Stone M., Goldbart P. Mathematics for Physics: A Guided Tour for Graduate Students

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4.4 Further exercises and problems 135

Here are three further problems involving the completeness of operators with a

continuous spectrum:

Problem 4.11: Missing state. In Problem 4.4.7 you will have found that the Schrödinger

equation



−

− 2 sech



ψ = E ψ

has eigensolutions

(x) = e

ikx

(−ik + tanh x)

with eigenvalue E = k

• For x large and positive ψ

(x) ≈ Ae

ikx

iη(k)

, while for x large and negative ψ

(x) ≈

ikx

−iη(k)

, the (complex) constant A being the same in both cases. Express the phase

shift η(k) as the inverse tangent of an algebraic expression in k.

• Impose periodic boundary conditions ψ(−L/2) = ψ(+L/2) where L  1. Find

the allowed values of k and hence an explicit expression for the k-space density,

ρ(k) =

, of the eigenstates.

• Compare your formula for ρ(k) with the corresponding expression, ρ

(k) = L/2π ,

for the eigenstate density of the zero-potential equation and compute the integral

N =



∞

−∞

{ρ(k) − ρ

(k)}dk.

• Deduce that one eigenfunction has gone missing from the continuum and becomes the

localized bound state ψ

(x) =

√

sech x.

Problem 4.12: Continuum completeness. Consider the differential operator

L =−

,0≤ x < ∞

with self-adjoint boundary conditions ψ(0)/ψ



(0) = tan θ for some fixed angle θ.

• Show that when tan θ<0 there is a single normalizable negative-eigenvalue eigen-

function localized near the origin, but none when tan θ>0.

• Show that there is a continuum of positive-eigenvalue eigenfunctions of the form

(x) = sin(kx + η(k)) where the phase shift η is found from

iη(k)

1 + ik tan θ

√

1 + k

tan

•

Write down (no justification required) the appropriate completeness relation

δ(x −x



) =



(x)ψ



) dk +



bound

(x)ψ



)

136 4 Linear differential operators

with an explicit expression for the product (not the separate factors) of the density of

states and the normalization constant N

, and with the correct limits on the integral

over k.

• Confirm that the ψ

continuum on its own, or together with the bound state when it

exists, form a complete set. You will do this by evaluating the integral

I(x, x



) =



∞

sin(kx + η(k)) sin(kx



+ η(k)) dk

and interpreting the result. You will need the following standard integral



∞

−∞

2π

ikx

1 + k

2|t|

−|x|/|t|

Take care! You should monitor how the bound state contribution switches on and off

as θ is varied. Keeping track of the modulus signs |...| in the standard integral is

essential for this.

Problem 4.13: One-dimensional scattering redux. Consider again the one-dimensional

Schrödinger equation from Chapter 3, Problem 3.4:

−

+ V (x)ψ = Eψ,

where V (x) is zero except in a finite interval [−a, a] near the origin (Figure 4.7).

For k > 0, consider solutions of the form

ψ(x) =



ikx

+ a

out

−ikx

, x ∈ L,

−ikx

+ a

out

ikx

, x ∈ R.

(a) Show that, in the notation of Problem 3.4, we have



out





(k) t

(−k)

(k) r

(−k)





aa

out

V(x)

out

Figure 4.7 Incoming and outgoing waves in Problem 4.13. The asymptotic regions L and R are

defined by L ={x < −a} and R ={x > a}.

4.4 Further exercises and problems 137

and show that the S-matrix

S(k) ≡



(k) t

(−k)

(k) r

(−k)



is unitary.

(b) By observing that complex conjugation interchanges the “in” and “out” waves, show

that it is natural to extend the definition of the transmission and reflection coefficients

to all real k by setting r

L,R

(k) = r

∗

L,R

(−k), t

L,R

(k) = t

∗

L,R

(−k).

(x) =



ikx

+ r

(k)e

−ikx

, x ∈ L,

(k)e

ikx

, x ∈ R,

k > 0,



(k)e

ikx

, x ∈ L,

ikx

+ r

(k)e

−ikx

, x ∈ R.

k < 0.

Show that, together with any bound states ψ

(x), these ψ

(x) satisfy the complete-

ness relation



bound

∗

(x)ψ



) +



∞

−∞

2π

∗

(x)ψ



) = δ(x − x



)

provided that

−



bound

∗

(x)ψ



) =



∞

−∞

2π

(k)e

−ik(x+x



)

, x, x



∈ L,



∞

−∞

2π

(k)e

−ik(x−x



)

, x ∈ L, x



∈ R,



∞

−∞

2π

(k)e

−ik(x−x



)

, x ∈ R, x



∈ L,



∞

−∞

2π

(k)e

−ik(x+x



)

, x, x



∈ R.

(d) Compute r

L,R

(k) and t

L,R

(k) for the potential V (x) = λδ(x −b), and verify that the

conditions in part (c) are satisfied.

