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Contents

The Hartree-Fock method is the underlying approximation to nearly all methods ofcomputational chemistry, both ab initio and semi-empirical. Therefore, a clear understanding ofwhere it comes from and what it means is essential for quantum chemists.These notes outline the Hartree-Fock method and derive several of the most useful HFequations. These notes are organized as follows:

1. General principles and definitions.1.1)Wavefunctions and Slater determinants:1.2) The Method of Lagrange Undetermined Multipliers:

2. Energy expression for a Slater determinant wave function2.1) One-electron terms:2.2) Two-electron terms2.3) Slater's rules

3. The general Hartree-Fock equations.

4. The Roothan Equations for molecules4.1) Closed shell HF equations4.2) Introduction of a basis set4.3) Open shell HF equations

5. Illustrating concepts: The hydrogen molecule, H2

5.1) Elliptical coordinates5.2) Overlap integral5.3) The Heitler-London Method for Hydrogen—Detailed Calculation

5.3a) The Restricted Hartree-Fock (RHF) Method for Hydrogen

5.3b) The Unrestricted Hartree-Fock (UHF) Method for Hydrogen

5.3c) The Full CI Method for Hydrogen

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1. General principles and definitions

1.1) Wavefunctions and Slater determinants:

In quantum chemistry, one is usually interested in solving the molecular Schrödinger equation:

HΨ = EΨ (Eq. 1)

where H is the non-relativistic, Born-Oppenheimer, molecular Hamiltonian (in atomic units):

H = −1

2∇i

2

i=1

N∑ −

ZA

riAA=1

M∑

i=1

N∑ +

1

rijj> i

N∑

i=1

N∑ −

ZAZB

rABB>A

M∑

B=1

M∑ (Eq. 2)

where i and j are electron indices, A and B are a nuclei indices, Z is the nuclear charge, N is thenumber of electrons, and M is the number of nuclei. The final term in (Eq. 2) is the nuclearrepulsion energy which is a function only of the nuclear coordinates and will be ignored in thesubsequent discussion. For convenience, H is usually divided into a one-electron term (h(i)) anda two-electron term (g(i,j)) defined as follows:

h i( ) = −1

2∇i

2 −ZA

riAA=1

M∑

g i , j( ) =1

rij

so that

H = − h i( )i=1

N∑ + g i , j( )

j> i

N∑

i=1

N∑ (Eq. 3)

In the Hartree-Fock method, the electronic wavefunction is usually approximated as a singleSlater determinant (vide infra):

ΨHFr x 1 ,

r x 2 ,...,

r x N( ) =

ψ1r x 1( ) ψ2

r x 1( ) L ψN

r x 1( )

ψ1r x 2( ) ψ2

r x 2( ) L ψ N

r x 2( )

M M O M

ψ1r x N( ) ψ2

r x N( ) L ψN

r x N( )

(Eq. 4)

which is abbreviated as:

ΨHF

r x 1 ,

r x 2 ,...,

r x N( ) =

1

N!det ψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( )

3

where each ψ is a orthonormal, single electron spin orbital which actually is a product of aspatial orbital φ(r) and a spin function σ(s)=α(s) or β(s). These functions have the followingproperties:

ψ i ψ j = δij

α s( ) α s( ) = β s( ) β s( ) = 1

α s( ) β s( ) = β s( ) α s( ) = 0

(Eq. 5)

The mathematical properties of the Slater determinant guarantee two closely-related propertiesthat are necessary for any multielectron wavefunction: antisymmetry and the Pauli principle.The first states that swapping the positions and spins of any two electron simply changes thesign of the wavefunction. This corresponds to swapping two rows of the determinant whichchanges its sign. The Pauli principle states that two electrons of identical spin cannot be at theexact same point in space. This would correspond to two rows of the determinant being equal,which would make the overall determinant equal zero, (hence meaning that the probability ofthis happening is zero.)

It is important to remember that the Slater determinant given in (eq. 4) is not a matrix, but adeterminant that can be expanded to yield a sum of products of spin orbitals. More precisely, thedeterminant expansion contains terms for all possible permutations of indices (xi) in the spin-orbitals:

Ψ =1

N!detψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( ) =

1

N!−1( )Pi Pi ψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( )( )

i=1

N!∑

=1

N!ψ1

r x 1( )ψ2

r x 2( )...ψ N

r x N( )( ) − ψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( )( )L{ }

(Eq. 6)

where Pi is the permutation operator that permutes the indices in all N! possible combinations.

1.2) The Method of Lagrange Undetermined Multipliers:A functional is just a function that takes a function as an argument and returns a value. Forexample, consider the functional F and the function g:

If F g x( )[ ] ≡ dxg x( )0

∞∫ and g x( ) ≡ e−x2

then F g x( )[ ] = 12

π

Functional derivatives can be defined, analogous to the derivative of regular functions, thatspecify how the value of a functional varies as its argument varies. These derivatives can beused to find the functions that minimize a particular functional.

Sometimes it is desirable to find the function that minimizes a functional, but with an additionalconstraint. For example, we might want to find the function, ƒ, that minimizes F[ƒ], but with theadditional condition that G[ƒ]=n (where G is another functional and n is a number). One methodfor accomplishing this is called the method of Lagrange undetermined multipliers. This involves

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constructing a function, L, which includes both functionals and a term l, which is theundetermined multiplier:

L f, λ( ) = F f[ ]− λ G f[ ] − n( )

Recall that the constraint on ƒ is that G[ƒ]=n, so the term in parenthesis is zero and therefore theminimum of L occurs for exactly the same ƒ as for F. Hence we simply needto set the first variation d of f to 0:

δL f, λ( ) = δF f[ ] − λ δG f[ ]( ) = 0

then solving for ƒ in terms of l and then solving for l by imposing the original constraint,G[ƒ]=n.

