chapter2 introduction to quantum mechanics

Microelectronics I

Chapter 2: Introduction to Quantum Mechanics

2.1 Principles of Quantum Mechanics

2.2 Schrodinger’s Wave Equation

2.3 Applications of Schrodinger’s Wave equation

2.4 Extensions of the Wave Theory to Atoms

Microelectronics I : Chapter 2

Control electron in the solid (crystal)

�Position

�Velocity � device’s speed

I ∝ n x q x vCurrent,

Introduction 1

Need to know electron behavior in the crystal and

the material (energy band, etc)

�Velocity � device’s speed

�No. of electronDensity of electron velocity

Introduction of quantum mechanics (Defines electron with wave function)

� Schrodinger equation


�Quantum mechanic becomes more significant as electronic device becomes smaller

Appearance of “quantum effect”

Introduction 2

Current, I

Voltage, V

Current, I

Width, W

I

Ohm’s law

“classical”

Smaller W I

?

Electron channel

V

“classical”

V

�Here, classical Physics no longer applicable !!


Objective:

To understand the basic of quantum mechanics

�Wave-particle duality

�Schrodinger equation

- equation

- physical meaning

�Application:

-quantized energy

-tunneling effect

Chapter 3: energy band theory of solids


propagates as wave (frequency, ν)Particle having energy, hν

*h, Planck's constant = 6.625 x 10-34 Js

light

1905, Einstein�Interference 1905, Einstein

“Photon (discrete packet)”

Explains the photoelectric effect

�Interference

�Refraction

�diffraction

light photoelectron

Max kinetic energy of photoelectron

Wave-Particle duality

frequencyνo


Wave-Particle duality of electron

�Electron: charged particle (q=1.6 x 10-19 C)

�De Broglie (1924)

I ∝ n x q x vex:

�De Broglie (1924)

Particle with momentum, p has wavelength, λ

λ = h/p P: Planck’s constant

Ex:

e

Velocity, v=105 m/s

Wave nature

e

λ = h/p =h/m.v

=6.625 x 10-34/(105 x 9.11 x 10-31)

= 7.27 nm


Electron gun

Double slit screen

e

Wave-particle duality : experiment

When electron hits the screen, a dot will appears

Particle naturee

① Shoot electron 4 times

Double slit screen

Particle nature

1st�Electron could go through the slit

Electron gun

e 2nd

3rd

4th�Electron could go through the slit�Position of electron was random


② Shoot electron many time

Electron gun

slit screen

e

http://www.hitachi.com/rd/research/em/doubleslit.html

e

slit screena: 8 electrons, b: 270 electrons, c: 2000 electrons,d: 160,000 electrons

�Interference pattern

Wave nature

d: 160,000 electrons

Interpretations


1. Electron propagates in space like wave2. Each electron pass through both two open slit at the same time3. Electron interfere with itself

Electron gun

Double slit screen

e

e

The experimental results confirm the wave-particle duality of electron


The Uncertainty principle (Heisenberg 1927)

∆p∆x ≥ ħ

①

Impossible to simultaneously describe with absolute accuracy the position and momentum of a particle

∆p∆x ≥ ħ

∆E∆t ≥ ħ

② Impossible to simultaneously describe with absolute accuracy the energy of particle and the instant of time the particle has this energy

∆p: uncertainty in momentum

∆x : uncertainty in position

∆E∆t ≥ ħ∆E: uncertainty in energy

∆t : uncertainty in time

�Cannot determine exact position of electron

use “probability”


Schrodinger Wave Equation

Describe and discuss electron behavior

Wave function� sin, cos

k : wave numberΨ(x,t)= a.cos(kx-ωt+ε))

Ψ(x,t)= exp ( j(kx-ωt))= exp( jkx)exp(-jωt)

k : wave numberω : angular momentumε : initial phase

Ψ(x,t)= a.cos(kx-ωt+ε))

Use “exp”

exp(jθ)=cosθ+jsinθ

ε=0

= exp( jkx)exp(-jωt)

