PHYS 3313 – Section 001
Lecture #20
Monday, April 10, 2017
Dr. Amir Farbin
•
•
•
•
•
Infinite Square-well Potential
Finite Square Well Potential
Penetration Depth
Degeneracy
Simple Harmonic Oscillator
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
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, Apr. 12
• Quiz 3 results
– Class average: 25.1
• Equivalent to: 50.2/100
• Previous quizzes: 21/100 and 57.6/100
– Top score: 47/50
• Reminder: Quadruple extra credit
– Colloquium on April 19, 2017
– Speaker: Dr. Nigel Lockyer, Director of Fermilab
– Make your arrangements to take advantage of this opportunity
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
2
Reminder: Special project #6
• Show that the Schrodinger equation
becomes Newton’s second law in the
classical limit. (15 points)
• Deadline Monday, Apr. 17, 2017
• You MUST have your own answers!
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
3
Infinite Square-Well Potential
• The simplest such system is that of a particle trapped in a
box with infinitely hard walls that the particle cannot
penetrate. This potential is called an infinite square well
and is given by
x £ 0, x ³ L
ì¥
V ( x) = í
0<x<L
î0
• The wave function must be zero where the potential is
infinite.
• Where the potential is zero inside the box, the time
d y ( x)
independent Schrödinger wave equation 2m dx + V ( x )y ( x ) = Ey ( x )
2
d
y
2mE
2
2 .
becomes
where
=
-k
y
k
=
2mE
=
y
2
2
2
2
2
dx
• The general solution is y ( x ) = Asin kx + Bcos kx .
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
4
Quantization
• Since the wave function must be continuous, the boundary conditions
of the potential dictate that the wave function must be zero at x = 0
and x = L. These yield valid solutions for B=0, and for integer values
of n such that kL = n  k=n/L
np x ö
æ
• The wave function now becomes y x = Asin ç
÷
n
• We normalize the wave function
ò
( )
è L ø
+¥
-¥
y * ( x )y n ( x )dx = 1
n
L
æ np x ö
2
2 æ np x ö
A ò sin ç
dx = A ò sin ç
dx = 1
÷
÷
-¥
0
è L ø
è L ø
2
+¥
2
• The normalized wave function becomes
y n ( x) =
2
æ np x ö
sin ç
÷ø
è
L
L
• These functions are identical to those obtained for a vibrating string
with fixed ends.
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
5
Quantized Energy
n
• The quantized wave number now becomes kn ( x ) = np = 2mE
2
L
• Solving for the energy yields
2 2
p
2
n
( n = 1,2, 3, )
En =
2
2mL
• Note that the energy depends on the integer values of n.
Hence the energy is quantized and nonzero.
• The special case of n = 1 is called the ground state energy.
y n ( x) =
9p 2 2
E3 =
= 9E1
2mL2
2
æ np x ö
sin ç
è L ÷ø
L
2p 2 2
E2 =
= 4E1
mL2
y n*y n = y n =
2
2 2 æ np x ö
sin ç
è L ÷ø
L
Monday, April 10, 2017
E1 =
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
p2
2
2mL2
6
How does this correspond to Classical Mech.?
• What is the probability of finding a particle in a box of length L?
• Bohr’s correspondence principle says that QM and CM must
correspond to each other! When?
1
L
– When n becomes large, the QM approaches to CM
• So when n∞, the probability of finding a particle in a
box of length L is
P ( x ) = y n ( x )y n ( x ) = y n ( x )
*
2
2 1 1
2
2
2 æ np x ö
2 æ np x ö
sin ç
sin ç
= × =
= lim
÷ø »
÷
n®¥
è
L 2 L
è L ø
L
L
L
• Which is identical to the CM probability!!
• One can also see this from the plot of P!
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
7
Expectation Values & Operators
• Expectation value for any function g(x)
+¥
g ( x ) = ò Y ( x,t ) g ( x ) Y ( x,t ) dx
*
-¥
• Position operator is the same as itself, x
• Momentum Operator
¶
p̂ = -i
¶x
• Energy Operator
¶
Ê = i
¶t
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
8
Ex 6.8: Expectation values inside a box
Determine the expectation values for x, x2, p and p2 of a particle in
an infinite square well for the first excited state.