In order to illustrate the method of Lagrange's undetermined multipliers let us consider thesimple case of a function of two variables z=f(x,y) for which extrema are seek and suppose thatthe two variables x and y are subjected to the condition: φ(x,y)=0. The function f(x,y) can berepresented by a surface and the boundary condition φ(x,y)=0 by a curve. This is shown in theFig. below:

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Fig. Extrema with boundary conditions

For the localisation of the extrema we are only interested in those values of f(x,y) which alsofulfil the condition φ(x,y) = 0, i.e. in those points located on the surface curve K Thus, ourproblem reduces to the localization of the extrema on K! For this purpose, let us consider thebunch of niveau-lines f(x,y) = c (= constant). One of these curve will touch K in exactly onepoint: the seek extremum. The projection of this niveau-line onto the (x,y)-plane will also touchthe curve φ(x,y) = 0 in exactly one point, and moreover, both curves will have the samederivative (common tangent). That is.

∂f

∂x:

∂f

∂y=

∂φ∂x

:∂φ∂y

or, in other words,

∂f

∂x∝

∂φ∂x

and∂f

∂y∝

∂φ∂y

Thus, introducing proportionality constants –λ – Lagrange’s multipliers – we get the equations:

φ(x,y)=0

z=f(x,y)

x

y

K

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∂f

∂x= −λ

∂φ∂x

and∂f

∂y= − λ

∂φ∂y

or

∂f

∂x+ λ

∂φ∂x

= 0 and∂f

∂y+ λ

∂φ∂y

= 0

Since the left hand side of these equations are nothing but the partial derivatives of.

F(x,y) = f(x,y) + λφ(x,y)

We yield the Method of Lagrange Undetermined Multipliers.

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2. Energy expression for Slater Determinant:

The quantum mechanical expectation value for the total molecular energy is given by:

E =Ψ H ΨΨ Ψ

(Eq. 7)

where H is the Hamiltonian given in Eq. 2. To calculate the energy of a single Slaterdeterminant wavefunction, we simply substitute Eq. 3 into this expression (note that thedenominator of Eq. 7 equals 1 since the Slater determinant is normalized):

E =1

N!det ψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( ) h i( ) 1

N!detψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( )

i=1

N∑

+1

N!det ψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( ) g i , j( ) 1

N!det ψ1

r x 1( )ψ 2

r x 2( )...ψN

r x N( )

j> i

N∑

i=1

N∑

(Eq. 8)

To evaluate Eq. 8, we will separately treat first the one-electron and then the two-electron termsin the Hamiltonian.

2.1) One-electron terms:

E1 =

1

N!det ψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( ) h i( ) 1

N!det ψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( )

i=1

N∑ (Eq. 9)

If we simply expand each Slater determinant, each integral will be a sum of N! x N! terms. Thekey to reducing this expression is to use the normalization conditions given in Eq. 5 and the factthat h(i) only operates on the i-th electron coordinate, so that all other spin orbitals can befactored out into separate integrations. Using the Greek subscripts to indicate an arbitrarypermutation of the spin orbitals, the value of any single term in the N! x N! expansion has thefollowing value (see next):

ψαr x 1( )ψβ

r x 2( )...ψγ

r x N( ) h i( ) ψχ

r x 1( )ψδ

r x 2( )...ψε

r x N( )

= ψµr x i( ) h i( ) ψν

r x i( ) ψα

r x 1( )ψβ

r x 2( )...ψγ

r x N( ) ψχ

r x 1( )ψδ

r x 2( )...ψ ε

r x N( )

where ψ µr x i( ) and ψ ν

r x i( ) do not appear in the second term above.

= ψµr x i( ) h i( ) ψν

r x i( ) ψα

r x 1( ) ψχ

r x 1( ) ψβ

r x 2( ) ψδ

r x 2( ) L ψγ

r x N( ) ψε

r x N( )

= ψµr x i( ) h i( ) ψν

r x i( ) δαχδβδL δγε

This means that the expression will be non - zero only when ψα and ψχ have the same electroncoordinate, etc. for all orbital overlaps. Hence, out of the N! x N! terms in an h matrix element,only (N - 1)! will be non-zero, since that's how many ways N - 1 coordinates can be distributedinto N - 1 overlap integrals.

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Thus, the h matrix element is:

ψαr x 1( )ψβ

r x 2( )...ψγ

r x N( ) h i( ) ψχ

r x 1( )ψδ

r x 2( )...ψε

r x N( )

=1

N!ψ1

r x i( ) h i( ) ψ1

r x i( ) + ψ2

r x i( ) h i( ) ψ2

r x i( ) + L + ψN

r x i( ) h i( ) ψN

r x i( )( )

This result is the same for each of the N h(i) matrix elements in the expression for E1 , so thatthe one-electron energy for a Slater determinant wavefunction is:

E1 = ψ i

r x 1( ) h 1( ) ψ i

r x 1( )

i=1

N∑ (Eq. 10)

(Note that the sum is now over spin orbitals rather than electron indices.)

2.2) Two-electron terms:

E2 =

1

N!det ψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( ) g i , j( ) 1

N!det ψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( )

j> i

N∑

i=1

N∑ (Eq. 11)

Now, following a very similar approach to that we took for the one-electron terms, we consideran arbitrary term from the N! x N! terms in the g(i,j) matrix element:

ψαr x 1( )ψβ

r x 2( )...ψγ

r x N( ) g i , j( ) ψχ

r x 1( )ψδ

r x 2( )...ψ ε

r x N( )

= ψµr x i( )ψλ

r x j( ) g i , j( ) ψ ν

r x i( )ψ σ

r x j( ) ψ α

r x 1( )ψβ

r x 2( )...ψγ

r x N( ) ψχ

r x 1( )ψδ

r x 2( )...ψε

r x N( )

= ψµr x i( )ψλ

r x j( ) g i , j( ) ψ ν

r x i( )ψ σ

r x j( ) ψα

r x 1( ) ψ χ

r x 1( ) ψβ

r x 2( ) ψδ

r x 2( ) L ψγ

r x N( ) ψε

r x N( )

= ψµr x i( )ψλ

r x j( ) g i , j( ) ψ ν

r x i( )ψ σ

r x j( ) δαχδβδL δγε

the product of delta functions will be non - zero in only two situations:

(i) m = n and l = s, or (ii) m = s and l = n, so:

= ψµ

r x i( )ψ ν

r x j( ) g i , j( ) ψµ

r x i( )ψ ν

r x j( ) δαχδβδL δγε

or

= ψµ

r x i( )ψ ν

r x j( ) g i , j( ) ψ ν

r x i( )ψµ

r x j( ) δαχδβδL δγε

In the N! x N! expansion for each g(i,j) above, there will be (N-1)! terms of type (i) and(N-1)! terms of type (ii).

