hartree-fock theory

Hartree-Fock Theory

Erin DahlkeDepartment of ChemistryUniversity of Minnesota

VLab TutorialMay 25, 2006

Elementary Quantum Mechanics

€

HΨ = EΨ

€

H = − h2

2me∇i

2 − h2

2Mα∇α

2 − Zα eriαα =1

M∑ + e2

rij+

Zα Zβrαββ >α

M∑

α =1

M∑

j<i

N∑

i=1

N∑

i=1

N∑

α =1

M∑

i=1

N∑

€

H = T + V

Kinetic energy Potential energy

• The Hamiltonian for a many-electron system can be written as :

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:

€

h=1 e =1

me =1 MA = Mαme

€

H = − 12∇i

2 − 12MA

∇A2 − ZA

riAA=1

M∑ + 1

rij+ ZAZB

rABB>A

M∑

A=1

M∑

j<i

N∑

i=1

N∑

i=1

N∑

A=1

M∑

i=1

N∑


• The Hamiltonian for a many-electron system can be written in atomic units as :

The Born-Oppenheimer Approximation

€

mpme

=1836.15152

To a good approximation one can assume that the electrons move in a field of fixed nuclei

€

Helec = − 12∇i

2 −i=1

N∑

ZAr1AA=1

M∑

i=1

N∑ + 1

riji=1

N∑

i=1

N∑

€

Helecφelec = Eelecφelec

€

Eelec = Eelec ( rA{ })

€

E tot = Eelec + ZAZBrABB>A

M∑

A=1

M∑


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.

€

i*Ψi∫ dτ =1

€

i*Ψ j∫ dτ = 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.

€

2 = Ψ*Ψ units of probability density

Properties of a Wave Function

1. The wave function must vanish at the boundaries of the system2. 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

€

φ(r1ω1,L ,riωi ,L ,rjω j ,K ,rNωN ) = −φ(r1ω1,L ,rjω j ,L ,riωi ,K ,rNωN )

Spin Orbitals and Spatial Orbitals

orbital - a wave function for a single electronspatial orbital - a wave function that depends on the position of the electron and describes its spatial distribution

spin orbital - a wave function that depends on both the position and spin of the electron

€

χ(x) =ψ (r)α (ω)

orψ (r)β (ω)

⎧ ⎨ ⎪

⎩ ⎪

€

ψi (r)2 dr

The probability of finding electron i in the volume element given by dr.

Hartree ProductsConsider a system of N electrons

€

H = h(i)i=1

N∑

€

Helec = − 12∇i

2 −i=1

N∑

ZAr1AA=1

M∑

i=1

N∑ + 1

riji=1

N∑

i=1

N∑

Consider a system of N non-interacting electrons

The simplest approach in solving the electronic Schrödinger equation is to ignore the electron-electron correlation.

Without this term the remaining terms are completely separable.

Hartree Products

€

H = h(i)i=1

N∑

€

h(i)χ j (xi ) = ε jχ j (xi )

€

HP(x1,x2,K ,xN ) = χ i (x1)χ j (x2)L χ k (xN )

€

HΨHP = EΨHP

€

E = εi + ε j +L + εk

Each of the one-electron Hamiltonians will satisfy a one-electron Schrödinger equation.

Because the Hamiltonian is separable the wave function for this system can be written as a product of the one-electron wave functions.

This would result in a solution to the Schrödinger equation

in which the total energy is simply a sum of the one-electron orbital energies

Hartree Method

€

Vi{ j} =ρ jrij

dr∫j≠i∑

Self Consistent Field (SCF) procedure1. Guess the wave function for all the occupied orbitals2. Construct the one-electron operators3. Solve the Schrödinger equation to get a new guess at the wave

function.

€

ρ j = Ψ j*Ψ j = Ψ j

2

€

h(i) = − 12∇i

2 − Zkrikk=1

M∑ + Vi{ j}

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:

where

Hartree Method

€

E = εii∑ − 1

2Ψi

2 Ψ j2

rijdridr j∫∫

i≠ j∑

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.

Sum of the one-electron orbital eigenvalues

One half the electron-electron repulsion energy

Hartree Products

€

HP(x1,x2,K ,xN )2

dx1L dxN = χ i (x1)χ j (x2)L χ k (xN )2

dx1L dxN

€

χi (x1) 2 dx1 χ j (x2)2

dx2L χ k (xN ) 2 dxN

€

HP(x1,x2,K ,xN ) = χ i (x1)χ j (x2)L χ k (xN )

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.

