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

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

AtomsMolecules

ClustersCongeries

Nano materials

Bulk materials

Geometric Structure

Sizemakes the difference

Electronic Structure

Structure determines propertiesProperties reflects 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 Sciencein Surface Science

fcc(100) fcc(111)

fcc(775) fcc(10 8 7)

Surface structures of Pt single crystal

(111)

(775)

(100)

(10 8 7)

Different surfaces do different chemistry

418

1

25

Surface Structure vs. Catalytic Activity

N2 + 3H2 ⎯→ 2NH3

Fe single crystal，20atm/700K

Role of Structural Chemistry Role of Structural Chemistry in Material Sciencein Material Science

Graphite & Diamond StructuresDiamond: Insulator or wide bandgap

semiconductor: →→→ →→→Graphite: Planar structure: →→→sp2 bonding ≈ 2d metal (in plane)

Other Carbon Crystal Structures“Buckyballs” (C60) →→→→→→“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 C50Cl10SCIENCE 304 (5671): 699-699 APR 30 2004

Role of Structural Chemistry Role of Structural Chemistry in Life Sciencein 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.

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 The basic knowledge of quantum mechanicsChapter 2 Atomic structureChapter 3 SymmetryChapter 4 Diatomic moleculesChapter 5/6 Polyatomic structuresChapter 7 Basics of CrystallographyChapter 8 Metals and AlloysChapter 9 Ionic compounds

OutlineOutline

Chapter 1 The basic knowledge of quantum mechanics

1.1 The failures of classical physics

• Classical physics: (prior to 1900)

Newtonian classical mechanicsMaxell’s theory of electromagnetic wavesThermodynamics and statistical physics

1.1.1 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)

1.1.2 The photoelectric effect

The photoelectric effect

The Photoelectric Effect

1. The kinetic energy of the ejected electrons depends 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).

– won the Nobel prize in 1921

1. Light consists of a stream of photons. The energy of a photon is proportional to its frequency.

ε = hν h = Planck’s constant

2. A photon has energy as well as mass. m= hν /c2

3. A photon has a definite momentum. p=mc= hν /c=h/λ

4. The intensity of light depends on the photon density

Therefore, the photon’s energy is equaled to the electron’s kinetic energy added to the electron’s binding energy

• Ephoton = E binding + E Kinetic energy

• hν=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 =hν = (6.626 x 10 -34Js)(9.73 x 109/s) = 6.447098 x 10 -24J

Ephoton = 6.45 x 10 - 24 J

Egamma ray =hν = ( 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 = hν).

Example II: Calculation of Energy from Wavelength

Problem: What is the photon energy of 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 = hν = (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=hν 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 = hν = (6.626 x10 - 34Js)(4.3 x1014 /s)E red = 2.8 x 10 - 19 J

• E blue = hν = (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 spetra

The Line Spectra of Several Elements

• The electron are like planets --- orbit the nucleus

• Light of energy E given off when electrons change orbits (i.e., different energies)

Why do the electrons not fall into the nucleus?

Why only discrete energies?

Planetary model:

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 = −RZ 2

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

21n2

2

Ephoton = -2.18 x 10 -18J - = - 4.09 x 10 -19J12 2

14 2

λ = = h x 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

Black-Body Radiation

Lesson 1 Summary

1.1 The failures of classical physics

The photoelectric effectA corpuscular theory of light (photons)ε = hν h = Planck’s constantp=h/λ

Atomic and molecular spetra

1.2 The characteristic of the motion of microscopic particles

1.2.1 The wave-particle duality of microscopic particles

In 1924 de Beoglie suggested that microscopic particles might have wave properties.

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

De Broglie considered that the wave-particle relationship in light is also applicable to particles of matter, i.e.

The wavelength of a particle could be determined by

λ= h/p = h/mv

E=hν

p=h/λ

h = Planck’s constant,

p = particle momentum,

λ = 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

λ = = h

mv6.626 x 10 - 34Js

( 9.11 x 10 - 31kg )( 1.00 x 106 m/s )

Wavelength λ = 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

−

−−

−

×=

××××

×=

==λ

1eV=1.602x10-19 J

The de Broglie Wavelengths of Several particles

Particles Mass (g) Speed (m/s) λ (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

• Wave (i.e., light)

- can be wave-like (diffraction)

- can be particle-like (p=h/λ)

• Particles

- can be wave-like (λ =h/p)

- can be particle-like (classical)

The wave-particle duality

For photon:

p=mc,

E=hν=h (c/λ) = pc = mc2

Discussions:

For particles:

p=mv,

E=(1/2) mv2 = p2/2m=pv/2

E=h ν

p = h/λ

λ = u / ν … u is not the moving speed of particles.