In order to determine the sign of the arbitrary matrix elements derived above, we need to

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look carefully at the signs of the individual terms in the Slater determinant expansion. The finalresult is that the first term above has a positive sign and the second a negative sign. Hence, theresulting terms for each g(i,j) are:

1

N!detψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( ) g i , j( ) 1

N!det ψ1

r x 1( )ψ2

r x 2( )...ψN

r x N( )

=1

N!

ψ1r x i( )ψ2

r x j( ) g i , j( ) ψ1

r x i( )ψ2

r x j( ) − ψ1

r x i( )ψ2

r x j( ) g i , j( ) ψ 2

r x i( )ψ1

r x j( ) +

ψ1r x i( )ψ3

r x j( ) g i , j( ) ψ1

r x i( )ψ3

r x j( ) − ψ1

r x i( )ψ3

r x j( ) g i , j( ) ψ3

r x i( )ψ1

r x j( ) + L

ψN−1r x i( )ψN

r x j( ) g i , j( ) ψ N−1

r x i( )ψN

r x j( ) − ψ N−1

r x i( )ψN

r x j( ) g i , j( ) ψ N

r x i( )ψ N−1

r x j( )

Thus, we get the following expression for the two-electron energy in terms of spin orbitals:

E2 =1

2

dr x 1d

r x 2ψ i

* r x 1( )ψ i

r x 1( ) 1

r12ψ j

* r x 2( )ψ j

r x 2( )∫

− dr x 1d

r x 2ψi

* r x 1( )ψ j

r x 1( ) 1

r12ψ j

* r x 2( )ψi

r x 2( )∫

j=1

N∑

i=1

N∑ (Eq. 12)

Note that the term in brackets is simply the difference between the two terms, not a vector.Combining Eq. 10 and Eq. 12, we get our completed energy expression for a Slater determinantwave function:

E ψi{ }[ ] = dr x 1ψ i

* r x 1( )h r

x 1( )ψ jr x 1( )∫

i=1

N∑ +

1

2

dr x 1d

r x 2ψi

* r x 1( )ψ i

r x 1( ) 1

r12ψ j

* r x 2( )ψ j

r x 2( )∫

− dr x 1d

r x 2ψi

* r x 1( )ψ j

r x 1( ) 1

r12ψ j

* r x 2( )ψi

r x 2( )∫

j=1

N∑

i =1

N∑

defining the following notation for integrations over spin orbitals:

i hj[ ] ≡ dr x 1ψ i

* r x 1( )h r

x 1( )ψ jr x 1( )∫

i j k l[ ] ≡ dr x 1d

r x 2ψ i

* r x 1( )ψ j

r x 1( ) 1

r12ψk

* r x 2( )ψl

r x 2( )∫

we get the following final expression for the energy of a Slater determinant of spin orbitals:

E ψi{ }[ ] = i h i[ ]i=1

N∑ +

1

2i i j j[ ]− ij ji[ ]

j=1

N∑

i=1

N∑ (Eq. 13)

2.3) Slater's rules:

This can be generalized for the calculation of matrix elements of one- and two-electronoperators between determinantal wave functions.

(i) One-electron operators

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Consider the matrix elements A F B of a one-electron operator F = f(i)i=1

N∑ between the two

Slater determinants: A = a1a2 ...ai ...aN* and B = b1b2...b i ...bN . We get, thus, the following

rule:

A F B =0 if A ≠ B b ymore than onespinorbital

± ak f bl if A ≠ B by exact ly1spinorbi ta lak a n d bl

ai f bii=1

N∑ if A = B

where + or – corresponds to the sign (parity) of the permutation needed to rearrange |B> asB = a1 ...ak−1b lak+1 ...aN with respect to A = a1 ...ak−1akak+1 ...aN .

(ii) Two-electrons operators

Consider the matrix elements A G B of a two-electrons operator G = g( i , j )j> i

N∑

i=1

N∑ between the

two Slater determinants: A = a1a2 ...ai ...aN* and B = b1b2...b i ...bN . We get, thus, the

following rule:

A G B =

0 if A ≠ B b ymore than two spinorbitals

± akbm g a lbn − a kal g bmbn[ ] if A ≠ B b yexactly 2spinorbitals aka l{ } and bmbn{ }± ak bt g ata t − a kat g a tbl[ ]

t∑ if A ≠ B b yexactly 1spinorbitalak and bl

ak ak g a ta t − akat g aka t[ ]k> t∑ if A = B

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3. Hartree-Fock Equations:

Given the final energy expression derived in section 2 (Eq. 13), we need to derive the equationsnecessary to find the optimal spin-orbitals to variationally minimize the energy. Starting with thefinal energy expression given above (using the [ij|kl] integral nomenclature defined in section2), we want to minimize E[{yi}], subject to the constraint that the spin-orbitals are orthonormal:

E ψi{ }[ ] = i h i[ ]i=1

N∑ +

1

2i i j j[ ]− ij ji[ ]

j=1

N∑

i=1

N∑

with the constraint: [i|j] = δij

Using the method of Lagrange undetermined multipliers for optimizing a functional withconstraints (see section I), we define:

L ψi{ }[ ] = E ψ i{ }[ ]− εij i j[ ] − δij( )j=1

N∑

i=1

N∑

where εij are the N2 undetermined multipliers. We want to find the {ψi} that minimize E, so wewant to find were δL = 0:

δL ψ i{ }[ ] = δE ψi{ }[ ] − ε ijδ i j[ ]j=1

N∑

i=1

N∑ = 0 (Eq. 14)

Expanding the variations in the two terms on the right hand side of equation 14:

δE ψ i{ }[ ] = δ i hi[ ]i=1

N∑

+ δ 1

2i i j j[ ]− ij ji[ ]

j=1

N∑

i=1

N∑

= δψi h ψ i[ ]+ ψi h δψ i[ ]i=1

N∑

+1

2δψ iψ i ψ jψ j[ ] + ψ iδψ i ψ jψ j[ ] + ψ iψi δψ jψ j[ ] + ψ iψi ψ jδψ j[ ]

j=1

N∑

i=1

N∑

−1

2δψ iψ j ψ jψ i[ ] + ψ iδψ j ψ jψ i[ ] + ψ iψ j δψ jψ i[ ] + ψ iψ j ψ jδψ i[ ]

j=1

N∑

i =1

N∑

From the definition of the integral nomenclature, we have the following equivalencies:

[i|j] = [j|i]*

[i|h|j] = [j|h|i]*

[ij|kl] = [kl|ij]

[ij|kl] = [ji|lk]*

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Using these relations, the expression for dE reduces to:

δE ψ i{ }[ ] = δψ i h ψ i[ ]i=1

N∑ +

1

2δψ iψ i ψ jψ j[ ] + δψ iψ j ψ jψ i[ ]

j=1

N∑

i=1

N∑

+ complex conjugate

Similarly, for the second term in Eq. 14:

ε ijδ i j[ ]j=1

N∑

i =1

N∑ = εij δi j[ ]

j=1

N∑

i=1

N∑ + ε ij i δj[ ]

j=1

N∑

i=1

N∑

= ε ij δi j[ ]j=1

N∑

i=1

N∑ + ε ij

* δi j[ ]*

j=1

N∑

i=1

N∑

= ε ij δi j[ ]j=1

N∑

i=1

N∑ + complex conjugate

Before combining the different terms in the variation, let’s define two operators forconvenience, note that they are defined in terms of their result when operating on a spin orbital.

J i 1( )ψ j 1( ) = dr x 2ψ i

* 2( ) 1

r12ψ i 2( )∫

ψ j 1( )

K i 1( )ψ j 1( ) = dr x 2ψ i

* 2( ) 1

r12

ψ j 2( )∫

ψ i 1( )

J(1) is known as the Coulomb operator and K(1) is known as the Exchange operator. Note thatK(1) is a “non-local” operator since the result of its operation on an orbital depends on the valueof that orbital throughout all of space, and not just at point x1.

Substituting these definitions and accumulating both terms in the expression for the variation,we get:

δE ψ i{ }[ ] = d

r x 1δψ i

* 1( ) h 1( )ψi 1( ) + J j 1( ) − K j 1( )( )ψ i 1( ) − ε ijψ j 1( )j=1

N∑

j=1

N∑

∫

i=1

N∑

+ complex conjugate = 0

The variation in the orbital, δψ, is completely arbitrary, so that in order for the above expressionto always be equal to zero, the term in the square brackets must be zero for all values of i.Slightly rearranging the terms in the brackets we get:

h 1( ) + J j 1( ) − K j 1( )( )j=1

N∑

ψ i 1( ) = − ε ijψ j 1( )

j=1

N∑ for i=1, 2, 3, … N

The term in square brackets in this expression is known as the Fock operator, ƒ(1). Substitutingthis we get a series of N integral equations known as the Hartree-Fock equations:

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f 1( )ψ i 1( ) = εijψ j 1( )j=1

N∑ for i=1, 2, 3, … N (Eq. 15)

or in matrix form:

f 1( ) r ψ 1( )† =

r ψ 1( )† ε (Eq. 16)

where r ψ 1( ) is a vector of N spin orbitals and ε is the matrix of Lagrange multipliers.

Note that we retain considerable flexibility in the choice of the Lagrange undeterminedmultipliers {εij} introduced to insure that the spin orbitals are orthonormal. To see this, considera new vector of spin orbitals

r θ related to

r ψ by a unitary transformation T:

r θ †T =

r ψ † .

Substituting into the Hartree-Fock equations:

f 1( ) r ψ 1( )† = ε

r ψ 1( )†

f 1( )r θ 1( )†T =

r θ 1( )†Tε

multiplying both sides of the equation by T† :

f 1( )r θ 1( )† TT† =

r θ 1( )†TεT†

f 1( )r θ 1( )† =

r θ 1( )†TεT† s i n c eTT† = 1( )

We have complete freedom to choose T, so that we can choose T to be the unitary matrixthat diagonalizes ε. Using this choice of T, leads us to one particular case of the Hartree-Fockequations, known as the Canonical Hartree-Fock Equations:

f 1( )ψ i 1( ) = εijψ j 1( )j=1

N∑ for i=1, 2, 3, … N

but, since we' re using a diagonalized e: (Eq. 17)

f 1( )ψ i 1( ) = εiiψ i 1( ) for i=1, 2, 3, … N

It is important to remember that the set of spin orbitals solved for in the above equation are onlyone set of an infinite number of valid spin orbitals that would arise from different choices of thetransformation matrix T, all yielding the same energy. Another point to remember, that will beclear when we derive the Roothan equations, is that the Fock operator, ƒ(1), is itself a functionof all N spin orbitals. This means that the Hartree-Fock equations must be solved iteratively.

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4. Roothan Equations for molecules:

The Hartree-Fock equations derived above make no assumption about the functional form of theindividual spin orbitals. In fact, for simple systems, these equations can be solved by standardmethods for numerically solving integral equations. However, a generally more useful approachis to recast these integral equations as matrix equations. The first step is to convert the canonicalHartree-Fock equations into a spin-independent form (for closed-shell molecules). After that,we’ll substitute as basis set expansion for the spatial part of the orbitals to generate the Roothanmatrix equations that are at the heart of virtually all Hartree-Fock programs.

4.1) Closed shell HF equations:In closed shell molecules the electrons are all paired into N/2 orbitals. The spatial portion of agiven pair of such orbitals is identical:

ψ i

r x ( ) =

φ ir r ( )α ω( )

φ ir r ( )β ω( )

where φ ir r ( ) are the spatial orbitals,

r r is the 3D spatial coordinate, and ω is the spin coordinate.

