We shall consider a very simple situation in which a particle
of mass m is restricted to moving in a circular path of radius
r in the xy plane. (The model is sometimes referred to as a
particle on a ring.)
Since there is no variation in the vertical displacement of
the particle (the z direction) its potential energy is constant.
We may set the value of the potential energy for the system
to zero, as we are concerned with variations in the energy of
the particle, and not the absolute value of its energy. Thus
the total energy is equal to the kinetic energy, and we can
use classical mechanics to write E = p2/2m.
We also make use of the classical mechanical expression for
the angular momentum, Jz, about the z axis, which
is Jz = ± pr, allowing us to manipulate the expression
for the energy to give E = Jz2/2mr2.
Since mr2 is the moment of inertia, I,
of a particle of mass m rotating around an axis at a distance
r, it follows that:

However, it is relatively simple to show that not all values
of the angular momentum are permitted in quantum mechanics,
and hence the angular momentum and rotational energy of this
are both quantised.
We have already used the classical result that Jz
= ± pr , and we may substitute for the momentum using the de
Broglie relation:

This gives us the following expression for the angular momentum:

The different signs correspond to different directions of rotation.
i.e. if the system is viewed from above, the particle may be
rotating either clockwise (conventionally a negative angular
momentum) or anticlockwise (a positive angular momentum).
We must now consider the wavelength of the particle, λ.
The circumference of the circle traced
out by the particle (the distance that the particle travels)
must be an integral number of wavelengths; this means that the
wavefunction will reproduce itself on successive orbits.
If the distance travelled by the particle is not an integral
number of wavelengths, then the wavefunction becomes out of
phase with itself on successive orbits. This is unacceptable
for two reasons: Firstly it means that the wavefunction is then
multivalued, and secondly it means that over an infinite number
of orbits the wavefunction will interfere with itself destructively
and the overall wavefunction will cancel to zero, indicating
a particle cannot exist under such conditions. We can express
this requirement mathematically as:

where ml can take values 0, ±1, ±2.... This definition
of ml allows us to write:

Using the classical expression for energy derived at the top
of the page, we can write:

Notice that in the expression for the energy, ml
occurs as its square. This indicates that the energy of rotation
is independent of the direction of rotation, as we would expect.
Thus the energy levels in this system are all doubly degenerate
(due to the fact that ml takes both positive and
negative integer values) except the level for which ml
= 0, which is non-degenerate. ml is a quantum number
for the system.
Note that since there is one boundary condition, one quantum
number is required to describe the system. (The single boundary
condition is the cyclic boundary condition that the wavefunction
must reproduce itself on successive circuits. This requirement
is commonly expressed as Ψml(θ)
= Ψml(θ + 2π), i.e. the
value of the function must be the same after one complete rotation
of 2π for a given value of ml.)
It will be noted that thus far in the treatment of this system,
we have not considered the Schrodinger equation at any point.
For this simple situation, it is easier to use the de Broglie
relation to manipulate classical results. This course of action
even allows us to generate the form of the acceptable energy
values for the particle.
Note however that the same results would have been obtained
by setting up and solving the Schrodinger equation for the system,
and for more complicated systems this more formal approach to
the problem is essential. This approach allows us to write down
the explicit forms of the wavefunctions for the system, but
as there is little to be gained from this we shall not consider
them here.
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