Lecture 2, April, 2012 QM-1: HF2.pdf · 2 1 2 ¦ ¦ 1 ¦ 2 ¦ ¦ ¦ A Aii j ij A B AB B A i R r Z Z Z M 1 2 1 1 2 H elΨ=EΨ For benzene we have 12 nuclear degrees of freedom (dof)

Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 1

Ch121a Atomic Level Simulations of Materials and

Molecules

William A. Goddard III, [email protected]

Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics,

California Institute of Technology

BI 115

Hours: Monday, Wednesday, Friday 2-3pm

Teaching Assistants Wei-Guang Liu, Fan Lu, Jose Mendozq

Lecture 2, April, 2012

QM-1: HF


CH121a Atomic Level Simulations of Materials and

Molecules

Instructor: William A. Goddard III

Prerequisites: some knowledge of quantum mechanics, classical

mechanics, thermodynamics, chemistry, Unix. At least at the Ch2a

level

Ch121a is meant to be a practical hands-on introduction to

expose students to the tools of modern computational

chemistry and computational materials science relevant to

atomistic descriptions of the structures and properties of

chemical, biological, and materials systems.

This course is aimed at experimentalists (and theorists) in

chemistry, materials science, chemical engineering, applied

physics, biochemistry, physics, geophysics, and mechanical

engineering with an interest in characterizing and designing

molecules, drugs, and materials.


Motivation: Design Materials, Catalysts, Pharma from 1st Principles so

can do design prior to experiment

Big breakthrough making FC simulations

practical:

reactive force fields based on QM Describes: chemistry,charge transfer, etc. For

metals, oxides, organics.

Accurate calculations for bulk phases

and molecules (EOS, bond dissociation)

Chemical Reactions (P-450 oxidation)

time

distance

hours

millisec

nanosec

picosec

femtosec

Å nm micron mm yards

MESO

Continuum

(FEM)

QM

MD

ELECTRONS ATOMS GRAINS GRIDS

Deformation and Failure

Protein Structure and Function

Micromechanical modeling

Protein clusters

simulations real devices

full cell (systems biology)

To connect 1st Principles (QM) to Macro work use an overlapping hierarchy of

methods (paradigms) (fine scale to coarse) so that parameters of coarse level

are determined by fine scale calculations.

Thus all simulations are first-principles based


Lectures

The lectures cover the basics of the fundamental methods:

quantum mechanics,

force fields,

molecular dynamics,

Monte Carlo,

statistical mechanics, etc.

required to understand the theoretical basis for the simulations

the homework applies these principles to practical problems

making use of modern generally available software.


Homework

First 6 weeks: The homework each week uses generally available

computer software implementing the basic methods on

applications aimed at exposing the students to understanding how

to use atomistic simulations to solve problems.

Each calculation requires making decisions on the specific

approaches and parameters relevant and how to analyze the

results.

Midterm: each student submits proposal for a project using the

methods of Ch120a to solve a research problem that can be

completed in 4 weeks.

The homework for the last 3 weeks is to turn in a one page report

on progress with the project

The final is a research report describing the calculations and

conclusions.


Methods to be covered in the lectures include:

Quantum Mechanics: Hartree Fock and Density Function

methods

Force Fields standard FF commonly used for simulations of

organic, biological, inorganic, metallic systems, reactions;

ReaxFF reactive force field: for describing chemical reactions,

shock decomposition, synthesis of films and nanotubes, catalysis

Molecular Dynamics: structure optimization, vibrations, phonons,

elastic moduli, Verlet, microcanonical, Nose, Gibbs

Monte Carlo and Statistical thermodynamics Growth

amorphous structures, Kubo relations, correlation functions, RIS,

CCBB, FH methods growth chains, Gauss coil, theta temp

Coarse grain approaches

eFF for electron dynamics

Tight Binding for electronic properties

solvation, diffusion,

mesoscale force fields


Applications will include prototype examples

involving such materials as:

Organic molecules (structures, reactions);

Semiconductors (IV, III-V, surface reconstruction)

Ceramics (BaTiO3, LaSrCuOx)

Metal alloys (crystalline, amorphous, plasticity)

Polymers (amorphous, crystalline, RIS theory, block);

Protein structure, ligand docking

DNA-structure, ligand docking


Outline Topic 1: QM: HF, DFT, basis sets, reactions, transition states, vibrations

Topic 2: Force Fields, nonbonds, hydrogen bonds, charges (QEq), QM FF

Topic 3: Molecular Dynamics: Verlet, NVE, NVT, NPT, Periodic Systems

Topic 4: Statistical mechanics: liquid simulations, entropy, nonequilibrium MD,

Green-Kubo, Monte Carlo, Grand Canonical MC, gas storage, surface tension

Topic 5: polymers: crystalline, amorphous, structure prediction,

Topic 6: ReaxFF Reactive Force Field and reactive Dynamics

Topic 7: PBC QM, band structure, phonons, elastic constants, Ab Initio MD,

Topic 8: surfaces: reconstruction, chemisorption, physisorption, solvation

Topic 9: eFF, Tight binding

Topic 9: applications:

Fuel Cell: oxygen reduction reaction, migration in Nafion solid oxides

Batteries: anode for Li battery, Solid electrolyte interface,

Nanotechnology: rotaxane molecular switches, carbon nanotube interfaces

Water Treatment: Dendrimers, captymers

Catalysis: alkane activation, ammoxidation

Metallic alloys: force fields, dislocations, plasticity, cavitation

Proteins: structures, ligand docking, GPCRs

DNA: A to B transition, Origami based nanotechnology


Topic 1: Practical Quantum Chemistry

Solve Schrödinger Equation

Hel =

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

HelΨ=EΨ

For benzene we have 12 nuclear

degrees of freedom (dof) and 42

electronic dof

For each set of nuclear dof, we solve HelΨ=EΨ to

calculate Ψ, the probability amplitudes for finding the 42

electrons at various locations

1st term: kinetic energy operator HOW MANY

2nd term: attraction of electrons to nuclei: HOW MANY

3rd term: electron-electron repulsion HOW MANY

4th term: what is missing?


The Schrödinger Equation: Kinetic Energy


Hel =

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

HelΨ=EΨ



electronic dof




1st term: kinetic energy operator: 42 terms





The Schrödinger Equation: Nuclear-Electron


Hel =

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

HelΨ=EΨ



electronic dof





2nd term: attraction of electrons to nuclei: 42*12= 504




The Schrödinger Equation: Electron-Electron


Hel =

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

HelΨ=EΨ



electronic dof






3rd term: electron-electron repulsion 42*41/2= 861 terms



The Schrödinger Equation: Nuclear-Nuclear


Hel =

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

HelΨ=EΨ

Missing is the nuclear-nuclear repulsion

Enn = SA/<Ψ|Ψ>=

Etotal = Eel + Enn

EES810-L1-2011 14

Homework and Research Project

First 8 weeks: The homework each week uses generally available

computer software implementing the basic methods on

applications aimed at exposing the students to understanding how

to use atomistic simulations to solve problems.

Each calculation requires making decisions on the specific

approaches and parameters relevant and how to analyze the

results.

Midterm: each student submits proposal for a project using the

methods of EEW80.810 to solve a research problem that can be

completed in the final 7 weeks.

The homework for the last 7 weeks is to turn in a one page report

on progress with the project

The final is a research report describing the calculations and

conclusions with a one hour oral describing the results

EES810-L1-2011 15

Methods to be covered in the lectures include:

Quantum Mechanics: Hartree Fock and Density Function

methods

Force Fields standard FF commonly used for simulations of

organic, biological, inorganic, metallic systems, reactions;

ReaxFF reactive force field: for describing chemical reactions,

shock decomposition, synthesis of films and nanotubes, catalysis

Molecular Dynamics: structure optimization, vibrations, phonons,

elastic moduli, Verlet, microcanonical, Nose, Gibbs

Monte Carlo and Statistical thermodynamics Growth

amorphous structures, Kubo relations, correlation functions, RIS,

CCBB, FH methods growth chains, Gauss coil, theta temp

Coarse grain approaches

eFF for electron dynamics

Tight Binding for electronic properties

solvation, diffusion,

mesoscale force fields

EES810-L1-2011 16

Applications will include prototype examples

involving such materials as:

Organic molecules (structures, reactions);

Semiconductors (IV, III-V, surface reconstruction)

Ceramics (BaTiO3, LaSrCuOx)

Metal alloys (crystalline, amorphous, plasticity)

Polymers (amorphous, crystalline, RIS theory, block);

Protein structure, ligand docking

DNA-structure, ligand docking

EES810-L1-2011 17

Outline Topic 1: QM: HF, DFT, basis sets, reactions, transition states, vibrations

Topic 2: Force Fields, nonbonds, hydrogen bonds, charges (QEq), QM FF

Topic 3: Molecular Dynamics: Verlet, NVE, NVT, NPT, Periodic Systems

Topic 4: Statistical mechanics: liquid simulations, entropy, nonequilibrium MD,

Green-Kubo, Monte Carlo, Grand Canonical MC, gas storage, surface tension

Topic 5: polymers: crystalline, amorphous, structure prediction,

Topic 6: ReaxFF Reactive Force Field and reactive Dynamics

Topic 7: PBC QM, band structure, phonons, elastic constants, Ab Initio MD,

Topic 8: surfaces: reconstruction, chemisorption, physisorption, solvation

Topic 9: eFF, Tight binding

Topic 9: applications:

Fuel Cell: oxygen reduction reaction, migration in Nafion solid oxides

Batteries: anode for Li battery, Solid electrolyte interface,

Nanotechnology: rotaxane molecular switches, carbon nanotube interfaces

Water Treatment: Dendrimers, captymers

Catalysis: alkane activation, ammoxidation

Metallic alloys: force fields, dislocations, plasticity, cavitation

Proteins: structures, ligand docking, GPCRs

DNA: A to B transition, Origami based nanotechnology

EES810-L1-2011 18

Topic 1: Practical Quantum Chemistry


Hel =

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

HelΨ=EΨ

For benzene we have 12 nuclei and

hence 3*12=36 degrees of freedom

(dof) and 42 electrons or 3*42 dof




1st term: kinetic energy operator HOW MANY




EES810-L1-2011 19

The Schrödinger Equation: Kinetic Energy


Hel =

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

HelΨ=EΨ











EES810-L1-2011 20

The Schrödinger Equation: Nuclear-Electron


Hel =

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

HelΨ=EΨ











EES810-L1-2011 21

The Schrödinger Equation: Electron-Electron


Hel =

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

HelΨ=EΨ






3rd term: electron-electron repulsion 42*41/2= 861 terms





EES810-L1-2011 22

The Schrödinger Equation: Nuclear-Nuclear


Hel =

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

HelΨ=EΨ

Missing is the nuclear-nuclear repulsion

Enn = SA/<Ψ|Ψ>=

Etotal = Eel + Enn

EES810-L1-2011 23

The stratospheric review of QM

You should have already been exposed to all the

material on the next xx slides

This is just a review to remind you of the key points

EES810-L1-2011 24

Quantum Mechanics – First postulate

The essential element of QM is that all properties that can

be known about the system are contained in the

wavefunction, Φ(x,y,z,t) (for one electron), where the

probability of finding the electron at position x,y,z at time t

is given by

P(x,y,z,t) = | Φ(x,y,z,t) |2 = Φ(x,y,z,t)* Φ(x,y,z,t)

Note that ∫Φ(x,y,z,t)* Φ(x,y,z,t) dxdydz = 1

since the total probability of finding the electron

somewhere is 1.

I write this as < Φ|Φ>=1, where it is understood that the

integral is over whatever the spatial coordinates of Φ are

EES810-L1-2011 25

Quantum Mechanics – Second postulate In QM the total energy can be written as

EQM = KEQM + PEQM

where for a system with a classical potential energy function,

V(x,y,z,t)

PEQM=∫Φ(x,y,z,t)*V(x,y,z,t)Φ(x,y,z,t)dxdydz ≡ < Φ| V|Φ>

Just like Classical mechanics except that V is weighted by P=|Φ|2

For the H atom

PEQM=< Φ| (-e2/r) |Φ> = -e2/

where is the average value of 1/r R _

R _

KEQM = (Ћ2/2me) <(Φ·Φ>

where <(Φ·Φ> ≡ ∫ [(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] dxdydz

In QM the KE is proportional to the average square of the gradient

or slope of the wavefunction

Thus KE wants smooth wavefunctions, no wiggles

EES810-L1-2011 26

Summary 2nd Postulate QM EQM = KEQM + PEQM

where for a system with a potential energy function, V(x,y,z,t)

PEQM= < Φ| V|Φ>=∫Φ(x,y,z,t)*V(x,y,z,t)Φ(x,y,z,t)dxdydz

Just like Classical mechanics except weight V with P=|Φ|2

KEQM = (Ћ2/2me) <(Φ·Φ>

where <(Φ·Φ> ≡ ∫ [(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] dxdydz

The stability of the H atom was explained by this KE (proportional

to the average square of the gradient of the wavefunction).

We will use the preference of KE for smooth wavefunctions to

explain the bonding in H2+ and H2.

However to actually solve for the wavefunctions requires the

Schrodinger Eqn., which we derive next.

We have assumed a normalized wavefunction, <Φ|Φ> = 1

EES810-L1-2011 27

3rd Postulate of QM, the variational principle

Consider that Φex is the exact wavefunction with energy

Eex = <Φ’|Ĥ|Φ’>/<Φ’|Φ’> and that

Φap = Φex + dΦ is some other approximate wavefunction.

Then Eap = <Φap|Ĥ|Φap>/<Φap|Φap> ≥ Eex

Eex Eap

E This means that for sufficiently small

dΦ, dE = 0 for all possible changes,

dΦ

We write dE/dΦ = 0 for all dΦ

This is called the variational principle.

For the ground state, d2E/dΦ ≥ 0 for all

possible changes

The ground state wavefunction is the system, Φ, has the lowest

possible energy out of all possible wavefunctions.

EES810-L1-2011 28

Write the energy of any approximate wavefunction, Φap, as

Eap = <Φap|Ĥ|Φap>/<Φap|Φap>

Ignoring terms 2nd order in dΦap, we obtain

<Φap|Ĥ|Φap> = Eex + <dΦ|Ĥ|Φex> + <Φex|Ĥ|dΦ>

= Eex + 2 Re[<dΦ|Ĥ|Φex>]

<Φap|Φap> = 1 + <dΦ|Φex> + <Φex|dΦ>

= 1 + 2 Re[<dΦ|Φex>]

where Re means the real part. To 1st order:

(a + db)/(1+dd) = [a /(1+dd)] + db = a+ db –a dd = a+ db –a dd

Thus Eap - Eex = 2 Re[<dΦ|Ĥ|Φex>]} - Eex{2 Re[<dΦ|Φex>]}

Eap - Eex = 2 Re[<dΦ|Ĥ-Eex|Φex>]} = 0 for all possible dΦ

But ∫ dΦ*[(Ĥ-Eex)Φex] = 0 for all possible dΦ [(Ĥ-Eex)Φex] = 0

This leads to the Schrödinger equation Ĥ Φex = EexΦex

Side comment: the next 4 slides Derive Schrödinger equation

from variational principle. You are not responsible for this

EES810-L1-2011 29

Derivation of Schrodinger Equation

Assume

EQM = {(Ћ2/2me)<(dΦ/dx)| (dΦ/dx)> + <Φ|V|Φ>}/<Φ|Φ>

Variational principle says that ground state Φ0 leads to the lowest

possible E, E0

Then starting with this optimum Φ0 , and making any change, dΦ

will increase E.

The first order change in E is

dE = (Ћ2/2me)<(d dΦ/dx)| (dΦ/dx)> + < dΦ| V|Φ> + CC

Integrate by parts

dE = -(Ћ2/2me)<(dΦ| (d2Φ/dx2)> + < dΦ| V|Φ> + CC

EES810-L1-2011 30

Derivation of Schrodinger Equation But even though <Φ0|Φ0> = 1, changing Φ0 by dΦ, might change

the normalization.

