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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

i(r) =K

=1

c i(r) (5.1)

where i is a spatial molecular orbital, is an atomic orbital, K is the number ofatomic orbitals, andc i is a coefficient.

For the H+2 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


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2 CHAPTER 5. HARTREE FOCK CALCULATIONS

and thus a Slater determinant for a molecular system can be thus constructed as

(1, , n) = 1n!

1(1) 2(1) n(1)

1(2) 2(2) n(2)...

... ...

1(n) 2(n) n(n)

(5.2)

wherei 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

Hcoreii =

d1i(1)

1

22i

N

I=1

ZIriI

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

Ecoretotal=n

i

Hcoreii (5.4)

Other type of terms in the expansion are the Columbic integrals

Jij =

d1d2i(1)j(2) 1

r12i(1)j(2) (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 representationij

1r12 ij

.


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5.1. TOTAL ENERGY CALCULATION USING MOLECULAR ORBITALS 3

to express it. Total electrostatic interaction between an electron in iand others inn 1spin orbitals is

ECoulombi =n

j=i

Jij (5.6)

Thus the total Coulomb electronic energy is

ECoulombtotal =n1

i

n

j>i

Jij (5.7)

The third type of integrals are the exchange integrals

Kij =

d1d2i(1)j(2) 1

r12i(2)j(1) (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 representationij

1r12ji

.

Exchange energy due to an electron in a spin orbital is

E

Exchange

i =

n

j=i Kij (5.9)

Thus the total exchange energy is given by

EExchangetotal =

n1

i

n

j>1

Kij (5.10)

Totalelectronicenergy is

Ecoretotal+ECoulombtotal +EExchangetotal

Total energy of aclosedshell1 system will be

E= 2n/2

i=1

Hcoreii +n/2

i=1

n/2

j=i+1

4Jij 2Kij

+

n/2

i=1

Jii (5.11)

1which 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.


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By using the identity that Jii =Kii,

E= 2n/2

i=1

Hcoreii +n/2

i=1

n/2

j=1

2JijKij

(5.12)

5.2 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:

Fii =ii

where Fi is a oneelectron operator, i is the atomic/molecular orbital, and i 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) = Hcore(1) +

n/2

j=1

2 Jj(1) Kj(1)

(5.14)

where,

Hcore(1) = 1221

N

I=1

ZIr1I

(5.15)

Jj(1) =

d2j(2) 1

r12j(2) (5.16)

Kj(1)i(1) =

d2j(2)

1

r12i(2)

j(1) (5.17)

To solve Eqn. 5.14, Jand

Khave to be constructed, which in turn requires{i}. Thus

aself consistent field(SCF) approach is used. One starts a calculation with an initial set of


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5.3. ROOTHAANHALL EQUATIONS 5

{i}, which is used to build F. By solving the eigenvalue equation, a new set of orbitalscan be obtained. This new set is then used to build improved F, which is again used

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 thesystem.

5.3 RoothaanHall Equations

When orbitals are expressed using the basis functions,

i =K

=1

ci

total energy minimization as required for SCF, can be carried out by varying the coeffi-

cients{ci}; at minimum,E

ci

=0

The Hartree Fock equation expressed using the basis functions

Fi(1)K

=1

ci(1) =iK

=1

ci(1) (5.18)

After multiplying the = Kci on the left side of the above equation, and inte-grating both sides, one can simplify the above equation, and arrives at

E=1

2

K

=1

K

=1

P Hcore +F (5.19)where

F =Hcore +

K

K

P

(|)1

2(|)

(5.20)

with

(|) =

d1d2 (1)(1) 1

r12(2)(2)

Here charge density matrix

P =2n/2i=1

cici


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and

Hcore =

d1 (1)

1

22

K

I

ZI|r1 RI|

(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 matrixP is only depen-

dent 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 Fis the Fock matrix whose elements are F. Fock matrix is a square matrix of

size KK. Here the coefficient matrix

C=

c1,1 c1,2 c1,K

c2,1 c2,2 c2,K...

... ...

cK,1 cK,2

cK,K

(5.22)

and the energy matrix

E=

1 0 0

0 2 0...

... ...

0 0 K

(5.23)


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5.3. ROOTHAANHALL EQUATIONS 7

You may remember the standard eigen value problems that you solved in mathemat-

ics 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 inFC = CEform, in-

stead of Eqn. 5.21. Such a form can be arrived ifS = 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

XTSX= I

HereXT means the transpose of the matrix X. We need to find X, which candiagonalize

the matrixS. This procedure is called diagonalization. There are many ways to do this

procedure. As

S1/2SS1/2 =I

we shall useS1/2 as X. Computing S1/2 involves the following steps:

1. findU, that diagonalizeS:

UTSU= D = diag(1 k)

2. find

D1/2 =diag(1/21 1/2k )

3. Then

S1/2 =UD1/2UT

By knowingS1/2, we can modify the Roothan-Hall equation as follows:


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1. MultiplyS1/2 both sides of Eqn. 5.21:

S1/2FC= S1/2SCE = S1/2CE

2. InsertingI = S1/2S1/2 into the above equation

S1/2F(S1/2S1/2)C= S1/2CE

F C=C E (5.24)

whereF =S1/2FS1/2, andC =S1/2C.

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

F EI= 0This is the secular determinant and the solutions are{k}. For a large matrix, theprocedure to find the solution of the secular determinent is by diagonalizing the matrix

F. Eigenvectors of this diagonalization procedure givesC and eigenvalues will be E.

KnowingC , we can obtainC from

C= S1/2C

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