
530 Classical Mechanics
© 2010 Taylor & Francis Group, LLC
Solution:
The operator T has the form
T
H
pq
H
pp t
p
mq
mq
p
ii ii
=
∂
∂
∂
∂
−
∂
∂
∂
∂
+
∂
∂
=
∂
∂
−
∂
∂
ω
2
.
Then,
(Tq)
t=0
= p
0
/m, (T
2
q)
t=0
= −ω
2
q, (T
3
q)
t=0
= −ω
2
p/m, (T
4
q)
t=0
= −ω
4
q
and so on. Substituting these into Equation 15.42, we obtain
qt qt
p
m
t
q
m
q
()
!
()
!
...=+ +− +−
+
=
0
0
2
2
0
3
2
0
23
00
22
0
33
0
1
−+
+−+
=
ω
ω
ω
ωtp
m
t
t
q
!
...
!
...
ccossinω
ω
ωt
p
m
t+
0
with a similar development for p(t).
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