Page 57 - 71 the abc of dft_opt

114 CHAPTER 14. DISCONTINUITIES
electron in an orbital should see an effective nuclear charge of Z − (N − 1), the N − 1
because there are N − 1 electrons close to the nucleus. Since v ext (r) = −Z/r, and at
large distances v H (r) = N/r, this implies v X (r) = −1/r. Since this effect will occur in
any wavefunction producing the exact density, this must also be true for v XC (r), so that the
Chapter 14 correlation contribution must decay more rapidly. In fact,
v X (r) → −1/r, v C (r) → −α(N − 1)/2r 4 (r → ∞) (14.3)
Discontinuities where α(N − 1) is the polarizability of the (N − 1)-electron species. The decay of the
correlation potential is so rapid that it has only ever been clearly identified in an exact
−
calculation for H .
14.1 Koopman’s theorem
Exercise 69 Asymptotic potentials
Derive the asymptotic condition on v X (r) from the exact decay of the density. How does
Since the large distance condition Eq. (5.18) is true for both the interacting wavefunction LDA
and an independent particle description, the following approximate theorem was first noted v X (r) behave at large distances?
by Koopmans (who later won a Nobel prize in economics). In a Hartree-Fock calculation, if As mentioned earlier, LDA does not provide very realistic looking exchange-correlation
you ignore relaxation, meaning the change in the self-consistent potential, when an electron potentials. Because the density decays exponentially in the tail region, so too will the LDA
is removed from the system, you find potential. In Fig. 14.1, we illustrate this effect. Of course, LDA also misses the shell

I HF = E HF (N − 1) − E HF (N) ≈ −& N (14.1) structure between the 1s and 2s and 2p electrons, just as smooth approximations for T S miss
shell structure. This is partially explained by the above argument: the long-range decay of
where & N is the eigenvalue of the highest occupied orbital. 0
-1
Exercise 67 Koopman’s theorem for 1d He -2
For the accurate HF 1-d He calculation, test Koopman’s theorem. v XC (r) -3
But, in exact density functional theory, Koopman’s theorem is exact. To see why, note -4
that Eq. (5.18) is true independent of the strength of the interaction. Consider the coupling -5
constant λ, keeping the density fixed. Since the density is fixed, its asympototic decay is the -6
Ne atom
same for any value of λ, i.e., the ionization potential is independent of λ. In particular, at -7 exact
LDA
λ = 0, its just −& HOMO , where we now refer to the HOMO of the KS potential. Thus -8
0 0.2 0.4 0.6 0.8 1
I = E(N − 1) − E(N) = −& HOMO . (14.2) r
We will see below that this condition is violated by all our approximate functionals. Figure 14.1: XC potential in the Ne atom, both exact and in LDA

Exercise 68 Asymptotic behavior of the density v X (r) is due to non-self-interaction in the exact theory, an effect missed by LDA. This has a
From the information in this and earlier chapters, deduce the limiting values of the asymptotes strong effect on the HOMO orbital energy (and all those above it), as will be discussed in
on the right of Fig. 5.2. more detail below.

14.2 Potentials 14.3 Derivative discontinuities
It is straightforward to argue for the exact asymptotic behaviour of the exchange-correlation We can understand the apparent differences between LDA and exact potentials in much more
potential for a Coulombic system. Consider first exchange. Far from the nucleus, and detail. Consider what happens when two distincct subsystems are brought a large distance
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