RP - INAOE · 2011-03-25 · algebraic methods by using diﬀerential operators when possible. The...

1

Differential Equations:An Operational Approach

Héctor Manuel Moya-Cessa Francisco Soto-Eguibar

RP Rinton Press

RP

Differential Equations:An Operational ApproachIn this short textbook on differential equations an alternativeapproach to the one that is usually found in textbooks ispresented. Original material that deals with the application ofoperational methods are developed. Particular attention is paidto algebraic methods by using differential operators whenapplicable. The methods presented in this book are useful inall applications of differential equations. These methods areused in quantum physics, where they have been developed intheir majority, but can be used in any other branch of physicsand engineering.

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Differential E

quations: An O

perational Approach M

oya-Cessa &

Soto-Eguibar

Differential Equations:

An Operational Approach

Héctor Manuel Moya-Cessa

Francisco Soto-Eguibar

RP

Rinton Press, Inc.

© 2011 Rinton Press, Inc.565 Edmund TerraceParamus, New Jersey 07652, [email protected]://www.rintonpress.com

All right reserved. No part of this bookcovered by the copyright hereon may bereproduced or used in any form or by anymeans-graphic, electronic, or mechanical,including photocopying, recording,taping, or information storage andretrieval system - without permission ofthe publisher.

Published by Rinton Press, Inc.

Printed in the United States of America

ISBN 978-1-58949-060-4

March 11, 2011 6:45 Rinton-Book9x6 Differential*equations.*An*operational*approach

Publishers’ page

iv


a Isabel y Leonardo

a Conchis, Sofi y Quique

v


vi


Preface

This short textbook, mainly on new methods to solve differential equations, is

the result from notes of the courses on special topics of mathematical methods

that we have taught, for several years, at Instituto Nacional de Astrofısica,

Optica y Electronica in Puebla, Mexico.

We present an alternative approach to the one that is usually found in text-

books. We have developed original material that is presented here, which deals

with the application of operational methods. Because of this, new material is

developed in five of the six chapters it contains, paying particular attention to

algebraic methods by using differential operators when possible. The methods

presented in this book are useful in all applications of differential equations.

These methods are used in quantum physics, where they have been developed in

their majority, but can be used in any other branch of physics and engineering.

The first Chapter is a review of linear algebra, as it is the basis of the rest

of the Chapters. The second Chapter is a survey of special functions, where we

introduce new operational techniques that will be used throughout the book. In

this very Chapter we show new material, in particular a sum of Hermite poly-

nomials of even order. The third Chapter is devoted to solve finite systems of

differential equations; we do this by using what we have called Vandermonde

methods. In the fourth and the fifth Chapters we solve infinite and semi-infinite

systems of ordinary differential equations using operational methods. Finally, in

Chapter six we solve some partial differential equations. The book is self con-

tained, as the methods established are new and they are completely developed.

The book is intended for undergraduate students and so we assume that the

reader has elementary knowledge on matrix algebra, determinants and a basic

working knowledge on ordinary differential equations. The reader acquainted

with linear algebra can go directly to Chapter 2, i.e. to the Special Functions

Chapter. This Chapter should not be skipped, even when the reader has been

vii


viii Preface

exposed to special functions, as the main notions used in the book are introduced

here. In some parts of the book, where we think the reader may not have much

experience, we provide explicit calculations in order to give details on how the

operational calculations have to be done.

We are very grateful to our colleagues Omar Lopez and Enrique Landgrave

for valuable criticisms and friendly recommendations. We would like to thank

several students for typing part of the lectures, in particular we are very grateful

to Juan Martınez-Carranza.

Hector Manuel Moya-Cessa and Francisco Soto-Eguibar

Santa Marıa Tonantzintla, Puebla, Mexico

March 2011


Contents

Preface vii

Chapter 1 Linear Algebra 1

1.1 Vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.2 The span of a set of vectors . . . . . . . . . . . . . . . . . . 3

1.1.3 Linear independence . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.4 Bases and dimension of a vector space . . . . . . . . . . . . . 6

1.1.5 Coordinate systems. Components of a vector in a given basis 7

1.2 The scalar product. Euclidian spaces . . . . . . . . . . . . . . . . . . 7

1.2.1 The norm in an Euclidian space . . . . . . . . . . . . . . . . 8

1.2.2 The concept of angle between vectors. Orthogonality . . . . . 8

1.3 Linear transformations . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3.1 The kernel of a linear transformation . . . . . . . . . . . . . . 12

1.3.2 The image of a linear transformation . . . . . . . . . . . . . . 13

1.3.3 Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.3.4 Linear transformations and matrices . . . . . . . . . . . . . 14

1.3.5 The product of linear transformations . . . . . . . . . . . . . 16

1.4 Eigenvalues and eigenvectors . . . . . . . . . . . . . . . . . . . . . . 20

1.4.1 The finite dimension case . . . . . . . . . . . . . . . . . . . . 24

1.4.2 Similar matrices . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.4.3 Diagonal matrices . . . . . . . . . . . . . . . . . . . . . . . . 28

1.4.3.1 Procedure to diagonalize a matrix . . . . . . . . . . 30

1.4.3.2 The Cayley-Hamilton theorem . . . . . . . . . . . . 30

1.5 Linear operators acting on Euclidian spaces . . . . . . . . . . . . . . 31

1.5.1 Adjoint operators . . . . . . . . . . . . . . . . . . . . . . . . 32

1.5.2 Hermitian and anti-Hermitian operators . . . . . . . . . . . . 33

ix


x Contents

1.5.3 Properties of the eigenvalues and eigenvectors of the Hermi-

tian operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

Chapter 2 Special functions 37

2.1 Hermite polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.1.1 Baker-Hausdorff formula . . . . . . . . . . . . . . . . . . . . 40

2.1.2 Series of even Hermite polynomials . . . . . . . . . . . . . . . 43

2.1.3 Addition formula . . . . . . . . . . . . . . . . . . . . . . . . . 48

2.2 Associated Laguerre polynomials . . . . . . . . . . . . . . . . . . . . 49

2.3 Chebyshev polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.3.1 Chebyshev polynomials of the first kind . . . . . . . . . . . . 52

2.3.2 Chebyshev polynomials of the second kind . . . . . . . . . . . 53

2.4 Bessel functions of the first kind of integer order . . . . . . . . . . . 55

2.4.1 Addition formula . . . . . . . . . . . . . . . . . . . . . . . . . 56

2.4.2 Series of the Bessel functions of the first kind of integer order 58

2.4.3 Relation between the Bessel functions of the first kind of in-

teger order and the Chebyshev polynomials of the second kind 60

Chapter 3 Finite systems of differential equations 63

3.1 Systems 2× 2 first . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.1.1 Eigenvalue equations . . . . . . . . . . . . . . . . . . . . . . . 64

3.1.2 Cayley-Hamilton theorem method . . . . . . . . . . . . . . . 69

3.1.2.1 Case A: λ1 = λ2 . . . . . . . . . . . . . . . . . . . . 70

3.1.2.2 Case B: λ1 = λ2 = λ . . . . . . . . . . . . . . . . . 72

3.2 Systems 4× 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

3.2.1 Case 1. All the eigenvalues are different (λ1 = λ2 = λ3 = λ4) 77

3.2.2 Case 2. λ1 = λ2 = λ3 = λ4 . . . . . . . . . . . . . . . . . . . 78

3.2.3 Case 3. λ1 = λ2 = λ3 = λ = λ4 . . . . . . . . . . . . . . . . . 79

3.2.4 Case 4. λ1 = λ2 = λ3 = λ4 = λ . . . . . . . . . . . . . . . . . 80

3.3 Systems n× n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

3.3.1 All the eigenvalues are distinct . . . . . . . . . . . . . . . . . 84

3.3.2 The eigenvalues are: λ1 = λ2 = λ3, λ4 = λ5, and the rest are

different . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

Chapter 4 Infinite systems of differential equations 91

4.1 Addition formula for the Bessel functions . . . . . . . . . . . . . . . 92

4.2 First neighbors interaction . . . . . . . . . . . . . . . . . . . . . . . . 94

4.3 Second neighbors interaction . . . . . . . . . . . . . . . . . . . . . . 96

4.4 First neighbors interaction with an extra interaction . . . . . . . . . 100

4.4.1 Interaction ωn . . . . . . . . . . . . . . . . . . . . . . . . . . 100


Contents xi

4.4.2 Interaction ω(−1)n . . . . . . . . . . . . . . . . . . . . . . . 107

Chapter 5 Semi-infinite systems of differential equations 115

5.1 First a semi-infinite system . . . . . . . . . . . . . . . . . . . . . . . 115

5.2 Semi-infinite system with first neighbors interaction . . . . . . . . . 119

5.3 Nonlinear system with first neighbors interaction . . . . . . . . . . . 126

Chapter 6 Partial differential equations 133

6.1 A simple partial differential equation . . . . . . . . . . . . . . . . . . 133

6.1.1 A Gaussian function as boundary condition . . . . . . . . . . 133

6.1.2 An arbitrary function as boundary condition . . . . . . . . . 134

6.2 Airy system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

6.2.1 Airy function as boundary condition . . . . . . . . . . . . . . 137

6.3 Harmonic oscillator system . . . . . . . . . . . . . . . . . . . . . . . 138

6.4 z-dependent harmonic oscillator . . . . . . . . . . . . . . . . . . . . . 139

Appendix A Dirac notation 141

Appendix B Inverse of the Vandermonde and Vandermonde con-

fluent matrices 143

B.1 The inverse of the Vandermonde matrix . . . . . . . . . . . . . . . . 145

B.2 The inverse of the confluent Vandermonde matrix . . . . . . . . . . . 146

Appendix C Tridiagonal matrices 149

C.1 Fibonacci system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

Bibliography 153

Index 155


Chapter 1

Linear Algebra

In this Chapter, and in the rest of the book, we will assume that the reader

is familiar with elementary matrix algebra and with determinants; however, in

special cases, and for the sake of clearness, we will define and review some con-

cepts. To review elementary matrix algebra and determinants, we recommend

the following books [Shores 07; Larson 09; Lang 86; Lang 87; Nicholson 90;

Poole 06].

1.1 Vector spaces

The concept of vector space is very important in mathematics. This concept

has applications in physics, engineering, chemistry, biology, the social sciences,

and other areas. It is an abstraction of the idea of geometrical vectors that we

acquire in elementary mathematics and physics, and even in our daily life. The

idea of vector is generalized to any size and to entirely different kinds of objects.

A vector space consists of a nonempty set V of objects, called vectors, that

has an addition + and a multiplication · by scalars (real or complex numbers).

The following ten axioms are assumed to hold.

Addition

(A1) Closure under addition: If u and v belong to V; then so does u+ v.

(A2) Commutative law for addition: If u and v belong to V, then u+ v = v+u.

(A3) Associative law for addition: If u, v, and w belong to V, then u+(v+w) =

(u+ v) + w.

(A4) There exists a zero vector in V, denoted by 0, such that for every vector u

in V, u+ 0 = u.

(A5) For every vector u in V there exists a vector −u, called the additive inverse

1


2 Linear Algebra

of u, such that u+ (−u) = 0.

Scalar Multiplication

(S1) Closure under scalar multiplication: If u belongs to V; then so α ·u for any

scalar α.

(S2) For any two scalars α and β and any vector u in V, α · (β · u) = (αβ) · u.(S3) For any vector u in V, 1 · u = u.

(S4) For any two scalars α and β and any vector u in V, (α+β) ·u = α ·u+β ·u.(S5) For any scalar α and any two vectors u and v in V, α · (u+v) = α ·u+α ·v.

From now on, for simplicity, we will denote the scalar multiplication of a scalar

α by a vector v as αv instead of α · v.

If the scalars are real numbers, we call the vector space a real vector space;

if the scalars are complex, we call the vector space a complex vector space.

