PHYS 3313 – Section 001
Lecture #12
Monday, Oct. 21, 2013
Dr. Jaehoon Yu
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The Schrödinger Wave Equation
Probability Density
Wave Function Normalization
Time-Independent Schrödinger Wave Equation
Expectation Values
Operators – Position, Momentum and Energy
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
1
Announcements
• Mid-term exam results
– Class average: 45.2/96
• Equivalent to 47/100
– Top score: 96/96
• Mid-term grade discussion moved to Monday, Oct. 28
• Reminder Homework #4
– End of chapter problems on CH5: 8, 10, 16, 24, 26, 36 and 47
– Due: This Wednesday, Oct. 23
• Reading assignments
– CH6.1 – 6.7 + the special topic
• Colloquia this week
– 4pm today, SH101, Dr. C. Guo of U. of Rochester
– 4pm Wednesday, SH101, Dr. X. Chu, U. of Colorado
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
2
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
3
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
4
Research Projects
1. Each of the 12 research groups picks one research topic
2. Study the topic as a group, looking up references
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Original theory or Original observation
Experimental proofs or Theoretical predictions + subsequent
experimental proofs
Importance and the impact of the theory/experiment
3. Each member of the group writes a 10 page report,
including figures (must not copy!!)
–
–
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10% of the total grade
Can share the theme and facts but you must write your own!
Due Mon., Nov. 25, 2013
4. The group presents a 10min power point talk
–
–
5% of the total grade
Date and time will be announced close to the end of the semester
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
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Group – Research Topic Association
Research Group Number
1
2
3
Research Topic
2
3
11
4
5
6
7
7
5
1
9
8
9
8
4
10
11
12
6
10
12
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
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1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Research Topics
Black body radiation
Michelson–Morley experiment
The Photoelectric effect
Special Relativity
The property of molecules, Browning Motion
Compton Effect
Discovery of the electron
Radioactivity
Rutherford Scattering
Super-conductivity
The Unification of Electromagnetic and Weak forces
The Discovery of the Higgs-like particle
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
7
Special Project #4
• 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. 28.
• You MUST have your own answers!
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
8
The Schrödinger Wave Equation
• Erwin Schrödinger and Werner Heinsenberg
proposed quantum theory in 1920
• The two proposed very different forms of equations
• Heinserberg: Matrix based framework
• Schrödinger: Wave mechanics, similar to the
classical wave equation
• Paul Dirac and Schrödinger later on proved that
the two give identical results
• The probabilistic nature of quantum theory is
contradictory to the direct cause and effect seen
in classical physics and makes it difficult to grasp!
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
9
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
¶Y ( x,t )
¶2 Y ( x,t )
i
=+ VY ( x,t )
2
¶t
2m ¶x
 The extension into three dimensions is
2
æ ¶2 Y ¶2 Y ¶2 Y ö
¶Y
i
=+ 2 + 2 ÷ + VY ( x, y, z,t )
2
ç
¶t
2m è ¶x
¶y
¶z ø
• where
Monday, Oct. 21, 2013
is an imaginary number
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
10
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.
2
¶Y1
¶ 2 Y1
Y = aY1 + bY2
i
=+ VY1
2
¶t
2m ¶x
¶Y ¶
¶Y
¶Y 2
= ( aY1 + bY 2 ) = a 1 + b
¶t ¶t
¶t
¶t
¶Y ¶
¶Y
¶Y 2
= ( aY1 + bY 2 ) = a 1 + b
¶x ¶x
¶x
¶x
2
¶Y
¶2 Y
i
=+ VY
2
¶t
2m ¶x
Rearrange terms
2
¶Y 2
¶2 Y 2
i
=+ VY 2
2
¶t
2m ¶x
¶2 Y ¶ æ ¶Y1
¶Y 2 ö
¶2 Y1
¶2 Y 2
= ça
+b
+b
÷ =a
¶x 2 ¶x è ¶x
¶x ø
¶x 2
¶x 2
2
2
æ ¶
ö
¶Y
¶2 Y
¶2
i
+
VY
=
i
+
V
çè ¶t 2m ¶x 2
÷ø Y = 0
¶t 2m ¶x 2
2
æ ¶ 2 Y1
¶Y
¶Y 2 ö
¶2 Y 2 ö
æ ¶Y1
i
= i ça
+b
a
+b
+ V ( aY1 + bY 2 )
÷ =è ¶t
¶t
¶t ø
2m çè ¶x 2
¶x 2 ÷ø
2
2
æ ¶Y1
ö
æ ¶Y 2
ö
¶ 2 Y1
¶2 Y 2
aç i
+
VY
=
-b
i
+
VY
1÷
2÷ = 0
çè
¶t
2m ¶x 2
è ¶t 2m ¶x 2
ø
ø
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
11
General Solution of the Schrödinger
Wave Equation
• The general form of the solution of the Schrödinger wave
equation is given by:
Y ( x,t ) = Ae
i( kx-w t )
= A éë cos ( kx - w t ) + isin ( kx - w t ) ùû
• which also describes a wave propagating 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. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
12
Ex 6.2: Solution for Wave Equation
Show that Aei(kx- t) satisfies the time-dependent Schrodinger wave Eq.
