With every physical rotation R, we postulate a quantum mechanical rotation operator D(R) which rotates quantum mechanical states.
|
α
⟩
R
=
D
(
R
)
|
α
⟩
In terms of the generators of rotation,
D
(
n
^
,
ϕ
)
=
exp
(
−
i
ϕ
n
^
⋅
J
ℏ
)
n
^
is rotation axis, and
J
is angular momentum.
The rotation operator
R
(
z
,
θ
)
, with the first argument
z
indicating the rotation axis and the second
θ
the rotation angle, can operate through the translation operator
T
(
a
)
for infinitesimal rotations as explained below. This is why, it is first shown how the translation operator is acting on a particle at position x (the particle is then in the state
|
x
⟩
according to Quantum Mechanics).
Translation of the particle at position x to position x+a:
T
(
a
)
|
x
⟩
=
|
x
+
a
⟩
Because a translation of 0 does not change the position of the particle, we have (with 1 meaning the identity operator, which does nothing):
T
(
0
)
=
1
T
(
a
)
T
(
d
a
)
|
x
⟩
=
T
(
a
)
|
x
+
d
a
⟩
=
|
x
+
a
+
d
a
⟩
=
T
(
a
+
d
a
)
|
x
⟩
⇒
T
(
a
)
T
(
d
a
)
=
T
(
a
+
d
a
)
Taylor development gives:
T
(
d
a
)
=
T
(
0
)
+
d
T
(
0
)
d
a
d
a
+
.
.
.
=
1
−
i
h
p
x
d
a
with
p
x
=
i
h
d
T
(
0
)
d
a
From that follows:
T
(
a
+
d
a
)
=
T
(
a
)
T
(
d
a
)
=
T
(
a
)
(
1
−
i
h
p
x
d
a
)
⇒
[
T
(
a
+
d
a
)
−
T
(
a
)
]
/
d
a
=
d
T
d
a
=
−
i
h
p
x
T
(
a
)
This is a differential equation with the solution
T
(
a
)
=
exp
(
−
i
h
p
x
a
)
.
Additionally, suppose a Hamiltonian
H
is independent of the
x
position. Because the translation operator can be written in terms of
p
x
, and
[
p
x
,
H
]
=
0
, we know that
[
H
,
T
(
a
)
]
=
0
. This result means that linear momentum for the system is conserved.
Classically we have for the angular momentum
l
=
r
×
p
. This is the same in quantum mechanics considering
r
and
p
as operators. Classically, an infinitesimal rotation
d
t
of the vector r=(x,y,z) about the z-axis to r'=(x',y',z) leaving z unchanged can be expressed by the following infinitesimal translations (using Taylor approximation):
x
′
=
r
cos
(
t
+
d
t
)
=
x
−
y
d
t
+
.
.
.
y
′
=
r
sin
(
t
+
d
t
)
=
y
+
x
d
t
+
.
.
.
From that follows for states:
R
(
z
,
d
t
)
|
r
⟩
=
R
(
z
,
d
t
)
|
x
,
y
,
z
⟩
=
|
x
−
y
d
t
,
y
+
x
d
t
,
z
⟩
=
T
x
(
−
y
d
t
)
T
y
(
x
d
t
)
|
x
,
y
,
z
⟩
=
T
x
(
−
y
d
t
)
T
y
(
x
d
t
)
|
r
⟩
And consequently:
R
(
z
,
d
t
)
=
T
x
(
−
y
d
t
)
T
y
(
x
d
t
)
Using
T
k
(
a
)
=
exp
(
−
i
h
p
k
a
)
from above with
k
=
x
,
y
and Taylor expansion we get:
R
(
z
,
d
t
)
=
exp
[
−
i
h
(
x
p
y
−
y
p
x
)
d
t
]
=
exp
(
−
i
h
l
z
d
t
)
=
1
−
i
h
l
z
d
t
+
.
.
.
with lz = x py - y px the z-component of the angular momentum according to the classical cross product.
To get a rotation for the angle
t
, we construct the following differential equation using the condition
R
(
z
,
0
)
=
1
:
R
(
z
,
t
+
d
t
)
=
R
(
z
,
t
)
R
(
z
,
d
t
)
⇒
[
R
(
z
,
t
+
d
t
)
−
R
(
z
,
t
)
]
/
d
t
=
d
R
/
d
t
=
R
(
z
,
t
)
[
R
(
z
,
d
t
)
−
1
]
/
d
t
=
−
i
h
l
z
R
(
z
,
t
)
⇒
R
(
z
,
t
)
=
exp
(
−
i
h
t
l
z
)
Similar to the translation operator, if we are given a Hamiltonian
H
which rotationally symmetric about the z axis,
[
l
z
,
H
]
=
0
implies
[
R
(
z
,
t
)
,
H
]
=
0
. This result means that angular momentum is conserved.
For the spin angular momentum about the y-axis we just replace
l
z
with
S
y
=
h
2
σ
y
and we get the spin rotation operator
D
(
y
,
t
)
=
exp
(
−
i
t
2
σ
y
)
.
Effect on the spin operator and quantum states
Operators can be represented by matrices. From linear algebra one knows that a certain matrix
A
can be represented in another basis through the transformation
A
′
=
P
A
P
−
1
where
P
is the basis transformation matrix. If the vectors
b
respectively
c
are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle
t
between them. The spin operator
S
b
in the first basis can then be transformed into the spin operator
S
c
of the other basis through the following transformation:
S
c
=
D
(
y
,
t
)
S
b
D
−
1
(
y
,
t
)
From standard quantum mechanics we have the known results
S
b
|
b
+
⟩
=
ℏ
2
|
b
+
⟩
and
S
c
|
c
+
⟩
=
ℏ
2
|
c
+
⟩
where
|
b
+
⟩
and
|
c
+
⟩
are the top spins in their corresponding bases. So we have:
ℏ
2
|
c
+
⟩
=
S
c
|
c
+
⟩
=
D
(
y
,
t
)
S
b
D
−
1
(
y
,
t
)
|
c
+
⟩
⇒
S
b
D
−
1
(
y
,
t
)
|
c
+
⟩
=
ℏ
2
D
−
1
(
y
,
t
)
|
c
+
⟩
Comparison with
S
b
|
b
+
⟩
=
ℏ
2
|
b
+
⟩
yields
|
b
+
⟩
=
D
−
1
(
y
,
t
)
|
c
+
⟩
.
This means that if the state
|
c
+
⟩
is rotated about the y-axis by an angle
t
, it becomes the state
|
b
+
⟩
, a result that can be generalized to arbitrary axes. It is important, for instance, in Sakurai's Bell inequality.