Example 1. The set of n-tuples of ordered real numbers,

(x1, x2, ..., xn)|xi ∈ R for all i,with the usual addition and multiplication operations, is the real vector space Rn.

Example 2. The set of all m × n matrices M with real entries, and with

the usual matrix addition and multiplication by a scalar, is the real vector space

M(m× n).

Example 3. The set of all real-valued functions on a set A is a real vector

space. Addition and scalar multiplication are defined so that f + g is the func-

tion x→ f(x) + g(x), and λf is the function x→ λf(x). Similarly, the space of

complex valued functions on A is a complex vector space.

Example 4. The set C(a, b) = f : (a, b) → C| f is a continuos function in (a, b),where C are the complex numbers, with the same operations of the preceding

example, is a complex vector space.

Example 5. The set of all polynomials of degree at most two with the standard

function addition and scalar multiplication forms a vector space.

We list, without demonstration, some properties of vector spaces.

Let v be a vector in some vector space V and let c be any scalar. Then

1. 0v = 0.

2. c0 = 0.


Vector spaces 3

3. (−c)v = c(−v) = −(cv).

4. If cv = 0, then v = 0 or c = 0.

5. A vector space has only one zero element.

6. Every vector has only one additive inverse.

1.1.1 Subspaces

A better understanding of the properties of vector spaces is obtained by intro-

ducing the concept of subsets and the bigger sets from which they can be subsets.

Of primordial importance is when the subsets are by themselves vector spaces;

in that case, the subset is called a subspace. We define subspace as follows:

If V is a vector space, a subset U of V is called a subspace of V if U is itself

a vector space, where U uses the vector addition and scalar multiplication of V.

Example 1. In the case of the vector space R2, all the straight lines that

goes through the origin, are subspaces.

Example 2. If we consider the vector space of real valued continuous func-

tions defined in a certain interval (a, b) in R, the subset of all polynomials of a

fixed degree is a subspace.

Let U be a subset of a vector space V. Then U is a subspace, if and only if,

it satisfies the following three conditions.

1. The zero vector 0 of V lies in U .2. If u and v lie in U , then u+ v lies in U ; i.e., the subset U is closed under the

addition.

3. If u lies in U , then α · u lies in U for all scalars; i.e., the subset U is closed

under the multiplication by scalars.

1.1.2 The span of a set of vectors

If v1, v2, ..., vn is any set of vectors in a vector space V, the set of all linear com-

binations of these vectors is called their span, and is denoted by spanv1, v2, ..., vn.

Example 1. In the vector space R2, the vector (1, 1) span the subspace that

consists of a straight line of slope 1 that goes through the origin.


4 Linear Algebra

Example 2. The set 1, x, x2, x3 spans the vector space of polynomials of

degree less or equal to 3.

Example 3. The three vectors (1, 0, 0), (0, 1, 0), (0, 0, 1) span the vector space

R3.

1.1.3 Linear independence

A set of vectors v1, v2, ..., vn is called linearly independent if any linear com-

bination of them equal to zero necessary implies that all the coefficients of the

combination are zero; i.e., if

α1v1 + α2v2 + ...+ αnvn = 0 (1.1)

then

α1 = α2 = ... = αn = 0. (1.2)

A set of vectors that is not linearly independent is said to be linearly depen-

dent.

Example 1. The vectors (1, 0, 1), (0, 1, 0) and (−1, 1, 1), of the vector space

R3, are linearly independent.

We take a linear combination of the three vectors,

r1(1, 0, 1) + r2(0, 1, 0) + r3(−1, 1, 1) = 0.

As

det

1 0 −1

0 1 1

1 0 1

= 2,

the only solution to the homogeneous system we have, is the trivial one; i.e.,

r1 = r2 = r3 = 0, and then the set is linearly independent.

Example 2. The subset B = sinx, sin 2x, sin 3x, ..., sinnx of the vector space

C(−π, π) = f : (−π, π) → R| f is a continuos function in (−π, π) is linearly

independent for all n.

Taking a linear combination of all the elements in the set B and equating it to


Vector spaces 5

zero, we get

n∑k=1

ck sin kx = 0.

Multiplying this equation by sinmx, integrating it from −π to π, and using the

integral∫ π

−πsinmx sin kxdx = πδmk, we get

πn∑

k=1

ckδkm = πcm = 0.

As m is an arbitrary integer from 1 to n, necessarily ck = 0 for all k from 1 to

n, and we have proved that the set B is linearly independent.

It is clear that a set v1, v2, ..., vn of vectors in a vector space V is linearly

dependent, if and only if, some vi is a linear combination of the others.

We know, from our hypothesis, that in the following equation

n∑k=1

ckvk = 0

at least one coefficient is no null, let say it is the m; then, we can write

cmvm +n∑

k=1,k =m

ckvk = 0.

As cm = 0, we get

vm = −n∑

k=1,k =m

ckcm

vk = 0,

as we wanted to show.

Example 3. In the vector space R2, the vectors (−1, 2) and (3,−6) are lin-

early dependent, as one is a multiple of the other, (3,−6) = −3(−1, 2).

If a vector space V can be spanned by n vectors, then in case that any other

set of m vectors in V is linearly independent, thus m ≤ n


6 Linear Algebra

1.1.4 Bases and dimension of a vector space

One of the main ideas of vector space theory is the notion of a basis. We already

know that a spanning set for a vector space V is a set of vectors v1, v2, ..., vnsuch that V = spanv1, v2, ..., vn . However, some spanning sets are better than

others because they have less elements. We know that a set of vectors has no

redundant vectors in it if and only if it is linearly independent. This observation

take us to the following definition.

A set e1, e2, ..., en of vectors in a vector space V is called a basis of V if it

satisfies the two conditions:

1. The set e1, e2, ..., en is linearly independent

2. The set e1, e2, ..., en spans the vector space V; i.e., V = spane1, e2, ..., en

The fact that a basis spans a vector space means that every vector in the space

has a representation as a linear combination of basis vectors. The linearly inde-

pendence of the basis implies that the representation is unique.

Example 1. In the vector space R3, the set of three vectors

(1, 0, 0), (0, 1, 0), (0, 0, 1) is a basis.

It is very easy to show that the three vectors are linearly independent.

Any vector (x, y, z) in R3, can be written as (x, y, z) = x(1, 0, 0) + y(0, 1, 0) +

z(0, 0, 1), then the set spans the space.

The number of vectors in two different bases of a vector space V must be the

same.

If e1, e2, ..., en is a basis of the nonzero vector space V, the number n of

vectors in the basis is called the dimension of V, and we write dimV.The zero vector space is defined to have dimension 0.

A vector space V is called finite dimensional if V = 0 or V has a finite basis.

Otherwise it is infinite-dimensional.

Let V be a vector space and assume that dimV = n > 0.

1. No set of more than n vectors in V can be linearly independent

2. No set of fewer than n vectors can span V

Let V be a vector space, and assume that dimV = n > 0.

1. Any set of n linearly independent vectors in V is a basis.

2. Any spanning set of n nonzero vectors in V is a basis.


The scalar product. Euclidian spaces 7

1.1.5 Coordinate systems. Components of a vector in a given

basis

A basis of a vector space V was defined to be an independent set of vectors

spanning V. The real significance of a basis lies in the fact that the vectors of

any basis of Cn or Rn may be regarded as the unit vectors of the space, under

a suitably chosen coordinate system, in virtue of the following theorem.

A given vector v, in a vector space V, can always be expanded as

v =n∑

k=1

αkek, (1.3)

where the set e1, e2, ..., en is a basis of the vector space.

The set of scalars α1, α2, ..., αn (complex or real) are called the components

or the coordinates of the vector v in the basis e1, e2, ..., en. It is clear that if

we have different bases, the components of the vector will be different; however

given a basis, this components are unique.

Example 1. In the real vector space R3, the standard basis is the set

(1, 0, 0), (0, 1, 0), (0, 0, 1). The components of the vector (−1, 2,−3) in this

basis are −1, 2 and −3. The set (1, 0, 1), (1, 1, 1), (1, 1, 0) is linearly indepen-

dent, then it is also a basis of R3; in this new basis, the components of the vector

(−1, 2,−3) are −3, 0 and 2.

1.2 The scalar product. Euclidian spaces

We say that a complex vector space V has a scalar product or dot product or an

internal product, if for any two elements u and v in V, a unique complex number

is associated, which will be denoted by u · v, and that association satisfies the

following four properties.

For any three vectors u, v and w in V, and for any scalar α,

1. Hermitian symmetry. u · v = (v · u)∗2. Distributivity or linearity. u · (v + w) = u · v + u · w3. Associativity or homogeneity. (αu) · v = α(u · v)4. Positivity. u · u > 0 if u = 0

In the case of a real vector space, property 1 simplifies to u · v = v ·u, the othersremain equal.


8 Linear Algebra

A vector space with a scalar product, is called an Euclidian space. If the scalars

are real, it is a real Euclidian space, if the scalars are complex, it is called a

complex Euclidian space or Unitary space. In what follows, we will think in

complex Euclidian spaces, and treat the real Euclidian spaces as a particular

case.

Example 1. In the vector space R3, the product

v · u = x1y1 + x2y2 + x3y3,

is a scalar product.

Example 2. If we consider the vector space

C(a, b) = f : (a, b) → C| f is a continuos function in (a, b),

the product f · g =∫ b

af(x)g∗(x)dx constitutes a scalar product.

1.2.1 The norm in an Euclidian space

With the introduction of the scalar product, we can define a norm (a ”size”) for

the elements of a vector space. Given a vector v in an Euclidian space V, wedefine the norm of the vector as

|v| =√v · v. (1.4)

Note that the norm of a vector is always a nonnegative integer.

Cauchy-Schwarz inequality. In a vector space V all the scalar products satisfy

the Cauchy-Schwarz inequality; i. e., given two vectors v and u in V,

|v · u|2 ≤ |v|2|u|2. (1.5)

The equality is satisfied, if and only if, the two vectors, v and u, are linearly

dependent.

1.2.2 The concept of angle between vectors. Orthogonality

In a real Euclidian space, and in analogy with the Euclidian space R3, it is also

possible to introduce the concept of angle between vectors. If we have v and u


The scalar product. Euclidian spaces 9

in V, the angle between them is defined as

θ = arccos(v · u|v||u|

). (1.6)

In a real Euclidian space V,a) Two vectors are orthogonal, if and only if, their scalar product is zero.

b) A vectorial subspace S, of a vector space V, is called orthogonal, if and only

if, v · u = 0 for any pair of unequal vectors in S.c) A vectorial subspace S, of a vector space V, is called orthonormal if in addi-

tion to be orthogonal all their vectors have norm 1.

Example 1. Let us consider the complex Euclidian space formed by the com-

plex vector space

C(a, b) = f : (a, b) → C| f is a continuos function in (a, b),with the scalar product f · g =

∫ b

af(x)g∗(x)dx. The subset 1, x, x2, ..., xn of

C(a, b) spans the subspace of all the polynomials of degree less than or equal to

n; however, this set is not orthogonal.

We have to calculate the scalar product of any two functions in the set:

xj · xk =∫ b

axjxkdx =

∫ b

axj+kdx = 1

j+k+1 [bj+k+1 − aj+k+1] = 0,

thus, no pair is orthogonal and the set is not orthogonal.

Example 2. The subset B = sinx, sin 2x, sin 3x, ..., sinnx of the real Euclid-

ian space C(−π, π) = f : (−π, π) → R| f is a continuos function in (−π, π)with the scalar product f · g =

∫ b

af(x)g(x)dx is an orthogonal set.

We already used the integral∫ π

−πsinmx sin kxdx = πδmk, so the scalar product

of any two members of the set is zero, then the set is orthogonal.

