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Rigid Rotations
Consider the rotation of two particles at a fixed distance R from one another:
m 2
m
1
r + r ≡ R
1 2
r 1
m r = m r center of mass (COM)
1 1 2 2
COM
r 2
These two particles could be an electron and a proton (in which case we’d be
looking at a hydrogen atom) or two nuclei (in which case we’d be looking at a
diatomic molecule. Classically, each of these rotating bodies has an angular
momentum L = I ω where ω is the angular velocity and I
i
is the moment of
i i
inertia I
i
= mr
i
2
for the particle. Note that, in the COM, the two bodies must
have the same angular frequency. The classical Hamiltonian for the particles is:
L
2
L
2
1 1 1
H =
1
2
= m r
2
ω
2
2
ω
2
=
2 2 2
m r + m r ω
1 1 2 2
( 1 1 2 2
)
2 I 2 I 2 2 2
1 2
Instead of thinking of this as two rotating particles, it would be really nice if we
could think of it as one effective particle rotating around the origin. We can do
this if we define the effective moment of inertia as:
m m
I = m
1
r
1
2
2
r
2
2
=μ r
0
2
μ =
1 2
m + m
1 2
where, in the second equality, we have noted that this two particle system
behaves as a single particle with a reduced mass μ rotating at a distance R
from the origin. Thus we have
1
2
L
2
H = I ω =
2 2 I
where, in the second equality,
we have defined the angular
momentum for this effective
particle, L = I ω. The problem
is now completely reduced to a
1 body problem with mass μ.
Similarly, if we have a group of
objects that are held in rigid
y
x
z
r
0
positions relative to one another – say the atoms in a crystal – and we rotate the
whole assembly with an angular velocity ω, about a given axis r then by a similar
method we can reduce the collective rotation of all of the objects to the
rotation of a single “effective” object with a moment of inertia I r
. In this
manner we can talk about rotations of a molecule (or a book or a pencil) without
having to think about the movement of every single electron and quark
individually.
It is important to realize, however, that even for a classical system, rotations
about different axes do not commute with each other. For example,
Hence , one gets different answers depending on what order the rotations are
Rotate 90 ˚
about x
R
x
Rotate 90 ˚
about y
R
y
x
y
z
Rotate 90 ˚
about y
R
y
Rotate 90 ˚
about x
R
x
performed in! Given our experience with quantum mechanics, we might define an
operator R
x
( R
y
) that rotates around x (y). Then we would write the above
experiment succinctly as: R R ≠ R R. This rather profound result has nothing
x y y x
to do with quantum mechanics – after all there is nothing quantum mechanical
about the box drawn above – but has everything to do with geometry. Thus, we
will find that, while linear momentum operators commute with one another
( p p
= p p
) the same will not be true for angular momenta because they relate
x y y x
to rotations ( L L
≠ L L ).
x y y x
Classically, angular momentum is given by L = r × p. This means the corresponding
quantum operator should be L = r × p
. This vector operator has three
components, which we identify as the angular momentum operators around each
of the three Cartesian axes, x,y and z:
2
z z
L L
,
⎯⎯⎯ L �
x → y
L
z
x
⎦
= i L
y
y → z
z → x
Because the angular momentum operators do not commute, the do not share a
common set of eigenfunctions. Thus, we arrive at the important conclusion that
a system cannot simultaneously have well defined angular momentum around
the x, y and z axes. The best we can hope for is well defined angular
momentum around 1 axis.
Returning to the problem of rigid rotations, we are interested in the
eigenstates of the Hamiltonian
H
ˆ
=
L
ˆ
2
1
2 2 2
=
(
L + L + L
)
2 I 2 I
x y z
It is relatively easy to show that while the x, y and z angular momentum
operators do not commute with each other, they do commute with L
2
L ,L
2
⎤
=
L , L
2
⎤
L , L
2
⎤
z
z x
z y
=
L , L
2
L , L
2
=
L ,L
L + L
L ,L
L ,L
L + L
L ,L
z x
z y
z x
⎦
x x
⎣
z x
⎦ ⎣
z y
y y
z y
=
(
−
) y
i L � L + L
( x
− i L � + L
y
)
x
x
i L � L
y
y
i L �
x
= 0
At this point, we make use of cyclic permutations to assert that
L
y
, L
2
= 0 and
L
, L
2
= 0 as well. Since L
z
commutes with L
2
, the two operators share
x
common eigenfunctions, so we can talk about a particle with a well defined z
component of angular momentum (call it m ) and a well defined total angular
momentum (call it l ). Thus, when we are looking for the eigenfunctions of the
rigid rotor Hamiltonian, we are looking for states indexed by two quantum
numbers l and m. This makes sense because we started with a 3D system that
would have had three quantum numbers but we’ve now restricted the motion to
the surface of a sphere, which is two dimensional. For such a 2D system, we
expect 2 quantum numbers. Based on the above analysis, we will denote the
angular momentum eigenstates by Y
l
m
, with m associated with the eigenvalue of
L
and l associated with L
2
z
We are now left with the fairly difficult problem of solving for the eigenvalues,
E l
, and eigenstates, Y
l
m
, of the rigid rotor. Notice that the eigenvalues of the
rigid rotor Hamiltonian will only depend on l because the Hamiltonian is
proportional to L
2
. In order to fully solve this differential equation, it is most
convenient to work in spherical polar coordinates. In most math textbooks, φ is
defined to be the angle relative to the z
z
axis while the vast majority of quantum
mechanics texts use θ in this capacity. We
will use the latter definition, but be careful
that any equations taken from other sources
use this same convention! Here are some
useful relations in spherical polar
coordinates:
x ≡ r cos φ sin θ
y ≡ r sin φ sin θ
z ≡ r cos θ
2
2
2
x
∇ = r + + sin θ
r
2
∂ r ∂ r r
2
sin
2
2
r
2
sin θ
In order to make progress, we need to express the angular momentum operators
in spherical polar coordinates, as well. This rearrangement turns out to be a
fairly tedious application of the chain rule and we will merely state the results:
⎛ ∂ ∂ ⎞
L
ˆ
x
= − i �
⎜
− sin φ − cot θ cos φ
⎟
⎝
∂θ ∂φ
⎠
⎛ ∂ ∂ ⎞
L
ˆ
y
= − i �
⎜
cos φ − cot θ sin φ
⎟
⎝
∂θ ∂φ
⎠
∂
L
ˆ
= − i �
z
∂φ
L
ˆ
2
= L
ˆ
2
x
ˆ
2
Y
ˆ
2
z
⇒ L
ˆ
2
= − �
2
⎢
⎡ 1 ∂
⎜
⎛
sin θ
∂
⎟
⎞
1
2
∂
2
2 ⎥
⎤
⎣
sin θ ∂θ
⎝
∂θ
⎠
sin θ ∂φ
⎦
As a result, the Schrödinger equation for the rigid rotor becomes
−�
2
⎡ 1 ∂ ⎛
∂ ⎞
1 ∂
2
⎤
m m
, )
= E Y (
θ φ, )
⎢
⎜
sin θ
⎟ 2 2 ⎥
l
θ φ
l l
2 I
⎣
sin θ ∂θ
⎝
∂θ
⎠
sin θ ∂φ
⎦
φ
θ
r
y
