Hartree-Fock Theory
Erin Dahlke
Department of Chemistry
University of Minnesota
VLab Tutorial
May 25, 2006
Elementary Quantum Mechanics
H  E

H T V
• The Hamiltonian for a many-electron system can be written as :
2

2
M
N M Z e N N e2
M M Z Z 
2
2

H 
i  
   
 
  
i1 2me
 1 2M
i1 1 ri
i1 ji rij  1   r
N
Kinetic energy
Potential energy
 where the first two terms represent the kinetic energy of the electrons and
nuclei, respectively, the third term is the nuclear-electron attraction, the
fourth term is the electron-electron repulsion, and the fifth term is the
nuclear-nuclear repulsion.
Elementary Quantum Mechanics
Atomic units:
1
e 1
me  1
MA 
M
me
• The Hamiltonian for a many-electron system can be written in atomic

units as :
N 1
M
N M Z
N N 1
M M Z Z
1
2
2
A
H    i  
A   
 
   A B
i1 2
A1 2M A
i1 A1 riA i1 ji rij A1 BA rAB
Kinetic energy

Potential energy
The Born-Oppenheimer Approximation
mp
me
 1836.15152
To a good approximation one can assume that the electrons move in a field
of fixed nuclei

N 1
N M Z
N N 1
2
A
Helec    i   
 
2
r
i1
i1 A1 1A i1 i1 rij
Kinetic energy
Potential energy
Helecelec  Eelecelec

E elec  E elec (rA )

M

M
Z A ZB
A1 B A rAB
E tot  E elec   
What is a Wave Function?
For every system there is a mathematical function of the coordinates of the
system  the wave function ()
This function contains within it all of the information of the system.
2
  *
units of probability density
*

 i i d  1

*

 i j d  Sij

In general,for a given molecular system, there are many different wave
functions that are eigenfunctions of the Hamiltonian operator, each with its
own eigenvalue, E.
Properties of a Wave Function
1. The wave function must vanish at the boundaries of the system
2. The wave function must be single-valued.
3. The wave function must be continuous.
A description of the spatial coordinates of an electron is not enough. We
must also take into account spin (). (Spin is a consequence of
relativity.)
= spin up
 = spin down
A many-electron wave function must be antisymmetric with respect to the
interchange of coordinate (both space and spin) of any two electrons. 
Pauli exclusion principle
 (r11, ,rii , ,rj j , ,rN  N )   (r11, ,rj j , ,rii , ,rN  N )
Spin Orbitals and Spatial Orbitals
orbital - a wave function for a single electron
spatial orbital - a wave function that depends on the position of the
electron and describes its spatial distribution
2
i (r) dr

The probability of finding
electron i in the volume
element given by dr.
spin orbital - a wave function that depends on both the position
and spin of the electron
 (r) ( )

 (x)   or
 (r) ( )

Hartree Products
Consider a system of N electrons
N 1
N M Z
N N 1
2
A
Helec    i   
 
i1 2
i1 A1 r1A i1 i1 rij
The simplest approach in solving the electronic Schrödinger equation is
 the electron-electron correlation.
to ignore
Without this term the remaining terms are completely separable.
Consider a system of N non-interacting electrons
N
H   h(i)
i1
Hartree Products
N
H   h(i)
i1
Each of the one-electron Hamiltonians will satisfy a one-electron
Schrödinger equation.

h(i)  j (x i )   j  j (x i )
Because the Hamiltonian is separable the wave function for this system
can be written as a product of the one-electron wave functions.

HP (x1,x 2, ,x N )  i (x1) j (x 2 )
 k (x N )
This would result in a solution to the Schrödinger equation

HHP  EHP
in which the total energy is simply a sum of the one-electron orbital
energies
E  i   j   k


Hartree Method
M
Zk
 Vi { j}
k1 rik
h(i)   12  i2  
Instead of completely ignoring the
electron-electron interactions we
consider each electron to be moving in a
field created by all the other electrons
From variation calculus you can show that:
j
Vi { j}    r dr
ij
ji
where
 j  *j j  j
2

Self Consistent Field (SCF) procedure
1. Guess the wave function for all the occupied orbitals

2. Construct the one-electron operators
3. Solve the Schrödinger equation to get a new guess at the wave
function.
Hartree Method
The one-electron Hamiltonian h(i) includes the repulsion of electron i with
electron j, but so does h(j). The electron-electron repulsion is being double
counted.
2
E
 i  12 
i
i j
2
i j
dri dr j

rij

Sum of the one-electron
orbital eigenvalues
One half the
electron-electron
repulsion energy
Hartree Products
What are the short comings of the Hartree Product wave function?
Electron motion is uncorrelated - the motion of any one electron
is completely independent of the motion of the other N-1
electrons.
HP