In a real Euclidian space V, any orthogonal set S of no null vectors is linearly

independent.

Let be S = v1, v2, ..., vk the orthogonal set. We take a linear combination of

these vectors and we make it equal to zero,

k∑j=1

λjvj = 0.

Now we take the scalar product of this equation with an arbitrary vector vm,

with m from 1 to k, and we use that vm · vj = δmj/(|vm||vj |), to write

k∑j=1

λjδmj

|vm||vj |=

λm|vm|2

= 0,


10 Linear Algebra

so λm = 0. As m is arbitrary, the necessary conclusion is that λj = 0 for all j

from 1 to k, and the set S is linearly independent.

In a real Euclidian space V of finite dimension n, any orthogonal set S of n

no null vectors is a basis.

Given an Euclidian space V of finite dimension n, and S = e1, e2, ..., en an

orthogonal basis of it, any vector v of V can be expanded as

v =

n∑k=1

ckek, (1.7)

where

ck =v · ek|ek|2

(1.8)

for k = 1, 2, 3..., n.

We take the scalar product of expression (1.7) with em, being m an arbitrary

integer from 1 to n, we use the properties 2 and 3 of the scalar product and we

use the fact that the basis is orthogonal (i.e., ei · ej = |ei||ej |δij), to write

v · em =n∑

k=1

ck(ek · em) =n∑

k=1

ck|em||ek|δmk = |em|2cm,

and from this equation (1.8) follows trivially.

Given an Euclidian space V of finite dimension n and S = e1, e2, ..., en an

orthonormal basis of it, any vector v of V can be expanded as

v =

n∑k=1

ckek, (1.9)

where

ck = v · ek (1.10)

for k = 1, 2, 3..., n.

Parseval formula. Let be V an Euclidian space of finite dimension n, and

S = e1, e2, ..., en an orthonormal basis of it. Then for any two vectors, v and


Linear transformations 11

u in V, the following formula is satisfied

v · u =n∑

k=1

(v · ek)(u · ek)∗. (1.11)

That implies that

v · u =

n∑k=1

vku∗k, (1.12)

where vk and uk are the components of the two vectors in the given basis.

In particular, if u = v,

|v|2 =n∑

k=1

v2k. (1.13)

In all Euclidian spaces is always possible to build an orthogonal basis, and

therefore an orthonormal basis. The construction processes is known as the

Gram-Schmidt orthogonalization process [Nicholson 90; Poole 06].

1.3 Linear transformations

A transformation is a function with vector spaces for its domain and codomain.

A transformation is linear if it preserves linear combinations; or more precisely,

if V and W are two vector spaces, a function T : V → W is called a linear

transformation in V, if it satisfies the following properties:

a) T (v + u) = T (v) + T (u) for all v and u in V.b) T (rv) = rT (v) for all v in V and all scalars r (real or complex).

The linear transformations are also called linear maps and linear operators.

However, we will use the denomination linear operator when the domain and

the codomain are the same vector space.

Example 1. The transformation from R3 to R2 given by the rule

(x, y, z) → (x+ y + z, x− 2y − z)

is a linear transformation.

Example 2. The differential operator acting on the vector space

C = f : (a, b) → R|f is a continuos function in (a, b)is a linear transformation.


12 Linear Algebra

A linear transformation always maps the zero vector of the domain in the

zero vector of the codomain. In other words, given a linear transformation T ,

from the vector space V to the vector space W, we have

T (0V) = 0W , (1.14)

where 0V is the zero vector in the vector space V and W is the zero vector in

the vector space W.

The second property of a linear transformation, T (rv) = rT (v), holds for any

vector v in V, and for any scalar r; in particular it holds for the zero scalar, and

then expression (1.14) follows trivially.

As the linear transformations are in fact functions from one vector space into

another, all the operations for functions applied. We can add linear transforma-

tions, and we can multiply linear transformations by a scalar. It is easy to show

that all these operations give again a linear transformation, then the space of

all linear transformations is itself a vector space.

1.3.1 The kernel of a linear transformation

Given a linear transformation T , from the vector space V to the vector space

W, we define the kernel of the transformation as the subset of the vector space

V of all the vectors that have as image the null vector in W; i.e., if we denote

the kernel of the linear transformation as ker(T ), we have

ker(T ) = vϵV|T (v) = 0. (1.15)

The kernel of a linear transformation is a vector subspace of the domain V.We already know that it is enough to show that the zero vector is in the kernel,

and that the kernel is closed under the addition and under the multiplication by

a scalar. First, the zero vector is in the kernel, as we also know that the zero

vector always goes to the zero vector under a linear map (see expression (1.14)).

Second, as the transformation is linear, if we have v, u in ker(T ) and two arbi-

trary scalars r and s, then T (rv + su) = rT (v) + sT (u) = r0 + s0 = 0, showing

that the kernel is closed under both operations.

The two following statements are completely equivalent:

1. The kernel of a linear transformation T is the set 0.2. If v and u are two elements of the vector space V such that T (v) = T (u),

then v = u.



We suppose 1 and show 2. If T (v) = T (u), using the linearity of T we get

T (v)− T (u) = T (v − u) = 0, but as by hypothesis the only vector that goes to

zero is zero, we obtain v − u = 0, or, as we wanted to show, v = u.

Now we suppose 2 and show 1. If we have a vector w in V such that T (w) = 0,

it is trivially true that T (w) = T (0) and then, by hypothesis, w = 0.

Example 1. The kernel of the linear transformation from R3 to itself, given by

the law, (x, y, z) → (x − y, x + y + z, 2x − 3y + z), is clearly the set with only

the vector zero, (0, 0, 0); any other vector goes to a nonzero vector in R3.

Example 2. The kernel of the linear operator d/dx, that acts in the vector

space C(a, b) = f : (a, b) → R| f is a continuos function in (a, b),is the set of constant functions in the interval, f : (a, b) → R| f = constant.

1.3.2 The image of a linear transformation

Given a linear transformation T , from the vector space V to the vector space W,

we define the image or range of the transformation as the subset of the vector

space W of all the vectors that are image of some vector of V; i.e., if we denote

the image of the linear transformation as im(T ), we have

im(T ) = wϵW| exists vϵV such that T (v) = w. (1.16)

The image of a linear transformation is a vector subspace of the domain W.

As we did for the kernel, we only need to show that the zero vector belongs to

the image, and that the image is closed under the two vector space operations.

First, as for a linear map zero goes to zero, the zero vector in W is in the image.

Second, if we have w1, w2 in im(T ), then exists v1 and v2 in V such that

T (v1) = w1 and T (v2) = w2. Then, rw1 + sw2 = rT (v1) + sT (v2) and as

the transformation T is linear rw1+ sw2 = rT (v1)+ sT (v2) = T (rv1+ sv2), and

clearly rw1 + sw2 ∈ im(T ).

Example 1. The image of a rotation of the plane R2 is the plane R2 itself.

1.3.3 Isomorphisms

A linear transformation T : V → W, is called injective if for all pair of vectors,

u and v in V, with u = v, we have T (u) = T (v).


14 Linear Algebra

With the results of the previous section, we can conclude that a linear trans-

formation is injective, if and only if, the only element that maps to zero is zero.

Example 1. A rotation in R2 is an injective mapping, as the only element

that ”remains” in place is the origin, and only the origin is mapped into the

origin.

Example 2. The linear map from R3 into R2, with the rule (x, y, z) → (x, y),

is not injective as maps all the straight line (0, 0, t), t ∈ R, into the vector (0, 0).

A linear transformation T : V → W, is called surjective if im(T ) = W.

Example 3. The linear map from R2 into R3, with the rule (x, y) → (x +

y, x − y, 3), is not surjective, as all the elements of the subset (r, s, t)|r, s, t ∈R with t = 3 of R3 do not have associated any vector in the domain R2.

If a linear transformation is injective and surjective, then it is called isomor-

phism. An isomorphism is then a map one-to-one.

Any finite dimensional nonzero vector space, over the field of the real num-

bers R or over the field of the complex numbers C, is isomorph to Rn or to Cn,

respectively .

To understand this one to one correspondence, we have to think in the set of n

numbers formed by the components of any vector in a given basis, and associate

this set with an element of Rn or Cn.

1.3.4 Linear transformations and matrices

An important and very useful result of linear algebra, is that exists a one-to-one

relationship between linear transformations, acting in finite dimensional vector

spaces, and matrices. That means that we can associate a linear transformation

with a matrix and that every linear transformation has associated a matrix. In

fact, there exists an isomorphism between linear transformations on finite di-

mensional vector spaces and matrices.

First, we prove the easy case, when a given matrix generates a linear trans-

formation.

Given a m× n complex matrix A, a linear transformation T from Cn to Cm is



generated with the rule

v → Av (1.17)

where v is a vector in the vector space Cn and Av is a vector in the vector space

Cm.

To show that the transformation generated is linear, we apply the rule (1.17) to

a linear combination of two arbitrary vectors v and u in Cn, getting A(rv+ sv);

but from the properties of the matrix algebra, it is trivial that A(rv + sv) =

rA(v) + sA()u, which is the linearity property of a transformation.

Now we show, without demonstration, what is the matrix that represents a

given linear transformation.

Given a linear transformation T : V → W, with dim(V) = n and dim(W) = m,

a m×n matrix A can be built. Let e1, e2, ..., en be a basis of the vector space

V and let f1, f2, ..., fm be a basis of the vector space W; the columns of the A

matrix, associated with the transformation T in these bases, are the transformed

vectors of the basis of V. As these last elements are in W, they must be written

in terms of the given basis; in other words,

A = (T (e1), T (e2), ..., T (en)). (1.18)

Example 1. Given the 3× 2 matrix

−2 1

−1 0

2 −3

, the linear transformation

from R2 to R3, with the rule (x, y) → (−2x+ y,−x, 2x− 3y), is generated.

Example 2. If we have the linear transformation T from R4 to R3, given

by the transformation law

(x1, x2, x3, x4) → (2x1 − 3x2 + 4x3,−x1 + 5x4, x2 + x3 + x4),

the associated matrix, in the standard basis, is given by(T (1, 0, 0, 0) T (0, 1, 0, 0) T (0, 0, 1, 0) T (0, 0, 0, 1)

);

or in other words,

2 −3 4 0

−1 0 0 5

0 1 1 1

.

Example 3. Let us consider the linear transformation T : R2 → R3, given by

T (x, y) = (x+y, x, y). Consider also en R2 the basis B = (1,−1), (0, 1) and the

basis B′ = (1, 1, 0), (0, 1, 1), (1, 0, 1) in R3. To find the matrix representation

of this linear transformation in these bases, we calculate the transforms of the

basis of R2; we have, T (1,−1) = (0, 1,−1) and T (0, 1) = (1, 0, 1). We write now,


16 Linear Algebra

(0, 1,−1) = (1, 1, 0) + 0(0, 1, 1) − (1, 0, 1) and (1, 0, 1) = 0(1, 1, 0) + 0(0, 1, 1) +

(1, 0, 1), then the matrix that represents T in these bases is

1 0

0 0

−1 1

.

Example 4. Let B = 1, x, ex, xex be a basis of the vector subspace W of

continuous functions, and let D be the differential operator on W. To calculate

the representation of this operator in this basis, we have to evaluate its action

on each of the elements of the basis; we get D(1) = 0, D(x) = 1, D(ex) =

ex, D(xex) = (1 + x)ex. We write now this functions in terms of the ba-

sis, 0 = 0(1) + 0(x) + 0(ex) + 0(xex), 1 = 1(1) + 0(x) + 0(ex) + 0(xex), ex =

0(1)+ 0(x)+ 1(ex)+ 0(xex), (1+x)ex = 0(1)+ 0(x)+ 1(ex)+ 1(xex). Thus, the

corresponding matrix is

0 1 0 0

0 0 0 0

0 0 1 1

0 0 0 1

.

