Lecture 1, April 6, 2020 Elements QM, stability H, H2+,H2...Ch120a-Goddard-L01 © copyright 2017 William A. Goddard III, all rights reserved 4 Applications: Organics: Resonance, strain,

© copyright 2017 William A. Goddard III, all rights reservedCh120a-Goddard-L01 1

Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic

chemistry, and energy

William A. Goddard, III, [email protected] Beckman Institute, x3093

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

California Institute of Technology

Teaching Assistant: Yalu Chen <[email protected]>

Lecture 1, April 6, 2020Elements QM, stability H, H2+,H2

Course number: Ch120aHours: 2-3pm Monday, Wednesday, Friday


OverviewThis 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 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.


Intended audienceThis 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 spanning the periodic table.


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.Hydrogen bonding, Hypervalent systems: XeFn, ClFn, IBX.Transition metal systems: organometallic reaction mechanisms. (oxidative-addition, reductive elimination, metathesis)Bioinorganics: Electronic states, reactions in heme molecules.Organometallic catalysts: CH4 à CH3OH, ROMP, MetallocenesMetal oxide catalysts: selective oxidation, ammoxidationMetals and metal alloys: chemisorption, Fuel cell catalysts Superconductors: mechanisms: organic and cuprate systems.


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

OK to collaborate on homework, but indicate who your partners were and write your own homework (no xerox from partners)

Exams: no collaboration, open book for everything distributed in course, no internet or computers except for course materials

Grade: Final 48%, Midterm 24%, Homework 28% (best 7 of 8)

No late homework or exams

Will have TA office hour in this room after the Friday lecture

lectures will often start with a review of the important stuff from previous lecture.

Ask questions during lectures


Course details

On line: old course notes, 16 chapters, from 1974-75; can download and print

Lectures this year on powerpoint, will be on line after the lecture

Also ppt from last year also on line

Schedule: MWF 2-3pm

Occasionally I will add an extra lecture from 3-4pm to make up for missing a lecture while on a trip


Need for quantum mechanics

Consider the classical description of the simplest atom, hydrogen with 1 proton of charge qp = +e and one electron with charge qe = –e separated by a distance R between them

PE = potential energy = qeqp/R = -e2/RKE = kinetic energy = mev2/2 = p2/2me where p = me vWhat is the lowest energy (ground state) of this system?

PE: R = 0 è PE = - ∞KE: p = 0 è KE = 0

Total Energy = E = KE + PE = - ∞ (cannot get any lower)

assume electron has velocity v(t) and that the proton is sitting still


Problem with classical mechanicsGround 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 H2 molecule

Similarly the carbon atom would have all electrons at the nucleus; thus no hydrocarbons, no amino acids, no DNA,

Thus no people.

This classical world would be a very dull universe with no room for us


Quantum Mechanics to the rescue

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.

We say that the wavefunction is normalized


PE of H atom, QMIn QM the total energy can be written as E = KE + PE where for the H atom PE = the average value of (-e2/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) (-e2/r) dxdydz or

PE = ∫Φ(x,y,z,t)* (-e2/r) Φ(x,y,z,t) dxdydz which we write asPE = < Φ| (-e2/r) |Φ> = -e2/Where is the average value of 1/rR

_ R_

Thus PE in QM is very similar to CM, just use average distance


PE of H atom, QMIn QM the total energy can be written as E = KE + PE where for the H atom PE = the average value of (-e2/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) (-e2/r) dxdydz or

PE = ∫Φ(x,y,z,t)* (-e2/r) Φ(x,y,z,t) dxdydz which we write asPE = < Φ| (-e2/r) |Φ> = -e2/Where is the average value of 1/rR

_ R_

Thus PE in QM is very similar to CM, just use average distance

© copyright 2017 William A. Goddard III, all rights reserved

Since PE = -Ze2/ , it is case c.The lowest PE is for a delta function the wavefunction with = 0 QM lowest PE = - ∞ just as for Classical Mechanics

Ch125a-Goddard-L01 12

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=0Which has the lowest PE?

R_R

_

For PE. QM is the same as CM, just average over P= |Φ(x,y,z,t)|2

PE scales as 1/ R_


What about KE?In Classical Mechanics the position and momentum of the electron can be specified independently, Thus the lowest energy had R=0 and v=0, But in QM both the KE and PE are derived from the SAME wavefunction.

