Chapter 5
Hartree Fock Calculations
5.1
Total Energy Calculation Using Molecular Orbitals
To generalize the LCAO method we have practiced for simple diatomic molecules, we
can express each molecular orbitals as
K
i (r) =
c i (r)
(5.1)
=1
where i is a spatial molecular orbital, is an atomic orbital, K is the number of
atomic orbitals, and c i is a coefficient.
For the H2+ problem, we used K = 2, and 1 = 1sA and 2 = 1sB . We found
that c11 = c12 = N+ for the ground state of this molecule. For a more accurate MO
calculation, we would like to use large number of basis functions.
For any molecule, we can use LCAO method to express the spatial part of the MO,
1
CHAPTER 5. HARTREE FOCK CALCULATIONS
and thus a Slater determinant for a molecular system can be thus constructed as
1 (1) 2 (1) n (1)
1 1 (2) 2 (2) n (2)
(1, , n) =
(5.2)
n!
..
..
..
.
.
.
1 ( n ) 2 ( n ) n ( n )
where i is a spin orbital, whose spatial part is expressed using LCAO. Note that spin
orbitals are orthonormal:
i | j = ij
Energy of the system can be now be computed using
| H | = E
When we substitute by the Slater determinant, we find three type of terms. The first
kind is
Hiicore =
d1 i (1)
N
Z
1
2i I
2
r
I =1 iI
!
i (1)
(5.3)
The above accounts for the kinetic+potential energy of an electron i moving in the
filed of the nuclei, computed using a molecular orbital i . The above integral is a
one-electron integral. Total contribution due to all the n electrons is
core
Etotal
=
Hiicore
(5.4)
Other type of terms in the expansion are the Columbic integrals
Jij =
Z Z
d1 d2 i (1) j (2)
1
(1) j (2)
r12 i
(5.5)
Jij is the Columbic electrostatic interaction between electrons in spin orbitals i and j.
This is a two-electron integral, and we will use a shorthand representation
1
ij ij .
r12
5.1. TOTAL ENERGY CALCULATION USING MOLECULAR ORBITALS
to express it. Total electrostatic interaction between an electron in i and others in n 1
spin orbitals is
n
Jij
EiCoulomb =
(5.6)
j 6 =i
Thus the total Coulomb electronic energy is
Coulomb
Etotal
n 1 n
Jij
(5.7)
j >i
The third type of integrals are the exchange integrals
Kij =
Z Z
d1 d2 i (1) j (2)
1
(2) j (1)
r12 i
(5.8)
The above integral is nonzero only if i and j have the same spin. This is also a
two-electron integral, and we will use a shorthand representation
1
ij ji .
r12
Exchange energy due to an electron in a spin orbital is
Exchange
Ei
Kij
(5.9)
j 6 =i
Thus the total exchange energy is given by
Exchange
Etotal
n 1 n
Kij
i
(5.10)
j >1
Total electronic energy is
Exchange
core
Coulomb
Etotal
+ Etotal
+ Etotal
Total energy of a closedshell1 system will be
n/2
E=2
i =1
Hiicore
n/2 n/2
i =1 j = i +1
n/2
4Jij 2Kij +
Jii
(5.11)
i =1
1 which means a fully paired electronic configuration representing the groundstate of a molecule/atom
with even number of electrons. Each (spatial) molecular orbital will be having two electrons of opposite
spins.
CHAPTER 5. HARTREE FOCK CALCULATIONS
4
By using the identity that Jii = Kii ,
n/2
E=2
Hiicore
i =1
5.2
n/2 n/2
2Jij Kij
(5.12)
i =1 j =1
The Hartree Fock Method
The Hartree Fock method involves obtaining oneelectron orbital functions (either
molecular orbital or atomic orbital) by solving a oneelectron eigen value equation:
Fi i = ei i
where Fi is a oneelectron operator, i is the atomic/molecular orbital, and ei is the
orbital energy. The effective oneelectron operator is constructed as
Fi = K.E. of electron i + nuclearelectron i attraction +
charge density of all other electron with electron i interaction (Columbic) +
exchange operator for electron i
(5.13)
The key point is in the meanfield approximation in building the effective oneelectron
Hamiltonian. The Fock operator for a closed-shell system is thus
Fi (1) = H core (1) +
n/2
2 Jj (1) K j (1)
(5.14)
j =1
where,
N
ZI
1
H core (1) = 21
2
r
I =1 1I
1
(2)
r12 j
Z
1
K j (1) i (1) =
d2 j (2) i (2) j (1)
r12
Jj (1) =
d2 j (2)
(5.15)
(5.16)
(5.17)
To solve Eqn. 5.14, J and K have to be constructed, which in turn requires {i }. Thus
a self consistent field (SCF) approach is used. One starts a calculation with an initial set of
5.3. ROOTHAANHALL EQUATIONS
By solving the eigenvalue equation, a new set of orbitals
{i }, which is used to build F.
which is again used
can be obtained. This new set is then used to build improved F,
to solve Eqn. 5.14 to get new set of orbitals, and the process continues. This iteration
is continued till one obtains a {i } that gives the lowest total electronic energy for the
system.
