Quantum operator
This article concerns the rotation operator , as it appears in quantum mechanics .
Quantum mechanical rotations
With every physical rotation
R
{\displaystyle R}
, we postulate a quantum mechanical rotation operator
D
(
R
)
{\displaystyle D(R)}
which rotates quantum mechanical states.
|
α α -->
⟩ ⟩ -->
R
=
D
(
R
)
|
α α -->
⟩ ⟩ -->
{\displaystyle |\alpha \rangle _{R}=D(R)|\alpha \rangle }
In terms of the generators of rotation,
D
(
n
^ ^ -->
,
ϕ ϕ -->
)
=
exp
-->
(
− − -->
i
ϕ ϕ -->
n
^ ^ -->
⋅ ⋅ -->
J
ℏ ℏ -->
)
,
{\displaystyle D(\mathbf {\hat {n}} ,\phi )=\exp \left(-i\phi {\frac {\mathbf {\hat {n}} \cdot \mathbf {J} }{\hbar }}\right),}
where
n
^ ^ -->
{\displaystyle \mathbf {\hat {n}} }
is rotation axis,
J
{\displaystyle \mathbf {J} }
is angular momentum , and
ℏ ℏ -->
{\displaystyle \hbar }
is the reduced Planck constant .
The translation operator
The rotation operator
R
-->
(
z
,
θ θ -->
)
{\displaystyle \operatorname {R} (z,\theta )}
, with the first argument
z
{\displaystyle z}
indicating the rotation axis and the second
θ θ -->
{\displaystyle \theta }
the rotation angle, can operate through the translation operator
T
-->
(
a
)
{\displaystyle \operatorname {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
⟩ ⟩ -->
{\displaystyle |x\rangle }
according to Quantum Mechanics ).
Translation of the particle at position
x
{\displaystyle x}
to position
x
+
a
{\displaystyle x+a}
:
T
-->
(
a
)
|
x
⟩ ⟩ -->
=
|
x
+
a
⟩ ⟩ -->
{\displaystyle \operatorname {T} (a)|x\rangle =|x+a\rangle }
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
{\displaystyle \operatorname {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
)
{\displaystyle \operatorname {T} (a)\operatorname {T} (da)|x\rangle =\operatorname {T} (a)|x+da\rangle =|x+a+da\rangle =\operatorname {T} (a+da)|x\rangle \Rightarrow \operatorname {T} (a)\operatorname {T} (da)=\operatorname {T} (a+da)}
Taylor development gives:
T
-->
(
d
a
)
=
T
-->
(
0
)
+
d
T
-->
(
0
)
d
a
d
a
+
⋯ ⋯ -->
=
1
− − -->
i
ℏ ℏ -->
p
x
d
a
{\displaystyle \operatorname {T} (da)=\operatorname {T} (0)+{\frac {d\operatorname {T} (0)}{da}}da+\cdots =1-{\frac {i}{\hbar }}p_{x}da}
with
p
x
=
i
ℏ ℏ -->
d
T
-->
(
0
)
d
a
{\displaystyle p_{x}=i\hbar {\frac {d\operatorname {T} (0)}{da}}}
From that follows:
T
-->
(
a
+
d
a
)
=
T
-->
(
a
)
T
-->
(
d
a
)
=
T
-->
(
a
)
(
1
− − -->
i
ℏ ℏ -->
p
x
d
a
)
⇒ ⇒ -->
T
-->
(
a
+
d
a
)
− − -->
T
-->
(
a
)
d
a
=
d
T
d
a
=
− − -->
i
ℏ ℏ -->
p
x
T
-->
(
a
)
{\displaystyle \operatorname {T} (a+da)=\operatorname {T} (a)\operatorname {T} (da)=\operatorname {T} (a)\left(1-{\frac {i}{\hbar }}p_{x}da\right)\Rightarrow {\frac {\operatorname {T} (a+da)-\operatorname {T} (a)}{da}}={\frac {d\operatorname {T} }{da}}=-{\frac {i}{\hbar }}p_{x}\operatorname {T} (a)}
This is a differential equation with the solution
T
-->
(
a
)
=
exp
-->
(
− − -->
i
ℏ ℏ -->
p
x
a
)
.
{\displaystyle \operatorname {T} (a)=\exp \left(-{\frac {i}{\hbar }}p_{x}a\right).}
Additionally, suppose a Hamiltonian
H
{\displaystyle H}
is independent of the
x
{\displaystyle x}
position. Because the translation operator can be written in terms of
p
x
{\displaystyle p_{x}}
, and
[
p
x
,
H
]
=
0
{\displaystyle [p_{x},H]=0}
, we know that
[
H
,
T
-->
(
a
)
]
=
0.