In CM, KE = p2/2me

In QM the KE for a one dimensional system is KEQM = <p2>/2me

In QM p =(Ћ/i) (dΦ/dx) ,* è complex conjugateThus KEQM= (Ћ2/2me)<(dΦ/dx)*| (dΦ/dx)> =Or KEQM=- (Ћ2/2me)<Φ*| (d2Φ/dx2)> =

Thus KE increases as the average slope increasesSince it depends on the slope squared KE will not allow the electron to localize at a point. For a periodic function, KE increases as the period decreases. Thus fast wiggles much higher energy than slow wiggles


Comment about KE

KE = (Ћ2/2me)<(dΦ/dx)| (dΦ/dx)>

This is not the usual form for KE. Most books writeKE = -(Ћ2/2me) ∫ Φ* (d2Φ/dx2) dx = ∫ Φ* [-(Ћ2/2me)(d2/dx2)] Φ

= ∫Φ*[px]2 Φ where px = (Ћ/i) (d/dx)This is the form that Schrodinger came up with in 1926 and is in essentially all QM books.

I consider that the fundamental form for the KE is

To compare to standard QM books, write u= (dΦ/dx) and dv= (dΦ/dx)* dx and integrate by parts ∫u dv = u(∞)v(∞) - u(-∞)v(-∞) - ∫v duThus since u and v must be 0 on the boundaries (otherwise get infinite total probability <Φ|Φ> rather than 1) the standard form converts to my form


Interpretation of QM form of KE

KE = (Ћ2/2me)<(dΦ/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 = (Ћ2/2me)<(ÑΦ ÑΦ> =

=(Ћ2/2me) ∫ [(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] dxdydz

Still same interpretation:

.

the KE is proportional to the average square of the gradient or slope of the wavefunction


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 è ∞ Leading to a ground state with KE = 0 just as for Classical Mechanics

R_


The compromise between PE and KE

How do PE and KE scale with , the average size of the orbital?

PE ~ -C1/

KE ~ +C2/ 2

Now lets find the optimum

R_

R_

R_

R_


R_

Consider very large Here PE is small and negative, while KE is (small)2 but positive, thus PE wins and the total energy is negative

Analysis for optimum R_


Now consider very small Here PE is large and negative, while KE is (large)2 but positive, thus KE wins and the total energy is positive

R_


R_

R_




R_


R_

Thus there must be some intermediate for which the total energy is most negativeThis is the for the optimum wavefunction

R_

R_




R_


R_

Thus there must be some intermediate for which the total energy is most negativeThis is the for the optimum wavefunction

R_

R_


Conclusion in QM the H atom has a finite size,


Discussion of KE

In QM KE wants to have a smooth wavefunction but electrostatics wants the electron concentrated at the nucleus.

Since KE ~ 1/R2 , 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

<(dx)(dp)> ≥ 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.


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

© copyright 2017 William A. Goddard III, all rights reserved 24

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+

For the atom, we use spherical coordinates r,Ө,Φ not x,y,z

N0 is the normalization constant, <Φ|Φ>=1

As the nuclear charge Z increases the orbital gets smaller (more attraction to nucleus) but taller, since it is normalized


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

Note: QM books sometimes plot vs. R not z. This leads to the Φ=0 at R=0


Hydrogen atom, Atomic Units

a0 = Ћ2/me2 = 0.529 A =0.053 nm is the Bohr radius

Z=1 for Hydrogen atom

R_

For Z ≠ 1, = a0/Z

Charge of nucleus = ZeCharge of electron = -eAverage distance = R

_

PE = - e (Ze)/ R_

For the atom, we use spherical coordinates r,Ө,Φ not x,y,z

Size of atom decreases as Z increases

E-Hatom = -(½) [me e4/ Ћ2] = -13.6058 eV1 Hartree = h0 =[me e4/ Ћ2] = 27.2116 eV = 627.51 kcal/mol

= 2625.5 kJ/mol

h0 = me e4/ Ћ2 = e2/a0 = thus E-Hatom=-(1/2)(e2/a0)=(1/2)PE


Atomic Units

Atomic units:me = 1, e = 1, Ћ = 1 leads to unit of length = a0 = 0.529 A =0.053 nm, the Bohr radiusunit of energy = h0= 27.2116 eV = 627.51 kcal/mol = 2625.5 kJ/mol, the Hartree

