Brook Abegaz, Tennessee Technological University, Fall 2013 1 - - PowerPoint PPT Presentation

Brook Abegaz, Tennessee Technological University, Fall 2013

Saturday, October 05, 2013 Tennessee Technological University

1

C hapt er 2 – I nt r oduct i on t o Q uant um M echani cs

• Quantum Mechanics:

 Used to understand the current – voltage

characteristics of materials.

 Explains the electron behavior when the material

is subject to various potential functions.

 The behavior and characteristics of sub-atomic

particles can be de described by the formulation

• f quantum mechanics called Wave Mechanics.

Saturday, October 05, 2013 Tennessee Technological University

2

• Principles of Quantum Mechanics

 Three basic principles for Quantum Mechanics.

1. The Principle of Energy Quanta,

• 2. The Principle of Wave-Particle Duality,

3. The Principle of Uncertainty

Saturday, October 05, 2013 Tennessee Technological University

3

Albert Einstein Louis de Broglie Werner Heisenberg

Saturday, October 05, 2013 Tennessee Technological University

4

1.

The Principle of Energy Quanta:

 Photo-Electric Effect experiment:

a.

if monochromatic light is incident on a clean surface

• f a material, electrons or photoelectrons are emitted

from the surface.

b.

At constant incident intensity, the maximum kinetic energy of the photoelectron varies linearly with frequency with a limiting frequency υ = υo below which no photoelectronsare produced.

c.

In 1900, Plank postulated that such thermal radiation is emitted from a heated surface in discrete packets of energy called Quanta whose energy is:

E= h* υ, where h = 6.625x10-34 J* sec,

υ = frequency of incident light

Saturday, October 05, 2013 Tennessee Technological University

5

Saturday, October 05, 2013 Tennessee Technological University

6

1.

The Principle of Energy Quanta:

 In 1905, Einstein interpreted the photoelectric results

by suggesting that the energy in a light wave is also contained in discrete packets or bundles called photon whose energy is E= h* υ.

 Work Function = minimum energy required to remove

an electron.

 If Incoming Energy > Work Function, excess photon

energy goes to the kinetic energy of the photoelectron: T = ½ mv2 = h* υ – Φ = h* υ - h* υo (υ ≥ υo) WhereT = maximum Kinetic Energy of a photoelectron

Φ = h* υo = work function

• 2. Wave-Particle Duality (de Broglie)

 Photo-electric effect = light waves behave as

particles.

 Compton effect = electromagnetic waves act as

particles.

 De Broglie’s experiment = particles also act like

waves with a wavelength λ = h/p where h = Plank’s constant, p = momentum of a particle.

 Wave-Particle Duality principle of Quantum

Mechanics applies to small particles such as electrons and for very large particles, the equations reduce to classical mechanics.

Saturday, October 05, 2013 Tennessee Technological University

7

• 3. The Uncertainty Principle (Heisenberg)

 It is not possible to describe the absolute accuracy

• f the behavior of small particles, including

[position and momentum] and [energy and time].

 It is impossible to simultaneously describe the

position and momentum of a particle accurately.

 It is impossible to accurately describe energy and

time for a particle simultaneously.

 ∆p∆x ≥ ħ where ∆p = uncertainty in momentum, ∆x = uncertainty in position.  ∆E∆t ≥ ħ where ∆E = uncertainty in energy, ∆t = uncertainty in time.

Saturday, October 05, 2013 Tennessee Technological University

8

• Exercise

1.

The uncertainty in position of an electron is ∆x =

8Å where 1Å =0.1nm.

a)

Determine the minimum uncertainty in momentum.

b)

If the nominal value of momentum is p = 1.2x10-

23kgms-1, determine the corresponding uncertainty

in Kinetic energy where the uncertainty in Kinetic energy is ∆E = (dE/dp)∆p = p ∆p/m.

Saturday, October 05, 2013 Tennessee Technological University

9

• Solution

a) Minimum uncertainty in momentum:

∆p∆x ≥ ħ, ∆p ≥ 6.625x10-34Js / (2π*8*0.1*10-9m) ∆p ≥ 1.319*10-25Kgm/s b) Corresponding uncertainty in Kinetic Energy for ∆p ≥ 1.2*10-

23Kgms-1,

∆E = p ∆p/me = 1.2x10-23 Kgms-1 * 1.319x10-25Kgms-1/me ∆E = 1.737x10-18Kgm2s-2 ∆E = (1.737x10-18Kgm2s-2)/(1.6x10-19J) = 10.86eV.

