We saw that Laplace™s equation in spherical polars is

ˆ

1 ‚ 2 (r•) L2

’ 2 •.

0= (8.19)

r ‚r2 r

222 CHAPTER 8. SPECIAL FUNCTIONS I

To solve this by the method of separation of variables, we factorize

• = R(r)Y (θ, φ), (8.20)

so that

1 d2 (rR) 1 1 ˆ2

’2 L Y = 0. (8.21)

Rr dr 2 rY

Taking the separation constant to be l(l + 1), we have

d(rR)

r2 ’ l(l + 1)(rR) = 0, (8.22)

dr 2

and

ˆ

L2 Y = l(l + 1)Y. (8.23)

The solution for R is rl or r ’l’1 . The equation for Y can be further decom-

ˆ

posed by setting Y = ˜(θ)¦(φ). Looking back at the de¬nition of L2 , we see

that we can take

¦(φ) = eimφ (8.24)

with m an integer to ensure single valuedness. The equation for ˜ is then

m2

1d d˜

’ ˜ = ’l(l + 1)˜.

sin θ (8.25)

sin2 θ

sin θ dθ dθ

It is convenient to set x = cos θ; then

m2

d 2d

(1 ’ x ) + l(l + 1) ’ ˜ = 0. (8.26)

1 ’ x2

dx dx

8.2.1 Legendre Polynomials

We ¬rst look at the axially symmetric case where m = 0. We are left with

d d

(1 ’ x2 ) + l(l + 1) ˜ = 0. (8.27)

dx dx

This is Legendre™s equation. We can think of it as an eigenvalue problem

d d

(1 ’ x2 )

’ ˜(x) = l(l + 1)˜(x), (8.28)

dx dx

8.2. SPHERICAL HARMONICS 223

on the interval ’1 ¤ x ¤ 1, this being the range of cos θ for real θ. Legendre™s

equation is of Sturm-Liouville form, but with regular singular points at x =

±1. Because the endpoints of the interval are singular, we cannot impose

as boundary conditions that ˜, ˜ , or some linear combination of these, be

zero there. We do need some boundary conditions, however, so as to have a

self-adjoint operator and hence a complete set of eigenfunctions.

Given one or more singular endpoints, one possible route to a well-de¬ned

eigenvalue problem is to demand solutions that are square-integrable, and so

normalizable. This works for the harmonic oscillator equation, for example,

and, as we will describe in detail later in the chpater, the oscillator equation™s

singular endpoints at x = ±∞ are in Weyl™s limit-point class. For Legen-

dre™s equation with l = 0, the two independent solutions are ˜(x) = 1 and

˜(x) = ln(1 + x) ’ ln(1 ’ x). Both of these solutions have ¬nite L2 [’1, 1]

norms, and this square integrability persists for all values of l. Thus, requir-

ing normalizability is not enough to select a unique boundary condition. This

means that both of the Legendre equation™s singular endpoints are in Weyl™s

limit-circle class, and there is therefore a family of boundary conditions all

of which give rise to self-adjoint operators. We therefore make the more re-

strictive demand that the allowed eigenfunctions be ¬nite at the endpoints.

Because the the north and south pole of the sphere are not special points,

this is a physically reasonable condition. If we start with a ¬nite ˜(x) at one

end of the interval and demand that the solution remain ¬nite at the other

end, we obtain a discrete spectrum of eigenvalues. When l is an integer,

then one of the solutions, Pl (x), becomes a polynomial, and so is ¬nite at

x = ±1. The second solution, Ql (x), is diveregent at both ends, and so is not

an allowed solution. When l is not an integer, neither solution is ¬nite. The

eigenvalues are therefore l(l + 1) with l zero or a positive integer. Despite its

unfamiliar form, the “¬nite” boundary condition makes the Legendre opera-

tor self-adjoint, and the eigenfunctions Pl (x) form a complete orthogonal set

for L2 [’1, 1].

The Pl (x) are the Legendre Polynomials. They can be expressed in closed

form as

1 dl 2

(x ’ 1)l .

Pl (x) = l (8.29)

l

2 l! dx

This is Rodriguez™ formula. The polynomials are here normalized in the

traditional way, so that

Pl (1) = 1. (8.30)

224 CHAPTER 8. SPECIAL FUNCTIONS I

They have simple symmetry properties

Pl (’x) = (’1)l Pl (x), (8.31)

and the ¬rst few are

P0 (x) = 1,

P1 (x) = x,

1

(3x2 ’ 1),

P2 (x) =

2

1

(5x3 ’ 3x3 ),

P3 (x) =

2

1

(35x4 ’ 30x2 + 3).

P4 (x) =

8

Being Sturm-Liouville eigenfunctions, the Pl for di¬erent n are orthogonal

2

1

Pl (x)Pm (x) dx = δlm . (8.32)

2l + 1

’1

Indeed, the Pl can be obtained by applying the Gram-Schmidt proceedure to

the sequence 1, x, x2 , . . . so as to obtain polynomials orthogonal with respect

to this inner product, and then ¬xing the normalization constant so that

Pl (1) = 1.

For us, the essential property of the Pl (x) is that the general axisymmetric

solution of 2 • = 0 can be expanded in terms of them as

∞

Al r l + Bl r ’l’1 Pl (cos θ).

•(r, θ) = (8.33)

l=0

You should memorize this formula. You should also know by heart the ex-

plicit expressions for the ¬rst four Pl (x), and the factor of 2/(2l + 1) in the

orthogonality formula.