Transcript Wednesday, Mar. 26, 2014

PHYS 3313 – Section 001
Lecture #17
Wednesday, Mar. 26, 2014
Dr. Jaehoon Yu
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Probability of Particle
Schrodinger Wave Equation and Solutions
Normalization and Probability
Time Independent Schrodinger Equation
Expectation Values
Momentum Operator
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
1
Announcements
• Research paper template has been uploaded onto
the class web page link to research
• Special colloquium on April 2, triple extra credit
• Colloquium this Wednesday at 4pm in SH101
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
2
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, Apr. 7.
• You MUST have your own answers!
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
3
Probability of the Particle
•
The probability of
observing the particle
between x and x + dx in
each state is
•
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed
in the Bohr model, has
surfaced in a natural
way by using waves.
•
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
4
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!
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
5
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
Wednesday, Mar. 26,
2014
is an imaginary number
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
6
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
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
7
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.
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
8
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.
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
9
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.
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
10
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
-¥
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
11
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
Wednesday, Mar. 26,
2014
+¥
( 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, Spring 2014
Dr. Jaehoon Yu
+¥
= 0+
0
A2
a
= 1
Y = ae
-a x
12
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 ∞?
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
13
Properties of Valid Wave Functions
Boundary conditions
To avoid infinite probabilities, the wave function must be finite
everywhere.
2) To avoid multiple values of the probability, the wave function must be
single valued.
3) For finite potentials, the wave function and its derivatives must be
continuous. This is required because the second-order derivative
term in the wave equation must be single valued. (There are
exceptions to this rule when V is infinite.)
4) In order to normalize the wave functions, they must approach zero as
x approaches infinity.
Solutions that do not satisfy these properties do not generally
correspond to physically realizable circumstances.
1)
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
14
Time-Independent Schrödinger Wave Equation
• The potential in many cases will not depend explicitly on time.
• The dependence on time and position can then be separated
in the Schrödinger wave equation. Let, Y ( x,t ) = y ( x ) f ( t )
2
¶ f (t )
f ( t ) ¶2y ( x )
which yields: i y ( x )
=+ V ( x )y ( x ) f ( t )
2
¶t
2m
¶x
2
1 ¶ f (t )
1 ¶2y ( x )
i
=+ V ( x)
2
f ( t ) ¶t
2m y ( x ) ¶x
Now divide by the wave function:
• The left side of this last equation depends only on time, and
the right side depends only on spatial coordinates. Hence each
side must be equal to a constant. The time dependent side is
1 df
i
=B
f dt
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
15
Time-Independent Schrödinger Wave Equation(con’t)

We integrate both sides and find: i
df
ò f = ò Bdt Þ i ln f = Bt + C
where C is an integration constant that we may choose to be 0.
Bt
Therefore
ln f =
i
This determines f to be by comparing it to the wave function of a free
particle
f ( t ) = eBt i = e-iBt = e-iw t Þ B = w Þ B = w = E
1 ¶ f (t )
i
=E
f ( t ) ¶t

This is known as the time-independent Schrödinger wave
equation, and it is a fundamental equation in quantum mechanics.
d 2y ( x )
+ V ( x )y ( x ) = Ey ( x )
2
2m dx
2
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
16
Stationary State
• Recalling the separation of variables: Y ( x,t ) = y ( x ) f ( t )
-iw t
and with f(t) = e
the wave function can be
-iw t
written as:
Y ( x,t ) = y ( x ) e
• The probability density becomes:
Y Y=y
*
2
( x )( e
iw t -iw t
e
) =y ( x )
2
• The probability distributions are constant in time.
This is a standing wave phenomena that is called the
stationary state.
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
17
Comparison of Classical and
Quantum Mechanics
•
•
•
Newton’s second law and Schrödinger’s wave
equation are both differential equations.
Newton’s second law can be derived from the
Schrödinger wave equation, so the latter is the more
fundamental.
Classical mechanics only appears to be more precise
because it deals with macroscopic phenomena. The
underlying uncertainties in macroscopic measurements
are just too small to be significant due to the small size
of the Planck’s constant
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
18
Expectation Values
• In quantum mechanics, measurements can only be expressed in
terms of average behaviors since precision measurement of each
event is impossible (what principle is this?)
• The expectation value is the expected result of the average of many
measurements of a given quantity. The expectation value of x is
denoted by <x>.
• Any measurable quantity for which we can calculate the expectation
value is called a physical observable. The expectation values of
physical observables (for example, position, linear momentum, angular
momentum, and energy) must be real, because the experimental
results of measurements are real.
N i xi
å
N
x
+
N
x
+
N
x
+
N
x
+
• The average value of x is x = 1 1
2 2
3 3
4 4
= i
N1 + N 2 + N 3 + N 4 +
åN
i
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
19
i
•
Continuous Expectation Values
+¥
We can change from discrete to
xP ( x ) dx
ò
-¥
continuous variables by using
the probability P(x,t) of
observing the particle at a
particular x.
• Using the wave function, the x =
expectation value is:
• The expectation value of any
function g(x) for a normalized
wave function:
+¥
x=
ò
+¥
-¥
+¥
ò
ò
xY ( x,t ) Y ( x,t ) dx
-¥
+¥
-¥
P ( x ) dx
*
Y ( x,t ) Y ( x,t ) dx
*
g ( x ) = ò Y ( x,t ) g ( x ) Y ( x,t ) dx
*
-¥
Wednesday, Mar. 26,
2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
20
Momentum Operator
• To find the expectation value of p, we first need to represent p in
terms of x and t. Consider the derivative of the wave function of a
free particle with respect to x:
¶Y ¶ i( kx-w t )
éë e
ùû = ikei( kx-w t ) = ikY
=
¶x ¶x
With k = p / ħ we have ¶Y = i p Y
¶x
¶Y ( x,t )
This yields
p éë Y ( x,t ) ùû = -i
¶x
• This suggests we define the momentum operator as
• The expectation value of the momentum is
p =
p̂ = -i.
¶
¶x
¶Y ( x,t )
ò-¥ Y ( x,t )p̂Y ( x,t ) dx = -i ò-¥ Y ( x,t ) ¶x dx
+¥
Wednesday, Mar. 26,
2014
+¥
*
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
*
21