Transcript Slide 1

Lecture 1, January 3, 2011
Elements QM, stability H, H2+
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]>
Caitlin Scott <[email protected]>
Ch120a-Goddard-L01
© copyright 2011 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 70 years.
We develop an atomistic QM-based understanding of the
structures and properties of chemical, biological, and materials
systems.
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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
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spanning
the periodic table.
Applications:
Organics: Resonance, strain, and group additivity. WoodwardHoffman 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: XeFn, ClFn, IBX chemistry.
Transition metal systems: organometallic reaction mechanisms.
(oxidative-addition, reductive elimination, metathesis)
Bioinorganics: Electronic states, reactions in heme molecules.
Organometallic catalysts: CH4  CH3OH, ROMP, Metallocenes
Metal oxide catalysts: selective oxidation, ammoxidation
Metals and metal alloys: chemisorption, Fuel cell catalysts
Ch120a-Goddard-L01
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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 homework, 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 except for course materials
Grade: Final 48%, Midterm 24%, Homework 28% (best 7)
No late homework or exams
TA budget cut by ¾ from usual level : 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
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Course details
On line: course notes, 16 chapters, can download and print
Lectures this year on powerpoint, will be on line after the lecture
Last years ppt 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
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6
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 = ?
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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
assume that the
proton is sitting still
If e is distance R from
proton the PE is
PE = potential energy = qeqp/R = -e2/R
KE = kinetic energy = ?
Ch120a-Goddard-L01
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8
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
(assume proton is
sitting still)
PE = potential energy = qeqp/R = -e2/R
KE = kinetic energy = ½ mev2 = p2/2me where p = me v
What is the lowest energy (ground state) of this system?
Ch120a-Goddard-L01
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9
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
assume electron has
velocity v(t) and that
the proton is sitting still
PE = potential energy = qeqp/R = -e2/R
KE = kinetic energy = mev2/2 = p2/2me where p = me v
What is the lowest energy (ground state) of this system?
PE: R = 0  PE = - ∞
KE: p = 0  KE = 0
TotalCh120a-Goddard-L01
Energy = E = KE ©+copyright
PE =2011
-∞
(cannot
get
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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)
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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 H2 molecule
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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 H2 molecule
Similarly the carbon atom would have all electrons at
the nucleus; thus no hydrocarbons, no amino acids,
no DNA,
Ch120a-Goddard-L01
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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 H2 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
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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 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 would be a very dull universe with no room for us
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15
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.
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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 (-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 as
_
2
2
PE = < Φ| (-e /r) |Φ> = -e / R
_
Where R is the average value of 1/r
Now what is the best value of PE in QM?
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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
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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?
_
Since PE = -e2/ R, it is case c.
_
Indeed the lowest PE is for a delta function with R = 0
Leading to a ground state with PE = - ∞ just as for Classical Mecha
2
For PE. QM is the
same
as
CM,
just
average
over
P=
|Φ(x,y,z,t)|
_
PE scales
as 1/ R
19
Ch120a-Goddard-L01
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What about KE?
In Classical Mechanics 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 = p2/2me
In QM the KE for a one dimensional system is
KE = (Ћ2/2me)<(dΦ/dx)| (dΦ/dx)>
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What about KE?
KE = (Ћ2/2me)<(dΦ/dx)| (dΦ/dx)>
This is not the usual form for KE. 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 du
Thus since u and v must be 0 on the boundaries (otherwise get
infinite total probability <Φ|Φ> rather than 1) we get
KE = -(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 and that is in
essentially all QM books.
Later we show that applying the variational principle to my form
leads directly to the Schrodinger equation.
Ch120a-Goddard-L01
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What about KE?
KE = (Ћ2/2me)<(dΦ/dx)| (dΦ/dx)>
This is not the usual form for KE. 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 du
Thus since u and v must be 0 on the boundaries (otherwise get
infinite total probability <Φ|Φ> rather than 1) we get
KE = -(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 Shrodinger came up with and that is in
essentially all QM books.
Later
we show
that
applying
the variational
to my form
Both forms
of the
KE
are correct,
since one principle
can be derived
from
leads
directly
to the Schrodinger
the other
by integrating
by parts equation.
I consider my form as more fundamental and more useful
Thus it makes it clear that KE is always positive and
decreasing the slopes decreases the KE
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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)<(Φ. Φ> =
=(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
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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
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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 R  ∞
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 R  ∞ whereas PE wants R = 0. Who wins?
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The compromise between PE and KE
_
How do PE and KE scale with R , the average size of the orbital?
