PHYS 507
Answers to Homework V
1.
The assumption
~μ
=
k
~
S
is not necessary.
We only need to use the commutation
relation
[
S
i
, μ
j
] =
i
¯
h
X
k
²
ijk
μ
k
,
which expresses the fact that
~μ
is a vector operator.
d
dt
h
S
i
i
t
=
i
¯
h
h
[
H, S
i
]
i
t
=

i
¯
h
X
j
B
j
h
[
μ
j
, S
i
]
i
t
=
i
¯
h
X
j
B
j
h
[
S
i
, μ
j
]
i
t
=

X
jk
B
j
²
ijk
h
μ
k
i
t
=

(
~
B
× h
~μ
i
t
)
i
=
+(
h
~μ
i
t
×
~
B
)
i
.
2.
We have calculated the rotation operator for a single spin 1/2 particle before. It is
D
(
θ,
ˆ
x
) = exp

i
2
θσ
x
¶
= cos
θ
2

i
sin
θ
2
σ
x
=
•
cos
θ
2

i
sin
θ
2

i
sin
θ
2
cos
θ
2
‚
As a result we have the following
D
(
θ,
ˆ
x
)
 ↑i
=
cos
θ
2
 ↑i 
i
sin
θ
2
 ↓i
,
D
(
θ,
ˆ
x
)
 ↓i
=

i
sin
θ
2
 ↑i
+ cos
θ
2
 ↓i
.
(a)
We will rotate the twoparticle state

0
,
0
i
by applying individual rotations to
each particle using the relations above. Let us start with
 ↑↓i
. We have
D
(
θ,
ˆ
x
)
 ↑↓i
=
(
D
1
(
θ,
ˆ
x
)
 ↑i
)
⊗
(
D
2
(
θ,
ˆ
x
)
 ↓i
)
=
cos
θ
2
 ↑i 
i
sin
θ
2
 ↓i
¶
⊗

i
sin
θ
2
 ↑i
+ cos
θ
2
 ↓i
¶
=
cos
2
θ
2
 ↑↓i 
i
sin
θ
2
cos
θ
2
(
 ↑↑i
+
 ↓↓i
)

sin
2
θ
2
 ↓↑i
.
Continue with
 ↓↑i
,
D
(
θ,
ˆ
x
)
 ↓↑i
=
(
D
1
(
θ,
ˆ
x
)
 ↓i
)
⊗
(
D
2
(
θ,
ˆ
x
)
 ↑i
)
=

i
sin
θ
2
 ↑i
+ cos
θ
2
 ↓i
¶
⊗
cos
θ
2
 ↑i 
i
sin
θ
2
 ↓i
¶
=
cos
2
θ
2
 ↓↑i 
i
sin
θ
2
cos
θ
2
(
 ↑↑i
+
 ↓↓i
)

sin
2
θ
2
 ↑↓i
.
1
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For the state

0
,
0
i
we have
D
(
θ,
ˆ
x
)

0
,
0
i
=
1
√
2
cos
2
θ
2
+ sin
2
θ
2
¶
(
 ↑↓i   ↓↑i
) =

0
,
0
i
.
(b)
Here we have
D
(
θ,
ˆ
x
)
 ↑↑i
=
(
D
1
(
θ,
ˆ
x
)
 ↑i
)
⊗
(
D
2
(
θ,
ˆ
x
)
 ↑i
)
=
cos
θ
2
 ↑i 
i
sin
θ
2
 ↓i
¶
⊗
cos
θ
2
 ↑i 
i
sin
θ
2
 ↓i
¶
=
cos
2
θ
2
 ↑↑i 
i
sin
θ
2
cos
θ
2
(
 ↑↓i
+
 ↓↑i
)

sin
2
θ
2
 ↓↓i
=
1
2
(1 + cos
θ
)

1
,
1
i 
i
2
sin
θ
·
√
2

1
,
0
i 
1
2
(1

cos
θ
)

1
,

1
i
.
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 Spring '11
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 Angular Momentum, Cos, Rotational symmetry, total angular momentum, Ly Sy

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