Thus we get an additional term

E+dE = E0/{<Φ0|Φ0> + <dΦ|Φ0> + CC} = E0 – E0{<dΦ|Φ0> + CC}

Thus

dE ={-(Ћ2/2me)<(dΦ| (d2Φ0/dx2)>+< dΦ|V|Φ0> -E0<dΦ|V|Φ0>} + CC

At a minimum the energy must increase for both +dΦ and –dΦ,

hence dE=0 = <(dΦ| {-(Ћ2/2me)(d2/dx2)+V -E0}|Φ>} + CC

Must get dE=0 for all possible dΦ, hence the coefficient of dΦ,

must be zero. Get

(H - E0)Φ=0 where H= {-(Ћ2/2me)(d2/dx2)+V} or HΦ= E0Φ

EES810-L1-2011 31

Summary deriviation of Schrödinger Equation

EQM = <Φ| | Φ> + < Φ| V|Φ> = <Φ| Ĥ | Φ>

where the Hamiltonian is Ĥ ≡ + V and = - (Ћ2/2me)2

And we assume a normalized wavefunction, <Φ|Φ> = 1

V(x,y,z,t) is the (classical) potential energy for the system

KE ^

KE ^

KE ^

Consider arbitrary Φap = Φex + dΦ and require that

dE= Eap – Eex = 0

Get <dΦ|Ĥ-Eex|Φex>] = ∫ dΦ*[(Ĥ-Eex)Φex] = 0 for all possible dΦ

This [(Ĥ-Eex)Φex] = 0 or the Schrödinger equation

Ĥ Φex = EexΦex

The exact ground state wavefunction is a solution of this equation

EES810-L1-2011 32

4th postulate of QM - Excited states The Schrödinger equation Ĥ Φk = EkΦk

Has an infinite number of solutions or eigenstates (German

for characteristic states), each corresponding to a possible

exact wavefunction for an excited state

For example H atom: 1s, 2s, 2px, 3dxy etc

Also the g and u states of H2+ and H2.

These states are orthogonal: <Φj|Φk> = djk= 1 if j=k

= 0 if j≠k

Note < Φj| Ĥ|Φk> = Ek < Φj|Φk> = Ek djk

We will denote the ground state as k=0

The set of all eigenstates of Ĥ is complete, thus any arbitrary

function Ө can be expanded as

Ө = Sk Ck Φk where <Φj| Ө>=Cj or Ө = Sk Φk <Φk| Ө>

EES810-L1-2011 33

Phase factor

Consider the exact eigenstate of a system

HΦ = EΦ

and multiply the Schrödinger equation by some CONSTANT

phase factor (independent of position and time)

exp(ia) = eia

eia HΦ = H (eia Φ) = E (eia Φ)

Thus Φ and (eia Φ) lead to identical properties and we

consider them to describe exactly the same state.

wavefunctions differing only by a constant phase factor

describe the same state

EES810-L1-2011 34

Configuration interaction

Consider a set of N-electron wavefunctions:

{i; i=1,2, ..M}

where < i|j> = dij {=1 if i=j and 0 if i ≠j)

Write approx = S (i=1 to M) Ci i

Then E = < approx|H|approx>/< approx|approx>

E= < Si Ci i |H| Sk Ck k >/ < Si Ci i | Si Ck k >

How choose optimum Ci?

Require dE=0 for all dCi get

Sk - Ei = 0 ,which we

write as ΣikHikCki = ΣikSikCkiEi

where Hjk = <j|H|k > and Sjk = < j|k >

Which we write as HCi = SCiEi in matrix notation

Ci is a column vector for the ith eigenstate

EES810-L1-2011 35

Configuration interaction upper bound theorem

Consider the M solutions of the CI equations

HCi = SCiEi ordered as i=1 lowest to i=M highest

Then the exact ground state energy of the system

Satisfies Eexact ≤ E1

Also the exact first excited state of the system satisfies

E1st excited ≤ E2

etc

This is called the Hylleraas-Unheim-McDonald Theorem

EES810-L1-2011 36

Electron spin, 5th postulate QM Consider application of a magnetic field

Our Hamiltonian has no terms dependent on the magnetic field.

Hence no effect.

But experimentally there is a huge effect. Namely

The ground state of H atom splits into two states

This leads to the 5th postulate of QM

In addition to the 3 spatial coordinates x,y,z each electron has

internal or spin coordinates that lead to a magnetic dipole aligned

either with the external magnetic field or opposite.

We label these as a for spin up and b for spin down. Thus the

ground states of H atom are φ(xyz)a(spin) and φ(xyz)b(spin)

B=0 Increasing B a

b

EES810-L1-2011 37

Permutational symmetry, summary Our Hamiltonian for H2,

H(1,2) =h(1) + h(2) + 1/r12 + 1/R

Does not involve spin

This it is invariant under 3 kinds of permutations

Space only: r1 r2

Spin only: s1 s2

Space and spin simultaneously: (r1,s1) (r2,s2)

Since doing any of these interchanges twice leads to the identity,

we know that

Ψ(2,1) = Ψ(1,2) symmetry for transposing spin and space coord

Φ(2,1) = Φ(1,2) symmetry for transposing space coord

Χ(2,1) = Χ(1,2) symmetry for transposing spin coord

EES810-L1-2011 38

Permutational symmetries for H2 and He

H2

He

Have 4 degenerate g

ground states for H2

Have 4 degenerate u

excited states for H2

Have 4 degenerate

ground state for He

EES810-L1-2011

H2

He

39

Permutational symmetries for H2 and He

the only states

observed are

those for

which the

wavefunction

changes sign

upon

transposing all

coordinates of

electron 1 and

2

Leads to the

6th postulate of

QM

EES810-L1-2011 40

The 6th postulate of QM: the Pauli Principle

For every eigenstate of an electronic system

H(1,2,…i…j…N)Ψ(1,2,…i…j…N) = EΨ(1,2,…i…j…N)

The electronic wavefunction Ψ(1,2,…i…j…N) changes

sign upon transposing the total (space and spin)

coordinates of any two electrons

Ψ(1,2,…j…i…N) = - Ψ(1,2,…i…j…N)

We can write this as

tij Ψ = - Ψ for all I and j

EES810-L1-2011 41

Consider H atom

The Hamiltonian has the form

h = - (Ћ2/2me)2 – Ze2/r

In atomic units: Ћ=1, me=1, e=1

h = - ½ 2 – Z/r

r

+Ze

We will consider one electron, but a nucleus with charge Z

φnlm = Rnl(r) Zlm(θ,φ)

Thus we want to solve

hφk = ekφk for the ground and excited states k

where Rnl(r) depends only on r and

Zlm(θ,φ) depends only on θ and φ

Assume φ10 = exp(-zr) E = ½ z2 – Z z

dE/dz = z – Z = 0 z = Z

EES810-L1-2011 42

The H atom ground state the ground state of H atom is

φ1s = N0 exp(-Zr/a0) ~ exp(-Zr) where we will ignore normalization

Line plot along z, through the origin

Maximum amplitude at z = 0

1

x = 0

Contour plot in the xz

plane, Maximum

amplitude at x,z = 0,0

EES810-L1-2011

We will use atomic units for which me = 1, e = 1, Ћ = 1

For H atom the average size of the 1s orbital is

a0 = Ћ2/ mee2 = 0.529 A =0.053 nm = 1 bohr is the unit of length

For H atom the energy of the 1s orbital []ionization potential (IP) of

H atom is

e1s = - ½ me e4/ Ћ2 = - ½ h0 = -13.61 eV = -313.75 kcal/mol

In atomic units the unit of energy is me e4/ Ћ2 = h0 = 1, denoted as

the Hartree

Note h0 = e2/a0 = 27.2116 eV = 627.51 kcal/mol = 2625.5 kJ/mol

The kinetic energy of the 1s state is KE1s = ½ and

the potential energy is PE1s = -1 = 1/R, where R = 1 a0 is the

average radius

43

Atomic units

EES810-L1-2011 44

The excited states of H atom - 1

The ground and excited states of a system can all be written as

hφk = ekφk, where <φk |φj> = dkj

Here dkj the Kronecker delta function is 0 when j=k, but it is

0 otherwise

We say that different excited states are orthogonal.

EES810-L1-2011

Nodal theorem

45

The ground state has no nodes (never changes sign),

like the 1s state for H atom

EES810-L1-2011 46


Use spherical polar coordinates, r, θ, φ

where z = rcosθ, x = rsinθcosφ, y = rsinθsinφ

2 = d2/dx2 + d2/dy2 + d2/dy2 transforms like

r2 = x2 + y2 + z2 so that it is independent of θ, φ

Thus h(r,θ,φ) = - ½ 2 – Z/r is independent of θ

and φ

x

z

y θ

φ

EES810-L1-2011 47


Use spherical polar coordinates, r, θ, φ

where z=r cosθ, x=r sinθ cosφ, y=r sinθ sinφ

2 = d2/dx2 + d2/dy2 + d2/dy2 transforms like

r2 = x2 + y2 + z2 so that it is independent of θ, φ

Thus h(r,θ,φ) = - ½ 2 – Z/r is independent of θ

and φ

x

z

y θ

φ

Consequently the eigenfunctions of h can be factored into Rnl(r)

depending only on r and Zlm(θ,φ) depending only on θ and φ

φnlm = Rnl(r) Zlm(θ,φ)

The reason for the numbers nlm will be apparent later

EES810-L1-2011 48

Excited radial functions

Consider excited states with Znl = 1; these are ns states with l=0

The lowest is R10 = 1s = exp(-Zr), the ground state.

All other radial functions must be orthogonal to 1s, and hence

must have one or more radial nodes.

The cross section is plotted along the z axis, but it would look

exactly the same along any other axis. Here

R20 = 2s = [Zr/2 – 1] exp(-Zr/2)

R30 = 3s = [2(Zr)2/27 – 2(Zr)/3 + 1] exp(-Zr/3)

Zr = 2 Zr = 1.9

Zr = 7.1

0 nodal

planes 1 spherical

nodal plane

2 spherical

nodal planes

EES810-L1-2011 49

Angularly excited states

Ground state 1s = φ100 = R10(r) Z00(θ,φ), where Z00 = 1 (constant)

Now consider excited states, φnlm = Rnl(r) Zlm(θ,φ), whose angular

functions, Zlm(θ,φ), are not constant, l ≠ 0.

Assume that the radial function is smooth, like R(r) = exp(-ar)

Then for the excited state to be orthogonal to the 1s state, we

must have

<Z00(θ,φ)|Zlm(θ,φ)> = 0

Thus Zlm must have nodal planes with equal positive and negative

regions.

The simplest cases are

rZ10 = z = r cosθ, which is zero in the xy plane

rZ11 = x = r sinθ cosφ, which is zero in the yz plane

rZ1,-1 = y = r sinθ sinφ, which is zero in the xz plane

These are denoted as angular momentum l=1 or p states

EES810-L1-2011 50

The p excited angular states of H atom φnlm = Rnl(r) Zlm(θ,φ)

Now lets consider excited angular functions, Zlm.

They must have nodal planes to be orthogonal to Z00

x

z

+

-

pz

The simplest would be Z10=z = r cosθ, which is

zero in the xy plane.

Exactly equivalent are

Z11=x = rsinθcosφ which is zero in the yz plane,

and

Z1-1=y = rsinθsinφ, which is zero in the xz plane

Also it is clear that these 3 angular functions

with one angular nodal plane are orthogonal to

each other. Thus <Z10|Z11> = <pz|px>=0 since

the integrand has nodes in both the xy and xz

planes, leading to a zero integral

x

z

+ -

px

x

z

+

-

pxpz -

+

pz

EES810-L1-2011 51

More p functions? So far we have the s angular function Z00 = 1 with no angular

nodal planes

And three p angular functions: Z10 =pz, Z11 =px, Z1-1 =py, each

with one angular nodal plane

Can we form any more angular functions with one nodal plane

orthogonal to all 4 of the above functions?

x

z

+

-

px’

a

x

z

+ -

pzpx’

a

+ -

For example we might rotate px by an angle a

about the y axis to form px’. However multiplying,

say by pz, leads to the integrand pzpx’ which

clearly does not integrate to zero

. Thus there are exactly three pi functions, Z1m,

with m=0,+1,-1, all of which have the same KE.

Since the p functions have nodes, they lead to a

higher KE than the s function (assuming no

radial nodes)

EES810-L1-2011 52

More angular functions?

So far we have the s angular function Z00 = 1 with no angular

nodal planes

And three p angular functions: Z10 =pz, Z11 =px, Z1-1 =py, each with

one angular nodal plane

Next in energy will be the d functions with two angular nodal

planes. We can easily construct three functions

x

z

+

-

dxz -

+

dxy = xy =r2 (sinθ)2 cosφ sinφ

dyz = yz =r2 (sinθ)(cosθ) sinφ

dzx = zx =r2 (sinθ)(cosθ) cosφ

where dxz is shown here

Each of these is orthogonal to each other (and to the s and the

three p functions). <dxy|dyz> = ʃ (x z y2) = 0, <px|dxz> = ʃ (z x2) = 0,

EES810-L1-2011 53

In addition we can construct three other functions with two

nodal planes

dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2]

dy2-z2 = y2 – z2 = r2 [(sinθ)2(sinφ)2 – (cosθ)2]

dz2-x2 = z2 – x2 = r2 [(cosθ)2 -(sinθ)2(cosφ)2]

where dz2-x2 is shown here,

Each of these three is orthogonal to the previous three d

functions (and to the s and the three p functions)

This leads to six functions with 2 nodal planes

More d angular functions?

x

z

+

-

dz2-x2

-

+

EES810-L1-2011 54

In addition we can construct three other functions with two

nodal planes

dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2]

dy2-z2 = y2 – z2 = r2 [(sinθ)2(sinφ)2 – (cosθ)2]

dz2-x2 = z2 – x2 = r2 [(cosθ)2 -(sinθ)2(cosφ)2]

where dz2-x2 is shown here,

Each of these three is orthogonal to the previous three d

functions (and to the s and the three p functions)

This leads to six functions with 2 nodal planes

More d angular functions?

x

z

+

-

dz2-x2

-

+

However adding these 3 (x2 – y2) + (y2 – z2) + (z2 – x2) = 0

Which indicates that there are only two independent such

functions. We combine the 2nd two as

(z2 – x2) - (y2 – z2) = [2 z2 – x2 - y2 ] = [3 z2 – x2 - y2 –z2] =

= [3 z2 – r2 ] which we denote as dz2

EES810-L1-2011 55

Summarizing the d angular functions

x

z

+

-

dz2

-

+

57º

Z20 = dz2 = [3 z2 – r2 ] m=0, ds

Z21 = dzx = zx =r2 (sinθ)(cosθ) cosφ

Z2-1 = dyz = yz =r2 (sinθ)(cosθ) sinφ

Z22 = dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2]

Z22 = dxy = xy =r2 (sinθ)2 cosφ sinφ

We find it useful to keep track of how often the

wavefunction changes sign as the φ coordinate is

increased by 2p = 360º

When this number, m=0 we call it a s function

When m=1 we call it a p function

When m=2 we call it a d function

When m=3 we call it a f function

m = 1, dp

m = 2, dd

EES810-L1-2011 56

Summarizing the angular functions

So far we have

•one s angular function (no angular nodes) called ℓ=0

•three p angular functions (one angular node) called ℓ=1

•five d angular functions (two angular nodes) called ℓ=2

Continuing we can form

•seven f angular functions (three angular nodes) called ℓ=3

•nine g angular functions (four angular nodes) called ℓ=4

where ℓ is referred to as the angular momentum quantum number

And there are (2ℓ+1) m values for each ℓ

EES810-L1-2011 57

real (Zlm) and complex (Ylm) ang. momentum fnctns

Here the bar over

m negative

EES810-L1-2011 58

Combination of radial and angular

nodal planes

Combining radial and angular functions gives the

various excited states of the H atom. They are

named as shown where the n quantum number is

the total number of nodal planes plus 1

The nodal theorem does not specify how 2s and

2p are related, but it turns out that the total

energy depends only on n.

