PHYS 3313 – Section 001
Lecture #14
Wednesday, Oct. 30, 2013
Dr. Jaehoon Yu
•
•
•
•
•
Infinite Square Well Potential
Finite Potential Well
Penetration Depth
Degeneracy
Simple Harmonic Oscillator
Wednesday, Oct. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
1
• Homework #5
Announcements
– CH6 end of chapter problems: 3, 5, 11, 14, 22, and 26
– Due on Wednesday, Nov. 6 in class
• Quiz #3
– At the beginning of the class Wed., Nov. 6
– Covers CH5.6 to what we cover Monday, Nov. 4
– Prepare your own formula sheet
• Research paper template is posted onto the research link
– Deadline for research paper submission is Monday, Nov. 25!!
• Colloquium today
– 4pm today, SH101, Dr. O. Auciello, UTD
• Colloquium coming week
– 4pm Monday, Nov. 4, SH101, Dr. Yujie Ding of Lehigh Univ.: Double extra credit
– 4pm Wednesday, Nov. 6, SH101, Dr. David Nygren of Lorentz Berkeley National
Laboratory, Triple extra credit
Wednesday, Oct. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
2
Wednesday, Oct. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
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Wednesday, Oct. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
4
Special project #5
Show that the Schrodinger equation
becomes Newton’s second law. (15 points)
Deadline Monday, Nov. 11, 2013
You MUST have your own answers!
Wednesday, Oct. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
5
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 derivative 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, Oct. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
6
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
2
d
y
2mE
2
becomes
where
.
k
=
2mE
=
-k
y
=
y
2
2
2
2
2
dx
• The general solution is y ( x ) = Asin kx + Bcos kx .
Wednesday, Oct. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
7
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
æ np x ö
• The wave function is now
y x = Asin
n
( )
çè
÷ø
L
• We normalize the wave function
ò
+¥
-¥
y ( x )y n ( x )dx = 1
*
n
A
2
ò
L
0
æ np x ö
sin ç
dx = 1
÷
è L ø
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.
Wednesday, Oct. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
8
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.
9p 2 2
E3 =
= 9E1
2mL2
y n ( x) =
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
E1 =
Wednesday, Oct. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
p2
2
2mL2
9
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?
– 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
æ np x ö 2
2 1 1
2 æ np x ö
lim sin 2 ç
»
sin
=
÷
× =
ç
÷
è L ø L
è L ø
L n®¥
L 2 L
• Which is identical to the CM probability!!
• One can also see this from the plot of P!
Wednesday, Oct. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
10
1
L
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, Oct. 30, 2013
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, Fall 2013
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
11
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, Oct. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
12