Lecture 1, January 4, 2010

Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number:

Ch120a Hours: 2-3pm Monday, Wednesday, Friday

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 Teaching Assistants: Wei-Guang Liu < [email protected]

> Ted Yu < [email protected]> Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 1

Overview

This course aims to provide a conceptual understanding of the chemical bond sufficient to predict semi-quantitatively the structures, properties, and reactivities of materials, without computations The philosophy is similar to that of Linus Pauling, who in the 1930’s revolutionized the teaching of chemistry by including the concepts from quantum mechanics (QM), but not its equations.

We now include the new understanding of chemistry and materials science that has resulted from QM studies over the last 50 years. We develop an atomistic QM-based understanding of the structures and properties of chemical, biological, and materials systems. Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 2

Intended audience

This course is aimed at experimentalists and theorists in

chemistry

,

materials science

,

chemical engineering

,

applied physics

,

biochemistry

,

physics

,

electrical engineering

, and

mechanical engineering

with an interest in characterizing and designing catalysts, materials, molecules, drugs, and systems for energy and nanoscale applications. Courses in QM too often focus more on applied mathematics rather than physical concepts. Instead, we start with the essential differences between quantum and classical mechanics (the description of kinetic energy) which is used to understand why atoms are stable and why chemical bonds exist. We then introduce the role of the Pauli Principle and spin and proceed to use these basic concepts to predict the structures and properties of various materials, including molecules and solids 3 Ch120a-Goddard-L01

Applications: Organics:

Resonance, strain, and group additivity. Woodward Hoffman rules, and reactions with dioxygen, and ozone.

Carbon Based systems

: bucky balls, carbon nanotubes, graphene; mechanical, electronic properties, nanotech appl.

Semiconductors, Surface Science:

Si and GaAs, donor and acceptor impurities, surface reconstruction, and surface reactions.

Ceramics:

Oxides, ionic materials, covalent vs. ionic bonding, concepts ionic radii, packing in determining structures and properties.

Examples

: silicates, perovskites, and cuprates.

Hypervalent systems:

XeF n

,

ClF n , IBX chemistry.

Transition metal systems:

organometallic reaction mechanisms. (oxidative-addition, reductive elimination, metathesis)

Bioinorganics:

Electronic states, reactions in heme molecules.

Organometallic catalysts

: CH 4  CH3OH, ROMP, Metallocenes

Metal oxide catalysts

: selective oxidation, ammoxidation

Metals and metal alloys:

chemisorption, Fuel cell catalysts

Superconductors:

mechanisms: organic and cuprate systems.

4

Course details

Homework every week, hand out on Wednesday 3pm, due Monday 2pm, graded and back 1pm Wednesday.

OK to collaborate on midterm, but indicate who your partners were and write your own homework (no xerox from partners) Exams: open book for everything distributed in course, no internet or computers outside of course materials Grade: Final 48%, Midterm 24%, Homework 28% (best 7) No late homework or exams TA budget cut in half this year: no TA office hours Each lecture will start with a review of the important stuff from previous lecture.

This is the time to ask questions Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 5

Course details

On line: course notes, 16 chapters, can download and print Lectures this year on powerpoint, will be on line after the lecture Schedule: MWF 2-3pm, I would like to change to two days a week 2-3:30pm on MWF (arranged in advance) Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 6

Need for quantum mechanics

Consider the classical description of the simplest atom, hydrogen with 1 proton of charge q p with charge q e = +e and one electron = –e separated by a distance R between them assume electron has velocity v(t) and that the proton is sitting still PE = potential energy = ?

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 7

Need for quantum mechanics

Consider the classical description of the simplest atom, hydrogen with 1 proton of charge q p with charge q e = +e and one electron = –e separated by a distance R between them assume electron has velocity v(t) and that the proton is sitting still PE = potential energy = q e q p /R = -e 2 /R KE = kinetic energy = ?

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 8

Need for quantum mechanics

Consider the classical description of the simplest atom, hydrogen with 1 proton of charge q p with charge q e = +e and one electron = –e separated by a distance R between them \ PE = potential energy = q e q p /R = -e 2 /R KE = kinetic energy = ½ m e v 2 (assume proton is sitting still) What is the lowest energy (ground state) of this system?

