PHYS 507
Answers to Homework IV
Assigned:
December 3, 2003, Wednesday.
Due:
December 12, 2003, Friday, at 5:00 pm.
1.
(a)
h
J
x
i
=
h
j, m

J
x

j, m
i
=
1
2
h
j, m

(
J
+
+
J

)

j, m
i
=
¯
h
2
‡
p
j
(
j
+ 1)

m
(
m
+ 1)
h
j, m

j, m
+ 1
i
+
p
j
(
j
+ 1)

m
(
m

1)
h
j, m

j, m

1
i
·
=
0
h
J
y
i
=
1
2
i
h
j, m

(
J
+

J

)

j, m
i
=
¯
h
2
i
‡
p
j
(
j
+ 1)

m
(
m
+ 1)
h
j, m

j, m
+ 1
i 
p
j
(
j
+ 1)

m
(
m

1)
h
j, m

j, m

1
i
·
=
0
Coupled with
h
J
z
i
=
m
¯
h
these relations can be written as
D
~
J
E
=
m
¯
h
ˆ
z
. This is a good
rule to keep in mind. In general if a state is an eigenstate of
J
n
, then
D
~
J
E
is a vector
along ˆ
n
.
For calculating
h
J
2
x
i
and
›
J
2
y
fi
let us use the following trick for a faster calculation
›
J
2
x
fi
=
h
j, m

J
2
x

j, m
i
=
h
φ

φ
i
where

φ
i
=
J
x

j, m
i
,
and similarly for
›
J
2
y
fi
. We have
J
x

j, m
i
=
1
2
(
J
+
+
J

)

j, m
i
=
¯
h
2
‡
p
j
(
j
+ 1)

m
(
m
+ 1)

j, m
+ 1
i
+
p
j
(
j
+ 1)

m
(
m

1)

j, m

1
i
·
J
y

j, m
i
=
1
2
i
(
J
+

J

)

j, m
i
=
¯
h
2
i
‡
p
j
(
j
+ 1)

m
(
m
+ 1)

j, m
+ 1
i 
p
j
(
j
+ 1)

m
(
m

1)

j, m

1
i
·
which gives
›
J
2
x
fi
=
›
J
2
y
fi
=
¯
h
2
2
(
j
(
j
+ 1)

m
2
)
This result is easy to understand.
We have
h
J
2
x
i
=
›
J
2
y
fi
by symmetry.
Since
J
2
=
J
2
x
+
J
2
y
+
J
2
z
we should have
›
J
2
x
fi
=
1
2
›
J
2

J
2
z
fi
=
¯
h
2
2
(
j
(
j
+ 1)

m
2
)
.
1
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Note that, again by symmetry, we should have
h
J
2
n
i
=
h
J
2
x
i
for any unit vector ˆ
n
on
the
xy
plane (which is proved in part (c)).
(b)
Start with
h
J
x
J
y
i
=
1
4
i
h
(
J
+
+
J

)(
J
+

J

)
i
=
1
4
i
›
J
2
+

J
2

+
J

J
+

J
+
J

fi
It can be seen easily that
›
J
2
+
fi
=
›
J
2

fi
= 0.
For the remaining terms, we either
calculate the commutator or use the application rules for
J
±
. Following the second
h
J

J
+
i
= ¯
h
2
(
j
(
j
+ 1)

m
(
m
+ 1))
,
h
J
+
J

i
= ¯
h
2
(
j
(
j
+ 1)

m
(
m

1))
.
The final result is
h
J
x
J
y
i
=
i
2
m
¯
h
2
.
For
h
J
y
J
x
i
we either follow the same procedure or use the relation (
J
x
J
y
)
†
=
J
y
J
x
.
h
J
y
J
x
i
=

i
2
m
¯
h
2
.
Since both of
J
x
J
y
and
J
y
J
x
is not hermitian, their expectation values do not have to
be real. Note that
J
x
J
y
+
J
y
J
x
is hermitian and we have
h
J
x
J
y
+
J
y
J
x
i
= 0
which is real.
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 Cartesian Coordinate System, René Descartes, Jy Jx, Jx Jy

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