2/9/11
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PHYS 360 Quantum Mechanics
Wed Feb 9, 2011
Lecture 13:
How can you get quanFzed energy levels from
an harmonic oscillator, which has no boundary
condiFons?
x
(
t
)
=
A
sin(
ω
t
)
+
B
cos(
ω
t
)
The Harmonic Oscillator
Consider the force exerted by a spring:
The soluFon has the form:
ω
≡
k
m
.
where
In quantum mechanics we rarely deal
with springs,
but many systems in equilibrium have
restoring forces that look like Hooke’s Law (above).
F
=
ma
=
m
d
2
x
dt
2
=
−
kx
2
The potenFal energy associated with a spring is….
F
=
−
kx
=
−
dV
dx
→
V
(
x
)
=
1
2
kx
2
=
1
2
m
ω
2
x
2
2
1
0
1
2
1
2
3
Put this potenFal into the Schrödinger EquaFon…..
3
V
(
x
)
=
1
2
m
ω
2
x
2
The harmonic potenFal:
−
2
2
m
d
2
ψ
n
dx
2
+
1
2
m
ω
2
x
2
ψ
n
=
E
n
ψ
n
We need to solve the Fme‐independent eigenvalue equaFon:
This is not as simple as it may look. There are two completely
different approaches: using operator algebra, or using power
series expansions.
We will focus now on the operator algebra method….
4
Raising and Lowering operators:
1
2
m
[ ˆ
p
2
+
(
m
ω
ˆ
x
)
2
]
ψ
n
=
E
n
ψ
n
Rewrite the Schrödinger
equaFon as:
We emphasize the operators:
ˆ
p
→
i
d
dx
,
ˆ
x
→
x
ˆ
H
=
1
2
m
[ ˆ
p
2
+
(
m
ω
ˆ
x
)
2
].
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 Spring '11
 DURBIN,STEPHEN
 Energy, Force, Trigraph, Mω, soluFon, a+ a− a+, a+ − a+

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