15.4 Derivation of the Hartree-Fock equations 505
15.4.2 Varying the single-particle wave functions
If we generalize the Euler-Lagrange equations to more variables and introduceN^2 Lagrange
multipliers which we denote byεμ ν, we can write the variational equation for the functional
of Eq. (15.23) as
δE−
N
∑
μ= 1
N
∑
ν= 1
εμ νδ
∫
ψμ∗ψν= 0. (15.23)
For the orthogonal wave functionsψμthis reduces to
δE−
N
∑
μ= 1
εμδ
∫
ψμ∗ψμ= 0. (15.24)
Variation with respect to the single-particle wave functionsψμyields then
N
∑
μ= 1
∫
δ ψμ∗hˆiψμdxi+^1
2
N
∑
μ= 1
N
∑
ν= 1
[∫
δ ψ∗μψν∗^1
ri j
ψμψνd(xixj)−
∫
δ ψμ∗ψν∗^1
ri j
ψνψμdridrj
]
+
N
∑
μ= 1
∫
ψμ∗hˆiδ ψμdri+
1
2
N
∑
μ= 1
N
∑
ν= 1
[∫
ψ∗μψν∗
1
ri j
δ ψμψνdridrj−
∫
ψμ∗ψν∗
1
ri j
ψνδ ψμdridrj
]
−
N
∑
μ= 1
Eμ
∫
δ ψ∗μψμdxi−
N
∑
μ= 1
Eμ
∫
ψμ∗δ ψμdri= 0.
(15.25)
Although the variationsδ ψandδ ψ∗are not independent, they may in fact be treated as
such, so that the terms dependent on eitherδ ψandδ ψ∗individually may be set equal to
zero. To see this, simply replace the arbitrary variationδ ψbyiδ ψ, so thatδ ψ∗is replaced by
−iδ ψ∗, and combine the two equations. We thus arrive at the Hartree-Fock equations
[
−
1
2
∇^2 i−
Z
ri
+
N
∑
ν= 1
∫
ψν∗(rj)
1
ri j
ψν(rj)drj
]
ψμ(xi)
−
[N
∑
ν= 1
∫
ψν∗(rj)
1
ri j
ψμ(rj)drj
]
ψν(ri) =εμψμ(ri).
(15.26)
Notice that the integration
∫
drjimplies an integration over the spatial coordinatesrjand
a summation over the spin-coordinate of electronj.
The two first terms are the one-body kinetic energy and the electron-nucleus potential.
The third ordirectterm is the averaged electronic repulsion of the other electrons. This
term includes the ’self-interaction’ of electrons wheni=j. The self-interaction is cancelled in
the fourth term, or theexchangeterm. The exchange term results from our inclusion of the
Pauli principle and the assumed determinantal form of the wave-function. The effect of the
exchange is for electrons of like-spin to avoid each other. Atheoretically convenient form of
the Hartree-Fock equation is to regard the direct and exchange operator defined through the
following operators
Vμd(ri) =
∫
ψμ∗(rj)^1
ri j
ψμ(rj)drj (15.27)
and
Vμex(ri)g(ri) =
(∫
ψ∗μ(rj)
1
ri jg(rj)drj
)
ψμ(ri), (15.28)