The corresponding operations in linear transformations are mapped into the

corresponding operations in matrices; i.e., if we have two linear transformations,

T and S, with corresponding matrices A and B, the associated matrix with the

linear transformation αT + βS will be αA + βB, being α and β two arbitrary

complex numbers (or real numbers, according to the case). Therefore, the sum of

linear transformations corresponds to the sum of matrices and the multiplication

of a linear transformation by a scalar, corresponds to the multiplication of ma-

trices by scalars. In the next subsection, we will see that matrix multiplication

corresponds to linear transformations composition or multiplication.

1.3.5 The product of linear transformations

Very important in mathematics and in physics is the composition of linear trans-

formations; it is also called multiplication. We define the composition and give

some of his properties in the rest of this subsection.

Given two linear transformations T : V → W and S : W → U , the composite

ST : V → U of T and S is defined by

ST (v) = S[T (v)], (1.19)

for all v in V.

The composition or multiplication of two linear transformations is a linear



transformation.

Let be v and u two arbitrary vectors in V, and α and β two arbitrary scalars

(real or complex). We make act the multiplication of the two linear transforma-

tions in the linear combination of our two arbitrary vectors, remembering that

both transformations are linear,

ST (αv + βu) = S[T (αv + βu)] = S[αT (v) + βT (u)] = αS[T (v)] + βS[T (u)]

that is exactly the linear property.

Example 1. Let be T : R4 → R3 such that (x1, x2, x3, x4) → (x1 + x2 −x3, x3, x2 − x4) and S : R3 → R2 such that (x1, x2, x3) → (2x1 − x3, x2), then

the multiplication of this two linear transformations is a linear transformation

with the rule (x1, x2, x3, x4) → (2x1+x2−2x3+x4, x3) acting from R4 into R2.

Example 2. Let us consider the real vector space of polynomials of degree equal

or less than 3 with real coefficients defined in the open interval (0, 1); i.e., the

set P = f : (0, 1) → R|f(x) = a0 + a1x+ a2x2 + a3x

3, with a0, a1, a2, a3 ∈ R.Let be the derivative operator D = d/dx acting on P. We can multiply D by

itself and get the linear operator that corresponds to the second derivative, and

so on, and get any order derivative. We can consider also the integral operator

acting on this same vector space; then if we compose the derivative operator

with the integral operator, we get the identity transformation.

It is very easy to see that not all pairs of linear transformations can be

composed. The codomain of the first linear transformation must be the domain

of the second. In the example 1 above, we make the product ST , but it is

clear that the product TS can not be done. As another example, if we have the

transformation T : R3 → R3 with the rule (x, y, z) → (x + y, y − z, x + y + z),

and the transformation S : R2 → R3 with the rule (x, y) → (x+ y, x− y, y), it isobvious that it is not possible to build the composition ST ; however the product

TS exists.

Moreover, and very important, even if TS and ST can both be formed, the new

linear transformations TS and ST need not be equal. The quantity

TS − ST (1.20)

is called the commutator of T and S and is denoted by the symbol [T, S].

Example 3. Let be R : R2 → R2 a rotation in an angle θ in the plane;

that means (x, y) → (x cos θ − y sin θ, x sin θ − y cos θ) and let be T : R2 → R2

the linear operator with the law (x, y) → (2x− y, x− 2y).


18 Linear Algebra

We have TR : R2 → R2 with the law

(x, y) →(2x cos θ − y cos θ − x sin θ − 2y sin θ, x cos θ − 2y cos θ − 2x sin θ − y sin θ)

and we have RT : R2 → R2 with the law

(x, y) →(2x cos θ − y cos θ − x sin θ + 2y sin θ, x cos θ − 2y cos θ + 2x sin θ − y sin θ).

The two new linear transformations are so similar that it is very easy to get

confused and think that they commute; however, we can see that they are not

equal, calculating the commutator, we obtain [T,R] : R2 → R2, with the rule

(x, y) → −4 sin θ(y, x).

Example 4. Two very important linear operators (linear transformations)

in quantum physics are x, multiplication by the independent variable, and

p = −id/dx, both acting on the vector space

C(a, b) = f : (a, b) → R| f is a continuos function in (a, b).It is obvious that xp[f(x)] = −ixdf(x)

dx and that px[f(x)] = −ixdf(x)dx − if(x), so

xp = px. In fact, we already showed that the commutator of these two operator

is i; i.e., [x, p] = i.

Linear transformations or operators whose commutator vanishes are called

commuting transformations. Any linear transformation commutes with the cor-

responding unit transformation, since

(TI)v = Tv, (1.21)

(IT )v = Tv. (1.22)

We shall consider as relevant the multiplication of linear transformations by

scalars, treating the equation

S = αT = Tα (1.23)

as equivalent to the equation

Sv = α(Tv) (1.24)

for any v.

With the definition of the product of two linear transformations, we can

define a new linear transformation raising a given one to a certain power. For

example, Tmv means

Tmv = TT · · ·Tv. (1.25)



Similarly, it is possible to define functions of linear transformations by their

formal power series expansions. One very useful and very important case is the

exponential function; the linear transformation eT formally means

eT = 1 + T +T 2

2!+T 3

3!+ · · · =

∞∑k=1

T k

k!. (1.26)

We will introduce now the very important concept of the inverse of a linear

transformation. For that we establish the following theorem.

Let V and W be finite dimensional vector spaces. The two following conditions

are equivalent for a linear transformation T : V → W:

1. T is an isomorphism.

2. There exists a linear transformation S : W → V, such that TS = IW and

ST = IV , where IW is the unit operator on the vector space W and IV is the

unit operator on the vector space V.The isomorphism S is called the inverse of T and is denoted by T−1.

At the end of the previous subsection, we already mentioned that the ma-

trix associated with the product of two linear transformations, acting in finite

dimensional vector spaces, is the product of the corresponding matrices; i.e.,

if we have two linear transformations, T and S, with corresponding matrices

A and B, the associated matrix with the linear transformation TS will be AB.

In this case the vector space of linear transformations is enriched with a new

operation, a multiplication, that has well defined properties; the set obtained is

now a non-commutative algebra.

Example 5. In example 1 of this subsection, we considered the linear transfor-

mations T : R4 → R3, such that (x1, x2, x3, x4) → (x1 + x2 − x3, x3, x2 − x4),

and S : R3 → R2, such that (x1, x2, x3) → (2x1 − x3, x2). We found that the

multiplication of this two linear transformations is a linear transformation with

the rule (x1, x2, x3, x4) → (2x1 + x2 − 2x3 + x4, x3), acting from R4 to R2.

If we call A to the matrix associated with the linear transformation T and

we call B to the matrix associated with the linear transformation S, we have

A =

1 1 −1 0

0 0 1 0

0 1 0 −1

and B =

(2 0 −1

0 1 0

).

The matrix associated with the composition ST is given by the product of matri-

ces BA; i.e., ST ⇒(

2 1 −2 1

0 0 1 0

). It is very easy to verify that effectively

that matrix corresponds to the matrix of the composition.


20 Linear Algebra

It is worth to note, that the product TS does not exists, and that corresponds

to the fact that the product of matrices AB can not be done.

Example 6. In the case of the linear operators acting on R2, that we stud-

ied in example 3, we have the matrix R =

(cos θ − sin θ

sin θ cos θ

)for the linear

operator R, and the matrix T =

(2 −1

1 −2

)for the linear operator T .

The matrix associated with the linear operator TR is then

TR =

(− sin θ + 2 cos θ −2 sin θ − cos θ

−2 sin θ + cos θ − sin θ − 2 cos θ

)and the matrix associated with

the linear operator RT is RT =

(− sin θ + 2 cos θ 2 sin θ − cos θ

2 sin θ + cos θ − sin θ − 2 cos θ

).

Thus, the matrix associated with the commutator of T and R is

[T,R] = −4 sin θ

(0 1

1 0

)and correctly correspond to the law

(x, y) → −4 sin θ(y, x) that we found in example 3.

1.4 Eigenvalues and eigenvectors

Let be V a vector space, S a subspace of V, and T a linear transformation from

S to V. A scalar λ is an eigenvalue of T , if there is a non-null vector v in S,such that

T (v) = λv. (1.27)

The vector v is called an eigenvector of T , corresponding to the eigenvalue λ.

Also the scalar λ is called the eigenvalue of the transformation T corresponding

to the eigenvector v.

The eigenvalues and eigenvectors are also called characteristic values and char-

acteristic vectors, respectively.

It is worth to emphasize, that even if the null vector 0 satisfies the equation

T (0) = λ0 for any scalar λ, it is not considered an eigenvector. It is also worth

to note, and very easy to prove, that there is only one eigenvalue for a given

eigenvector.

Example 1. Consider the linear operator T , acting on R3 and defined by

the rule (x, y, z) = (x−2y+ z, 0, y+ z). The vector (−3,−1, 1) is an eigenvector

of T with eigenvalue 0 and the vector (1, 0, 0) is an eigenvector of T with eigen-


Eigenvalues and eigenvectors 21

value 1.

In terms of matrices, the linear transformation is 1 −2 1

0 0 0

0 1 1

x

y

z

=

x− 2y + z

0

y + z

and the two ”eigenequations” are 1 −2 1

0 0 0

0 1 1

−3

−1

1

= 0

−3

−1

1

and

1 −2 1

0 0 0

0 1 1

1

0

0

=

1

0

0

.

Example 2. The eigenfunctions of the derivative operator D = d/dx, acting

on the vector space

C(a, b) = f : (a, b) → R| f is a continuos function in (a, b)are the solutions of the differential equation

df(x)

dx= λf(x);

they are the functions f(x) = aeλx, with a an arbitrary real constant. The

corresponding eigenvalue is λ.

We will give now several very useful definitions:

Let be V a vector space, S a subspace of V, and T a linear transformation

from S to V. A subspace U of S is called invariant under the T transformation,

if T maps every element of U into U .

If λ is an eigenvalue of T : S ⊂ V → V, the set

Eλ = Eλ(T ) = v|vϵS, T (v) = λv (1.28)

is a vector space, a subspace of S, called the eigenspace associated with λ.

It is clear that the subspace generated by an eigenvector is invariant under

the transformation.

Let be V a vector space, S a subspace of V, and T a linear transformation

from S to V. If we have k eigenvectors v1, v2, ..., vk with k different eigenvalues

λ1, λ2, ..., λk, then the corresponding eigenvectors v1, v2, ..., vk constitute a

linearly independent set.


22 Linear Algebra

We prove this theorem by mathematical induction [Bronshtein 07, page 5].

a) We prove the theorem for k = 1.

We take a linear combination of the vectors in the set (just v1) and make it equal

to zero; i.e., c1v1 = 0. As by definition, v1 is non-null, necessarily c1 = 0.

b) We suppose now the theorem true for k − 1 and we prove it for k.

We want to prove that∑k

i=1 civi = 0 implies that ci = 0 for all i from 1 to k.

As T is a linear map and vi its eigenvectors

T (k∑

i=1

civi) =k∑

i=1

ciT (vi) =k∑

i=1

ciλivi,

then

T (k∑

i=1

civi)− λk

k∑i=1

civi =k∑

i=1

ciλivi − λk

k∑i=1

civi =

= ckλkvk +k−1∑i=1

ciλivi − ckλkvk − λk

k−1∑i=1

civi =k−1∑i=1

ci(λi − λk)vi = 0,

but our induction assumption is that the set v1, v2, ..., vk−1 is linearly inde-

pendent, then ci(λi−λk) = 0 for i = 1, 2, 3, ..., k−1. However, in the hypothesis

of the theorem we supposed that λi = λj for i, j = 1, 2, 3, ..., k, and then neces-

sarily ci = 0 for i = 1, 2, 3, ..., k − 1.

It only remains to prove that ck is also zero. We take again the original linear

combination,∑k

i=1 civi, we use that ci = 0 for i = 1, 2, 3, ..., k − 1 and that all

the eigenvectors are different of zero, and arrive to ck = 0.

The inverse of this theorem is not true. If a linear transformation T has a

set of k linearly independent eigenvectors, then the corresponding eigenvalues

do not have to be different. We can have a linear transformation with a set of

eigenvectors linearly independent and all the eigenvalues be the same, as with

the identity transformation.

Example 3. Let be the linear operator acting on R3, whose associated ma-

trix is 2 3 1

0 −1 2

0 0 3

.

In example 1 of the next subsection we will demonstrate that their eigenvalues

and eigenvectors are



2 ↔

1

0

0

, −1 ↔

1

−1

0

, 3 ↔

5

1

2

.

As the determinant

det

1 1 5

0 −1 1

0 0 2

= −2

is different from zero, the three vectors are linearly independent.

Let be V a finite dimension vector space, S a subspace of V, and T a linear

transformation from S to V. If the dimension of V is n, then the maximum

number of eigenvalues is n.

If the linear transformation T has exactly n different eigenvalues, then the corre-

sponding eigenvectors constitute a basis for the vector space V, and the matrix

A, associated to the linear transformation T , is a diagonal matrix with the eigen-

values as diagonal elements.

The existence of n different eigenvalues is a sufficient condition to have a

diagonal representation of the associated matrix, but it is not necessary. There

are linear transformations with less than n eigenvalues and with a diagonal rep-

resentation.

The existence of n linearly independent eigenvectors is a necessary and suffi-

cient condition for the linear transformation to have a diagonal representation.

Example 4. Again we will analyze a linear operator acting on R3, whose

associated matrix is

−2 2 −3

2 1 −6

−1 −2 0

.

Their eigenvalues and eigenvectors are

5 ↔

1

2

−1

, −3 ↔

−2

1

0

, −3 ↔

3

0

1

.

The eigenvalue −3 has multiplicity 2. However, the three eigenvectors are lin-


24 Linear Algebra

early independent. To prove that we calculate the determinant

det

1 −2 3

2 1 0

−1 0 1

= 8

and as it is different from zero, the three vectors are linearly independent. It is

clear then, that the condition that all the eigenvalues be different is sufficient

but not necessary.

1.4.1 The finite dimension case

In the rest of this section, we will develop methods to find the eigenvalues and

eigenvectors of linear transformations between vector spaces of finite dimension.

If T is a linear transformation T : S ⊂ V → V, with dim(V) = n, we want to

find the scalars λ such that the equation T (v) = λv has nontrivial solutions; in

other words, we want to find the nontrivial solutions to the equation

(λI − T )v = 0 (1.29)

with v = 0 and with I the identity transformation.

If A is the matrix associated with the linear transformation T : S ⊂ V → V,then the equation

(λI−A)x = 0 (1.30)

has a nontrivial solution, if and only if, the matrix λI−A is singular, or in other

words, it does not have an inverse or also that det(A) = 0.

Then, if λ is an eigenvalue of the linear transformation T , it must satisfy the

equation

det(λI−A) = 0. (1.31)

Inversely, if λ satisfies the above equation, then it is an eigenvalue of the trans-

formation T .

If we define the function

p(λ) = det(λI−A), (1.32)

then

a. The function p(λ) is a polynomial of degree n.

b. The highest degree term is λn.



c. The constant term, p(0), is det(−A) = (−1)n detA.

The function p(λ) is called the characteristic polynomial of the matrix A.

If the vector space is over the real numbers, then the eigenvalues of the linear

transformation are the real roots of the characteristic polynomial.

Example 1. As promised in the example 2 of the previous subsection, we

will demonstrate now that the eigenvalues and eigenvectors of the matrix 2 3 1

0 −1 2

0 0 3

are

2 ↔

1

0

0

, −1 ↔

1

−1

0

, 3 ↔

5

1

2

.

a) We calculate first the characteristic polynomial.

p(λ) = det(λI−A) = λ3 − 4λ2 + λ+ 6.

It is easy to find [Spiegel 98, Chapter 20, page 212] that their roots (and then

also the eigenvalues) are 2,−1 and 3.

To find the eigenvectors, we have to solve the equation Av = λv, for each

eigenvalue. For the eigenvalue 2, we obtain, after all reductions, the system of

equations

y = 0 and z = 0.

The invariant subspace corresponding to the eigenvalue 2 is the X axis . The

parametric equation is then (x, y, z) = (1, 0, 0)t and we take as eigenvector

(1, 0, 0).

For the eigenvalue -1 we obtain the system of linear equations

x+ y = 0 and z = 0.

Thus, the invariant subspace is the straight line (x, y, z) = (1,−1, 0)t and we

take as eigenvector (1,−1, 0).

Finally, for the eigenvalue 3, we obtain the system of linear equations

x− 3y − z = 0 and 2y − z = 0.

Thus, the invariant subspace is the straight line (x, y, z) = (5, 1, 2)t and we take

as eigenvector (5, 1, 2).

Example 2. We will calculate the eigenvalues and eigenvectors of the 3 × 3


26 Linear Algebra

matrix, A =

7√3 −6√

3 5 6√3

−6 6√3 4

.

a) The characteristic polynomial is

p (λ) = det (λI −A) = det

λ1 0 0

0 1 0

0 0 1

−

7√3 −6√

3 5 6√3

−6 6√3 4

=

= λ3 − 16λ2 − 64λ+ 1024.

b) To calculate the eigenvalues, we factorize the characteristic polynomial as

λ3−16λ2−64λ+ 1024 = (λ− 8) (λ+ 8) (λ− 16)

and then the eigenvalues of this matrix are λ1 = 8, λ2 = −8, λ3 = 16.

c) To find the eigenvectors, we have to solve the linear system Avi = λivi for

each of the already known eigenvalues.

We start with λ = 8, and we get the equation 7√3 −6√

3 5 6√3

−6 6√3 4

x

y

z

= 8

x

y

z

.

The solution to this system is

x =√3y,

z = 0.

The intersection of these two planes in the space R3, is a straight line with

parametric equation (x, y, z) =(√

3, 1, 0)t, where t ∈ R − 0. We have to

exclude the zero vector, as it is not considered an eigenvector, even if satisfies

the eigenvalue equation. We can take as eigenvector any one in this invariant

subspace. One possibility is to take a normalized eigenvector; for simplicity we

take v1 = (√3, 1, 0).

For the other two eigenvalues, the procedure is identical, and the invariant sub-

spaces are the straight line (x, y, z) =(1,−

√3, 2)t, with t ∈ R − 0 for the

eigenvalue −8, and (x, y, z) =(−1,

√3, 2)t, with t ∈ R− 0 for the eigenvalue

16.

Resuming, the eigenvalues and the eigenvectors are

8 ↔

√3

1

0

, −8 ↔

1

−√3

2

, 16 ↔

−1√3

2

.



1.4.2 Similar matrices

Two n× n matrices A and B are called similar if

B = P−1AP (1.33)

for some invertible n× n matrix P.

Similar matrices represent the same linear transformation under two differ-

ent bases. If the matrix A represents the linear transformation T in some basis

and the matrix B represents the same linear transformation in another basis,

then B = P−1AP, with P being the change of bases matrix. The matrix P is

sometimes called a similarity transformation.

It is worth to note, that this condition is sufficient and necessary; then two ma-

trices are similar, if and only if, they represent the same linear transformation.

Similarity is an equivalence relation on the space of square matrices. That

is, all the matrix representations of a linear operator T form an equivalence class

of similar matrices.

Similar matrices share many properties:

• The determinant.

• The trace.

• The eigenvalues.

• The characteristic polynomial.

Example 1. Consider the linear operator T on R2 defined by T (x, y) =

(5x + y, 3x − 2y) and the following two bases B = (1, 2), (2, 3) and B′ =

(1, 3), (1, 4). We want to find the representation of the linear operator on

both bases. We also want the change-of basis matrix, and finally we want to

show how the two representations are related.

We calculate the representations of T . To find the first column of the matrix A

representing the linear operator T in the basis B = (1, 2), (2, 3), we have to

calculate T (1, 2) = (7,−1) and represent it in the basis B = (1, 2), (2, 3); weget (7,−1) = −23(1, 2) + 15(2, 3).

To evaluate the second column, we do the same with the other element of the ba-

sis, T (2, 3) = (13, 0) and (13, 0) = −39(1, 2) + 26(2, 3), so A =

(−23 −39

15 26

).

For the matrixB, representing the linear operator T in the basis B = (1, 3), (1, 4),

we do exactly the same, getting B =

(35 41

−27 −32

).


28 Linear Algebra

Now we will find the change-of-basis matrix P. As (1, 3) = 3(1, 2) − (2, 3) and

(1, 4) = 5(1, 2)− 2(2, 3), we have P =

(3 5

−1 −2

).

Finally, we can verify that

B = P−1AP

=

(3 5

−1 −2

)−1( −23 −39

15 26

)(3 5

−1 −2

)=

(35 41

−27 −32

).

1.4.3 Diagonal matrices

If T is a linear transformation T : S ⊂ V → V, with dim(V) = n, and all the roots

λ1, λ2, ..., λn of the characteristic polynomial are different in the corresponding

field of scalars:

a) The corresponding eigenvectors u1, u2, ..., un constitute a basis for V.b) The matrix that represents the linear transformation T , with respect to the

ordered basis u1, u2, ..., un, is the diagonal matrix Λ = diag(λ1, λ2, ..., λn).

c) If A is the matrix that represents the linear transformation T with respect to

other basis e1, e2, ..., en, then

Λ = C−1AC, (1.34)

where C is the matrix that relates the two bases; i.e.,

U = EC. (1.35)

Example 2 In the case of the matrix of the example 1 above, the diagonal ma-

trix is A =

8 0 0

0 −8 0

0 0 16

. in the basis formed by the eigenvectors,

√3

1

0

,

1

−√3

2

,

−1√3

2

.

The similarity matrix or the matrix that makes the conversion from the original

basis to the eigenvector basis, where the matrix has a diagonal representation,

is

√3 1 −1

1 −√3

√3

0 2 2

.

If the eigenvalues are not all different, that does not mean that there is not



a diagonal representation. We will have a diagonal representation, if and only

if, there are k linearly independent eigenvectors associated with each eigenvalue

of multiplicity k.

Example 3 We will calculate the eigenvalues and eigenvectors of the 3 × 3

matrix, A =

−13 −8 −4

12 7 4

24 16 7

and we will give a diagonal representation of

it. Also the similarity matrix will be presented.

a) The characteristic polynomial is

f (λ) = det (λI −A) = det

λ+ 13 8 4

−12 λ− 7 −4

−24 −16 λ− 7

=

= λ3 − λ2 − 5λ− 3 = (λ− 3) (λ+ 1)2.

b) The eigenvalues, which are the roots of the characteristic polynomial, are −1,

with multiplicity 2, and 3.

c) In order to find the eigenvectors, we have to solve now for each eigenvalue the

equation Av = λv.

For the eigenvalue −1, reducing the system with the Gauss-Jordan procedure[Larson 09; Shores 07], we get the equation 3x + 2y + z = 0; that is a plane

that goes through the origin. Then, the invariant subspace corresponding to the

eigenvalue −1, with multiplicity 2, is two dimensional, it is a plane. Any pair

of linearly independent vectors in that plane works as eigenvectors. The easy

way to choose them, is to write the parametric equations of the plane; taking

as parameters y and z, and renaming them as s and t, respectively; we get,

(x, y, z) = (−23s−

13 t, s, t) = (−2

3 , 1, 0)s+(− 13 , 0, 1)t. To avoid fractions, we take

as eigenvectors (−2, 3, 0) and (−1, 0, 3).

For the eigenvalue 3, the reduced system of equations is 2x + z = 0 and

−2y+z = 0; i.e., the intersection of two planes, a straight line that goes through

the origin. The parametric equation of that line is (x, y, z) = t (−1, 1, 2) with

t ∈ R and t = 0. Then any vector in this invariant subspace works fine as eigen-

vector, we get the vector (−1, 1, 2).

Resuming, the eigenvalues and the eigenvectors are

−1 ↔

−2

3

0

, −1 ↔

−1

0

3

, 3 ↔

−1

1

2

.


30 Linear Algebra

Finally the diagonal matrix is

−1 0 0

0 −1 0

0 0 3

and the similarity matrix is

−2 −1 −1

3 0 1

0 3 2

.

Example 4. In this example, we will try to find a diagonal representation

for the 2× 2 matrix, A =

(3 2

−2 −1

).

The characteristic polynomial is

det(λI−A) = (λ− 1)2

and so, we have just one eigenvalue, 1, but with multiplicity 2.

To find the corresponding eigenvectors we have to solve the equation(3 2

−2 −1

)(x

y

)=

(x

y

),

which after the normal procedures it is reduced to x = y. Thus, the invariant

subspace is a straight line that goes through the origin with an slope of 45

degrees. But we already know that there will be a diagonal representation, if

and only if, there are 2 linearly independent eigenvectors associated with our

eigenvalue 1; then matrix A does not have a diagonal representation.

1.4.3.1 Procedure to diagonalize a matrix

We present a summary of what we have studied until now about the method to

diagonalize a matrix.

1. Calculate all the eigenvalues of the matrix A.

2. Calculate all the corresponding eigenvectors.

3. Build the C matrix with the eigenvectors as columns.

4. Apply the similarity transformation C−1AC.

1.4.3.2 The Cayley-Hamilton theorem

Functions of linear transformations are of fundamental importance in theory and

in practice; specially to solve algebraic and differential systems of equations is

the exponential function of a linear operator. In the calculation of that kind

of functions, is central the Cayley-Hamilton theorem that we expose in what

follows.

Let T be a linear operator on a finite dimensional vector space V . If p is the


Linear operators acting on Euclidian spaces 31

characteristic polynomial for T , then p(T ) = 0.

In other words, every square matrix satisfies its own characteristic equation.

The Cayley-Hamilton theorem always provides a relationship between the pow-

ers of the matrix A, associated to the linear operator T , which allows one to

simplify expressions involving such powers, and evaluate them without having

to compute the power An or any higher powers of A.

Example 1. Given the 2 × 2 matrix A =

(−1 2

3 1

), we have, pA = λ2 − 7,

and then

A2 − 7I =

(−1 2

3 1

)2

− 7

(1 0

0 1

)= 0.

Hence, we get

A2 = 7I, A3 = A(7I) = 7A, A4 = AA3 = A× 7A = 7A2 = 49I

and so on. In fact, A2j = 7jI and A2j+1 = 7jA, for j = 0, 1, 2, ... .

If we want to calculate etA, we have

etA =

∞∑k=0

tk

k!Ak =

= I∞∑k=0

(√7t)2k

(2k)!+

A√7

∞∑k=0

(√7t)2k+1

(2k + 1)!= cosh(

√7t)I+

sinh(√7t)√

7A,

or explicitly

etA =

(cosh(

√7t)−

√77 sinh(

√7t) 2

√7

7 sinh(√7t)

3√7

7 sinh(√7t) cosh(

√7t) +

√77 sinh(

√7t)

).

1.5 Linear operators acting on Euclidian spaces

We already said that the words linear transformation, linear maps and linear

operators are synonymous; however, it is usual that when a linear transformation

acts from a vector space into itself, it is called linear operator, or just operator;

we will use that convention in all this section. Also in this section, E will denote

an Euclidian space over the complex numbers; i.e., a complex vector space with

a specific scalar product.


32 Linear Algebra

1.5.1 Adjoint operators

Definition of the adjoint operator. Let T be a linear operator on an Euclidian

space E . Then, we say that T has an adjoint on E if there exists a linear operator

T ′ on E such that (T (v), u) = (v, T ′(u)) for all v and u in E .

Let T be a linear operator on a finite dimensional Euclidian space E . Then

there exists a unique linear operator T ′ on E , such that for all v, u in E , we have

(T (v), u) = (v, T ′(u)). (1.36)

Then every linear operator on a finite-dimensional Euclidian space E has an ad-

joint. In the infinite-dimensional case this is not always true. When this linear

operator exists, we call it the adjoint of T , and we denote it T †.

Two comments should be made about the finite-dimensional case.

1. When it exists, the adjoint T † is unique.

2. The adjoint of T depends not only on T , but on the inner product as well.

The adjoint has the following properties.

1. (T + S)† = T † + S†.

2. (TS)† = S†T †.

3. (rT )† = r∗T † where r is an arbitrary scalar.

4. [(T )†]† = T .

Example 1. In the usual Euclidian space R2, consider the linear operator, T :

(x, y) → (2x−3y, 5x+y). The adjoint operator is T † : (x, y) → (2x+5y,−3x+y).

Given two arbitrary vectors, v1 = (x1, y1) and v2 = (x2, y2), in R2, we have that

(T (v1), v2) = ((2x1 − 3y1, 5x1 + y1), (x2, y2))

= (2x1 − 3y1)x2 + (5x1 + y1)y2 = 2x1x2 − 3y1x2 + 5x1y2 + y1y2

and

(v1, T†(v2)) = ((x1, y1), (2x2 + 5y2,−3x2 + y2))

= x1(2x2 + 5y2) + y1(−3x2 + y2) = 2x1x2 − 3y1x2 + 5x1y2 + y1y2.

Thus, (T (v1), v2) = (v1, T†(v2)), and effectively T † is the adjoint operator of T .

Let T be a linear operator on a finite dimensional Euclidian space E , and let

be A the matrix associated with it in a given basis. Then the matrix associated

with the adjoint linear operator is the adjoint matrix A†, i.e., the transpose and

conjugated matrix.



Example 2. Consider the linear operator of example 1 above. The matrix

associated with the linear operator T is

(2 −3

5 1

), and the matrix associated

with the adjoint linear operator T † is

(2 5

−3 1

).

1.5.2 Hermitian and anti-Hermitian operators

If T is a linear operator on E and (T (v), u) = (v, T (u)) for all v and all u in E ,the linear transformation is called Hermitian or self-adjoint; in other words, a

linear operator is Hermitian or self-adjoint if T † = T .

Hermitian transformations are very important, since they play for operators the

role that real numbers play for normal functions. In quantum mechanics, all

dynamical variables have associated an operator, and this operator must be a

Hermitian operator.

Example 1. In the Euclidian space, formed by the vector space

A = f : (a, b) → C| f is a continuos function in (a, b) and f(a) = f(b) = 0,

with the scalar product f · g =∫ b

af(x)g∗(x)dx, the operator p = −id/dx is a

Hermitian operator.

If we take any two functions, f and g in A, and we integrate by parts

(f, p(g)) = (f,−i dgdx

) =

∫ b

a

f(x)[−idg(x)dx

]∗dx =

∫ b

a

[−idf(x)dx

]g∗(x)dx = (p(f), g),

so the operator satisfies the Hermitian condition.

If T is a linear operator on E and (T (v), u) = −(v, T (u)) for all v and all u

en E , the linear transformation is called anti-Hermitian or anti-adjoint; in other

words, a linear operator is anti-Hermitian or anti-adjoint if T † = −T .Anti-Hermitian operators are equivalent to the pure imaginary numbers.

In the case of real Euclidian spaces, the Hermitian and anti-Hermitian linear

operators area called symmetric and anti-symmetric linear operators, respec-

tively.

If T is a linear operator on E and e1, e2, ..., en is a basis of E , thena) The linear operator T is Hermitian, if and only if, (T (ej), ei) = (ej , T (ei)) for

all i and all j from 1 to n.


34 Linear Algebra

b) The linear operator T is anti-Hermitian, if and only if, (T (ej), ei) = −(ej , T (ei))

for all i and all j from 1 to n.

These properties follow directly from the linearity of the operator T and from

the fact that in E all vectors can be written as linear combinations of the basis

vectors.

If T is a linear operator on E , e1, e2, ..., en is an orthonormal basis of E ,and A = (aij) is the matrix associated with the linear transformation T in the

given basis, then

a) The linear operator T is Hermitian, if and only if, the matrix A is self-adjoint

or Hermitian; i.e., A† = A.

b) The linear operator T is anti-Hermitian, if and only if, the matrix A is skew-

adjoint or anti-Hermitian; i.e., A† = −A.

1.5.3 Properties of the eigenvalues and eigenvectors of the Her-

mitian operators

If T is a linear operator on the Euclidian space E , λ is one of its eigenvalues and

v is the corresponding eigenvector, then

λ =(T (v), v)

(v, v). (1.37)

As v is an eigenvector of the linear operator T with eigenvalue λ, we have that

Tv = λv, then, from the properties of the scalar product, (T (v), v) = (λv, v) =

λ(v, v), and expression (1.37) follows immediately.

Also

λ∗ =(v, T (v))

(v, v). (1.38)

It is clear that an eigenvalue is real, if and only if,

(T (v), v) = (v, T (v)) (1.39)

for all vϵE .This condition is trivially satisfied in a real Euclidian space.

In case that

(T (v), v) = −(v, T (v)) (1.40)



for all vϵE , the corresponding eigenvalue is pure imaginary.

Summarizing, if T is a linear operator on E and λ is an eigenvalue of it, then

a) If the linear operator T is Hermitian, λ is real.

b) If the linear operator T is anti-Hermitian, λ is pure imaginary.

If T is a Hermitian linear operator on the Euclidian space E and λ and µ are

different eigenvalues with v and u the corresponding eigenvectors, then v and u

are orthogonal; i.e., v · u = 0.

As v and u are eigenvalues of T , we have that (T (v), u) = (λv, u) = λ(v, u)

and also that (v, T (u)) = (v, µu) = µ∗(v, u) then as T is Hermitian, (T (v), u) =

(v, T (u)) and λ(v, u) = µ∗(v, u) or (λ − µ∗)(v, u) = 0. By hypothesis, all the

eigenvalues are different, then necessarily (v, u) = 0.

Exactly the same thing happens in the case of the anti-Hermitian transforma-

tions; or in other words, the eigenvectors corresponding to different eigenvalues

are orthogonal.

If T is a Hermitian linear operator on E and dimE = n, then there are n

eigenvectors v1, v2, ..., vn of T that form an orthonormal basis of E .

The matrix corresponding to T in the basis v1, v2, ..., vn is a diagonal matrix,

Λ = diag(λ1, λ2, ..., λn), where λk is the eigenvalue corresponding to the eigen-

vector vk, for k from 1 to n.

If the transformation is anti-Hermitian a similar thing happens; i.e., also

there are n eigenvectors of T that form an orthonormal basis of E .

Any square matrix A = (aij), Hermitian or anti-Hermitian, is similar to the

diagonal matrix Λ = diag(λ1, λ2, ..., λn) of the eigenvalues. Exists then, a matrix

P, such that Λ = P−1AP.

The similarity matrixP is the eigenvectors matrix, i.e., the matrix whose columns

are the eigenvectors. The similarity matrix P is non-singular and unitary; i.e.,

P−1 = P†.


36 Linear Algebra


Chapter 2

Special functions

In this Chapter we will revise briefly a few properties of some special functions

that will be used to solve later some differential equations.

