1 HartreeFock Approximation
We have a system of
N
interacting fermions with a Hamiltonian
H
=
N
i
=1
h
(
i
) +
N
i<j
V
int
(
i,j
)
.
(1)Here,
h
(
i
) is the “singleparticle Hamiltonian” for the particle
i
and
V
int
(
i,j
)is the part of the Hamiltonian describing the interaction between particles
i
and
j
. By the “singleparticle Hamiltonian” we mean the terms that wouldbe present if there were only one fermion in the system. For the case of
N
electrons in an atom with proton number
Z
at the nucleus, these terms wouldbe
h
(
i
) =
p
2
i
2
m
−
Ze
2
r
i
,V
int
(
i,j
) =
e
2

r
i
−
r
j

.
Again we ignored the rest of the terms like spinorbit interaction which wouldintroduce spindependent terms into the Hamiltonian. Note that in Eq. (1),the interaction part is summed over each particle pair only once. As a result,there are
N
2
=
N
(
N
−
1)2terms in the last sum.Such problems cannot be solved exactly in general (there are a few uninteresting exceptions). As a result, we need a general approximation procedurefor solving such manybody equations. Obviously, what we want to calculate is important in choosing the approximation procedure. Hence, thereare various methods developed to cope with diﬀerent aspects of such kindof problems. The very ﬁrst step should be obtaining a “feel” for the energylevels for those problems. That certainly means that we don’t need accuratenumerical values.Independent particle picture is this ﬁrst step in visualizing the energylevels of most of the manybody problems. This is the language used indescribing the electronic states of atoms in Atomic Physics courses. Whenwe are saying that in Lithium, there are two electrons in the 1s orbital and1
one electron in the 2s orbital, we are using that language. However, we say, inthose atoms, the orbital wavefunctions and their energies are diﬀerent thanthose of the hydrogenic atoms. Our purpose in here is to formalize thosekind of statements.Therefore, the question we would like to answer is “What is the bestindependentparticle approximation for an interacting system of fermions?”And the answer can be provided by the variational theory. We should forma manybody wavefunction of the atom as a Slater determinant using someoneparticle states
ϕ
1
,...,ϕ
N
. And, we should choose these oneparticlestates as the ones that minimize the expectation value of the Hamiltonian inEq. (1). This is the HartreeFock approximation.
2 Expectation Values
Let us suppose that we have
N
arbitrary oneparticle states
ϕ
1
,...,ϕ
N
. Weshould choose them as an orthonormal set, since this will make the calculationof expectation values easier. So, we have
ϕ
i

ϕ
j
=
δ
i,j
i,j
= 1
,...,N.
(2)As a result, we have
N
2
equations giving us a restriction on the oneparticlestates. (Some may notice that, there are only
N
(
N
+ 1)
/
2 equations inEq. (2), but the number of real quantities ﬁxed to a certain value is
N
2
. So,there are
N
2
real equations.)We form the Slater determinant from these
N
states asΨ(1
,
2
,...,N
) = 1
√
N
!
ϕ
1
(1)
ϕ
1
(2)
... ϕ
1
(
N
)
ϕ
2
(1)
ϕ
2
(2)
... ϕ
2
(
N
)... ... ... ...
ϕ
N
(1)
ϕ
N
(2)
... ϕ
N
(
N
)
.
By using the manybody wavefunction thus formed, we will calculate theexpectation value of the Hamiltonian,
H
, as
Ψ

H

Ψ
=
N
Ψ

h
(1)

Ψ
+
N
2
Ψ

V
int
(1
,
2)

Ψ
.
These quantities were calculated before, so we just reproduce the results inhere
Ψ

H

Ψ
=
N
i
=1
E
i
+
i<j
(
C
ij
−
E
ij
)
,
(3)2
where
E
i
=
ϕ
i

h

ϕ
i
,C
ij
=
ϕ
i
ϕ
j

V
int

ϕ
i
ϕ
j
,E
ij
=
ϕ
j
ϕ
i

V
int

ϕ
i
ϕ
j
.
Here,
C
ij
is usually called as the Coulomb integral for the orbitals
ϕ
i
and
ϕ
j
(speciﬁcally considering electrons in atoms and molecules), and
E
ij
iscalled the exchange integral. We note that in Eq. (3) sum over Coulomband exchange integrals is over pairs of states. That sum can be written inanother way. We just need to note that
C
ij
=
C
ji
and
E
ij
=
E
ji
. It canalso be noticed that, by the deﬁnition of Coulomb and exchange integrals,we have the relation
C
ii
=
E
ii
. Keep in mind that
C
ii
is not physicallymeaningful (interaction of an electron with itself?). But, in order to simplifythe Eq. (3), those terms are frequently used. By using these relations, Eq. (3)can be rewritten as,
Ψ

H

Ψ
=
N
i
=1
E
i
+ 12
N
i
=1
N
j
=1
(
C
ij
−
E
ij
)
.
(4)
3 Eﬀective Hamiltonian
Now, all we have to do is to minimize the expectation value in Eq. (4) withthe restrictions in Eq. (2). We can use the Lagrange multiplier method tocarry out this minimization. As a result the quantity,
A
=
Ψ

H

Ψ
−
N
i
=1
N
j
=1
Λ
ij
ϕ
i

ϕ
j
(5)should be minimized, where Λ
ij
are the Lagrange multipliers. The Lagrangemultipliers in this expression can take on complex values, but you shouldnot be concerned about this. Obviously, only real valued functions can beminimized, hence the expression in Eq. (5) should be real valued. This impliesthat the Λ matrix is hermitian. If you are concerned about the presenceof complex numbers in a realvalued expression, you should derive Eq. (5)starting from the realvalued forms of the expressions in Eq. (2).3
Now, we can consider the variations in each element of the oneparticlestate set
ϕ
1
,...,ϕ
N
. In eﬀect, what we will be doing is evaluating the functional derivatives and equating them to zero. Since we are using the Lagrangemultiplier method, we can vary each
ϕ
i
in an arbitrary way. That means,we will make sure that the replacement

ϕ
i
→
ϕ
i
+

δϕ
i
, where

δϕ
i
is an arbitrary inﬁnitesimal state, will not change the expression in Eq. (4)up to ﬁrst order. (To be precise, we will set

δϕ
i
=

ψ
where

ψ
isan arbitrary oneparticle state and we will equate the order
term of thevariation in Eq. (4) to 0.)Now, the ﬁrst order variation of realvalued expressions that depend ona quantum mechanical state (say

ϕ
i
) are of the form
δϕ
i

α
+
α

δϕ
i
for some (perhaps complicated) “state”

α
. This expression is then setequal to 0 at the extrema. Since

δϕ
i
is arbitrary, this will only imply that

α
= 0 (show). If we keep this in mind, we can quickly ﬁnd the conditionsfor minimum in our calculation. As a result we ﬁnd,
δA
=
N
i
=1
δϕ
i

h

ϕ
i
+
N
i,j
=1
δϕ
i
ϕ
j

V
int

ϕ
i
ϕ
j
−
δϕ
i
ϕ
j

V
int

ϕ
j
ϕ
i
−
N
i,j
=1
Λ
ij
δϕ
i

ϕ
j
+
complexconjugate
.
Now, if arbitrary inﬁnitesimal values of

δϕ
i
makes the expression abovevanish, the the following equation should be satisﬁed for the
i
th
state:(
h
+
V
Coul
+
V
Exch
)

ϕ
i
=
N
j
=1
Λ
ij

ϕ
j
,
(6)where
V
Coul
and
V
Exch
are operators which are a little bit complicated. In theabstract case that we are using in here, we will deﬁne them by their matrix4