Ψ(x,t)= Φ(t)φ(x)

position-dependenttime-dependent

Hφ(x)=Eφ(x)

Time-independent Schrodinger equation

Hamiltonian

Schrodinger Wave Equation2


Energy of electronHamiltonian

Hamiltonian: total energy operator

H = kinetic energy + potential energy

)(2 2

22

xVxm

H +∂

∂−=h

Energy of electron

)()()(2 2

22

xExxVxm

ϕϕ =

+

∂

∂−h


Physical meaning of wave equation

φ(x): wave function

|φ(x)|2: probability of existence of electron at x

position of electron cannot be determined precisely

∫∫+∞

∞−

+∞

∞−

== 1)()(|)(| *2dxxxdxx ϕϕϕ

total probability=1

φ*(x): complex conjugate function

total probability=1


Major quantities: energy, momentum, position of electron

)(

)(

xxi

xH

ϕ

ϕ

∂

∂h

Energy,

Momentum, Result :equation

)(

)(

xx

xxi

ϕ

ϕ∂

Momentum,

Position,

Value of major quantities given by expected value in probability theory

∫+∞

∂* ϕϕh

∫+∞

ϕϕ ∫+∞

dxxxx )()(* ϕϕ

energy momentum position

∫

∫∞+

∞−

∞−

∂

∂

dxxx

dxxxi

x

)()(

)()(

*

*

ϕϕ

ϕϕh

∫

∫∞+

∞−

∞−

dxxx

dxxHx

)()(

)()(

*

*

ϕϕ

ϕϕ

∫

∫∞+

∞−

∞−

dxxx

dxxxx

)()(

)()(

*

*

ϕϕ

ϕϕ

Result :real number


Region I Region II

Boundary Condition

x=a

Condition 1: φ(x) must be finite, single-valued, and continuousCondition 1: φ(x) must be finite, single-valued, and continuousCondition 2: ∂φ(x)/∂x must be finite, single-valued and continuous

ax

II

ax

I

III

xx

aa

==∂

∂=

∂

∂

=

ϕϕ

ϕϕ )()( Condition 1

Condition 2


Basic solution of Schrodinger equation

Consider V: constant

)()(2 2

22

xExVxm

ϕϕ =

+

∂

∂−h

0)()(2)(

22

2

=−

−∂

∂x

EVm

x

xϕ

ϕ

h

--eq.1

Second order differential equation

∂x h

constant


1. if, E < V

0)(2 2

2>=

−α

h

EVm

eq.1ϕαϕ 2"

0=−

Solution: xx

BeAeααϕ −+= A,B: Coefficient

ϕαϕ

ϕαϕ2"

0

=

=−

∞∞

Condition 1( φ must be finite)

xBe

αϕ −=

1. if, E > V

0)(2 2

2>−=

−− β

h

EVm

eq.1ϕβϕ =+


Solution: xixiDeCe

ββϕ −+= C,D: Coefficient

eq.1

ϕβϕ

ϕβϕ2"

2"0

−=

=+

�Wave function is given by the combination of the two type solution


Application 1: Potential well

∞ ∞

region I region II region III

region I and III

V(x)=∞

x=Lx=0

region I region II region IIIV(x)=∞

Electron cannot exist in the regions

region II (0<x<L)

−∂ ϕ

φ=0

……Time-independent equation

V=0

0)(2)(

22

2

=+∂

∂x

mE

x

xϕ

ϕ

h

---eq. 1

0)()(2)(

22

2

=−

−∂

∂x

EVm

x

xϕ

ϕ

h

eq. 1


)(2)(

22

2

xmE

x

xϕ

ϕ

h−=

∂

∂

ϕβϕ 2" −= 2

2 2

h

mE=βϕβϕ −= 2

h

Solution ;xixi

BeAeββϕ −+=

Boundary condition;