• Exercise
• 2. A proton’s energy is measured with an

uncertainty of 0.8eV.

a) Determine the minimum uncertainty in time

• ver which this energy is measured.

b) Compute the same problem for an electron.

Saturday, October 05, 2013 Tennessee Technological University

10

• Solution

a) Minimum uncertainty in time:

∆E∆t ≥ ħ, ∆t ≥ 6.625x10-34Js / (2π*0.8*1.6*10-19C) ∆t ≥ 8.25*10-16s b) Corresponding uncertainty in time for an electron, ∆t ≥ 8.25*10-16s. (It is the same!)

• Shrödinger’sWave Equation

Incorporates the principle of Quanta introduced by Plank and the Wave-Particle Duality principle Introduced by De Broglie to formulate Wave Mechanics.

• Wave Mechanics = a formulation
• f

Quantum Mechanics that describes the behavior and characterization

• f

electron movement and subatomic particles in materials and devices.

• Wave Equation = a one-dimensional, non-relativistic

Shrödinger’swave equation: (1)

where (x,t) is the wave function, V(x) is the potential function assumed to be independent of time, m is the mass

• f the particle and j is√(-1).

Saturday, October 05, 2013 Tennessee Technological University

11

t t x j t x x V x t x m        ) , ( ) , ( ) ( ) , ( 2

2 2 2

    

• Shrödinger’sWave Equation
• From Wave-Particle Duality Principles

Saturday, October 05, 2013 Tennessee Technological University

12

Therefore, we would like to split equation 1to

a)

A left side that isindependent of time and

b)

A right side that isindependent of position. Dividing by the total wave function gives:

The left side of the equation isa function of position only and the right side

• f the equation is a function of time t, each side must be equal to a constant η:

(2)

) ( ) ( ) , ( t x t x    

t t x j t x x V x x t m        ) ( ) ( ) ( ) ( ) ( ) ( ) ( 2

2 2 2

        t t t j x V x x x m        ) ( ) ( 1 ) ( ) ( ) ( 1 2

2 2 2

      t t t j    ) ( . ) ( 1 .    

• Shrödinger’sWave Equation
• The solution of such equation is:

Saturday, October 05, 2013 Tennessee Technological University

13

Substituting in (2)

(3)

where Eis the total energy of the particle, V(x) is the potential experienced by the particle and m is the mass of the particle.

t j

e t

 





 ) (

t E j

e t

 

 ) (    2 h hv E     E     E x V x x x m      ) ( ) ( . ) ( 1 . 2

2 2 2

   ) ( )) ( ( 2 ) (

2 2 2

     x x V E m x x   

• Physical Meaning of theWave Equation
• The total wave equation is the product of position-dependent, time-

independent function and a time-dependent, position-independent function.

Saturday, October 05, 2013 Tennessee Technological University

14

Max Born in 1926 postulated that the probability of finding a particle between x and x+dx at a given time t is: And

Therefore, the probability density function is Independent of Time. Main Difference between Classical Physics and Quantum Mechanics:  Position of a particle can be determined precisely in Classical Physics;  But in Quantum Mechanics, it can be done so only with a

probability.

t E j

e x t x t x

) (

) ( ) ( ) ( ) , (

 

      ) , ( ). , ( | ) , ( |

* 2

t x t x t x    

t E j

e x t x

) ( * *

). ( ) , (

 

 

2 * 2

| ) ( | ) , ( ). , ( | ) , ( | x t x t x t x      

• Boundary Conditions
• Using the probability density function for a single particle.

Saturday, October 05, 2013 Tennessee Technological University

15

Two additional conditions: Condition 1:Ψ(x) must be finite, single valued and continuous. Condition 2: d Ψ(x) /dx must be finite, single valued and continuous. These conditions are important to have a finite energy E and a finite potential V(x).

 Potential Functions and Corresponding Wave Function solutions

• Fig. a) When the potential function is Finite everywhere,
• Fig. b) When the potential function is Infinite in some regions.

