PHYS 3313 – Section 001
Lecture # 20
Wednesday, April 15, 2015
Dr. Jaehoon Yu
•
•
•
•
•
•
Finite Square Well Potential
Penetration Depth
Degeneracy
Simple Harmonic Oscillator
Barriers and Tunneling
Alpha Particle Decay
Wednesday, April 15,
2015
PHYS 3313-001, Spring 2015
Dr. Jaehoon Yu
1
Announcements
• Research paper deadline is Monday, May 4
• Research presentation deadline is Sunday, May 3
• Homework #5
– CH6 end of chapter problems: 34, 39, 46, 62 and 65
– Due Wednesday, Apr. 22
• Reading assignments
– CH7.6 and the entire CH8
• Bring out the special project #5
• No Colloquium today
Wednesday, April 15,
2015
PHYS 3313-001, Spring 2015
Dr. Jaehoon Yu
2
Wednesday, April 15,
2015
PHYS 3313-001, Spring 2015
Dr. Jaehoon Yu
3
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.
Wednesday, April 15,
2015
PHYS 3313-001, Spring 2015
Dr. Jaehoon Yu
4
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?
Wednesday, April 15,
2015
PHYS 3313-001, Spring 2015
Dr. Jaehoon Yu
5
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.
Wednesday, April 15,
2015
PHYS 3313-001, Spring 2015
Dr. Jaehoon Yu
6
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
Wednesday, April 15,
2015
¶y
¶z ø
PHYS 3313-001, Spring 2015
Dr. Jaehoon Yu
2m
7
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
Wednesday, April 15,
2015
n1p
k1 =
L1
k2 =
n2p
L2
PHYS 3313-001, Spring 2015
Dr. Jaehoon Yu
k3 =
n3p
L3
8
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
Wednesday, April 15,
2015
PHYS 3313-001, Spring 2015
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?
9
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
Wednesday, April 15,
2015
PHYS 3313-001, Spring 2015
Dr. Jaehoon Yu
10