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课程教学组 : 吕鑫 教授 Tel: 2181600 (mobile/office) email: [email protected] 化学楼 236 室 http://pcss.xmu.edu.cn/users/xlu/ 教辅: 王雪 [email protected] 张嘉伟 [email protected] 课程主页 : http://pcss.xmu.edu.cn/users/xlu/group/courses/structurechem/index.html. 参考书 : - PowerPoint PPT Presentation
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Structural Chemistry参考书 :
<< 结构化学 >> 厦门大学化学系物构组编
<<Physical Chemistry>> P.W. Atkins
<< 结构化学基础 >> 周公度编著 , 北京大学出版社
《化学键的本质》鲍林
上海科学技术出版社
《结构化学习题解析》周公度 段连运课程 QQ 群 : 8675318 ( 加入时请注明个人学号 )
课程教学组 : 吕鑫 教授 Tel: 2181600 (mobile/office) email: [email protected] 化学楼 236 室 http://pcss.xmu.edu.cn/users/xlu/
教辅:王雪 [email protected] 张嘉伟 [email protected] 课程主页 :http://pcss.xmu.edu.cn/users/xlu/group/courses/structurechem/index.html
两点说明:
• 作业要求 (交作业时间、理科作业规范)
• 考勤、考试方式与成绩 (期中 / 期末 / 随堂考试,期平成绩等)
What is ChemistryWhat is Chemistry
The branch of natural science that deals with composition, structure, properties of substances and the changes they undergo.
Types of substancesTypes of substances
Nano materials
Bulk materials
Geometric Structure
Size makes the difference
Electronic Structure
AtomsMolecules
ClustersCongeries
Structure determines propertiesProperties reflect structures
Structure vs. PropertiesStructure vs. Properties
Structural ChemistryStructural Chemistry
Inorganic ChemistryOrganic ChemistryCatalysisElectrochemistryBio-chemistryetc.
Material ScienceSurface ScienceLife ScienceEnergy ScienceEnvironmental Scienceetc.
Role of Structural Chemistry Role of Structural Chemistry in Surface Science in Surface Science
fcc(100) fcc(111)
fcc(775) fcc(10 8 7)
Surface structures of Pt single crystal
Low-index surface :NCS
LIS < NCB.
High-index surface :•Abundant edge sites.•NCes < NCS
LIS < NCB
•Lower NC ~ higher reactivity.
(111)
(775)
(100)
(10 8 7)
Different surfaces do different chemistry.
Structure-sensitive Catalysis!
418
1
25
Surface Structure vs. Catalytic Activity
N2 + 3H2 2NH3
Fe single crystal ,20atm/700K
Another example of Structure-sensitive Catalysis
Role of Structural Chemistry Role of Structural Chemistry in Material Science in Material Science
Graphite & Diamond StructuresDiamond: Insulator or wide bandgap
semiconductor: Graphite: Planar structure: sp2 bonding 2d metal (in plane)
Other Carbon allotropes“Buckyballs” (C60, C70 etc) “Buckytubes” (nanotubes),
other fullerenes
C Crystal StructuresC Crystal Structures
Structure makes the difference
Zheng LS ( 郑兰荪 ), et al.
Capturing the labile fullerene[50] as C50Cl10
SCIENCE 304 (5671): 699-699 APR 30 2004
The pentagon-pentagon fusions in pristine C50-D5h are sterically strained and highly reactive. Perchlorination of these active sites stabilizes the labile C50-D5h.
Nature Materials, 2008, 7, 790.
Chlorofullerenes featuring triple
sequentially fused pentagons
Xie, S.Y., Lu, X., Zheng, L.S. et al
Nature Chemistry, in
press. #540C54Cl8 (a), #864C56Cl12 (b),
#4,169C66Cl6 (c), #4,169C66Cl10 (d).
Endohedral Metallofullerene:Sc4@C82 (C3v) vs. Sc4C2@C80 (Ih)
• (Sc2+)4@C828-
Shinohara, H. Rep. Prog. Phys. 2000, 63, 843.
Proposed QM-predicted
Sc4@C82
E = 28.8 kcal/mol
E = 0.0 kcal/mol
C26-@(Sc3+)4@C80
6--Ih
A Russian-Doll endofullerene
X.Lu, J. Phys. Chem. B. 2006, 110, 11098; Lu & Wang, J. Am. Chem. Soc. 2009, 131, 16646.Highlighted by C&EN and Nat. Chem.
Role of Structural Chemistry Role of Structural Chemistry in Life Science in Life Science
What do proteins do ?Proteins are the basis of how biology gets things
done.
• As enzymes, they are the driving force behind all of the biochemical reactions which makes biology work.
• As structural elements, they are the main constituents of our bones, muscles, hair, skin and blood vessels.
• As antibodies, they recognize invading elements and allow the immune system to get rid of the unwanted invaders.
