Orbital approximation

• Assign the electrons to an atomic orbital and a spin• Construct an antisymmetrized wave function using a Slater

determinant• evaluate the energy with the Hamiltonian that includes the

electron-electron interactions

| |HE

Ψ Ψ=

Ψ Ψ

1 1 1 2 1

1 1 1 2 11 2

1 2

( ) ( ) ( )( ) ( ) ( )1( , , , )

!( ) ( ) ( )

Z Z Zs s s NZ Z Zs s s N

N

Z Z ZN N N N

r r rr r r

r r rN

r r r

φ φ φφ φ φ

φ φ φ

↑ ↑ ↑↓ ↓ ↓

Ψ =

↓ ↓ ↓

Lithium

The antisymmetric 3 electron wavefunction can be written

1 1 1 2 1 3

1 2 3 1 1 2 1 1 1 2 1 3

2 1 2 2 2 3

( ) ( ) ( )1( , , ) , , ( ) ( ) ( )3! ( ) ( ) ( )

Li Li Lis s s

Li Li Li Li Li Lis s s s s s

Li Li Lis s s

r r rr r r r r r

r r r

φ φ φφ φ φ φ φ φ

φ φ φ

↑ ↑ ↑Ψ = ↑ ↓ ↑ = ↓ ↓ ↓

↑ ↑ ↑

Construct the Hamiltonian matrix to see which has the lowest energy

There are two possible configurations

1 1 2, ,Li Li Lis s sφ φ φ↑ ↓ ↑ and 1 1 2, ,Li Li Li

s s sφ φ φ↑ ↓ ↓

Berylium

The antisymmetric 4 electron wave function can be written,

1 2 3 4 1 1 2 2

1 1 1 2 1 3 1 4

1 1 1 2 1 3 1 4

2 1 2 2 2 3 2 4

2 1 2

( , , , ) , , ,

( ) ( ) ( ) ( )( ) ( ) ( ) ( )1( ) ( ) ( ) ( )4!( ) (

Be Be Be Bes s s s

Be Be Be Bes s s sBe Be Be Bes s s sBe Be Be Bes s s s

Be Bes s

r r r r

r r r rr r r rr r r rr

φ φ φ φ

φ φ φ φφ φ φ φφ φ φ φφ φ

Ψ = ↑ ↓ ↑ ↓

↑ ↑ ↑ ↑↓ ↓ ↓ ↓

=↑ ↑ ↑ ↑↓ ↓

2 2 3 2 4) ( ) ( )Be Bes sr r rφ φ↓ ↓

Gold

The antisymmetric 79 electron wave function can be written,

79 79 791 2 79 1 1 1 2 6 79( , , , ) ( ), ( ), , ( )s s sr r r r r rφ φ φΨ = ↑ ↓ ↑

The determinant of an N×N matrix has N! terms.

79! = 8.95 × 10116

(more than the atoms in the observable universe)

Start with the valence electrons. Stop adding electrons when the matrix gets too big.

Pauli exclusion

The sign of the wave function must change when two electrons are exchanged

( )1 1 2 11 2 1 1 2 2 1 2 2 1

1 2 2 2

( ) ( )1 1( , ) ( ) ( ) ( ) ( )( ) ( )2 2r r

r r r r r rr r

φ φφ φ φ φ

φ φΨ = = −

The Pauli exclusion principle only holds in the noninteracting electron approximation when the many electron wave function can be written as a product of single electron wave functions, only one electron can occupy each single electron state.

If two columns are exchanged, the wave function changes sign.

If two columns are the same, the determinant is zero.

Atomic physics summary

Solutions to the Schrödinger equation accurately describe the observed energy levels in atoms.

We know the equation that needs to be solved but it is intractable.

A common first approximation is the orbital approximation: Assign the electrons to an atomic orbital and a spin antisymmetrized product of spin orbitals.

The energy is then evaluated including the electron-electron interactions.

Sometimes assumptions have to be made about the core wave functions and the wave functions of the valence electrons are determined numerically.

| |HE

Ψ Ψ=

Ψ Ψ

The full Hamiltonian of a molecule

Everything you can know about the molecule is contained in the Hamiltonian.

This explains life, the universe, and everything!

