brook abegaz, tennessee technological...

Brook Abegaz, Tennessee Technological University, Fall 2013

Saturday, October 05, 2013Tennessee Technological University 1

Chapter 2 – Introduction to Quantum Mechanics 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 of quantum mechanics called Wave Mechanics.


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


Albert Einstein Louis de Broglie Werner Heisenberg

1. The Principle of Energy Quanta: Photo-Electric Effect experiment:a. if monochromatic light is incident on a clean surface

of a material, electrons or photoelectrons are emittedfrom the surface.

b. At constant incident intensity, the maximum kineticenergy of the photoelectron varies linearly withfrequency with a limiting frequency υ = υo belowwhich no photoelectrons are produced.

c. In 1900, Plank postulated that such thermal radiationis emitted from a heated surface in discrete packets ofenergy called Quanta whose energy is:

E = h*υ, where h = 6.625x10-34 J*sec,υ = frequency of incident light



1. The Principle of Energy Quanta:

In 1905, Einstein interpreted the photoelectric resultsby suggesting that the energy in a light wave is alsocontained in discrete packets or bundles called photonwhose energy is E = h*υ.

Work Function = minimum energy required to removean electron.

If Incoming Energy > Work Function, excess photonenergy goes to the kinetic energy of the photoelectron:

T = ½ mv2 = h*υ – Φ = h*υ - h*υo (υ ≥ υo)Where T = 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.


3. The Uncertainty Principle (Heisenberg)

It is not possible to describe the absolute accuracy of 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.


Exercise1. 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.


Solutiona) 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.

Exercise2. A proton’s energy is measured with an

uncertainty of 0.8eV.a) Determine the minimum uncertainty in time

over which this energy is measured.b) Compute the same problem for an electron.


Solutiona) 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’s Wave 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 of Quantum

Mechanics that describes the behavior andcharacterization of electron movement andsubatomic particles in materials and devices.

Wave Equation = a one-dimensional, non-relativisticShrödinger’s wave equation:

(1)

where (x,t) is the wave function, V(x) is the potentialfunction assumed to be independent of time, m is the massof the particle and j is √(-1).


ttxjtxxV

xtx

m

),(),()(),(2 2

22

Shrödinger’s Wave Equation From Wave-Particle Duality Principles


Therefore, we would like to split equation 1 toa) A left side that is independent of time andb) A right side that is independent of position.

Dividing by the total wave function gives:

The left side of the equation is a function of position only and the right sideof the equation is a function of time t, each side must be equal to a constant η:

(2)

)()(),( txtx

ttxjtxxV

xxt

m

)()()()()()()(2 2

22

tt

tjxV

xx

xm

)()(

1)()()(

12 2

22

tt

tj

)(.

)(1.

Shrödinger’s Wave Equation The solution of such equation is:


Substituting in (2)

(3)

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

tjet

)(tEj

et

)(

2

hhvE

E

ExVx

xxm

)()(.)(

1.2 2

22

0)())((2)(22

2

xxVEmx

x

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

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


Max Born in 1926 postulated that the probability of finding a particle between xand 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.

tEjextxtx

)()()()(),(

),().,(|),(| *2 txtxtx

tEjextx

)(** ).(),(

2*2 |)(|),().,(|),(| xtxtxtx

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


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

∞∞

1|)(| 2

dxx

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.)


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

examples using various potential functions1. Electron in Free Space• No force acting on the particle.• The potential function, v(x) = 0.

whose solution is:

wherek = √(2mE/ħ2)

which is called the Wave Number.


0)(2)(22

2

xmEx

x

]2exp[]2exp[)(

mEjxBmEjxAx

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.


tEjet

)()(

)]2(exp[)]2(exp[),( EtmExjBEtmExjAtx

)](exp[),( tkxjAtx

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.


∞∞

0)())((2)(22

2

xxVEmx

x

Infinite Potential Well• In region II• V = 0.• Time independent wave equation is:

• A solution to these equation is:

• Using Boundary Conditions


0)(2)(22

2

xmEx

x

0)())((2)(22

2

xxVEmx

x

KxAKxAx sincos)( 21

0)()0( axx

2

2

mEK

01 A

KaAax sin0)( 2

nKa a

nK

Infinite Potential Well Since we know that:

from 0 to a:

Solution of the integral

The time-independent wave solution is:


1)()( * dxxx

a

KxdxA0

222 1sin

aA 2

2

,...3,2,1

)sin(2)(

na

xna

x

Infinite Potential Well Since we have:

The total energy is then:

The wave function is then:

is a standing wave solution that represents theelectron in the infinite potential well, where K musthave discrete values meaning the total energy onlyhas discrete values.