What is the wave function of the first excited state? n=? 2
y n=2 ( x ) =
L
2 L
2 æ 2p x ö
= ò y ( x ) xy n=2 ( x ) = ò x sin ç
dx =
n=2
è L ÷ø
-¥
2
L 0
+¥
x
n=2
x2
p
=
n=2
n=2
2
æ 2p x ö
sin ç
è L ÷ø
L
*
2 L 2 2 æ 2p x ö
2
x
sin
dx
=
0.32L
ç
÷
è L ø
L ò0
2 L æ 2p x ö
= ò sin ç
÷ø ( -i
0
è
L
L
2 L æ 2p x ö
p n=2 = ò sin ç
÷ø ( -i
0
è
L
L
2
p n=2
4p 2 2
E2 =
=
2
2m
2mL
2
Monday, April 10, 2017
2 2p
¶ é
np x ù
) êsin æçè ö÷ø údx = -i L L
¶x ë
L û
)
2
¶2 é æ 2p x ö ù
sin ç
÷ø údx =
2 ê
è
¶x ë
L û
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
2
ò
L
0
æ 2p x ö
æ 2p x ö
sin ç
cos
çè
÷ dx = 0
è L ÷ø
L ø
2 æ 2p ö
ç ÷
Lè L ø
2
ò
L
0
4p 2
æ 2p x ö
sin ç
dx =
L2
è L ÷ø
2
9
2
Ex 6.9: Proton Transition Energy
A typical diameter of a nucleus is about 10-14m. Use the infinite square-well
potential to calculate the transition energy from the first excited state to the
ground state for a proton confined to the nucleus.
The energy of the state n is En = n 2
p2
2
2mL2
What is n for the ground state? n=1
2
2
15
2
p 2 2 p 2 2c2
p
×
197.3eV
×
nm
1
1.92
´10
eV
(
)
E =
=
=
=
= 2.0MeV
1
2mL2
2mc 2 L2
mc 2
2 × (10 -5 nm )
938.3 ´10 6 eV
What is n for the 1st excited state? n=2
E2 = 2
2
p2
2
2
2mL
= 8.0MeV
So the proton transition energy is
DE = E2 - E1 = 6.0MeV
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
10
Finite Square-Well Potential
• The finite square-well potential is
ìV0
ï
V ( x ) = í0
ïV
î 0
x £ 0,
0<x<L
x³L
• The Schrödinger equation outside the finite well in regions I and III is
2
1 d 2y
= ( E - V0 ) for regions I and III, or using a 2 = 2m (V0 - E ) 2
2
2m y dx
d 2y
yields dx 2 = a 2y . The solution to this differential has exponentials of the
form eαx and e-αx. In the region x > L, we reject the positive exponential
and in the region x < 0, we reject the negative exponential. Why?
y I ( x ) = Aea x
region I, x < 0
y III ( x ) = Ae-a x region III, x > L
This is because the wave function
should be 0 as x±infinity.
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
11
Finite Square-Well Solution
•
Inside the square well, where the potential V is zero and the particle is free, the
2
d
y
2
wave equation becomes
where k = 2mE
=
-k
y
dx 2
•
Instead of a sinusoidal solution we can write
•
The boundary conditions require that
•
•
•
2
y II ( x ) = Ceikx + De-ikx region II, 0<x < L
y I = y II at x = 0 and y II = y III at x = L
and the wave function must be smooth where the regions meet.
Note that the
wave function is
nonzero outside
of the box.
Non-zero at the
boundary either..
What would the
energy look like?
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
12
Penetration Depth
• The penetration depth is the distance outside the
potential well where the probability significantly
decreases. It is given by
dx »
1
a
=
2m (V0 - E )
• It should not be surprising to find that the penetration
distance that violates classical physics is
proportional to Planck’s constant.
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
13
Three-Dimensional Infinite-Potential Well
• The wave function must be a function of all three spatial coordinates.
2
p
• We begin with the conservation of energy E = K +V =
+V
2m
• Multiply this by the wave function to get
p2
æ p2
ö
y + Vy
Ey = ç
+ V ÷y =
2m
è 2m
ø
• Now consider momentum as an operator acting on the wave function.