2 22 2 22 2

, 0 0 02 2 4 4 4A A B

i Ai A i A i j A Be A iA ij AB

Z e Z Z eeHm m r r rπε πε πε< <

= − ∇ − ∇ − + +∑ ∑ ∑ ∑ ∑

Born Oppenheimer approximation

Fix the positions of the nuclei and consider the many electron Hamiltonian.

This is still too difficult. Neglect the electron-electron interactions.

2 22 22

, 0 0 02 4 4 4A A B

elec ii i A i j A Be iA ij AB

Z e Z Z eeHm r r rπε πε πε< <

= − ∇ − + +∑ ∑ ∑ ∑

Separation of variables (Trennung der Veränderlichen)

The Schrödinger equation can be solved by the separation of variables if the total Hamiltonian can be written as a sum of Hamiltonians each depending on only one variable.

Ht(r1,r2,...rn) = H1(r1) + H2(r2) + ... Hn(rn)

22 22

, 0 02 4 4A

elec ii i A i je iA ij

Z e eHm r rπε πε<

= − ∇ − +∑ ∑ ∑

222

_02 4

Aelec red i MO

i A ie iA

Z eH Hm rπε

−= ∇ − =

∑ ∑ ∑

Molecular orbital Hamiltonian

A B

1

C

+

-

+rAB

r1C

rAC

rBC

r1Br1A

+

Fix the positions of the nuclei. Solve the Schrödinger equation for one electron. Find the ground state and the excited states.

2221

0 12 4A

MOAe A

Z eHm r rπε−

= ∇ −−∑

Molecular orbitals

Molecular orbitals of a molecule are like the atomic orbitals of an atom.

1

2

3

4

( )( )( )( )

MO

MO

MO

MO

rrrr

ψψψψ

1

2

2

2

( )( )( )

( )x

y

s

s

p

p

rrr

r

φφφ

φ

You can put two electrons, spin up and spin down, in each molecular orbital.

Molecular orbitals

The first approximation for the many electron wave function of a molecule is an antisymmetrized product of molecular orbitals. The energy of this wave function should be evaluated using the electronic Hamiltonian.

1 1 1 1 , 1

1 21

1 ,

( ) ( ) ( )( )1( , )

!( ) ( )

MO MO MO N

MON

MO N MO N N

r r rr

r rN

r r

ψ ψ ψψ

ψ ψ

↑ ↓ ↑↑

Ψ =

↑ ↑

Molecular orbitals

Calculate the energy of a molecular orbital using the electronic Hamiltonian.

1 1 1 2 2( , ) , , , ,N MO MO MO MOr r ψ ψ ψ ψΨ = ↑ ↓ ↑ ↓

2 22 22

, 0 0 02 4 4 4A A B

elec ii i A i j A Be iA ij AB

Z e Z Z eeHm r r rπε πε πε< <

= − ∇ − + +∑ ∑ ∑ ∑

elecHE

Ψ Ψ=

Ψ Ψ

Bond potentials

Calculate the energies for different atomic distances.The minimum yields the bond length and bond strength.

elecHE

Ψ Ψ=

Ψ Ψ

XPS

Bond angles

Find the angle that minimizes the energy.

elecHE

Ψ Ψ=

Ψ Ψ

Shape of a molecule

2 22 22

, 0 0 02 4 4 4A A B

elec ii i A i j A Be iA ij AB

Z e Z Z eeHm r r rπε πε πε< <

= − ∇ − + +∑ ∑ ∑ ∑

Shape of a molecule

http://lampx.tugraz.at/~hadley/ss1/molecules/moleculeviewer/viewer.php

In Jmol, double click to start and stop a measurement.

Chemical reactions

Calculate the energy at every stage of the reaction.

elecHE

Ψ Ψ=

Ψ Ψ

http://en.wikipedia.org/w

iki/File:Walden-inversion-3D

-balls.png

To calculate the speed of a chemical reaction, solve the time-dependent Schrödinger equation.

Chemical reactions

It is possible to calculate if the reaction is endothermic or exothermic.

elecHE

Ψ Ψ=

Ψ Ψ

C3H8 + 5 O2 → 3 CO2 + 4 H2O + Energy

+

Propane Oxygen Carbon dioxide

Water

+ →

The molecular orbital can be found by:Linear Combination of Atomic Orbitals

Look for a solution to the molecular orbital Hamiltonian,

of the form,

Here φn are atomic orbitals.

