PHYS 3313 – Section 001
Lecture #19
Wednesday, Apr. 2, 2014
Dr. Jaehoon Yu
•
•
•
•
Infinite Square Well Potential
Finite Square Well Potential
Penetration Depth
3D Infinite Potential Well
Wednesday, Apr. 2, 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
• Homework #4
– End of chapter problems on CH5: 8, 10, 16, 24, 26, 36 and 47
– Due Wednesday, Apr. 9
• Quiz #3 at the beginning of the class Monday, Apr. 7
– Covers CH 5.1 through CH6.4
• Special colloquium at 4pm, today for triple extra credit
– Dr. Gerald Blazey of the White House
Wednesday, Apr. 2, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
2
Reminder: 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, Apr. 2, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
4
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!
Wednesday, Apr. 2, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
5
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!
Wednesday, Apr. 2, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
6
Details of the computation
2
2 1
2 æ np x ö
sin ç
= ×
÷
è L ø
L
L 2p
ò
2p
0
2
sin
( y)dy
np x
Let y =
L
2 1
= ×
L 2p
ò
2p
0
1
(1- cos 2y )dy
2
2p
2 1 1é 1
ù = 2 × 1 × 1 2p - 0 - 0
= ×
× ê y - sin 2y ú
[
]
L 2p 2 ë 2
û 0 L 2p 2
2 1 1
2 1 1
= ×
× ×2p =
× =
L 2p 2
L 2 L
Wednesday, Apr. 2, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
7
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
Wednesday, Apr. 2, 2014
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 2014
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
8
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
Wednesday, Apr. 2, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
9
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
2
d
yields y2 = a 2y . The solution to this differential has exponentials of the
form
dx
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, Apr. 2, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
10
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, Apr. 2, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
11
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, Apr. 2, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
12