Using the nomenclature:

φ i = φir r ( )α ω( )

φ i = φir r ( )β ω( )

The closed shell ground state can be written:

Ψ0 = ψ1ψ2 ...ψN−1ψN = φ1φ 1...φN 2φ N 2

We want the convert the N general HF equation to N/2 equations for the spatial orbitals {φ}. Todo this we must integrate the equations over the spin coordinate ω. Consider, a single one of thegeneral HF equations:

fr x 1( )ψ i

r x 1( ) = ε iψi

r x 1( )

assuming that ψi is an a spin orbital:

fr x 1( )φ i

r r 1( )α ω1( ) = ε iφ i

r r 1( )α ω1( ) (Eq. 18)

multiplying on the left by a α* ω1( ) and integrating over w1 :

dω1∫ α* ω1( )f r

x 1( )α ω1( )[ ]φ ir r 1( ) = εiφi

r r 1( )

Evaluating the closed-shell spin-independent Fock operator ƒ(r1):

15

dω1∫ α* ω1( )f r x 1( )α ω1( )[ ]φ i

r r 1( ) =

dω1∫ α* ω1( ) hr r 1( ) + J j

r r 1( ) − K j

r r 1( )( )

j=1

N∑

α ω1( )

φi

r r 1( ) =

dω1∫ α* ω1( )h r r 1( )α ω1( )[ ]φi

r r 1( ) + dω1∫ α* ω1( ) J j

r r 1( ) − Kj

r r 1( )( )

j=1

N∑ α ω1( )

φi

r r 1( )

We go ahead and integrate the one - electron term and separate the sums in the two electron terminto sums over N/2 α and β electrons:

fr r 1( )φ i

r r 1( ) = h

r r 1( )φi

r r 1( )

+ dω1dω2dr r 2∫ α* ω1( )φ j

* r r 2( )α* ω2( ) 1

r12φ j

r r 2( )α ω2( )α ω1( )φi

r r 1( )

j=1

N 2∑

+ dω1dω2dr r 2∫ α* ω1( )φ j

* r r 2( )β* ω2( ) 1

r12φ j

r r 2( )β ω2( )α ω1( )φi

r r 1( )

j=1

N 2∑

− dω1dω2dr r 2∫ α* ω1( )φ j

* r r 2( )α* ω2( ) 1

r12φ j

r r 1( )α ω1( )α ω2( )φi

r r 2( )

j=1

N 2∑

− dω1dω2dr r 2∫ α* ω1( )φ j

* r r 2( )β* ω2( ) 1

r12φ j

r r 1( )β ω1( )α ω2( )φi

r r 2( )

j=1

N 2∑

Now we can go ahead and do the spin integrations. Note that all of these integrations equal 1,except for the last term, which integrates to zero. This leaves us with the following expressionfor the closed shell, spin-independent Fock matrix:

fr r 1( )φ i

r r 1( ) = h

r r 1( )φi

r r 1( )

+2 dr r 2∫ φ j

* r r 2( ) 1

r12φ j

r r 2( )φ i

r r 1( )

j=1

N 2∑ − d

r r 2∫ φ j

* r r 2( ) 1

r12φ j

r r 1( )φi

r r 2( )

j=1

N 2∑

we can rewrite this as:

f

r r 1( )φ i

r r 1( ) = h

r r 1( )φi

r r 1( ) + 2

) J j

r r 1( ) −

) K j

r r 1( )( )φi

r r 1( )

j=1

N 2∑

where ) J and

) K are the closed shell coulomb and exchange operators:

) J j

r r 1( )φi

r r 1( ) = d

r r 2φ j

* r r 2( ) 1

r12φ j

r r 2( )∫

φi

r r 1( ) and

) K j

r r 1( )φ i

r r 1( ) = d

r r 2φ j

* r r 2( ) 1

r12φ i

r r 2( )∫

φ j

r r 1( )

Now, combining this result with the original right hand side of the HF equations (in Eq. 18), weget the following closed shell HF equations:

fr r 1( )φ i

r r 1( ) = ε iφi

r r 1( ) for i = 1, 2,3,…, N/2 (Eq. 19)

16

Following the same procedure, we can rewrite the HF energy expression for a closed shellmolecule:

E = 2 i( h i)i=1

N 2∑ + 2 ii jj) − ij ji)((

j=1

N 2∑

i=1

N 2∑

= 2 hiii=1

N 2∑ + 2J ij − K ij

j=1

N 2∑

i =1

N 2∑

where (Eq. 20)

J ij ≡ dr r 2φi

* r r 2( )

) J j

r r 2( )φi

r r 2( )∫

K ij ≡ dr r 2φi

* r r 2( ) )

K jr r 2( )φi

r r 2( )∫

4.2) Introduction of a basis set:

In order to make the closed shell HF equations solvable using matrix operations, it’s necessaryto introduce a basis set to approximate the individual spin orbitals. This basis set {χ} is usually aset of “cartesian gaussian” functions (see lecture notes), but the equations below are general forany type of basis function:

φ i

r r ( ) = Caiχa

r r ( )

a=1

Nbasis

∑

Substituting this expansion into the closed shell HF equations (see next):

fr r 1( )φ i

r r 1( ) = ε iφi

r r 1( ) ⇒

fr r 1( ) Cνi χν

r r 1( )

ν=1

Nbasis

∑ = ε i Cνiχ νr r 1( )

ν=1

Nbasis

∑

To convert to a matrix equation, multiply on the left by χµr r 1( ) and integrate:

Cνi d

r r 1χµ

r r 1( )f r

r 1( )χνr r 1( )∫

ν=1

Nbasis

∑ = ε i Cν i dr r 1χµ

r r 1( )χν

r r 1( )∫

ν=1

Nbasis

∑

Now, defining the Fock matrix F, and the overlap matrix S:

Fµν = dr r 1χµ

r r 1( )f

r r 1( )χν

r r 1( )∫ and Sµν = d

r r 1χµ

r r 1( )χν

r r 1( )∫

we get the Roothan matrix equations for closed shell molecules:

FµνCν iν=1

N basis

∑ = ε i SµνCνiν=1

Nbasis

∑ for i = 1,2,...,Nbasis

17

or in pure matrix form: FC = SCε

where ε is a Nbasis × Nbasis diagonal matrix.