Molecular orbitals of H2

The molecular orbital Hamiltonian for H2 is,

What about spin?

Ar

and Br

are the positions of the protons.

1 1 , 2 1 , 3 2 , 4 2 ,H H H H

mo s A s B s A s Bc c c cψ φ φ φ φ= + + + +

Molecular orbitals of H2

The time independent Schrödinger equation,

Multiply from the left by

mo moH Eψ ψ=

1 ,Hs Aφ

( )1 1 , 1 , 2 1 , 1 , 1 1 , 1 , 2 1 , 1 ,H H H H H H H Hs A mo s A s A mo s B s A s A s A s Bc H c H E c cφ φ φ φ φ φ φ φ+ + = + +

Multiply from the left by 1 ,Hs Bφ

( )1 1 , 1 , 2 1 , 1 , 1 1 , 1 , 2 1 , 1 ,H H H H H H H Hs B mo s A s B mo s B s B s A s B s Bc H c H E c cφ φ φ φ φ φ φ φ+ + = + +

Two equations with two unknowns: c1 and c2

Molecular orbitals of H2

Hamiltonian matrix Overlap matrix

1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 ,1 1

2 21 , 1 , 1 , 1 , 1 , 1 , 1 , 1 ,

H H H H H H H Hs A mo s A s A mo s B s A s A s A s B

H H H H H H H Hs B mo s A s B mo s B s B s A s B s B

H H c cE

c cH H

φ φ φ φ φ φ φ φ

φ φ φ φ φ φ φ φ

=

1 1* *

2 2

AA AB AA AB

AB BB AB BB

H H c S S cE

H H c S S c

=

Roothaan equations:

1 00 1

S ≈

AA BBH H=

Molecular orbitals of H2

1 1*

2 2

AA AB

AB AA

H H c cE

H H c c

=

The eigenvalues and eigenfunctions are:

AA ABE H H± = ±

( ) ( ) ( )( )1 112

H HsA sBr r rψ φ φ± = ±

Both HAA and HAB are negative E+ < E-

Molecular orbitals of H2

( ) ( )1 11 2 1 2

2 2

( ) ( )1 1, ( ) ( )( ) ( )2 2r r

r r r rr r

ψ ψψ ψ

ψ ψ+ +

+ ++ +

↑ ↓Ψ = = ↑↓ − ↓↑

↑ ↓

In the ground state, both electrons occupy the lower energy symmetric orbital (bonding orbital).

( ) ( )( )1 2 1 1 1 2 1 1 1 2 1 2 1 1 1 1 1 21, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2H H H H H H H HsA sA sA sB sA sB sB sBr r r r r r r r r rφ φ φ φ φ φ φ φΨ = + + + ↑↓ − ↓↑

Use this wave function including the electron-electron interaction to calculate the bond potential.

Student project: Draw the Bond potential of H2

http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/hmol.html

elecHE

Ψ Ψ=

Ψ Ψ

Student project Henögl / Pranter 2014

r0 = 0.869 AD0 = 3.18 eV

exact valuesr0 = 0.74 AD0 = 4.52 eV

student project: Lang / Röthel

r0 = 0.85 AD0 = 3.63 eV

exact valuesr0 = 0.74 AD0 = 4.52 eV

Homonuclear diatomic molecules

All homonuclear diatomic molecules use the molecular orbitals of H2.

1 1 , 2 1 , 3 2 , 4 2 , 5 2 , 6 2 ,x x

Z Z Z Z Z Zmo s A s B s A s B p A p Bc c c c c cψ φ φ φ φ φ φ= + + + + + +

The Hamiltonian matrix is as large as the number of atomic orbitals in the molecular orbital sum.

H2, N2, O2, ...

Homonuclear diatomic molecules

All homonuclear diatomic molecules use the molecular orbitals of H2.

from: Blinder, Introduction to Quantum Mechanics

g → inversion symmetry

1σg < 1σu < 2σg < 2σu < 3σg ~ 1πu < 1πg < 3σu

Separtion of variablesnumber of electron pairs shared

from

: Blin

der,

Intro

duct

ion

to Q

uant

um M

echa

nics