Wave function is not antisymmetric with respect to the interchange of two particles

Electrons are not indistinguishable

€

χi (x1)L χ j (xα ),χ k (xβ ),L χ l (xN ) = χ i (x1)L χ j (xβ ),χ k (xα ),L χ l (xN )

Consider a 2 electron system with two spin orbitals. There are two ways to put two electrons into two spin orbitals.

Slater Determinants

€

12HP (x1,x2) = χ i (x1)χ j (x2)

Ψ21HP (x1,x2) = χ i (x2)χ j (x1)

€

(x1,x2) = 12

(χ i (x1)χ j (x2) − χ i (x2)χ j (x1))

€

(x1,x2) = χ i (x1)χ j (x2) ± χ i (x2)χ 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

Slater Determinants

€

(x1,x2) = 12

(χ i (x1)χ j (x2) − χ i (x2)χ j (x1))

= − 12

(χ i (x2)χ j (x1) − χ i (x1)χ j (x2)) = −Ψ(x2,x1)

€

(x1,x2) = 12

χ i (x1) χ j (x1)χ i (x2) χ j (x2)

€

(x1,x2) = 12

(χ i (x1)χ j (x2) + χ i (x2)χ j (x1))

= 12

(χ i (x2)χ j (x1) + χ i (x1)χ j (x2)) ≠ −Ψ(x2,x1)

Try the linear combination with the addition.

This wave function is not antisymmetric!

What about the linear combination with the subtraction.

This wave function is antisymmetric!

An alternative way to write this wave function would be:

Slater Determinants

For an N electron system

€

(x1,x2,K ,xN ) = 1N!

χ i (x1) χ j (x1) L χ k (x1)χ i (x2) χ j (x2) L χ k (x2)

M M O Mχ i (xN ) χ j (xN ) L χ k (xN )

Wave function is not antisymmetric with respect to the interchange of two particles

Electrons are not indistinguishable

Slater Determinant

rows correspond to electrons, columns correspond to orbitals

Electron motion is uncorrelated ?

Slater DeterminantsIs the electron motion correlated?

Consider a Helium atom in the singlet state:

€

He : 1s2

€

s = 12

1s(r1)α (ω1) 1s(r1)β (ω1)1s(r2)α (ω2) 1s(r2)β (ω2)

€

s = 12

(1s(r1)α (ω1)1s(r2)β (ω2) −1s(r1)β (ω1)1s(r2)α (ω2))€

s = 12

1s(r1)α (ω1) 1s(r1)β (ω1)1s(r2)α (ω2) 1s(r2)β (ω2)

Slater Determinants

€

s2dr1dr2dω1dω2 ⇒ probability distribution∫∫∫∫

€

s2 dr1dr2dω1dω2 =∫∫∫∫

12

(1s(r1)α (ω1)1s(r2)β (ω2) −1s(r1)β (ω1)1s(r2)α (ω2))∫∫∫∫2dr1dr2dω1dω2

€

2 = Ψ*Ψ

Slater Determinants

What happens to the spin terms?

€

s2dr1dr2dω1dω2∫∫∫∫ =

12 [ 1s*(r1)1s(r1)1s*(r2)1s(r2)dr1∫∫ dr2 α *(ω1)α (ω1)∫ dω1 β *(ω2)β (ω2)∫ dω2

- 1s*(r1)1s(r1)1s*(r2)1s(r2)dr1dr2∫∫ α *(ω1)β (ω1)∫ dω1 β *(ω2)α (ω2)∫ dω2

- 1s*(r1)1s(r1)1s*(r2)1s(r2)dr1∫∫ dr2 β *(ω1)α (ω1)∫ dω1 α *(ω2)β (ω2)∫ dω2

+ 1s*(r1)1s(r1)1s*(r2)1s(r2)dr1∫∫ dr2 β *(ω1)β (ω1)∫ dω1 α *(ω2)α (ω2)∫ dω2

Slater Determinants

€

*(ω)∫ α (ω)dω = β *(ω)∫ β (ω)dω =1

α *(ω)∫ β (ω)dω = β *(ω)∫ α (ω)dω = 0

€






Slater Determinants

€

*(ω)∫ α (ω)dω = β *(ω)∫ β (ω)dω =1

α *(ω)∫ β (ω)dω = β *(ω)∫ α (ω)dω = 0

1

0

0

1

€






Slater Determinants

€


12 [ 1s*(r1)1s(r1)dr1 1s*(r2)1s(r2)dr2 +∫∫ 1s*(r1)1s(r1)dr1 1s*(r2)1s(r2)dr2]∫∫

€

s2 dr1dr2dω1dω2∫∫∫∫ = 1s(r1)∫ 2dr1 1s(r2)∫ 2dr2

€

s* 1

r12Ψsdr1dr2dω1dω2∫∫∫∫ = 1s(r1)∫ 2∫ 1

r121s(r2)∫ 2dr1dr2 = J ⇒ Coulomb repulsion

After integrating out the spin coordinates, we’re left with only two terms in the integral:

Coulomb RepulsionClassically the Coulomb repulsion between two point charges is given by:

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.