In this case, the operator must act twice on each dimension. Given:
¶y p̂ y = -i ¶y p̂ y = -i ¶y
2
2
2
2
p = px + py + pz
p̂xy = -i
y
z
¶y
¶z
¶x
• The three dimensional Schrödinger wave equation is
2
2
æ ¶2y ¶2y ¶2y ö
2
Rewrite
+
+
+
V
y
=
E
y
Ñ
y + Vy = Ey
2
2 ÷
ç 2
2m è ¶x
Monday, April 10, 2017
¶y
¶z ø
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
2m
14
Ex 6.10: Expectation values inside a box
Consider a free particle inside a box with lengths L1, L2 and L3 along the x, y, and z axes,
respectively, as shown in the figure. The particle is constrained to be inside the box. Find
the wave functions and energies. Then find the ground energy and wave function and the
energy of the first excited state for a cube of sides L.
What are the boundary conditions for this situation?
Particle is free, so x, y and z wave functions are independent from each other!
Each wave function must be 0 at the wall! Inside the box, potential V is 0.
-
2
2m
Ñ2y + Vy = Ey Þ -
A reasonable solution is
2
2m
Ñ2y = Ey
y ( x, y, z ) = Asin ( k1x ) sin ( k2 y ) sin ( k3z )
Using the boundary
condition
y = 0 at x = L1 Þ k1L1 = n1p Þ k1 = n1p L1
So the wave numbers are
Monday, April 10, 2017
n1p
k1 =
L1
k2 =
n2p
L2
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
k3 =
n3p
L3
15
Ex 6.10: Expectation values inside a box
Consider a free particle inside a box with lengths L1, L2 and L3 along the x, y, and z axes,
respectively, as shown in figure. The particle is constrained to be inside the box. Find the
wave functions and energies. Then find the round energy and wave function and the energy
of the first excited state for a cube of sides L.
The energy can be obtained through the Schrödinger equation
2
2
2
æ
ö
¶
¶
¶
2
+ 2 + 2 ÷ y = Ey
Ñy =2
ç
2m è ¶x ¶y ¶z ø
2m
2
2
¶y
¶
= ( Asin ( k1 x ) sin ( k2 y ) sin ( k3z )) = k1Acos ( k1x ) sin ( k2 y ) sin ( k3z )
¶x ¶x
¶2y
¶2
2
2
=
Asin
k
x
sin
k
y
sin
k
z
=
-k
Asin
k
x
sin
k
y
sin
k
z
=
-k
y
(
)
(
)
(
)
(
)
(
)
(
)
(
)
1
2
3
1
1
2
3
1
2
2
¶x
¶x
2
æ ¶2
¶2
¶2 ö
2
2
2
+
+
y
=
k
+
k
+
k
(
2
2
2
ç
÷
1
2
3 )y = Ey
2m è ¶x ¶y ¶z ø
2m
2
p 2 2 æ n12 n22 n32 ö
2
2
2
E=
k1 + k2 + k3 ) =
+ 2 + 2÷
(
2
ç
2m
2m è L1 L2 L3 ø
2
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
What is the ground state energy?
E1,1,1 when n1=n2=n3=1, how much?
When are the energies the same
for different combinations of ni?
16
Degeneracy*
• Analysis of the Schrödinger wave equation in three
dimensions introduces three quantum numbers that
quantize the energy.
• A quantum state is degenerate when there is more
than one wave function for a given energy.
• Degeneracy results from particular properties of the
potential energy function that describes the system.
A perturbation of the potential energy, such as the
spin under a B field, can remove the degeneracy.
*Mirriam-webster: having two or more states or subdivisions
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
17
The Simple Harmonic Oscillator
•
Simple harmonic oscillators describe many physical situations: springs, diatomic molecules
and atomic lattices.
F = -k ( x - x0 )
•
Consider the Taylor expansion of a potential function:
1
2
V ( x ) = V0 + V1 ( x - x0 ) + V2 ( x - x0 ) +
2
The minimum potential at x=x0, so dV/dx=0 and V1=0; and the zero potential V0=0, we
have
1
2
V ( x ) = V2 ( x - x0 )
2
Substituting this into the wave equation:
æ 2m
d 2y
2m æ
k x2 ö
mk x 2 ö
= - 2 çEy = ç - 2 E + 2 ÷y
2
÷
dx
2 ø
è
è
ø
2
dy
2 2
2mE
mk
2
=
a
x - b )y
(
Let a = 2 and b = 2 which yields dx 2
.
