CHAPTER
5
.
1.
We are given
dx
(
A
Ψ
(
x
))*
Ψ
(
x
)
=
∞
∞
∫
dx
Ψ
(
x
)*
A
Ψ
(
x
)
∞
∞
∫
Now let
Ψ
(
x
)
=φ
(
x
)
+λψ
(
x
)
, where
λ
is an arbitrary complex number. Substitution
into the above equation yields, on the l.h.s.
dx
(
A
φ
(
x
)
+ λ
A
ψ
(
x
))*
(
(
x
)
+ λψ
(
x
))
∞
∞
∫
=
dx
(
A
)*
+ λ
(
A
)*
+ λ
*(
A
)*
+


2
(
A
)*
[ ]
∞
∞
∫
On the r.h.s. we get
dx
(
(
x
)
+ λψ
(
x
))*(
A
(
x
)
+ λ
A
(
x
))
∞
∞
∫
=
dx
*
A
+ λ
*
*
A
+ λφ
*
A
+


2
*
A
[ ]
∞
∞
∫
Because of the hermiticity of
A
, the first and fourth terms on each side are equal. For the
rest, sine
is an arbitrary complex number, the coefficients of
and
* are independent ,
and we may therefore identify these on the two sides of the equation. If we consider
,
for example, we get
dx
(
A
(
x
))*
(
x
)
=
∞
∞
∫
dx
(
x
)*
A
(
x
)
∞
∞
∫
the desired result.
2.
We have
A
+
=
A
and
B
+
=
B
, therefore (
A
+
B
)
+
= (
A
+
B
). Let us call (
A
+
B
) =
X
.
We have shown that
X
is hermitian. Consider now
(
X
+
)
n
=
X
+
X
+
X
+
…
X
+
=
X
X
X
…
X
= (
X
)
n
which was to be proved.
3.
We have
⟨
A
2
⟩=
dx
*
(
x
)
A
2
∞
∞
∫
(
x
)
Now define
A
(
x
) =
(
x
). Then the above relation can be rewritten as
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A
2
⟩=
dx
ψ
(
x
)
A
φ
(
x
)
=
dx
∞
∞
∫
∞
∞
∫
(
A
(
x
))*
(
x
)
=
dx
∞
∞
∫
(
A
(
x
))*
A
(
x
)
≥
0
4.
Let
U
=
e
iH
=
i
n
H
n
n
!
n
=
0
∞
∑
. Then
U
+
=
(

i
)
n
(
H
n
)
+
n
!
n
=
0
∞
∑
=
(

i
)
n
(
H
n
)
n
!
n
=
0
∞
∑
=
e

iH
, and thus
the hermitian conjugate of
e
iH
is
e
iH
provided
H
=
H
+.
.
5.
We need to show that
e
iH
e

iH
=
i
n
n
!
n
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 Spring '05
 mokhtari
 Equations, Trigraph, Elementary algebra, arbitrary complex number

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