o

d2

Hψ = ’ 2 ’ l(l + 1) sech2 x ψ = Eψ (4.150)

dx

in the homework. When l is an integer, the potential in this Schr¨dinger

o

equation has the special property that it is re¬‚ectionless.

The simplest non-trivial example is l = 1. In this case, H has a single

discrete bound state at E0 = ’1. The normalized eigenfunction is

1

ψ0 (x) = √ sech x. (4.151)

2

The rest of the spectrum consists of a continuum of unbound states with

eigenvalues E(k) = k2 and eigenfunctions

1

eikx (’ik + tanh x).

ψk (x) = √ (4.152)

1 + k2

Here, k is any real number. The normalization of ψk (x) has been chosen so

that, at large |x|, where tanh x ’ ±1, we have

ψk (x)ψk (x ) ’ e’ik(x’x ) .

—

(4.153)

The measure in the completeness integral must therefore be dk/2π, the same

as that for a free particle.

114 CHAPTER 4. LINEAR DIFFERENTIAL OPERATORS

Let us compute the di¬erence

dk —

∞

ψ (x)ψk (x ) ’ δ(x ’ x )

I=

2π k

’∞

dk —

∞

ψk (x)ψk (x ) ’ e’ik(x’x )

=

2π

’∞

dk ’ik(x’x ) ik(tanh x ’ tanh x ) + tanh x tanh x ’ 1

∞

= e .

1 + k2

2π

’∞

(4.154)

We use the standard result,

dk ’ik(x’x ) 1 1

∞

= e’|x’x | ,

e (4.155)

1 + k2

2π 2

’∞

together with its x derivative,

dk ’ik(x’x ) ik 1

∞

= sgn (x ’ x ) e’|x’x | ,

e (4.156)

1 + k2

2π 2

’∞

to ¬nd

1

sgn (x ’ x )(tanh x ’ tanh x ) + tanh x tanh x ’ 1 e’|x’x | . (4.157)

I=

2

Assume, without loss of generality, that x > x ; then this reduces to

1 1

’ (1 + tanh x)(1 ’ tanh x )e’(x’x ) = ’ sech x sech x

2 2

= ’ψ0 (x)ψ0 (x ). (4.158)

Thus, the expected completeness relation

dk —

∞

ψ (x)ψk (x ) = δ(x ’ x ),

ψ0 (x)ψ0 (x ) + (4.159)

2π k

’∞

is con¬rmed.

Chapter 5

Green Functions

In this chapter we will study strategies for solving the inhomogeneous linear

di¬erential equation Ly = f . The tool we use is the Green function, which

is an integral kernel representing the inverse operator L’1 . Apart from their

use in solving inhomogeneous equations, Green functions play an important

role in many areas of physics.

5.1 Inhomogeneous Linear equations

We wish to solve Ly = f for y. Before we set about doing this, we should

ask ourselves whether a solution exists, and, if it does, whether it is unique.

The answers to these questions are summarized by the Fredholm alternative.

5.1.1 Fredholm Alternative

The Fredholm alternative for operators on a ¬nite-dimensional vector space

is discussed in detail in the appendix on linear algebra. You will want to

make sure that you have read and understood this material. Here, we merely

restate the results.

Let V be ¬nite-dimensional vector space, and A be a linear operator

A : V ’ V on this space. Then

I. Either

i) Ax = b has a unique solution,

or

ii) Ax = 0 has a non-trivial solution.

115

116 CHAPTER 5. GREEN FUNCTIONS

II. If Ax = 0 has n linearly independent solutions, then so does A† x = 0.

III. If alternative ii) holds, then Ax = b has no solution unless b is perpen-

dicular to all solutions of A† x = 0.

What is important for us in the present chapter is that this result continues

to hold for linear di¬erential operators L on a ¬nite interval ” provided that

we de¬ne L† as in the previous chapter, and provided the number of boundary

conditions is equal to the order of the equation.

If the number of boundary conditions is not equal to the order of the

equation then the number of solutions to Ly = 0 and L† y = 0 will di¬er in

general. It is still true, however, that Ly = f has no solution unless f is

perpendicular to all solutions of L† y = 0.

Example: Let

dy

Ly = , y(0) = y(1) = 0. (5.1)

dx

Clearly Ly = 0 has only the trivial solution y ≡ 0. If a solution to Ly = f

exists, therefore, it will be unique.

dy

We know that L† = ’ dx , with no boundary conditions on the functions

in its domain. The equation L† y = 0 therefore has the non-trivial solution

y = 1. This means that there is no solution to Ly = f unless

1

1, f = f dx = 0. (5.2)