≠ p2/2m =(1/2) mv2

≠ pv

1.2.2 The uncertainty principle

hpx

hpxD

hD

hDp

ppp

pp

DDOAOC

APOP

xo

x

≥ΔΔ

=ΔΔ

===

==Δ=

===

=−

//

)0(sin

sin

/2

1/

2

1/sin

2

1

λλ

λ

θθ

λλθ

λ

Include higher order,

y

e OA

P

Q

C

x

e OA

P

Q

C

x

e OA

P

Q

C

x

psinθ

A

O

Cθθ

P

θ psinθ

A

O

Cθθ

P

θ psinθ

A

O

Cθθ

P

θ

h2

1

4or

hpx

π≥ΔΔ

A quantitative version

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.

ExampleThe 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

Δp = p x 0.001% = 9.11 x 10-33 kg m/s

Δx = h / Δp = 6.626 x 10-34 / (9.11 x 10-33 )

= 7.27 x 10-2 (m)

ExampleThe 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

Δp = p x 0.001% = 1 x 10-4 kg m/s

Δx = h / Δp = 6.626 x 10-34 / (1 x 10-4 )

= 6.626 x 10-30 (m)

ExampleThe average time that an electron exists in an excited

state is 10-8 s. What is the minimum uncertainty in energy of that state?

h ≥ΔΔ 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

−

−

−

×=××

==

=Δ=Δ tE h

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 exhibita 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”.

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 continuousvalues 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).

Lesson 2 Summary

• Wave (i.e., light)

- can be wave-like (diffraction)

- can be particle-like (p=h/λ)

• Particles

- can be wave-like (λ =h/p)

- can be particle-like (classical)

The wave-particle duality

The uncertainty principle

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 ψ(r, t) that depends onthe coordinates of the particle and on the time.

))(/2exp[( EtxphiA x −= πψ

The wavefunction Ψ for a single particle of 1-D motion is:

For example:

)/(2exp[ vtxiA −= λπψThe wavefunction of plane monochromatic light:

A wave function must satisfy 3 mathematical conditions:

1. Single-value2. Continuous3. Quadratically integrable.

dxdydztrtr ),(),(* ψψ The probability that the particle lies in the volume elementdxdydz, located at r, at time t.

The probability

1 ),(),(* =∫ ∫ ∫∞

∞−

∞

∞−

∞

∞−

dxdydztrtr ψψ

To be generally normalized

Postulate 2. For every observable mechanical quantity of a microscopic system, there is a corresponding linear Hermitian operator associated with it.

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 corrdinate 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 Âwith eigenvalue k.

kf(x)f(x) ≡A

2x2x 2e(d/dx)e =

Eigen is a German word meaning characteristic.

Examples

τψψτψψτψψ

τψψτψψτψψ

dxdxdx

dxdxdx

∫∫∫∫∫∫

∞∞−

∞∞−

∞∞−

∞∞−

∞∞−

∞∞−

==

===

*12

*12

*12

2*12

*12

*1

ˆ)(

ˆˆ

Examples for calculation

operatorsHermitonNotdx

d

x

operatorsHermitonpx x

,

ˆ,ˆ

∂∂

*1

*1

*21

*22

*1

2*12

*12

*1

ˆ]|[

)(ˆ

ψψτψψτψψψψ

τψψτψψτψψ

x

x

pdx

hidx

hi

dx

hidx

hidp

∫∫∫

∫∫∫∞

∞−∞

∞−∞

∞−∞

∞−

∞∞−

∞∞−

∞∞−

=∂∂

=∂∂

−−

=∂∂

−=∂∂

−=

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 ∇=

∂∂

+∂∂

+∂∂

=ππhh

V )z

y

x

(m8

- H2

2

2

2

2

2

2

2

+∂∂

+∂∂

+∂∂

=πh

xx2-p

∂∂

−=∂∂

= hiih

π

)x

-y

(2

-Mz ∂∂

∂∂

= yxih

π

xx =

VV =

If a system is in a state described by a normalized wave function ψ, then the average value of the observable corresponding to A is given by –

τdAa ΨΨ= ∫∞

∞−ˆ *

The average value of the physical observable

Exercise :

Suppose a particle in a box is in a state -

Note that the wave function ψ(x) is not an eigenfunction for a particle in a box. Sketch ψ(x) vs. x and show that y(x) isnormalized. Calculate the average energy associated with thisstate. (Assume V = 0).

otherwise 0

ax 0 )()a

30( )( 2

1

5

=

≤≤−=Ψ xaxx

22

2

4

5

ma

h

π

xm8

-V xm8

- H2

2

2

2

2

2

2

2

d

dh

d

dh

ππ=+=

dxHEa

ΨΨ= ∫0

* ˆ

0

-

-

*

*

*

*

22

222

2

22

==

=

=

ΨΨ=

=

ΨΨ=

ΨΨ=

ΨΨ=

∫

∫∫∫

∞

∞−

∞

∞−

∞

∞−

∞

∞−

nn

a

n

nn

n

nnn

nnn

nn

aa

aa

a

dAa

a

da

da

dAa

σ

τ

τ

τ

τ

Thus the only value we measure is the value an.

