lecture 1, january 4, 2016 quantum mechanics-1:...

Lecture 1-Ch121a-Goddard-L01 © copyright 2016 William A. Goddard III, all rights reserved\ 1

Ch121a Atomic Level Simulations of Materials and

Molecules

William A. Goddard III, [email protected]

316 Beckman Institute, x3093

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

California Institute of Technology

Lecture 1, January 4, 2016

Quantum Mechanics-1: HF

Special Instructor: Julius Su <[email protected]>

Teaching Assistants:

Daniel Brooks [email protected]

Jin Qian [email protected]

Room BI 115

Lecture: Monday, Wednesday 2-3pm

Lab Session: Friday 2-3pm

mailto:[email protected]

mailto:[email protected]


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.

3

CATALYSTS ALKANE OXIDATION, AMMOXIDATION : VOPO, MoVNbTeOx

CATALYSTS for METHANE TO LIQUID : Ir, Os, Rh, Ru organometallic (220C)

ARTIFICIAL PHOTOSYNTHESIS (JCAP) H2O +h H2+O2, CO2 +h fuels

FUEL CELL CATALYST: Oxygen Reduction Reaction (Pt alloy, nonPGM)

BATTERIES: organic cathode (green), Li air-CO2.

BIOTECHNOLGY: GPCR Membrane Proteins, Pharma, Novel Amino Acids

NANOSYSTEMS: Nanomanufacturing, DNA based assembly

SEMICONDUCTORS: damage free etching

THERMOELECTRICS: (high ZT at low T)

SOLAR ENERGY: dye sensitized solar cells, CuInGaSe (CIGS/CdS) cells

GAS STORAGE (H2, CH4, CO2) : MOFs, COFs, metal alloys, nanoclusters

CERAMICS: FC electrodes, membranes, Ferroelectrics, Superconductors

POLYMERS: Higher Temperature Fuel Cell PEM (Replace Nafion)

WATER: Captymers for Selective removal metals and anions

ENERGETIC MATERIALS: PETN, RDX, HMX, TATB, TATP, Propellants

We develop methods and software simultaneously with

Applications to most challenging problems. Goddard Focus

MultiParadigm Strategy enables application of 1st principles

to complex systems

Lecture 1-Ch121a-Goddard-L01 © copyright 2016 William A. Goddard III, all rights reserved\

Motivation: Design Materials, Catalysts, Pharma from 1st

Principles so can do design prior to experiment

4

Big breakthrough making FC simulations

practical:

reactive force fields based on QMDescribes: 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

Ch121a

5

1:Quantum Mechanics

Challenge: increased accuracy

• New Functionals DFT (dispersion)

• Quantum Monte Carlo methods

• Tunneling thru molecules (I/V)

2:Force Fields

Challenge: chemical reactions

• ReaxFF- Describe Chemical

Reaction processes, Phase

Transitions, for Mixed Metal,

Ceramic, Polymer systems

• Electron Force Field (eFF)

describe plasma processing

3:Molecular Dynamics

Challenge: Extract properties

essential to materials design

• Non-Equilibrium Dynamics

– Viscosity, rheology

– Thermal Conductivity

• Solvation Forces (continuum Solv)

– surface tension, contact angles

• Hybrid QM/MD

• Plasticity, Dislocations, Crack

• Interfacial Energies

• Reaction Kinetics

• Entropies, Free energies

4:Biological Predictions

1st principles structures GPCRs

1st principles Ligand Binding

5:MesoScale Dynamics

Coarse Grained FF

Hybrid MD and Meso Dynamics

6: Integration: Computational

Materials Design Facility (CMDF)

•Seamless across the hierarchies of

simulations using Python-based scripts

Materials Design Requires Improvements in Methods

to Achieve Required Accuracy. Our Focus:

Essential to develop new methods for theory to

help solve problems in energy and environment


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 and Research Project

First 5 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 Ch121a to solve a research problem that can be

completed in the final 5 weeks.

The homework for each of the last 5 weeks is to turn in a one

page report on progress with the project

The final is a research report ~ 5 page 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

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


The stratospheric review of QM

10

You should have already been exposed to much of

this material

This overview to remind you of the key points

Overview of Quantum Mechanics, Hydrogen Atom, etc

Please review again to make sure that you are comfortable

with the concepts, which you should have seen before


Classical Mechanics

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

attractionElectron-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

Not consistent with real world. Solution? Quantum mechanics

Ab Initio, quantum mechanics

Quantum mechanics

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

Schrodinger Equation

Too complicated to solve

exactly. What do we do?


