unbound states

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Unbound States Unbound States 1. 1. A review about the A review about the discussions we have had so discussions we have had so far on the Schrödinger far on the Schrödinger equation. equation. 2. 2. Quiz 10.21 Quiz 10.21 3. 3. Topics in Unbound States: Topics in Unbound States: The potential step. The potential step. Two steps: The potential Two steps: The potential barrier and tunneling. barrier and tunneling. Real-life examples: Alpha Real-life examples: Alpha decay and other applications. decay and other applications. A summary: Particle-wave A summary: Particle-wave today Thurs.

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Unbound States. today. A review about the discussions we have had so far on the Schrödinger equation. Quiz 10.21 Topics in Unbound States: The potential step. Two steps: The potential barrier and tunneling. Real-life examples: Alpha decay and other applications. - PowerPoint PPT Presentation

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Page 1: Unbound States

Unbound StatesUnbound States1.1. A review about the discussions A review about the discussions

we have had so far on the we have had so far on the Schrödinger equation.Schrödinger equation.

2.2. Quiz 10.21Quiz 10.213.3. Topics in Unbound States:Topics in Unbound States:

The potential step. The potential step. Two steps: The potential barrier Two steps: The potential barrier

and tunneling. and tunneling. Real-life examples: Alpha decay Real-life examples: Alpha decay

and other applications.and other applications. A summary: Particle-wave A summary: Particle-wave

propagation.propagation.

today

Thurs.

Page 2: Unbound States

Review: The Schrödinger equationReview: The Schrödinger equation

2 2

2 Ψ Ψ2

x,t i x,tm x t

The free particle Schrödinger Equation:

22

2

Ψ ΨΨ

2x,t x,t

U x x,t im x t

From

Ψ i kx tx,t Ae

Ψ ΨKE x,t E x,t

Ψ ΨKE U x x,t E x,t

The Schrödinger Equation

The plane wave solution

And this leads to the equation that adds an external potential

Solve for with the knowledge of , for problems in QM. Ψ x,t U x

The momentum of this particle: p kcompletely defined. The location of this particle: 2 2Ψ i kx t i kx tx,t Ae Ae A

2

2pKEm

and p k E

We understand this equation as energy accounting

undefined.

Page 3: Unbound States

KE U x x E x

Review: the time independent Review: the time independent Schrödinger equation and the two Schrödinger equation and the two conditions for the wave functionconditions for the wave function

When the wave function can be expressed as

2 2

22d U x x E x

m dx

Ψ x,t x t

and the time independent Schrödinger Equation:

We have found

The solution of this equation is the stationary states becauseThe probability of finding a particle does not depend on time:

22 2

Ψ i E tx,t x e x

i E tt e

Normalization:

2

all phase space

Ψ 1x,t dx

The wave function be smooth the continuity of the wave function and its first order derivative.

Two conditions

22

22d x

U x x E xm dx

Page 4: Unbound States

Review: Solving the Schrödinger Review: Solving the Schrödinger equation. equation.

Case 1: The infinite potential wellCase 1: The infinite potential well

0 0 or

2 sin 0

x x Lx n x x L

L L

Equation and Solution: Energy and probability 2 2

2 , 1 2 32

E n n , , ,...mL

2 22 sin nx xL L

0 x L

1. Standing wave. 2. The QM ground-

state. A bound state particle cannot be stationary, although its wave function is stationary.

3. Energy ratio at each level: n2.

4. With very large n, QM CM.

22

2

0 0 or

from: 02

x x Lx d x

E x x Lm dx

Page 5: Unbound States

Solving the Schrödinger equation. Solving the Schrödinger equation. Case 2: The finite potential wellCase 2: The finite potential well

0U U L

x0 L

E KE

x0 L

E KE

00U U 00U U

0 00x L

U xx ,x L

0

0 00x L

U xU x ,x L

22

2

220

02

0 0 20

2

d xE xx L m dxU x

U x ,x L d xU x E x

m dx

The change

The change

Equations:

Page 6: Unbound States

Review: Solving the Schrödinger Review: Solving the Schrödinger equation. equation.

Case 2: The finite potential wellCase 2: The finite potential well 00U U

x0 L

E KE

00U U

2

2sin cos mEx A kx B kx , k

2

0 22 2

2d x m U Ex x

dx

0x x

x x

x Ce De , xx Fe Ge , x L

0

sin cos 0

x

x

Ce xx A kx B kx x L

Ge x L

Solutions: 2cot kkLk

Energy quantization:

0

12m U E

Penetration depth:

22

22d x

E xm dx

22

022d x

U x E xm dx

Page 7: Unbound States

Review: Solving the Schrödinger Review: Solving the Schrödinger equation. equation.

Case 3: The simple harmonic oscillatorCase 3: The simple harmonic oscillator

212

U x x

This model is a good approximation of particles oscillate about an equilibrium position, like the bond between two atoms in a di-atomic molecule.

22

22

12 2d x

x x E xm dx

Solve for wave function and energy level

Page 8: Unbound States

Review Solving the Schrödinger Review Solving the Schrödinger equation. equation.