∞ ∞

1 | ) ( |

2





  

dx x 

• Reading Assignment
• Text Book: Semiconductor Physics and

Devices, Basic Principles, Donald A. Neamen

• Finish Reading Chapter 2:

“Introduction to Quantum Mechanics”

• Discussion on that topic is on Friday, 9/6/13.
• A Homework per Two Chapters covered

(expect a homework after next class, Its due date would be in two weeks.)

Saturday, October 05, 2013 Tennessee Technological University

16

• Applications of Schrödinger’s Wave Equation
• Applying the wave equation in several

examples using various potential functions

• 1. Electron in Free Space
• No force acting on the particle.
• The potential function, v(x) = 0.

whose solution is: where

k = √(2mE/ħ2)

which is called the Wave Number.

Saturday, October 05, 2013 Tennessee Technological University

17

) ( 2 ) (

2 2 2

    x mE x x    ] 2 exp[ ] 2 exp[ ) (   mE jx B mE jx A x    

• Applications of Schrödinger’s Wave Equation
• Time dependent portion of the solution is:
• The total solution for the wave function is:

which is a traveling wave in +x and –x direction.

• For a +x direction traveling wave:
• where

k = √(2mE/ħ2) = √(p2/ħ2) = p/ħ. => p = ħk

• Since de Broglie’s wavelength is:

λ = h/p = 2πħ/p. λ = 2π/ k. k = 2π/ λ.

Saturday, October 05, 2013 Tennessee Technological University

18 t E j

e t

) (

) (

 

  )] 2 ( exp[ )] 2 ( exp[ ) , ( Et mE x j B Et mE x j A t x        

)] ( exp[ ) , ( t kx j A t x    

• 2. Infinite Potential Well
• In region I & III
• V is in finite, E is finite, the wave function must

be zero.

• A particle cannot penetrate the infinite

potential barriers.

• The probability of finding a particle in those

regions is zero.

Saturday, October 05, 2013 Tennessee Technological University

19

∞ ∞

) ( )) ( ( 2 ) (

2 2 2

     x x V E m x x   

• Infinite P
• tential Well
• In region II
• V = 0.
• Time independent wave equation is:
• A solution to these equation is:
• Using Boundary Conditions

Saturday, October 05, 2013 Tennessee Technological University

20

) ( 2 ) (

2 2 2

    x mE x x    ) ( )) ( ( 2 ) (

2 2 2

     x x V E m x x    Kx A Kx A x sin cos ) (

2 1

   ) ( ) (     a x x  

2

2  mE K 

1 

A Ka A a x sin ) (

2

     n Ka  a n K  

• Infinite P
• tential Well
• S

ince we know that:

• from 0 to a:
• S
• lution of the integral
• The time-independent wave solution is:

Saturday, October 05, 2013 Tennessee Technological University

21



  

 1 ) ( ) (

*

dx x x  





a

Kxdx A

2 2 2

1 sin a A 2

2 

,... 3 , 2 , 1 ) sin( 2 ) (   n a x n a x  

• Infinite P
• tential Well
• S

ince we have:

• The total energy is then:
• The wave function is then:

is a standing wave s

• lution that

repres ents the electron in the infinite potential well, where K mus t have dis crete values meaning the total energy only hasdis crete values.

• Energy is Quantized!

Saturday, October 05, 2013 Tennessee Technological University

22

2

2  mE K 

a n K  

2 2 2 2

2 a n mE   

,... 3 , 2 , 1 2

2 2 2 2

   n ma n E E

n

 

Kx a x sin 2 ) (   ,... 3 , 2 , 1 sin 2 ) (         n a x n a x  

• Exercise

1 . Calculate the first three energy levels of an

electron in an infinite potential well. Consider an electron in an infinite potential well of width 5Å.

Saturday, October 05, 2013 Tennessee Technological University

23

• S
• lution

We know for an electron in an infinite potential well: E

n = n2 (1.054*10-34)2 (3.14)2

= n2(1.51) eV (2*(9.11*10-31)*(5*10-1

0)2 )

Therefore, the first three energy levels (n= 1, n= 2, n= 3) are: E

1 = 1.5eV, E 2 = 6.04eV, E 3 = 13.59eV

• Exercise
• 2. The width of an infinite potential well is

12Å. Determine the first three allowed energy levels in (eV) for a) an electron and b) a proton.

