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There is another method to prove the Question 1:
Let the L.H.S. of the relation to be a function of ξ as
() ( ) ( )
exp
exp
f
AB
A
ξ
ξξ
=−
,
(1)
then we could calculate that
()
(
)
(
)
(
)
(
)
(
)
[] ()
(
)
1
exp
exp
exp
exp
exp
exp
exp
exp
exp
exp
exp
,
exp
exp
exp
d
fA
B
A
d
A
A
ABA
A
A
BA
A
A AB
A
A
A
AC
A
′
⎡⎤
⎣⎦
=− ⋅
−
+
−
⋅ ⋅
⋅
⋅
−−
⋅
⋅
⋅
⋅
≡−
⋅
⋅
,
(2)
where we have defined C
1
= [B, A] and used the fact that A commutes with exp (ξA) or exp (ξA).
Similarly, we have
(
)
(
)
(
)
(
)
1
11
1
2
exp
exp
exp
exp
exp
exp
exp
,
exp
xp
,
,
exp
exp
exp
d
C
A
d
A CA
A
A AC
A
ACA
A
A
A
A
A
′′
⋅
⋅
⋅
⋅
⋅
⋅
=−⋅
⋅
⋅
⋅
,
(3)
in which C
2
= [C
1
, A] = [[B, A], A].
Go on as this, we will easily have
(
)
(
)
1
exp
exp
exp
exp
n
n
n
d
C
A
d
A
−
⋅
⋅
(4)
where C
n
= [C
n1
, A] = […[[B, A], A], …A] (there are n A`s).
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This note was uploaded on 12/10/2011 for the course PHYS 4221 taught by Professor Cflo during the Spring '11 term at CUHK.
 Spring '11
 CFLO

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