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 9

Need for quantum mechanics

Consider the classical description of the simplest atom, hydrogen with 1 proton of charge q p with charge q e = +e and one electron = –e separated by a distance R between them assume electron has velocity v(t) and that the proton is sitting still PE = potential energy = q e q p /R = -e 2 /R KE = kinetic energy = m e v 2 /2 = p 2 /2m e where p = m e v What is the lowest energy (ground state) of this system?

PE: R = 0  PE = -

∞

KE: p = 0  KE = 0 Ch120a-Goddard-L01

∞

10

Problem with classical mechanics

Ground state for H atom has the electron sitting on the proton (R=0) with v=0.

Thus electron and proton move together Since their charges cancel there is no interaction of the H atom with anything else in the universe (except gravity) Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 11

Problem with classical mechanics

Ground state for H atom has the electron sitting on the proton (R=0) with v=0.

Thus electron and proton move together Since their charges cancel there is no interaction of the H atom with anything else in the universe (except gravity) Thus there is no H 2 molecule Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 12

Problem with classical mechanics

Ground state for H atom has the electron sitting on the proton (R=0) with v=0.

Thus electron and proton move together Since their charges cancel there is no interaction of the H atom with anything else in the universe (except gravity) Thus there is no H 2 molecule Similarly the carbon atom would have all electrons at the nucleus; thus no hydrocarbons, no amino acids, no DNA, Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 13

Problem with classical mechanics

Ground state for H atom has the electron sitting on the proton (R=0) with v=0.

Thus electron and proton move together Since their charges cancel there is no interaction of the H atom with anything else in the universe (except gravity) Thus there is no H 2 molecule Similarly the carbon atom would have all electrons at the nucleus; thus no hydrocarbons, no amino acids, no DNA, Thus no people.

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 14

Problem with classical mechanics

Ground state for H atom has the electron sitting on the proton (R=0) with v=0.

Thus electron and proton move together Since their charges cancel there is no interaction of the H atom with anything else in the universe (except gravity) Thus there is no H 2 molecule Similarly the carbon atom would have all electrons at the nucleus; thus no hydrocarbons, no amino acids, no DNA, Thus no people.

This would be a very dull universe with no room for us Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 15

Quantum Mechanics to the rescue

The essential element of QM is that all properties that can be known about the system is 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.

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 16

PE of H atom, QM

In QM the total energy can be written as E = KE + PE where for the H atom PE = the average value of (-e 2 /r) over all positions of the electron. Since the probability of the electron at xyz is P(x,y,z,t) = | Φ(x,y,z,t) | 2 = Φ(x,y,z,t) * Φ(x,y,z,t) We can write PE =

∫

Φ(x,y,z,t) * Φ(x,y,z,t) (-e 2 /r) dxdydz or PE =

∫

Φ(x,y,z,t) * (-e 2 /r) Φ(x,y,z,t) dxdydz which we write as _ PE = < Φ| (-e 2 /r) | Φ> = -e 2 / _ R Now what is the best value of PE in QM?

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 17

Best value for PE in QM of H atom

Consider possible shapes of wavefunction Φ(x,y,z,t) * of H atom We plot the wavefunction along the z axis with the proton at z=0 Which has the lowest PE?

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 18

Best value for PE in QM of H atom

Consider possible shapes of wavefunction Φ(x,y,z,t) * of H atom We plot the wavefunction along the z axis with the proton at z=0 Since PE = -e 2 / , it is case c.

Indeed the lowest PE is for a delta function with = 0 Leading to a ground state with PE = -

∞

_ R just as for Classical Mechanics For PE QM is the same as CM, just average over P= | Φ(x,y,z,t)| 2 PE scales as 1/ Ch120a-Goddard-L01 R © copyright 2010 William A. Goddard III, all rights reserved 19

What about KE?

In CM the position and momentum of the electron can be specified independently, R=0 and v=0, but in QM both the KE and PE are derived from the SAME wavefunction.