The Roothan equation derived above cannot be solved as a simple matrix eigenvalue equation(AX=λX) because the overlap matrix S, appears on the right hand side. We can get around thisproblem by making a substitution for the C matrix in the Roothan matrix equation:

Starting with the matrix Roothan equations:

FC = SCε

make the following substitution:

C' = XC C = X-1 C'

to yield: FXC' = SXC'ε

Multiplying on the left by X† :

X†FXC' = X†SXC'ε

If we choose X to be a unitary matrix that diagonalizes S, such as S1/2, then

X†SX = 1 (1 := unit matrix), so that the equations become:

F'C' = C'ε (defining F' = X†FX)

This final expression for the closed shell Roothan matrix equations are what are actuallyimplemented in conventional HF programs. This equation shows how to derive a new C matrixfrom the Fock matrix by solving a matrix eigenvector problem. But, as we stated earlier, theFock matrix is itself a function of the orbitals with means that it is a function of C. Hence westill require a convenient expression for constructing the Fock matrix from C.

Fµν = dr r 1χµ

r r 1( )f

r r 1( )χν

r r 1( )∫

= dr r 1χµ

r r 1( )h r

r 1( )χνr r 1( )∫ + d

r r 1χµ

r r 1( ) 2Ji

r r 1( ) − Ki

r r 1( )[ ]χν

r r 1( )∫

i =1

N 2∑

Substituting the basis set expansion into the definitions of J and K and renaming the oneelectron portion of the Fock matrix:

= Hµνcore + CλiCσi 2 µνλσ )( − µ σ λ ν)([ ]

σ

Nbasis

∑λ

Nbasis

∑i

N 2∑

= Hµνcore + Pλσ 2 µνλσ )( − µ σ λ ν)([ ]

σ

Nbasis

∑λ

Nbasis

∑

where:

18

Pλσ = CλiCσii

N 2∑

and:

µ ν λ σ)( ≡ d

r r 1d

r r 2χµ

* r r 1( )χν

* r r 1( ) 1

r12χλ

r r 2( )χσ

r r 2( )∫∫

This final set of definitions, combined with the matrix eigenvalue equation, F´C´ = εC´, are acomplete set of expression for writing a closed shell Hartree Fock program. Note that the twosets of integrals defined above, Hcore , and (µν|λσ) are calculated using numerical integrationschemes that depend on the exact form of the individual basis functions {χ}.Specific implementations of these numerical integration methods, and many complete HFpackages are available from the commercial vendors and from the QCPE.

4.3) Open shell HF equations

Alors que pour la grande majorité des molécules organiques ou des molécules minérales sansélément de transition le terme fondamental est totalement symétrique (du type lA1), cela n'estplus vrai pour les complexes des éléments de transition qui possèdent fréquemment uneconfiguration électronique à couches ouvertes (open shells). Par exemple de tous les ions dn

(0<n<10) en champ octaédrique, seuls les ions d6 dans la limite du champ fort ont pour termefondamental lA1g. Ainsi est-il particulièrement important de savoir étendre les méthodes abinitio aux composés à couches ouvertes.

La méthode la plus utilisée, mais non la seule, consiste à exprimer la fonction d'onde par unsimple déterminant construit à partir de deux ensembles indépendants d'OM, l'un φα décrivantles électrons de spin α, l'autre φβ décrivant les électrons de spin α. Si nous avons p spins α et qspins β (p>q), la diagonale principale du déterminant prend la forme:

Ψ0 = φ1α 1( )α 1( )φ1

β 2( )β 2( )...φqα 2q −1( )α 2q −1( )φq

β 2q( )β 2q( )φq+1α 2q +1( )α 2q + 1( )...φq+p

α 2q + p( )α 2q + p( )(Eq. 21)

La matrice F se scinde également en deux matrices Fα et Fβ dont les éléments matriciels sont:

Fµνα = Hµν

coeur + Pλσ(µν|λσ) - Pλσα (µλ|νσ)∑

λ,σ

(Eq. 22)

avec

Pλσα = Ciλ

α*∑i

OM occ.spin α

Ciσα (Eq. 23)

et une expression équivalente pour Fµνβ ,. Les équations à résoudre s'écrivent matriciellement:

19

Cα Fα = Eα Cα S

Cβ Fβ = Eβ Cβ S (Eq. 24)

où la signification des différentes matrices Cα , Cβ, Eα et Eβ est évidente. C'est à nouveau paritérations successives qu'on peut résoudre les équations (Eq. 24).

Une fonction d'onde du type (Eq. 21) à laquelle conduisent les équations (Eq. 24) possède ungrave défaut: elle n'est pas fonction propre de l'opérateur S2, c'est-à-dire qu'elle n'a pas lesbonnes propriétés de symétrie. Cette fonction d'onde décrivant un terme dont la multiplicité despin est 2(p—q)+1 est contaminée par des fonctions décrivant des termes de plus hautemultiplicité de spin. Seule une combinaison linéaire de fonctions déterminantales (Eq. 21)pourrait être fonction propre de S2.

Outre ce défaut, la méthode conduit à deux ensembles d'OM φα et φβ qui n'ont pas deux à deuxla même énergie, ce qui ne correspond pas aux diagrammes qualitatifs. Il existe une solutionévidente à cette difficulté qui consiste à imposer aux q OM de plus basses énergies de spin βd'être identiques aux q OM de plus basse énergie de spin α. La diagonale principale dudéterminant est alors:

ψ = ϕ1(1)α(1)⋅ ϕ1(2)β(2)⋅ ... ⋅ ϕq(2q-1)α(2q-1)⋅ ϕq(2q)β(2q)⋅ ϕq+1α (2q+1)α(2q+1)⋅ ... ⋅ϕ p

α(p+q)α(p+q)

(Eq. 25)

Ce déterminant est dit restricted par opposition au déterminant (Eq. 21) dit unrestricted. Cetterestriction imposée à la fonction d'onde se ressent sur l'énergie qui est plus élevée donc moinsbonne avec un déterminant restricted qu'avec un déterminant unrestricted. En revanche, lafonction d'onde restricted possède les bonnes propriétés de symétrie; elle est fonction propre deS2.

Une dernière difficulté, mais de toute autre nature, rencontrée dans le calcul des fonctions d'ondeassociées à un terme non totalement symétrique provient de ce qu'on ne peut introduire dans leséquations de Roothan le couplage spin-orbite. Nous avons vu que celui-ci peut partiellementlever la dégénérescence des termes dégénérés orbitalement et de spin. Lorsque ce couplage spin-orbite est important, il conduit à un état fondamental abaissé par rapport au terme fondamental,ce dont le calcul ab initio ne tient généralement pas compte.