Energy is Quantized!


22

mEK a

nK

2

22

22

anmE

,...3,2,12 2

222

nmanEE n

Kxa

x sin2)( ,...3,2,1

sin2)(

na

xna

x

Exercise1. 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.


SolutionWe know for an electron in an infinite potential well:En = n2 (1.054*10-34)2 (3.14)2 = n2(1.51) eV

(2*(9.11*10-31)*(5*10-10)2 ) Therefore, the first three energy levels (n=1, n=2,

n=3) are:E1 = 1.5eV, E2 = 6.04eV, E3 = 13.59eV

Exercise2. 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.


Solutiona) 0.261eV, 1.045eV, 2.351eVb) 1.425*10-4 eV, 5.70*10-4eV, 1.28*10-3eV

Ene = ħ2n2π2/(2mea2) = n2(1.11*10-68)(9.86)Ene = n2(1.054*10-34)2(3.14)2 = 4.179x10-20J

2(9.11*10-31)(12*10-10)2Ene = 0.2612(n2) eV.

The Step Potential Function A case when a flux of particles are incident

on a potential barrier. Region 1: V = 0 T.I.S.W.E.

General Solution

where the constant K is:


0)(2)(122

12

xmEx

x

)0()( 11111 xeBeAx xjKxjK

212

mEK

The Step Potential Function A case when a flux of particles are incident

on a potential barrier. Region 2: V = V0

T.I.S.W.E.

General Solution

where But since the wave equation must remain

finite:


0)()(2)(2022

22

xEVmx

x

)0()( 22222 xeBeAx xKxK

20

2)(2

EVmK

)0()( 222 xeAx xK 02 B

The Step Potential Function Using Boundary Conditions at x = 0:

And, since potential difference is finiteeverywhere, first derivative of wavefunction should be continuous.


)0()0( 21

211 ABA

0

2

0

1

xx xx

The Step Potential Function Solving the partial differential:

Solving together:

Reflected probability density function is:

Vr = reflected velocity and Vi = incident velocity.


221111 AKBjKAjK

)()2(

21

22

12

12122

1 KKAKKjKKB

)()(2

21

22

12112 KK

AjKKKA

221

22

*11

2121

22

2121

22*

11 )(.)2).(2(.

KKAAKKjKKKKjKKBB

*11

*11

..

..AABBR

i

r

The Step Potential 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 or transmitted.

Region I is similar to Classical Physics!


2

21 mvT

mvvmmvmK

2

222

21 212

1.Km

v i

1.Km

v r

The Step Potential Function Region 2 Since: for the case of E < V0 , A2≠ 0 The Probability Density Function is not zero,

thus the particle may penetrate potentialbarrier 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 theReflection Coefficient was 1.


xKeAx 222 )(

0)(.)( *22 xx

Exercise1. 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.


SolutionRegion 1: V(x) = 0.E = T = ½ mv2 = ½ (9.11x10-31Kg)(1x105m/s)2

=4.56x10-21J = 2.85x10-2eVRegion 2: V(x) = 2E. xKeAx 2

22 )( 20

2)(2

EVmK

Exercise1. 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.


SolutionSince k2d = 1 gives a decay to (1/e), thepenetration distance is:

22)(2)2(21

EmdEEmd

mxxx

xmE

d 102131

342

106.11)1056.4)(1011.9(2

10054.12

Exercise2. Consider an electron traveling in Region 2 at a velocity of 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.


SolutionV0 = 3E.d1 = 10Å = 1x10-9m = 1x10-8m.1) First calculate energyE = ½ mv2 = ½ (9.11x10-31Kg)(105m/s)2 = 4.56x10-21J1) 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 Solution1.7 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?


Solutiona) 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 Solution1.7 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?


Solutionc) 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.

Picture Credits BBC Science

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

Visual Science Bloghttp://3dciencia.com/blog/?p=278

Fifth Solvay Conferencehttp://www.purephysics.net/2012/09/colorized-fifth-solvay-conference-of.html

Max Plankhttp://en.wikipedia.org/wiki/Max_Planck

Werner Heisenberghttp://www.thelaunchcomplex.com/Heisenberg.php

Louis De Brogliehttp://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/quantum_theory_origins/

Albert Einsteinhttp://www.forbes.com/sites/moneybuilder/2012/12/19/albert-einsteins-philosophies-for-

growing-wealth/

Erwin Schrödingerhttp://www.freegreatpicture.com/news-and-events/the-physics-behind-schrdingers-cat-

paradox-45944


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