In CM, KE = p 2 /2m e In QM the KE for a one dimensional system is KE = [

∫

( p Φ)

†

( p Φ) dx]/2m e = [

∫

( Ћ = ( Ћ 2 /2m e )

∫

(d Φ/dx) * (d Φ/dx) dx d Φ/idx) * ( Ћ d Φ/idx) dx]/2m e Thus in QM KE =

(

Ћ

2 /2m e )<(d Φ/dx)| (dΦ/dx)>

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 20

Interpretation of QM form of KE

KE =

(

Ћ

2 /2m e )<(d Φ/dx)| (dΦ/dx)>

KE proportional to the average square of the gradient or slope of the wavefunction Thus the KE in QM prefers smooooth wavefunctions In 3-dimensions KE =

(

Ћ

2 /2m e )<(



Φ

.



Φ> = =(

Ћ

2 /2m e ) ∫ [

(d Φ/dx) 2 + (d Φ/dx) 2 + (d Φ/dx) 2 ] dxdydz Still same interpretation: the KE is proportional to the average square of the gradient or slope of the wavefunction Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 21

My form for KE differs from usual one in QM

In usual QM, write KE operator = - ( Ћ 2 /2m e ) (d 2 /dx 2 ) One dimensional, usual form KE =

<

Φ| - ( Ћ 2 /2m e ) (d 2 /dx 2 )| Φ>= - ( Ћ 2 /2m e ) = - ( Ћ 2 /2m e )

<

Φ|(d 2 /dx 2 )| Φ>

∫

( Φ * )(d 2 Φ/dx 2 ) dx Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 22

My form for KE differs from usual one in QM

In usual QM, write KE operator = - ( Ћ 2 /2m e ) (d 2 /dx 2 ) One dimensional, usual form KE =

<

Φ| - ( Ћ 2 /2m e ) (d 2 /dx 2 )| Φ>= - ( Ћ 2 /2m e ) = - ( Ћ 2 /2m e )

<

Φ|(d 2 /dx 2 )| Φ>

∫

( Φ * )(d 2 Φ/dx 2 ) dx Now integrate by parts: -

∫

u(dv/dx)dx =

∫

(du/dx)(v)dx if u,v  Let u = Φ * and dv/dx = d 2 Φ/dx 2 here then get Goddard form 0 at boundaries KE = ( Ћ 2 /2m e )<(d Φ/dx)| (dΦ/dx)> Both forms of the KE are correct, I consider the second form more fundamental and more useful It can be used to derive the 1 st form by integrating by parts Clearly KE is always positive and decreasing the slopes decreases the KE Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 23

Best value for KE in QM of H atom

Consider possible shapes of wavefunction Φ(x,y,z,t) * of H atom Which has the lowest KE?

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 24

Best value for KE in QM of H atom

Consider possible shapes of wavefunction Φ(x,y,z,t) * of H atom Which has the lowest KE?

clearly it is case a.

_ Indeed the lowest KE is for a wavefunction with 

∞

Leading to a ground state with KE = 0 just as for Classical Mechanics But in QM the same wavefunction must be used for KE and PE _ KE wants 

∞

_ whereas PE wants = 0. Who wins?

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 25

The compromise between PE and KE

_ PE ~ -C 1 / R _ KE ~ +C 2 /

2

_ Now lets find the optimum R _ Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 26

_

_ R

Analysis for optimum R

Here PE is small and negative, but KE is (small) 2 but positive, thus PE wins and the total energy is negative _ Here PE is large and negative, but KE is (large) 2 but positive, thus KE wins and the total energy is positive _ total energy is most negative _ Conclusion in QM the H atom has a finite size, Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 27

Discussion of KE

In QM KE wants to have a smooth wavefunction but electrostatics wants the electron concentrated at the nucleus. Since KE ~ 1/R 2 , KE always keeps the wavefunction finite, leading to the finite size of H and other atoms. This allows the formation of molecules and hence to existence of life In QM it is not possible to form a wavefunction in which the position is exactly specified simultaneous with the momentum being exactly specified. The minimum value is <( d x)( d p> >

Ћ

/2 (The Heisenberg uncertainty principle) Sometimes it is claimed that this has something to do with the finite size of the atom. It does but I consider this too hand-wavy. Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 28

Implications

QM leads to a finite size for the H atom and for C and other atoms This allows formation of bonds to form H2, benzene, amino acids, DNA, etc.