Enlm = - Z2/2n2

The potential energy is given by

PE = - Z2/n2 = -Z/ , where =n2/Z

Thus Enlm = - Z/2

1s 0 0 0 1.0

2s 1 1 0 4.0

2p 1 0 1 4.0

3s 2 2 0 9.0

3p 2 1 1 9.0

3d 2 0 2 9.0

4s 3 3 0 16.0

4p 3 2 1 16.0

4d 3 1 2 16.0

4f 3 0 3 16.0

nam

e

tota

l nodal pla

nes

rad

ial n

od

al p

lan

es

angula

r nodal pla

nes

ˉ R ˉ R

ˉ R

Siz

e (

a0)

This is all you need to remember

EES810-L1-2011 59

Sizes hydrogen orbitals

Hydrogen orbitals 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f

Angstrom (0.1nm) 0.53, 2.12, 4.77, 8.48

H--H C

0.74

H

H

H

H

1.7

H H

H H

H H

4.8

=a0 n2/Z ˉ R Where a0 = bohr = 0.529A=0.053 nm = 52.9 pm

EES810-L1-2011 60

Hydrogen atom excited states

1s -0.5 h0 = -13.6 eV

2s -0.125 h0 = -3.4 eV

2p

3s -0.056 h0 = -1.5 eV

3p 3d

4s -0.033 h0 = -0.9 eV

4p 4d 4f

Energy zero

Enlm = - Z/2 ˉ R = - Z2/2n2

EES810-L1-2011 61

Plotting of orbitals:

line cross-section vs.

contour

contour plot

in yz plane

line plot

along z axis

EES810-L1-2011 62

Contour plots of 1s, 2s, 2p hydrogenic orbitals

EES810-L1-2011 63

Contour plots of 3s, 3p, 3d hydrogenic

orbitals

EES810-L1-2011 64

Contour plots of 4s, 4p, 4d hydrogenic

orbtitals

EES810-L1-2011 65

Contour plots of hydrogenic 4f orbitals

EES810-L1-2011 66

He+ atom

Next lets review the energy for He+.

Writing Φ1s = exp(-zr) we determine the optimum z for He+ as

follows

<1s|KE|1s> = + ½ z2 (goes as the square of 1/size)

<1s|PE|1s> = - Zz (linear in 1/size)

E(He+) = + ½ z2 - Zz

Applying the variational principle, the optimum z must satisfy

dE/dz = z - Z = 0 leading to z = Z,

KE = ½ Z2, PE = -Z2, E=-Z2/2 = -2 h0.

writing PE=-Z/R0, we see that the average radius is R0=1/z = 1/2

So that the He+ orbital is ½ the size of the H orbital

EES810-L1-2011 67

Estimate J1s,1s, the electron repulsion

energy of 2 electrons in He+ 1s orbitals

How can we estimate J1s,1s

Assume that each electron moves on a sphere

With the average radius R0 = 1/z =1/2

And assume that e1 at along the z axis (θ=0)

Neglecting correlation in the electron motions, e2 will on the

average have θ=90º so that the average e1-e2 distance is

~sqrt(2)*R0

Thus J1s,1s ~ 1/[sqrt(2)*R0] = 0.707 z

R0

e1

e2

Now consider He atom: EHe = 2(½ z2) – 2Zz J1s,1s

EES810-L1-2011 68

Estimate J1s,1s, the electron repulsion

energy of 2 electrons in He+ 1s orbitals

How can we estimate J1s,1s

Assume that each electron moves on a sphere

With the average radius R0 = 1/z =1/2

And assume that e1 at along the z axis (θ=0)

Neglecting correlation in the electron motions, e2 will on the

average have θ=90º so that the average e1-e2 distance is

~sqrt(2)*R0

Thus J1s,1s ~ 1/[sqrt(2)*R0] = 0.707 z

A rigorous calculation gives

J1s,1s = (5/8) z = 0.625 z = (5/16) h0 = 8.5036 eV = 196.1 kcal/mol

R0

e1

e2

Now consider He atom: EHe = 2(½ z2) – 2Zz J1s,1s

Since e1s = -Z2/2 = -2 h0 = 54.43 eV = 1,254.8 kcal/mol the

electron repulsion increases the energy (less attractive) by 15.6%

EES810-L1-2011 69

The optimum atomic orbital for He atom

He atom: EHe = 2(½ z2) – 2Zz (5/8)z


dE/dz = 0 leading to

2z - 2Z + (5/8) = 0

Thus z = (Z – 5/16) = 1.6875

KE = 2(½ z2) = z2

PE = - 2Zz (5/8)z = -2 z2

E= - z2 = -2.8477 h0

EES810-L1-2011 70

The optimum atomic orbital for He atom

He atom: EHe = 2(½ z2) – 2Zz (5/8)z


dE/dz = 0 leading to

2z - 2Z + (5/8) = 0

Thus z = (Z – 5/16) = 1.6875

KE = 2(½ z2) = z2

PE = - 2Zz (5/8)z = -2 z2

E= - z2 = -2.8477 h0

Ignoring e-e interactions the energy would have been E = -4 h0

The exact energy is E = -2.9037 h0 (from memory, TA please

check).

Thus this simple approximation of assuming that each electron is

in a 1s orbital and optimizing the size accounts for 98.1% of the

exact result.

EES810-L1-2011 71

Interpretation: The optimum atomic orbital for He atom

Assume He(1,2) = Φ1s(1)Φ1s(2) with Φ1s = exp(-zr)

We find that the optimum z = (Z – 5/16) = Zeff = 1.6875

With this value of z, the total energy is E= - z2 = -2.8477 h0

This wavefunction can be interpreted as two electrons moving

independently in the orbital Φ1s = exp(-zr) which has been

adjusted to account for the average shielding due to the other

electron in this orbital.

On the average this other electron is closer to the nucleus about

31.25% of the time so that the effective charge seen by each

electron is Zeff = 2 - 0.3125=1.6875

The total energy is just the sum of the individual energies,

E = -2 (Zeff2/2) = -2.8477 h0

Ionizing the 2nd electron, the 1st electron readjusts to z = Z = 2

with E(He+) = -Z2/2 = - 2 h0.

Thus the ionization potential (IP) is 0.8477 h0 = 23.1 eV (exact

value = 24.6 eV)

EES810-L1-2011 72

Now lets add a 3rd electron to form Li

ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb)(Φ1sg)]

Problem: with either g = a or g = b, we get ΨLi(1,2,3) = 0

Since there are two electrons in the same spinorbital

This is an essential result of the Pauli principle

Thus the 3rd electron must go into an excited orbital

ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2sa)]

or

ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py)

EES810-L1-2011 73

First consider Li+

First consider Li+ with ΨLi(1,2) = A[(Φ1sa)(Φ1sb)]

Here Φ1s = exp(-zr) with z = Z-0.3125 = 2.69 and

E = -z2 = -7.2226 h0.

For Li2+ we get E =-Z2/2=-4.5 h0

Thus the IP of Li+ is IP = 2.7226 h0 = 74.1 eV

The size of the 1s orbital for Li+ is

R0 = 1/z = 0.372 a0 = 0.2A

EES810-L1-2011 74

Consider adding the 3rd electron to the 2p

orbital ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py)

Since the 2p orbital goes to zero at z=0, there is very

little shielding so that the 2p electron sees an effective

charge of

Zeff = 3 – 2 = 1, leading to

a size of R2p = n2/Zeff = 4 a0 = 2.12A

And an energy of e = -(Zeff)2/2n2 = -1/8 h0 = 3.40 eV

0.2A

1s

2.12A

2p

EES810-L1-2011 75

Consider adding the 3rd electron to the 2s

orbital ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py)

The 2s orbital must be orthogonal to the 1s, which means that

it must have a spherical nodal surface below ~ 0.2A, the size

of the 1s orbital. Thus the 2s has a nonzero amplitude at z=0

so that it is not completely shielded by the 1s orbitals.

The result is Zeff2s = 3 – 1.72 = 1.28

This leads to a size of R2s = n2/Zeff = 3.1 a0 = 1.65A

And an energy of e = -(Zeff)2/2n2 = -0.205 h0 = 5.57 eV

0.2A

1s

2.12A 2s

R~0.2A

EES810-L1-2011 76

Li atom excited states

1s

2s

-0.125 h0 = -3.4 eV 2p

Energy

zero

-0.205 h0 = -5.6 eV

-2.723 h0 = -74.1 eV

MO picture State picture

(1s)2(2s)

(1s)2(2p)

DE = 2.2 eV

17700 cm-1

564 nm

Ground

state

1st

excited

state

Exper

671 nm

DE = 1.9 eV

EES810-L1-2011 77

Aufbau principle for atoms

1s

2s

2p

3s

3p 3d

4s

4p 4d

4f

Energy

2

2

6

2

6 2

6

10

10 14

He, 2

Ne, 10

Ar, 18

Zn, 30

Kr, 36

Get generalized energy

spectrum for filling in the

electrons to explain the

periodic table.

Full shells at 2, 10, 18, 30,

36 electrons

EES810-L1-2011 78

He, 2

Ne, 10

Ar, 18

Zn, 30

Kr, 36

EES810-L1-2011 79

Many-electron configurations

General

aufbau

ordering

Particularly stable

EES810-L1-2011 80

General trends along a row of the periodic

table As we fill a shell, say B(2s)2(2p)1 to Ne (2s)2(2p)6

we add one more proton to the nucleus and one more electron to

the valence shell

But the valence electrons only partially shield each other.

Thus Zeff increases, leading to a decrease in the radius ~ n2/Zeff

And an increase in the IP ~ (Zeff)2/2n2

Example Zeff2s=

1.28 Li, 1.92 Be, 2.28 B, 2.64 C, 3.00 N, 3.36 O, 4.00 F, 4.64 Ne

Thus (2s Li)/(2s Ne) ~ 4.64/1.28 = 3.6

EES810-L1-2011 81

General trends along a column of the

periodic table As we go down a column

Li [He}(2s) to Na [Ne]3s to K [Ar]4s to Rb [Kr]5s to Cs[Xe]6s

We expect that the radius ~ n2/Zeff

And the IP ~ (Zeff)2/2n2

But the Zeff tends to increase, partially compensating for

the change in n so that the atomic sizes increase only

slowly as we go down the periodic table and

The IP decrease only slowly (in eV):

5.39 Li, 5.14 Na, 4.34 K, 4.18 Rb, 3.89 Cs

(13.6 H), 17.4 F, 13.0 Cl, 11.8 Br, 10.5 I, 9.5 At

24.5 He, 21.6 Ne, 15.8 Ar, 14.0 Kr,12.1 Xe, 10.7 Rn

EES810-L1-2011 82

Plot of rφ(r) for

the outer s

valence orbital

EES810-L1-2011 83

Plot of rφ(r) for

the outer s and

p valence

orbitals

Note for C row

2s and 2p have

similar size, but

for other rows

ns much

smaller than np

EES810-L1-2011 84

Plot of rφ(r) for the

outer s and p valence

orbitals

Note for C row 2s

and 2p have similar

size, but for other

rows ns much

smaller than np

EES810-L1-2011 85

Transition metals; consider [Ar] + 1 electron [IP4s = (Zeff

4s )2/2n2 = 4.34 eV Zeff

4s = 2.26; 4s<4p<3d

IP4p = (Zeff4p )

2/2n2 = 2.73 eV Zeff4p = 1.79;

IP3d = (Zeff3d )

2/2n2 = 1.67 eV Zeff3d = 1.05;

IP4s = (Zeff4s )

2/2n2 = 11.87 eV Zeff4s = 3.74; 4s<3d<4p

IP3d = (Zeff3d )

2/2n2 = 10.17 eV Zeff3d = 2.59;

IP4p = (Zeff4p )

2/2n2 = 8.73 eV Zeff4p = 3.20;

IP3d = (Zeff3d )

2/2n2 = 24.75 eV Zeff3d = 4.05; 3d<4s<4p

IP4s = (Zeff4s )

2/2n2 = 21.58 eV Zeff4s = 5.04;

IP4p = (Zeff4p )

2/2n2 = 17.01 eV Zeff4p = 4.47;

K

Ca+

Sc++

As the net charge increases the differential shielding for 4s vs 3d

is less important than the difference in n quantum number 3 vs 4

Thus charged system prefers 3d vs 4s

EES810-L1-2011 86

Transition metals; consider Sc0, Sc+, Sc2+

3d: IP3d = (Zeff3d )

2/2n2 = 24.75 eV Zeff3d = 4.05;

4s: IP4s = (Zeff4s )

2/2n2 = 21.58 eV Zeff4s = 5.04;

4p: IP4p = (Zeff4p )

2/2n2 = 17.01 eV Zeff4p = 4.47;

Sc++

As increase net charge increases, the differential shielding for 4s

vs 3d is less important than the difference in n quantum number 3

vs 4. Thus M2+ transition metals always have all valence

electrons in d orbitals

(3d)(4s): IP4s = (Zeff4s )

2/2n2 = 12.89 eV Zeff4s = 3.89;

(3d)2: IP3d = (Zeff3d )

2/2n2 = 12.28 eV Zeff3d = 2.85;

(3d)(4p): IP4p = (Zeff4p )

2/2n2 = 9.66 eV Zeff4p = 3.37;

Sc+

(3d)(4s)2: IP4s = (Zeff4s )

2/2n2 = 6.56 eV Zeff4s = 2.78;

(4s)(3d)2: IP3d = (Zeff3d )

2/2n2 = 5.12 eV Zeff3d = 1.84;

(3d)(4s)(4p): IP4p = (Zeff4p )

2/2n2 = 4.59 eV Zeff4p = 2.32;

Sc

EES810-L1-2011 87

Implications for transition metals

The simple Aufbau principle puts 4s below 3d

But increasing the charge tends to prefers 3d vs 4s.

Thus Ground state of Sc 2+ , Ti 2+ …..Zn 2+ are all (3d)n

For all neutral elements K through Zn the 4s orbital is

easiest to ionize.

This is because of increase in relative stability of 3d for

higher ions

EES810-L1-2011 88

Transtion metal valence ns and (n-1)d orbitals

EES810-L1-2011 89

Review over, back to quantum mechanics

Stopped Lecture 1

EES810-L1-2011 90

Excited states The Schrödinger equation Ĥ Φk = EkΦk

Has an infinite number of solutions or eigenstates (German

for characteristic states), each corresponding to a possible

exact wavefunction for an excited state

For example H atom: 1s, 2s, 2px, 3dxy etc

Also the g and u states of H2+ and H2.