In atomic units: KE= <ÑΦ.ÑΦ>/2 (leave off Ћ2/me)PE = <Φ|-Z/r|Φ> (leave off e2)


E = -0.5 h0 Classical turning point r = a0/√2

Local PE and KE of H atom

Local KE positive Local KE negativeLocal KE negative

Local PE, negativeLocal KE, positive

PE = -Ze2/ R_

0 energy

Energy


Bring a proton up to an H atom to form H2+

Is the molecule bound?That is does it have a lower energy at finite R than at R = ∞

Now consider H2+ molecule

L R

rLrR

e

Actually get two states at for finite R One is bound and one is unbound

This makes sense since can have two states H and H+Or H+ and H


Several possibilitiesElectron is on the left proton,orElectron is on the right protonOr we could combine them

Now consider what is the wavefunction for H2+ molecule

L R

rLrR

e

At R = ∞ these are have the same energy, but not for finite RIn QM we always want the wavefunction with the lowest energy. Question: which combination is lowest?


Combine Atomic Orbitals for H2+ molecule

Two extreme possibilitiesSymmetric combination

the above equations ignore normalizationDg = Sqrt[2(1+S)] and Du = Sqrt[2(1-S)] factors ensure that <Φg|Φg> =∫ Φg|Φg dxdydz = 1 (normalized)<Φu|Φu> =∫ Φu|Φu dxdydz = 1 (normalized)

Which is best (lowest energy)?

Antisymmetric combination


Energies of of H2+ Molecule

LCAO = Linear Combination of Atomic Orbitals

Good state: g

Ungood state: u

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


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,

Why is it wrong

+ +-


But WHY is the g state bound?

Moving charge to the bond midpoint decreases density near atoms, thus electrons move from attractive region near nuclei to less attractive region near bond midpoint, this INCREASES the PE

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

+ +-

PE(r) = -1/ra – 1/rb


The change in electron density for molecular orbitals

The densities rg and ru for the g and u LCAO wavefunctions of H2

+ compared to superposition of rL + rRatomic densities (all densities add up to one electron)

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 squared to get probability density


Compare change in density with local PE function

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 nucleusThus the Φg = cL + cR wavefunction moves charge to the bond region AT THE EXPENSE of the charge near the nuclei, causing an increase in the PE, and opposing bonding

PE(r) = -1/ra – 1/rb


The PE of H2+ for g and u states

The total PE of H2+ for the and wavefunctions (relative to the

values of Vg = Vu = -1 h0 at R = ∞)

Φg = cL + cR

Φu = cL - cR

Φu moves charge from bond midpoint to nuclei thus gets better as decrease R; But Φg gets worse


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

The shape of the Φg = cL + cR and Φu = cL - cR wavefunctions compared to the pure atomic orbital (all normalized to a total probability of one).

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

This decrease arises only in the bond region.

It is this decrease in KE that is responsible for the bonding in H2

+


The KE of g and u wavefunctions for H2+

The change in the KE as a function of distance for the g and u wavefunctions of H2

+

(relative to the value at R=∞ of KEg=KEu=+0.5 h0)

Use top part of 2-7

Comparison of the g and u wavefunctions of H2

+ (near the optimum bond distance for the g state), showing why g is so bonding and u is so antibonding


R too short leads to a big decrease in slope but over a very short region, èlittle bondingR is too large leads to a decrease in slope over a long region, but the change in slope is very small è little bondingOptimum bonding occurs when there is a large region where both atomic orbitals have large slopes in the opposite directions (contragradient). This leads to optimum bonding

Why does KEg has an optimum?


KE dominates PE

Changes in the total KE and PE for the g and u wavefunctions of H2

+ (relative to values at R=∞ of KE :+0.5 h0PE: -1.0 h0E: -0.5 h0

The g state is bound between R~1.5 a0 and ∞(starting the atoms at any distance in this range leads to atoms vibrating forth and back. Exciting to the u state leads to dissociation

Good state: g

Ungood state: u


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


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 H2

+ is the ground state (not the u state)


The nodal Theorem 1DSchrodinger equation, H Φk = Ek ΦkOne dimensional: H =- ½ d2/dx2 + V(x)Consider the best possible eigenstate of Hwith a node, Φ1 and construct a nonnegative function Ө0 =|Φ1| as in bFor every value of x, V(x)[Φ1]2 = V(x)[Ө0]2 so that V0 = ∫ [Ө0]*V(x)[Ө0] = ∫ [Φ1]*V(x)[Φ1]2 = V1

a

b

Φ1

Ө0

c Φ0

Also |dӨ0/dx|2 = |dΦ1/dx|2 for every value of x except the single point at which the node occurs (does not count since no length)

Thus T0 = ½ ∫ |dӨ0/dx|2 = ½ ∫ |dΦ1/dx|2 = T1. Hence E0 = T0 + V0 = T1 + V1 = E1.


