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Orbital Angular Momentum Eigenfunctions
Michael Fowler 1/11/08
Introduction
In
the last lecture , we established that the operators 
have a common set of
eigenkets 
, 
where j, m
are integers or half odd integers, and we found the matrix elements of 
(and hence those of Jx, Jy
) between these eigenkets. This purely formal structure, therefore, nails down
the allowed values of total angular momentum and of any measured
component. But there are other things we
need to know: for example, how is an electron in a particular angular momentum
state in an atom affected by an external field?
To compute that, we need to know the wave function 
If
a system has spherical symmetry, such as an electron in the Coulomb field of a
hydrogen nucleus, then the Hamiltonian H
and the operators 
have a common set of
eigenkets 
. The spherically symmetric
Hamiltonian is unchanged by rotation, so must commute with any rotation
operator, 
. Recall that
commuting Hermitian operators can be diagonalized simultaneously—and
therefore have a common set of eigenkets.
Fortunately, many systems of interest do have spherical symmetry, at least to a good approximation, the basic example of course being the hydrogen atom, so the natural set of basis states is the common eigenkets of energy and angular momentum. It turns out that even when the spherical symmetry is broken, the angular momentum eigenkets may still be a useful starting point, with the symmetry breaking treated using perturbation theory.
Two-Dimensional Models
As a warm-up exercise for the complications of the three-dimensional spherically symmetric model, it is worth analyzing a two-dimensional circularly symmetric model, that is,
(In this section, we’ll denote the particle mass by M, to avoid confusion with the angular momentum quantum number m – but be warned you are often going to find m used for both in the same discussion!)
The two-dimensional angular momentum operator is
It is a straightforward exercise to check that for the circularly-symmetric Hamiltonian above,
Exercise: check this.
To take advantage of the circular symmetry, we switch to
circular variables 
where
Transforming the Hamiltonian and angular momentum into 
coordinates,
and
Exercise: check these results!
The angular momentum eigenfunctions 
satisfy
equivalent to 
. So 
, and for this to be a single-valued wave function, m must be an integer. (This also ensures the hermiticity of the
operator—the integration-by-parts check has canceling contributions from 
and 
)
Notice this means that any
function of r multiplied by 
is an eigenfunction of
angular momentum with eigenvalue 
, and in fact any eigenfunction of L with eigenvalue must be of this
form. So we can factor out the r-dependence, and write a complete set
of orthonormal eigenfunctions of L,
normalized by integrating around the circle:
It is interesting to note that this would be a complete set
of wave functions for a particle confined to a ring—rather like the original
Bohr orbits. In fact, nanotech rings in
which electrons have wave functions like this can now be manufactured. Note also that in such rings one can also
have real wave functions 
, which are still energy eigenstates, but not angular momentum eigenstates, since
they are standing waves, linear superpositions of waves going around the ring
in opposite directions.
The common eigenstates of the Hamiltonian and the angular momentum evidently have the form
We should emphasize that although the angular part of the
wave function does not depend on the radial potential, the radial component 
does depend on the angular momentum m. This becomes obvious on
putting this 
into the version of Schrödinger’s equation,
,
noting that 
, and canceling out the common factor 
to give
.
In this
one-dimensional equation for the radial wave function , the angular momentum term 
evidently is
equivalent to a repulsive potential.
It’s called the “centrifugal barrier” and is easy to understand from
classical mechanics. To see this,
consider a classical particle bound (in two dimensions) by an attractive
central force V(r). Split the momentum into
a radial component pr and
a component in the direction perpendicular to the radius, 
The angular momentum 
and is constant (since the force is central). The energy
substituting 
. Since 
, the angular part is exactly equivalent to the above
Schrödinger equation.
But what about the radial part? Why isn’t 
, and 
equal to 
? We know the more
complicated differentiation with respect to r
in the Schrödinger equation above must be correct, because it came from 
and 
To see why 
is incorrect, even though it satisfies 
, recall what happens in x-space.
We argued there that 
for a plane wave
because from the photon analogy, acting on the plane wave state 
this operator gives
the rate of change of phase and therefore the momentum. But a radial
wave is a little different: picture a photon wave coming out of a single narrow
slit, that is, a slit having width far smaller than the photon wavelength. The photon wave will radiate outwards with
equal amplitude in all directions (180°) but the wave amplitude will decrease with
distance from the slit to conserve probability.
For a long (narrow) slit, this is essentially a two-dimensional problem,
so the wave function will be 
. We know that if we
measure the momentum of photons at
different distances from the slit we’ll get the same result. The wavelength determines the photon’s
momentum, and it isn’t changing. The
color stays the same. However, 
operating on 
doesn’t just give pr: it picks up an extra term from differentiating the 
, so it is obviously not
giving us the right momentum.
Fortunately, this is easy to fix: we define the operator
which eliminates the extra term, and still satisfies .
However, there is still a small problem. If we substitute
this 
in the classical expression for the energy, following the
procedure we used successfully to find Schrödinger’s equation in Cartesian
coordinates, we find
This is almost—but not quite—the same as the equation we
found by transforming from Cartesian coordinates. The difference is the term 
. So which is right?
Actually our first one was right—this second one, derived directly from the
classical Hamiltonian, does give the same result in the classical limit,
because the difference between them vanishes for 
. We conclude that
beginning with the classical Hamiltonian, and replacing dynamical variables
with the appropriate quantum operators, cannot guarantee that we get the
correct quantum Hamiltonian: it might be off by some term of order 
. This would become
evident in predicting properties of truly quantum systems, such as atomic
energy levels. Problems of this kind are
common in constructing quantum theories starting from a classical theory:
essentially, in a classical theory, the order of variables in an expression is
irrelevant, but in the quantum theory there can only be one correct order of noncommuting variables such as 
and r in any
expression.
What can we say about the radial wave function ? If both the energy
and the potential at the origin are finite, then for small r 
. However, the wave
function cannot be discontinuous, so 
. To make further
progress in finding the wave function, we need to know the potential. Specific examples will be analyzed in due
course. It is interesting to note that
the allowed wave functions, proportional to 
, are the complex functions 
if the two-dimensional space is mapped into the complex
plane. Representing many-electron wave functions in the plane in this way was a
key to understanding the quantum Hall effect.