What are proteins made of ?
• Proteins are necklaces of amino acids, i.e. long chain molecules.
Definition of Structural Chemistry• It is a subject to study the microscopic
structures of matters at the atomic/molecular level using Chemical Bond Theory.
• Chemical bonds structures properties.
Objective of Structural ChemistryObjective of Structural Chemistry
1) Determining the structure of a known substance
2) Understanding the structure-property relationship
3) Predicting a substance with specific structure and property
Chapter 1 basics of quantum mechanics 4Chapter 2 Atomic structure 4Chapter 3 Symmetry 3Chapter 4 Diatomic molecules 3 Chapter 5/6 Polyatomic structures 5Chapter 7 Basics of Crystallography 4 Chapter 8 Metals and Alloys 1Chapter 9 Ionic compounds 3Special talk
OutlineOutline
Chapter 1 The basic knowledge of quantum mechanics
1.1 The origin of quantum mechanics--- The failures of classical physics
Black-body radiation, Photoelectric effect, Atomic and molecular spectra
• Classical physics: (prior to 1900) Newtonian classical mechanics
Maxell’s theory of electromagnetic waves
Thermodynamics and statistical physics
1.1.1 Black-body radiation
Device for the experimentation of black-body radiation.
It can not be explained by classical thermodynamics and statistical mechanics.
Black-Body Radiation
Classical Solution:
Rayleigh-Jeans Law
(high energy, Low T)
Wien Approximation
(long wave length)Wave length / m
A number of experiments revealed the temperature-dependence of max and independence on the substance made of the black-body device.
The dawn of quantum mechanics
1.1.2 The photoelectric effect
The photoelectric effect
The Photoelectric Effect
1. The kinetic energy of the ejected electrons depends exclusively and linearly on the frequency of the light. 2. There is a particular threshold frequency for each metal. 3. The increase of the intensity of the light results in the increase of the number of photoelectrons.
Classical physics: The energy of light wave should be directly proportional to intensity and not be affected by frequency.
Explaining the Photoelectric Effect• Albert Einstein
– Proposed a corpuscular theory of light (photons) in 1905.
– won the Nobel prize in 1921
1. Light is consisted of a stream of photons. The energy of a photon is proportional to its frequency.
= h h = Planck’s constant
2. A photon has energy as well as mass. Mass-energy relationship: = mc2 m= hc2
3. A photon has a definite momentum. p=mc= hc=h/
4. The intensity of light depends on the photon density.
Therefore, the photon’s energy is the sum of the photoelectron’s kinetic energy and the binding energy of the electron in metal.
• Ephoton = E binding + E Kinetic energy
• h=W + Ek
Explaining the Photoelectric Effect
Example I: Calculation Energy from Frequency
Problem: What is the energy of a photon of electromagnetic radiation emitted by an FM radio station at 97.3 x 108 cycles/sec?What is the energy of a gamma ray emitted by Cs137 if it has a frequencyof 1.60 x 1020/s?
Ephoton =h = (6.626 x 10 -34Js)(9.73 x 109/s) = 6.447098 x 10 -24J
Ephoton = 6.45 x 10 - 24 J
Egamma ray =h = ( 6.626 x 10-34Js )( 1.60 x 1020/s ) = 1.06 x 10 -13J
Egamma ray = 1.06 x 10 - 13J
Solution:
Plan: Use the relationship between energy and frequency to obtain the energy of the electromagnetic radiation (E = h).
Example II: Calculation of Energy from Wavelength
Problem: What is the photon energy of electromagnetic radiationthat is used in microwave ovens for cooking, if the wavelength of theradiation is 122 mm ?
wavelength = 122 mm = 1.22 x 10 -1m
frequency = = = 2.46 x 1010 /s3.00 x108 m/s1.22 x 10 -1m
cwavelength
Energy = E = h = (6.626 x 10 -34Js)(2.46 x 1010/s) = 1.63 x 10 - 23 J
Plan: Convert the wavelength into meters, then the frequency can becalculated using the relationship;wavelength x frequency = c (where cis the speed of light), then using E=h to calculate the energy.Solution:
Example III: Photoelectric Effect
• The energy to remove an electron from potassium metal is 3.7 x 10 -19J. Will photons of frequencies of 4.3 x 1014/s (red light) and 7.5 x 1014 /s (blue light) trigger the photoelectric effect?
• E red = h = (6.626 x10 - 34Js)(4.3 x1014 /s)
E red = 2.8 x 10 - 19 J
• E blue = h = (6.626 x10 - 34Js)(7.5x1014 /s)
E blue = 5.0 x 10 - 19 J
• The binding energy of potassium is = 3.7 x 10 - 19 J
• The red light will not have enough energy to knock an electron out of the potassium, but the blue light will eject an electron !