2.1 Hermite polynomials

The generating function for the Hermite polynomials is

e−α2+2αx =∞∑

n=0

Hn(x)αn

n!. (2.1)

The Hermite polynomials may be obtained from Rodrigues’ formula

Hn(x) = (−1)nex2 dn

dxne−x2

, (2.2)

the first ones are

H0(x) = 1,

H1(x) = 2x,

H2(x) = 4x2 − 2,

H3(x) = 8x3 − 12x,

H4(x) = 16x4 − 48x2 + 12; (2.3)

some of them are shown in Figure 2.1.

From the recurrence relations

Hn+1(x) = 2xHn(x)− 2nHn−1(x), (2.4)

37


38 Special functions

Fig. 2.1 Some Hermite polynomials.

and

dHn(x)

dx= 2nHn−1(x), (2.5)

we can generate all the Hermite polynomials.

From the above recurrence relations, we can also prove that, if we define the

functions

ψn(x) =π−1/4

√2nn!

e−x2/2Hn(x), (2.6)

then ([Arfken 05])

A†ψn(x) ≡1√2

(x− d

dx

)ψn(x) =

√n+ 1ψn+1(x) (2.7)

and

Aψn(x) ≡1√2

(x+

d

dx

)ψn(x) =

√nψn−1(x). (2.8)

The functions (2.6) constitute a complete orthonormal set for the space of square


Hermite polynomials 39

integrable functions, then we can expand any function in that space as

f(x) =

∞∑n=0

cnψn(x), (2.9)

where

cn =

∫ ∞

−∞dxf(x)ψn(x). (2.10)

Hermite polynomials are also solutions of the second order ordinary differ-

ential equation

y′′ − 2xy′ + 2ny = 0. (2.11)

Let us define the differential operator p as

p = −i ddx, (2.12)

then we have that

dn

dxn=(−pi

)n,

and we can rewrite (2.2) in the form

Hn(x) = (−i)nex2

pne−x2

= (−i)n(ex

2

pe−x2)n

. (2.13)

The operator inside the parenthesis above has the form

eξABe−ξA, (2.14)

that will appear frequently when we solve differential equations in the following

Chapters.

We can obtain an expression for this type of operators by developing the expo-

nentials in a Taylor series , to obtain ([Arfken 05; Orszag 08])

Hadamard lemma: Given two linear operators A and B then

eξABe−ξA = B + ξAB − ξBA+ξ2

2!A2B − ξ2ABA+

ξ2

2!BA2 + · · ·

= B + ξ [A,B] +ξ2

2![A, [A,B]] +

ξ3

3![A, [A, [A,B]]] + · · · ,

(2.15)

where [A,B] ≡ AB−BA is the commutator of operators A and

B.



We are now in the position to obtain an expression for

Hn(x) = (−i)n(ex

2

pe−x2)n

,

formula (2.13) developed above.

We identify

ξ = 1, A = x2, B = p,

in equation (2.15), and we get,

ex2

pe−x2

= p+ 1[x2, p

]+

1

2!

[x2,[x2, p

]]+

1

3!

[x2,[x2,[x2, p

]]]+ · · · (2.16)

To calculate the first commutator[x2, p

], we use the general property

[AB,C] = A[B,C] + [A,C]B (2.17)

of commutators, and that

[x, p]f(x) ≡ −i(x d

dx− d

dxx)f(x) = −xf ′(x) + xf ′(x) + if(x) = if(x);

i.e. [x, p] = i, to get the commutation relation[x2, p

]= 2ix.

It is obvious that all the other commutators in (2.16) are zero, and we finally

get (see [Arfken 05], problem 13.1.5)

Hn(x) = (−i)n (p+ 2ix)n1. (2.18)

This last expression can be used to obtain the generating function. We have,

∞∑n=0

Hn(x)αn

n!=

∞∑n=0

αn

n!(−i)n (p+ 2ix)

n1 = e−iα(p+2ix)1. (2.19)

We obtained the exponential of the sum of two quantities that do not commute.

The above exponential can be factorized in the product of exponentials via the[Louisell 90]:

2.1.1 Baker-Hausdorff formula

Given two operators A and B that obey

[[A,B] , A] = [[A,B] , B] = 0, (2.20)



then

eA+B = e−12 [A,B]eAeB . (2.21)

Demonstration: We define

F (λ) = eλ(A+B) = eg(λ)[A,B ]eλAeλB . (2.22)

By deriving the second part of the above equation with respect to λ, we have

dF (λ)

dλ= (A+B) eλ(A+B) = (A+B)F (λ), (2.23)

and deriving the last part of (2.22) with respect to λ, we have

dF (λ)

dλ= g′(λ) [A,B]F (λ) + eg(λ)[A,B]AeλAeλB + eg(λ)[A,B]eλABeλB .

Using equation (2.15) and the hypothesis of the problem, [[A,B] , A] = [[A,B] , B] =

0, it is very easy to show that

eg(λ)[A,B]A = Aeg(λ)[A,B]

and that

eλABe−λA = B + λ [A,B] ,

so

dF (λ)

dλ= g′(λ) [A,B] +A+B + λ [A,B]F (λ). (2.24)

By comparing (2.23) and (2.24), we obtain the following differential equation

[g′(λ) + λ] [A,B] = 0,

with solution

g(λ) = −λ2

2,

where the initial condition g (λ = 0) = 0 has been used, because F (λ = 0) = 1.

Now we evaluate in λ = 1, and we get the Baker-Hausdorff formula

eA+B = e−[A,B]

2 eAeB .



We are now ready to apply the Baker-Hausdorff formula, expression (2.21),

to the formula

∞∑n=0

Hn(x)αn

n!= e−iα(p+2ix)1,

derived above (equation(2.19)), and obtain the generating function. We identify

A = 2αx,

B = −iαp,

and then

[A,B] = −2iα2 [x, p] = 2α2,

such that

∞∑n=0

Hn(x)αn

n!= e−α2

e2αxe−iαp1.

Using now the obvious fact that e−iαp1 = 1, we get

∞∑n=0

Hn(x)αn

n!= e−α2+2αx, (2.25)

that is the generating function for Hermite polynomials.

The Hermite polynomials also can be calculated as the determinant of the

matrix

Ai,j =

2x if i = j√2i if j = i+ 1√

2(i− 1) if j = i− 1

0 otherwise;

(2.26)

or in other words



Hn(x) =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

2x√2 0 0 0 · · · · · · 0

√2 2x

√4 0 0 · · · · · · 0

0√4 2x

√6 0 · · · · · · 0

0 0√6 2x

√8 0 · · · 0

0 0 0√8 2x

√10 · · · 0

· · · · · · · · · · · ·√10 2x

. . . · · ·

· · · · · · · · · · · · · · ·. . .

. . .√

2(n− 1)

0 0 0 0 0 · · ·√2(n− 1) 2x

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

.

(2.27)

2.1.2 Series of even Hermite polynomials

In order to show the power of the operator methods, we calculate now the value

of the following even Hermite polynomials series,

F (t) =∞∑

n=0

tn

n!H2n (x) . (2.28)

From (2.18), we get

H2n (x) = (−1)n(p+ 2ix)

2n1.

Therefore,

F (t) =

∞∑n=0

tn

n!H2n (x) =

∞∑n=0

tn

n!(−1)

n(p+ 2ix)

2n1

=

∞∑n=0

tn

n!(−1)

n[(p+ 2ix)

2]n

1 = exp[−t (p+ 2ix)

2]1.

Expanding the square in the exponential, we get

F (t) =∞∑

n=0

tn

n!H2n (x) = exp

−t[p2 − 4x2 + 2i (xp+ px)

]1. (2.29)



The operators in the exponential in this last expression do not satisfy the con-

ditions of the Baker-Hausdorff formula, equation (2.21); so we need another

method to understand the action of the full operator that appears in the right

side of expression (2.29). What we do is propose the ansatz (it is a German

word, and in physics and mathematics, it is an educated guess that is verified

later by its results),

F (t) = exp[f (t)x2

]exp [g (t) (xp+ px)] exp

[h (t) p2

]1, (2.30)

where f(t), g(t) and h(t) are functions that we have to determine. Deriving this

expression with respect to t, and dropping the explicit dependence of f(t), g(t)

and h(t) on t,

dF (t)

dt=

=df

dtx2F (t) +

dg

dtexp

(fx2

)(xp+ px) exp [g (xp+ px)] exp

(hp2)1

+dh

dtexp

(fx2

)exp [g (xp+ px)] p2 exp

(hp2)1.

Introducing an ”smart” 1 in the second and third term, we get

dF (t)

dt=df

dtx2F (t) +

dg

dtefx

2

(xp+ px) e−fx2

F (t)

+dh

dtefx

2

exp [g (xp+ px)] p2 exp [−g (xp+ px)] e−fx2

F (t).

(2.31)

We work then with the operator in the second term; we have to use the very

useful expression, already introduced (expression 2.15 on page 39),

eξABe−ξA = B + ξ [A,B] +ξ2

2![A, [A,B]] +

ξ3

3![A, [A, [A,B]]] + · · · ,

to get

exp(fx2

)(xp+ px) exp

(−fx2

)= xp+ px+ f

[x2, xp+ px

]+f2

2!

[x2,[x2, xp+ px

]]+f3

3!

[x2,[x2,[x2, xp+ px

]]]+ · · · .

(2.32)

The first commutator that appears in the above expression is easily calculated,

[x2, xp+ px] = 4ix2,

and so all the others commutators are zero. Substituting back in (2.32), we get

exp(fx2

)(xp+ px) exp

(−fx2

)= xp+ px+ 4ifx2.



We analyze now the third term in expression (2.31), i.e.

exp(−fx2) exp [g (xp+ px)] p2 exp [−g (xp+ px)] exp(−fx2).

We study first only a part of it, exp [g (xp+ px)] p2 exp [−g (xp+ px)]. Using

again (2.15),

exp [g (xp+ px)] p2 exp [−g (xp+ px)] =

= p2 + g[xp+ px, p2

]+g2

2!

[xp+ px,

[xp+ px, p2

]]+g3

3!

[xp+ px,

[xp+ px,

[xp+ px, p2

]]]+ · · · .

Calculating the first commutators,[xp+ px, p2

]= 4ip2,

[xp+ px,

[xp+ px, p2

]]= −16p2,

[xp+ px,

[xp+ px,

[xp+ px, p2

]]]= −64ip2,

and so on. It is clear that

exp [g (xp+ px)] p2 exp [−g (xp+ px)] = p2∞∑j=0

(4i)j g

j

j!= p2 exp (4ig) .

We proceed now to complete the study of the third term in expression (2.31).

Until now we have

exp(−fx2) exp [g (xp+ px)] p2 exp [−g (xp+ px)] exp(−fx2) == exp(−fx2)p2 exp (4ig) exp(−fx2).

We use once more formula (2.15), to write

exp(fx2

)p2 exp

(−fx2

)=

= p2 + f[x2, p2

]+f2

2!

[x2,[x2, p2

]]+f3

3!

[x2,[x2,[x2, p2

]]]+ · · ·

The first commutator gives [x2, p2

]= 2i(xp+ px),

the second one gives [x2,[x2, p2

]]= −8x2,



and the third one [x2,[x2,[x2, p2

]]]= 0;

such that all the other commutators are zero, and

exp(fx2

)p2 exp

(−fx2

)=

= p2 + 2if (xp+ px) +f2

2!

(−8x2

)= p2 + 2if (xp+ px)− 4f2x2.

Finally, we can write a reduced expression for the derivative of the original series

F (t),

dF (t)

dt=dfdtx2 +

dg

dt

(xp+ px+ 4ifx2

)+

+ exp (4ig)dh

dt

[p2 + 2if (xp+ px)− 4f2x2

] F (t)

and rearranging terms

dF (t)

dt=

=[df

dt+ 4if

dg

dt− 4f2 exp (4ig)

dh

dt

]x2+[

dg

dt+ 2if exp (4ig)

dh

dt

](xp+ px) + exp (4ig)

dh

dtp2F (t).