Ignore electron-electron interactions

Independent Particle Approximation

14

Prob(1,2,3,4, ..N-1,N) =|Ψ|2= Ψ*Ψ

Ψ*(1,2,3,4, ..N-1,N)Ψ(1,2,3,4, ..N-1,N)

= ψa*(1) ψb

*(2) ψc*(3) ---ψN

*(N)ψa(1) ψb(2) ψc(3) ---ψN(N)

=|ψa*(1) ψa(1)| |ψb

*(2) ψb(2)| ----

=P(1) P(2)---

Solve N different 1-electron problems: h(1) ψa(1) = ea ψa(1)

Total wavefunction is the product of 1-e orbitals

Ψ(1,2,3,4, ..N-1,N) =ψa(1) ψb(2) ψc(3) ----ψN(N)

With the product wavefunction the

probability of finding e1 at position (x1,y1,z1)

is independent of the probability of finding e2 at position (x2,y2,z2)

The electrons are independent of each other

(no correlation of their motions)

This wavefunction is called the Hartree approximation


Simplest wave function satisfying PP for 2 electrons:

Ψ(1,2) = ψa(1) ψb(2) - ψb(1)ψa(2)

Thus Ψ(1,2) = ψb(1) ψa(2) - ψa(1)ψb(2) = -[ψa(1) ψb(2) - ψb(1)ψa(2)]

We write this as a Slater Determinant

Ψ(1,2) = ψb(1) ψa(2) - ψa(1)ψb(2)= =

Pauli Principle

15

Ψ(1,2,4,3, ..N-1,N) = - Ψ(1,2,3,4, ..N-1,N)

Interchanging any two electrons changes the sign of the total

wavefunction

Hartree wavefunction not satisfy PP

2 electrons: ΨH(2,1) = ψa(2) ψb(1) = ψb(1)ψa(2) ǂ -ψa(1) ψb(2)

3 electrons


General case – independent electrons –

Slater determinant

16

Ψ(1,2,3,4, ..N-1,N) = A[ψa(1) ψb(2) ψc(3) ----ψN(N)]

Properties of determinants:

Get zero if any two rows or columns are identical (simple Pauli

exclusion principle from old QM)

Interchange any two rows or columns changes the sign

Every row or column can be taken as orthogonal to every other

row or column

Adding some amount of one column to any other column leaves

determinant unchanged

Even if they do not interact, PP requires a determinant

wavefunction

Consider the product wavefunction

Ψ(1,2) = ψa(1) ψb(2)

And the Hamiltonian

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>

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

Represent 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 Hartree Product Wavefunction

consider 2 electron case

Very simple: the

one-electron

energy for ψa(1)

and for ψa(2) plus

the e-e repulsion

between them (and

the 0 electron

nuclear energy

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 terms

Energy for Hartree wavefunction N electrons

20

ΨH(1,2,3,4, ..N-1,N) =ψa(1) ψb(2) ψc(3) ----ψN(N)

EH= Sa=1,N haa + Sa<b=1,N Jab + 1/R

N one-electron terms, N*(N-1)/2 Coulomb terms

The 2 electron exchange term

22

The 4th term leads to-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 sign of the exchange energy

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

Discussion of 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



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

Thus we can write the energy as

E = Sa=1,2 haa + <ψa(1)| [ Jb (1)-Kb (1)] |ψa(1)>+ 1/R

Total energy for N electron systems

26

Thus the total energy for the 4e wavefunction Aψaψbψcψd is

E = Sj=1-4 hjj + Sj<k=1-4 (Jjk –Kjk) + 1/R

Here there are 4*3/2=6 electron repulsion terms

Now we can generalize to the case of N electrons

Aψaψbψc---ψN

E = Sj=1-N hjj + Sj<k=1-N (Jjk –Kjk) + 1/R

Note we can add the self-term with j=k, since Jjj –Kjj =0

Thus the energy can be written as

E = Sj=1-N hjj + (1/2) Sj,k=1-N (Jjk –Kjk) + 1/R

Here we can write the number Jjk as

Jjk = <ψj (1)| Jk(1)|ψj(1)> where Jk(1) = ʃ [ψk(2)]2/r12 is the

Kjk = <ψj(1)|Kk(1)|ψj(1)> so that the E becomes

E = Sj=1-N <ψj |h|ψj> + (1/2)Sj,k=1-N <ψj| Jk-Kk|ψj> + 1/R

This would appear to have N2/2 two-e interactions, but N/2 are

zero, so we still get N(N-1/2 two-e terms

Lecture 1-Ch121a-Goddard-L01 © copyright 2016 William A. Goddard III, all rights reserved\ Ch120a-