• E Total = E Binding Energy + EKinetic Energy of Electron
• E Electron = ETotal - E Binding Energy
• E Electron = 5.0 x 10 - 19J - 3.7 x 10 - 19 J
= 1.3 x 10 - 19Joules
1.1.3 Atomic and molecular spectra
The Line Spectra of Several Elements
An atom can emit lights of discrete/specific frequencies upon electric/photo-stimulation.
• First proposed by Rutherford in 1911.
•The electrons are like planets of the solar system --- orbit the nucleus (the Sun).
• Light of energy E given off when electrons change orbits of different energies.
Why do the electrons not fall into the nucleus?
Why are they in discrete energies?
Planetary model:
Based on classical physics, the electrons would be attracted by the nucleus and eventually fall into the nucleus by continuously emitting energy/light!!
Bohr’s atomic model • Niels Bohr, a Danish physicist, combined the Plank’s
quanta idea, Einstein’s photon theory and Rutherford’s Planetary model, and first introduced the idea of electronic energy level into atomic model.
• Quantum Theory of Energy.• The energy levels in atoms can be
pictured as orbits in which electrons travel at definite distances from the nucleus.
• These he called “quantized energy levels”, also known as principal energy levels.
n=1n=2n=3n=4
n : principal quantum number
The electron in H atom can be promoted to higher energy levels by photons or electricity.
The Energy States of the Hydrogen Atom
Bohr derived the energy for a system consisting of a nucleus plus a single electron eg.
He predicted a set of quantized energy levels given by :
- R is called the Rydberg constant (2.18 x 10-18 J)- n is a quantum number- Z is the nuclear charge
n 1,2,3.. .En RZ2
n2
H He Li2
Problem: Find the energy change when an electron changes from then=4 level to the n=2 level in the hydrogen atom? What is the wavelengthof this photon?
Plan: Use the Rydberg equation to calculate the energy change, then calculate the wavelength using the relationship of the speed of light.Solution:
Ephoton = -2.18 x10 -18J -1n1
2
1n2
2
Ephoton = -2.18 x 10 -18J - = 4.09 x 10 -19J14 2
12 2
= = h c E
(6.626 x 10 -34Js)( 3.00 x 108 m/s)
4.09 x 10 -19J= 4.87 x 10 -7 m = 487 nm
1.1.1 Black-Body Radiation
Planck’s quanta idea E= nh
Lesson 1 Summary
1.1 The failures of classical physics
1.1.2 The photoelectric effect
A corpuscular theory of light (photons) = h h = Planck’s constantp=h/
1.1.3 Atomic and molecular spectra Planetary model: orbits of electrons around the nucleusBohr’s atomic model: quantized energy levels of orbits
En RZ2
n2
1.2 The characteristic of the motion of microscopic particles
1.2.1 The wave-particle duality of microscopic particles
• In classical physics, waves and particles behave differently and can be described by rather different theories.
• As mentioned above, Einstein’s Corpuscular Theory of Lights (photons) to explain the photoelectric effect for the first time introduced the wave-particle duality of photon.
• In 1924 de Broglie suggested that microscopic particles such as electron and proton might also have wave properties in addition to their partical properties.
De Broglie considered that the wave-particle relationship in light is also applicable to particles of matter, i.e.
•The latter equation clearly reflects the wave-particle duality by involving both a particle property (momentum) and a wave property (wavelength)!
•The wavelength of a particle could be determined by
= h/p = h/mv (v: velocity, m: mass)
E=h
p=h/
h = Planck’s constant,
p = particle momentum,
= de Broglie wavelength
de Broglie Wavelength
Example: Calculate the de Broglie wavelength of an electron with speed 3.00 x 106 m/s.
electron mass = 9.11 x 10 -31 kg
velocity = 3.00 x 106 m/s
= =
h mv
6.626 x 10 - 34Js( 9.11 x 10 - 31kg )( 1.00 x 106 m/s )
Wavelength = 2.42 x 10 -10 m = 0.242 nm
J = kg m2
s2hence
The moving speed of an electron is determined by the potential difference of the electric field (V)
eVm 2v2
1
If the unit of V is volt, then the wavelength is:
)(10226.1
1
10602.110110.92
10626.6
2v/
9
1931
34
mV
V
Vme
hmh
1eV=1.602x10-19 J
The de Broglie Wavelengths of Several particles
Particles Mass (g) Speed (m/s) (m)
Slow electron 9 x 10 - 28 1.0 7 x 10 - 4
Fast electron 9 x 10 - 28 5.9 x 106 1 x 10 -10
Alpha particle 6.6 x 10 - 24 1.5 x 107 7 x 10 -15
One-gram mass 1.0 0.01 7 x 10 - 29
Baseball 142 25.0 2 x 10 - 34
Earth 6.0 x 1027 3.0 x 104 4 x 10 - 63
Different Behaviors of Waves and Particles
Si Crystal
Electron beam (50eV)
The diffraction of electrons
STM image of Si(111) 7x7 surface
Spatial image of the confined electron states of a quantum corral. The corral was built by arranging 48 Fe atoms on the Cu(111) surface by means of the STM tip. Rep. Prog. Phys. 59(1996) 1737
Electron as waves
• Wave (i.e., light)
- can be wave-like (diffraction)
- can be particle-like (p=h/)• Particles
- can be wave-like ( =h/p)
- can be particle-like (classical)
The wave-particle duality
•A wave of microscopic particles is a probability wave, neither like the macroscopic mechanical wave nor like the normal electromagnetic wave!
