PHYS 3313 – Section 001
Lecture #20
Monday, Apr. 7, 2014
Dr. Jaehoon Yu
•
•
•
•
3D Infinite Potential Well
Degeneracy
Simple Harmonic Oscillator
Barriers and Tunneling
Monday, Apr. 7, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
1
Announcements
• Research paper deadline is Monday, Apr. 28
• Research presentation deadline is Sunday, Apr. 27
• Reminder: Homework #4
– End of chapter problems on CH5: 8, 10, 16, 24, 26, 36 and 47
– Due this Wednesday, Apr. 9
• Bring out special project #4 at the end of class
Monday, Apr. 7, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
2
Reminder: Special project #5
Show that the Schrodinger equation becomes
Newton’s second law in the classical limit. (15
points)
Deadline Monday, Apr. 21, 2014
You MUST have your own answers!
Monday, Apr. 7, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
3
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, Apr. 7, 2014
¶y
¶z ø
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
2m
4
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, Apr. 7, 2014
n1p
k1 =
L1
k2 =
n2p
L2
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
k3 =
n3p
L3
5
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 Fire. 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, Apr. 7, 2014
PHYS 3313-001, Spring 2014
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?
6
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, Apr. 7, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
7