Numerical Methods in Quantum Mechanics - Dipartimento di Fisica

(the variations of all other normalization integers will be zero) and, using the
hermiticity of H as in Sec.4.1.1,
∫
δ〈ψ|H|ψ〉 = δφ ∗ k(1)f 1 φ k (1) dv 1 + c.c.
+ ∑ ∫
δφ ∗ k(1)φ ∗ j(2)g 12 φ k (1)φ j (2) dv 1 dv 2 + c.c. (6.12)
j≠k
This result is obtained by noticing that the only involved terms of Eq.(6.8) are
those with i = k or j = k, and that each pair is counted only once. For instance,
for 4 electrons the pairs are 12, 13, 14, 23, 24, 34; if I choose k = 3 the only
contributions come from 13, 23, 34, i.e. ∑ j≠k (since g is a symmetric operator,
the order of indices in a pair is irrelevant)
Thus the variational principle takes the form
⎡
⎤
∫
∫
δφ ∗ k(1)
φ ∗ j(2)g 12 φ k (1)φ j (2) dv 2 − ɛ k φ k (1) ⎦ dv 1 + c.c. = 0
⎣f 1 φ k (1) + ∑ j≠k
i.e. the term between square brackets (and its complex conjugate) must vanish
so that the above equation is satisfied for any variation:
f 1 φ k (1) + ∑ ∫
φ ∗ j(2)g 12 φ k (1)φ j (2) dv 2 = ɛ k φ k (1) (6.13)
j≠k
These are the Hartree equations (k = 1, . . . , N). It is useful to write down
explicitly the operators:
⎡
⎤
− ¯h2 ∇ 2
2m
1φ k (1) − Zq2 e
φ k (1) + ⎣ ∑ ∫
φ ∗
e r
j(2) q2 e
φ j (2) dv 2 ⎦ φ k (1) = ɛ k φ k (1)
1 r
j≠k
12
(6.14)
We remark that each of them looks like a Schrödinger equation in which in
addition to the Coulomb potential there is a Hartree potential:
where we have used
∫
V H (r 1 ) =
ρ k (2) q2 e
r 12
dv 2 (6.15)
ρ k (2) = ∑ j≠k
φ ∗ j(2)φ j (2) (6.16)
ρ j is the charge density due to all electrons differing from the one for which we
are writing the equation.
6.2.1 Eigenvalues and Hartree energy
Let us multiply Hartree equation, Eq(6.13), by φ ∗ k (1), integrate and sum over
orbitals: we obtain
∑
ɛ k = ∑ ∫
φ ∗ k(1)f 1 φ k (1)dv 1 + ∑ ∑
∫
φ ∗ k(1)φ k (1)g 12 φ ∗ j(2)φ j (2)dv 1 dv 2 .
k k
k j≠k
(6.17)
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