Case 3: The simple harmonic oscillatorCase 3: The simple harmonic oscillator

102

00 1 2 3

E n

n , , , ,..., m

Energy are equally spaced, characteristic of an oscillator

Wave function at each energy level

Gaussian

Page 9: Unbound States

Compare Case 1, Case 2 and 3: Compare Case 1, Case 2 and 3:

0 0 or

2 sin 0

x x Lx n x x L

L L

2 22

2 , 1 2 32

E n n , , ,...mL

Energy levels:

Wave function:

0

sin cos 0

x

x

Ce xx A kx B kx x L

Ge x L

2cot kkLk

0

12m U E

Penetration depth:

Wave function:

Energy levels: 1

02

00 1 2 3

E n

n , , , ,..., m

Energy levels:

Page 10: Unbound States

New cases, unbound states: the New cases, unbound states: the potential steppotential step

From potential well to a one-side well, or a step:

0x

KE E0KE E U

E

00U x U

0 0U x

Free particle with energy E. Standing waves do not form and energy is no quantized.

Ψ i kx tx,t Ae We discussed about free particle wave function before. Which is: Ψ ikx i t ikx i tx,t Ae Ae e Re-write it in the form

Right moving. Why?Right moving (along

positive x-axis) wave of free particle:

ikxx Ae

Left moving wave of free particle:

ikxx Ae

x

2 0k p k

Page 11: Unbound States

The potential step: solve the The potential step: solve the equationequationKE E

00U x U

0x

0KE E U E

0 0U x

Initial condition: free particles moving from left to right.

x

0U UWhen

22

22d x

U x x E xm dx

The Schrödinger Equation:

When 0U

2

22 2

2d x mE x k xdx

2

0 22 2

2d x m E Ux k' x

dx

ikx ikxx Ae Be Solution:

Inc.

Refl.

Trans.

Why no reflection here? Ans: no 2nd potential step on the right.

Are we done? What do we learn here? Any other conditions to apply to the solutions?

Normalization and wave function smoothness

0E UWhen

ik' xx Ce

Page 12: Unbound States

The potential step: apply The potential step: apply conditionsconditions

0x x

Smoothness requires:

2

Inc.*A A 2

Refl*

.B B 2

Trans*

.C C

2 2# particles # particles distancetime distance time

v k

0 0 00 00 0 : ik ik ik'

x x Ae Be Ce

A B C

0 0 00 0

0 0

: ik ik ik'x x

x x

d d ikAe ikBe ik ' Cedx dx

k A B k' C

Transmission probability:

20Trans.

2 20Inc.

# trans. time 4 incident time

*x

*x

k C Ck' kk 'T# A Akk k k '

Reflection probability:

2 20Refl.

2 20Inc.

# Refl. time 1 incident time

*x

*x

k k k 'B BR T# A Ak k k'

Express B and C in terms of A:

B k k' k k ' A 2C k k k' A

Page 13: Unbound States

The potential step: transmission The potential step: transmission and reflectionand reflection

0x x

Now using the definitions of k and k’:

02

2m E Uk'

2

2mEk

nair

Reference:for normal incidence, light transmission probability:

2

41nT

n

2

2

1

1

nR

n

Reflection probability:

Transmission probability:

2

4kk 'Tk k '

Reflection probability:

2

2

k k'R

k k '

0

2

0

4E E U

TE E U

2

0

2

0

E E UR

E E U

Page 14: Unbound States

The potential step: solve the The potential step: solve the equationequationKE E

00U x U

0x

0 0KE E U

E

0 0U x

Initial condition: free particles moving from left to right.

x

22

22d x

U x x E xm dx

The Schrödinger Equation:

0U UWhen

20 2

2 2

2d x m U Ex x

dx

ikx ikxx Ae Be Solution:

Inc.

Refl.

When 0U

2

22 2

2d x mE x k xdx

0E UWhen

xx Ce

0E U

Page 15: Unbound States

The potential step: apply The potential step: apply conditionsconditions

Smoothness requires: 0 0 0

0 00 0 : ik ikx x Ae Be Ce

A B C

0 0 00 0

0 0

: ik ikx x

x x

d d ikAe ikBe Cedx dx

k A B C

Transmission probability:

1 0T R

Reflection probability:

1*

*

B BRA A

Express B in terms of A: B ik ik A

Penetration depth: 0

12m U E

* *B B A AOne can prove:

Page 16: Unbound States

Review questionsReview questions The plane wave solution of a free The plane wave solution of a free

particle Schrödinger Equation isparticle Schrödinger Equation isCan you normalize this wave function? Can you normalize this wave function?

Try to solve for the wave function and Try to solve for the wave function and discuss about transmission and discuss about transmission and reflection for this situation: reflection for this situation:

Ψ i kx tx,t Ae

KE E 00U x U

0x

0KE E U E

x 0 0U x

Page 17: Unbound States

Preview for the next class Preview for the next class (10/23)(10/23)

Text to be read:Text to be read: In chapter 6:In chapter 6:

Section 6.2Section 6.2 Section 6.3Section 6.3 Section 6.4Section 6.4

Questions:Questions: How much do you know about radioactive How much do you know about radioactive

decays of isotopes? Have you heard of alpha decays of isotopes? Have you heard of alpha decay, beta decay and gamma sources? decay, beta decay and gamma sources?

Have you heard of the “tunneling effect” in the Have you heard of the “tunneling effect” in the EE department (only for EE students)? EE department (only for EE students)?

What is a wave phase velocity? What is a wave What is a wave phase velocity? What is a wave group velocity? group velocity?

Page 18: Unbound States

Homework 8, due by 10/28Homework 8, due by 10/281.1. Problem 25 on page 187.Problem 25 on page 187.2.2. Problem 34 on page 188.Problem 34 on page 188.3.3. Problem 15 on page 224.Problem 15 on page 224.4.4. Problem 18 on page 224.Problem 18 on page 224.