We just showed that for the best possible eigenfunction of H with a node, H Φ1 = E1 Φ1

Ө0 =|Φ1| has the same energy as Φ1E0 = T0 + V0 = T1 + V1 = E1. However Ө0 is just a special case of a nodeless wavefunction that happens to go to 0 at one point. Thus we could smooth out Ө0 in the region of the node as in c, decreasing the KE and lowering the energy. Thus the optimum nodeless wavefunction Φ0leads to E0 < E1. Only for a potential so repulsive at some point, that all wavefunctions are 0, do we get E0 = E1

The nodal Theorem 1D a

b

Φ1

Ө0

c Φ0


The nodal Theorem for excited states in 1D – skip thisFor one-dimensional finite systems, we can order all eigenstates by the number of nodes

E0 < E1 < E2 .... En < En+1

(where only a sufficiently singular potential can lead to an = sign )

The argument is the same as for the ground state.

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

Show that En-1 = En

But Өn-1 is not the best with n-1 sign changes.

Thus we can smooth out Өn-1 in the region of the extra node to decrease the KE and lower the energy for the Φn-1,. Thus the optimum n-1 node wavefunction leads to En-1 < En.


The nodal Theorem 3DIn 2D a wavefunction that changes size once will have a line of points with Φ1=0 (a nodal line) For 3D there will be a 2D nodal surface with Φ1=0. In 3D the same argument as for 1D shows that the ground state is nodeless. We start with Φ1 the best possible eigenstate with a nodal surface and construct a nonnegative function Ө0 =|Φ1|For every value of x,y,z, V(x,y,z)[Φ1]2 = V(x,y.z)[Ө0]2 so that V0 = ∫ [Ө0]*V(xyz)[Ө0] = ∫ [Φ1]*V(xyz)[Φ1]2 = V1

Also |ÑӨ0|2 = |ÑΦ1|2 everywhere except along a 2D plane Thus T0 = ½ ∫ |ÑӨ0|2 = ½ ∫ |ÑΦ1|2 = T1. Hence E0 = T0 + V0 = T1 + V1 = E1. As before E1 is the best possible energy for an eigenstate with a nodal plane. However Ө0 can be improved by smoothing Thus the optimum nodeless wavefunction Φ0 leads to E0 < E1.


The nodal Theorem for excited states in 3D skip thisFor 2D and 3D, one cannot order all eigenstates by the number of nodes. Thus consider the 2D wavefunctions

+ Φ00 Φ10

Φ01 Φ20+-

+-

+-+

It is easy to show as in the earlier analysis that

E00 < E10 < E20< E21

E00 < E01 < E11 < E21

But the nodal argument does not indicate the relative energies of E10 and E20 versus E01

- Φ11+-

++-+

+- -Φ21


Back to H2+

Nodal theorem è The ground state must be the g wavefunction

g state u state


H2 molecule, independent atomsStart with non interacting H atoms, electron 1 on H on earth, cE(1)the other electron 2 on the moon, cM(2)What is the total wavefunction, Ψ(1,2)?

Maybe Ψ(1,2) = cE(1) + cM(2) ? NO

Since the motions of the two electrons are completely independent, we expect that the probability of finding electron 1 at some location on earth to be independent of the probability of finding electron 2 at some location on the moon.

Thus P(1,2) = PE(1)*PM(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) = PE(1)*PM(2)?© copyright 2009 William A. Goddard III, all rights reserved


Ψ(1,2) = cE(1)cM(2) leads to P(1,2) = |Ψ(1,2)|2 = Ψ(1,2)* Ψ(1,2) =

= [cE(1)cM(2)]* [cE(1)cM(2)] =

= [cE(1)* cE(1)] [cM(2)* cM(2)] =

= PE(1) PM(2)

Answer: product of amplitudes

Conclusion the wavefunction for independent electrons is the product of the independent orbitals for each electron

Back to H2, ΨEM(1,2) = cE(1)cM(2)

But ΨME(1,2) = cM(1)cE(2) is equally good since the electrons are identicalAlso we could combine these wavefunctions cE(1)cM(2) ± cE(1)cM(2)

Which is best?