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
18
Parabolic Potential Well
•
•
•
If the lowest energy level is zero, this violates the uncertainty principle.
-a x 2 2
y n = H n ( x ) e where Hn(x) are Hermite
The wave function solutions are
polynomial function of order n.
In contrast to the particle in a box, where the oscillatory wave function is a sinusoidal
curve, in this case the oscillatory behavior is due to the polynomial, which dominates
at small x. The exponential tail is provided by the Gaussian function, which
dominates at large x.
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
19
Analysis of the Parabolic Potential Well
•
•
The energy levels are given by
1ö
æ
En = ç n + ÷
è
2ø
•
Monday, April 10, 2017
1ö
2
The zero point energy is called the Heisenberg limit:
E0 =
•
æ
k m = çn+ ÷ w
è
ø
1
w
2
Classically, the probability of finding the mass is
greatest at the ends of motion’s range and smallest at
the center (that is, proportional to the amount of time
the mass spends at each position).
Contrary to the classical one, the largest probability for
this lowest energy state is for the particle to be at the
center.
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
20
Ex. 6.12: Harmonic Oscillator stuff
• Normalize the ground state wave function  0 for the
simple harmonic oscillator and find the expectation
values <x> and <x2>.
y n ( x) = Hn ( x) e
ò
+¥
-¥
-a x 2 2
Þy 0 ( x ) = H 0 ( x ) e
+¥
-a x 2 2
+¥
= Ae
-a x 2 2
æ1 p ö
=1
÷
è2 aø
y *y 0 dx = ò A2 e-a x dx = 2A2 ò e-a x dx = 2A2 ç
2
-¥
0
a
æaö
A2 =
ÞA=ç ÷
èpø
p
0
14
æaö
Þ H0 ( x) = ç ÷
èpø
a
x = ò y xy 0 dx =
-¥
p
+¥
*
0
+¥
x 2 = ò y *0 x 2y 0 dx =
-¥
x2 =
2 mk
ò
+¥
-¥
a
p
ò
xe
+¥
-¥
-a x2
14
æaö
Þy 0 ( x ) = ç ÷
èpø
14
e-a x
2
2
dx = 0
x 2 e-a x dx = 2
2
Þ w = k m Þ x2 =
Monday, April 10, 2017
2
a
p
ò
+¥
0
x 2 e-a x dx = 2
2
aæ p ö 1
=
32÷
ç
p è 4a ø 2a
2mw
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
21
Barriers and Tunneling
•
•
•
Consider a particle of energy E approaching a potential barrier of height V0 and the
potential everywhere else is zero.
We will first consider the case when the energy is greater than the potential barrier.
In regions I and III the wave numbers are: kI = kIII = 2mE
•
In the barrier region we have
Monday, April 10, 2017
kII =
2m ( E - V0 )
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
where V = V0
22
Reflection and Transmission
•
•
The wave function will consist of an incident wave, a reflected wave, and a transmitted
wave.
The potentials and the Schrödinger wave equation for the three regions are as
follows:
d 2y I 2m
Region I ( x < 0 )
V =0
+ 2 Ey I = 0
2
dx
Region II ( 0 < x < L ) V = V0
Region III( x > L )
•
The corresponding solutions are:
V =0
Region I ( x < 0 )
d 2y II 2m
+ 2 ( E - V0 )y II = 0
dx 2
d 2y III 2m
+ 2 Ey III = 0
dx 2
I
I
Region II ( 0 < x < L ) y II = CeikII x + De-ikII x
Region III( x > L )
•
y I = Aeik x + Be-ik x
y III = Feik x + Ge-ik x
I
I
As the wave moves from left to right, we can simplify the wave functions to:
Incident wave
y I (incident) = AeikI x
Reflected wave
y I (reflected) = Be-ikI x
Transmitted wave y III (transmitted) = FeikI x
Monday, April 10, 2017
PHYS 3313-001, Spring 2017
Dr. Jaehoon Yu
23