•It reflects the statistic probability of particle presenting.
For photon:
p=mc,
E=h=h (c/) = pc = mc2
The differences between photon and microscopic particles
For particles:
p=mv,
E=(1/2) mv2 = p2/2m=pv/2
E=h
p = h/
= u / … u is not the moving speed of particles.
p2/2m =(1/2) mv2
pv
Question: what is the exact expression of u?
1.2.2 The uncertainty principle
•In classical Physics, the position and momentum of a macroscopic particle (a body) can be certainly determined at a give time.
•This is not the case for a microscopic particle.
•In the diffraction experiments to reveal the wave-particle duality of electrons, the observed wave pattern is just a statistic distribution of electron motion. The exact position and momentum of an electron at a given time remains uncertain.
The experiments of electron beam diffraction revealed that the narrow the slit is, the larger is the central area of the diffraction pattern.
hpx
hpxD
hD
hDp
ppp
pp
DDOAOC
APOP
xo
x
//
)0(sin
sin
/2
1/
2
1/sin
2
1
Include higher order,
y
psin
A
O
C
P
A
O
P
A
O
P
2
1
4or
hpx
A quantitative version
electrons. ofy uncertaintposition
slit). theof(width 2OADx
e OA
P
Q
C
x
e OA
P
Q
C
x
e OA
P
Q
C
x
Example
The speed of an electron is measured to be 1000 m/s to an accuracy of 0.001%. Find the uncertainty in the position of this electron.
The momentum is
p = mv = (9.11 x 10-31 kg) (1 x 103 m/s)
= 9.11 x 10-28 kg.m/s
p = p x 0.001% = 9.11 x 10-33 kg m/s
x = h / p = 6.626 x 10-34 / (9.11 x 10-33 )
= 7.27 x 10-2 (m)
Example
The speed of a bullet of mass of 0.01 kg is measured to be 1000 m/s to an accuracy of 0.001%. Find the uncertainty in the position of this bullet.
The momentum is
p = mv = (0.01 kg) (1 x 103 m/s) = 10 kg.m/s
p = p x 0.001% = 1 x 10-4 kg m/s
x = h / p = 6.626 x 10-34 / (1 x 10-4 )
= 6.626 x 10-30 (m)
Example
The average time that an electron exists in an excited state is 10-8 s. What is the minimum uncertainty in energy of that state?
tE
eV 100.66
eV 106.1
1006.1 J 10 x 1.06
s 10Js/ 10 x 1.06 /
7
19
26 26-
-8-34min
tE
Measurement
•Classical: the error in the measurement depends on the precision of the apparatus, could be arbitrarily small.
•Quantum: it is physically impossible to measure simultaneously the exact position and the exact velocity of a particle.
CLASSICAL vs QUANTUM MECHANICS
Macroscopic matter - Matter is particulate, energy varies continuously. The motion of a group of particles can be predicted knowing their positions, their velocities and the forces acting between them.
Microscopic particles - microscopic particles such as electrons exhibit a wave-particle “duality”, showing both particle-like and wave-like characteristics. The energy level is discrete. …
The description of the behavior of electrons in atoms requires a completely new “quantum theory”.
Quantum mechanical description of Electron
• Quantum mechanics is basically statistical in nature.
• Quantum mechanics does not say that an electron is distributed over a large region of space as a wave is distributed.
• Rather it is the probability patterns used to describe the electron’s motion that behave like waves.
1.3 The basic assumptions (postulates) of quantum
mechanics
Postulate 1. The state of a system is described by a wave function of the coordinates and the time.
In CM (classical mechanics), the state of a system of N particlesis specified totally by giving 3N spatial coordinates (Xi, Yi, Zi)and 3N velocity coordinates (Vxi, Vyi, Vzi).
In QM, the wave function takes the form (r, t) that depends onthe coordinates of the particles and on the time.