2
(x1,x 2 , ,x N ) dx1
2
dx N   i (x1)  j (x 2 )
2
i (x1) dx1  j (x 2 ) dx 2
2
 k (x N ) dx1
dx N
2
 k (x N ) dx N

Electrons are not indistinguishable

HP (x1,x 2, ,x N )  i (x1) j (x 2 )
 k (x N )
Wave function is not antisymmetric with respect to the interchange of

two particles
i (x1)
 j (x ),  k (x  ),
 l (x N )  i (x1 )
 j (x  ),  k (x ),
 l (x N )
Slater Determinants
Consider a 2 electron system with two spin orbitals. There are two ways
to put two electrons into two spin orbitals.
HP
12
(x1,x 2 )  i (x1) j (x 2 )
HP
21
(x1,x 2 )  i (x 2 ) j (x1 )
Neither of these two wave functions are antisymmetric with respect
to interchange of two particles, nor do they account for the fact that

electrons are indistinguishable…
Try taking a linear combination of these two possibilities
(x1,x 2 )  i (x1)  j (x 2 )  i (x 2 ) j (x1 )

(x1,x 2 )  1 (i (x1) j (x 2 )  i (x 2 ) j (x1))
2
Slater Determinants
Try the linear combination with the addition.
(x1,x 2 )  1 (i (x1 ) j (x 2 )  i (x 2 )  j (x1))
2
 1 (i (x 2 ) j (x1)  i (x1)  j (x 2 ))  (x 2,x1)
2
This wave function
is not antisymmetric!
What about the linear combination with the subtraction.
(x1,x 2 )  1 (i (x1 ) j (x 2 )  i (x 2 )  j (x1))
2
  1 (i (x 2 ) j (x1 )  i (x1)  j (x 2 ))  (x 2 ,x1)
2
An alternative way to write this wave function would be:
(x1,x 2 )  1
2
i (x1 )  j (x1)
i (x 2 )  j (x 2 )
This wave function
is antisymmetric!
Slater Determinants
For an N electron system
(x1,x 2 , ,x N ) 
1
N!
i (x1 )
i (x 2 )
 j (x1)
 j (x 2 )
 k (x1 )
 k (x 2 )
i (x N )  j (x N )
 k (x N )
rows correspond
to electrons,
columns
correspond to
orbitals
Slater Determinant
Electrons are not indistinguishable

Wave function is not antisymmetric with respect to the interchange of
two particles

Electron motion is uncorrelated
?
Slater Determinants
Is the electron motion correlated?
Consider a Helium atom in the singlet state:

He : 1s2
s 
1
2
1s(r1 ) (1 ) 1s(r1 ) (1)
1s(r2 ) (
2 ) 1s(r2 ) ( 2 )
s 
1
2
1s(r1 ) (1 ) 1s(r1 ) (1)
1s(r2 ) ( 2 ) 1s(r2 ) ( 2 )
  1 (1s(r ) ( )1s(r ) ( ) 1s(r ) ( )1s(r ) ( ))

s
1
1
2
2
1
1
2
2
2

Slater Determinants
2
    s dr1dr2d1d2  probability distribution
2
    s dr1dr2 d1d 2 


1 (1s(r ) ( )1s(r ) ( ) 1s(r ) ( )1s(r ) ( ))
1
1
2
2
1
1
2
2
2
2
  *

2
dr1dr2 d1d 2
Slater Determinants
2
    s dr1dr2 d1d 2 
1 [   1s* (r )1s(r )1s* (r )1s(r )dr dr   * ( ) ( )d   * ( ) ( )d
1
1
2
2
1 2
1
1
1
2
2
2
2
-   1s* (r1 )1s(r1 )1s* (r2 )1s(r2 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
-   1s* (r1 )1s(r1 )1s* (r2 )1s(r2 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
+   1s* (r1 )1s(r1 )1s* (r2 )1s(r2 )dr1dr2   * (1 ) (1 )d1   * ( 2 ) ( 2 )d 2
What happens to the spin terms?
Slater Determinants
2
    s dr1dr2 d1d 2 
1 [   1s* (r )1s(r )1s* (r )1s(r )dr dr   * ( ) ( )d   * ( ) ( )d
1
1
2
2
1 2
1
1
1
2
2
2
2
-   1s* (r1 )1s(r1 )1s* (r2 )1s(r2 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
-   1s* (r1 )1s(r1 )1s* (r2 )1s(r2 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
+   1s* (r1 )1s(r1 )1s* (r2 )1s(r2 )dr1dr2   * (1 ) (1 )d1   * ( 2 ) ( 2 )d 2
  * ( ) ( )d    * ( ) ( )d  1
  * ( ) ( )d    * ( ) ( )d  0

