Transcript Precursors to Modern Physics

Unbound States
1.
2.
3.
today
A review about the discussions we have
had so far on the Schrödinger equation.
Quiz 10.21
Topics in Unbound States:


Thurs.


The potential step.
Two steps: The potential barrier and
tunneling.
Real-life examples: Alpha decay and other
applications.
A summary: Particle-wave propagation.
Review: The Schrödinger equation
2


Ψ
x,t

i
Ψ  x,t 


2m x 2
t
2
The free particle Schrödinger Equation:
The plane wave solution Ψ  x,t   Aeikxt 
The momentum of this particle: p  k completely defined.
2
The location of this particle: Ψ  x,t   Aei kxt  Aeikxt   A2 undefined.
From
p2
KE 
2m
and
p k
E 
We understand this equation as energy accounting KEΨ  x,t   EΨ  x,t 
And this leads to the equation that adds an external potential
 KE  U  x   Ψ  x,t   EΨ  x,t 
The Schrödinger Equation 
Solve for
Ψ  x,t 

2
 2Ψ  x,t 
2m
with the knowledge of
x 2
U  x
 U  x  Ψ  x,t   i
Ψ  x,t 
t
, for problems in QM.
Review: the time independent Schrödinger equation
and the two conditions for the wave function
When the wave function can be expressed as Ψ  x,t    x  t 
i E t
We have found   t   e  
and the time independent Schrödinger Equation:
 KE  U  x    x   E  x 
2


d2


U
x



  x   E  x 
2
2
m
dx




2
2m
d 2  x 
dx
2
 U  x   x   E  x 
The solution of this equation is the stationary states because
2
The probability of finding a
2
2
i  E t
   x
particle does not depend on time: Ψ  x,t     x  e
Ψ  x,t  dx  1
Normalization

all phase space
:
The wave function be smooth  the continuity of
the wave function and its first order derivative.
2
Two
conditions
Review: Solving the Schrödinger equation.
Case 1: The infinite potential well
Equation and Solution:
Energy and probability
0
x  0 or x  L


2
  x  
d 2  x 
 E  x 
0 x L
from: 
2m dx 2

0


  x    2  n
 L sin  L


x  0 or x  L

x

E
2
2mL
  x 
2
0 xL
1. Standing wave.
2. The QM groundstate. 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.
2
n 2 , n  1, 2,3,...
2 2  n
sin 
L
 L

x 0  x  L

Solving the Schrödinger equation.
Case 2: The finite potential well
U  0  
U  L  
E  KE
0
U  0  U0
E  KE
The change 
L
0 0  x  L
U  x  
 x  0,x  L
Equations:
U  0  U0
0
x
x
 0 0 x L
U  x  
U 0 x  0,x  L
The change 
 0 0 x L
U  x  

U 0 x  0,x  L
L


2
2m
2
2m
d 2  x 
dx
2
d 2  x 
dx
2
 E  x 
 U 0  x   E  x 
Review: Solving the Schrödinger equation.
Case 2: The finite potential well
U  0  U0
U  0  U0
Penetration depth:
E  KE

0


2
2m
d 2  x 
dx 2
2m
d 2  x 
dx 2
 E  x 
 U 0  x   E  x 


2m U 0  E 
x
L
2
1
  x   Asin  kx   Bcos  kx  , k 
d 2  x 
dx 2

2m U 0  E 
2
  x     x

Ce x
x0
Solutions:   x    Asin  kx   Bcos  kx  0  x  L

Ge x
xL

2
2mE
2
  x   Ce x  De x , x  0
  x   Fe x  Ge x , x  L
Energy quantization: 2cot  kL   k  

k
Review: Solving the Schrödinger equation.
Case 3: The simple harmonic oscillator
This model is a good approximation of
particles oscillate about an equilibrium
position, like the bond between two atoms
in a di-atomic molecule.
1
U  x    x2
2

2
2m
d 2  x 
dx 2
1 2
  x   x   E  x 
2
Solve for wave function and energy level
Review Solving the Schrödinger equation.
Case 3: The simple harmonic oscillator
Wave function at each energy level
Gaussian
E   n  12  0
Energy are equally
spaced, characteristic
n  0,1, 2,3,...,0   m of an oscillator
Compare Case 1, Case 2 and 3:
Wave function:
0


  x    2  n
 L sin  L


x  0 or x  L

x

0 xL
Energy levels:
E
2
2
n , n  1, 2,3,...
2
2
2mL
Wave function:

Ce x
x0

  x    Asin  kx   Bcos  kx  0  x  L

Ge x
xL

Energy levels:
2cot  kL  
k



E   n  12  0
k
n  0,1, 2,3,...,0   m
Penetration depth:

1


Energy levels:
2m U 0  E 
New cases, unbound states: the potential step
From potential well to a one-side well, or a step:
E
KE  E  U 0
U  x  0  U0
KE  E
U  x  0  0
x
x0
We discussed about free particle
wave function before. Which is:
Re-write it in the form
Free particle with energy E.
Standing waves do not form
and energy is no quantized.
Ψ  x,t   Ae
i  kx t 
Ψ  x,t   Ae
ikxit
ikx it
 Ae
Right moving (along positive xaxis) wave of free particle:
  x   Aeikx
Left moving wave of free particle:
  x   Ae
ikx
e
Right moving. Why?
k  2   0
p k
The potential step: solve the equation
Initial condition: free
particles moving
from left to right.
E
KE  E  U 0
U  x  0  0
x
x0
The Schrödinger Equation: 
2
2m
d 2  x 
dx 2
 U  x   x   E  x 
When U  0
d 2  x 
dx 2

2mE
2
U  x  0  U0
KE  E
When U  U 0
  x   k   x 
2
d 2  x 
dx 2

2m  E  U 0 
2
  x   k' 2   x 
When E  U 0
Solution:
  x   Aeikx  Beikx
Inc.
Refl.
  x   Ceik' x
Trans.
Are we done? What do we learn here? Any Normalization and wave
other conditions to apply to the solutions? function smoothness
Why no
reflection here?
Ans: no 2nd
potential step on
the right.
The potential step: apply conditions

x0
x
2
Inc.
 A* A

*

B
B
Refl .
2

*

C
C
Trans .
2
# particles # particles distance
2
2

 v k
time
distance time
Smoothness
 x0  0  x0  0: Aeik 0  Beik 0  Ceik' 0  A  B  C
requires:
d x0
d x0

: ikAeik 0  ikBeik 0  ik' Ceik' 0  k  A  B   k' C
dx x0
dx x0
Express B and C
B   k  k'   k  k'   A C   2k  k  k '   A
in terms of A:
Transmission
probability:
Reflection
probability:

2
*
k
# trans. time
C
Ck'
4kk'
x

0
T
 Trans.


# incident time  2 k x0
A* Ak  k  k' 2
Inc.
 Refl. k x0 B* B  k  k' 
# Refl. time
R

 * 
 1 T
2
2
# incident time  k x0 A A  k  k' 
Inc.
2
2
The potential step: transmission and reflection
Now using the
definitions of k and
k’:
x0
x
n
k' 
2
Transmission T  4kk'
T 4

2
probability:
 k  k' 
Reflection
probability:
air
k
2mE
 k  k' 
R
2
 k  k' 

R

2

Reference:
for normal incidence, light T  4n 2
 n  1
transmission probability:
Reflection
probability:
 n  1
2
 n  1
2
R

2m  E  U 0 
2
E  E U0 
E  E U0
E  E U0
E  E U0



2
2
2
The potential step: solve the equation
Initial condition: free
particles moving
from left to right.
E
KE  E
U  x  0  0
KE  E  U0  0
x
x0
The Schrödinger Equation: 
2
2m
d 2  x 
dx 2
 U  x   x   E  x 
When U  0
d 2  x 
dx 2

2mE
2
U  x  0  U0
When U  U 0
  x   k   x 
2
d 2  x 
dx 2

2m U 0  E 
2
  x    2  x 
 U0
When E 
Solution:
  x   Aeikx  Beikx
Inc.
Refl.
  x   Ce x
The potential step: apply conditions
Smoothness requires:
 x0  0  x0  0: Aeik 0  Beik 0  Ce 0
 A B  C
d x0
dx


x 0
d x 0
dx
: ikAeik 0  ikBeik 0   Ce 0
x 0
Express B in
B     ik    ik   A
terms of A:
One can
*
*
B
B

A
A
prove:
Reflection
probability:
B* B
R  * 1
AA
Transmission T  1  R  0
probability:
Penetration
depth:

1


2m U 0  E 
k  A  B  C
Review questions

The plane wave solution of a free particle
Schrödinger Equation is Ψ  x,t   Ae  
Can you normalize this wave function?
Try to solve for the wave function and discuss
about transmission and reflection for this
situation:
i kx t

E
KE  E  U 0
U  x  0  U0
KE  E
x0
x U  x  0  0
Preview for the next class (10/23)

Text to be read:

In chapter 6:




Section 6.2
Section 6.3
Section 6.4
Questions:



How much do you know about radioactive decays of isotopes?
Have you heard of alpha decay, beta decay and gamma
sources?
Have you heard of the “tunneling effect” in the EE
department (only for EE students)?
What is a wave phase velocity? What is a wave group velocity?
Homework 8, due by 10/28
1.
2.
3.
4.
Problem 25 on page 187.
Problem 34 on page 188.
Problem 15 on page 224.
Problem 18 on page 224.