Page 21 - 71 the abc of dft_opt

4.2. HARTREE-FOCK 41 42 CHAPTER 4. TWO ELECTRONS
To improve on our estimate, and restore the variational principle, we should evaluate the where
expectation value of V ee on our wavefunction: v H (x) = δU = 1 ∞ dx n(x )δ(x − x ) = n(x) (4.16)
#
#
#
δn(x) −∞
1 1 1 Z
∞ ∞ 2 # 2 2 ∞
#
V ee = dx dx |φ(x)| |φ(x )| δ(x − x ) = Z dx exp(−4Z|x|) = . (4.10) is the Hartree potential. This is also known as the classical or electrostatic potential, as it
#
−∞ −∞ −∞ 2
is the electrostatic potential due to the charge distribution in classical electrostatics (if the
For Helium (Z=2), V ee = +1 so that the ground state energy becomes E = −4 + 1 = −3. electrostatic interaction were a δ-function, for this case). (Compare with Eq. (1.6) of the
Because we evaluated all parts of the Hamiltonian on our trial wavefunction, we know that introduction.) This equation is the Hartree-Fock equation for this problem.
the true E ≤ −3.
Exercise 15 Hartree-Fock equations for two electrons
However, using the variational principle, we see that, with almost no extra work, we can
Derive the Hartree-Fock equation for two electrons, Eq. (4.15), by minimizing the energy as
do better still. We chose the length scale of our orbital to be that of the V ee = 0 problem,
i.e., α = Z, the nuclear charge. But we can treat this instead as an adjustable parameter, a functional of the orbital, Eq. (4.14), keeping the orbital normalized, using the techniques
of Chapter ??.
and ask what value minimizes the energy. Inserting this orbital into the components of the
energy, we find: There are several important aspects of this equation that require comment. First, note the
( 2 )
T s = 2 α /2 V ext = −2Zα V ee = α/2 (4.11) factor of 2 dividing the Hartree potential in the equation. This is due to exchange effects:
The Hartree potential is the electrostatic potential generated by the total charge density due
so that the total energy, as a function of α, is
to all the electrons. A single electron should only see the potential due to the other electron,
2
E(α) = α − 2α(Z − 1/4). (4.12) hence the division by two.
Next, note that these are self-consistent equations, since the potential depends on the
Minimizing this, we find α min = Z − 1/4 and thus
density, which in turn depends on the orbital, which is the solution of the equation, etc.
< = 2 These can be solved in practice by the iterative method described in the introduction.
7
2
E min = −α 2 = −(Z − 1/4) = − = −3.0625. (4.13)
min To get an idea of what the effective potential looks like, we construct it for our approximate
4
solution above. There α = 1.75 for He, so the potential is a delta function with an added
We have lowered the energy by 1/16 of a Hartree, which may not seem like much, but is Hartree potential of 4 exp(−7|x|). This contribution is positive, pushing the electron farther
about 1.7 eV or 40 kcal/mol. Chemical accuracy requires errors of about 1 or 2 kcal/mol.
away from the nucleus. We say the other electron is screening the nucleus. The effective
The best solution to this problem is to find the orbital that produces the lowest energy.
nuclear charge is no longer Z, but rather Z − 1/4.
We could do this by including many variational parameters in a trial orbital, and minimize It is straightforward to numerically solve the orbital equation Eq. (4.15) on a grid, essen-
the energy with respect to each of them, giving up when addition of further parameters
tially exactly. But there is an underlying analytic solution in this case. Write
has negligible effect. A systematic approach to this would be to consider an infinite set
of functions, usually of increasing kinetic energy, and to include more and more of them. φ(x) = γ cosech (γ |x| + β) (4.17)
However, having learned some functional calculus, we can provide a more direct scheme.
where coth(β) = Z/γ, and γ = Z − 1/2. Insertion into Eq. (4.15) yields
The energy, as a functional of the orbital, is:
2
& = −γ /2, (4.18)
1 1 ∞ 2
E[φ] = 2T S [φ] + 2V ext [φ] + U[φ]/2, U[φ] = dx n (x) (4.14) 2
2 −∞ but this does not mean E = −γ . The energy of the effective non-interacting system is not
2
where n(x) = 2|φ(x)| , T S and V ext are the one-electron functionals mentioned in chapter 3, the energy of the original interacting system. In this case, from the derivation, we see E =
2
and U is the 1-d equivalent of the Hartree energy (discussed more below). For two electrons 2&−U/2 (since & = T S +V ext +U/2). We find U/2 = Z/2−1/6, or E = −Z +Z/2−1/12 as
the exact HF energy, slightly lower than our crude estimate using hydrogenic wavefunctions.
in HF, V ee = U/2, and we discuss this much more later. If we simply minimize this functional
But this slight difference is 1/48, or 0.6 eV or 13 kcal/mol, which is still a significant error
of the orbital, subject to the restriction that the orbital is normalized, we find:
by modern standards.
 
 1 d 2 1  In Fig. 4.1, we plot both the exact orbital and the scaled Hydrogenic orbital, showing there
− − Zδ(x) + v H (x) φ(x) = &φ(x), (4.15)
2 dx 2 is very little difference.
 2 