€

J = q1q2r12

€

J = ψ1(r1)∫ 2∫ 1r12

ψ2(r2)∫ 2dr1dr2

This is the expression for the Coulomb repulsion between an electron in orbital i with an electron in orbital j. Since is always positive, as is the Coulomb energy is always a positive term and causes destabilization.

€

1r12

€

2

Slater Determinants

Consider a Helium atom in the triplet state:

€

He : 1s12s1

€

t = 12

1s(r1)α (ω1) 2s(r1)α (ω1)1s(r2)α (ω2) 2s(r2)α (ω2)

€

t = 12

(1s(r1)α (ω1)2s(r2)α (ω2) − 2s(r1)α (ω1)1s(r2)α (ω2))

€

t2dr1dr2dω1dω2 =∫∫∫∫

12

(1s(r1)α (ω1)2s(r2)α (ω2) − 2s(r1)α (ω1)1s(r2)α (ω2))2

dr1dr2dω1dω2

Slater Determinants

€

t2dr1dr2dω1dω2∫∫∫∫ =

12 [ 1s*(r1)1s(r1)2s*(r2)2s(r2)dr1∫∫ dr2 α *(ω1)α (ω1)∫ dω1 α *(ω2)α (ω2)∫ dω2

- 1s*(r1)2s(r1)2s*(r2)1s(r2)dr1∫∫ dr2 α *(ω1)α (ω1)∫ dω1 α *(ω2)α (ω2)∫ dω2


+ 2s*(r1)2s1s*(r2)1s(r2)(r1)dr1∫∫ dr2 α *(ω1)α (ω1)∫ dω1 α *(ω2)α (ω2)∫ dω2

€

t2dr1dr2dω1dω2 =∫∫∫∫

12

(1s(r1)α (ω1)2s(r2)α (ω2) − 2s(r1)α (ω1)1s(r2)α (ω2))2

dr1dr2dω1dω2

Slater Determinants

1

1

1

1

€


12 [ 1s*(r1)1s(r1)2s*(r2)2s(r2)dr1∫∫ dr2 α *(ω1)α (ω1)∫ dω1 α *(ω2)α (ω2)∫ dω2



+ 2s*(r1)2s1s*(r2)1s(r2)(r1)dr1∫∫ dr2 α *(ω1)α (ω1)∫ dω1 α *(ω2)α (ω2)∫ dω2

Slater Determinants

€

t2 dr1dr2dω1dω2∫∫∫∫ =

12 [ 1s*(r1)1s(r1)2s*(r2)2s(r2)dr1∫∫ dr2 − 1s*(r1)2s(r1)2s*(r2)1s(r2)dr1∫∫ dr2

- 2s*(r1)1s(r1)1s*(r2)2s(r2)dr1dr2 + 2s*(r1)2s(r1)1s*(r2)1s(r2)dr1dr2]∫∫∫∫

€

t2dr1dr2dω1dω2∫∫∫∫ ≠ 1s(r1)∫ 2dr1 1s(r2)∫ 2dr2

€

t* 1

r12Ψtdr1dr2dω1dω2∫∫∫∫ = 1

2 (J − 2 1s*(∫ r1)2s(r1) 1r12

1s(r2)2s*(r2)dr1dr2 + J)

exchange integral

€

=J − K

The exchange integral has no classical analog.

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%).

Exchange Integral

€

1s*(∫ r1)2s(r1) 1r12

1s(r2)2s*(r2)dr1dr2 = 1s*(∫ r1)2s*(r2) 1r12

1s(r2)2s(r1)dr1dr2

Slater Determinants

€

hen r1 = r2 Ψt2 dr1dr2dω1dω2∫∫∫∫ = 0

So no two electrons with the same spin can occupy the same point in space the Pauli Exclusion principle

€


12 [ 1s*(r1)1s(r1)dr1 2s*(r2)2s(r2)dr2 −∫∫ 1s*(r1)2s(r1)dr1 2s*(r2)1s(r2)dr2]∫∫

- 2s*(r1)1s(r1)dr1 1s*(r2)2s(r2)dr2 + 2s*(r1)2s(r1)dr1 1s*(r2)1s(r2)dr2]∫∫∫∫

Hartree–Fock Theory

What if we apply Hartree theory to a Slater determinant wave function?

