6Quantum mechanics in three dimensions
IB Quantum Mechanics
6.3 Angular momentum
Recall that in IA Dynamics and Relativity, we also solved a spherically symmetric
system in classical dynamics. We obtained beautiful, exact solutions for the
orbits of particle in a gravitational potential in terms of conic sections.
To do so, we came up with a conserved quantity known as the angular
momentum. Using the fact that this is conserved, we were able to make a lot of
calculations and derive many results. We can also use this conserved quantity to
understand, say, why the Earth doesn’t just fall into the sun directly.
To understand spherically symmetric potentials in quantum mechanics, it is
also helpful to understand the angular momentum.
Definitions
Definition
(Angular momentum)
.
The angular momentum is a vector of oper-
ators
L =
ˆ
x ∧
ˆ
p = −i~x ∧∇.
In components, this is given by
L
i
= ε
ijk
ˆx
j
ˆp
k
= −i~ε
ijk
x
j
∂
∂x
k
.
For example, we have
L
3
= ˆx
1
ˆp
2
− ˆx
2
ˆp
1
= −i~
x
1
∂
∂x
2
− x
2
∂
∂x
1
.
Note that this is just the same definition as in classical dynamics, but with
everything promoted to operators.
These operators are Hermitian, i.e.
L
†
= L,
since
ˆx
i
and
ˆp
j
are themselves Hermitian, and noting the fact that
ˆx
i
and
ˆp
j
commute whenever i 6= j.
Each component of
L
is the angular momentum in one direction. We can
also consider the length of L, which is the total angular momentum.
Definition
(Total angular momentum)
.
The total angular momentum operator
is
L
2
= L
i
L
i
= L
2
1
+ L
2
2
+ L
3
3
.
Again, this is Hermitian, and hence an observable.
Commutation relations
We have the following important commutation relations for angular momentum:
[L
i
, L
j
] = i~ε
ijk
L
k
. (i)
For example,
[L
1
, L
2
] = i~L
3
.
Recall that in classical dynamics, an important result is that the angular mo-
mentum is conserved in all directions. However, we know we can’t do this in
quantum mechanics, since the angular momentum operators do not commute,
and we cannot measure all of them. This is why we have this
L
2
. It captures the
total angular momentum, and commutes with the angular momentum operators:
[L
2
, L
i
] = 0 (ii)
for all i.
Finally, we also have the following commutation relations.
[L
i
, ˆx
j
] = i~ε
ijk
ˆx
k
, [L
i
, ˆp
j
] = i~ε
ijk
ˆp
k
. (iii)
These are rather important results.
L
i
and
L
2
are observables, but (i) implies we
cannot simultaneously measure, say
L
1
and
L
2
. The best that can be done is to
measure
L
2
and
L
3
, say. This is possible by (ii). Finally, (iii) allows computation
of the commutator of L
i
with any function of ˆx
i
and ˆp
i
.
First, we want to prove these commutation relations.
Proposition.
(i) [L
i
, L
j
] = i~ε
ijk
L
k
.
(ii) [L
2
, L
i
] = 0.
(iii) [L
i
, ˆx
j
] = i~ε
ijk
ˆx
k
and [L
i
, ˆp
j
] = i~ε
ijk
ˆp
k
Proof.
(i)
We first do a direct approach, and look at specific indices instead of general
i and j. We have
L
1
L
2
= (−i~)
2
x
2
∂
∂x
3
− x
3
∂
∂x
2
x
3
∂
∂x
1
− x
1
∂
∂x
3
= −~
2
x
2
∂
∂x
3
x
3
∂
∂x
1
− x
1
x
2
∂
2
∂x
2
3
− x
2
3
∂
2
∂x
2
∂x
1
+ x
3
x
1
∂
2
∂x
2
∂x
3
Now note that
x
2
∂
∂x
3
x
3
∂
∂x
1
= x
2
x
3
∂
2
∂x
3
x
1
+ x
2
∂
∂x
1
.
Similarly, we can compute L
2
L
1
and find
[L
1
, L
2
] = −~
2
x
2
∂
∂x
1
− x
1
∂
∂x
2
= i~L
3
,
where all the double derivatives cancel. So done.