Saturday, October 05, 2013 Tennessee Technological University

24

• S
• lution

a) 0.261eV, 1.045eV, 2.351eV b) 1.425*10-4 eV, 5.70*10-4eV, 1.28*10-3eV

E

ne = ħ2n2π2/ (2mea2) = n2(1.11*10- 68)(9.86)

E

ne = n2(1.054*10- 34)2(3.14)2

= 4.179x10- 20J 2(9.11*10- 31)(12*10- 10)2

E

ne = 0.2612(n2) eV.

• The S

tep P

• tential Function
• A case when a flux of particles are incident
• n a potential barrier.
• Region 1: V = 0
• T

.I.S .W.E.

• General S
• lution
• where the constant K is:

Saturday, October 05, 2013 Tennessee Technological University

25

) ( 2 ) (

1 2 2 1 2

    x mE x x   

) ( ) (

1 1

1 1 1

  



x e B e A x

x jK x jK



2 1

2  mE K 

• The S

tep P

• tential Function
• A case when a flux of particles are incident
• n a potential barrier.
• Region 2: V = V0
• T

.I.S .W.E.

• General S
• lution
• where
• But since the wave equation must remain

finite:

Saturday, October 05, 2013 Tennessee Technological University

26

) ( ) ( 2 ) (

2 2 2 2 2

     x E V m x x    ) ( ) (

2 2

2 2 2

  

 

x e B e A x

x K x K



2 2

) ( 2  E V m K   ) ( ) (

2

2 2

 



x e A x

x K



2 

B

• The S

tep P

• tential Function
• Using Boundary Conditions at x = 0:
• And,

since potential difference is finite everywhere, first derivative

• f

wave function should be continuous .

Saturday, October 05, 2013 Tennessee Technological University

27

) ( ) (

2 1

  

2 1 1

A B A  

2 1  

    

x x

x x  

• The S

tep P

• tential Function
• S
• lving the partial differential:
• S
• lving together:
• Reflected probability density function is:
• Vr = reflected velocity and

Vi = incident velocity.

Saturday, October 05, 2013 Tennessee Technological University

28

2 2 1 1 1 1

A K B jK A jK   

) ( ) 2 (

2 1 2 2 1 2 1 2 1 2 2 1

K K A K K jK K B     

) ( ) ( 2

2 1 2 2 1 2 1 1 2

K K A jK K K A   

2 2 1 2 2 * 1 1 2 1 2 1 2 2 2 1 2 1 2 2 * 1 1

) ( . ) 2 ).( 2 ( . K K A A K K jK K K K jK K B B      

* 1 1 * 1 1

. . . . A A B B R

i r

  

• The S

tep P

• tential Function
• Region 1
• V = 0, E = T where T = Kinetic Energy of

particle.

• Therefore:-
• Incident velocity is also:
• Then Reflection Coefficient:
• Meaning: All incident particles are later

reflected and thus particle are not absorbed

• r transmitted.
• Region I is similar to Classical Physics!

Saturday, October 05, 2013 Tennessee Technological University

29

2

2 1 mv T     mv v m mv m K         

2 2 2 2 2 1

2 1 2

1

.K m vi  

1

.K m vr  

• The S

tep P

• tential Function
• Region 2
• S

ince:

• for the case of E < V0 , A2≠ 0
• The Probability Density Function is not zero,

thus the particle may penetrate potential barrier and exist in Region 2.

• Region 2 is different from Classical Physics!
• However, the particle must eventually go

back and return to Region 1 where the Reflection Coefficient was 1.

Saturday, October 05, 2013 Tennessee Technological University

30

x K

e A x

2

2 2

) (



  ) ( . ) (

* 2 2

 x x  

• Exercise
• 1. In region 1, consider an incident electron

traveling at a velocity of 1x105m/s. Calculate the penetration depth of a particle impinging on a potential barrier that is twice as large as the total energy of the incident particle.