Slater Determinants
2
    s dr1dr2 d1d 2 
1 [   1s* (r )1s(r )1s* (r )1s(r )dr dr   * ( ) ( )d   * ( ) ( )d
1
1
2
2
1 2
1
1
1
2
2
2
2
-   1s* (r1 )1s(r1 )1s* (r2 )1s(r2 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
-   1s* (r1 )1s(r1 )1s* (r2 )1s(r2 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
+   1s* (r1 )1s(r1 )1s* (r2 )1s(r2 )dr1dr2   * (1 ) (1 )d1   * ( 2 ) ( 2 )d 2
  * ( ) ( )d    * ( ) ( )d  1
  * ( ) ( )d    * ( ) ( )d  0

1
0
0
1
Slater Determinants
After integrating out the spin coordinates, we’re left with only
two terms in the integral:
2
    s dr1dr2 d1d 2 
1 [  1s* (r )1s(r )dr  1s* (r )1s(r )dr   1s* (r )1s(r )dr  1s* (r )1s(r )dr ]
1
1 1
2
2
2
1
1 1
2
2
2
2
2
2
2
    s dr1dr2d1d2   1s(r1) dr1  1s(r2 ) dr2
2
2
  
  s* r1 s dr1dr2 d1d 2    1s(r1 ) r1  1s(r2 ) dr1dr2  J  Coulomb repulsion
12
12
Coulomb Repulsion
Classically the Coulomb repulsion between two point charges is given by:
qq
J 1 2
r12
where q1 is the charge on particle one, q2 is the charge on particle 2, and
r12 is the distance between the two point charges.

An electron is not a point charge. Its position is delocalized, and described
by its wave function. So to describe its position we need to use the wave
function for the particle, and integrate its square modulus. (in atomic units
the charge of an electron (e) is set equal to one.
2
2
J    1 (r1 ) r1   2 (r2 ) dr1dr2
12
This is the expression for the Coulomb repulsion between an electron
1
in orbital i with an electron in orbital j. Since r12 is always positive, as
2

is  the Coulomb energy is always a positive term and causes
destabilization.

Slater Determinants
Consider a Helium atom in the triplet state:
He : 1s12s1

1s(r1 ) (1 ) 2s(r1 ) (1 )
1
t 
2 1s(r2 ) ( 2 ) 2s(r2 ) ( 2 )
 1
t 
(1s(r1) (1)2s(r2 ) (2 )  2s(r1) (1)1s(r2 ) (2 ))
2

Slater Determinants
2
    t dr1dr2 d1d 2 
1 (1s(r ) ( )2s(r ) ( )  2s(r ) ( )1s(r ) ( ))
1
1
2
2
1
1
2
2
2
2
dr1dr2 d1d 2
2
    t dr1dr2 d1d 2 
1 [   1s* (r )1s(r )2s* (r )2s(r )dr dr   * ( ) ( )d   * ( ) ( )d
1
1
2
2
1 2
1
1
1
2
2
2
2
-   1s* (r1)2s(r1 )2s* (r2 )1s(r2 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
-   2s* (r1)1s(r1 )1s* (r2 )2s(r2 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
+   2s* (r1)2s1s* (r2 )1s(r2 )(r1 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
Slater Determinants
2
    t dr1dr2 d1d 2 
1 (1s(r ) ( )2s(r ) ( )  2s(r ) ( )1s(r ) ( ))
1
1
2
2
1
1
2
2
2
2
dr1dr2 d1d 2
2
    t dr1dr2 d1d 2 
1 [   1s* (r )1s(r )2s* (r )2s(r )dr dr   * ( ) ( )d   * ( ) ( )d
1
1
2
2
1 2
1
1
1
2
2
2
2
-   1s* (r1)2s(r1 )2s* (r2 )1s(r2 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
-   2s* (r1)1s(r1 )1s* (r2 )2s(r2 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
+   2s* (r1)2s1s* (r2 )1s(r2 )(r1 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
Slater Determinants
2
    t dr1dr2 d1d 2 
1 (1s(r ) ( )2s(r ) ( )  2s(r ) ( )1s(r ) ( ))
1
1
2
2
1
1
2
2
2
2
dr1dr2 d1d 2
2
    t dr1dr2 d1d 2 
1 [   1s* (r )1s(r )2s* (r )2s(r )dr dr   * ( ) ( )d   * ( ) ( )d
1
1
2
2
1 2
1
1
1
2
2
2
2
-   1s* (r1)2s(r1 )2s* (r2 )1s(r2 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
-   2s* (r1)1s(r1 )1s* (r2 )2s(r2 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
+   2s* (r1)2s1s* (r2 )1s(r2 )(r1 )dr1dr2   * (1) (1)d1   * ( 2 ) ( 2 )d 2
1
1
1
1
Slater Determinants
2
    t dr1dr2 d1d 2 
1[
2
 1s* (r1 )1s(r1 )2s* (r2 )2s(r2 )dr1dr2    1s* (r1 )2s(r1 )2s* (r2 )1s(r2 )dr1dr2
-   2s* (r1 )1s(r1 )1s* (r2 )2s(r2 )dr1dr2    2s* (r1 )2s(r1 )1s* (r2 )1s(r2 )dr1dr2 ]
2
2
2
    t dr1dr2d1d2   1s(r1) dr1  1s(r2 ) dr2