5.3
RoothaanHall Equations
When orbitals are expressed using the basis functions,
K
i =
ci
=1
total energy minimization as required for SCF, can be carried out by varying the coefficients {ci }; at minimum,
E
=0
ci
The Hartree Fock equation expressed using the basis functions
Fi (1)
ci (1) = ei
=1
ci (1)
(5.18)
=1
After multiplying the = K ci on the left side of the above equation, and integrating both sides, one can simplify the above equation, and arrives at
1 K K
core
E = P H + F
2 =1 =1
(5.19)
where
F =
core
H
+ P
1
(|) (|)
2
with
(|) =
Z Z
d1 d2 (1) (1)
Here charge density matrix
n/2
P = 2 ci ci
i =1
1
(2) (2)
r12
(5.20)
CHAPTER 5. HARTREE FOCK CALCULATIONS
6
and
core
H
=
#
K
1 2
ZI
(1)
d1 (1)
2
r RI |
I | 1
"
It may be noted that , , , and indicate basis functions, centered on same or
different atoms. The above equations are called the RoothaanHall equations for the
HartreeFock calculations. Note here that all the integrals as required for the energy
calculation is now casted using the basis functions. The density matrix P is only dependent on the coefficients. Thus computationally, this is advantageous, as the integrals
can be programmed for the basis functions, and the only quantity that change during
the SCF is the density matrix elements.
But the main advantage is that, we can construct matrix equations for carrying out
the above calculations:
FC = SCE
(5.21)
where F is the Fock matrix whose elements are F . Fock matrix is a square matrix of
size K K. Here the coefficient matrix
c1,1 c1,2
c2,1 c2,2
C=
.
..
..
.
cK,1 cK,2
c1,K
c2,K
..
.
cK,K
(5.22)
and the energy matrix
E=
e1
e2
..
.
..
.
..
.
eK
(5.23)
5.3. ROOTHAANHALL EQUATIONS
You may remember the standard eigen value problems that you solved in mathematics lectures in the matrix forms: the same mathematics can now be used for solving the
Hartree Fock equations. In order to do that we need the equation in FC = CE form, instead of Eqn. 5.21. Such a form can be arrived if S = I, which is a diagonal matrix with
all the diagonal elements equal 1. This however requires that all the basis functions be
mutually orthonormal, which is not obviously the case (for e.g. because basis functions
can be on different atoms).
We have a way out if we can convert the non-orthogonal basis to an orthogonal basis
for solving the above eigenvalue equation: Let X be a matrix such that
XT SX = I
Here XT means the transpose of the matrix X. We need to find X, which can diagonalize
the matrix S. This procedure is called diagonalization. There are many ways to do this
procedure. As
S1/2 SS1/2 = I
we shall use S1/2 as X. Computing S1/2 involves the following steps:
1. find U, that diagonalize S:
UT SU = D = diag(1 k )
2. find
1/2
D1/2 = diag(11/2
)
k
3. Then
S1/2 = UD1/2 UT
By knowing S1/2 , we can modify the Roothan-Hall equation as follows:
CHAPTER 5. HARTREE FOCK CALCULATIONS
1. Multiply S1/2 both sides of Eqn. 5.21:
S1/2 FC = S1/2 SCE = S1/2 CE
2. Inserting I = S1/2 S1/2 into the above equation
S1/2 F(S1/2 S1/2 )C = S1/2 CE
F0 C0 = C0 E
(5.24)
where F0 = S1/2 FS1/2 , and C0 = S1/2 C.
Eqn. 5.24 is in the form we would like to have. Now, we can solve Eqn. 5.24: as we
know, this equation is true only for
0
F EI = 0
This is the secular determinant and the solutions are {ek }. For a large matrix, the
procedure to find the solution of the secular determinent is by diagonalizing the matrix
F0 . Eigenvectors of this diagonalization procedure gives C0 and eigenvalues will be E.
Knowing C0 , we can obtain C from
C = S1/2 C0