Consider H2 at R=∞

ΦRL

Two equivalent wavefunctions

At R=∞ these are all the same, what is best for finite R

The wavefunction for H2 at long R

ΦLR

ΦLR

ΦLR

ΦRL

ΦRL


Plot of two-electron

wavefunctions along molecular

axis

z

x

LR wavefunction

L

LR

L

R

R


Plot of two-electron wavefunctions along molecular axis

z

x

LR wavefunction RL wavefunction

L

L

R

R

L

L

R

R


Combine the LR and RL

wavefunctions

z

x

Lower KE (good)

Higher PE (bad)

EE poor

higher KE (bad)

Lower PE (good)

EE good

L

L

R

R

L

L

R

R

LR+RL LR-RL

Who wins? LR+RL is best because no nodesWe can see the reason here. The KE is decreased for the plus combination.


Valence Bond wavefuntion of H2

Valence bond: start with ground state at R=∞ and build molecule by combining best state of atoms

Nodal theorem says ground state is nodeless, thus Fg is better than Fu

LR+RL

LR-RL


Details

E(H2+)1 = KE1 + PE1a + PE1b + 1/Rab for electron 1

E(H2+)2 = KE2 + PE2a + PE2b + 1/Rab for electron 2

E(H2) = (KE1 + PE1a + PE1b) + (KE2 + PE2a+PE2b) + 1/Rab+EE

EE = <Φ(1,2)|1/r12| Φ(1,2)> assuming <Φ(1,2)|Φ(1,2)>=1

Note that EE = ∫dx1dy1dz1 ∫dx2dy2dz2[| Φ(1,2)|2/r12] > 0 since all terms in integrand > 0


Summary: bonding in H2 is due to decrease in KE as add the L(1)R(2) and R(1)L(2) two electron

wavefunctions

Ch125a-Goddard-L02 58

Stability of H atom, bonding in H2+ and bonding in H2 all due to KESmooth wavefunctions better than wiggly wavefunctions


Ch120a-Goddard-L02

Molecular Orbitals: Alternative way to view states of H2

Valence bond: start with ground state at R=∞ and buildmolecule by bonding atoms

Molecular orbitals (MO): start with optimum orbitals of one electron molecule at R=Re and add electronsH2: two MO’s

g: bonding

u: antibonding

15

Low E

High E

g:bonding

u:antibonding


Put 2 electrons into the 2 MO’s, get four states

Best

worst

In between

First lets look at the ground state, g(1)g(2)

60Ch120a-Goddard-L02


Compare ground state MO and VB near Re

KE: best possible

EE: bad, too much ionic character

KE: pretty good

EE: excellent

Now consider large R:61Ch120a-Goddard-L02

e1 and e2 on same

center, very bad

e1 and e2 on different

centers, good

z1

z2

z1

z2


compare ground state from VB and MO

MO wavefunction is halfcovalent and half ionic

Good for R=Requilbrium = Re,terrible for R = ∞


VB wavefunction is fullcovalent and no ionic

Great for R = ∞

Good for Re


Compare bond dissociation for VB and MO

Covalent limit H + H

MO limit half covalent and half

ionic H- + H+



Put 2 electrons into the 2 MO’s, get four states

Best

worst

In between

Now do the excited states, gu and ug



ddard III, all rights reserved 21

Ch120a-Goddard-L02

Analyze Φgu = φg(1)φu(2)wavefunction in 2-electron space

φg(1)

φu(2)

L

L

R

R

L

L

R

R

L

L

R

R

L R


Analyze guand ug statesin 2 electron

space

Both have exactlythe same KE andsame PE

gu ug

Now combine into gu+ug and gu – ug

Which is better?


all rights reserved 23

Ch120a-Goddard-L02

Analyze guand ug statesin 2 electron

spaceAll four have one nodal plane and lead to same KE and same PE

However the electron-electron repulsion is different

<Φ(1,2)|1/r12| Φ(1,2)>

Worst case is for z1=z2, along diagonal Never have

z1=z2Maximum at z1=z2terrible EE

gu ug

gu-ug gu+ug


Ground and excited states from MO analysis

Thus gu-ug is best excited state because it has very low electron repulsion. The electrons have zero chance of both being at the same spot



VB and MO descriptions of u excited states

Pure covalent, but antibonding

Pure ionic, but bonding

MO VB

MO VB



Energies for H2 states based on atomic orbitals(z=1)

covalent

ionic


© copyright 2017 William A. Goddard III, all rights reserved 28

3rd Postulate of QM, the variational principle

EexThis is called the variational principle.

