is a reasonable approximation to the true wavefunction, and yields roughly accurate
numbers:
V ee = 0 E X = −U E C = 0 (N = 1) (11.2) HF
Φ (r 1 , r 2 ) = φ(r 1 )φ(r 2 ) ≈ Ψ(r 1 , r 2 ) (11.5)
This is one of the most difficult properties for local and semilocal density functionals to get However, when the bond is stretched to large distances, the correct wavefunction becomes
right, because the exact Hartree energy is a non-local functional. The error made for one- a linear combination of two Slater determinants (the Heitler-London wavefunction), one for
electron systems is called the self-interaction error, as it can be thought of as arising from each electron on each H-atom:
the interaction of the charge density with itself in U. By spin-scaling, we find 1
Ψ(r 1 , r 2 ) = √ (φ A (r 1 )φ B (r 2 ) + φ B (r 1 )φ A (r 2 )) , (11.6)
E X = −U/2 (N = 2, unpolarized) (11.3) 2
for two spin-unpolarized electrons. LSD does have a self-interaction error, as can be seen for where φ A is an atomic orbital centered on nucleus A and similarly for B. The total density is
the H atom results in the tables. just n(r) = n A (r) + n B (r), where n I (r) is the hydrogen atom density centered on nucleus I.
1 c !2000 by Kieron Burke. All rights reserved. But note that the density remains unpolarized.
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