))(/2exp[( EtxphiA x
The wavefunction for a single particle of 1-D motion is:
For example:
)/(2exp[ vtxiA The wavefunction of plane monochromatic light:
A wave function must satisfy 3 mathematical conditions:
1. Single-value; 2. Continuous; 3. Quadratically integrable.
a) The product of wave function (r,t) and its complex conjugate (r,t)* represents the probability distribution function of the system. (Physical meaning of wave function!)
b) The wave function (r,t) must be continuous in space. Otherwise its second derivative would not be attainable.
c) The wave function of a system must be quadratically integrable so as to evaluate the statistical average values of its properties.
dxdydztrtr ),(),(* The probability that the particle lies in the volume element dxdydz, located at r, at time t.
The probability
1 ),(),(*
dxdydztrtr
To be generally normalized
0 ),(),(
dxdydztrtr ji
Wave functions of different states for a given system must be generally orthogonal
Postulate 2. Each observable mechanical quantity of a microscopic system is associated respectively with a linear Hermitian operator.
To find this operator, write down the classical-mechanical expression for the observable in terms of Cartesian coordinates and corresponding linear-momentum, and then replace each coordinate x by the operator, and each momentum component px by the operator –iћ/x.
An operator is a rule that transforms a given function into another function. E.g. d/dx, sin, log
C)BA()CB(A
Operators obey the associative law of multiplication:
d/dxD 5)( 3 xxf23 3)'5()(ˆ xxxfD
f(x)Bf(x)A)f(x)BA(
f(x)Bf(x)A)f(x)BA(
f(x)]B[Af(x)BA
• A linear operator means
• A Hermitian operator means
*111
*1 )ψA(ψψAψ *
122*1 )ψA(ψψAψ
• A Hermitian operator ensures that the eigenvalue of the operator is a real number
2121 ψAψA)ψ(ψA
ψAcψA c
Eigenfunctions and Eigenvalues
Suppose that the effect of operating on some function f(x) with the operator  is simply to multiply f(x) by a certain constant k. We then say that f(x) is an eigenfunction of  with eigenvalue k.
kf(x)f(x)A
2x2x 2e(d/dx)e
Eigen is a German word meaning characteristic.
To every physical observable there corresponds a linear Hermitian operator. To find this operator, write down the classical-mechanical expression for the observable in terms of Cartesian coordinates and corresponding linear-momentum components, and then replace each coordinate x by the operator x. and each momentum component px by the operator
iћ/x.
Mechanical quantities and their Operators
Position x
Momentum (x) px
Angular Momentum (z) Mz=xpy-ypx
Kinetic Energy T=p2/2m
Potential Energy V
Total Energy E =T+V
Some Mechanical quantities and their Operators
Mechanical quantities Methematical Operator
22
2
2
2
2
2
2
2
2
2
m8-)
z
y
x(
m8-T
hh
V )z
y
x
(m8
- H2
2
2
2
2
2
2
2
h
xx2-px
iih
)x
-y
(2
-Mz
yxih
xx
VV
Hamiltonian
If a system is in a state described by a normalized wave function , then the average value of the observable A corresponding to operator  is given by –
dAa
ˆ *
The average value of the physical observable
n
nnn
nnn
nn
a
da
da
dAa
*
*
ˆ*
Thus the only value we measure is the value an.
If the wave function is an eigenfunction of Â, with eigenvalue an,
then a measurement of the observable corresponding to Â
will give the value an with certainty.
n
nn
a
dAa
2
22
ˆ*
0
-
-
22
222
nn
a
aa
aa
• When two operators are commutable, their corresponding mechanical quantities can be measured simultaneously.
Commuted operators ( 对易算符 )
0FG-GF]G ,F[
Poisson bracket
Postulate 3: The wave-function of a system evolves in time according to the time-dependent Schrödinger equation
Assumption 3: The wave-function of a system evolves in time according to the time-dependent Schrödinger equation -
ttzyxH
i ),,,(ˆ
In general the Hamiltonian H is not a function of t, so we canapply the method of separation of variables.
f(t) z)y,(x, ),,,( tzyx
t
zyxtzyx-iE
e ),,( ),,,(
Edt
tdf
f(t)
H
)(1i
z)y,(x,
z)y,(x,ˆ
dt
tdff(t)H
)(z)y,(x,i z)y,(x,ˆ
z)y,(x,z)y,(x,ˆ EH
Edt
tdf
f(t)
)(1i
Time-independent Schrödinger’s Equation
V )z
y
x
(m8
- H2
2
2
2
2
2
2
2
h
VTH ˆˆˆ
xx2-Px
iih
m
PT
2
ˆˆ
2
2
04
ZeV
r
e.g. H atom
EH ˆ Eigenvalue equation
2z
2y
2x
2 PPPP
m
pmvT
22
1 22
Spherical polar coordinates
2
2
2222
22
sin
1)(sin
sin
1)(
1
rrr
rrr
The Schrödinger equation EH ˆ
e-n2
e2
2
V m8
- H h
2
2
22
2
2
2
2
2 12
rrrryxa
2
2
22
2
2
2 1
ayx
Ema 2
2
2
2
2 ,3,2,1,
8
)(1
),(1
22
22
21
21
nma
nhEE
nCosnSin
The Schrödinger’s Equation is an eigenequation.
aψψ A
I. The eigenvalue of a Hermitian operator is a real number.
****A ψaψ
ψψaψψ **A
**** )A( ψψaψψ *aa
Proof:
Quantum mechanical operators have to have real eigenvalues
In any measurement of the observable associated with the operator A, the only values that will ever be observed are the eigenvalues a, which satisfy the eigenvalue equation.