{\displaystyle [H,\operatorname {T} (a)]=0.}
This result means that linear momentum for the system is conserved.
In relation to the orbital angular momentum
Classically we have for the angular momentum
L
=
r
× × -->
p
.
{\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} .}
This is the same in quantum mechanics considering
r
{\displaystyle \mathbf {r} }
and
p
{\displaystyle \mathbf {p} }
as operators. Classically, an infinitesimal rotation
d
t
{\displaystyle dt}
of the vector
r
=
(
x
,
y
,
z
)
{\displaystyle \mathbf {r} =(x,y,z)}
about the
z
{\displaystyle z}
-axis to
r
′
=
(
x
′
,
y
′
,
z
)
{\displaystyle \mathbf {r} '=(x',y',z)}
leaving
z
{\displaystyle 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
+
⋯ ⋯ -->
{\displaystyle {\begin{aligned}x'&=r\cos(t+dt)=x-y\,dt+\cdots \\y'&=r\sin(t+dt)=y+x\,dt+\cdots \end{aligned}}}
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
⟩ ⟩ -->
{\displaystyle \operatorname {R} (z,dt)|r\rangle =\operatorname {R} (z,dt)|x,y,z\rangle =|x-y\,dt,y+x\,dt,z\rangle =\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)|x,y,z\rangle =\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)|r\rangle }
And consequently:
R
-->
(
z
,
d
t
)
=
T
x
-->
(
− − -->
y
d
t
)
T
y
-->
(
x
d
t
)
{\displaystyle \operatorname {R} (z,dt)=\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)}
Using
T
k
(
a
)
=
exp
-->
(
− − -->
i
ℏ ℏ -->
p
k
a
)
{\displaystyle T_{k}(a)=\exp \left(-{\frac {i}{\hbar }}p_{k}a\right)}
from above with
k
=
x
,
y
{\displaystyle k=x,y}
and Taylor expansion we get:
R
-->
(
z
,
d
t
)
=
exp
-->
[
− − -->
i
ℏ ℏ -->
(
x
p
y
− − -->
y
p
x
)
d
t
]
=
exp
-->
(
− − -->
i
ℏ ℏ -->
L
z
d
t
)
=
1
− − -->
i
ℏ ℏ -->
L
z
d
t
+
⋯ ⋯ -->
{\displaystyle \operatorname {R} (z,dt)=\exp \left[-{\frac {i}{\hbar }}\left(xp_{y}-yp_{x}\right)dt\right]=\exp \left(-{\frac {i}{\hbar }}L_{z}dt\right)=1-{\frac {i}{\hbar }}L_{z}dt+\cdots }
with
L
z
=
x
p
y
− − -->
y
p
x
{\displaystyle L_{z}=xp_{y}-yp_{x}}
the
z
{\displaystyle z}
-component of the angular momentum according to the classical cross product .
To get a rotation for the angle
t
{\displaystyle t}
, we construct the following differential equation using the condition
R
-->
(
z
,
0
)
=
1
{\displaystyle \operatorname {R} (z,0)=1}
:
R
-->
(
z
,
t
+
d
t
)
=
R
-->
(
z
,
t
)
R
-->
(
z
,
d
t
)
⇒ ⇒ -->
d
R
d
t
=
R
-->
(
z
,
t
+
d
t
)
− − -->
R
-->
(
z
,
t
)
d
t
=
R
-->
(
z
,
t
)
R
-->
(
z
,
d
t
)
− − -->
1
d
t
=
− − -->
i
ℏ ℏ -->
L
z
R
-->
(
z
,
t
)
⇒ ⇒ -->
R
-->
(
z
,
t
)
=
exp
-->
(
− − -->
i
ℏ ℏ -->
t
L
z
)
{\displaystyle {\begin{aligned}&\operatorname {R} (z,t+dt)=\operatorname {R} (z,t)\operatorname {R} (z,dt)\\[1.1ex]\Rightarrow {}&{\frac {d\operatorname {R} }{dt}}={\frac {\operatorname {R} (z,t+dt)-\operatorname {R} (z,t)}{dt}}=\operatorname {R} (z,t){\frac {\operatorname {R} (z,dt)-1}{dt}}=-{\frac {i}{\hbar }}L_{z}\operatorname {R} (z,t)\\[1.1ex]\Rightarrow {}&\operatorname {R} (z,t)=\exp \left(-{\frac {i}{\hbar }}\,t\,L_{z}\right)\end{aligned}}}
Similar to the translation operator, if we are given a Hamiltonian
H
{\displaystyle H}
which rotationally symmetric about the
z
{\displaystyle z}
-axis,
[
L
z
,
H
]
=
0
{\displaystyle [L_{z},H]=0}
implies
[
R
-->
(
z
,
t
)
,
H
]
=
0
{\displaystyle [\operatorname {R} (z,t),H]=0}
. This result means that angular momentum is conserved.