    t* r1 t dr1dr2d1d 2  12 (J  2  1s* (r1)2s(r1) r1 1s(r2 )2s* (r2 )dr1dr2  J)
12
12
exchange integral
= J K
Exchange Integral

The exchange integral has no classical analog.
 1s* (r1)2s(r1 ) r1 1s(r2 )2s* (r2 )dr1dr2   1s* (r1)2s* (r2 ) r1 1s(r2 )2s(r1 )dr1dr2
12
12
The exchange integral gets its name from the fact that the two electrons
exchange their positions as you go from the left to the right in the
integrand.
The exchange integrals are there to correct the Coulomb integrals, so
that they take into account the antisymmetrization of the wave function.
Electrons of like spin tend to avoid each other more than electrons of
different spin, so the destabilization predicted by the Coulomb integrals is
too high. The exchange integrals, which are always positive, lower the
overall repulsion of the electron (~25%).
Slater Determinants
2
    t dr1dr2 d1d 2 
1 [  1s* (r )1s(r )dr  2s* (r )2s(r )dr   1s* (r )2s(r )dr  2s* (r )1s(r )dr ]
1
1 1
2
2
2
1
1 1
2
2
2
2
-  2s* (r1 )1s(r1)dr1  1s* (r2 )2s(r2 )dr2   2s* (r1 )2s(r1 )dr1  1s* (r2 )1s(r2 )dr2 ]
when r1 = r2
2
    t dr1dr2d1d2  0
So 
no two electrons with the same spin can occupy the same point
in space  the Pauli Exclusion principle
Hartree–Fock Theory
What if we apply Hartree theory to a Slater determinant wave function?
f (i)   12  i2
nuclei Z
 
k
k
rik
 ViHF { j}
one-electron Fock operator
ViHF { j}  J(i)  K(i)

Coulomboperator :
J(i)     *j (x 2 ) r1  j (x 2 )dx 2
ij
ji
Exchange operator :

K(i)     *j (x 2 ) r1  j (x1)dx 2
ij
ji
Hartree–Fock Theory
Coulomb operator:
J(i)     *j (x 2 ) r1  j (x 2 )dx 2   J j (i)
ij
j
j
 i* (x1 )J(i)i (x1 )dx1     i* (x1 )i (x1 ) r1  *j (x 2 )  j (x 2 )dx1dx 2
12
j
Exchange operator:
K(i)     *j (x 2 ) r1  j (x1)dx 2   K j (i)
ij
j
j
 i* (x1 )K(i)i (x 2 )dx1     i* (x1 ) *j (x 2 ) r1 i (x 2 )  j (x1)dx1dx 2
12
j
Self Interaction
J(i)     *j (x 2 ) r1  j (x 2 )dx 2   J j (i)
ij
j

j
K(i)     *j (x 2 ) r1  j (x1)dx 2   K j (i)
ij
j
j

•The sums in these equations run over all values of j, including j = i.
• Each of these terms contains a self-interaction term
- a Coulomb integral between an electron and itself
- an exchange integral between an electron and itself
• Since both J and K contain the self-interaction term, and since we’re
subtracting them from each other, the self-interaction cancels.
What is a Wave Function?
For every system there is a mathematical function of the coordinates of the
system  the wave function ()
This function contains within it all of the information of the system.
2
  *
units of probability density
*

 i i d  1

*

 i j d  Sij

In general,for a given molecular system, there are many different wave
functions that are eigenfunctions of the Hamiltonian operator, each with its
own eigenvalue, E.
Hartree–Fock Theory
For a molecular system we don’t know what the true wave function is.
In general in order to approximate it we make the assumption that the
true wave function is a linear combination of one-electron orbitals.
b
Let i (x)   c yi  y (x)
y1
b
b
y1
y1
f (i)i (x)  f (i)  c yi  y (x)   c yi f (i) y (x)

b
b
nuclei Z
b

2
k
1
 J(i)  K(i)] y (x)  i  c yi  y (x)
 c yi f (i) y (x)   c yi [ 2 i  
r
ik
y1
y1
k
y1