These states are orthogonal: <Φj|Φk> = djk= 1 if j=k

= 0 if j≠k

Note < Φj| Ĥ|Φk> = Ek < Φj|Φk> = Ek djk

We will denote the ground state as k=0

The set of all eigenstates of Ĥ is complete, thus any arbitrary

function Ө can be expanded as

Ө = Sk Ck Φk where <Φj| Ө>=Cj or Ө = Sk Φk <Φk| Ө>


Quick fix to satisfy the Pauli Principle

Combine the product wavefunctions to form a symmetric

combination

Ψs(1,2)= ψe(1) ψm(2) + ψm(1) ψe(2)

And an antisymmetric combination

Ψa(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)

We see that

t12 Ψs(1,2) = Ψs(2,1) = Ψs(1,2) (boson symmetry)

t12 Ψa(1,2) = Ψa(2,1) = -Ψa(1,2) (Fermion symmetry)

Thus for electrons, the Pauli Principle only allows the

antisymmetric combination for two independent

electrons


Implications of the Pauli Principle

Consider two independent electrons,

1 on the earth described by ψe(1)

and 2 on the moon described by ψm(2)

Ψ(1,2)= ψe(1) ψm(2)

And test whether this satisfies the Pauli Principle

Ψ(2,1)= ψm(1) ψe(2) ≠ - ψe(1) ψm(2)

Thus the Pauli Principle does NOT allow

the simple product wavefunction for two

independent electrons


Consider some simple cases: identical spinorbitals

Ψ(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)

Identical spinorbitals: assume that ψm = ψe

Then Ψ(1,2)= ψe(1) ψe(2) - ψe(1) ψe(2) = 0

Thus two electrons cannot be in identical spinorbitals

Note that if ψm = eia ψe where a is a constant phase

factor, we still get zero


Consider some simple cases: orthogonality

Consider the wavefunction

Ψold(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)

where the spinorbitals ψm and ψe are orthogonal

hence <ψm|ψe> = 0

Define a new spinorbital θm = ψm + l ψe (ignore normalization)

That is NOT orthogonal to ψe. Then

Ψnew(1,2)= ψe(1) θm(2) - θm(1) ψe(2) =

=ψe(1) θm(2) + l ψe(1) ψe(2) - θm(1) ψe(2) - l ψe(1) ψe(2)

= ψe(1) ψm(2) - ψm(1) ψe(2) =Ψold(1,2)

Thus the Pauli Principle leads to orthogonality of

spinorbitals for different electrons, <ψi|ψj> = dij = 1 if i=j

=0 if i≠j


Consider some simple cases: nonuniqueness

Starting with the wavefunction

Ψold(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)

Consider the new spinorbitals θm and θe where

θm = (cosa) ψm + (sina) ψe

θe = (cosa) ψe - (sina) ψm Note that <θi|θj> = dij

Then Ψnew(1,2)= θe(1) θm(2) - θm(1) θe(2) =

+(cosa)2 ψe(1)ψm(2) +(cosa)(sina) ψe(1)ψe(2)

-(sina)(cosa) ψm(1) ψm(2) - (sina)2 ψm(1) ψe(2)

-(cosa)2 ψm(1) ψe(2) +(cosa)(sina) ψm(1) ψm(2)

-(sina)(cosa) ψe(1) ψe(2) +(sina)2 ψe(1) ψm(2)

[(cosa)2+(sina)2] [ψe(1)ψm(2) - ψm(1) ψe(2)] =Ψold(1,2)

Thus linear combinations of the spinorbitals do not change Ψ(1,2)

ψe

ψm

θe

θm

a

a a


Determinants

The determinant of a matrix is defined as

The determinant is zero if any two columns (or rows) are identical

Adding some amount of any one column to any other column

leaves the determinant unchanged.

Thus each column can be made orthogonal to all other

columns.(and the same for rows) The above properties are just those of the Pauli Principle

Thus we will take determinants of our wavefunctions.


The antisymmetrized wavefunction

Where the antisymmetrizer can be thought of as the

determinant operator.

Similarly starting with the 3!=6 product wavefunctions of the form

Now put the spinorbitals into the matrix and take the determinant

The only combination satisfying the Pauil Principle is


Example:

From the properties of determinants we know that interchanging

any two columns (or rows), that is interchanging any two

spinorbitals, merely changes the sign of the wavefunction

Interchanging electrons 1 and 3 leads to

Guaranteeing that the Pauli Principle is always satisfied


Consider the product wavefunction

Ψ(1,2) = ψa(1) ψb(2)

And the Hamiltonian for H2 molecule

H(1,2) = h(1) + h(2) +1/r12 + 1/R

In the details slides next, we derive

E = < Ψ(1,2)| H(1,2)|Ψ(1,2)>/ <Ψ(1,2)|Ψ(1,2)>

E = haa + hbb + Jab + 1/R

where haa =<a|h|a>, hbb =<b|h|b> are just the 1 electron energies

Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>=ʃ [ψa(1)]2 [ψb(1)]2/r12

represents the total Coulomb interaction between the electron

density ra(1)=| ψa(1)|2 and rb(2)=| ψb(2)|2

Since the integrand ra(1) rb(2)/r12 is positive for all positions of 1

and 2, the integral is positive, Jab > 0

Energy for 2-electron product wavefunction


Details in deriving energy: normalization

First, the normalization term is

<Ψ(1,2)|Ψ(1,2)>=<ψa(1)|ψa(1)><ψb(2) ψb(2)>

Which from now on we will write as

<Ψ|Ψ> = <ψa|ψa><ψb|ψb> = 1 since the ψi are normalized

Here our convention is that a two-electron function such as

<Ψ(1,2)|Ψ(1,2)> is always over both electrons so we need not put

in the (1,2) while one-electron functions such as <ψa(1)|ψa(1)> or

<ψb(2) ψb(2)> are assumed to be over just one electron and we

ignore the labels 1 or 2


Using H(1,2) = h(1) + h(2) +1/r12 + 1/R

We partition the energy E = <Ψ| H|Ψ> as

E = <Ψ|h(1)|Ψ> + <Ψ|h(2)|Ψ> + <Ψ|1/R|Ψ> + <Ψ|1/r12|Ψ>

Here <Ψ|1/R|Ψ> = <Ψ|Ψ>/R = 1/R since R is a constant

<Ψ|h(1)|Ψ> = <ψa(1)ψb(2) |h(1)|ψa(1)ψb(2)> =

= <ψa(1)|h(1)|ψa(1)><ψb(2)|ψb(2)> = <a|h|a><b|b> =

≡ haa

Where haa≡ <a|h|a> ≡ <ψa|h|ψa>

Similarly <Ψ|h(2)|Ψ> = <ψa(1)ψb(2) |h(2)|ψa(1)ψb(2)> =

= <ψa(1)|ψa(1)><ψb(2)|h(2)|ψb(2)> = <a|a><b|h|b> =

≡ hbb

The remaining term we denote as

Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)> so that the total energy is


Details of deriving energy: one electron termss


The energy for an antisymmetrized product, A ψaψb

The total energy is that of the product plus the exchange term

which is negative with 4 parts

Eex=-< ψaψb|h(1)|ψb ψa >-< ψaψb|h(2)|ψb ψa >-< ψaψb|1/R|ψb ψa >

- < ψaψb|1/r12|ψb ψa >

The first 3 terms lead to < ψa|h(1)|ψb><ψbψa >+

<ψa|ψb><ψb|h(2)|ψa >+ <ψa|ψb><ψb|ψa>/R

But <ψb|ψa>=0

Thus all are zero

Thus the only nonzero term is the 4th term:

-Kab=- < ψaψb|1/r12|ψb ψa > which is called the exchange energy

(or the 2-electron exchange) since it arises from the exchange

term due to the antisymmetrizer.

Summarizing, the energy of the Aψaψb wavefunction for H2 is

E = haa + hbb + (Jab –Kab) + 1/R


The energy of the antisymmetrized wavefunction

The total electron-electron repulsion part of the energy for any

wavefunction Ψ(1,2) must be positive

Eee =∫ (d3r1)((d3r2)|Ψ(1,2)|2/r12 > 0

This follows since the integrand is positive for all positions of r1

and r2 then

We derived that the energy of the A ψa ψb wavefunction is


Where the Eee = (Jab –Kab) > 0

Since we have already established that Jab > 0 we can conclude

that

Jab > Kab > 0


Separate the spinorbital into orbital and spin parts

Since the Hamiltonian does not contain spin the spinorbitals can

be factored into spatial and spin terms.

For 2 electrons there are two possibilities:

Both electrons have the same spin

ψa(1)ψb(2)=[Φa(1)a(1)][Φb(2)a(2)]= [Φa(1)Φb(2)][a(1)a(2)]

So that the antisymmetrized wavefunction is

Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)]=

=[Φa(1)Φb(2)- Φb(1)Φa(2)][a(1)a(2)]

Also, similar results for both spins down

Aψa(1)ψb(2)= A[Φa(1)Φb(2)][b(1)b(2)]=

=[Φa(1)Φb(2)- Φb(1)Φa(2)][b(1)b(2)]

Since <ψa|ψb>= 0 = < Φa| Φb><a|a> = < Φa| Φb>

We see that the spatial orbitals for same spin must be orthogonal


Energy for 2 electrons with same spin

The total energy becomes


where haa ≡ <Φa|h|Φa> and hbb ≡ <Φb|h|Φb> where Jab= <Φa(1)Φb(2) |1/r12 |Φa(1)Φb(2)>

We derived the exchange term for spin orbitals with same spin as

follows

Kab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)>

`````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><a(1)|a(1)><a(2)|a(2)>

≡ Kab

where Kab ≡ <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)>

Involves only spatial coordinates.


Now consider the exchange term for spin orbitals with opposite

spin

Kab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)>

`````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><a(1)|b(1)><b(2)|a(2)>

= 0

Since <a(1)|b(1)> = 0.

Energy for 2 electrons with opposite spin

Thus the total energy is

Eab = haa + hbb + Jab + 1/R

With no exchange term unless the spins are the same

Since <ψa|ψb>= 0 = < Φa| Φb><a|b>

There is no orthogonality condition of the spatial orbitals for

opposite spin electrons

In general we can have <Φa|Φb> =S, where the overlap S ≠ 0


Summarizing: Energy for 2 electrons

When the spinorbitals have the same spin,

Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)]

The total energy is

Eaa = haa + hbb + (Jab –Kab) + 1/R

But when the spinorbitals have the opposite spin,

Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)b(2)]=

The total energy is

Eab = haa + hbb + Jab + 1/R

With no exchange term

Thus exchange energies arise only for the case in

which both electrons have the same spin


Consider further the case for spinorbtials with opposite spin

The wavefunction

[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)]

Leads directly to 3Eab = haa + hbb + (Jab –Kab) + 1/R

Exactly the same as for

[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)a(2)]

[Φa(1)Φb(2)-Φb(1)Φa(2)][b(1)b(2)]

These three states are collectively referred to as the triplet state

and denoted as having spin S=1


Consider further the case for spinorbtials with opposite spin

The other combination leads to one state, referred to as the

singlet state and denoted as having spin S=0

[Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)]

For the ground state of a 2-electron system, Φa=Φb so we get

[Φa(1)Φa(2)][a(1)b(2)-b(1)a(2)] = A[Φa(1)a(1)] [Φa(2)b(2)]

Leading directly to 1Eaa = 2haa + Jaa + 1/R

This state is referred to as the closed shell single state and

denoted as having spin S=0


Re-examine He atom with spin and the Pauli Principle

Ψ(1,2) = A[(φ1s a) (φ1s b)]

E= 2 <1s|h|1s> + J1s,1s

Which is exactly what we assumed above when we

ignore spin and the Pauli Principle

So for 2 electrons nothing changes


Consider the product wavefunction

Ψ(1,2) = ψa(1) ψb(2)

And the Hamiltonian for H2 molecule

H(1,2) = h(1) + h(2) +1/r12 + 1/R

In the details slides next, we derive

E = < Ψ(1,2)| H(1,2)|Ψ(1,2)>/ <Ψ(1,2)|Ψ(1,2)>


where haa =<a|h|a>, hbb =<b|h|b> are just the 1 electron energies

Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>=ʃ d3r1[ψa(1)]2 ʃd3r2[ψb(2)]2/r12 =

= ʃ [ψa(1)]2 Jb (1) = <ψa(1)| Jb (1)|ψa(1)>

Where Jb (1) = ʃ [ψb(2)]2/r12 is the Coulomb potential at 1 due to

the density distribution [ψb(2)]2

Energy for 2-electron product wavefunction

Jab is the Coulomb repulsion between densities ra=[ψa(1)]2 and rb


The energy for an antisymmetrized product,

Aψaψb

The total energy is that of the product wavefunction plus the new

terms arising from exchange term which is negative with 4 parts

Eex=-< ψaψb|h(1)|ψb ψa >-< ψaψb|h(2)|ψb ψa >-< ψaψb|1/R|ψb ψa >

- < ψaψb|1/r12|ψb ψa >

The first 3 terms lead to < ψa|h(1)|ψb><ψbψa >+

<ψa|ψb><ψb|h(2)|ψa >+ <ψa|ψb><ψb|ψa>/R

But <ψb|ψa>=0

Thus all are zero

Thus the only nonzero term is the 4th term:

-Kab=- < ψaψb|1/r12|ψb ψa > which is called the exchange energy

(or the 2-electron exchange) since it arises from the exchange

term due to the antisymmetrizer.

Summarizing, the energy of the Aψaψb wavefunction for H2 is


One new term from the antisymmetrizer


Summary electron-electron energies

Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>=<ψa(1)| Jb (1)|ψa(1)>

is the total Coulomb interaction between the electron density

ra(1)=| ψa(1)|2 and rb(2)=| ψb(2)|2

Since the integrand ra(1) rb(2)/r12 is positive for all positions of 1

and 2, the integral is positive, Jab > 0

Here Jb (1) = ʃ [ψb(2)]2/r12 is the potential at 1 due to the density

distribution [ψb(2)]2

Kab=< ψaψb|1/r12|ψb ψa >= ʃ d3r1[ψa(1)ψb(1)] ] ʃd3r2[ψb(2) ψa(2)]]2/r12

= <ψa(1)| Kb (1)|ψa(1)>

Where Kb (1) ψa(1)] ] = ψb(1) ʃ [ψb(2)ψa(2)]2/r12 is an integral

operator that puts Kab into a form that looks like Jab. The

difference is that Jb (1) is a function but Kb (1) is an operator


Alternative form for electron-electron energies

It is commen to rewrite Jab as

Jab ≡ [ψa(1) ψa(1)|ψb(2)ψb(2)] where all the electron 1 stuff is on

the left and all the electron 2 stuff is on the right. Note that the

1/r12 is now understood

Similarly Kab= [ψa(1)ψb(1)|ψb(2)ψa(2)]

Here the numbers 1 and 2 are superflous so we write

Jab ≡ [ψaψa|ψbψb] = [aa|bb] since only the orbital names are

significant

Siimilarly

Kab ≡ [ψaψb|ψbψa] = [ab|ba]

Thus the total 2 electron energy is

Jab - Kab = [aa|bb] - [ab|ba]

But if a and b have opposite spin then the exchange term is zero


Consider the case of 4 e in 2 orbitals, a,b

Ψ(1,2,3,4) = A[(aa)(ab)(ba)(bb)]

E = 2 haa + 2 hbb + Enn + [2Jaa-Kaa] +2[2Jab-Kab] + [2Jbb-Kbb]

= 2 haa + 2 hbb + Enn + 2(aa|aa)-(aa|aa)+4(aa|bb)-2(ab|ba)

+2(bb|bb)-(bb|bb)

Where we see that the self-Coulomb and self-exchange can

cancel.

Now change φ1 to φ1 + dφ1 the change in the energy is

dE = 4<dφ1|h|φ1> + 4 <dφ1|2J1-K1|φ1> + 4 <dφ1|2J2-K2|φ1>

= 4 <dφ1|HHF|φ1>

Where HHF = h + Σj=1,2 [2Jj-Kj] is called the HF Hamiltonian

In the above expression we assume that φ1 was normalized,

<φ1|h|φ1> = 1.