0 * ijji d
II. The eigenfunctions of Hermitian operators are orthogonal
II. The eigenfunctions of Hermitian operators are orthogonal
Consider these two eigen equations
mmm ψaψ A
nnn ψaψ A
Multiply the left of the 1st eqn by m* and integrate, then take the complex conjugate of eqn 2, multiply by n and integrate
a ˆ *n
* ddA nmnm
a ˆ **m
** ddA mnmn
There are 2 cases, n = m, or n m
0**)a - (a
**ˆ ˆ*
mn
d
dAdA
nm
mnnm
0 **)a - (a mn dnm
Subtracting these two equations gives -
If n = m, the integral = 1, by normalization, so an = an*
If n m, and the system is nondegenerate (i.e.different eigenfunctions do not have the same eigenvalues, an am ), then
0 *)a - (a mn dnm
0 * dnm
The eigenfunctions of Hermitian operators are orthogonal
0 * ijji d
Example:
)2(32
1)(
0
2/
30
20
a
re
aH ar
s
oar
s ea
H /
30
1
1)(
0 02 / / 2 2
1 2 3 0 0 000
1(2 ) sin
4 2r a r a
s s
rd e e r drd d
aa
dra
rre
aa
r
0
0
22
3
30
)2(24
40
0
0
32
3
0
22
3
30
00[2
1dr
a
redrre
aa
r
a
r
0 0
32 ]81
16
27
16[
2
1dyyedyye yy
ya
r
02
3
1 16 16[ (3) (4)]27 812
0]!381
16!2
27
16[
2
1
Postulate 4 : Superposition Principle ( 态叠加原理 ) If 1, 2,… n are the possible states of a microscopic system (a complete set), then the linear combination of these states is also a possible state of the system.
i
iinn ccccc 332211
The coefficient ci reflects the contribution of wavefunction i to .
Postulate 5 : Pauli’s principle( 泡利不相容原理 ).
Every atomic or molecular orbital can only contain a maximum of two electrons with opposite spins.
Energy level diagram for He. Electron configuration: 1s2
N
S
Hparamagnetic – one (more) unpaired electrons
N
S
Hediamagnetic – all paired electrons
Ene
rgy
1
2
3
0 1 2
n
l
ms = spin magnetic electron spin
ms = ±½ (-½ = ) (+½ = )
The complete wavefunction for the description of electronic motion should include a spin parameter in addition to its spatial coordinates.
•Two electrons in the same orbital must have opposite spins.
•Electron spin is a purely quantum mechanical concept.
Pauli exclusion principle:
Each electron must have a unique set of quantum numbers.
),(),,( sl msmln
The complete wavefunction for the description of electronic motion should include a spin parameter in addition to its spatial coordinates.
Fermions
•Particles that do obey the Pauli Exclusion Principle.
Bosons
•Particles that do not obey the Pauli Exclusion Principle
+ symmetry(Bosons)
- antisymmetry(Fermions))()(
)()(
)(
)(
1,22,1
2
1,2
2
2,1
2,1
qqqq
qqqq
Heatomelectrontwofor
1.4 Solution of free particle in a box – a simple application of Quantum Mechanics
1.4.1 The free particle in a one dimensional box
V(x)= V(x)=
V(x)= 0
x=0 x=l
I
II
III
1. The Schrödinger’s Equation and its solution
V xm8
- H2
2
2
2
d
dh
In area I, III:
EVxm
h
2
2
2
2
8
08 2
2
2
2
mV
h
x
VEVVVh
m
x
)(08
2
2
2
2
1D box
(meaning no particle in area I and III).