Alternatively, we can do this in an abstract way. We have
L
i
L
j
= ε
iar
ˆx
a
ˆp
r
ε
jbs
ˆx
b
ˆp
s
= ε
iar
ε
jbs
(ˆx
a
ˆp
r
ˆx
b
ˆp
s
)
= ε
iar
ε
jbs
(ˆx
a
(ˆx
b
ˆp
r
− [ˆp
r
, ˆx
b
])ˆp
s
)
= ε
iar
ε
jbs
(ˆx
a
ˆx
b
ˆp
r
ˆp
s
− i~δ
br
ˆx
a
ˆp
s
)
Similarly, we have
L
j
L
i
= ε
iar
ε
jbs
(ˆx
b
ˆx
a
ˆp
s
ˆp
r
− i~δ
as
ˆx
b
ˆp
r
)
Then the commutator is
L
i
L
j
− L
j
L
i
= −i~ε
iar
ε
jbs
(δ
br
ˆx
a
ˆp
s
− δ
as
ˆx
b
ˆp
r
)
= −i~(ε
iab
ε
jbs
ˆx
a
ˆp
s
− ε
iar
ε
jba
ˆx
b
ˆp
r
)
= −i~((δ
is
δ
ja
− δ
ij
δ
as
)ˆx
a
ˆp
s
− (δ
ib
δ
rj
− δ
ij
δ
rb
)ˆx
b
ˆp
r
)
= i~(ˆx
i
ˆp
j
− ˆx
j
ˆp
i
)
= i~ε
ijk
L
k
.
So done.
(ii) This follows from (i) using the Leibnitz property:
[A, BC] = [A, B]C + B[A, C].
This property can be proved by directly expanding both sides, and the
proof is uninteresting.
Using this, we get
[L
i
, L
2
] = [L
i
, L
j
L
j
]
= [L
i
, L
j
]L
j
+ L
j
[L
i
, L
j
]
= i~ε
ijk
(L
k
L
j
+ L
j
L
k
)
= 0
where we get 0 since we are contracting the antisymmetric tensor
ε
ijk
with
the symmetric tensor L
k
L
j
+ L
j
L
k
.
(iii)
We will use the Leibnitz property again, but this time we have it the other
way round:
[AB, C] = [A, C]B + A[B, C].
This follows immediately from the previous version since [
A, B
] =
−
[
B, A
].
Then we can compute
[L
i
, ˆx
j
] = ε
iab
[ˆx
a
ˆp
b
, ˆx
j
]
= ε
iab
([ˆx
a
, ˆx
j
]ˆp
b
+ ˆx
a
[ˆp
b
, ˆx
j
])
= ε
iab
ˆx
a
(−i~δ
bj
)
= i~ε
ija
ˆx
a
as claimed.
We also have
[L
i
, ˆp
j
] = ε
iab
[ˆx
a
ˆp
b
, ˆp
j
]
= ε
iab
([ˆx
a
, ˆp
j
]ˆp
b
+ ˆx
a
[ˆp
b
, ˆp
j
])
= ε
iab
(i~δ
aj
ˆp
b
)
= i~ε
ijb
ˆp
b
.
Spherical polars and spherical harmonics
Recall that angular momentum is something coming from rotation. So let’s
work with something with spherical symmetry. We will exploit the symmetry
and work with spherical polar coordinates. To do so, we first need to convert
everything we had so far from Cartesian coordinates to polar coordinates.
We define our spherical polar coordinates (r, θ, ϕ) by the usual relations
x
1
= r sin θ cos ϕ
x
2
= r sin θ sin ϕ
x
3
= r cos θ.
If we express our angular momentum operators as differential operators, then we
can write them entirely in terms of
r, θ
and
ϕ
using the chain rule. The formula
for
L
3
will be rather simple, since
x
3
is our axis of rotation. However, those for
L
1
and
L
2
will be much more complicated. Instead of writing them out directly,
we instead write down the formula for
L
±
=
L
1
± iL
2
. A routine application of
the chain rule gives
L
3
= −i~
∂
∂ϕ
L
±
= L
1
± iL
2
= ±~e
±iϕ
∂
∂θ
± i cot θ
∂
∂ϕ
L
2
= −~
2
1
sin θ
∂
∂θ
sin θ
∂
∂θ
+
1
sin
2
θ
∂
2
∂ϕ
2
.
Note these operators involve only
θ
and
ϕ
. Furthermore, the expression for
L
2
is something we have all seen before — we have
∇
2
=
1
r
∂
2
∂r
2
r −
1
~
2
r
2
L
2
.
Since we know
[L
3
, L
2
] = 0,
there are simultaneous eigenfunctions of these operators, which we shall call
Y
`m
(
θ, ϕ
), with
`
= 0
,
1
,
2
, ···
and
m
= 0
, ±
1
, ±
2
, ··· , ±`
. These have eigenval-
ues ~m for L
3
and ~
2
`(` + 1) for L
2
.
In general, we have
Y
`m
= const e
imϕ
P
m
`
(cos θ),
where
P
m
`
is the associated Legendre function. For the simplest case
m
= 0, we
have
Y
`0
= const P
`
(cos θ),
where P
`
is the Legendre polynomial.
Note that these details are not important. The important thing to take away
is that there are solutions
Y
`m
(
θ, ϕ
), which we can happily use and be confident
that someone out there understands these well.