_
PE ~ -C1/ R
_
KE ~ +C2/ R 2
_
Now lets find the optimum R
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_
Analysis for optimum R
_
Consider very large R
Here PE is small and negative, while KE is (small)2 but
positive, thus PE wins and the total energy is negative
_
Now consider very small R
Here PE is large and negative, while KE is (large)2 but
positive, thus KE wins and the total energy is positive
_
Thus there must be some intermediate R for which the
total energy is most negative
_
This is the R for the optimum wavefunction
Conclusion in QM the H atom has
a finite size,
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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.
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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
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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
Here a0 = Ћ2/me2 = 0.529 A =0.053_ nm is the Bohr radius (the
average size of the H atom),
=aR
0/Z
Here
is the normalization constant, <Φ|Φ>=1
_
_
2
The PE =
while KE= Ze / 2 R
R
_
2
Thus the total energy E = -Ze /(2R) = PE/2 = -KE
-Ze2/
ThisCh120a-Goddard-L01
is called the Virial ©Theorem
general
molecules
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reserved
30
Atomic Units
Z=1 for Hydrogen atom
a0 = Ћ2/me2_= = 0.529 A =0.053 nm is the Bohr radius
For Z ≠ 1, R = a0/Z
_
E = -Ze2/2 R = - Z2 e2/2a0 = - meZ2 e4/2 Ћ2 = -Z2 h0/2
where h0 = e2/a0 = me e4/ Ћ2 = Hartree = 27.2116 eV
= 627.51 kcal/mol = 2625.5 kJ/mol
Atomic units:
me = 1, e = 1, Ћ = 1 leads to
unit of length = a0 and unit of energy = h0
In atomic units: KE= <Φ.Φ>/2 (leave off Ћ2/me)
PE = <Φ|-1/r|Φ>/2 (leave off e2)
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Local PE and KE of H atom
Local KE negative Local KE positive Local KE negative
E = -0.5 h0
Local PE, negative
Local KE, positive
Classical turning point
r = a0/√2
_
PE =
R
_
KE= Ze2/ 2 R
-Ze2/
Φ(z)
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32
2 ways to plot orbitals
1-dimensional Iine plot of
orbital along z axis
Ch120a-Goddard-L01
2-dimensional contour plot of
orbital in xz plane, adjacent
contours differ by 0.05 au
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33
Now consider H2+ molecule
Bring a proton up to an H atom to form H2+
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?
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Combine Atomic Orbitals for H2+ molecule
Symmetric
combination
Two extreme
possibilities
Antisymmetric combination
Which is best (lowest energy)?
the Dg = Sqrt[2(1+S)] and Du = Sqrt[2(1-S)] factors above are the
constants needed to ensure that
<Φg|Φg> =∫ Φg|Φg dxdydz = 1 (normalized)
<Φu|Φu> =∫ Φu|Φu dxdydz = 1 (normalized)
I will usually eschew writing such factors, leaving them to be
understood
Ch120a-Goddard-L01
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35
Energies of of H2+ 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
LCAO
= Linear Combination
of Atomic Orbitals
Ch120a-Goddard-L01
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36
But WHY is the g state bound?
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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
-
+
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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
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39
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 + rR
atomic 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
40
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2011 William A. Goddard III, all rights reserved
squared
to get probability
density
Compare change in density with local PE function
PE(r) = -1/ra – 1/rb
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 Φg = L + R wavefunction moves charge to the bond
region AT THE EXPENSE of the charge near the nuclei, causing
an increase
in the PE, ©and
opposing
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41
The PE of H2+ for g and u states
The total PE of H2+ for the Φg = L + R and Φu = L - R
wavefunctions (relative to the values of Vg = Vu = -1 h0 at R = ∞)
Ch120a-Goddard-L01
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42
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 Φg = L + R and
Φu = L - R wavefunctions
compared to the pure atomic
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 H2+
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43
The KE of g and u wavefunctions for H2+
Use top part of 2-7
Ch120a-Goddard-L01
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)
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
44
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rights
reserved
isallso
antibonding
Why does KEg has an optimum?
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).
Ch120a-Goddard-L01
This leads to optimum bonding
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45
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 h0
PE: -1.0 h0
E: -0.5 h0
Ch120a-Goddard-L01
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
46
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III, allto
rights
reserved
leads
dissociation
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
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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 H2+ is the ground state (not the u
state)
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48
a
The nodal Theorem 1D
Schrodinger equation, H Φk = Ek Φk
One dimensional: H =- ½ d2/dx2 + V(x)
Consider the best possible eigenstate of H
with a node, Φ1 and construct a
nonnegative function Ө0 =|Φ1| as in b
For every value of x, V(x)[Φ1]2 = V(x)[Ө0]2 so that
V0 = ∫ [Ө0
]*V(x)[Ө
0] = ∫
[Φ1]*V(x)[Φ1]2
= V1
Φ1
b
Ө0
c
Φ0
Also |dӨ0/dx|2 = |dΦ1/dx|2 for every value of x
except the single point at which the node occurs.