II: V=0
Boundary condition and continuous condition: (0)=0, (l)=0
Hence, (0) =Acos0+Bsin0 = A + 0 = 0 => A =0
=Bsinx (B0)
(l) =Bsinx =Bsin l=0, Thus, l=n, =n/l (n=1,2,…)
0
8
2
2
2
2
2
2
xd
d
h
mEset
xBxA
eBeA xixi
sincos
or ''
xm8
- V xm8
- H2
2
2
2
2
2
2
2
d
dh
d
dh
H E
E xm8
-2
2
2
2
d
dh0
82
2
2
2
E
h
m
xd
d
...)3,2,1(8
8
2
22
2
22
2
22
nml
hnE
l
n
h
mE
xl
nB
sin
Normalization of wave-function:
xxdxx 2sin4
1
2
1sin 2
xl
n
l
sin2
lB
21
2
1)(
2
1 20
2 ln
l
l
nBx
n
l
l
nB l
1sin0
22 dxxl
nB
l
1 2
0 dx
l
2. The properties of the solutions
a. The particle can exist in many states
b. quantization energy
c. The existence of zero-point energy. minimum energy (h2/8ml2)
d. There is no trajectory but only probability distribution
e. The presence of nodes
2
2
1 8ml
hE
l
x
l
sin2
1
2
2
2 8
4
ml
hE l
x
l
2sin
22
2
2
3 8
9
ml
hE
l
x
l
3sin
23
n=1
n=2
n=3
Ene
rgy
E2
E3
E4
E1
n=2
• In the ground state (n=1), the highest probability of the particle occurs at the location l/2.
• In the first excited state (n=2), the highest probability of the particle occurs at the locations l/4 and 3l/4, the lowest probability at the location l/2.
nodenode
Discussion:
i. Normalization and orthogonality
0sinsin2
)()(00
ll
mn dxl
xm
l
xn
ldxxx
ii. Average value
2sinsin
20
ldx
l
xnx
l
xn
lx
l
3sinsin
2 2
0
22 ldx
l
xnx
l
xn
lx
l
0ˆ0
*
0
* l
nn
l
nnx dxdx
didxPp
2
222
2
22
2
22
22
44sin
2
4
ˆ
l
hncp
l
hn
l
xn
ll
hn
cdx
dicP
iii. Uncertainty
3222 l
xxx
l
nhppp
222
34232
nh
l
nhlpx
2
)(1
hpx
stategroundnwhen
a. Obtain the potential energy functions followed by deriving the Hamiltonian operator and Schrödinger equation.
b. Solve the Schrödinger equation. (obtain n and En)
c. Study the characteristics of the distributions of n (e.g., boundary conditions)
d. Deduce the values of the various physical quantities of each corresponding state.
The general steps in the quantum mechanical treatment:
3. Quantum leaks --- tunneling
V
x0 l
I II III),0(
8
)0(8
2
2
2
2
2
2
2
2
lxxExm
h
and
lxEVxm
h
The probability of penetration is given by
lEVmeVEVEP
)(22
)]/(1)[/(4
When E<V
• In classical mechanics, all particles with E > V can pass through the barrier of potential V, whereas the particles with E<V can not pass through the potential .
• This is not the case in quantum mechanics.
P decreases exponentially with increasing V and its width AWA m!!
Tunneling
CLASSICAL MECHANICS
QUANTUM MECHANICS
Tunneling in the “real world”
• Tunneling is widely exploited:
- for the operation of many microelectronic devices (tunneling diodes, flash memory, …)
- for advanced analytical techniques (scanning tunneling microscope, STM)
• Responsible for radioactivity (e.g. alpha particles)
Tip
Piezo-Tube
STM System
Mode: Constant Current mode, Constant height mode
zEmKeP
)(22
C28H588
Free electrons of metals can tunnel between the surfaces of two metals of atomic distances (~ 1 nm) driven by a suitable bias voltage.
1.4.2 The free particle in a three dimensional box
Let = (x, y, z)= X (x) Y (y) Z (z) (separation of variables)Substituting into 3-D Schrödinger equation
Out of the box, V(x, y, z) = ∞ In the box, V(x, y, z) =0
Particle in a 3-D box of dimensions a, b, c
zz
by
ax
0
0
0
E
m
h 2
2
2
8
Ezyxm
h
)(8 2
2
2
2
2
2
2
2
EXYZXYZzyxm
h
Ezyxm
h
)(8
)(8
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
(separation of variables)
XEXxm
hx
2
2
2
2
8YEY
ym
hy
2
2
2
2
8ZEZ
zm
hz
2
2
2
2
8
E=Ex+Ey+Ez
Let Ez = E – (Ex+Ey)
EXYZz
ZXY
y
YXZ
x
XYZ
m
h
)(8 2
2
2
2
2
2
2
2
xEzZ
Z
yY
Y
m
hE
xX
X
m
h
)(88 2
2
2
2
2
2
2
2
2
2
c
zn
b
yn
a
xn
abcXYZ zyx sinsinsin
8
)(8 2
2
2
2
2
22
c
n
b
n
a
n
m
hEEEE zyx
zyx
b
yn
byY ysin
2)(
a
xn
axX xsin
2)(
c
zn
czZ zsin
2)(
The solution is:
c
zn
b
yn
a
xn
abcXYZ zyx sinsinsin
8
)(8 2
2
2
2
2
22
c
n
b
n
a
n
m
hEEEE zyx
zyx
2
22
8=
a
n
m
hE x
x
2
22
8=
b
n
m
hE
y
y
2
22
8=
c
n
m
hE z
z
( n= 1, 2, ……)
Multiply degenerate energy level when the box is cubic
(a = b = c)
)(8
)(8
2222
2
2
2
2
2
2
22
zyxzyx
zyx nnnma
h
c
n
b
n
a
n
m
hEEEE
The ground state: nx=ny=nz=1 2
2
8
3
ma
hE
The first excited state: ni=nj=1, nk=2
The wave-functions are called degenerate (triply degenerate)
2
2
8
6
ma
hE
112
121
211
E
m-dimensional box (m=1-3)• Wave functions
)1,...( sin2
,1
i mil
xn
l i
ii
i
m
ii
)1,...( 8
;2
2
1
miml
hnEEE
i
ii
m
ii
• Energy
1.4.3 Simple applications of a one-dimensional potential box model
Example 1: The delocalization effect of 1,3-butadiene
Four electron form two localized bonds
Four electron form a 44
delocalized bonds
>
l l l
E1
C CC C
E4/9
E1/9
C CC C
E=2×2 ×h2/8ml2=4E1E=2×h2/8m(3l )2+ 2×22 ×
h2/8m(3l )2 =(10/9)E1
Example 2: The adsorption spectrum of cyanines
R2N-(CH=CH-)m-CH=NR2+
..
