Physics 731
Assignment #3, Solutions
1. An operator
L
is selfadjoint if
f

Lg
w
=
Lf

g
w
(1)
for all
f, g
on the interval
[
a, b
]
with weight function
w
(
x
)
. For the Legendre operator on the interval
[

1
,
1]
with
w
(
x
) = 1
and functions that are finite at the endpoints, the LHS of Eq. (
1
) is
I
≡
f

Lg
=
1

1
f
*
(
x
)

(1

x
2
)
d
2
dx
2
+ 2
x
d
dx
g
(
x
)
dx.
(2)
Integrating the first term by parts, we have
1

1
f
*
(
x
)(

(1

x
2
))
d
2
g
dx
2
dx
=
f
*
(
x
)(

(1

x
2
))
dg
dx
1

1

1

1
df
*
dx
(

(1

x
2
)) + 2
xf
*
dg
dx
dx.
(3)
The boundary term is zero, since
f
and
g
are finite at
x
=
±
1
. In the remaining integral, note that the
second term is equal in magnitude and opposite in sign to the second term in Eq. (
2
). Therefore,
I
=
1

1
df
*
dx
(1

x
2
)
dg
dx
dx.
(4)
Integrating again by parts, we have
I
=
df
*
dx
(1

x
2
)
g
1

1

1

1
d
2
f
*
dx
2
(1

x
2
)
g

2
x
df
*
dx
g
dx.
(5)
Again, the boundary term vanishes.
I
is then given by
I
=
1

1
d
2
f
*
dx
2
(

(1

x
2
)) + 2
x
df
*
dx
gdx
=
Lf

g ,
(6)
which is what we wanted to prove.
2. The even and odd parity eigenfunctions of the infinite square well can be summarized as follows:
ψ
n
(
x
) =
1
a
sin
nπ
(
x
+
a
)
2
a
,
n
= 1
,
2
,
3
, . . .
(7)
Therefore,
x
=
1
a
a

a
x
sin
2
nπ
(
x
+
a
)
2
a
dx
= 0
(8)
x
2
=
1
a
a

a
x
2
sin
2
nπ
(
x
+
a
)
2
a
dx
=
a
2
1
3

2
n
2
π
2
,
(9)
and
p
=

i
¯
h
a
a

a
sin
nπ
(
x
+
a
)
2
a
d
dx
sin
nπ
(
x
+
a
)
2
a
dx
= 0
(10)
p
2
=

¯
h
2
a
a

a
sin
nπ
(
x
+
a
)
2
a
d
2
dx
2
sin
nπ
(
x
+
a
)
2
a
dx
=
¯
h
2
n
2
π
2
4
a
2
.
(11)
1
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Thus, the
x

p
uncertainty product and the generalized uncertainty principle take the form
(Δ
x
)
2
(Δ
p
)
2
=
¯
h
2
4
π
2
n
2
3

2
≥
1
4

[
X, P
]

2
=
¯
h
2
4
.
(12)
For
n
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 Fall '09
 Everett
 Trigraph, dx, dx dx dx, dx dx

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