Transcript Ring

Quantum
Physics II
Particle on a Ring
An introduction to Angular
Momentum
Recommended Reading:
Harris, Chapter 6
Particle confined to a circular ring
In this problem we consider a particle of mass m confined to move in a
horizontal circle of radius. This problem has important applications in
the spectroscopy of molecules and is a good way to introduce the
concept of ANGULAR MOMENTUM in quantum mechanics.
Can specify the position of the
particle at any time by giving its x
m
and y coordinates.
BUT this is a one dimensional
problem and so we only need one
coordinate to specify the position
of the particle at any time t.
r

The position is specified by the
angle (t), the angle made by the
vector r with the horizontal.
If orbit is horizontal then there potential energy of the particle is
constant and can be taken as zero: U = 0
What is the potential if the orbit is vertical? Then we must include
gravity, much more difficult problem.
Classical Treatment
m
r

s
v
m = mass of particle
v = instantaneous velocity = ds/dt
r = radius vector
s = arc length
 = angle = s/r
Can then define angular velocity 
d 1 ds v


dt r dt r
The kinetic energy is constant and equal to ½mv2

1 2 2
mr 
2
1
The quantity I = mr2 is the moment of inertia of the particle, then E  I2
2
The angular momentum L of the particle is defined as
In terms of the angular velocity this can be expressed as E 
L  mr  v  mr 2  I
Can write the energy of the particle in terms of the angular momentum
1 2 L2
E  I 
2
2I
Note that the angular momentum L is perpendicular to both r and v
(since L = mr  v ).
can have two directions for the velocity, clockwise or anti-clockwise,
L
v
r
Magnitude of L is the same in
both situations, but direction is
different. Must remember that L is
a vector
r
v
L
TISE for particle confined to a circular ring
The time independent Schrodinger equation is (in (x,y) coordinates)
x = r.cos(), y = r.sin()
(1) or
tan() = y/x
If we express this in terms of the angle  then
 2  d2  d2  


 0  E
2
2


2m  dx
dy 
 d2  d2   1 d2 



 dx2 dy2  r 2 dφ 2


(2) and Schrodinger’s equation becomes
 2  1 d2  

 E
2
2


2m  r dφ 
but mr2 = I, the moment of inertia of
the particle, So TISE is
 2 d2 

 Eψ
2
2Ι dφ
(3)
(4)
want to solve this equation for the
wavefunctions and the allowed energy
levels
Solutions to TISE for particle on a circular ring
d2 
rearrange (4)

2IE
0 
d2 
2  0
 mL
d2  2
d2
A general solution of this differential equation is
    AexpimL    Bexp- imL  
(5)
(6) Check this
1
 2IE  2
where
(7)

mL  
 2 
The wavefunction must be single valued for all values of .
In particular if we rotate through an angle 2 the wavefunction
must return back to the same value it started with:
       2π   exp(imL )  expimL (  2π)
from (7)
we then
get

exp2 πmL    1

mL  int eger  0,  1,  2,  3
E
22
mL
2I

22
mL
2mr 2
(8)
Energies are quantised.
Allowed values are given
by eqn (8)
Energy Spectrum
 2  2
m
E
 2I  L


where mL =0, 1, 2, 3, ….
mL =5
E2 =25
mL =4
E2 =16
mL =3
E2 =9
mL =2
mL =1
mL =0
E2 =4
E1 =1
E0 =0
Energies in units
of
2
  


 2I 


Energy Spectrum
 2  2
m
E
 2I  L


where mL =0, 1, 2, 3, ….
- mLc
+ mL
Energy
mL =5
E2 =25
mL
mL =4
E2 =16
mL =3
E2 =9
mL =2
mL =1
mL =0
E2 =4
Energies in units
of
2
  


 2I 


E1 =1
E0 =0
all states are doubly degenerate except the ground state (m = 0)
which is singly degenerate.
Degeneracy of Solutions
Equation (6) shows that there are two solutions for each value of
mL, (except mL = 0) [ exp(imL) and exp(-imL)]  doubly degenerate
system.
The two solutions correspond to particles with the same energy but
rotating in opposite directions
r
m

ψφ   Aexp- imL φ 
r
m

ψφ   AexpimL φ 
Compare this with linear motion of a free particle, where the
solutions are also doubly degenerate; Aexp(ikx) (+x direction) and
Aexp(-ikx) (-x direction)
Normalisation
From normalisation condition
2π
 ψ * ψ dφ  1
0
2π
  A exp imLφ A expimLφ  dφ  1
0
2π
  A2 dφ  1  2 πA2  1
0
Hence
1
 1 2
A 
 2π
(9)
And the normalised wavefunctions are
1
 1 2


2
π


expimL φ 
and
1
 1 2


2
π


exp- imL φ  (10)
Probability Distribution
What is it probability P, that the particle will be found at a particular
angle  ?
P  ψφ   ψ * φ ψφ   

2
π


Note that this is
independent of the angle 
and the quantum number
mL  equal probability of
finding the particle
anywhere on the ring no
matter what state it is in!
However, note that the
wavefunctions are complex.
They have a real and an
imaginary part
1
 1 2
exp- imL φ  

2
π


expimL φ  
1/2
Probability
2
1
 1 2
0
Angle 
2
1
2π
The Wavefunctions
eiθ  cosθ  isinθ
Recall that
and e -iθ  cosθ  isinθ
so the wavefunctions can be written as
1  imL φ
1
 cos mL φ  i sin mLφ 
e

2π
2π
Real part of the wavefunction = cos(mL)
Imaginary part of the wavefunction = sin(mL)
and again we see that
2
P  ψφ   ψ * φ ψφ 
1
cosmL φ  i sinmL φ  1 cosmL φ  i sinmL φ 
2π
2π
1
1

cos2 mL φ  sin2 mL φ 
2π
2π



We can visualise the real and imaginary parts of the wavefunction
as follows.
1
sin(1.)
cos(1.)
1
0
Angle 
2
2
sin(2.)
cos(2.)
0
Angle 
2
3
sin(3.)
cos(3.)
0
Angle 
2
sin(1.)
1
cos(1.)
2