€

f (i) = − 12∇i

2 − Zkrik

+ ViHF{ j}

k

nuclei∑

€

Coulom opeρtoρ: J(i) = χ j*(x2)∫

j=i∑ 1

rijχ j (x2)dx2

€

Exχhnge opeρtoρ: K(i) = χ j*(x2)∫

j=i∑ 1

rijχ j (x1)dx2

one-electron Fock operator

€

ViHF{ j} = J(i) − K(i)


€

J(i) = χ j*(x2)∫

j∑ 1

rijχ j (x2)dx2 = J j (i)

j∑

€

K(i) = χ j*(x2)∫

j∑ 1

rijχ j (x1)dx2 = K j (i)

j∑€

χi*(∫ x1)J(i)χ i (x1)dx1 = χ i

*(x1)χ i (x1)∫ 1r12

∫j

∑ χ j*(x2)χ j (x2)dx1dx2

€

χi*(∫ x1)K(i)χ i (x2)dx1 = χ i

*(x1)χ j*(x2)∫ 1

r12∫

j∑ χ i (x2)χ j (x1)dx1dx2

Coulomb operator:

Exchange operator:

Self Interaction

•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.

€

J(i) = χ j*(x2)∫

j∑ 1

rijχ j (x2)dx2 = J j (i)

j∑

€

K(i) = χ j*(x2)∫

j∑ 1

rijχ j (x1)dx2 = K j (i)

j∑

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.

€

i*Ψi∫ dτ =1

€

i*Ψ j∫ dτ = 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.

€

2 = Ψ*Ψ units of probability density


€

Let φi (x) = cyiχ y (x)y=1

b∑

€

cyi f (i)χ y (x)y=1

b∑ = cyi[− 1

2∇i2 − Zk

rik+ J(i) − K(i)]

k

nuclei∑

y=1

b∑ χ y (x) = εi cyiχ y (x)

y=1

b∑

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.

€

f (i)φi (x) = f (i) cyiχ y (y=1

b∑ x) = cyi f (i)χ y (

y=1

b∑ x)


€

J j (x1) = cmj* cnj χ m

* (x2) 1r12

χ n (∫n=1

b∑

m=1

b∑ x2)dx2

€

K j (x1) = cmj* cnj χ m

* (x2) 1r12

χ n (∫n=1

b∑

m=1

b∑ x1)dx2

€

Pmn = cmj* cnj

j∑ We call Pmn the density matrix

What happens to our Coulomb and exchange operators??


€

cyi χ z*∫ f (i)χ y

y=1

b∑ dτ = εi cyi χ z

*∫ χ yy=1

b∑ dτ

€

Fzy = χ z*∫ f (i)χ ydτ and Szy = χ z

*∫ χ ydτ

€

cyi (Fzy −εiSzy )y=1

b∑ = 0

One can write the one-electron Schrödinger equation as:

Where we can define the following matrix elements:

Fock matrix overlap matrix

We can then rearrange the one-electron Schrödinger equation to get:

€

cyi χ z*∫ f (i)χ y

y=1

b∑ dτ −εi cyi χ z

*∫ χ yy=1

b∑ dτ = 0

€

cyiFzy −y=1

b∑ εi cyiSzy

y=1

b∑ = 0

Where we will have one such equation for each electron is our system.


€

det(Fzy −εiSzy ) =

F11 −εiS11 F12 −εiS12 L F1b −εiS1bF21 −εiS21 F22 −εiS22 L F2b −εiS2b

M M O MFb1 −εiSb1 Fb2 −εiSb2 M Fbb −εiSbb

In order find a non-trivial solution to this set of equations one can set up and solve the secular determinant:

Solution of the secular determinant determines the coefficients cyi which can, in turn be used to solve for the one-electron energy eigenvalues, ei.

€

EHF = εii∑ − 1

2 J j (i) − K j (i)j

∑

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

Is new density matrix P(n) sufficiently similar to old density matrix P(n-1)

Replace P(n-1) with P(n)

Output data for given geometry

no

yes

Flow chart of the implementation of Hartree–Fock Theory

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.

hartree-fock theory

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