If the wave function is an eigenfunction of A, with eigenvalue an,

then the a measurement of the observable corresponding to A

will give the value an with certainty.

• When the two operators commute, their corresponded mechanical quantities can be measured simultaneously.

ˆ ˆ, 0F G FG GF⎡ ⎤ = − =⎣ ⎦) )) )

Commuted operators

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

∂Ψ∂

=Ψ hi ),,,(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, ),,,( ψ=Ψ tzyx

h

t

zyxtzyx-iE

e ),,( ),,,( ψψ =

Edt

tdf

f(t)

H==

)(1i

z)y,(x,

z)y,(x,ˆh

ψψ

dt

tdff(t)H

)(z)y,(x,i z)y,(x,ˆ ψψ h=⋅

z)y,(x,z)y,(x,ˆ ψψ EH =

Edt

tdf

f(t)=

)(1ih

Time-independent Schrödinger’s Equation

V )z

y

x

(m8

- H2

2

2

2

2

2

2

2

+∂∂

+∂∂

+∂∂

=πh

VTH ˆˆˆ += xx2-p

∂∂

−=∂∂

= hiih

πm

pmvT

22

1 22 ==

2

04

ZeV

rπε

Λ

= e.g. H atom

ψψ EH =ˆ Eigenvalue equation

The Schrödinger’s Equation is eigenvalue equation.

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 theeigenvalues a, which satisfy the eigenvalue equation.

ijji dx δψψ * =∫

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 ψm* and integrate, then take the complex conjugate of eqn 2, multiply by ψn and integrate

∫∫ = a ˆ *n

* dxdxA nmnm ψψψψ

∫∫ = a ˆ **m

** dxdxA mnmn ψψψψ

There are 2 cases, n = m, or n ≠ m

0**)a - (a

**ˆ ˆ*

mn ==

−

∫∫∫

dx

dxAdxA

nm

mnnm

ψψ

ψψψψ

0 **)a - (a mn =∫ dxnm ψψ ≠

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 have the same eigenvalues, an ≠ am ), then

0 *)a - (a mn =∫ dxnm ψψ

0 * =∫ dxnm ψψ

The eigenfunctions of Hermitian operators are orthogonal

ijji dx δψψ * =∫

Example:

)2(32

1

0

2/

30

20

a

re

a

ars −= −

πϕoar

s ea

H /

30

1

1)( −=

πϕ

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

π πϕ ϕ τ θ θ φ

π

+∞ ∞ − −

−∞= −∫ ∫ ∫ ∫

dra

rre

aa

r

∫∞ −

−=0

0

22

3

30

)2(24

40

ππ

∫∫∞ −∞ −

−=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

= ⋅ Γ − Γ 0]!381

16!2

27

16[

2

1=⋅−⋅=

Postulate 4 : If ψ1, ψ2,… ψn 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.

Assumption 4 : If ψ1, ψ2,… ψn 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.

∑=+⋅⋅⋅++=Ψi

iinn ccccc ψψψψψ 332211

Postulate 5 : Pauli’s principle. Every atomic or molecular orbital can only contain a maximum of two electrons with opposite spins.

N

S

H

N

S

He

Energy level diagram for He. Electron configuration: 1s2

paramagnetic – one (more) unpaired electrons

diamagnetic – all paired electrons

Ene

rgy

1

2

3

0 1 2

n

l

ms = spin magnetic → electron spin

ms = ±½ (-½ = α) (+½ = β)

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 χ⋅Ψ=Φ

+ symmetry(Bosons)

- antisymmetry(Fermions)

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

)()(

)()(

)(

)(

1,22,1

2

1,2

2

2,1

2,1

qqqq

qqqq

qq

Heatomelectrontwofor

φφ

φφ

φ

±=

=

−

Lesson 3 Summary

The basic assumptions (postulates) of quantum mechanics

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

π

I, III:

ψψψπ

EVxm

h=+

∂∂

−2

2

2

2

8

08 2

2

2

2

=⋅∂∂

=mV

h

x πψψ

VEVVVh

m

x=−∴∞==−

∂∂

)(08

2

2

2

2

Qψπψ

II: V=0

08

2

2

2

2

=+∂∂ ψπψ

Eh

m

x

xixe

xixexi

xi

αα

ααα

α

sincos

sincos

−=

+=−

Boundary condition and continuous condition: ψ(0)=0, ψ(l)=0

Hence, ψ(0) =Acos0+Bsin0

A=0, B≠0 ψ=Bsinαx

ψ (l) =Bsinαx =Bsin αl=0, Thus, αl=nπ,

0

8

2

2

2

2

2

2

=+∂∂

=

ψαψ

απ

x

h

mEset

xBxA

eBeA xixi

ααψψ αα

sincos

''

+=+= −

xixixi

xixixi

edx

dei

dx

de

edx

dei

dx

de

ααα

ααα

αα

αα

−−− −=Ψ

−=Ψ

=Ψ

−=Ψ

=Ψ

=Ψ

22

22

22

22

12

11

...)3,2,1(8

8

2

22

2

22

2

22

==

==

nml

hnE

l

n

h

mE παπx

l

nB

πψ sin=

Normalization of wave-function:

xxdxx 2sin4

1

2

1sin2 −=∫

xl

n

l

πψ sin2

=

lB

2=1

2

1)(

2

1 20

2 =⋅⋅⋅=⋅ ln

l

l

nBx

n

l

l

nB l

ππ

ππ

1sin0

22 =∫ dxxl

nB

l π

12

0=∫ dx

lψ

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

πψ sin2

1 =

2

2

2 8

4

ml

hE = l

x

l

πψ 2sin

22 =

2

2

3 8

9

ml

hE = LL

l

x

l

πψ 3sin

23 =

n=1

n=2

n=3

Discussion:

i. Normalization and orthogonality

0sinsin2

)()(00

== ∫∫ll

mn dxl

xm

l

xn

ldxxx

ππψψ

ii. Average value

2sinsin

20

ldx

l

xnx

l

xn

lx

l=>=< ∫

ππ

3sinsin

2 2

0

22 ldx

l

xnx

l

xn

lx

l=>=< ∫

ππ

0>=< p

2

222

4l

hnp >=<

ada

dadAa

*

* ˆ*

=ΨΨ=

ΨΨ=ΨΨ=

∫∫∫

∞

∞−

∞

∞−

∞

∞−

τ

ττ

If the wave function is aneigenfunction of Â

iii. Uncertainty

3222 l

xxx =><−><=Δ

l

nhppp

222 =><−><=Δ

34232

nh

l

nhlpx ==ΔΔ

π2

)(1

hpx

stategroundnwhen

≈ΔΔ

=

a. Obtain the potential energy functions followed by deriving the Hamiltonian operator and Schrödinger equation.

b. Solve the Schrödinger equation. (obtain ψn and En)

c. Study the characteristics of the distributions of ψn.

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

><=∂∂

−

<<=+∂∂

−

ψψπ

ψψψπ

The probability of penertration is given by

lEVmeVEVEP

)(22

)]/(1)[/(4−−

−≈ h When E<V

Tunneling

CLASSICAL MECHANICS

QUANTUM MECHANICS

Tunneling in the “real world”

• Tunneling is used:

- 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 high mode

zEmKeP

)(22

−Φ−≈ h

C28H588

1.4.2 The free particle in a three dimensional box

Let ψ = ψ(x, y, z)= X (x) Y (y) Z (z) (separation of variables)Substituting into 3-D Schroedinger 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

0ψψ

πE

m

h=∇− 2

2

2

8

ψψπ

Ezyxm

h=

∂∂

+∂∂

+∂∂

− )(8 2

2

2

2

2

2

2

2

xEXYZXYZzyxm

h

Ezyxm

h

=∂∂

+∂∂

+∂∂

−

=∂∂

+∂∂

+∂∂

−

)(8

)(8

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

π

ψψπ

(separation of variables)

XEXxm

hx=

∂∂

−2

2

2

2

8πYEY

ym

hy=

∂∂

−2

2

2

2

8πZEZ

zm

hz=

∂∂

−2

2

2

2

8π

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

π

xEzZ

Z

yY

Y

m

hE

xX

X

m

h=

∂∂

+∂

∂+=

∂∂

− )(88 2

2

2

2

2

2

2

2

2

2

ππ

c

zn

b

yn

a

xn

abcXYZ zyx πππψ sinsinsin

8==

)(8 2

2

2

2

2

22

c

n

b

n

a

n

m

hEEEE zyx

zyx ++=++=

b

yn

byY yπsin

2)( =

a

xn

axX xπsin

2)( =

c

zn

czZ zπsin

2)( =

The solution is:

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

1.4.3 Simple applications of a one-dimensional potential box model

Example 1: The delocalization effect of 1,3-butadiene

Four π electron form two π localized bonds

Four π electron form a π44

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-)mCH=NR2

+..

Total π 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

+=+−+=Δ 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 λmax (calc) / nm λmax (expt) /nm

1 311.6 309.0

2 412.8 409.0

3 514.6 511.0

)52(8

])2()3[(8 2

222

+=+−+=Δ

= mlm

hmm

lm

h

h

Ev

ee

)(565248 pmml +=)(52

30.3

)52(

8 22

pmm

l

mh

clme

+=

+=λ