Imposing this constraint (a Lagrange multiplier) leads to

<dφ1|HHF – l1|φ1> = 0 and <dφ2|H

HF – l2|φ2> = 0

Thus the optimum orbitals satisfy HHFφk = lk φk the HF equations


Starting point for First Principles QM

Energy = Kinetic energy + Potential energy

Kinetic energy =

Potential energy =

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

Nucleus-Nucleus

repulsion

Nucleus-Electron

attraction Electron-Electron

repulsion

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

atoms electrons

p p + +

Classical Mechanics

Can optimize electron coordinates and momenta separately,

thus lowest energy: all p=0 KE =0

All electrons on nuclei: PE = - infinity

Makes for dull world


Ab Initio, quantum mechanics

Starting point for First Principles

Energy = < Ψ|KE operator|Ψ> + < Ψ|PE operator|Ψ>

Kinetic energy op =

Potential energy =

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

atoms electrons

Optimize Ψ, get HelΨ=EΨ

Hel =

The wavefunction Ψ(r1,r2,…,rN) contains all

information of system determine KE and PE

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22


HelΨ=EΨ

Hel =

Schrodinger Equation

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

A

A

A rR

Z

R

ZZ

M

1

2

1

2

1 22

Solving SE gives exact properties of molecules, solids,

enzymes, etc

History

H atom, Schrodinger 1925-26

H2 Simple (Valence bond) 1927, accurate 1937

C2H6 simple 1963, accurate 1980’s

2008: can get accurate wavefunctions for ~100-200

atoms


Closed shell Hartree Fock (HF)

For benzene with 42 electrons, the ground state HF wavefunction

has 21 doubly occupied orbitals, φ1,.. φi,.. φ21 And we want to determine the optimum shape and energy for

these orbitals

First consider the componets of the total energy

Σ i=1,21< φi|h|φi> from the 21 up spin orbitals

Σ i=1,21< φi|h|φi> from the 21 down spin orbitals

Σ I<j=1,21 [Jij – Kij] interactions between the 21 up spin orbitals

Σ I<j=1,21 [Jij – Kij] interactions between the 21 down spin orbitals

Σ I≠j=1,21 [Jij] interactions of the 21 up spin orbitals with the 21

down spin orbitals

Enn = Σ A<B=1,12 ZAZB/RAB nuclear-nuclear repulsion

Combining these terms leads to

E = Σ i=1,21 2< φi|h|φi> + Enn + Σ I≠j=1,21 2[2Jij-Kij] + Σ I=1,21 [2Jii]

But Jii = Kii so we can rewrite this as


The energy expression for closed shell HF

E = Σ i=1,21 2< φi|h|φi> + Enn + Σ I<j=1,21 2[2Jij-Kij] + Σ I=1,21 [Jii]

This says for any two different orbitals we get 4 coulomb

interactions and 2 exchange interactions, but the two electrons in

the same orbital only lead to a single Coulomb term

Since Jii = Kii (self coulomb = self exchange) we can write

E = Σ i=1,21 2< φi|h|φi> + Enn + Σ I≠j=1,21 [2Jij-Kij] + Σ I=1,21 [2Jii-Kii]

and hence

E = Σ i=1,21 2< φi|h|φi> + Enn + Σ I,j=1,21 [2Jij-Kij]

which is the final expression for Closed Shell HF

Now we need to apply the variational principle to find the

equations determining the optimum orbitals, the HF orbitals


Consider the case of 4 electrons in 2 orbitals

E = 2<φ1|h|φ1> + 2< φ2|h|φ2> + Enn

+ [2J11-K11] +2[2J12-K12] + [2J22-K22]

Here we can write Jij = (ii|jj) where the first two indices go with

electron 1 and the other two with electron 2

Also we write Jij = (ii|jj) = <i|Jj|i>, where Jj is the coulomb potetial

seen by electron 1 due to the electron in orbital j.

Thus if we change φ1 to φ1 + dφ1 the change in the energy is

dE = 4<dφ1|h|φ1> + 4 <dφ1|2J1-K1|φ1> + 4 <dφ1|2J2-K2|φ1>

= 4 <dφ1|HHF|φ1>

Where HHF = h + Σj=1,2 [2Jj-Kj] is called the HF Hamiltonian

In the above expression we assume that φ1 was normalized,

<φ1|h|φ1> = 1.

Imposing this constraint (a Lagrange multiplier) leads to

<dφ1|HHF – l1|φ1> = 0 and <dφ2|H

HF – l2|φ2> = 0

Thus the optimum orbitals satisfy HHFφk = lk φk the HF equations


The general case of 2M electrons

For the general case the HF closed shell equations are

HHFφk = lk φk where we solve for k=1,M occupied orbitals

HHF = h + Σj=1,M [2Jj-Kj]

This is the same as the Hamiltonian for a one electron system

moving in the average electrostatic and exchange potential, 2Jj-Kj

due to the other N-1 = 2M-1 electrons

Problem: sum over 2Jj leads to 2M Coulomb terms, not 2M-1

This is because we added the self Coulomb and exchange terms

But (2Jk-Kk) φk = (Jk) φk so that these self terms cancel.


Analyze HF equations

The optimum orbitals for the 4 electron closed shell wavefunction

Ψ(1,2,3,4) = A[(aa)(ab)(ba)(bb)]

Are eigenstate of the HF equations

HHFφk = lk φk for k=1,2

where HHF = h + Σj=1,2 [2Jj-Kj]

This looks like a one-electron Hamiltonian but it involves the

average Coulomb potential of 2 electrons in φa plus 2 electrons in

φb plus exchange interactions with one electron in φa plus one

electron in φb

It seems wrong that there should be 4 coulomb interactions

whereas each electron sees only 3 other electrons and that there

are two exchange interactions whereas each electron sees only

one other with the same spin.

This arises because we added and subtracted a self term in the

total energy

Since (Jk-Kk)φk = 0 there spurious terms cancel.


Main practical applications of QM

Determine the Optimum geometric structure and

energies of molecules and solids

Determine geometric structure and energies of

reaction intermediates and transition states for

various reaction steps

Determine properties of the optimized

geometries: bond lengths, energies,

frequencies, electronic spectra, charges

Determine reaction mechanism: detailed

sequence of steps from reactants to products


The Matrix HF equations

The HF equations are actually quite complicated because Kj

is an integral operator, Kj φk(1) = φj(1) ʃ d3r2 [φj(2) φk(2)/r12]

The practical solution involves expanding the orbitals in terms

of a basis set consisting of atomic-like orbitals,

φk(1) = Σm Cm Xm, where the basis functions, {Xm, m=1, MBF}

are chosen as atomic like functions on the various centers

As a result the HF equations HHFφk = lk φk

Reduce to a set of Matrix equations

ΣjmHjmCmk = ΣjmSjmCmkEk

This is still complicated since the Hjm operator includes

Exchange terms


Minimal Basis set – STO-3G

For benzene the smallest possible basis set is to use a 1s-like

single exponential function, exp(-zr) called a Slater function,

centered on each the 6 H atoms and

C1s, C2s, C2pz, C2py, C2pz functions on each of the 6 C atoms

This leads to 42 basis functions to describe the 21 occupied MOs

and is refered to as a minimal basis set.

In practice the use of exponetial functions, such as exp(-zr),

leads to huge computational costs for multicenter molecules and

we replace these by an expansion in terms of Gaussian basis

functions, such as exp(-ar2).

The most popular MBS is the STO-3G set of Pople in which 3

gaussian functions are combined to describe each Slater function


Double zeta + polarization Basis sets – 6-31G**

To allow the atomic orbitals to contract as atoms are brought

together to form bonds, we introduce 2 basis functions of the

same character as each of the atomic orbitals:

Thus 2 each of 1s, 2s, 2px, 2py, and 2pz for C

This is referred to as double zeta. If properly chosen this leads to

a good description of the contraction as bonds form.

Often only a single function is used for the C1s, called split

valence

In addition it is necessary to provide one level higher angular

momentum atomic orbitals to describe the polarization involved in

bonding

Thus add a set of 2p basis functions to each H and a set of 3d

functions to each C.

The most popular such basis is referred to as 6-31G**


6-31G** and 6-311G**

6-31G** means that the 1s is described with 6 Gaussians,

the two valence basis functions use 3 gaussians for the

inner one and 1 Gaussian for the outer function

The first * use of a single d set on each heavy atom

(C,O etc)

The second * use of a single set of p functions on each

H

The 6-311G** is similar but allows 3 valence-like functions

on each atom.

There are addition basis sets including diffuse functions (+)

and additional polarization function (2d, f) (3d,2f,g), but

these will not be relvent to EES810


Effective Core Potentials (ECP, psuedopotentials)

For very heavy atoms, say starting with Sc, it is computationally

convenient and accurate to replace the inner core electrons

with effective core potentials

For example one might describe:

• Si with just the 4 valence orbitals, replacing the Ne core with

an ECP or

• Ge with just 4 electrons, replacing the Ni core

• Alternatively, Ge might be described with 14 electrons with the

ECP replacing the Ar core. This leads to increased accuracy

because the

• For transition metal atoms, Fe might be described with 8

electrons replacing the Ar core with the ECP.

• But much more accurate is to use the small Ne core, explicitly

treating the (3s)2(3p)6 along with the 3d and 4s electrons


Software packages

Jaguar: Good for organometallics

QChem: very fast for organics

Gaussian: many analysis tools

GAMESS

HyperChem

ADF

Spartan/Titan


Results for Benzene

The energy of the C1s orbital is ~ - Zeff2/2

where Zeff = 6 – 0.3125 = 5.6875

Thus e1s ~ -16.1738 h0 = - 440.12 eV.

This leads to 6 orbitals all with very similar energies.

This lowest has the + combination of all 6 1s orbitals,

while the highest alternates with 3 nodal planes.

There are 6 CH bonds and 6 CC bonds that are

symmetric with respect to the benzene plane, leading to

12 sigma MOs

The highest MOs involve the p electrons. Here there are

6 electrons and 6 pp atomic orbitals leading to 3 doubly

occupied and 3 empty orbitals with the pattern


Pi orbitals of benzene

Top view


The HF orbitals of

N2

With 14 electrons we

get M=7 doubly

occupied HF orbitals

We can visualize this

as a triple NN bond

plus valence lone

pairs


The energy diagram for N2

TAs put energies of 7

occupied orbitals plus

lowest 2 unoccupied

orbitals, use correct

symmetry notation


The HF orbitals of H2O



lowest 2 unoccupied


symmetry notation

Show orbitals


The HF orbitals of ethylene



lowest 2 unoccupied


symmetry notation

Show orbitals


The HF orbitals of benzene

TAs put energies of

21 occupied orbitals

plus lowest 4

unoccupied orbitals,

use correct symmetry

notation

Show orbitals


HF wavefunctions

Good distances, geometries, vibrational levels

But

breaking bonds is described extremely poorly

energies of virtual orbitals not good description of

excitation energies

cost scales as 4th power of the size of the

system.


Electron correlation

In fact when the electrons are close (rij small), the electrons

correlate their motions to avoid a large electrostatic repulsion,

1/rij

Thus the error in the HF equation is called electron correlation

For He atom

E = - 2.8477 h0 assuming a hydrogenic orbital exp(-zr)

E = -2.86xx h0 exact HF (TA look up the energy)

E = -2.9037 h0 exact

Thus the elecgtron correlation energy for He atom is 0.04xx h0

= 1.x eV = 24.x kcal/mol.

Thus HF accounts for 98.6% of the total energy


Configuration interaction

Consider a set of N-electron wavefunctions: {i;

i=1,2, ..M} where < i|j> = dij {=1 if i=j and 0 if i ≠j)

Write approx = S (i=1 to M) Ci i

Then E = < approx|H|approx>/< approx|approx>

E= < Si Ci i |H| Si Cj j >/ < Si Ci i | Si Cj j >

How choose optimum Ci?

Require dE=0 for all dCi get

Sj - Ei = 0 ,which we

write as

HCi = SCiEi in matrix notation, ie ΣjkHjkCki = ΣjkSjkCkiEi

where Hjk = <j|H|k > and Sjk = < j|k > and Ci is a

column vector for the ith eigenstate


Configuration interaction upper bound theorm

Consider the M solutions of the CI equations

HCi = SCiEi ordered as i=1 lowest to i=M highest

Then the exact ground state energy of the system

Satisfies Eexact ≤ E1

Also the exact first excited state of the system

satisfies

E1st excited ≤ E2

etc

This is called the Hylleraas-Unheim-McDonald

Theorem


Alternative to Hartree-Fork,

Density Functional Theory

Walter Kohn’s dream:

replace the 3N electronic degrees of freedom needed to define

the N-electron wavefunction Ψ(1,2,…N) with

just the 3 degrees of freedom for the electron density r(x,y,z).

It is not obvious that this would be possible but

P. Hohenberg and W. Kohn Phys. Rev. B 76, 6062 (1964).

Showed that there exists some functional of the density

that gives the exact energy of the system

=

rrr

VFV

HK ][rep-

min

Kohn did not specify the nature or form of this functional,

but research over the last 46 years has provided

increasingly accurate approximations to it. Walter Kohn (1923-)

Nobel Prize Chemistry 1998


The Hohenberg-Kohn theorem

The Hohenberg-Kohn theorem states that if N interacting

electrons move in an external potential, Vext(1..N), the

ground-state electron density r(xyz)=r(r) minimizes the

functional

E[r] = F[r] + ʃ r(r) Vext(r) d3r

where F[r] is a universal functional of r and the minimum

value of the functional, E, is E0, the exact ground-state

electronic energy.

Here we take Vext(1..N) = Si=1,..N SA=1..Z [-ZA/rAi], which is the

electron-nuclear attraction part of our Hamiltonian. HK do

NOT tell us what the form of this universal functional, only of

its existence


Proof of the Hohenberg-Kohn theorem

Mel Levy provided a particularly simple proof of Hohenberg-Kohn

theorem {M. Levy, Proc. Nat. Acad. Sci. 76, 6062 (1979)}.

Define the functional O as O[r(r)] = min <Ψ|O|Ψ>

|Ψ>r(r)

where we consider all wavefunctions Ψ that lead to the same

density, r(r), and select the one leading to the lowest expectation

value for <Ψ|O|Ψ>.

F[r] is defined as F[r(r)] = min <Ψ|F|Ψ>

|Ψ>r(r)

where F = Si [- ½ i2] + ½ Si≠k [1/rik].

Thus the usual Hamiltonian is H = F + Vext

Now consider a trial function Ψapp that leads to the density r(r)

and which minimizes <Ψ|F|Ψ>

Then E[r] = F[r] + ʃ r(r) Vext(r) d3r = <Ψ|F +Vext|Ψ> = <Ψ|H|Ψ>

Thus E[r] ≥ E0 the exact ground state energy.


The Kohn-Sham equations

Walter Kohn and Lou J. Sham. Phys. Rev. 140, A1133 (1965).

Provided a practical methodology to calculate DFT wavefunctions

They partitioned the functional E[r] into parts

E[r] = KE0 + ½ ʃʃd3r1 d3r2 [r(1) r(2)/r12 + ʃd3r r(r) Vext(r) + Exc[r(r)]

Where

KE0 = Si <φi| [- ½ i2 | φi> is the KE of a non-interacting electron

gas having density r(r). This is NOT the KE of the real system.

The 2nd term is the total electrostatic energy for the density r(r).

Note that this includes the self interaction of an electron with itself.

The 3rd term is the total electron-nuclear attraction term

The 4th term contains all the unknown aspects of the Density

Functional


Solving the Kohn-Sham equations

Requiring that ʃ d3r r(r) = N the total number of electrons and

applying the variational principle leads to

[d/dr(r)] [E[r] – m ʃ d3r r(r) ] = 0

where the Lagrange multiplier m = dE[r]/dr = the chemical

potential

Here the notation [d/dr(r)] means a functional derivative inside

the integral.

To calculate the ground state wavefunction we solve

HKS φi = [- ½ i2 + Veff(r)] φi = ei φi

self consistently with r(r) = S i=1,N <φi|φi>

where Veff (r) = Vext (r) + Jr(r) + Vxc(r) and Vxc(r) = dEXC[r]/dr

Thus HKS looks quite analogous to HHF


The Local Density Approximation (LDA)

We approximate Exc[r(r)] as

ExcLDA[r(r)] = ʃ d3r eXC(r(r)) r(r)

where eXC(r(r)) is derived from Quantum Monte Carlo

calculations for the uniform electron gas {DM Ceperley and BJ

Alder, Phys.Rev.Lett. 45, 566 (1980)}

It is argued that LDA is accurate for simple metals and simple

semiconductors, where it generally gives good lattice

parameters

It is clearly very poor for molecular complexes (dominated by

London attraction), and hydrogen bonding


Generalized gradient approximations

The errors in LDA derive from the assumption that the density

varies very slowly with distance.

This is clearly very bad near the nuclei and the error will depend

on the interatomic distances

As the basis of improving over LDA a powerful approach has been

to consider the scaled Hamiltonian

cxxc EEE = ] drρ(r),...ρ(r)ρ(r),εE xx =


LDA exchange

( ) ( )3

1

x

LDA

x rρAρε = xA = -3

1

π

3

4

3

.

Here we say that in LDA each electron interacts with all N

electrons but should be N-1. The exchange term cancels this

extra term. If density is uniform then error is proportional to 1/N.

since electron density is r = N/V


Generalized gradient approximations

cxxc EEE =

] drρ(r),...ρ(r)ρ(r),εE xx =

( ) ( )sFερρ,ε LDA

x

GGA

x =

( ) 3

4

3

12 ρπ24

ρs

=

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

0.0 5.0 10.0

S

F(S

)

B88

PW91

new(mix)

PBE

Becke 88

X3LYP

PBE

PW91

s

F(s) GGA functionals

( )( )

( )2

1

1

2

32

1

188B

sasinhsa1

sasasinhsa1sF

=

( )( ) ( )

( ) d

52

1

1

2s100

432

1

191PW

sasasinhsa1

seaasasinhsa1sF

2

=

Here ( )3

12

2 π48a = , 21 βa6a = , βA2

aa

x

3/1

2

23 = , 34 a

81

10a = ,

x

3/1

64

25

A2

10aa

= , and d = 4.

