5.61 Fall 2007 Rigid Rotor page 1
Rigid Rotations
Consider the rotation of two particles at a fixed distance R from one another:
ω
r1 + r2 ≡ R
m1
r1 m1r1 = m2 r2 center of mass (COM)
COM r2 m2
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 Li = I iω where ω is the angular velocity and Ii is the moment of
inertia Ii = mri 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:
L12 L2 2 1 1 1
H= + = m1r12ω 2 + m2 r22ω 2 = ( m1r12 + m2 r22 ) ω 2
2I1 2I 2 2 2 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:
m1 m2
I = m1r12 + m2 r22 = µ r02 µ=
m1 + m2
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
z
1 2 L2
H = Iω =
2 2I
where, in the second equality,
we have defined the angular µ
r0
momentum for this effective θ
particle, L = Iω . The problem
is now completely reduced to a
y
1body problem with mass µ. φ
Similarly, if we have a group of
objects that are held in rigid x
, 5.61 Fall 2007 Rigid Rotor page 2
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 Ir. 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,
Rotate 90˚ Rotate 90˚
about x about y
Rx Ry
z
y
x
Rotate 90˚ Rotate 90˚
about y about x
Ry Rx
Hence,one gets different answers depending on what order the rotations are
performed in! Given our experience with quantum mechanics, we might define an
operator R̂x ( R̂y ) that rotates around x (y). Then we would write the above
experiment succinctly as: Rˆ Rˆ ≠ Rˆ Rˆ . 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
( pˆ x pˆ y = pˆ y pˆ x ) the same will not be true for angular momenta because they relate
to rotations ( Lˆ Lˆ ≠ Lˆ Lˆ ).
x y y x
Classically, angular momentum is given by L = r × p . This means the corresponding
quantum operator should be Lˆ = rˆ × pˆ . 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:
Rigid Rotations
Consider the rotation of two particles at a fixed distance R from one another:
ω
r1 + r2 ≡ R
m1
r1 m1r1 = m2 r2 center of mass (COM)
COM r2 m2
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 Li = I iω where ω is the angular velocity and Ii is the moment of
inertia Ii = mri 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:
L12 L2 2 1 1 1
H= + = m1r12ω 2 + m2 r22ω 2 = ( m1r12 + m2 r22 ) ω 2
2I1 2I 2 2 2 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:
m1 m2
I = m1r12 + m2 r22 = µ r02 µ=
m1 + m2
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
z
1 2 L2
H = Iω =
2 2I
where, in the second equality,
we have defined the angular µ
r0
momentum for this effective θ
particle, L = Iω . The problem
is now completely reduced to a
y
1body problem with mass µ. φ
Similarly, if we have a group of
objects that are held in rigid x
, 5.61 Fall 2007 Rigid Rotor page 2
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 Ir. 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,
Rotate 90˚ Rotate 90˚
about x about y
Rx Ry
z
y
x
Rotate 90˚ Rotate 90˚
about y about x
Ry Rx
Hence,one gets different answers depending on what order the rotations are
performed in! Given our experience with quantum mechanics, we might define an
operator R̂x ( R̂y ) that rotates around x (y). Then we would write the above
experiment succinctly as: Rˆ Rˆ ≠ Rˆ Rˆ . 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
( pˆ x pˆ y = pˆ y pˆ x ) the same will not be true for angular momenta because they relate
to rotations ( Lˆ Lˆ ≠ Lˆ Lˆ ).
x y y x
Classically, angular momentum is given by L = r × p . This means the corresponding
quantum operator should be Lˆ = rˆ × pˆ . 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:
