Monday, Oct. 15, 2012

```PHYS 3313 – Section 001
Lecture #12
Monday, Oct. 15, 2012
Dr. Jaehoon Yu
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The Schrödinger Wave Equation
Time-Independent Schrödinger Wave Equation
Probability Density
Wave Function Normalization
Expectation Values
Operators – Position, Momentum and Energy
Monday, Oct. 15, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
1
Announcements
• Reminder Homework #4
– End of chapter problems on CH5: 8, 10, 16, 24, 26, 36
and 47
– Due: This Wednesday, Oct. 17
– CH6.1 – 6.7 + the special topic
• Colloquium this week
– 4pm, Wednesday, Oct. 17, SH101
– Drs. Musielak and Fry of UTA
Monday, Oct. 15, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
2
Monday, Oct. 15, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
3
Special project #5
Prove that the wave function =A[sin(kx t)+icos(kx- t)] is a good solution for the timedependent Schrödinger wave equation. Do NOT
use the exponential expression of the wave
function. (10 points)
Determine whether or not the wave function =Ae |x| satisfy the time-dependent Schrödinger wave
equation. (10 points)
Due for this special project is Monday, Oct. 22.
Monday, Oct. 15, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
4
The Schrödinger Wave Equation
 The Schrödinger wave equation in its time-dependent
form for a particle of energy E moving in a potential V in
one dimension is
2
2
  x, t 
h    x, t 
ih
 
 V   x, t 
2
t
2m
x
 The extension into three dimensions is
2
2
2
2
h  
 
 
ih
 


 V   x, y, z, t 
2
2
2 

t
2 m  x
y
z 

• where
Monday, Oct. 15, 2012
is an imaginary number
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
5
Ex 6.1: Wave equation and Superposition
The wave equation must be linear so that we can use the superposition principle to. Prove
that the wave function in Schrodinger equation is linear by showing that it is satisfied for
the wave equation  x,t1 x,t2 x,t where a and b are constants and 1
x,t and 2 x,t describe two waves each satisfying the Schrodinger Eq.
  a 1  b 2

t

x
ih



t

t

x
ih
 1
t
 
a 1  b 2   a
 1
a 1  b 2   a
 1
 
h
2
 
t
x
h
 1
2
2
2 m x
b
b
2
 V 1
2 m x
V
t
 
h
2
 2
2
2 m x
2
 V2
 2
t
 2
 
x
x
2
2
2
ih
 2
Rearrange terms
ih

t

2
   1
 2 
 1
 2

a

b

a

b


2
2
x 
x
x 
x
x
2
h
2
 
2
2 m x
2
2
2
2



h 
 V    ih


V
2
  0
 t 2 m x

2
2
2
 2 
h   1
 2 
  1
ih
 ih  a
b
a
b
 V a 1  b 2 
  
2
2

t
t
t 
2 m  x
x 

2
2
2
2
  1

  2

h  1
h  2
a  ih


V



b
ih


V

 0
1
2
2
2

t
2 m x
t
2 m x




Monday, Oct. 15, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
6
General Solution of the Schrödinger Wave
Equation
 The general form of the solution of the Schrödinger wave
equation is given by:
  x, t   Ae
i  kx   t 
 A  cos  kx   t   i sin  kx   t 
• which also describes a wave propergating in the x direction. In
general the amplitude may also be complex. This is called the
wave function of the particle.
 The wave function is also not restricted to being real. Only the
physically measurable quantities (or observables) must be
real. These include the probability, momentum and energy.
Monday, Oct. 15, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
7
Ex 6.2: Solution for wave equation
Show that Aei(kx- t) satisfies the time-dependent Schrodinger wave Eq.
  Ae


x
 
2
x
ih
2


t

t

x

x

i  kx   t 
Ae
i  kx   t 

ik    ik
 iA ke

x
i  kx   t 



Ae

t
i kx   t 
h
i kx   t 
  i 
 ik 
   ik iAke i kx   t   
 ih  i    h   
   iA e 
2
k

2m
2
 Ak e
2
i  kx   t 
2
2 2


h k



V
  0


2m



  V
2


p
 E 
V  0
2m


  
T he E nergy: E  hf  h 
  h
 2 
2
2
2 p
p
The w ave num ber: k 




h p
h
h
From the energy conservation: E  K  V 
 k 
T he m om entum : p  hk
p
2
2m
V
E
So Aei(kx- t) is a good solution and satisfies Schrodinger Eq.
Monday, Oct. 15, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
p
2
V  0
2m
8
Ex 6.3: Bad solution for wave equation
Determine  x,tAsin(kx- t) is an acceptable solution for the timedependent Schrodinger wave Eq.
  A sin  kx   t 

x
 

2
x
2


x

x

t


t
A sin  kx   t    A cos  kx   t 
A sin  kx   t   kA cos  kx   t 
kA cos  kx   t    k A sin  kx   t    k 
ih   cos  kx   t   
2
h
2
k

2m
2
2

sin  kx   t   V sin  kx   t 
 h2 k 2

 ih  cos  kx   t   
 V  sin  kx   t 
 2m

 p2

 iE cos  kx   t   
 V  sin  kx   t 
 2m

This is not true in all x and t. So  (x,t)=Asin(kx- t) is not an acceptable
solution for Schrodinger Eq.
Monday, Oct. 15, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
9
Normalization and Probability
 The probability P(x) dx of a particle being between x
and X + dx was given in the equation
P  x  dx  
 x, t    x, t  dx
• Here * denotes the complex conjugate of 
 The probability of the particle being between x1 and x2
is given by
x
*
P 

2
*
  dx
x1
 The wave function must also be normalized so that the
probability of the particle being somewhere on the x
axis is 1.

*

  x , t   x , t d x  1
Monday, Oct. 15, 2012
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
10
Ex 6.4: Normalization
Consider a wave packet formed by using the wave function that Ae- x,
where A is a constant to be determined by normalization. Normalize this
wave function and find the probabilities of the particle being between 0 and
1/ , and between 1/ and 2/ .
  Ae
Probabilit
y density





2
A e


2  x
A
 x
  dx 
*
dx  2 

Monday, Oct. 15, 2012

 A e


2
A e
0
 x
2  x
dx 
 A e
*
2A
 x
d x   A e

*
 x


2
2 
e
2  x
Normalized Wave Function
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu

0
0
A

 
A e
 x
d x 
2
 1
e
 x
11
Ex 6.4: Normalization, cont’d
Using the wave function, we can compute the probability for a particle to be
with 0 to 1/ and 1/ to 2/ .
e
 
 x
For 0 to 1/ :
P 

1
  dx 
*
0

1 
e
2  x
dx 
0

2 
1
e
2  x
 
0
e

2
1
2

 1  0.432
For 1/ to 2/ :
P 

2 
1
  dx 
*

2 
1
e
2  x
dx 

 2
2 
e
2  x
 
1
e

2
1
4
e
2
  0.059