Thus T0 = ½ ∫ |dӨ0/dx|2 = ½ ∫ |dΦ1/dx|2 = T1.
Hence E0 = T0 + V0 = T1 + V1 = E1.
Ch120a-Goddard-L01
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49
The nodal Theorem 1D
a
Φ1
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 Φ1
E0 = 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 Φ0
leads to E0 < E1.
Only for a potential so repulsive at some point,
that all wavefunctions are 0, do we get E0 = E1
Ch120a-Goddard-L01
b
Ө0
c
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Φ0
50
The nodal Theorem for excited states in 1D
For one-dimensional finite systems, we can order all eigenstates
by the number of nodes
E0 < E1 < E2 .... En < En+1
(where 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-1III, <
En. reserved
Ch120a-Goddard-L01
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51
The nodal Theorem 3D
In 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 |Ө0|2 = |Φ1|2 everywhere except along a 2D plane
Thus T0 = ½ ∫ |Ө0|2 = ½ ∫ |Φ1|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.
Ch120a-Goddard-L01
© copyright 2011 William A. Goddard III, all rights reserved
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
+
Φ00 Φ10
+
-
Φ01 Φ20
+
-
+
Φ11 Φ21
+
-
+
+
-
+
+
-
-
+
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
Ch120a-Goddard-L01
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53
Back to H2+
g state
u state
Nodal theorem  The ground state must be the g wavefunction
Ch120a-Goddard-L01
© copyright 2011 William A. Goddard III, all rights reserved
54
H2 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)?
Ch120a-Goddard-L01
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55
H2 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) ?
Ch120a-Goddard-L01
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56
H2 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
Ch120a-Goddard-L01
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57
H2 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) = 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)?
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58
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)M(2)] =
= [E(1)* E(1)] [M(2)* M(2)] =
= PE(1) PM(2)
Conclusion the wavefunction for independent electrons is the
product of the independent orbitals for each electron
Ch120a-Goddard-L01
© copyright 2011 William A. Goddard III, all rights reserved
59
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)M(2)] =
= [E(1)* E(1)] [M(2)* M(2)] =
= PE(1) PM(2)
Conclusion the wavefunction for independent electrons is the
product of the independent orbitals for each electron
Back to H2, ΨEM(1,2) = E(1)M(2)
But ΨME(1,2) = M(1)E(2) is equally good since the electrons
are identical
Ch120a-Goddard-L01
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60
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)M(2)] =
= [E(1)* E(1)] [M(2)* M(2)] =
= PE(1) PM(2)
Conclusion the wavefunction for independent electrons is the
product of the independent orbitals for each electron
Back to H2, ΨEM(1,2) = E(1)M(2)
But ΨME(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
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61
The wavefunction for H2 at long R
Consider H2 at R=∞
Two equivalent
wavefunctions
ΦLR
ΦRL
At R=∞ these are all the same, what is best for finite R
Ch120a-Goddard-L01
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62
Plot of twoelectron
wavefunctions
along molecular
axis
LR
R
LR
L
L
x
R
z
Ch120a-Goddard-L01
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63
Plot of two-electron wavefunctions along molecular axis
R
R
LR
RL
L
L
L
R
L
R
x
z
Ch120a-Goddard-L01
© copyright 2011 William A. Goddard III, all rights reserved
64
Plot two-electron
wavefunctions
along molecular
axis
x
LR+RL
R
R
L
L
L
z
Ch120a-Goddard-L01
R
LR-RL
L
R
Lower KE (good)
higher KE (bad)
Higher PE (bad)
Lower PE (good)
EE poor
EE good
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65
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
Ch120a-Goddard-L01
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66
Valence Bond
wavefuntion of H2
LR-RL
Nodal theorem says ground
state is nodeless, thus g is
better than u
LR+RL
Valence bond: start with ground state at R=∞ and build
molecule
by combining© best
state
of atoms
Ch120a-Goddard-L01
copyright
2011 William
A. Goddard III, all rights reserved
67
End of Lecture 1
Ch120a-Goddard-L01
© copyright 2011 William A. Goddard III, all rights reserved
68