Becke9 b = 0.0042 a4 and a5 zero

Here ( )3

12

2 π48a = , 21 βa6a = , βA2

aa

x

3/1

2

23 = , 34 a

81

10a = ,

x

3/1

64

25

A2

10aa

= , and d = 4.


adiabatic connection formalism

The adiabatic connection formalism provides a rigorous way to define Exc.

It assumes an adiabatic path between the fictitious non-interacting KS system (λ =

0) and the physical system (λ = 1) while holding the electron density r fixed at its

physical λ = 1 value for all λ of a family of partially interacting N-electron systems:

] ]1

,0

xc xcE U dlr r l= is the exchange-correlation energy at intermediate coupling strength λ.

The only problem is that the exact integrand is unknown.

Becke, A.D. J. Chem. Phys. (1993), 98, 5648-5652.

Langreth, D.C. and Perdew, J. P. Phys. Rev. (1977), B 15, 2884-2902.

Gunnarsson, O. and Lundqvist, B. Phys. Rev. (1976), B 13, 4274-4298.

Kurth, S. and Perdew, J. P. Phys. Rev. (1999), B 59, 10461-10468.


Perdew, J.P. Ernzerhof, M. and Burke, K. J. Chem. Phys. (1996), 105, 9982-

9985.

Mori-Sanchez, P., Cohen, A.J. and Yang, W.T. J. Chem. Phys. (2006), 124,

091102-1-4.


Becke half and half functional

assume a linear model ,xcU a bl l=

take , 0

exact

xc xU El= = the exact exchange of the KS orbitals

approximate , 1 , 1

LDA

xc xcU Ul l= =

partition LDA LDA LDA

xc x cE E E=

set ;exact LDA exact

x xc xa E b E E= = ;exact LDA exact

x xc xa E b E E= =

Get half-and-half functional ] ( )

1 1

2 2

exact LDA LDA

xc x x cE E E Er =

Becke, A.D. J. Chem. Phys. (1993), 98, 1372-1377


Becke 3 parameter functional

] ( )B3

1 2 3

LDA exact LDA GGA GGA

xc xc x x x cE E c E E c E c Er = D D

Empirically modify half-and-half

where GGA

xED is the gradient-containing correction terms to the LDA exchange

GGA

cED is the gradient-containing correction to the LDA correlation,

1 2 3, ,c c c are constants fitted against selected experimental thermochemical data.

The success of B3LYP in achieving high accuracy demonstrates that errors of for

covalent bonding arise principally from the λ 0 or exchange limit, making it important

to introduce some portion of exact exchange

DFT

xcE



Perdew, J.P. Ernzerhof, M. and Burke, K. J. Chem. Phys. (1996), 105, 9982-

9985.

Mori-Sanchez, P., Cohen, A.J. and Yang, W.T. J. Chem. Phys. (2006), 124,

091102-1-4.


LDA: Slater exchange

Vosko-Wilk-Nusair correlation, etc

GGA: Exchange: B88, PW91, PBE, OPTX, HCTH, etc

Correlations: LYP, P86, PW91, PBE, HCTH, etc

Hybrid GGA: B3LYP, B3PW91, B3P86, PBE0,

B97-1, B97-2, B98, O3LYP, etc

Meta-GGA: VSXC, PKZB, TPSS, etc

Hybrid meta-GGA: tHCTHh, TPSSh, BMK, etc

Some popular DFT functionals


Truhlar’s DFT functionals

MPW3LYP, X1B95, MPW1B95, PW6B95, TPSS1KCIS, PBE1KCIS, MPW1KCIS, BB1K, MPW1K, XB1K, MPWB1K, PWB6K, MPWKCIS1K MPWLYP1w,PBE1w,PBELYP1w, TPSSLYP1w G96HLYP, MPWLYP1M , MOHLYP M05, M05-2x M06, M06-2x, M06-l, M06-HF

Hybrid meta-GGA M06 = HF tPBE + VSXC

156

Fundamental problem in standard DFT methods

Use QM calculations on small systems ~100 atoms get

accurate energies, geometries, stiffness

Fit QM to force field to describe big systems (104 -107 atoms)

Fit to obtain parameters for continuum systems

macroscopic properties based on first principles (QM)

Can predict novel materials where no empirical data available.

General Problem with DFT: bad

description of vdw attraction

(London dispersion)

Invalidates multiscale

paradigm


XYG3 approach to include London Dispersion in DFT

Görling-Levy coupling-constant perturbation expansion

] ]1

,0

xc xcE U dlr r l= Take initial slope as the 2nd order correlation energy:

, 2

, 0

0

2xc GL

xc c

UU E

l

l

ll

=

=

= =

where

22

2

ˆˆˆ1

4

i xi j eeGL

c

ij ii j i

fE

aa b

ab aa b a

e e e e e e

=

where is the electron-electron repulsion operator, is the local exchange operator,

and is the Fock-like, non-local exchange operator.

ˆee ˆ

x

f̂

,xcU a bl l= Substitute into with 22 GL

cb E= ;exact LDA exact

x xc xa E b E E= = or

Combine both approaches (2 choices for b) ( )2

1 2

GL DFT exact

c xc xb b E b E E=

] ( ) ( )R5 2

1 2 3 4

LDA exact LDA GGA PT LDA GGA

xc xc x x x c c cE E c E E c E c E E c Er = D D

a double hybrid DFT that mixes some exact exchange into while also introducing a

certain portion of into

DFT

xE2PT

cEDFT

cE

contains the double-excitation parts of 2PT

cE

22

2

ˆˆˆ1

4

i xi j eeGL

c

ij ii j i

fE

aa b

ab aa b a

e e e e e e

=

This is a fifth-rung functional (R5) using information from both occupied and virtual KS

orbitals. In principle can now describe dispersion

Sum over virtual orbtials


Final form of XYG3 DFT

] ( ) ( )R5 2

1 2 3 4

LDA exact LDA GGA PT LDA GGA

xc xc x x x c c cE E c E E c E c E E c Er = D D

we adopt the LYP correlation functional but constrain c4 = (1 – c3) to exclude

compensation from the LDA correlation term.

This constraint is not necessary, but it eliminates one fitting parameter.

Determine the final three parameters {c1, c2, c3} empirically by fitting only to the

thermochemical experimental data in the G3/99 set of 223 molecules:

Get {c1 = 0.8033, c2 = 0.2107, c3 = 0.3211} and c4 = (1 – c3) = 0.6789

Use 6-311+G(3df,2p) basis set

XYG3 leads to mean absolute deviation (MAD) =1.81 kcal/mol,

B3LYP: MAD = 4.74 kcal/mol.

M06: MAD = 4.17 kcal/mol

M06-2x: MAD = 2.93 kcal/mol

M06-L: MAD = 5.82 kcal/mol .

G3 ab initio (with one empirical parameter): MAD = 1.05

G2 ab initio (with one empirical parameter): MAD = 1.88 kcal/mol

but G2 and G3 involve far higher computational cost.


Thermochemical accuracy with size

G3/99 set has 223 molecules:

G2-1: 56 molecules having up to 3 heavy atoms,

G2-2: 92 additional molecules up to 6 heavy atoms

G3-3: 75 additional molecules up to 10 heavy atoms.

B3LYP: MAD = 2.12 kcal/mol (G2-1), 3.69 (G2-2), and 8.97 (G3-3) leads to

errors that increase dramatically with size

B2PLYP MAD = 1.85 kcal/mol (G2-1), 3.70 (G2-2) and 7.83 (G3-3) does not

improve over B3LYP

M06-L MAD = 3.76 kcal/mol (G2-1), 5.71 (G2-2) and 7.50 (G3-3).

M06-2x MAD = 1.89 kcal/mol (G2-1), 3.22 (G2-2), and 3.36 (G3-3).

XYG3, MAD = 1.52 kcal/mol (G2-1), 1.79 (G2-2), and 2.06 (G3-3), leading to

the best description for larger molecules.


Accuracy (kcal/mol) of various QM methods for

predicting standard enthalpies of formation Functional MAD Max(+) Max(-)

DFT

XYG3 a 1.81 16.67 (SF6) -6.28 (BCl3)

M06-2x a 2.93 20.77 (O3) -17.39 (P4)

M06 a 4.17 11.25 (O3) -25.89 (C2F6)

B2PLYP a 4.63 20.37(n-octane) -8.01(C2F4)

B3LYP a 4.74 19.22 (SF6) -8.03 (BeH)

M06-L a 5.82 14.75 (PF5) -27.13 (C2Cl4)

BLYP b 9.49 41.0 (C8H18) -28.1 (NO2)

PBE b 22.22 10.8 (Si2H6) -79.7 (azulene)

LDA b 121.85 0.4 (Li2) -347.5 (azulene)

Ab initio

HFa 211.48 582.72(n-octane) -0.46 (BeH)

MP2a 10.93 29.21(Si(CH3)4) -48.34 (C2F6)

QCISD(T) c 15.22 42.78(n-octane) -1.44 (Na2)

G2(1 empirical parm) 1.88 7.2 (SiF4) -9.4 (C2F6)

G3(1 empirical parm) 1.05 7.1 (PF5) -4.9 (C2F4)


-5.00

0.00

5.00

10.00

15.00

20.00

25.00

30.00

-2.00 -1.50 -1.00 -0.50 0.00 0.50 1.00 1.50 2.00 2.50

Reaction coordinate

En

erg

y (

kca

l/m

ol)

HF

HF_PT2

XYG3

CCSD(T)

B3LYP

BLYP

SVWN

HF

HF_PT2 SVWN B3LYP

BLYP

XYG3 CCSD(T)

SVWN

H + CH4 H2 + CH3

Reaction Coordinate: R(CH)-R(HH) (in Å)

Energ

y (

kcal/m

ol)

Comparison of QM methods for reaction surface of

H + CH4 H2 + CH3


Reaction

barrier

heights

19 hydrogen transfer (HT) reactions,

6 heavy-atom transfer (HAT) reactions,

8 nucleophilic substitution (NS) reactions and

5 unimolecular and association (UM) reactions.

Functional All (76) HT38 HAT12 NS16 UM10

DFT

XYG3 1.02 0.75 1.38 1.42 0.98

M06-2x a 1.20 1.13 1.61 1.22 0.92

B2PLYP 1.94 1.81 3.06 2.16 0.73

M06 a 2.13 2.00 3.38 1.78 1.69

M06-La 3.88 4.16 5.93 3.58 1.86

B3LYP 4.28 4.23 8.49 3.25 2.02

BLYP a 8.23 7.52 14.66 8.40 3.51

PBEa 8.71 9.32 14.93 6.97 3.35

LDAb 14.88 17.72 23.38 8.50 5.90

Ab initio

HFb 11.28 13.66 16.87 6.67 3.82

MP2 b 4.57 4.14 11.76 0.74 5.44

QCISD(T) b 1.10 1.24 1.21 1.08 0.53

Zhao and Truhlar

compiled benchmarks

of accurate barrier

heights in 2004

includes forward and

reverse barrier heights

for

Note: no reaction

barrier heights used

in fitting the 3

parameters in

XYG3)


(A)

-15.00

-10.00

-5.00

0.00

5.00

10.00

15.00

20.00

25.00

30.00

3.0 4.0 5.0 6.0

Intermolecular distance

En

erg

y (

kca

l/m

ol)

BLYP

B3LYP

XYG3

CCSD(T)

SVWN

HF_PT2

(C)

-12.00

-9.00

-6.00

-3.00

0.00

Ec_VWN

Ec_B3LYP

Ec_LYP

Ec_XYG3

Ec_CCSD(T)

Ec_PT2

(B)

-5.00

0.00

5.00

10.00

15.00

20.00

25.00

30.00

3.0 4.0 5.0 6.0

Ex_B

Ex_B3LYP

Ex_XYG3

Ex_HF

Ex_S

HF

HF_PT2

B3LYP

BLYP

CCSD(T)

LDA

(SVWN)

A. Total Energy (kcal/mol)

Distance (A)

XYG3

B. Exchange Energy (kcal/mol)

C. Correlation Energy (kcal/mol)

B

S

B3LYP

XYG3

PT2

B3LYP

LYP CCSD(T)

VWN

XYG3

Distance (A)

Conclusion: XYG3 provides excellent accuracy for London dispersion, as good as

CCSD(T)

Test for

London

Dispersion


Accuracy of QM methods for noncovalent interactions.

Functional Total HB6/04 CT7/04 DI6/04 WI7/05 PPS5/05

DFT

M06-2x b 0.30 0.45 0.36 0.25 0.17 0.26

XYG3 a 0.32 0.38 0.64 0.19 0.12 0.25

M06 b 0.43 0.26 1.11 0.26 0.20 0.21

M06-L b 0.58 0.21 1.80 0.32 0.19 0.17

B2PLYP 0.75 0.35 0.75 0.30 0.12 2.68

B3LYP 0.97 0.60 0.71 0.78 0.31 2.95

PBE c 1.17 0.45 2.95 0.46 0.13 1.86

BLYP c 1.48 1.18 1.67 1.00 0.45 3.58

LDA c 3.12 4.64 6.78 2.93 0.30 0.35

Ab initio

HF 2.08 2.25 3.61 2.17 0.29 2.11

MP2c 0.64 0.99 0.47 0.29 0.08 1.69

QCISD(T) c 0.57 0.90 0.62 0.47 0.07 0.95

HB: 6 hydrogen bond

complexes,

CT 7 charge-transfer

complexes

DI: 6 dipole

interaction complexes,

WI:7 weak interaction

complexes,

PPS: 5 pp stacking

complexes. WI and PPS dominated

by London dispersion.

Note: no

noncovalent

complexes used

in fitting the 3

parameters in

XYG3)


Problem

] ]1

,0

xc xcE U dlr r l= Take initial slope as the 2nd order correlation energy:

, 2

, 0

0

2xc GL

xc c

UU E

l

l

ll

=

=

= =

where

22

2

ˆˆˆ1

4

i xi j eeGL

c

ij ii j i

fE

aa b

ab aa b a

e e e e e e

=

where is the electron-electron repulsion operator, is the local exchange operator,

and is the Fock-like, non-local exchange operator.