20

5. Illustrating concepts: The hydrogen molecule, H2

H2 belongs to the cylindrical group D∞h and is therefore best expressed in terms of ellipticalcoordinates.

5.1 Elliptical coordinates

The elliptical coordinates µ and λ are defined as follows (cf. fig. 1):

y

A z z

x x

rA rB

R

y

B

φ

φ

θΑ θΒ

Fig. 1 Elliptical coordinates

Definitions:

λ =rA + rB

R(1a)

µ =rA − rB

R(1b)

dτ =R3

8λ2 − µ2( )dλdµ dϕ (2)

1 ≤ λ ≤ ∞ -1 ≤ µ ≤ 1 0 ≤ Φ ≤ 2π

R is the distance AB. Further,

21

zA = rA cos θa( ) =R

21+ λµ( )

zB = rB cos θb( ) =R

21 − λµ( )

rA sin θa( ) = rB sin θb( ) =R

2λ2 −1( ) µ2 −1( )[ ]1 2

Fig. 2 below displays contours of constant µ- and λ-values

Fig. 2: Contours of constant λ = 1.01, 2, 3, 4 and 5 (—)and of constant µ = -1, -0,75, -0.5, -0.25, 0, 0.25, 0.5, 0.75 and 1 (---)

5.2 Overlap integral

22

Problem: Calculate the overlap integral, S(1s,1s), between two 1s orbitals, centred at adistance R

from each other. What is the value of S(1s, 1s) for R → O and ∞ ?

Use the following, normalised, 1s(A)=a( r r ) and 1s(B)=b(

r r ) Slater Type Orbitals (STO) given by

a(1) =α3

πexp −αrA( ) b(1) =

α3

πexp −αrB( ) (3)

It is advantageous to express the answer, using the so-called A integrals, defined as

An α( ) = xne−αxdx =n!e −α

αn+11

∞∫

αk

k!k= 0

n∑ (4)

Thus, we obtain:

a 1( ) b 1( ) = S =α3

πexp −αrA( )exp −αrB( )dV∫

=α3

πR3

8dφ

0

2π∫ λ2 − µ2( )exp −α λ + µ( ) R

2

exp −α λ − µ( ) R

2

dµdλ

1

∞∫

−1

1

∫

=α3

πR3

82π dµ λ2 exp −αλR[ ]dλ

1

∞∫

−1

1

∫ −α3

πR3

82π µ2 dµ exp −αλR[ ]dλ

1

∞∫

−1

1

∫ (5)

=αR( )3

63A2 α( ) − A0 α( )[ ]

= e−αR 1+ αR +αR( )2

3

5.3 The Heitler-London Method for Hydrogen—Detailed Calculation.

First we must normalise the functions. For convenience we shall assume that the atomic orbitalsa and b are properly normalised, and that the overlap integral ∫a(l)b(l) dv = S, as above. Then wefind that the function

1

2 1± S2( )a 1( )b 2( ) ± a 2( )b 1( )[ ] (6)

is normalised. We recall that the overlap integral S was computed previously. This was done forthe atomic orbital 1s(A)=a(

r r ) and 1s(B)=b(

r r ) with a parameter α which would equal unity for a

hydrogen ls orbital. We shall carry this parameter through in our present calculation as well, sothat we can vary it at the end, and improve the original Heitler-London calculation, in which itwas assumed that α = 1.We must now find the diagonal matrix element of the Hamiltonian with respect to these twofunctions. The Hamilton operator reads

23

−∇12 − ∇2

2 −2

r1a−

2

r2a−

2

r1b−

2

r2b+

2

r12+

2

R(7)

where rla is the distance from the first electron to the nucleus a, etc., r12 is the distance betweenthe electrons, and R is the distance between the nuclei. For all the terms of the Hamiltonianexcept the repulsion 2/rl2 between electrons, we can use the methods of calculation already usedfor the calculation of the overlap integral S. For the repulsive terms, the calculations are moreinvolved, and have been made by Heitler and London. These integrals are discussed in one ofSlater's books.We now give in Table 1 the various integrals met in the evaluation of the energy.

24

Table l Integrals needed for energy calculation, Heitler-London method for H2. The atomicorbitals a(l) and b(l) refer to the normalised functions centred on the atoms a and b. The quantityα is a variation parameter. Energies are expressed in Rydbergs, distances in atomic units. Thequantity w = αR, where R is the internuclear distance.

a 1( ) ∇12 a 1( ) := a 1( ) ∇1

2( )a 1( )dv1∫ = α2

a 1( ) −2

r1aa 1( ) = −2 α

a 1( ) b 1( ) = S = e−w 1 + w +w2

3

a 1( ) 2

r1ba 1( ) = αJ = α −

2

w+ e−2w 2 +

2

w

a 1( ) 2

r1bb 1( ) = αK = −α e−2w 2 +

2

w

a 1( ) ∇12 b 1( ) = −α2 K + S( ) = α2 1+ w −

w2

3

aa bb = a2 (1)2

r12b2 (1)dv1 dv2∫ = αJ' = α

2

w− e−2w 2

w+

11

4+

3w

2+

w2

3

ab ab = a(1)b(1)2

r12

a ( 2 ) b ( 2 ) d v1 dv2∫ = αK' = 2

5α −e2w − 25

8+ 23w

4+ 3w2 + w3

3

+6

wS2 C + l n w( ) + S'2 Ei −4w( ) − 2SS'Ei −2w( )[ ]

ab bb = a(1)b(1)2

r12b ( 2 ) b ( 2 ) d v1 dv2∫ = αL = α e−w 2w +

1

4+

5

8w

− e−3w 1

4+

5

8w

where S' = ew 1− w +w2

3

; C = Euler's constant = 0.57722

Ei −x( ) = −e−t

tdt

x

∞∫ ; x > 0 is the integral logarithm

25

Matlab script:

The MATLAB script below will calculate all the integrals needed for a complete SCF LCAOcalculation of H2 as a function of the interatomic distance R and the orbital exponent α.