Hartree–Fock Theory
What happens to our Coulomb and exchange operators??
b
b
J j (x1 )    c *mj c nj   *m (x 2 ) r1  n (x 2 )dx 2
12
m1 n1
b
K j (x1 )    c *mj c nj   *m (x 2 ) r1  n (x1 )dx 2
12

m1 n1

Pmn   c *mj c nj
j

b
We call Pmn the density matrix
Hartree–Fock Theory
One can write the one-electron Schrödinger equation as:
b
b
 c yi   z f (i)  y d  i  c yi   *z  y d
*
y1
y1
Where we can define the following matrix elements:

Fzy    *z f (i)  y d
and
Fock matrix
Szy    *z  y d
overlap matrix
We can then rearrange the one-electron Schrödinger equation to get:

b
b
 c yi   z f (i)  y d  i  c yi   *z  y d  0
*
y1
y1
b
b
b
 c yi Fzy i  c yi Szy  0
 c yi (Fzy  i Szy )  0
y1

y1
y1
Where we will have one such equation for each electron is our system.
Hartree–Fock Theory
In order find a non-trivial solution to this set of equations one can set up
and solve the secular determinant:
det(Fzy  i Szy ) 
F11  i S11
F21  i S21
F12  i S12
F22  i S22
F1b  i S1b
F2b  i S2b
Fb1  i Sb1
Fb2  i Sb2
Fbb  i Sbb
Solution of the secular determinant determines the coefficients cyi which
can, in turn be used to solve for the one-electron energy eigenvalues, i.
EHF  i  12  J j (i)  K j (i)
i
j
Flow chart of the
implementation of
Hartree–Fock Theory
Choose a basis set
Choose a molecular geometry
Compute and store all
overlap, one-electron, and
two-electron integrals
Guess initial density matrix
Construct and solve the
Hartree–Fock secular
determinant
Construct density matrix
from occupied orbitals
Replace P(n-1) with P(n)
no
Output data for given geometry
yes
Is new density matrix P(n)
sufficiently similar to old
density matrix P(n-1)
Limitations of Hartree–Fock Theory - Energetics
•Hartree–Fock theory ignores electron correlation
- cannot be used (accurately) in any process in which the
total number of paired electrons changes.
•Even if the total number of paired electrons stays the same, if the
nature of the bonds changes drastically HF theory can have serious
problems. (i.e., isomerization reactions)
•Does well for protonation/deprotonation reactions.
•Can be used to compute ionization potentials and electron affinities.
•Will not do well for describing systems in which there are dispersion
interactions, as they are completely due to electron correlation effects,
except by cancellation of errors.
•HF charge distributions tend to be over polarized which give
electrostatic interactions which are too large.
Limitations of Hartree–Fock Theory - Geometries
•HF theory tends to predict bonds to be too short, especially as you
increase the basis set size.
•Bad for transition state structures due to the large correlation associated
with the making and breaking of partial bonds.
•Nonbonded complexes tend to be far too loose, as HF theory does not
account for dispersion interactions.
Limitations of Hartree–Fock Theory - Charge Distributions
•The magnitude of dipole moments is typically overestimated by 10–25%
for medium sized basis sets.
•Results are erratic with smaller basis sets.
Summary
• Hartree–Fock theory is an approximate solution to the electronic
Schrödinger equation which assumes that each individual electron i,
moves in a field created by all the other electrons.
• Introduces the concept of exchange energy through the use of a Slater
determinant wave function.
• Ignores all other electron correlation.
• Contains a self-interaction term which cancels itself out.
• Often underbinds complexes.
• Predicts bond lengths which are too short.
References
Szabo, A.; Ostlund, N. S. Modern Quantum Chemistry. An
Introduction to Advanced Electronic Structure Theory. Dover
Publications: Mineola, NY; 1996.
Pilar, F. L. Elementary Quantum Mechanics, 2nd Ed. Dover
Publications: Mineola, NY; 2001.
Cramer, C. J. Essentials of Computational Chemistry. Wiley:
Chichester; 2002.