ˆee ˆ

x

f̂

Sum over virtual orbtials

XYG3 approach to include London Dispersion in DFT

Görling-Levy coupling-constant perturbation expansion

EGL2 involves double excitations to virtuals, scales as N5 with size

MP2 has same critical step

Yousung Jung (KAIST) has figured out how to get linear scaling for MP2

XYGJ-OS and XYGJ-OS


A solution: XYGJ-OS: include excitations to virtual orbitals in

order to describe London Dispersion in DFT

Goes beyond using just density (occupied orbitals)

Scales as (size)**3 just as B3LYP (CCSD scales as (size)**7

] ( ) ( )XYGJ- OS 2

2 ,1

HF S VWN LYP PT

xc x x x x VWN c LYP c PT c osE e E e E e E e E e E

lr =

Get {ex, eVWN, eLYP, ePT2} ={0.7731,0.2309, 0.2754, 0.4364}.

include only opposite spin & only local contributions N**3 scaling

A fast doubly hybrid density functional method close

to chemical accuracy: XYGJ-OS

Igor Ying Zhang, Xin Xu,

Yousung Jung, WAG

PNAS (2011) in press

Xin Xu


XYG4-OS and XYG4-LOS timings

0.0

40.0

80.0

120.0

160.0

200.0

0 20 40 60 80 100 120

alkane chain length

CP

U (

ho

urs)

XYG4-LOS

XYG4-OS

B3LYP

XYG3

Timings XYGJ-OS and XYGJ-LOS for long alkanes


0.0

40.0

80.0

120.0

160.0

200.0

0 20 40 60 80 100 120

alkane chain length

CP

U (

ho

urs)

XYG4-LOS

XYG4-OS

B3LYP

XYG3


0.0

40.0

80.0

120.0

160.0

200.0

0 20 40 60 80 100 120

alkane chain length

CP

U (

ho

urs)

XYG4-LOS

XYG4-OS

B3LYP

XYG3


0.0

40.0

80.0

120.0

160.0

200.0

0 20 40 60 80 100 120

alkane chain length

CP

U (

ho

urs)

XYG4-LOS

XYG4-OS

B3LYP

XYG3


0.0

40.0

80.0

120.0

160.0

200.0

0 20 40 60 80 100 120

alkane chain lengthC

PU

(h

ou

rs)

XYG4-LOS

XYG4-OS

B3LYP

XYG3

XYGJ-OS

XYGJ-LOS

XYGJ-LOS

XYGJ-OS


Accuracy of Methods (Mean absolute deviations MAD, in eV) HOF IP EA PA BDE NHTBH HTBH NCIE

All Time Methods

(223) (38) (25) (8) (92) (38) (38) (31) (493) C100H202 C100H100

DFT methods

SPL (LDA) 5.484 0.255 0.311 0.276 0.754 0.542 0.775 0.140 2.771

BLYP 0.412 0.200 0.105 0.080 0.292 0.376 0.337 0.063 0.322

PBE 0.987 0.161 0.102 0.072 0.177 0.371 0.413 0.052 0.562

TPSS 0.276 0.173 0.104 0.071 0.245 0.391 0.344 0.049 0.250

B3LYP 0.206 0.162 0.106 0.061 0.226 0.202 0.192 0.041 0.187 2.8 12.3

PBE0 0.300 0.165 0.128 0.057 0.155 0.154 0.193 0.031 0.213

M06-2X 0.127 0.130 0.103 0.092 0.069 0.056 0.055 0.013 0.096

XYG3 0.078 0.057 0.080 0.070 0.068 0.056 0.033 0.014 0.065 200.0 81.4

XYGJ-lOS 0.072 0.055 0.084 0.067 0.033 0.049 0.038 0.015 0.056 7.8 46.4

MC3BB 0.165 0.120 0.175 0.046 0.111 0.062 0.036 0.023 0.123

B2PLYP 0.201 0.109 0.090 0.067 0.124 0.090 0.078 0.023 0.143

Wavefunction based methods

HF 9.171 1.005 1.148 0.133 0.104 0.397 0.582 0.098 4.387

MP2 0.474 0.163 0.166 0.084 0.363 0.249 0.166 0.028 0.338

G2 0.082 0.042 0.057 0.058 0.078 0.042 0.054 0.025 0.068

G3 0.046 0.055 0.049 0.046 0.047 0.042 0.054 0.025 0.046

HOF = heat of formation; IP = ionization potential,

EA = electron affinity, PA = proton affinity,

BDE = bond dissociation energy,

NHTBH, HTBH = barrier heights for reactions,

NCIE = the binding in molecular clusters


Comparison of speeds

HOF IP EA PA BDE NHTBH HTBH NCIE All Time

Methods

(223) (38) (25) (8) (92) (38) (38) (31) (493) C100H202 C100H100

DFT methods

SPL (LDA) 5.484 0.255 0.311 0.276 0.754 0.542 0.775 0.140 2.771

BLYP 0.412 0.200 0.105 0.080 0.292 0.376 0.337 0.063 0.322

PBE 0.987 0.161 0.102 0.072 0.177 0.371 0.413 0.052 0.562

TPSS 0.276 0.173 0.104 0.071 0.245 0.391 0.344 0.049 0.250

B3LYP 0.206 0.162 0.106 0.061 0.226 0.202 0.192 0.041 0.187 2.8 12.3

PBE0 0.300 0.165 0.128 0.057 0.155 0.154 0.193 0.031 0.213

M06-2X 0.127 0.130 0.103 0.092 0.069 0.056 0.055 0.013 0.096

XYG3 0.078 0.057 0.080 0.070 0.068 0.056 0.033 0.014 0.065 200.0 81.4

XYGJ-lOS 0.072 0.055 0.084 0.067 0.033 0.049 0.038 0.015 0.056 7.8 46.4

MC3BB 0.165 0.120 0.175 0.046 0.111 0.062 0.036 0.023 0.123

B2PLYP 0.201 0.109 0.090 0.067 0.124 0.090 0.078 0.023 0.143


HF 9.171 1.005 1.148 0.133 0.104 0.397 0.582 0.098 4.387

MP2 0.474 0.163 0.166 0.084 0.363 0.249 0.166 0.028 0.338

G2 0.082 0.042 0.057 0.058 0.078 0.042 0.054 0.025 0.068

G3 0.046 0.055 0.049 0.046 0.047 0.042 0.054 0.025 0.046

HOF IP EA PA BDE NHTBH HTBH NCIE All Time

Methods

(223) (38) (25) (8) (92) (38) (38) (31) (493) C100H202 C100H100

DFT methods

SPL (LDA) 5.484 0.255 0.311 0.276 0.754 0.542 0.775 0.140 2.771

BLYP 0.412 0.200 0.105 0.080 0.292 0.376 0.337 0.063 0.322

PBE 0.987 0.161 0.102 0.072 0.177 0.371 0.413 0.052 0.562

TPSS 0.276 0.173 0.104 0.071 0.245 0.391 0.344 0.049 0.250

B3LYP 0.206 0.162 0.106 0.061 0.226 0.202 0.192 0.041 0.187 2.8 12.3

PBE0 0.300 0.165 0.128 0.057 0.155 0.154 0.193 0.031 0.213

M06-2X 0.127 0.130 0.103 0.092 0.069 0.056 0.055 0.013 0.096

XYG3 0.078 0.057 0.080 0.070 0.068 0.056 0.033 0.014 0.065 200.0 81.4

XYGJ-lOS 0.072 0.055 0.084 0.067 0.033 0.049 0.038 0.015 0.056 7.8 46.4

MC3BB 0.165 0.120 0.175 0.046 0.111 0.062 0.036 0.023 0.123

B2PLYP 0.201 0.109 0.090 0.067 0.124 0.090 0.078 0.023 0.143


HF 9.171 1.005 1.148 0.133 0.104 0.397 0.582 0.098 4.387

MP2 0.474 0.163 0.166 0.084 0.363 0.249 0.166 0.028 0.338

G2 0.082 0.042 0.057 0.058 0.078 0.042 0.054 0.025 0.068

G3 0.046 0.055 0.049 0.046 0.047 0.042 0.054 0.025 0.046

Lecture 1Ch121a-Goddard-L02 © copyright 2011 William A. Goddard III, all rights reserved\

Density Functional Theory errors kcal/mol)

170

LDA 130.88 15.2

Include density gradient (GGA)

BLYP 10.16 7.9

PW91 22.04 9.3

PBE 20.71 9.1

Hybrid: include HF exchange

B3LYP 6.08 4.5

PBE0 5.64 3.9

Include KE functional fit to barriers and complexes

M06-L 5.20 4.1

M06 3.37 2.2

M06-2X 2.26 1.3

atomize barrier

Popular with physicists

Popular with physicists

Popular with chemists

Include excitations to virtuals

XYGJ-OS 1.81 1.0

G3 (cc) 1.06 0.9

Accuracy needed for predictions

More rigorous foundation

No exact exchange, fast wag uses for catalysis

Does well not well founded

G3 CC (4 semiempirical parameters)

cannot be used for potential curves


0.01.02.03.04.05.06.07.08.09.0

10.0

B3LYP

M06

M06-2x

M06-L

B2PLYP

XYG3

XYG4-OS G2 G3

MAD (kcal/mol)

G2-1

G2-2

G3-3

Heats of formation (kcal/mol)

Large molecules

small molecules


0.0

5.0

10.0

15.0

20.0

25.0

B3LYP

BLYP PBE

LDA HF

MP2

QCISD(T)

XYG3

XYG4-OS

MAD (kcal/mol)

HAT12

NS16

UM10

HT38

Reaction barrier heights (kcal/mol)

Truhlar NHTBH38/04 set and HTBH38/04 set


0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

B3LYP

BLYP PBE

LDA HF MP2

QCISD(T)

XYG3

XYG4-OS

MAD (kcal/mol)

HB6

CT7

DI6

WI7

PPS5

Nonbonded interaction (kcal/mol)

Truhlar NCIE31/05 set


-5.00

0.00

5.00

10.00

15.00

20.00

25.00

30.00

-2.00 -1.50 -1.00 -0.50 0.00 0.50 1.00 1.50 2.00 2.50

Reaction coordinate

En

erg

y (

kca

l/m

ol)

HF

HF_PT2

XYG3

CCSD(T)

B3LYP

BLYP

SVWN

HF

HF_PT2 SVWN B3LYP

BLYP

XYG3 CCSD(T)

SVWN

H + CH4 H2 + CH3

Reaction Coordinate: R(CH)-R(HH) (in Å)

Energ

y (

kcal/m

ol)

Comparison of QM methods for reaction

surface of H + CH4 H2 + CH3


DFT-ℓg for accurate Dispersive Interactions for Full

Periodic Table

Hyungjun Kim, Jeong-Mo Choi, William A. Goddard, III 1Materials and Process Simulation Center, Caltech

2Center for Materials Simulations and Design, KAIST


Current challenge in DFT calculation for energetic materials

• Current implementations of DFT describe well strongly bound geometries and energies, but fail to describe the long range van der Waals (vdW) interactions.

• Get volumes ~ 10% too large

• XYGJ-lOS solves this problem but much slower than standard methods

• DFT-low gradient (DFT-lg) model accurate description of the long-range1/R6 attraction of the London dispersion but at same cost as standard DFT

Nlg,

lg 6 6,

-ij

ij i j ij eij

CE

r dR

=

DFT D DFT dispE E E =

C6 single parameter from QM-CC

d =1

Reik = Rei + Rek (UFF vdW radii)


PBE-lg for benzene dimer

T-shaped Sandwich Parallel-displaced

PBE-lg parameters

Nlg,

lg 6 6,

-ij

ij i j ij eij

CE

r dR

=

Clg-CC=586.8, Clg-HH=31.14, Clg-HH=8.691

RC = 1.925 (UFF), RH = 1.44 (UFF)

First-Principles-Based Dispersion Augmented Density Functional Theory: From

Molecules to Crystals’ Yi Liu and wag; J. Phys. Chem. Lett., 2010, 1 (17), pp

2550–2555


DFT-lg description for benzene

PBE-lg predicted the EOS of benzene crystal (orthorhombic phase I) in good agreement with

corrected experimental EOS at 0 K (dashed line).

Pressure at zero K geometry: PBE: 1.43 Gpa; PBE-lg: 0.11 Gpa

Zero pressure volume change: PBE: 35.0%; PBE-lg: 2.8%

Heat of sublimation at 0 K: Exp:11.295 kcal/mol; PBE: 0.913; PBE-lg: 6.762


DFT-lg description for graphite

graphite has AB stacking (also show AA eclipsed graphite)

Exper E

0.8, 1.0, 1.2

Exper c 6.556

PBE-lg

PBE

Bin

din

g e

ne

rgy (

kca

l/m

ol)

c lattice constant (A)


Universal PBE-ℓg Method

UFF, a Full Periodic Table Force Field for Molecular Mechanics and Molecular

Dynamics Simulations; A. K. Rappé, C. J. Casewit, K. S. Colwell, W. A. Goddard

III, and W. M. Skiff; J. Am. Chem. Soc. 114, 10024 (1992)

Derived C6/R6 parameters from scaled atomic polarizabilities for Z=1-103 (H-

Lr) and derived Dvdw from combining atomic IP and C6

Universal PBE-lg: use same Re, C6, and De as UFF, add a single new

parameter slg


blg Parameter Modifies Short-range Interactions

blg =1.0 blg =0.7

12-6 LJ potential (UFF parameter)

lg potential lg potential

When blg =0.6966, ELJ(r=1.1R0) = Elg(r=1.1R0)


DFT D DFT dispE E E =

Reik = Rei + Rek (UFF vdW radii)

Problem cannot yet do XYGJ-OS for crystals

Solution: use XYGJ-OS or CCSD to get accurate London Dispersion on

small vdW clusters.

Use to modify PBE for doing crystals by adding low gradient correction

(PBE-lg) (also B3LYP-lg) for accurate description of the long-range

1/R6 attraction of the London dispersion


Universal low gradient (ulg) method for DFT-ulg

Universal force field (UFF): Rappé, Goddard JACS 114, 10024 (1992)

Generic approach to force fields for whole periodic table (to Z=103 Lr)

For each atom: 6 rule based parameters 618 to describe all molecules

for all atoms up to Z=103

UFF has two vdw parameters: D0 and R0 per atom based on

• atomic polarizability from HF QM

• ionization potential from experiment

• atom size from experiment

Problem with DFT-lg: need a C6 parameter for every pair of atoms.

Can get from XYGJ-OS or CCSD calculation on small <100 atom

complexes, but for atoms up to Lr (Z=103) would need 5356 parameters,

far too tedious

ulg strategy: base C6 term in DFT-ulg on the C6 from UFF

wag962. Universal Correction of Density Functional Theory to Include London Dispersion

(up to Lr, Element 103); HJ Kim, JM Choi, wag; J. Phys. Chem. Lett. 2012, 3, 360−363


Universal low gradient (ulg) method

1. Match long R

2. Match at mid-range regime (r = 1.1R0):

3. Then introduce a single general scaling parameter for

whole periodic table (slg),

ulg method: use van der Waal’s parameters from Universal Force-Field

With 1 parameter, DFT-ulg

defined for Z=1 to 103

UFF vdw terms (up to Lr, Z=103) DFT-ulg


Determine the single parameter in DFT-ulg

from Benzene dimer interactions

J-M Choi, HJ Kim, WAG

DFT-ulg fit a single

parameter slg to benzene

dimer CCSD(T)

Get Slg = 0.7012


186

Validation: C6 parameters for lg fit to PBE for

benzene dimer does excellent job on crystals

Molecules PBE PBE-ℓg Exp.

Benzene 1.051 12.808 11.295

Naphthalene 2.723 20.755 20.095

Anthracene 4.308 28.356 27.042

Molecules PBE PBE-ℓg Exp.

Benzene 511.81 452.09 461.11

Naphthalene 380.23 344.41 338.79

Anthracene 515.49 451.55 451.59

Sublimation energy (kcal/mol/molecule)

Cell volume (angstrom3/cell) PBE-lg 0 to 2% too small,

thermal expansion

PBE-lg 3 to 5% too high

(zero point energy)


Crystals: Polyaromatic Hydrocarbons

V0 (Å3) Sublimation

E (kcal/mol)

Compres.

B0 (GPa)

PBE 511.8 1.05 1.3

PBE-ulg 452.1 12.81 8.8

PBE-Grimme 420.3 13.33 10

Exp. 461.8 11.3 ~8

Heat Vapor. PBE PBE-ulg Exp.

Naphthalene 0.89 18.93 18.4-23.5

Anthracene 1.75 25.80 24.6-30.0

Phenantracene 1.52 24.39 23.6-26.5

Benzene crystal:

Volume PBE PBE-ulg Exp.

Naphthalene 380.2 344.4 342.3

Anthracene 515.5 451.6 455.2

Phenantracene 524.5 461.7 459.5


Equation of States of Benzene Crystal

PBE-ulg predicts the

correct cold-

compression curve.


Hobza S22 database

Twenty-two prototypical small molecular complexes for non-covalent

interactions in biological molecules (h-bonded, dispersion dominated,

and mixed)

7 hydrogen bonded Mean average error (MAE)

PBE-ulg: 0.53 kcal/mol

PBE-Grimme: 1.01 kcal/mol

vdw-DF: 0.59 kcal/mol (lundqvist, PRL 2004)

8 dispersion dominated MAE



vdw-DF: 1.86 kcal/mol Overall:



vdw-DF: 1.20 kcal/mol

XYGJ-OS: 0.46 kcal/mol

Mean average error (MAE)



vdw-DF: 1.06 kcal/mol


Big Challenge for DFT

190

Proper description of spin states

Organometallic reaction barriers depend strongly on spin

Antiferromagnets

Cuprate superconductors

Ground states of Mn, Fe, Co, Ni metals

Current optimization of DFT methods focus mainly on 1st

and 2nd row compounds (H-Ar) but applications involve

transition metals, lanthanides, actinides where local d and f

orbitals can lead to magnetically complex systems

Example of the challenge:

Group 10:

s2d8 (3F) vs. s1d9 (3D) vs. s0d10 (1S)


Ground state configurations for group 10

Ni Pd Pt

wag206-Theoretical Studies of Oxidative Addition and Reductive Elimination. II.