Definition of variables:Input:

r: interatomic distance in Bohra: orbital exponent in Bohr-1

Output (The energy units are Rydbergs):s: overlap integralj = Jk = Kjp = J'kp = K'l = L

Matlab script:w = α*re = exp(w); tow = 2/w;e2 = e*e;g = 0.577215664901533;s = (1+w+w*w/3)/e;sp = (1-w+w*w/3)*e;j = -tow+(2+tow)/e2;k = -(2+w+w)/e;jp = tow-(tow+11/4+1.5*w+w*w/3)/e2;kp = 0.4*(-(-25/8+23*w/4+3*w*w+w*w*w/3)/e2 + 6*(s*s*(g+log(w))- sp*sp*expint(4*w)+2*s*sp*expint(w+w))/w);l = (w+w+0.25+0.625/w)/e+(-0.25-0.625/w)/e2/e;

5.3a) The Restricted Hartree-Fock (RHF) Method for HydrogenThe qualification "Restricted" indicates that the same one-electron function (m.o.) is used forboth the α- and the β-spin components, i.e.:

ψ ± 1σ( ) =1

2 1 ± S2( )a 1σ( )± b 1σ( )[ ] (8)

Recall the definition of the Fock operator for closed sell configurations:

) f 1( ) =

) h 1( ) + 2

) J i 1( ) −

) K i 1( )[ ]

i

N / 2∑ (9)

where the 1st term involves the one-electron operators representing the kinetic energy and thenuclear attraction of an electron; the 2nd term is the two-electron part representing the

26

electrostatic repulsion of an electron in the averaged potential of the (N-1) residual electrons. In

the basis φµ{ } this reads

Fµν = Hµνcore + Pµν µ ν( λσ) −

1

2µλ( νσ)

λσ∑ (10)

And the total energy Etot is given by:

E tot =ZAZB

RABA<B∑ + 2 h ii

i =1

N 2∑ + 2Jij − Kij[ ]

j

N / 2∑

i

N / 2∑ (11)

where the matrix elements in the eqn. above are expressed in the basis of the m.o.'si = Cµiφµ

µ∑ .

Fig. 3 below displays the potential energy curve of H2 as a function of the interatomic distanceR for an RHF calculation. It is seen that the bond energy and the equilibrium bond length isacceptable, but the dissociation of H2 does not go to the limit of two isolated H atom as R goesto infinity. This surprising result is totally incorrect and is not specific for H2.

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-2.2

-2.15

-2.1

-2.05

-2

-1.95

-1.9

-1.85

-1.8

-1.75

-1.7

Fig. 3 RHF potential curve for STO (α=1) H2 in Rydberg (1Ry=13.6eV) as a function of theinteratomic distance R in Bohr. Note: The energy of two isolated H atoms is -2Ry.

27

As we shall see in the next two paragraphs this incorrect behaviour can be cured as follows.

5.3b The Unrestricted Hartree-Fock (UHF) Method for HydrogenAs in the precious paragraph, the qualification "Unrestricted" indicates here that two differentone-electron function (m.o.) are used for both the α- and the β-spin components respectively,i.e.:

ψ1α 1α( ) =1

1 + 2qS + q2a 1α( )+ q ⋅ b 1α( )[ ] and ψ1β 1β( ) =

1

1 + 2qS + q2q ⋅a 1β( )+ b 1β( )[ ]

(12)

where 0 ≤ q ≤ 1 is an undetermined parameter to be optimised variationnally. The result of suchan UHF calculation for H2 is shown in Fig. 4. It is seen that the dissociation of H2 does now goto the correct limit of two isolated H atom as R goes to infinity. However, cf. Fig. 5, the groundstate vawefunction is no longer an eigenfunction of S2. The reason for this is that the additionaldegree of freedom "parameter q" does in fact admix into the ground state singlet a certainamount of the excited triplet state. Hence, the wavefunction thus obtained is no longer a puresinglet. In order to obtain both, a correct dissociation curve for large bond distances and a properspin eigenfunction, it is necessary to perform a configuration interaction calculation. This isdone in the following paragraph.

28

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-2.2

-2.15

-2.1

-2.05

-2

-1.95

-1.9

-1.85

-1.8

-1.75

Fig. 4 UHF potential curve for STO (α=1) H2 in Rydberg as a function of the interatomicdistance R in Bohr. Note: The energy of two isolated H atoms is -2Ry.

29

0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Fig. 5 <S2> for a UHF calculation of H2 as a function of the interatomic distance R in Bohr

5.3c The Full CI Method for Hydrogen

Here we consider the full CI matrix obtained with the manyfold of all 4

2

= 6 single

determinants:

ψ1α 1( )ψ1β 2( ) , ψ1α 1( )ψ2α 2( ) , ψ1β 1( )ψ 2α 2( ) , ψ1α 1( )ψ2β 2( ) , ψ1β 1( )ψ2β 2( ) and ψ2α 1( )ψ2β 2( )

It is useful to adapt these N-electrons wavefunctions in spin — the spatial part is alreadysymmetry adapted — yielding two 1Σg. and each one 3Σu and one 1Σu.

1Σg1 = ψ1α 1( )ψ1β 2( )

30

3Σu =

ψ1α 1( )ψ2α 2( )1

2ψ1β 1( )ψ2α 2( ) +

1

2ψ1α 1( )ψ2β 2( )

ψ1β 1( )ψ2β 2( )

1Σu =1

2ψ1β 1( )ψ2α 2( ) −

1

2ψ1α 1( )ψ2β 2( )

1Σg2 = ψ2α 1( )ψ2β 2( ) (13)

With this transformation only the 1Σg. do interact and hence, the whole CI problem reducessimply to the diagonalisation of a 2x2 matrix. The full CI potential energy curve for ground stateand all excited states i.e. 1Σg1, 3Σu, 1Σu and 1Σg2. of H2 as a function of the interatomicdistance R, is given in Fig. 6.

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

Fig. 6 Full CI potential curve for STO (α=1) H2 in Rydberg as a function of the interatomic

distance R in Bohr; where: — denotes 1Σg1, --- 3Σu, .-.- 1Σu and ... 1Σg2. Note: The energy oftwo isolated H atoms is -2Ry.