Lecture 11
Particle on a ring
(c) So Hirata, Department of Chemistry, University of Illinois at Urbana-Champaign. This material has
been developed and made available online by work supported jointly by University of Illinois, the
National Science Foundation under Grant CHE-1118616 (CAREER), and the Camille & Henry Dreyfus
Foundation, Inc. through the Camille Dreyfus Teacher-Scholar program. Any opinions, findings, and
conclusions or recommendations expressed in this material are those of the author(s) and do not
necessarily reflect the views of the sponsoring agencies.
The particle on a ring



The particle on a ring is the particle in a box
curved with its both ends tied together.
This is a model for molecular rotation, a part
of an atom’s electronic wave function, and
even a part of a crystal’s electronic wave
function.
We introduce a new operator, angular
momentum operator, and a new boundary
condition, cyclic boundary condition.
Rotational motion

We review the relevant classical physics and
mathematical concepts:



Angular momentum
Cylindrical coordinates
We then switch to quantum mechanics:




Angular momentum operator
Hamiltonian in cylindrical coordinates
Cyclic boundary condition
The particle on a ring
The angular momentum



Which is the most effective
way of applying force to
θ
rotate the gear?
a
(b); the right angle
b
to the radius
c
(a) and (c) are wasteful
d
because the effective
component is |F| sin θ; (d) is
not using leverage.
The angular momentum


For a given r (position)
and p (momentum), its
rotational motion is
proportional to |r| and
|p| sin θ.
Angular momentum J
(or l) is a vector vertical
to the plane formed by r
and p, with length
|r||p|sinθ.
p
r
θ
The angular momentum

The mathematical definition of the angular
momentum is a vector outer product:
l  r p
(lx , l y , lz )  (ry pz  rz p y , rz px  rx pz , rx p y  ry px )

Without losing generality, we can assume r =
(rx, 0, 0) and p = (px, py, 0). Then l = (0, 0, lz).
(0, 0, lz )  (0, 0, rx p y )  (0, 0,| r || p | sin  )
Note that the angular momentum is a vector parallel to z-axis
Cylindrical coordinates
x  r cos 
y  r sin 
zz
The angular momentum operator

Classical angular momentum

The quantum mechanical operator can be obtained
by the conversion
lz = rx py - ry px = xpy - ypx

p x  mvx  i
x

Therefore
ˆl = r̂ p̂ - r̂ p̂ = -i æ x ¶ - y ¶ ö = -i ¶ x  r cos 
z
x y
y x
çè ¶y
¶x ÷ø
¶f
y  r sin 

 sin  
 cos  
x
r
r 

 cos  
 sin  
y
r
r 
zz
The Hamiltonian
x  r cos 
y  r sin 
zz
2 æ 2
(mv) 2 p 2
¶
¶2 ö
E=
=
Û Ĥ = + 2÷
ç
2
2m
2m
2m è ¶x
¶y ø
2
(rmv) 2
l2
¶2
E=
=
Û2
2
2
2
2mr
2mr
2mr ¶f
The particle on a ring
¶
Y(
f
)
=
EY(
f
)
2
2
2mr0 ¶f
2
2
Cyclic boundary condition



In the cylindrical coordinates, the wave function is
Ψ(r,φ,z) at fixed r and z.
The wave function must satisfy the cyclic boundary
condition: Y(j + 2p ) = Y(j )
Y(2p ) = Y(0)
After a complete revolution, the function must have
the same value and also must trace the function in
the previous cycle. Otherwise it would not be singlevalued.
The particle on a ring
2 2

( )  E( )
2
2
2mr 
d 2 e kx
2 kx



k
e
2
dx
d 2 eikx
2 ikx


k
e
2
dx
This will have problems with the
cyclic boundary condition
Promising
 2  2 iml ml2 2 iml

e 
e
2
2
2
2mr 
2mr
Energy
The particle on a ring

The cyclic boundary condition
iml (j +2 p )
e
imlj
=e
i 2 p ml
®e
=1
ml  0,1,2,




Unlike the particle in a box, we have m = 0.
Unlike the particle in a box, we have negative m’s.
m
Energy 2mr
is doubly degenerate for ml ≥ 1.
No zero-point energy; no violation of uncertainty.
2
l
2
2
The particle on a ring
Box
Energy
differences in
the microwave
range for
molecular
rotations
No zeropoint energy
Ring
Doubly
degenerate
Cyclic boundary condition

Imagine bending the particle in a box and connect
the both ends to make a ring. Notice the difference
between the boundary conditions used in two
problems.
Quantum in nature
Rotation of
H2O
How does a
microwave
oven work?
Quantum in nature
Rotation of
electrons in
chlorophyll
Why is grass
green?
The particle on a ring

Normalization
2
N    e
 0
iml 2
d 

1 / 2
2
   1d 
 0

1 / 2
1

2
The angular momentum operator

Classical angular momentum

The quantum mechanical operator can be obtained
by the conversion
lz = rx py - px ry = xpy - ypx

p x  mvx  i
x

Therefore
ˆl = -i æ x ¶ - y ¶ ö = -i ¶
z
çè ¶y
¶x ÷ø
¶f

 sin  
 cos  
x
r
r 

 cos  
 sin  
y
r
r 
x  r cos 
y  r sin 
zz
The angular momentum

Let us act the angular momentum
operator on the wave function
imlf
imlf
e
ˆl Y = -i ¶ e
= ml
z
ml
¶f 2p
2p


The wave function is also an
eigenfunction of angular momentum
operator (H and lz commute and have
the simultaneous eigenfunctions).
Double degeneracy comes from the
two senses of rotations.
The uncertainty principle

The complementary observables
x̂
tˆ
ˆ
pˆ x  i

x
E  i

t

lˆx  i

The angle and angular moment are the
complementary observables.
The particle on a ring

For a state with a definite angular momentum,
angle must be completely unknown.

The probability density along the ring is,
 iml
iml
e
e
1
*
ml ml 

2
2

2

completely uniform regardless of m .
l
Cyclic boundary
condition and crystals
The crystalline solids have
repeated chemical units. We
impose cyclic boundary
condition on the wave
functions of crystals because
infinite periodic crystals are
mathematically isomorphic to
rings of large radius.
Crystals’ wave functions as a
result have particle on a ring
eigenfunctions as their part.
Summary


In the particle on a ring, we have learned a
number of important concepts: angular
momentum operator, cylindrical
coordinates, cyclic boundary condition.
Similarities and differences from the particle
in a box: the ring has no zero-point energy,
doubly degenerate (|ml| ≥ 1), uniform
probability density.