- D.14.1 Derivation from the eigenvalue problem
- D.14.2 Parity
- D.14.3 Solutions of the Laplace equation
- D.14.4 Orthogonal integrals
- D.14.5 Another way to find the spherical harmonics
- D.14.6 Still another way to find them

D.14 The spherical harmonics

This note derives and lists properties of the spherical harmonics.

D.14.1 Derivation from the eigenvalue problem

This analysis will derive the spherical harmonics from the eigenvalue problem of square angular momentum of chapter 4.2.3. It will use similar techniques as for the harmonic oscillator solution, {D.12}.

The imposed additional requirement that the spherical harmonics
are eigenfunctions of means that they are of the form
where function
is still to be determined. (There is also an arbitrary dependence on
the radius , but it does not have anything to do with angular
momentum, hence is ignored when people define the spherical
harmonics.) Substitution into with
as in (4.22) yields an ODE (ordinary differential equation)
for :

It is convenient define a scaled square angular momentum by so that you can divide away the from the ODE.

More importantly, recognize that the solutions will likely be in terms
of cosines and sines of , because they should be
periodic if changes by . If you want to use
power-series solution procedures again, these transcendental functions
are bad news, so switch to a new variable
. At the very least, that will reduce things to
algebraic functions, since is in terms of
equal to . Converting the ODE to the
new variable , you get

As you may guess from looking at this ODE, the solutions are likely to be problematic near , (physically, near the -axis where is zero.) If you examine the solution near those points by defining a local coordinate as in , and then deduce the leading term in the power series solutions with respect to , you find that it is either or , (in the special case that 0, that second solution turns out to be .) Either way, the second possibility is not acceptable, since it physically would have infinite derivatives at the -axis and a resulting expectation value of square momentum, as defined in chapter 4.4.3, that is infinite. You need to have that behaves as at each end, so in terms of it must have a factor near 1 and near 1. The two factors multiply to and so can be written as where must have finite values at 1 and 1.

If you substitute into the ODE
for , you get an ODE for :

Plug in a power series, , to get, after clean up,

Using similar arguments as for the harmonic oscillator, you see that the starting power will be zero or one, leading to basic solutions that are again odd or even. And just like for the harmonic oscillator, you must again have that the power series terminates; even in the least case that 0, the series for at 1 is like that of and will not converge to the finite value stipulated. (For rigor, use Gauss’s test.)

To get the series to terminate at some final power , you must have according to the above equation that , and if you decide to call the azimuthal quantum number , you have where since and , like any power , is greater or equal to zero.

The rest is just a matter of table books, because with , the ODE for is just the -th derivative of the differential equation for the Legendre polynomial, [40, 28.1], so the must be just the -th derivative of those polynomials. In fact, you can now recognize that the ODE for the is just Legendre's associated differential equation [40, 28.49], and that the solutions that you need are the associated Legendre functions of the first kind [40, 28.50].

To normalize the eigenfunctions on the surface area of the unit
sphere, find the corresponding integral in a table book, like
[40, 28.63]. As mentioned at the start of this long and
still very condensed story, to include negative values of ,
just replace by . There is one additional issue,
though, the sign pattern. In order to simplify some more advanced
analysis, physicists like the sign pattern to vary with according
to the so-called ladder operators.

That requires,
{D.65}, that starting from 0, the spherical
harmonics for 0 have the alternating sign pattern of the
ladder-up operator,

and those for 0 the
unvarying sign of the ladder-down operator.

Physicists
will still allow you to select your own sign for the 0
state, bless them.

where is the normal Legendre polynomial. Needless to say, some other authors use different definitions, potentially putting in a factor .

D.14.2 Parity

One special property of the spherical harmonics is often of interest: their “parity.” The parity of a wave function is 1, or even, if the wave function stays the same if you replace by . The parity is 1, or odd, if the wave function stays the same save for a sign change when you replace by . It turns out that the parity of the spherical harmonics is ; so it is 1, odd, if the azimuthal quantum number is odd, and 1, even, if is even.

To see why, note that replacing by means in spherical coordinates that changes into and into . According to trig, the first changes into . That leaves unchanged for even , since is then a symmetric function, but it changes the sign of for odd . So the sign change is . The value of has no effect, since while the factor in the spherical harmonics produces a factor under the change in , also puts derivatives on , and each derivative produces a compensating change of sign in .

D.14.3 Solutions of the Laplace equation

The Laplace equation

is

Solutions to this equation are called “harmonic functions.” In spherical coordinates, the Laplace equation has solutions of the form

This is a complete set of solutions for the Laplace equation inside a sphere. Any solution of the Laplace equation inside a sphere is a linear combination of these solutions.

As you can see in table 4.3, each solution above is a power series in terms of Cartesian coordinates.

For the Laplace equation outside a sphere, replace by 1 in the solutions above. Note that these solutions are not acceptable inside the sphere because they blow up at the origin.

To check that these are indeed solutions of the Laplace equation, plug them in, using the Laplacian in spherical coordinates given in (N.5). Note here that the angular derivatives can be simplified using the eigenvalue problem of square angular momentum, chapter 4.2.3.

D.14.4 Orthogonal integrals

The spherical harmonics are orthonormal on the unit sphere:

Further

See the notations for more on spherical coordinates and .

To verify the above expression, integrate the first term in the
integral by parts with respect to and the second term with
respect to to get

and then apply the eigenvalue problem of chapter 4.2.3.

D.14.5 Another way to find the spherical harmonics

There is a more intuitive way to derive the spherical harmonics: they
define the power series solutions to the Laplace equation. In
particular, each is a different power series solution
of the Laplace equation 0 in Cartesian coordinates.
Each takes the form

where the coefficients are such as to make the Laplacian zero.

Even more specifically, the spherical harmonics are of the form

where the coordinates and serve to simplify the Laplacian. That these are the basic power series solutions of the Laplace equation is readily checked.

To get from those power series solutions back to the equation for the spherical harmonics, one has to do an inverse separation of variables argument for the solution of the Laplace equation in a sphere in spherical coordinates (compare also the derivation of the hydrogen atom.) Also, one would have to accept on faith that the solution of the Laplace equation is just a power series, as it is in 2D, with no additional nonpower terms, to settle completeness. In other words, you must assume that the solution is analytic.

D.14.6 Still another way to find them

The simplest way of getting the spherical harmonics is probably the one given later in derivation {D.65}.