on the interval [0, R]. Here p(r) q(r) and w(r) are all supposed real so the

equation is formally self-adjoint with respect to the inner product

R

wu— v dr.

u, v = (8.168)

w

0

The endpoint x = 0 is singular if p(0) = 0. When this is so, we will not be

able to impose self-adjoint boundary conditions of our accustomed form

ay(0) + by (0) = 0 (8.169)

because one or both of y(r) and y (r) will diverge at r = 0. The various

possibilites are ennumerated by by Weyl™s theorem:

Theorem (Weyl, 1910): Suppose that r = 0 is a singular point and r = R a

regular point of the di¬erential equation (8.167). Then

I. Either:

a) Limit-circle case: There exists a »0 such that both solutions of

(8.167) have convergent w norm in the vicinity of r = 0. In this

case both solutions have convergent w norm for all values of ».

Or

b) limit-point case : No more than one solution has convergent w

norm for any ».

II. In either case, whenever Im » = 0, there is at least one ¬nite-norm

solution. When » lies on the real axis there may or may not exist a

¬nite norm solution.

248 CHAPTER 8. SPECIAL FUNCTIONS I

We will not attempt to prove Weyl™s theorem. The proof is not di¬cult and

may be found in many standard texts3 , but it is just a little more technical

than the level of this text. We will instead illustrate it with enough examples

to make the result plausible, and its practical consequences clear.

When we come to construct the Green function G(r, r ; ») obeying

’[pG] + (q ’ »w)G = δ(r ’ r ) (8.170)

we are obliged to choose a normalizable function for the r < r solution,

because otherwise the range of G will not be in L2 [0, R]. When we are in the

limit point case, and Im » = 0 there is a unique choice for this function, a

unique Green function, and hence a unique self-adjoint operator of which G is

the inverse. When » is on the real axis then there may be no such functions,

and G cannot exist. This will occur only when » is in the continuous spectrum

of the di¬erential operator.

When we have the limit-circle case there is more than one choice and

hence more than one way of obtaining a self-adjoint operator. How do we

characterize the boundary conditions to which these corespond?

Suppose that the two normalizable solutions for » = »0 are y1 (r) and

y2 (r). The proof of Weyl™s theorem reveals that once we are su¬ciently close

to r = 0 all solutions behave as a linear combination of these two, and we

can therefore impose as a boundary condition that the allowed solutions be

proportional to a speci¬ed real linear combination

y(r) ∼ ay1 (r) + by2 (r), r ’ 0. (8.171)

This is a natural generalization of the regular case, where we have solutions

y1 (r), y2 (r) with boundary conditions y1 (0) = 1, y1 (0) = 0, so y1 (r) ∼ 1,

and y2 (0) = 0, y2 (0) = 1, so y2 (r) ∼ r. The regular self-adjoint boundary

condition

au(0) + bu (0) = 0 (8.172)

with real a, b then forces y(r) to behave as

y(r) ∼ by1 (r) ’ ay2 (r) ∼ b 1 ’ a r, r ’ 0. (8.173)

Example: Consider the radial equation that arises when we separate the

Laplacian in spherical polar coordinates.

d 2 dψ

+ l(l + 1)ψ = k 2 r 2 ψ.

’ r (8.174)

dr dr

3

For example: Ivar Stackgold Boundary Value Problems of Mathematical Physics, Vol-

ume I (SIAM 2000).

8.4. SINGULAR ENDPOINTS 249

When k = 0 this has solutions ψ = rl , r ’l’1. For non-zero l only the ¬rst of

the normalization integrals

R R

2l 2

r ’2l’2 r 2 dr

r r dr, (8.175)

0 0

is ¬nite. Thus, for for l = 0, we are in the limit-point case, and the boundary

condition at the origin is uniquely determined by the requirement that the

solution be normalizable.

When l = 0, however, the two solutions are ψ1 (r) = 1 and ψ2 (r) = 1/r.

Both integrals

R R

2

r ’2 r 2 dr

r dr, (8.176)

0 0

converge and we are in the limit-circle case.

For l = 0 and general k, the solutions can be taken to be

sin kr cos kr

ψ2,k (r) = ’kn0 (kr) =

ψ1,k (r) = j0 (kr) = , (8.177)

kr r

and ψ1,k ∼ 1 and ψ2,k ∼ 1/r near r = 0. This is the same behaviour as the

k = 0 solutions, and so both remain normalizable in conformity with Weyl™s

theorem.

We obtain a self-adjoint operator if we choose a constant as and demand

that all functions in the domain be proportional to

as

ψ(r) ∼ 1 ’ (8.178)

r

when we are su¬ciently close to r = 0. If we write the solution with this

boundary condition as

sin(kr + δ) sin(kr) cos(kr)

ψk (r) = = cos δ + tan δ

r r r

tan δ

∼ k cos δ 1 + , (8.179)

kr

we read o¬ the phase shift δ as

tan δ(k) = ’kas . (8.180)

These boundary conditions arise in quantum mechanics when we study

the potential scattering of particles whose de Broglie wavelength is much

250 CHAPTER 8. SPECIAL FUNCTIONS I

larger than the range of the potential. The incident wave is unable to resolve

any of the internal structure of the potential and perceives it only as a singular

point at the origin. In this context the constant as is called the scattering

length. This physical model explains why only the l = 0 partial waves have a

choice of boundary condition: particles with angular momentum l = 0 miss

the origin by a distance rmin = l/k and never see the potential.

Example: Consider the radial part of the Laplace eigenvalue problem in two

dimensions.

d2 ψ 1 dψ m2 ψ

+ 2 = k 2 ψ.

’ 2’ (8.181)

dr r dr r

When k2 = 0, the m = 0 equation has solutions ψ1 (r) = 1 and ψ2 (r) = ln r.

Both these are normalizable, and we are in the limit-circle case at r = 0.

When k2 > 0 the solutions are

1

J0 (kr) = 1 ’ (kr)2 + · · · .

4

2

[ln(kr/2) + γ] + · · · ,

N0 (kr) = (8.182)

π

and again the short distance behaviour of the general solution coincides with

that of the k 2 = 0 solution. The self-adjoint boundary conditions at r ’ r

are therefore that all allowed functions be proportional to

1 + ± ln r (8.183)

with ± a real constant.

Exercise: Two-dimensional delta-function potential. Consider the quantum