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Chem 452 – Exam II
Oct 15 2014
For a 1D harmonic oscillator
N
v
2
=
1
α
π
2
v
v
!
H
v
(y): Hermite polynomial
H
0
= 1,
H
1
= 2y,
H
2
= 4y
2
‐2 …
H
v+1
‐ 2yH
v
+ 2vH
v‐1
=0
H
v
−∞
∞
∫
(
y
)
H
v
'
(
y
)
e
−
y
2
dy
=
0
if v
'
≠
v
π
2
v
v
!
if v
'
=
v
'
(
)
*
)
For a 3D rotor
only when
l
” =
l
+ 1 and m
l
” = m
l
’ + m
ˆ
L
2
=
−
2
Λ
2
=
−
2
1
sin
2
θ
∂
2
∂φ
2
+
1
sin
θ
∂
∂θ
sin
θ
∂
∂θ
#
$
%
&
'
(
For a classical rotor :
Reduced mass
:
Trigonometric identity :
sin
2
x
+ cos
2
x
= 1
cos
2
θ
a
b
∫
sin
θ
d
θ
=
1
3
cos
3
θ
"
#
$
%
&
'
a
b
Coordinates conversion:
x
= rsinθcosϕ ,
y
= rsinθsinϕ ,
z
= rcosθ
∇
2
=
∂
2
∂
x
2
+
∂
2
∂
y
2
+
∂
2
∂
y
2
=
∂
2
∂
r
2
+
2
r
∂
∂
r
+
1
r
2
Λ
2
,
Λ
2
=
1
sin
2
θ
∂
2
∂φ
2
+
1
sin
θ
∂
∂θ
sin
θ
∂
∂θ
ψ
v
(
x
)
=
N
v
H
v
(
y
)
e
−
y
2
2
,
y
=
x
α
,
α
=
2
k
f
m
%
&
'
'
(
)
*
*
1
4
E
v
=
(
v
+
1
2
)
ω
,
v
=
0, 1, 2...
ψ
l
,
m
l
(
θ
,
φ
)
=
Y
l
,
m
l
(
θ
,
φ
)
E
l
=
l
(
l
+
1)
2
2
I
,
I
=
mr
2
J
=
l
(
l
+
1)
,
J
z
=
m
l
l
=
0, 1, 2...
m
l
=
0,
±
1,
±
2...,
±
l
Y
l
',
m
l
'
*
∫
Y
l
,
m
l
d
τ
=
Y
l
',
m
l
'
*
Y
l
,
m
l
0
π
∫
0
2
π
∫
sin
θ
d
θ
d
φ
=
δ
l
'
l
δ
m
l
'
m
l
Y
l
",
m
l
"
*
Y
l
',
m
l
'
0
π
∫
0
2
π
∫
Y
l
,
m
l
sin
θ
d
θ
d
φ
≠
0
ˆ
l
z
=
−
i
∂
∂φ
E
=
J
2
2
I
I
=
mr
2
μ
=
m
1
m
2
m
1
+
m
2
e
cx
dx
a
b
∫
=
1
c
e
cx
[
]
a
b
sin
cxdx
a
b
∫
=
1
c
−
cos
cx
[
]
a
b
cos
cxdx
a
b
∫
=
1
c
sin
cx
[
]
a
b
d
τ
−∞
∞
∫
=
r
2
dr
sin
θ
d
θ
d
φ
0
∞
∫
0
π
∫
0
2
π
∫
h
= 6.626*10
–34
Js
= h/2
π
k
= 1.38*10
‐23
J/K
m
p
= 1.67*10
‐27
kg
m
e
= 9.109*10
–31
kg
Point Total
Problem
Possible Score
Your Score
1
8
2
12
3
16
4
14
5
10
6
10
7
12
8
18
Total
100
Name:
1/7
1.
(8 pts) Write the Hamiltonian operators for (a) a harmonic oscillator with mass m and
spring constant
k
f
and (b) an electron in a hydrogen atom separated by a fixed distance
r (i.e. r is a constant).
Hint: the restoration force exerted on a harmonic oscillator is
linearly proportional to the displacement from the equilibrium position. The potential
energy of a negative charge separated by r from a positive charge is ‐1/r.
(
a
)
−
2
2
m
∂
2
∂
x
2
+
1
2
k
f
x
2
(
b
)
−
2
2
μ
r
2
1
sin
2
θ
∂
2
∂φ
2
+
1
sin
θ
∂
∂θ
sin
θ
∂
∂θ
#
$
%
&
'
(
−
1
r
;
μ
=
m
e
m
p
m
e
+
m
p
≈
m
e
2.
(12 pts) Sketch the first four (v = 0 ~ 3) wavefunctions (Ψ
v
) and the probability
densities (Ψ
v

2
) of a 1D harmonic oscillator and mark approximate classical turning points.
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