Goddard-

27

Fundametal Postulate of QM, the variational principle

To find the best Φap we change parameters

in Φap until we get the minimum energy.

For this best approximate wavefunction

Φbest it must be that dEap = 0 for all possible

changes, dΦ.

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

possible energy out of all possible wavefunctions.

For the ground state,

the curvature d2E/dΦ

must be positive for all

possible changes,

d2E/dΦ ≥ 0

EexEap

E

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> ≥ Eexact

Thus the condition for the best Φap is

dEap/dΦ = 0 for all dΦ from Φap

This is called the variational principle.

The Hartree-Fock equations (for spinorbitals)

Find the best ψj for a Slater Determinant productif we change ψj to ψj + dψj the change in the energy is

dE=<dψj |h|ψj>+<ψj |h|dψj>+(1/2)[<dψj| Jj-Kj|ψj>+<ψj| Jj-Kj|dψj>]

S k=1-N,k≠j [<dψj| Jk-Kk|ψj> + <ψj| Jk-Kk|dψj>]= <dψj |Hj

HF|ψj> + <ψj |HjHF|dψj> = 2Re <dψj |Hj

HF|ψj>

where HjHF = h + Σk=1-N [Jk-Kk] is called the HF Hamiltonian

In the above expression we assume that φ1 was normalized,

<φ1|h|φ1> = 1.

Including this explicitly, we have

dE = <ψj |HjHF|ψj>/[<dψj |ψj> + <ψj |dψj>]

=-{<ψj |HjHF|ψj>/[<ψj |ψj>]2}{<dψj |ψj> + <ψj |dψj>]=

= – ej{<dψj |ψj> + <ψj |dψj>]

Where ej ={<ψj |HjHF|ψj>

Thus the total dE is 2Re<dψj |HjHF – ej|ψj> = 0 for all possible dψj

Thus the coefficient of dψj must be zero

Hence the optimum orbitals, the HF orbitals satisfy HjHFψj = ej ψj

which is like a one-electron Schrodinger Eqn.


The canonical HF equations

29

HHFψj = ej ψj where j=1..N

where HHF = h + Σk=1-N [Jk-Kk]

Thus each HF spinorbital is the optimum state for this

electron moveing in the field of the nuclei plus the

average interaction with the other N-1 electrons

Thus these equations must be solved self-consistently.

Now the operator seems to have N Coulomb and N

exchange terms whereas it should be N-1 terms

This is ok because

(Jj-Kj)ψj = 0

So that the self term cancels.


The closed shell Hartree Fock Equations

30

General concept: there are an infinite number of possible orbitals

for the electrons. For a system with 2M electrons we will put the

electrons into the M lowest orbitals, with two electrons in each

orbital (one up or a spin, the other down or b spin)

M occ orb

2M elect

In the above derivation we allowed each

electron to have a different spinorbital but

for the ground state of normal molecules we

can replace each spinorbital with a product

of a spatial function and a spin function

φaa or φab

Often the ground state is a close shell wavefunction in which

there are an even number of electrons, N=2M

each occupied orbital has two electrons, one with up or a spin

and the other with down or b spin

The closed shell wavefunction is written as

Ψ(1,2,3,4, ..N-1,N) = A[(φaa)(φab)(φba)(φbb)---------(φza)(φzb)]

Where the A is the antisymmetrizer or determinant operator

where the 1st column is φaa(1), φaa(2), φaa(3), etc

The 2nd column is φab(1), φab(2), φab(3), etc

Thus there are N! terms

This guarantees that the wavefunction changes sign if any 2

electrons are interchanged (Pauli Principle)

Properties of determinant: if two columns are identical get zero.

Thus can never have 2 electrons in same orbital with same spin

Can take every column to be orthogonal; thus <φa|φb>=0

Also can recombine any two orbitals and the wavefunction does

a = (cos) φa + (sin) φb not change

b = (-sin) φa + (cos) φb

The Hartree Fock Equations

Closed shell

31

M occ orb

N=2M elect

1 2 N

ab

z

The energy (closed shell)

32

Hel (1,2,---N) = Si h(i) + Si<j 1/rij

where h(i) = - ½ 2 + Si ZA/Rai is the interaction of all nuclei A

with electron 1 plus the kinetic energy, a total of N terms

and the other term is the Coulomb interaction between each pair

of electrons, a total of N(N-1)/2 terms

If we ignore the antisymmetrizer, so that the wavefunction is a

Hartree product

Ψ(1,2,3,4, ..N-1,N) = [(φaa)(φab)(φba)(φbb)---------(φza)(φzb)]

Then the energy is

Eproduct = Sa 2<a|h|a> + Sa Jaa + Sa<b 4Jab

Thus N=2M 1e terms and

M+2M(M-1)=M(2M-2+1)= N(N-1)/2 2e terms

The electronic Hamiltonian is

M=N/2 terms M=N/2 terms M(M-1)/2 terms

The Coulomb energy

33

Jab= <Φa(1)Φb(2) |1/r12 |Φa(1)Φb(2)>

Jab = ʃ1,2 [a(1)]2 [b(2)]2/r12

Jab =ʃ1 [a(1)]2 Jb(1)

where Jb(1) = ʃ [b(2)]2/r12 is the coulomb potential evaluated at

point 1 due to the charge density [b(2)]2 integrated over all

space

Thus Jab is the total Coulomb interaction between the electron

density ra(1)=|a(1)|2 and rb(2)=|b(2)|2

Consider the effect of the Antisymmetrizer

34

Two electrons with same spin

Ψ(1,2)) = A[(φaa)(φba)]= (φaa)(φba) - (φba)(φaa)

1 2 1 2

New term in energy is the exchange term

-<(φaa)(φba)|Hel(1,2)|(φba)(φaa)> is a sum of 3 terms

1. <(φaa)(φba)|h(1)|(φba)(φ1a)> = <φaa|h(1)|φba><φba|φaa>

2. <(φaa)(φba)|h(2)|(φba)(φ1a)>=<φaa|φba><φba|h(2)|φaa>

3. <(φaa)(φba)|1/r12|(φba)(φ1a)>=Kab

Thus the only new term is -Kab note that it is negative because

one side is exchanged but not the other

Thus the total energy becomes

E = <a|h|a> + <b|h|b> + Jab – Kab

0

0

The Exchange energy

35

Kab= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)>

Kab = ʃ1 [a(1)b(1)] ʃ2 [b(2)a(2)]/r12

Kab = ʃ1 [a(1) {P12 b(2)] ʃ2[b(2)]/r12 } a(1)] = ʃ1 [a(1) Kb(1) a(1)]

No simple classical interpretation, but we have written it in terms of

an integral operator Kb(1) so that is looks similar to the Coulomb

case

Relationship between Jab and Kab

36


wavefunction Ψ(1,2) = A[(φaa)(φba)] must be positive

Eee =∫ (d3r1)((d3r2)|Ψ(1,2)|2/r12 > 0


and r2 then

Thus Jab – Kab > 0 and hence Jab > Kab > 0

Thus the exchange energy is positive but smaller than the

Coulomb energy

Note that Kaa = <Φa(1)Φa(2) |1/r12 |Φa(1)Φa(2)> = Jaa


Interpret the Exchange term

37

For the gu excited states H2

we found two states:3S = [φgφu - φuφg][abba] 1S = [φgφu + φuφg][abba]

The energy of these two

wavefunctions is 3E=Jgu – Kgu1E=Jgu + Kgu

In 2-e space this leads to

electron correlation as

The biggest contribution is

when both electrons are at

the same spot, say z1=z2

Which is along the upper

diagonal.

2Kgu is just the difference

of these 2 cases

E = Sa 2<a|h|a> + Sa [2Jaa – Kaa] + Sa + Sa,b (2Jab – Kab)

There are M2 terms, so it appears that the number of Jij is

2M2 = 2(N/2)(N/2) = N2/2 terms,

but we should have N(N-1)/2 = N2 –N/2

This is because we added N/2 fake terms, Jaa that must be

cancelled by the N/2 fake Kaa terms.