Allowing life to form Thus we owe our lives to QM The essence of QM is that wavefunctions want to be smooth, wiggles are bad, because they increase KE Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 29

The wavefunction for H atom

In this course we are not interested in solving for wavefunctions, rather we want to deduce the important properties of the wavefunctions without actually solving any equations However it is useful to know the analytic form. The ground state of H atom has the form (here Z=1 for H atom, Z=6 for C 5+ Here we use r, Ө,Φ spherical coordinates rather than x,y,z (the average size of the H atom), =a 0 /Z Here is the normalization constant, <

Φ|Φ>=1

_ _ The PE = -Ze 2 / while KE= Ze 2 R _ Thus the total energy E = -Ze 2 /2 = PE/2 = -KE Ch120a-Goddard-L01 30

Atomic Units

Z=1 for Hydrogen atom For Z ≠ 1, _ R _ E = -Ze 2 /2 R = 0.529 A =0.053 nm is the Bohr radius = a 0 /Z = - Z 2 e 2 /2a 0 = - m e Z 2 e 4 /2

Ћ

2

= -Z 2 h 0 /2 where h 0 = e 2 /a 0 = m e e 4 /

Ћ

2

= Hartree = 27.2116 eV = 627.51 kcal/mol = 2625.5 kJ/mol Atomic units: m e = 1, e = 1, Ћ = 1 leads to unit of length = a 0 and unit of energy = h 0 In atomic units: KE=

<



Φ

.



Φ>/2

(leave off Ћ 2 /m e )

PE = <

Φ|

-1/r|

Φ

>/2

(leave off e 2 ) Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 31

Local PE and KE of H atom

Local KE negative Local KE positive Local KE negative E = -0.5 h 0 Local PE, negative Local KE, positive Classical turning point r = a 0 / √2 Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 32

2 ways to plot orbitals

1-dimensional Iine plot of orbital along z axis 2-dimensional contour plot of orbital in xz plane, adjacent contours differ by 0.05 au Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 33

Now consider H

2 +

molecule

Bring a proton up to an H atom to form H 2 + Is the molecule bound? That is, do we get a lower energy at finite R than at R = ∞ Two possibilities Electron is on the left proton Electron is on the right proton Or we could combine them At R = ∞, these are all the same, but not for finite R In QM we always want the wavefunction with the lowest energy Question: which combination is lowest?

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 34

Combine Atomic Orbitals for H

2 +

molecule

Antisymmetric combination Which is best (lowest energy)?

the D g = Sqrt[2(1+S)] and D u = Sqrt[2(1-S)] factors above are the constants needed to ensure that < Φ g | Φ g > =

∫

Φ g | Φ g dxdydz = 1 (normalized) < Φ u | Φ u > =

∫

Φ u | Φ u dxdydz = 1 (normalized) I will usually eschew writing such factors, leaving them to be 35 © copyright 2010 William A. Goddard III, all rights reserved

Ch120a-Goddard-L01

Energies of of H

2 +

Molecule

g state is bound since starting the atoms at any distance between arrows, the molecule will stay bonded, with atoms vibrating forth and back Ungood state: u Good state: g 36

But WHY is the g state bound?

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 37

But WHY is the g state bound?

Common rational : Superimposing two orbitals and squaring to get the probability leads to moving charge into the bond region. This negative charge in the bond region attracts the two positive nuclei + + Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 38

But WHY is the g state bound?

Common rational : Superimposing two orbitals and squaring to get the probability leads to moving charge into the bond region. This negative charge in the bond region attracts the two positive nuclei + + Sounds reasonable, but increasing the density in bond region  decrease density near atoms, thus moves electrons from very attractive region near nuclei to less attractive region near bond midpoint, this INCREASES the PE Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 39

The change in electron density for molecular orbitals

The densities r g and r u wavefunctions of H 2 + for the g and u LCAO compared to superposition of r L atomic densities (all densities add up to one electron) + r R Adding the two atomic orbitals to form the g molecular orbital increases the electron density in the bonding region, as expected. This is because in QM, the amplitudes are added and then 40 Ch120a-Goddard-L01

Compare change in density with local PE function

PE(r) = -1/r a – 1/r b The local PE for the electron is lowered at the bond midpoint from the value of a single atom But the best local PE is still near the nucleus Thus the Φ g =  L +  R wavefunction moves charge to the bond region AT THE EXPENSE of the charge near the nuclei, causing 41 Ch120a-Goddard-L01

If the bonding is not due to the PE, then it must be KE

We see a dramatic decrease in the slope of the g orbital along the bond axis compared to the atomic orbital. This leads to a dramatic decrease in KE compared to the atomic orbital The shape of the Φ g =  L Φ u =  L  R + wavefunctions  R compared to the pure atomic and orbital (all normalized to a total probability of one).

