dr

on the interval [0, R], and with boundary conditions ψ(0) = 0 = ψ(R). This

problem arises when we solve the Schr¨dinger equation for a central potential

o

in spherical polar coordinates, and assume that the wavefunction is a function

of r only (i.e. S-wave, or l = 0). Again, we want the boundary at R to be

in¬nitely far away, but we will start with R at a large but ¬nite distance,

and then take the R ’ ∞ limit. Let us ¬rst deal with the simple case that

V (r) ≡ 0; then the solutions are

ψk (r) ∝ sin kr, (4.124)

6

Peierls was justifying why the phonon contribution to the speci¬c heat of a crystal

could be calculated by using periodic boundary conditions. Some sceptics thought that

his calculation might be wrong by factors of two.

108 CHAPTER 4. LINEAR DIFFERENTIAL OPERATORS

with eigenvalue E = k2 , and with the allowed values of being given by

kn R = nπ. Since

R

R

sin2 (kn r) dr = , (4.125)

2

0

the normalized wavefunctions are

2

ψk = sin kr, (4.126)

R

and completeness reads

∞

2

sin(kn r) sin(kn r ) = δ(r ’ r ). (4.127)

R

n=1

As R becomes large, this sum goes over to an integral:

∞

2 2

∞

sin(kn r) sin(kn r ) ’ dn sin(kr) sin(kr ),

R R

0

n=1

Rdk 2

∞

= sin(kr) sin(kr ).(4.128)

π R

0

Thus,

2 ∞

dk sin(kr) sin(kr ) = δ(r ’ r ). (4.129)

π 0

As before, the large distance, here R, no longer appears.

Now consider the more interesting problem which has the potential V (r)

included. We will assume, for simplicity, that there is an R0 such that V (r)

is zero for r > R0 . In this case, we know that the solution for r > R0 is of

the form

ψk (r) = Nk sin (kr + δ(k)) , (4.130)

where the phase shift δ(k) is a functional of the potential V . The eigenvalue

is still E = k2 .

Example: A delta-function shell. We take V (r) = »δ(r ’ a).

4.3. COMPLETENESS OF EIGENFUNCTIONS 109

ψ

»δ (r’a)

x

a

Delta function shell potential.

A solution with eigenvalue E = k2 and satisfying the boundary condition at

r = 0 is

A sin(kr), r < a,

ψ(r) = (4.131)

sin(kr + δ), r > a.

The conditions to be satis¬ed at r = a are:

i) continuity, ψ(a ’ ) = ψ(a + ) ≡ ψ(a), and

ii) jump in slope, ’ψ (a + ) + ψ (a ’ ) + »ψ(a) = 0.

Therefore,

ψ (a + ) ψ (a ’ )

’ = », (4.132)

ψ(a) ψ(a)

or

k cos(ka + δ) k cos(ka)

’ = ». (4.133)

sin(ka + δ) sin(ka)

Thus,

»

cot(ka + δ) ’ cot(ka) = , (4.134)

k

and

»

δ(k) = ’ka + cot’1 + cot ka . (4.135)

k

110 CHAPTER 4. LINEAR DIFFERENTIAL OPERATORS

δ(k)

π 2π 4π

3π

ka

’π

Phase shift as a function of k.

The graph of δ(k) is shown in the ¬gure. The allowed values of k are

required by the boundary condition

sin(kR + δ(k)) = 0 (4.136)

to satisfy

kR + δ(k) = nπ. (4.137)

This is a transcendental equation for k, and so ¬nding the individual solutions

kn is not simple. We can, however, write

1

n= kR + δ(k) (4.138)

π

and observe that, when R becomes large, only an in¬nitesimal change in k

is required to make n increment by unity. We may therefore regard n as a

“continuous” variable which we can di¬erentiate with respect to k to ¬nd

dn 1 ‚δ