Monday, Oct. 15, 2012
Dr.
Jaehoon Yu
• 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

• Reminder Homework #4
– End of chapter problems on CH5: 8, 10, 16, 24, 26, 36 and 47
– Due: This Wednesday, Oct. 17
• Reading assignments
– 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.
 You MUST have your own answers!
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
   i h
 t
  h 2
2 m
 2   
 x
2

V
  
 The extension into three dimensions is i h

 t
  h 2

2 m

 2 
 x
2

 2 
 y
2

 2  
 z
2 

V
  x , y , z , t

• where is an imaginary number
Monday, Oct. 15, 2012 PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
5

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
x,t and 
2
1
x,t
2
x,t where a and b are constants and
x,t describe two waves each satisfying the Schrodinger Eq.

1
  a

1

 t


 t

 b

2 a

1
 b

2
 i h
 a

 t
1

1
 t
 
 b h 2
2 m

2
 t
 2 
1
 x
2

V

1

 x


 x
 a

1
 b

2

 a

1
 x
 b

2
 x i h

2
 t h 2
 
2 m
 2 
2
 x
2

V

2
 2 
 x
2


 x

 a

1
 x
 b

2
 x


 a
 2 
1
 x
2
 b
 2 
2
 x
2 i h

 t h 2
 
2 m
 2 
 x
2

V

Rearrange terms i h

 t
 h 2
2 m
 2 
 x
2

V
 

 i h

 t h 2

2 m
 2
 x
2

V


 
0 i h

 t
 i h

 a

1
 t
 b

2
 t

 h
2
 
2 m

 a
 2 
1
 x
2
 b
 2 
2
 x
2





1
 b

2


 h

1
 t h 2

2 m
 2 
1
 x
2

V

1


 

 h

2
 t h 2

2 m
 2 
2
 x
2

V

2



0
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:
   
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

Show that Ae i(kx t) satisfies the time-dependent Schrodinger wave Eq.
 
Ae i kx

 x


 x
  t
 
 t

Ae
   t


 iAke i kx
  t



 t

Ae
   t


  iA
 e i kx
  t

 ik

  i
  i
 2 
 x 2 h

 t



 x i h
 
  i


  ik

The Energy: E
 hf
 h
The wave number: k


 x h

2




 
  

2





2
 h p ik iAke h 2
2 m
 h



 k
2 
2
 p h


 p h

 t



 
V

Ak
2 e i kx
  t



  k
2 
 



E
 h 2 p k
2 m
2
2 m
2

V
The momentum: p
 h k
2
From the energy conservation: E

K

V
 p

V
2 m
So Ae i(kx t) is a good solution and satisfies Schrodinger Eq.
E


V p




2
2 m
 
0

0

V

0
Monday, Oct. 15, 2012 PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
8

Determine  x,t Asin(kx t) is an acceptable solution for the timedependent Schrodinger wave Eq.
 

 x

A

 x sin

 kx
  t

A sin
 kx
  t


 
A sin
 kx

 t
 t
 
 kA cos
 kx
  t

 t
 
 
A
 cos
 kx
  t
 i
 2 
 x 2 h

 


 x cos
 i h
 cos
 kx
  t

 kA kx
 

 cos
 h 2
 k
2 m
2 t

 kx

V




 
 t h 2
2 m sin
 kx

 

 t

 

 iE k
2 k
2 sin cos
A
 sin
 kx kx

 kx

 t


 t





 t

 p
 
V
2
2 m k sin
2


V
 kx


 
 t
 sin kx

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
 t


 The probability P(x) dx of a particle being between x and X + dx was given in the equation
  dx
  * x , t
 x , t dx
• Here  * denotes the complex conjugate of 
 The probability of the particle being between x
1 is given by
P
  x x
1
2  *  dx and x
2
 The wave function must also be normalized so that the probability of the particle being somewhere on the x axis is 1.
 

 *
     dx

1
Monday, Oct. 15, 2012 PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
10

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
  x
Probabilit y density
 

 *  dx

 


Ae
  x

*

Ae
  x
 dx

 


A
* e
  x
 
Ae
  x
 dx

 

A
2 e


2
 x dx
 
0

2 A
2 e

2
 x dx

A
 
Monday, Oct. 15, 2012
2 A
2

2
 e

2
 x
0


0

A
2


1
Normalized Wave Function
PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
   e
  x
11

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
 
0
1

 *  dx
 
0
1

 e

2
 x dx



2
 e

2
 x
1

0
 
1
2
 e

2

1


0.432
For
1/
 to 2/  :
2

P
  *
1

 2
   dx

1

 e

2
 x dx



2
 e

2
 x
2

1

 
1
2
 e

4  e

2


0.059
How about
2/

: to ∞?
Monday, Oct. 15, 2012 PHYS 3313-001, Fall 2012
Dr. Jaehoon Yu
12