PHYS 3313 – Section 001
Lecture # 19
Monday, April 13, 2015
Dr. Jaehoon Yu
•
Refresher: Infinite Square-well Potential
–
–
•
•
Energy quantization
Expectation value computations
Finite Square Well Potential
Penetration Depth
Monday, April 13, 2015
PHYS 3313-001, Spring 2015
Dr. Jaehoon Yu
1
Announcements
• Quiz #4 at the beginning of the class this
Wednesday, Apr. 15
– The class web site was down so I will give you an
extension..
– Covers CH 5.4 through what we finish today
– BFOF
• Colloquium 4pm Wednesday, SH101
Monday, April 13, 2015
PHYS 3313-001, Spring 2015
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 this Wednesday, Apr. 15, 2015
• You MUST have your own answers!
Monday, April 13, 2015
PHYS 3313-001, Spring 2015
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 .
Wednesday, April 8, 2015
PHYS 3313-001, Spring 2015
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
æ np x ö
• The wave function is now
y x = Asin
n
• We normalize the wave function
( )
ò
çè
÷ø
L
+¥
-¥
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.
Wednesday, April 8, 2015
PHYS 3313-001, Spring 2015
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 13, 2015
E1 =
PHYS 3313-001, Spring 2015
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 13, 2015
PHYS 3313-001, Spring 2015
Dr. Jaehoon Yu
7
Expectation Value & 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
Wednesday, April 8, 2015
PHYS 3313-001, Spring 2015
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 13, 2015
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 2015
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 13, 2015
PHYS 3313-001, Spring 2015
Dr. Jaehoon Yu
10