i( kx-w t )
¶Y ¶
=
Aei( kx-w t ) = -iAw ei( kx-w t ) = -iwY
¶t ¶t
(
Y = Ae
)
¶Y ¶
=
Aei( kx-w t ) = iAkei( kx-w t ) = ikY
¶x ¶x
(
)
¶2 Y ¶
¶
i( kx-w t )
2 i( kx-w t )
2
=
ikY
=
ik
Y
=
ik
iAke
=
-Ak
e
=
-k
Y
(
)
(
)
2
¶x
¶x
¶x
2 2
æ
ö
k
2
¶Y
w
V
Y=0
2
ç
÷
i
= i ( -iwY ) = wY = -k
Y
+
VY
( )
2m
è
ø
¶t
2m
æ
ö
p2
æw ö
=
E
V
=0
The Energy: E = hf = h ç ÷ = w
ç
÷
2m
è
ø
è 2p ø
2p 2p 2p p p
The momentum: p = k
The wave number: k =
=
=
=
l h p
h
2
2
(
)
p
From the energy conservation: E = K + V =
+V
2m
p
E-V = 0
2m
So Aei(kx- t) is a good solution and satisfies Schrodinger Eq.
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
13
Ex 6.3: Bad Solution for Wave Equation
Determine  (x,tAsin(kx- t) is an acceptable solution for the timedependent Schrodinger wave Eq.
Y = Asin ( kx - w t )
¶Y ¶
= ( Asin ( kx - w t )) = -Aw cos ( kx - w t )
¶t ¶t
¶Y ¶
= ( Asin ( kx - w t )) = kAcos ( kx - w t )
¶x ¶x
¶2 Y ¶
2
2
=
kAcos
kx
w
t
=
-k
Asin
kx
w
t
=
-k
Y
(
)
(
)
(
)
2
¶x
¶x
2
i ( -w cos ( kx - w t )) = -k 2 sin ( kx - w t )) + V sin ( kx - w t )
(
2m
æ 2k2
ö
-i w cos ( kx - w t ) = ç
+ V ÷ sin ( kx - w t )
è 2m
ø
æ p2
ö
-iE cos ( kx - w t ) = ç
+ V ÷ sin ( kx - w t )
è 2m
ø
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. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
14
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 = Y* ( x,t ) Y ( x,t ) dx
• Here * denotes the complex conjugate of 
• The probability of the particle being between x1 and x2
is given by
x2
P = ò Y*Y dx
x1
• The wave function must also be normalized so that the
probability of the particle being somewhere on the x
axis is 1.
+¥
*
Y
ò ( x,t ) Y ( x,t ) dx = 1
-¥
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
15
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/ .
Y = Ae
Probabilit
y density
ò
+¥
-¥
-a x
Y Ydx = ò
*
-¥
+¥
+¥
-¥
0
= ò A2 e-2a x dx = 2 ò
A= a
Monday, Oct. 21, 2013
+¥
( Ae ) ( Ae )dx = ò ( A e )( Ae )dx =
-a x
*
+¥
-a x
* -a x
-a x
-¥
2
2A -2a x
2 -2a x
e
Ae
dx =
-2a
Normalized Wave Function
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
+¥
= 0+
0
A2
a
= 1
Y = ae
-a x
16
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/ .
Y = ae
-a x
For 0 to 1/ :
P=ò
1a
0
Y Ydx = ò
*
1a
0
a -2a x
-2a x
e
a e dx =
-2a
1a
0
1 -2
= - ( e -1) » 0.432
2
For 1/ to 2/ :
P=ò
2a
1a
Y Ydx =
*
2a
ò a ae
1
2a
-2a x
a -2a x
1 -4 -2
e
=
e - e ) » 0.059
dx =
(
-2a
2
1a
How about 2/ :to ∞?
Monday, Oct. 21, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
17