Saturday, October 05, 2013 Tennessee Technological University

31

• Solution

Region 1: V(x) = 0. E = T = ½ mv2 = ½ (9.11x10-31Kg)(1x105m/s)2 =4.56x10-21J = 2.85x10-2eV Region 2: V(x) = 2E.

x K

e A x

2

2 2

) (



 

2 2

) ( 2  E V m K  

• Exercise
• 1. In region 1, consider an incident electron

traveling at a velocity of 1x105m/s. Calculate the penetration depth of a particle impinging on a potential barrier that is twice as large as the total energy of the incident particle.

Saturday, October 05, 2013 Tennessee Technological University

32

• Solution

Since k2d = 1 gives a decay to (1/e), the penetration distance is:

2 2

) ( 2 ) 2 ( 2 1   E m d E E m d    m x x x x mE d

10 21 31 34 2

10 6 . 11 ) 10 56 . 4 )( 10 11 . 9 ( 2 10 054 . 1 2

   

   

• Exercise
• 2. Consider an electron traveling in Region 2 at a velocity
• f 105m/s incident on a potential barrier whose height is

3 times the Kinetic energy of the electron. Calculate the probability of finding an electron at a distance ‘d’ in Region 2 compared with x = 0 where ‘d’ is: a) 10Å and b) 100Å into the potential barrier.

Saturday, October 05, 2013 Tennessee Technological University

33

• Solution

V0 = 3E. d1= 10Å = 1x10-9m = 1x10-8m.

1)

First calculate energy E = ½ mv2 = ½ (9.11x10-31Kg)(105m/s)2 = 4.56x10-21J

1)

Then calculate “k2” k2 = (2m(V-E)/(ħ)2)1/2 = (2(9.11x10-31)(3E-E)/(1.054x10- 34)2)1/2 k2 = 12.23x108 Probability at 10Å = e-2Kd = e -2(12.23x108)(1x10-9) = 0.0866 => 8.66 % Probability at 100Å = e-2Kd = e -2(12.23x108)(1x10-8) = 2.38x10-11 => 2.38x10-9 %

• Take Home Solution

1.7 Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95Å. The atom is placed in a) Simple Cubic, b) FCC, c) BCC, d) Diamond Lattice. Assuming the nearest atoms are touching each other, what is the lattice constant of each lattice?

Saturday, October 05, 2013 Tennessee Technological University

34

• Solution

a) Simple Cubic (SC)

a = 2r = 2(1.95Å) = 3.9Å.

b) FCC

4r = (a2 + a2)1/2 = (2a2)1/2 4r = (2)1/2a => a = 2(2)1/2r a = 2(2)1/2r = 2(2)1/2(1.95Å) = 5.515Å

• Take Home Solution

1.7 Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95Å. The atom is placed in a) Simple Cubic, b) FCC, c) BCC, d) Diamond Lattice. Assuming the nearest atoms are touching each other, what is the lattice constant of each lattice?

Saturday, October 05, 2013 Tennessee Technological University

35

• Solution

c)

BCC

Hypotenus= (2a2)1/2 . l = 4r. (l)2 = ((a2 + a2)1/2)2+ (a) 2 (4r)2 = ((2a2)1/2)2+ (a) 2 = 3a2 a = 4r/(3)1/2 =4(1.95Å)/(3)1/2= 4.503Å

d)

Diamond Lattice

(((a/2)2+(a/2)2)1/2 )2 + (a/2)2 = (4r)2 2a2/4 + a2/4 = (4r)2 3a2 = 64r2 a = 8r/(3)1/2 = 8(1.95Å)/(3)1/2 = 9.01Å.

Pi ct ur e C r edi t s

• BBC Science

http://www.bbc.co.uk/science/space/universe/questions_and_ideas/quantum_mechanics

• Visual Science Blog

http://3dciencia.com/blog/?p=278

• Fifth Solvay Conference

http://www.purephysics.net/2012/09/colorized-fifth-solvay-conference-of.html

• Max Plank

http://en.wikipedia.org/wiki/Max_Planck

• Werner Heisenberg

http://www.thelaunchcomplex.com/Heisenberg.php

• Louis De Broglie

http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/quantum_theory_origins/

• Albert Einstein

http://www.forbes.com/sites/moneybuilder/2012/12/19/albert-einsteins-philosophies-for- growing-wealth/

• Erwin Schrödinger

http://www.freegreatpicture.com/news-and-events/the-physics-behind-schrdingers-cat- paradox-45944

Saturday, October 05, 2013 Tennessee Technological University

36