Also note Sa,b 2Jab = (½)ʃ1,2 [r(1)] [r(2)]2/r12

where r(1)= Sa [Φa(1)]2 is the total electron density, the classical

electrostatic energy for this charge density. This includes Jii

The final energy for closed shell wavefunction

38

The total energy is

E = Sa 2<a|h|a> + Sa Jaa + Sa<b (4Jab – 2Kab)

One from aa and one from bb

[2Jaa – Kaa]

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


The Hartree Fock Equations

40

Variational principle: Require that each orbital be the best

possible (leading to the lowest energy) leads to

HHF(1)φa(1)= ea φa(1)

where we solve for the occupied orbital, φa, to be occupied by

both electron 1 and electron 2

Here HHF(1)= h(1) + Sb [2Jb(1) - Kb(1)]

This looks like the Hamiltonian for a one-electron system in which

the Hamiltonian has the form it would have for the average

potential due the electron in all other orbital orbitals

Thus the two-electron problem is factored into M=N/2 one-

electron problems, which we can solve to get φa, φb, etc


Self consistency

41

However to solve for φa(1) we need to know Sb [2Jb(1) - Kb(1)]

which depends on all M orbitals

Thus the HHFφa= ea φa equation must be solved iteratively until

it is self consistent

But after the equations are solved self consistently, we can

consider each orbital as the optimum orbital moving in the

average field of all the other electrons

In fact the motions between these electrons would tend to be

correlated so that the electrons remain farther apart than in this

average field

Thus the error in the HF energy is called the correlation

energy


He atom one slater orbital

42

If one approximates each orbital as φ1s = N0 exp(-zr) a Slater orbital then it is

only necessary to optimize the scale parameter z

In this case

He atom: EHe = 2(½ z2) – 2Zz (5/8)z

Applying the variational principle, the optimum z must satisfy

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 Thus this simple

approximation of assuming that each electron is in a H1s orbital

and optimizing the size accounts for 98.1% of the exact result.

The Koopmans orbital energy

43

The next question is the meaning of the one-electron energy, ea

in the HF equation

HHF(1)φa(1)= ea φa(1)

Multiplying each side by φa(1) and integrating leads to

ea <a|a> = <a|HHF|a> = <a|h|a> + 2Sb<a|Jb|a> - Sb<a|Kb|a>

= <a|h|a> + Jaa + Sb≠a<a|2Jb-Kb|a>

Thus in the approximation that the remaining electron does not

change shape,

ea corresponds to the energy to ionize an electron from the a

orbital to obtain the N-1 electron system

Sometimes this is referred to as the Koopmans theorem

(pronounced with a long o).

It is not really Koopmans theorem, which we will discuss later,

but we will use the term anyway


The ionization potential

44

There are two errors in using the ea to approximate the IP

IPKT ~ - ea

First the remaining N-1 electrons should be allowed to relax to the optimum

orbital of the positive ion, which would make the Koopmans IP too large

However the energy of the HF description is leads to a total energy less

negative than the exact energy,

Exact = EHF – Ecorr

Where Ecorr is called the electron correlation energy (since HF does NOT

allow correlation of the electron motions. Each electron sees the average

potential of the other)

which would make the Koopmans IP too small

These effects tend to cancel so that the ea from the HF wavefunction leads

to a reasonable estimate of IP

(N-1)e

exactHF from Ne

Ne

exactHFexact IP

Koopmans IP

Open shell wavefunctions

45

Re-examine the Slater determinant wavefunction

Aψa(1)ψb(2) = A[Φa(1)Φb(2)][a(1)b(2)]=

when the spinorbitals have the opposite spin

Since <ψa|ψb>= 0 = < Φa| Φb><a|b>

There is no orthogonality condition for the spatial orbitals for

opposite spin electrons

In general < Φa| Φb> =Sab, where the overlap Sab ≠ 0

This wavefunction has spin projection of MS=0, but if Sab ≠ 1

It is not a singlet stateThus S+ A[Φa(1)Φb(2)][a(1)b(2)]= A[Φa(1)Φb(2)][a(1)a(2)]

= [ΦaΦb - ΦbΦa]aa ≠ 0 unless Φa=Φb

Thus this wavefunction A[Φa(1)Φb(2)][a(1)b(2)] has a mixture of

singlet and triplet character

Why would we ever consider such a strange wavefunction?