If you are familiar with complex variable methods, look ahead to Chapter 18 where

Problem 18.22 shows you how to use complex variable methods to evaluate the Fourier

transforms in part (c), and so confirm that the bound state ψ

(x) and the ψ

(x) together

constitute a complete set of eigenfunctions.

138 4 Linear differential operators

Problem 4.14: Levinson’s theorem and the Friedel sum rule. The interaction between an

attractive impurity and (S-wave, and ignoring spin) electrons in a metal can be modelled

by a one-dimensional Schrödinger equation

−

+ V (r)χ = k

χ.

Here r is the distance away from the impurity, V (r) is the (spherically symmetric) impu-

rity potential and χ(r) =

√

4πrψ(r) where ψ(r) is the three-dimensional wavefunction.

The impurity attracts electrons to its vicinity. Let χ

(r) = sin(kr) denote the unperturbed

wavefunction, and χ

(r) denote the perturbed wavefunction that beyond the range of

impurity potential becomes sin(kr + η(k)). We fix the 2nπ ambiguity in the definition

of η(k) by taking η(∞) to be zero, and requiring η(k) to be a continuous function of

• Show that the continuous-spectrum contribution to the change in the number of

electrons within a sphere of radius R surrounding the impurity is given by









|χ

(x)|

−|χ

(x)|





dk =

[

η(k

) − η(0)

]

+ oscillations.

Here k

is the Fermi momentum, and “oscillations” refers to Friedel oscillations

≈ cos(2(k

R + η)). You should write down an explicit expression for the Friedel

oscillation term, and recognize it as the Fourier transform of a function ∝ k

−1

sin η(k).

• Appeal to the Riemann–Lebesgue lemma to argue that the Friedel density oscillations

make no contribution to the accumulated electron number in the limit R →∞.

(Hint: you may want to look ahead to the next part of the problem in order to show

that k

−1

sin η(k) remains finite as k → 0.)

The impurity-induced change in the number of unbound electrons in the interval [0, R]

is generically some fraction of an electron, and, in the case of an attractive potential,

can be negative – the phase shift being positive and decreasing steadily to zero as k

increases to infinity. This should not be surprising. Each electron in the Fermi sea speeds

up as it enters an attractive potential well, spends less time there, and so makes a smaller

contribution to the average local density than it would in the absence of the potential.

We would, however, surely expect an attractive potential to accumulate a net positive

number of electrons.

• Show that a negative continuous-spectrum contribution to the accumulated electron

number is more than compensated for by a positive number

bound



∞

(ρ

(k) − ρ(k))dk =−



∞

∂η

∂k

dk =

η(0)

4.4 Further exercises and problems 139

of electrons bound to the potential. After accounting for these bound electrons, show

that the total number of electrons accumulated near the impurity is

tot

η(k

This formula (together with its higher angular momentum versions) is known as the

Friedel sum rule. The relation between η(0) and the number of bound states is called

Levinson’s theorem. A more rigorous derivation of this theorem would show that

η(0) may take the value (n + 1/2)π when there is a non-normalizable zero-energy

“half-bound” state. In this exceptional case the accumulated charge will depend on R.

Green functions

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

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 finite-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 finite-dimensional vector space equipped with an inner product, and let A

be a linear operator A : V → V on this space. Then

I. Either

(i) Ax = b has a unique solution,

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

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 perpendicular 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 differential operators L on a finite interval – provided that we define 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 differ in general. It is still true, however,

that Ly = f has no solution unless f is perpendicular to all solutions of L

†

y = 0.

140

5.2 Constructing Green functions 141

Example: As an illustration of what happens when an equation possesses too many

boundary conditions, consider

Ly =

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

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

therefore, it will be unique.

We know that L

†

=−d/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 should be no solution to Ly = f unless

1, f =



fdx= 0. (5.2)

If this condition is satisfied then

y (x) =



f (x) dx (5.3)

satisfies both the differential equation and the boundary conditions at x = 0, 1. If the

condition is not satisfied, y(x) is not a solution, because y(1) = 0.

Initially we only solve Ly = f for homogeneous boundary conditions. After we have

understood how to do this, we will extend our methods to deal with differential equations

with inhomogeneous boundary conditions.

5.2 Constructing Green functions

We will solve Ly = f , a differential equation with homogeneous boundary conditions,

by finding an inverse operator L

−1

, so that y = L

−1

f . This inverse operator L

−1

will be

represented by an integral kernel

−1

)

x,ξ

= G(x, ξ), (5.4)

with the property

G(x, ξ) = δ(x − ξ). (5.5)

Here, the subscript x on L indicates that L acts on the first argument, x,ofG. Then

y (x) =



G(x, ξ)f (ξ) dξ (5.6)

will obey

y =



G(x, ξ)f (ξ) dξ =



δ(x −ξ)f (ξ ) dξ = f (x). (5.7)

142 5 Green functions

The problem is how to construct G(x, ξ). There are three necessary ingredients:

• the function χ(x) ≡ G(x, ξ) must have some discontinuous behaviour at x = ξ in

order to generate the delta function;

• away from x = ξ , the function χ(x) must obey Lχ = 0;

• the function χ(x) must obey the homogeneous boundary conditions required of y at

the ends of the interval.