Ch120a-Goddard-L01 This decrease arises only in the bond region.

It is this decrease in KE that is responsible for the bonding in H 2 + © copyright 2010 William A. Goddard III, all rights reserved 42

The PE of H

2 +

for g and u states

The total PE of H wavefunctions 2 + for the Φ g =  L +  R and (relative to the values of V g Φ = V u u =  L = -1 h  R 0 at R = ∞) Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 43

The KE of g and u wavefunctions for H

2 + Use top part of 2-7 The change in the KE as a function of distance for the g and u wavefunctions of H 2 + (relative to the value at R= ∞ of KE g =KE u =+0.5 h 0 ) Comparison of the g and u wavefunctions of H 2 + (near the optimum bond distance for the g state), showing why g is so bonding and u 44 Ch120a-Goddard-L01

Why does KEg has an optimum?

Ch120a-Goddard-L01 R too short leads to a big decrease in slope but over a very short region,  little bonding R is too large leads to a decrease in slope over a long region, but the change in slope is very small  little bonding Optimum bonding occurs when there is a large region where both atomic orbitals have large slopes in the opposite directions (contragradient). 45

Ch120a-Goddard-L01

KE dominates PE

Changes in the total KE and PE for the g and u wavefunctions of H 2 + (relative to values at R= ∞ of KE :+0.5 h 0 PE: -1.0 h 0 E: -0.5 h 0 The g state is bound between R~1.5 a 0 and ∞ (starting the atoms at any distance in this range leads to atoms vibrating forth and back. Exciting to the u state 46

KE dominates PE, leading to g as ground state

Calculations show this, but how could we have predicted that g is better than u without calculations?

Answer: the nodal theorem: The ground state of a QM systems has no nodes. Thus g state lower E than u state Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 47

The nodal Theorem

The ground state of a system has no nodes (more properly, the ground state never changes sign). This is often quite useful in reasoning about wavefunctions. For example the nodal theorem immediately implies that the g wavefunction for H 2 + state) is the ground state (not the u Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 48

The nodal Theorem 1D

Schrodinger equation,

H

Φ k = E k Φ k One dimensional: Ĥ =- ½ d 2 /dx 2 + V(x) Consider the best possible eigenstate of Ĥ with a node, Φ 1 and construct a nonnegative function Ө 0 =| Φ 1 | as in b For every value of x, V(x)[ Φ 1 ] 2 V 0 = ∫ [ Ө 0 ] * V(x)[ Ө 0 ] = ∫ = V(x)[ Ө 0 ] 2 [ Φ 1 ] * V(x)[ Φ 1 ] 2 = V 1 so that a b c Also |d Ө 0 /dx| 2 = |d Φ 1 /dx| 2 for every value of x except the single point at which the node occurs. Φ 1 Ө 0 Φ 0 Thus T 0 = ½ ∫ |d Ө 0 /dx| 2 = ½ ∫ |d Φ 1 /dx| 2 = T 1 . Hence E 0 = T 0 + V 0 = T 1 + V 1 = E 1 . Here E 1 is the best possible energy for an eigenstate with a node. However Ө 0 is not. Thus we could smooth out Ө 0 in the region of the node as in c, decreasing the KE and lowering the energy. Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 49

The nodal Theorem 1D

a Φ 1 Best possible eigenfunction of

H

node,

H

Φ 1 = E 1 Φ 1 with a Ө 0 =| Φ 1 | has the same energy as Φ 1 E 0 = T 0 + V 0 However Ө 0 = T wavefunction 1 + V 1 = E 1 . is not the best Thus we could smooth out Ө 0 in the region of the node as in c, decreasing the KE and lowering the energy. Thus the optimum nodeless wavefunction Φ 0 leads to E 0 < E 1 . Only for a very singular potential at some point, so repulsive that all wavefunctions are 0, do we get E 0 = E 1 Ch120a-Goddard-L01 b c © copyright 2010 William A. Goddard III, all rights reserved Ө 0 Φ 0 50

The nodal Theorem for excited states in 1D

For one dimensional finite systems we can order all eigenstates by the number of nodes E 0 < E 1 < E 2 .... E n < E n+1 (where a sufficiently singular potential can lead to an = sign ) The argument is the same as for the ground state.