Total electrons: 2m+4
In the ground state, these electrons occupy m+2 molecular orbitals
The adsorption spectrum correspond to excitation of electrons from the highest occupied (m+2) orbital to the lowest unoccupied (m+3) orbital.
)52(8
])2()3[(8 2
222
2
2
mlm
hmm
lm
hE
ee
The general formula of the cyanine dye:
E
n=1n=2
n=m+2n=m+3
Table 1. The absorption spectrum of the cyanine dye
R2N-(CH=CH-)mCH=NR2+
..
m max (calc) / nm max (expt) /nm
1 311.6 309.0
2 412.8 409.0
3 514.6 511.0
)52(8
])2()3[(8 2
222
mlm
hmm
lm
h
h
Ev
ee
)( 565248 pmml )(52
3.3
)52(
8 22
pmm
l
mh
clme
More Examples
1. Are eim and cos(m) eigenfunctions of operator id/d ? If so, please determine the eigenvalue.
imimim meimiee
d
di
So eim is an eigenfunction of operator id/d with an eigenvalue of –m.
mcmimmmimd
di cossin)sin(cos
So cosm is not an eigenfunction of operator id/d.
2. For the -conjugate molecule CH2CHCHCHCHCHCHCH2, its UV-vis spectrum shows the first long-wavelength absorption at 460 nm. Please estimate the length of its carbon chain using the 1D box model.
)8...3,2,1(8
2
22
nml
hnEn
hc
ml
h
ml
hEEE
2
2
2
222
45 8
9
8
)45(
1-4 are occupied and 5-8 are unoccupied in its ground state. The energy of the first excitation is
Ans: It has 8 p-electrons forming a delocalized 88 bond.
Energies of -MOs given by 1D box model are
pmmc
h
mhc
hl 1120
8
9
8
9 2
3. Does the following function represent a state of particle in a 1-dimensional box?
a
x
aa
x
ax
2sin
23sin
22)(
If yes, does it have a certain value of energy and what is the energy of this state? If not, determine the average energy.
Ans: Two wave functions of a particle in a 1-d box of length a are
a
x
ax
a
x
ax
2sin
2 )( ;sin
2)( 21
)(3)(2)( 21 xxx
So (x) represents a possible state of a 1d-box particle, according to the superposition principle.
)()(3)(2
)(ˆ3)(ˆ2)](3)(2[ˆ)(ˆ
2211
2121
xcxExE
xHxHxxHxH
So (x) is not an eigenfunction of Hamiltonian and has not a certain energy.
To evaluate <E>, (x) must be normalized.
)()(' xcx Suppose , then
1132
sin2
3sin2
2
)()()('
2
0
2
2
0
22
0
2
0
2
cdxa
x
aa
x
ac
dxxcdxxcdxx
a
aaa
13
12 c
2
2
2
22
0
2
2
2
2
2
0
13
55
2sin
23sin
22ˆ2
sin2
3sin2
2
8ˆ ,)('ˆ)'*(
ma
h
ma
hc
dxa
x
aa
x
aH
a
x
aa
x
ac
dx
d
m
hHdxxHxE
a
a
What is Quantum Mechanics?
QM is the theory of the behavior of very small objects (e.g. molecules, atoms, nuclei, elementary particles, quantum fields, etc.)
One of the essential differences between classical and quantum mechanics is that physical variables that can take on continuous values in classical mechanics (e.g. energy, angular momentum) can only take on discrete (or quantized) values in quantum mechanics (e.g. the energy levels of electrons in atoms, or the spins of elementary particles, etc).