5.61 Fall 2007
Rigid Rotor
page
1±
Rigid
Rotations
Consider
the
rotation
of
two
particles
at
a
fixed
distance
R
from
one
another:
m
2
m
1
ω
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
+
m r
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
+
m
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
1body
problem
with
mass
µ
.
Similarly,
if
we
have
a
group
of
objects
that
are
held
in
rigid
y
x
z
θ
r
0
µ
φ
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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
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
ˆ
x
(
R
ˆ
y
)
that
rotates
around
x
(y).
Then
we
would
write
the
above
ˆ
ˆ
ˆ
ˆ
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
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 Spring '08
 greenwood
 Calculus, Angular Momentum, Ly

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