0)0()0( ==III

ϕϕ 0)()( == LLIIIII

ϕϕ

0=+ BA 0=+ − LiLiBeAe

ββ...eq. 2 ...eq. 3


From eq. 2 & 3

0)( =− − LiLieeA

ββ

A≠0

0=− −ee

LiLi ββregion II

0)sin(2

0

=

=− −

Li

eeLiLi

β

ββ

n; integer

2nL

mE

nL

=

=

π

πβ

x=Lx=0

En=1

En=2

En=3

2

22

2n

LmE

nL

=

=

π

π

h

hx=Lx=0

The energy of particle is quantized

Particular discrete values

Classical; continuous values


Wave function

xixiBeAe

ββϕ −+=

= x

L

niA

πsin2

=

xL

nC

L

πsin

normalization

1sin0

22 =

∫ dxx

L

nC

L

πTotal probability=1

2

LC

2=

= x

L

n

L

πϕ sin

2n=1,2,3,4……


Corresponding probability functionsWave functions

x=L x=L


Application 2: Potential well


V0

x=Lx=0

eRegion I

0)(2)(

22

2

=+∂

∂x

mE

x

x

I

I ϕϕ

h

Region II Region III

0)(2)(

22

2

=+∂

∂x

mE

x

x

III

III ϕϕ

h0)(

)(2)(2

0

2

2

=−

−∂

∂x

EVm

x

x

II

II ϕϕ

h


Consider

E<V0, ,)(2

2

02

h

EVm −=α 2

2 2

h

mE=β

Solution; wave functionSolution; wave function

xixi

IBeAe

ββϕ −+=

xixi

IIIFeEe

ββϕ −+=

xx

IIDeCe

ααϕ −+=


V0

A

B

C

D

E

F

x=Lx=0F=0

xi

IIIEe

βϕ =


Boundary condition;

)()(

)0()0(

LLIIIII

III

ϕϕ

ϕϕ

=

=Continuous wave function

)()( LLIIIII

ϕϕ =

)()(

)0()0(

''

''

LL IIIII

III

ϕϕ

ϕϕ

=

=Continuous first derivative

4 equation

Can solve for the 4 coefficients B, C, D, E in term of A


Parameter of interest; Transmission coefficient, T


V0

A E *

*

AA

EET

⋅

⋅=

x=Lx=0

AA ⋅

)2exp(116 LEE

T β−

−

≈ )2exp(11600

LV

E

V

ET β−

−

≈

�T is not zero Electron penetrate the barrier

“tunneling”


L

Wave function through the potential barrier


Extensions of the wave Theory to Atoms

Potential function (coulomb attraction)

e2−

=

+Nucleus; positively charged

r

erV

0

2

4)(

πε

−=

“Quantized

approximation

+Quantum well

Expected results

+Quantum well

“Quantized

energy”

= x

L

n

L

πϕ sin

2

n=1,2,3,…


Solving Schrodinger equation

0),,())((2

),,(2

02 =−+∇ φθψφθψ rrVEm

rh

Wave function & energy of electron in the atomWave function & energy of electron in the atom

result

222

0

4

0 1

2)4( n

emE

n

hπε

−=

n; 1, 2, 3,…

(principal quantum number)

1. Energy of atom is quantized. The value is determined by a quantum number,n

2. Wave function of electron also determined by quantum numbers (n, l, m)


�n: principal quantum number

(determine total electron energy)

Quantum number

Quantum states of electron

(determine total electron energy)

N=1, 2, 3,….

�l: azimuthal quantum number

(specifies the shape of atomic orbital)

l= n-1, n-2,……,0 (s,p,d,..)

�m: magnetic quantum number

( direction)

|m|=l,l-1,…0 S- orbital

z

y

x

z

y

xz

z

y

x

|m|=l,l-1,…0

�s: spin quantum number

( spin of electron)

S=1/2, -1/2

S- orbital

y

x

p- orbital


electron

energyex: C (no. of electron: 14)

l=2p+ n=1

n=2

l=1s

l=2sl=2p

n=1

n=2

As n increases, energy of

quantum state increases

m

+

+ n=1

Pauli Exclusion Principle

No two electrons may occupy the same quantum state

quantum state increases

chapter2 introduction to quantum mechanics

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