The last ingredient ensures that the resulting solution, y(x ), obeys the boundary condi-

tions. It also ensures that the range of the integral operator G lies within the domain of

L, a prerequisite if the product LG = I is to make sense. The manner in which these

ingredients are assembled to construct G(x, ξ) is best explained through examples.

5.2.1 Sturm–Liouville equation

We begin by constructing the solution to the equation

(p(x)y



)



+ q(x)y(x) = f (x) (5.8)

on the finite interval [a, b] with homogeneous self-adjoint boundary conditions



(a)

y (a)

= tan θ



(b)

y (b)

= tan θ

. (5.9)

We therefore seek a function G(x, ξ) such that χ(x) = G(x, ξ) obeys

Lχ = (pχ



)



+ qχ = δ(x − ξ). (5.10)

The function χ(x) must also obey the homogeneous boundary conditions we

require of y(x).

Now (5.10) tells us that χ(x) must be continuous at x = ξ . For if not, the two

differentiationsapplied to a jump function would give us the derivative of a delta function,

and we want only a plain δ(x − ξ). If we write

G(x, ξ) = χ(x) =



(x)y

(ξ), x <ξ,

(ξ)y

(x), x >ξ,

(5.11)

then χ(x) is automatically continuous at x = ξ . We take y

(x) to be a solution of Ly = 0,

chosen to satisfy the boundary condition at the left-hand end of the interval. Similarly

(x) should solve Ly = 0 and satisfy the boundary condition at the right-hand end.

With these choices we satisfy (5.10) at all points away from x = ξ.

To work out how to satisfy the equation exactly at the location of the delta function,

we integrate (5.10) from ξ − ε to ξ + ε and find that

p(ξ )[χ



(ξ + ε) − χ



(ξ − ε)]=1. (5.12)

5.2 Constructing Green functions 143

With our product form for χ(x), this jump condition becomes

Ap(ξ )



(ξ)y



(ξ) − y



(ξ)y

(ξ)

= 1 (5.13)

and determines the constant A. We recognize the Wronskian W (y

, y

; ξ)on the left-hand

side of this equation. We therefore have A = 1/(pW ) and

G(x, ξ) =



(x)y

(ξ), x <ξ,

(ξ)y

(x), x >ξ.

(5.14)

For the Sturm–Liouville equation the product pW is constant. This fact follows from

Liouville’s formula,

W (x) = W (0) exp



−







dξ



, (5.15)

and from p

= p



= p



in the Sturm–Liouville equation. Thus

W (x) = W (0) exp



−ln[p(x)/p(0)]

= W (0)

p(0)

p(x)

. (5.16)

The constancy of pW means that G(x, ξ) is symmetric:

G(x, ξ) = G(ξ , x). (5.17)

This is as it should be. The inverse of a symmetric matrix (and the real, self-adjoint,

Sturm–Liouville operator is the function-space analogue of a real symmetric matrix) is

itself symmetric.

The solution to

Ly = (p



)



+ qy = f (x) (5.18)

is therefore

y (x) =



(x)



(ξ)f (ξ ) dξ + y

(x)



(ξ)f (ξ ) dξ



. (5.19)

Take care to understand the ranges of integration in this formula. In the first integral

ξ>x and we use G(x, ξ) ∝ y

(x)y

(ξ). In the second integral ξ<x and we use

G(x, ξ) ∝ y

(ξ)y

(x). It is easy to get these the wrong way round.

Because we must divide by it in constructing G(x, ξ), it is necessary that the Wronskian

W (y

, y

) not be zero. This is reasonable. If W were zero then y

∝ y

, and the single

function y

satisfies both Ly

= 0 and the boundary conditions. This means that the

differential operator L has y

as a zero-mode, so there can be no unique solution to

Ly = f .

144 5 Green functions



Figure 5.1 The function χ(x) = G(x, ξ).

Example: Solve

−∂

y = f (x), y(0) = y(1) = 0. (5.20)

We have

= x

= 1 − x



⇒ y



− y



≡ 1. (5.21)

We find that (Figure 5.1)

G(x, ξ) =



x(1 − ξ), x <ξ,

ξ(1 − x), x >ξ,

(5.22)

and

y (x) = (1 − x)



ξf (ξ ) dξ + x



(1 − ξ)f (ξ ) dξ . (5.23)

5.2.2 Initial value problems

Initial value problems are those boundary value problems where all boundary conditions

are imposed at one end of the interval, instead of some conditions at one end and some

at the other. The same ingredients go into constructing the Green function, though.

Consider the problem

− Q(t)y = F(t), y(0) = 0. (5.24)

We seek a Green function such that

G(t, t



) ≡



− Q(t)



G(t, t



) = δ(t − t



) (5.25)

and G(0, t



) = 0.