For the spin angular momentum about for example the
y
{\displaystyle y}
-axis we just replace
L
z
{\displaystyle L_{z}}
with
S
y
=
ℏ ℏ -->
2
σ σ -->
y
{\textstyle S_{y}={\frac {\hbar }{2}}\sigma _{y}}
(where
σ σ -->
y
{\displaystyle \sigma _{y}}
is the Pauli Y matrix ) and we get the spin rotation operator
D
-->
(
y
,
t
)
=
exp
-->
(
− − -->
i
t
2
σ σ -->
y
)
.
{\displaystyle \operatorname {D} (y,t)=\exp \left(-i{\frac {t}{2}}\sigma _{y}\right).}
Effect on the spin operator and quantum states
Operators can be represented by matrices . From linear algebra one knows that a certain matrix
A
{\displaystyle A}
can be represented in another basis through the transformation
A
′
=
P
A
P
− − -->
1
{\displaystyle A'=PAP^{-1}}
where
P
{\displaystyle P}
is the basis transformation matrix. If the vectors
b
{\displaystyle b}
respectively
c
{\displaystyle c}
are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle
t
{\displaystyle t}
between them. The spin operator
S
b
{\displaystyle S_{b}}
in the first basis can then be transformed into the spin operator
S
c
{\displaystyle S_{c}}
of the other basis through the following transformation:
S
c
=
D
-->
(
y
,
t
)
S
b
D
− − -->
1
-->
(
y
,
t
)
{\displaystyle S_{c}=\operatorname {D} (y,t)S_{b}\operatorname {D} ^{-1}(y,t)}
From standard quantum mechanics we have the known results
S
b
|
b
+
⟩ ⟩ -->
=
ℏ ℏ -->
2
|
b
+
⟩ ⟩ -->
{\textstyle S_{b}|b+\rangle ={\frac {\hbar }{2}}|b+\rangle }
and
S
c
|
c
+
⟩ ⟩ -->
=
ℏ ℏ -->
2
|
c
+
⟩ ⟩ -->
{\textstyle S_{c}|c+\rangle ={\frac {\hbar }{2}}|c+\rangle }
where
|
b
+
⟩ ⟩ -->
{\displaystyle |b+\rangle }
and
|
c
+
⟩ ⟩ -->
{\displaystyle |c+\rangle }
are the top spins in their corresponding bases. So we have:
ℏ ℏ -->
2
|
c
+
⟩ ⟩ -->
=
S
c
|
c
+
⟩ ⟩ -->
=
D
-->
(
y
,
t
)
S
b
D
− − -->
1
-->
(
y
,
t
)
|
c
+
⟩ ⟩ -->
⇒ ⇒ -->
{\displaystyle {\frac {\hbar }{2}}|c+\rangle =S_{c}|c+\rangle =\operatorname {D} (y,t)S_{b}\operatorname {D} ^{-1}(y,t)|c+\rangle \Rightarrow }
S
b
D
− − -->
1
-->
(
y
,
t
)
|
c
+
⟩ ⟩ -->
=
ℏ ℏ -->
2
D
− − -->
1
-->
(
y
,
t
)
|
c
+
⟩ ⟩ -->
{\displaystyle S_{b}\operatorname {D} ^{-1}(y,t)|c+\rangle ={\frac {\hbar }{2}}\operatorname {D} ^{-1}(y,t)|c+\rangle }
Comparison with
S
b
|
b
+
⟩ ⟩ -->
=
ℏ ℏ -->
2
|
b
+
⟩ ⟩ -->
{\textstyle S_{b}|b+\rangle ={\frac {\hbar }{2}}|b+\rangle }
yields
|
b
+
⟩ ⟩ -->
=
D
− − -->
1
(
y
,
t
)
|
c
+
⟩ ⟩ -->
{\displaystyle |b+\rangle =D^{-1}(y,t)|c+\rangle }
.
This means that if the state
|
c
+
⟩ ⟩ -->
{\displaystyle |c+\rangle }
is rotated about the
y
{\displaystyle y}
-axis by an angle
t
{\displaystyle t}
, it becomes the state
|
b
+
⟩ ⟩ -->
{\displaystyle |b+\rangle }
, a result that can be generalized to arbitrary axes.
See also
References
L.D. Landau and E.M. Lifshitz: Quantum Mechanics: Non-Relativistic Theory , Pergamon Press, 1985
P.A.M. Dirac: The Principles of Quantum Mechanics , Oxford University Press, 1958
R.P. Feynman, R.B. Leighton and M. Sands: The Feynman Lectures on Physics , Addison-Wesley, 1965
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