For the ground state, d2E/dΦ ≥ 0 for allpossible changes

The ground state wavefunction is the system, Φ, has the lowest possible energy out of all possible wavefunctions.

Consider that Φex is the exact wavefunction with energy Eex = <Φex|Ĥ|Φex>/<Φex|Φex> and thatΦap = Φex + dΦ is some other approximate wavefunction.

Then Eap = <Φap|Ĥ|Φap>/<Φap|Φap>

Eap

EThis means that for sufficiently small dΦ,dE = 0 for all possible changes, dΦ

We write dE/dΦ = 0 for all dΦ

Then Eap ≥ Eex


Ch120a-Goddard-L02 © copyright 2011

William A. Goddard III, all rights reserved29

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

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

dE = <(dΦ| H|Φ> + CC

Where H = -(Ћ2/2me) (d2/dx2) + V The QM Hamiltonian

Write QM energy as

Extra detail, ok to skipDerivation of Schrodinger Equation

Variational principle says that ground state Φ0 leads to the lowest possible E, E0

Then starting with this optimum Φ0 , and making any change, dΦwill increase E.

The first order change in E isdE = (Ћ2/2me)<(d dΦ/dx)| (dΦ/dx)> + < dΦ| V|Φ>+CC

Complex conjugate

Integrate by parts

© copyright 2017 William A. Goddard III, all rights reservedCh120a-Goddard-L02 © copyright 2011 William A. Goddard III, all rights reserved 30

Extra detail, ok to skipDerivation of Schrodinger Equation

But even though <Φ0|Φ0> = 1, changing Φ0 by dΦ, might changethe normalization.This leads to an additional termE+dE = E0/{<Φ0|Φ0> + <dΦ|Φ0> + CC} = E0 – E0{<dΦ|Φ0> + CC}ThusdE =<(dΦ|H|Φ0> -E0<dΦ|Φ0>} + CC = =<(dΦ|(H-E0)|Φ0> +CCAt the minimum the energy must increase for both +dΦ and –dΦ, hence dE=0 = <(dΦ|(H - E0}|Φ>} + CC for all possible dΦThus the coefficient of dΦ, must be zero. Get(H - E0)Φ=0 orwhere H= {-(Ћ2/2me)(d2/dx2)+V}

HΦ= E0Φ Schrodinger EquationThe Hamiltonian

Thus starting with EQM = {(Ћ2/2me)<(dΦ/dx)| (dΦ/dx)> + <Φ|V|Φ>}/<Φ|Φ>and minimizing the energy leads to the Schrodinger Equation

I consider minimization of energy as the fundamental Concept


Summary deriviation of Schrödinger Equation

|EQM = <Φ K̂E | Φ> + < Φ| V|Φ> = <Φ| Ĥ | Φ>where the Hamiltonian is Ĥ ≡ K̂E + V and K̂E = - (Ћ2/2me)Ñ2

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

V(x,y,z,t) is the (classical) potential energy for the system Consider arbitrary approximate wavefunction

Φap = Φex + dΦ and require that

dE= Eap – Eex = 0

Get <dΦ|Ĥ-Eex|Φex>] = ∫ dΦ*[(Ĥ-Eex)Φex] = 0 for all possible dΦThis è [(Ĥ-Eex)Φex] = 0 or the Schrödinger equation

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Ch120a-Goddard-L02

Ĥ Φex = EexΦex

The exact ground state wavefunction is a solution of this equation


Extra detail, ok to skip, Excited statesThe 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 stateFor example H atom: 1s, 2s, 2px, 3dxy etcAlso 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=0The set of all eigenstates of Ĥ is complete, thus any arbitraryfunction Ө can be expanded as

Ө = Sk Ck Φk where <Φj| Ө>=Cj or Ө = Sk Φk <Φk| Ө> 75

Ch120a-Goddard-L02

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Lecture 1, April 6, 2020 Elements QM, stability H, H2+,H2...Ch120a-Goddard-L01 © copyright 2017 William A. Goddard III, all rights reserved 4 Applications: Organics: Resonance, strain,