Reductive Coupling of H-H, H-C, and C-C Bonds from Pd and Pt Complexes

J. J. Low and W. A. Goddard III; Organometallics 5, 609 (1986)

Exper GVB-CI HF Exper GVB-CI HF Exper GVB-CI HF


Ab initio methods Ni atom (all electron)

192

method s1d9 (3D) s0d10 (1s) Ni atom

exper -0.69 39.43

HF(wag 1986) 15.30 114.80

HF(G3 basis, Yu 2012) 1.72 55.17

HF (numerical nonrelativistic) 29.29 126.14 Cowan-Griffin

HF (numerical relativistic) 37.59 139.29 Cowan-Griffin

GVB-CI (wag 1986) -14.20 26.20

MP2(G3 basis, Yu 2012) -30.92 -44.60 using s2d8 state

CCSD(G3 basis, Yu 2012) not conv 47.07 using s2d8 state

Basis set issues

Basis set issues:

G3 basis: contraction of 3s and 3p core functions (overlaps 3d)

Reference state issues for MP2 and CCSD: used s2d8


Compare DFT methods Ni atom (all electron)

193

method s1d9 (3D) s0d10 (1s)

exper -0.69 39.43

HF(G3 basis, Yu 2012) 1.72 55.17

GVB-CI (wag 1986) -14.20 26.20

PBE(G3 basis, Yu 2012) -12.30 29.37

PBE0(G3 basis, Yu 2012) -9.18 85.06

B3LYP(G3 basis, Yu 2012) -9.11 22.65

M06-L(G3 basis, Yu 2012) 36.92 -51.31

M06(G3 basis, Yu 2012) -10.33 19.43

M06-HF(G3 basis, Yu 2012) -14.01 49.11

M06-2X(G3 basis, Yu 2012) -3.59 48.36

XYGJ-OS (G3 basis, Yu 2012) 0.03 -2.12

Highlight

s1d9 < 3 kcal/mol

s0d10 <10 kcal/mol

Need to use multiple spin states in DFT optimization


Pt atom using LANL Core Effective Potential

194

method s0d10 (1s) s2d8 (3F) Pt atom

exper 11.07 14.76

HF(wag 1986) 31.40 8.40

HF(Yu 2012) 25.41 5.95

HF (numerical nonrelativistic) -32.52 75.64 Cowan-Griffin


GVB-CI (wag 1986) 12.20 14.20

PBE(Yu 2012) 14.39 -0.05

PBE0(Yu 2012) 15.03 9.08

B3LYP(Yu 2012) 14.67 6.84

M06-L(Yu 2012) 14.01 0.86

M06(Yu 2012) 0.40 19.17

M06-HF(Yu 2012) 24.95 21.07

M06-2X(Yu 2012) 11.72 15.24

XYGJ-OS na na Yu basis: LACV3P**++f,

Highlight

s2d8 < 2 kcal/mol

s0d10 <2 kcal/mol

Ground state s1d9 (3D)


method s1d9 (3D) s2d8 (3F) Pd atom

exper 21.91 77.94

HF(wag 1986) -12.70 41.80

HF(Yu 2012) 1.72 55.17

HF (numerical nonrelativistic) -17.29 86.71 Cowan-Griffin


GVB-CI (wag 1986) 19.60 82.20

MP2 13.41 5.43

CCSD 16.83 0.92

PBE( Yu 2012) 9.43 67.17

PBE0(Yu 2012) 19.99 88.40

B3LYP(Yu 2012) 19.98 84.79

M06-L(Yu 2012) 30.66 97.18

M06( Yu 2012) 38.63 114.45

M06-HF(Yu 2012) 10.09 89.50

M06-2X(Yu 2012) 26.07 97.61

XYGJ-OS na na

Yu basis:LACVP in Qchem

Pd atom

using LANL

Core

Effective

Potential

Highlight

s1d9 < 3 kcal/mol

s2d8 <5 kcal/mol


States for Ni d8 atom (real orbitals)

196

Hole type exp HF PBE PBE0 B3LYP M06-L M06 M06-HF M06-2X

z2, x2-y2 σδ 0 0 0 0 0 0 0 0 0

xy, z2 σδ 0 0 0.08 0.06 0.05 -0.42 -0.27 -0.53 -0.83

xz, yz ππ 9.004 11.3 10.2 10.4 10.25 16.22 10.4 -7.93 4.13

xz, x2-y2 πδ 9.004 11.6 9.39 9.79 9.8 13.17 7.5 0.52 6.05

yz, x2-y2 πδ 9.004 11.6 9.39 9.79 9.8 13.17 7.5 0.52 6.05

xz, xy πδ 9.004 11.6 10.4 10.5 10.42 16.38 10.5 -7.7 4.29

yz, xy πδ 9.004 11.6 10.4 10.5 10.42 16.38 10.5 -7.7 4.29

xz, z2 σπ 27.01 34.3 26.6 28.2 28.33 35.85 18.1 10.28 19.9

yz, z2 σπ 27.01 34.3 26.6 28.2 28.33 35.85 18.1 10.28 19.9

xy, x2-y2 δδ 36.02 45.4 36.7 38.3 38.44 48.36 23.8 14.84 26.92

ok exc exc exc ok

Bottom line: for transition metal systems, current levels of DFT

based on foundation of sand: must address in next generation DFT


Parameter Optimization

Implemented in VASP 5.2.11

0.7012

0.6966

Chem 121 - Applied Quantum Chemistry

Lecture 1Lecture 2 198

Method:

· Semi-Empirical, used for very big systems, or for rough approximations of

geometry (extended Huckel theory, CNDO/INDO, AM1, MNDO)

· HF (Hartree Fock). Simplest Ab Initio method. Very cheap, fairly inaccurate

· MP2 (Moeller-Plasset 2). Advanced version of HF. Usually not as cheap or as

accurate as B3LYP, but can function as a complement.

· CASSCF (Complete Active Space, Self Consisting Field). Advanced version

of HF, incorporating excited states. Mainly used for jobs where photochemistry is

important. Medium cost, Medium Accuracy. Quite complicated to run…

· QCISD (Quadratic Configuration Interaction Singles Doubles). Very

advanced version of HF. Very Expensive, Very accurate. Can only be used on

systems smaller than 10 heavy atoms.

· CCSD (Coupled Cluster Singles Doubles). Very much like QCISD.

Density Functional Theory

LDA (local density approximation)

PW91, PBE

· B3LYP (density functional theory). Cheap, Accurate.

Generally, B3LYP is the method of choice. If the system allows it, QCISD or CCSD

can be used. HF and/or MP2 can be used to verify the B3LYP results.



Basis Set: What mathematical expressions are used to describe orbitals. In

general, the more advanced the mathematical expression, the more accurate

the wavefunction, but also more expensive calculation.

· STO-3G - The ‘minimal basis set’. Not particularly accurate, but cheap and

robust.

· 3-21G - Smallest practical Basis Set.

· 6-31G - More advanced, i.e. more functions for both core and valence.

· 6-31G** - As above, but with ‘polarized functions’ added. Essentially

makes the orbitals look more like ‘real’ ones. This is the standard basis set

used, as it gives fairly good results with low cost.

· 6-31++G - As above, but with ‘diffuse functions’ added. Makes the orbitals

stretch out in space. Important to add if there is hydrogen bonding, pi-pi

interactions, anions etc present.

· 6-311++G** - As above, with even more functions added on… The more

stuff, the more accurate… But also more expensive. Seldom used, as the

increase in accuracy usually is very small, while the cost increases drastically.

· Frozen Core: Basis sets used for higher row elements, where all the core

electrons are treated as one big frozen chunk. Only the valence electrons are

treated explicitly



• Software packages

– Jaguar

– GAMESS

– TurboMol

– Gaussian

– Spartan/Titan

– HyperChem

– ADF



Running an actual calculation

– Determine the starting geometry of the

molecule you wish to study

– Determine what you’d like to find out

– Determine what methods are suitable and/or

affordable for the above calculation

– Prepare input file

– Run job

– Evaluate result



Example: Good ol’ water

Starting geometry: water is bent, (~104º), a normal

O-H bond is ~0.96 Å. For illustration, however, we’ll

start with a pretty bad guess.

Simple Z-matrix:

O1

H2 O1 1.00

H3 O1 1.00 H2 110.00

1.00 Å 1.00 Å

110º



What do we wish to find out?

How about the IR spectra?

What is a suitable method for this calculation?

Well, any, really, since it is so small. But 99% of the

time the answer to this question is “B3LYP/6-

31G**” – a variant of density functional theory that

is the main workhorse of applied quantum

chemistry, with a standard basis set. Let’s go with

that.



Actual jaguar input:

&gen

igeopt=1

ifreq=1

dftname=b3lyp

basis=6-31g**

&

&zmat

O1

H2 O1 0.95

H3 O1 0.95 H2 120.00

&



Running time!

Jaguar calculates the wave function for the

atomic coordinates we provided

From the wave function it determines the

energy and the forces on the current geometry

Based on this, it determines in what direction it

should move the atoms to reach a better

geometry, i.e. a geometry with a lower energy



1.00 Å 1.00 Å

110º

0.96 Å 0.96 Å

104º

Our horrible guess Target geometry

Think elastic springs:

The bonds are too long,

so there will be a force

towards shorter bonds

Forces



Optimization –

minimization of the

forces. When all forces

are zero the energy will

not change and we

have the resting

geometry

O1

H2 O1 0.9500000000

H3 O1 0.9500000000 H2 120.0000000000

SCF energy: -76.41367730925

--

O1

H2 O1 0.9566666804

H3 O1 0.9566666820 H2 106.8986301461

SCF energy: -76.41937497895

--

O1

H2 O1 0.9653619358

H3 O1 0.9653619375 H2 103.0739287925

SCF energy: -76.41969584939

--

O1

H2 O1 0.9653155294

H3 O1 0.9653155310 H2 103.6688074046

SCF energy: -76.41970381840

--



0.9653155294 Å

103.6688074046º

Computer accuracy

0.96 Å 0.96 Å

103.7º

“actual” accuracy

Accuracy

0.9653155294 Å

Accuracy is a relative concept



frequencies 1666.01 3801.19 3912.97

No negative frequencies!

(Compare IR spectra for gas-phase water)



Vibrational levels

“zero” level

Zero Point Energy (ZPE)

Zero Point Energies

Optimized energy is at the zero level, but in reality the molecule has a higher

energy due to populated vibrational levels.

At 0 K, all molecules populate the lowest vibrational level, and so the

difference between the “zero” level and the first vibrational level is the Zero

Point Energy (ZPE)

From our calculation:

The zero point energy (ZPE): 13.410 kcal/mol



Thermodynamic data at higher temperatures

T = 298.15 K

U Cv S H G

--------- --------- --------- --------- ---------

trans. 0.889 2.981 34.609 1.481 -8.837

rot. 0.889 2.981 10.503 0.889 -2.243

vib. 0.002 0.041 0.006 0.002 0.000

elec. 0.000 0.000 0.000 0.000 0.000

total 1.779 6.003 45.117 2.371 -11.080

Most thermodynamic data can be computed with very good

accuracy in the gas phase. Temperature dependant



Transition states

Reactant Product

Transition State (TS)

CH3Br + Cl- CH3Cl + Br- TS

Reaction coordinate

Line represents the

reacting coordinate, in this

case the forming C-Cl and

breaking C-Br bonds

Stationary points:

points on the surface

where the derivative

of the energy = 0



CH3Br + Cl- CH3Cl + Br- TS

Reaction coordinate

Not a hill, but a

mountain pass

Transition state =

stationary point where all forces

except one is at a minimum.

The exception is at its maximum



Reactant Product

TS

Derivative of the energy = 0

Second derivative:

For a minimum > 0

For a maximum < 0

So a TS should have a

negative second derivative of

the energy

Second derivative of the

energy = force



A transition state should have one

negative (imaginary) frequency!!!

(and ONLY one)



Reactant Product

TS

Optimizing transition states:

Simultaneously optimize all

modes (forces) towards their

minimum, except the reacting

mode

But for the computer to know

which mode is the reacting

mode, you must have one

imaginary frequency in your

starting point

Inflection points

Region with

imaginary frequency

Must start with a good guess!!!



Example:

CH3Br + Cl- CH3Cl + Br-

What do we know about this reaction? It’s an SN2

reaction, so the Cl- must come in from the backside of

the CH3Br. The C-Cl forms at the same time as the C-

Br forms. The transition state should be five

coordinate



2.0 2.2 Cl

Br

H H

H

C

Initial guess: C-Cl = 2.0 Å, C-Br = 2.2 Å

Single point frequency on the above geometry:

frequencies 98.64 99.58 109.11 310.66 1339.10 1348.64

frequencies 1349.46 1428.45 1428.73 2838.52 3017.70 3017.93

No negative frequencies! Bad initial guess



Refinement :

Initial guess most likely wrong because of erronous C-

Br and C-Cl bond lengths

Let the computer optimize the five-coordinate structure

Frozen optimizations:

Just like a normal optimization, but with one or more

geometry parameters frozen

In this case, we optimize the structure with all the H-C-

Cl angles frozen at 90º



Result:

2.32 2.62 Cl

Br

C-Cl and C-Br bonds quite a bit longer in the new structure

Frequency calculation: frequencies -286.26 168.54 173.32 173.43 874.16 874.76

frequencies 976.23 1413.99 1414.65 3220.91 3420.84 3421.80

One negative frequency! Good initial guess



Time for the actual optimization:

Jaguar follows the negative frequency towards the maximum

Geometry optimization 1: SCF Energy = -513.35042353681






Final energy higher than starting energy (although only 0.5 kcal/mol)

Frequency calculation

frequencies -268.67 162.64 174.22 174.31 848.15 848.24

frequencies 960.97 1415.75 1415.96 3220.77 3420.80 3421.15

One negative frequency! We found a true transition state



2.46 2.51 Cl

Br

Final geometry: C-Cl = 2.46 Å

C-Br = 2.51 Å

Cl-C-H = 88.7º

Br-C-H = 91.3º

Structure not quite symmetric, the

hydrogens are bending a little bit away

from the Br.



Solvation calculations

Explicit solvents:

Calculations where solvent molecules

are added as part of the calculation

Implicit solvents:

Calculations where solvation effects

are added as electrostatic interactions

between the molecule and a virtual

continuum of “solvent”.



Reaction energetics and barrier heights

Collect the absolute energies of the reactants, products and transition states

CH3Br + Cl- TS CH3Cl + Br- -53.078938 + -460.248741 -513.349681 -500.108371 + -13.237607

Sum each term

CH3Br + Cl- TS CH3Cl + Br- -513.327679 -513.349681 -513.345978

Define reactants as “0”, and deduct the reactant energy from all terms

CH3Br + Cl- TS CH3Cl + Br- 0 -.022002 -.018299

Convert to kcal/mol (1 hartree = 627.51 kcal/mol)




Convert to kcal/mol (1 hartree = 627.51 kcal/mol)

CH3Br + Cl- TS CH3Cl + Br- 0 -13.8 -11.5

But this doesn’t make sense




CH3Br + Cl- TS CH3Cl + Br- 0 -13.8 -11.5

Solvation not included!

Include solvation corrections!

CH3Br + Cl- TS CH3Cl + Br- 0 9.2 -6.4

Documents

Lecture 2, April, 2012 QM-1: HF2.pdf · 2 1 2 ¦ ¦ 1 ¦ 2 ¦ ¦ ¦ A Aii j ij A B AB B A i R r Z Z Z M 1 2 1 1 2 H elΨ=EΨ For benzene we have 12 nuclear degrees of freedom (dof)