Energies for H2

46

The closed shell HF wavefunction

for H2 φHF(1)φHF(2)[abba]

does ok near Re but goes to the

wrong limit as R>3 bohr~1.6A

(it is nearly as bad as MO)

HF

GVBexact

Consider now the wavefunction the

Unrestricted HF wavefunction (UHF) , where

we do not require Φa=Φb

A[Φa(1)Φb(2)][a(1)b(2)]

For R< ~2.2 bohr the optimum is Φa=Φb

But for R> ~2.2 bohr Φa localizes more on the

left while Φb(2) localizes more on the right to

that it goes to atomic orbitals at R = ∞ (similar

to the GVB orbitals). The energy curve has

~1/5 the bonding of GVB, starting at ~2.2

bohr. Of course it has the wrong spin.


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 = ΣjmSjmCmklk

This is still complicated since the Hjm operator includes exchange

terms

We still refer to this as solving the HF equations

This was worked out by Clemens Roothaan (pronounced with a

long o and short a) in the RS Mulliken group at Chicago in the

1950s. They called them LCAO SCF equations


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


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 HF orbitals of H2O

TAs put energies of 5

occupied orbitals plus

lowest 2 unoccupied

orbitals, use correct

symmetry notation

Show orbitals


The HF orbitals of ethylene



lowest 2 unoccupied


symmetry notation

Show orbitals


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


The HF orbitals of benzene

TAs put energies of

21 occupied orbitals

plus lowest 4

unoccupied orbitals,

use correct symmetry

notation

Show orbitals


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



lowest 2 unoccupied


symmetry notation


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


HF wavefunctions

61

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

62

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

63

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


Configuration interaction upper bound theorm

64

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

Transition states

65

ReactantProduct

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, which would

lead to an imaginary

frequency

Transition state is the stationary point, where all forces are zero,

but for which the force is at a minimum for all coordinates but

one, where it is at a maximum


ReactantProduct

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!!!


Stop Lecture 1

67

The following slides are supplementary material

They may give you more insight

Quantum Mechanics – First postulate

68

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

Quantum Mechanics – Second postulate

69

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/rR_

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

Summary 2nd Postulate QM

70

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

3rd Postulate of QM, the variational principle

71

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

EexEap

EThis 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.

Side comment: the next 4 slides Derive Schrödinger equation

from variational principle. You are not responsible for this

72

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

Extra: Derivation of Schrodinger Equation

73

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

Extra: Derivation of Schrodinger Equation

74

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

Extra Summary deriviation of Schrödinger

Equation

75

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

4th postulate of QM - Excited states

76

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>


Phase factor

77

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

Extra: Configuration interaction

78

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


Extra: Configuration interaction upper bound

theorem

79

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


Electron spin, 5th postulate QM

80

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 Ba

b


Permutational symmetry, summary

81

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


Permutational symmetries for H2 and He

82

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


H2

He

Permutational symmetries for H2 and He

83

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


The 6th postulate of QM: the Pauli Principle

84

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


Consider H atom

85

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


The H atom ground state

86

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


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

Atomic units

87

The excited states of H atom - 1

88

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

1 when j=k, but it is

0 otherwise

We say that different excited states are orthogonal.


Nodal theorem

89

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

like the 1s state for H atom



90

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

yθ

φ



91

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

yθ

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


The reason for the numbers nlm will be apparent later


Excited radial functions

92

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

planes1 spherical

nodal plane

2 spherical

nodal planes

Angularly excited states

93

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

The p excited angular states of H atom

94


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


More p functions?

95

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)

More angular functions?

96

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,


More d angular functions?

97

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

x

z

+

-

dz2-x2

-

+


More d angular functions?

98

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

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


Summarizing the d angular functions

99

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


Summarizing the angular functions

100

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


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

101

Here the bar over

m negative


Combination of radial and angular nodal

planes

102

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 n

od

al p

lane

s

radia

l nodal pla

nes

angula

r nodal pla

nes

R̄ R̄

R̄

Siz

e (

a0)

This is all you need to remember


Sizes hydrogen orbitals

103

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/ZR̄ Where a0 = bohr = 0.529A=0.053 nm = 52.9 pm


Hydrogen atom excited states

1041s

-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 R̄ = - Z2/2n2


Plotting of orbitals:

line cross-section vs. contour

105

contour plot

in yz plane

line plot

along z axis


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

106


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

107


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

108


Contour plots of hydrogenic 4f orbitals

109

He+ atom

110

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


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


Estimate J1s,1s, the electron repulsion energy of 2

electrons in He+ 1s orbitals

111

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


Estimate J1s,1s, the electron repulsion energy of 2

electrons in He+ 1s orbitals

112

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%


The optimum atomic orbital for He atom

113

He atom: EHe = 2(½ z2) – 2Zz (5/8)z



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


The optimum atomic orbital for He atom

114

He atom: EHe = 2(½ z2) – 2Zz (5/8)z



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.