Consider best wavefunction Φ n with n nodes and flip the sign at one node to get a wavefunction Ө n-1 that changes sign only n-1 times.

Calculate that E n-1 = E n But Ө n-1 is not the best with n-1 sign changes. Thus we can smooth out Ө n-1 in the region of the extra node to decrease the KE and lower the energy for the Φ n-1 ,. Thus the optimum n-1 node wavefunction leads to E Ch120a-Goddard-L01 n-1 < E n . 51

The nodal Theorem 3D

In 2D a wavefunction that changes size once will have a line of points with Φ 1 =0 (a nodal line) For 3D there will be a 2D nodal surface with Φ 1 =0. In 3D the same argument as for 1D shows that the ground state is nodeless. We start with Φ 1 the best possible eigenstate with a nodal surface and construct a nonnegative function Ө 0 =| Φ 1 | For every value of x,y,z, V(x,y,z)[ Φ 1 ] 2 V 0 = ∫ [ Ө 0 ] * V(xyz)[ Ө 0 ] = ∫ = V(x,y.z)[ Ө 0 ] 2 [ Φ 1 ] * V(xyz)[ Φ 1 ] 2 = V 1 so that Also |  Ө 0 | 2 Thus T 0 = |  Φ 1 | 2 = ½ ∫ |  Ө 0 | 2 everywhere except along a 2D plane = ½ ∫ |  Φ 1 | 2 = T 1 . Hence E 0 = T 0 + V 0 = T 1 + V 1 = E 1 . As before E 1 is the best possible energy for an eigenstate with a nodal plane. However Ө 0 can be improved by smoothing Thus the optimum nodeless wavefunction Φ 0 Ch120a-Goddard-L01 leads to E 0 © copyright 2010 William A. Goddard III, all rights reserved < E 1 . 52

The nodal Theorem for excited states in 3D For 2D and 3D, one cannot order all eigenstates by the number of nodes. Thus consider the 2D wavefunctions + Φ 00 Φ 10 + + Φ 01 Φ 20 + + + + Φ 11 Φ 21 + + + It is easy to show as in the earlier analysis that E 00 < E 10 < E 20 < E 21 E 00 < E 01 < E 11 < E 21 But the nodal argument does not indicate the relative energies of E 10 and E 20 versus E 01 Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 53

Back to H

2 + Nodal theorem  The ground state must be the g wavefunction Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 54

H

2

molecule, independent atoms

Start with non interacting H atoms, electron 1 on H on earth,  E (1) the other electron 2 on the moon,  M (2) What is the total wavefunction, Ψ(1,2)?

Maybe Ψ(1,2) =  E (1) +  M (2) ?

Since the motions of the two electrons are completely independent, we expect that the probability of finding electron 1 somewhere to be independent of the probability of finding electron 2 somewhere.

Thus P(1,2) = P E (1)*P M (2) This is analogous to the joint probability, say of rolling snake eyes (two ones) in dice P(snake eyes)=P(1 for die 1)*P(1 for die 2)=(1/6)*(1/6) = 1/36 Question what wavefunction Ψ(1,2) leads to P(1,2) = P E (1)*P M 55 (2)?

Answer: product of amplitudes

Ψ(1,2) =  E (1) M (2) leads to P(1,2) = | Ψ(1,2)| 2 = Ψ(1,2) * Ψ(1,2) = = [  E (1) M (2)] * = [ E (1) *  E [ (1)] [ E M (1) (2) * M  (2)] = M (2)] = = P E (1) P M (2) Conclusion the wavefunction for independent electrons is the product of the independent orbitals for each electron Back to H 2 , Ψ EM (1,2) =  E (1) M (2) But Ψ ME (1,2) =  M (1) E (2) is equally good since the electrons are identical Also we could combine these wavefunctions  E (1) M (2)   E (1) M (2) Which is best? Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 56

End of Lecture 1

Ch120a-Goddard-L01 © copyright 2010 William A. Goddard III, all rights reserved 57