Interpretation: The optimum atomic orbital for He atom

115

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)


Now lets add a 3rd electron to form Li

116

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)


First consider Li+

117

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


Consider adding the 3rd electron to the 2p orbital

118


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


Consider adding the 3rd electron to the 2s orbital

119


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

2.12A2s

R~0.2A


Li atom excited states

1201s

2s

-0.125 h0 = -3.4 eV2p

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


Aufbau principle for atoms

1211s

2s

2p

3s

3p3d

4s

4p4d

4f

Energy

2

2

6

2

62

6

10

1014

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


He, 2

Ne, 10

Ar, 18

Zn, 30

Kr, 36


Many-electron configurations

123

General

aufbau

ordering

Particularly stable


General trends along a row of the periodic table

124

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


General trends along a column of the periodic

table

125

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


Plot of rφ(r) for

the outer s

valence orbital

126


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


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

Transition metals; consider [Ar] + 1 electron

129

[IP4s = (Zeff4s )2/2n2 = 4.34 eV Zeff

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

IP4p = (Zeff4p )2/2n2 = 2.73 eV Zeff

4p = 1.79;

IP3d = (Zeff3d )2/2n2 = 1.67 eV Zeff

3d = 1.05;

IP4s = (Zeff4s )2/2n2 = 11.87 eV Zeff

4s = 3.74; 4s<3d<4p


3d = 2.59;


4p = 3.20;


3d = 4.05; 3d<4s<4p

IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff

4s = 5.04;


4p = 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


Transition metals; consider Sc0, Sc+, Sc2+

130

3d: IP3d = (Zeff3d )2/2n2 = 24.75 eV Zeff

3d = 4.05;

4s: IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff

4s = 5.04;

4p: IP4p = (Zeff4p )2/2n2 = 17.01 eV Zeff

4p = 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 Zeff

4s = 3.89;

(3d)2: IP3d = (Zeff3d )2/2n2 = 12.28 eV Zeff

3d = 2.85;

(3d)(4p): IP4p = (Zeff4p )2/2n2 = 9.66 eV Zeff

4p = 3.37;

Sc+

(3d)(4s)2: IP4s = (Zeff4s )2/2n2 = 6.56 eV Zeff

4s = 2.78;

(4s)(3d)2: IP3d = (Zeff3d )2/2n2 = 5.12 eV Zeff

3d = 1.84;

(3d)(4s)(4p): IP4p = (Zeff4p )2/2n2 = 4.59 eV Zeff

4p = 2.32;

Sc


Implications for transition metals

131

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


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

132


Review over, back to quantum mechanics

133

Stopped Lecture 1


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

136

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: 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

aa


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

Ψ(1,2) = ψa(1) ψb(2)

And the Hamiltonian for H2 molecule

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


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 [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



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





But <ψb|ψa>=0

Thus all are zero


-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.


The energy of the antisymmetrized wavefunction


wavefunction Ψ(1,2) must be positive

Eee =∫ (d3r1)((d3r2)|Ψ(1,2)|2/r12 > 0


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

Ψ(1,2) = ψa(1) ψb(2)

And the Hamiltonian for H2 molecule

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


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





But <ψb|ψa>=0

Thus all are zero


-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.



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



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


Topic 1: Practical Quantum Chemistry

160

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 nuclei and

hence 3*12=36 degrees of freedom

(dof) and 42 electrons or 3*42 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

161


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




1st term: kinetic energy operator: 42 terms

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







The Schrödinger Equation: Nuclear-Electron

162


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





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







The Schrödinger Equation: Electron-Electron

163


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





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

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





The Schrödinger Equation: Nuclear-Nuclear

164


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

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

lecture 1, january 4, 2016 quantum mechanics-1:...

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