We get back to the original expression for the series

F (t) = exp−t[p2 − 4x2 + 2i (xp+ px)

]1

and take the derivative with respect to t,

dF (t)

dt=

= −[p2 − 4x2 + 2i (xp+ px)

]exp

−t[p2 − 4x2 + 2i (xp+ px)

]1

=[−p2 + 4x2 − 2i (xp+ px)

]F.

Comparing now both expressions, we get the system of differential equations

df

dt+ 4if

dg

dt− 4f2 exp (4ig)

dh

dt= 4,

dg

dt+ 2if exp (4ig)

dh

dt= −2i, (2.33)

exp (4ig)dh

dt= −1.



The initial conditions that we must set on these equations can be easily under-

stood from equation (2.29) and from the ansatz (2.30); as for t = 0 the operator

in the right side of (2.29) is the identity, we must impose f(0) = g(0) = h(0) = 0.

We outline now the procedure to solve the system (2.33). In the last equation

the value of dhdt is found, and substituted in the second equation, where the value

of dgdt is also found, and then both values are substituted in the first equation.

The differential equation so obtained is solved for f , with the initial condition

f(0) = 0, substituting it in the second equation the value of function g is ob-

tained, and finally both values are substituted in the third one, and the value of

h obtained. The result that we get is

f =4t

4t+ 1,

g = − i

2ln (4t+ 1) ,

h = − t

4t+ 1.

We calculate now explicitly

F (t) = exp(fx2

)exp [g (xp+ px)] exp

(hp2)1.

Remembering the definition of p (expression 2.12) it is very easy to see that

exp(hp2)1 = 1

and then

F (t) = exp(fx2

)exp [g (xp+ px)] 1.

Using now the [x, p] = i, we write

exp [g (xp+ px)] 1 = exp [g (2xp− i)] 1 = exp (−ig) exp (2gxp) 1

and it is also clear that

exp (2gxp) 1 =

[1 + gxp+

g2

2(xp)

2+ ...

]1 =

= 1 + gxp1 +g2

2(xp)

21 + ... = 1

and also that

exp [g (xp+ px)] 1 = exp (−ig) .



We have then

F (t) = exp(fx2 − ig

).

Substituting the functions f and g,

F (t) = exp

[(4t

4t+ 1

)x2 − 1

2[ln (4t+ 1)]

]=

1√4t+ 1

exp

(4tx2

4t+ 1

),

and finally we get the formula we were looking for

∞∑n=0

tn

n!H2n (x) =

1√4t+ 1

exp

(4tx2

4t+ 1

). (2.34)

It is interesting to look for some special values. For example, for t = 0, it is

obvious that F (t) = 1, as correctly predicts formula (2.34).

As H2n(0) = (−1)n (2n)!n!

[Arfken 05], we have that for t = 1 and x = 0, formula

(2.34) give us

∞∑n=0

(−1)n(2n)!

n!2=

√5

5. (2.35)

2.1.3 Addition formula

We want to apply the form obtained in (2.18), Hn(x) = (−i)n (p+ 2ix)n1, to

evaluate the quantity

Hn(x+ y).

We write it as

Hn(x+ y) = (−i)n[−i d

d(x+ y)+ 2i(x+ y)]n, (2.36)

by using the chain rule we have

d

d(x+ y)=

1

2

(∂

∂x+

∂

∂y

),

so that we may re-express (2.36) in the form

Hn(x+ y) =

(−i√2

)n

(−i ∂

∂√2x

+ 2i√2x− i

∂

∂√2y

+ 2i√2y)n.


Associated Laguerre polynomials 49

By defining

pX = −i ∂∂X

, pY = −i ∂∂Y

with X =√2x and Y =

√2y, we obtain

Hn(x+ y) =1

2n/2

n∑k=0

(n

k

)(−i)k(pX + 2iX)k(−i)n−k(pY + 2iY )n−k,

that by using Hj(x) = (−i)n(p+ 2ix)j1 adds to

Hn(x+ y) =1

2n/2

n∑k=0

(n

k

)Hk(

√2x)Hn−k(

√2y), (2.37)

which is the addition formula we were looking for.

2.2 Associated Laguerre polynomials

The generating function for the associated Laguerre polynomials is [Arfken 05]

∞∑n=0

Lαn(x)t

n =1

(1− t)α+1 exp

(−xt1− t

), |t| < 1. (2.38)

The associated Laguerre polynomials may be obtained from the correspond-

ing Rodrigues’ formula

Lαn (x) =

1

n!x−αex

dn

dxn(e−xxn+α

), (2.39)

being the first ones,

Lα0 (x) = 1,

Lα1 (x) = (1 + α)− x,

Lα2 (x) =

1

2

(2 + 3α+ α2

)− (2 + α)x+

x2

2,

Lα3 (x) =

(1 +

11α

6+ α2 +

α3

6

)−(3 +

5α

2+α2

2

)x+

1

2(3 + α)x2 − x3

6.

(2.40)

The associated Laguerre polynomials satisfy several recurrence relations.

One very useful, when extracting properties of the wave functions of the hy-

drogen atom, is

(n+ 1)Lαn+1(x) = (2n+ α+ 1− x)Lα

n(x)− (n+ α)Lαn−1(x). (2.41)



We will use the operator method outlined above for the Hermite polynomials,

to derive the usual explicit expression for the associated Laguerre polynomials.

We rewrite expression (2.39) as

Lαn (x) =

1

n!x−αex (ip)

ne−xxn+α,

where again the operator p = −id/dx, defined in (2.12), was used.

We notice that ex (ip)ne−x = [ex (ip) e−x]n and that, using (2.15), expe−x =

(p+ i), so

Lαn(x) =

in

n!x−α (p+ i)

nxn+α,

or writing explicitly the operator p,

Lαn(x) =

1

n!x−α

(d

dx− 1

)n

xn+α. (2.42)

Using the binomial expansion,

Lαn(x) =

1

n!x−α

n∑m=0

(n

m

)(−1)n−m dm

dxmxn+α,

and because

dm

dxmxn+α =

(n+ α)!

(n+ α−m)!xn+α−m,

we obtain the usual form for associated Laguerre polynomials,

Lαn(x) =

n∑k=0

(n+ α

n− k

)(−1)k

xk

k!. (2.43)

The second order ordinary differential equation, from which the associated

Laguerre polynomials are solution, is

xy′′ + (α+ 1− x)y′ + ny = 0. (2.44)

The associated Laguerre polynomials are orthogonal in the interval [0,∞)

with the weight function xαexp(−x); i.e.∫ ∞

0

dx xαexp(−x)Lm(x)Ln(x) =(n+ k)!

n!δmn, (2.45)


Associated Laguerre polynomials 51

Fig. 2.2 Some Laguerre polynomials.

where δmn is the Kroenecker delta.

In the special case when α = 0, we get the Laguerre polynomials. The La-

guerre polynomials are denoted without the super index α.

The graphics of some of these first Laguerre polynomials are presented in Figure

2.2.

From the recurrence relation (2.41), it is possible to prove that the Laguerre

polynomials can be calculated as the determinant of the matrix

Ai,j =

1 + 2(i− 1)− x if i = j

i if j = i+ 1 and j = i− 1

0 otherwise;

(2.46)



or in other words

Ln(x) =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1− x 1 0 · · · · · · 0

1 3− x 2 0 · · · 0

0 2 5− x 3 · · · 0

0 0 3 7− x · · · · · ·

· · · · · · · · · · · ·. . . n− 1

0 0 0 · · · n− 1 1 + 2(n− 1)− x

∣∣∣∣∣∣∣∣∣∣∣∣∣∣. (2.47)

2.3 Chebyshev polynomials

Again, we will give here the main features of Chebyshev polynomials. We first

look at the polynomials of the first kind.

2.3.1 Chebyshev polynomials of the first kind

The generating function of the Chebyshev polynomials of the first kind, Tn(x),

is given by

g(t, x) =1− xt

1− 2xt+ t2=

∞∑n=0

Tn(x)tn, (2.48)

for |x| < 1 and |t| < 1.

They may be written in several forms, but one convenient for later purposes,

is

Tn(x) =

[n/2]∑m=0

(n

2m

)xn−2m(x2 − 1)m, (2.49)

where [n/2] is the so-called floor function, also called the greatest integer function

or integer value, and gives the largest integer less than or equal to n/2.

Another form to write them, that allows for easy calculation of their roots,

is

Tn(x) = 2n−1n∏

k=1

x− cos

[(2k − 1)π

2n

]. (2.50)

A few Chebyshev polynomials of the first kind are

T0(x) = 1,

T1(x) = x,

T2(x) = 2x2 − 1,

(2.51)


Chebyshev polynomials 53

Fig. 2.3 Some Chebyshev polynomials of the first kind.

and with the recurrence relations

Tn+1(x) = 2xTn(x)− Tn−1(x), (2.52)

we can find the rest.

In Figure 2.3, we plot some of the first Chebyshev polynomials of the first kind.

2.3.2 Chebyshev polynomials of the second kind

The generating function for the Chebyshev polynomials of the second kind,

Un(x), is given by

g(t, x) =1

1− 2xt+ t2=

∞∑n=0

Un(x)tn, (2.53)

for |x| < 1 and |t| < 1.

A few Chebyshev polynomials of the second kind are

U0(x) = 1,

U1(x) = 2x,

U2(x) = 4x2 − 1;

(2.54)



Fig. 2.4 Some Chebyshev polynomials of the second kind.

and with the recurrence relations

Un+1(x) = 2xUn(x)− Un−1(x), (2.55)

we can find the rest.

In figure 2.4, we plot some Chebyshev polynomials of the second kind.

As with the Chebyshev polynomials of the first kind, we can write them as

the sum

Un(x) =

[n/2]∑m=0

(n+ 1

2m+ 1

)xn−2m(x2 − 1)m, (2.56)

and the form that allows easy calculation of the roots is

Un(x) = 2nn∏

k=1

x− cos

[kπ

n+ 1

]. (2.57)

It also may be written as

Un(x) =sin[(n+ 1) cos−1 x]

sin[cos−1 x]. (2.58)


Bessel functions of the first kind of integer order 55

Both expressions above show that the roots of the Chebyshev polynomials of the

second kind are

xk = cos(k

n+ 1π). (2.59)

A nice way to obtain the Chebyshev polynomials of the second kind is via

the determinant

Un(x) =

∣∣∣∣∣∣∣∣∣∣∣∣∣

2x 1 0 . . . 0 0

1 2x 1 0 . . . 0

0 1 2x 1 . . . 0

. . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . .

0 0 0 . . . 1 2x

∣∣∣∣∣∣∣∣∣∣∣∣∣. (2.60)

2.4 Bessel functions of the first kind of integer order

Bessel functions of the first kind of integer order, Jn(x), are solutions of the

Bessel differential equation

x2y′′ + xy′ + (x2 − n2)y = 0, (2.61)

where n is an integer. They may be obtained from the generating function

exp

[x

2

(t− 1

t

)]=

∞∑n=−∞

tnJn(x), (2.62)

and also from the following recurrence relations

2n

xJn(x) = Jn−1(x) + Jn+1(x). (2.63)

In Figure 2.5, we plot some of the first Bessel functions of the first kind of

integer order.

Bessel functions of the first kind of integer order may be written as

Jn(x) =∞∑

m=0

(−1)mx2m+n

22m+nm!(m+ n)!, (2.64)

and also the following integral representation is very useful

Jn (x) =1

2π

∫ π

−π

e−i(nτ−x sin τ)dτ. (2.65)



Fig. 2.5 First four Bessel functions of the first kind of integer order.

Another important relations for the Bessel functions of the first kind are the

Jacobi-Anger expansions:

eix cos y =∞∑

n=−∞inJn(x)e

iny (2.66)

and

eix sin y =∞∑

n=−∞Jn(x)e

iny. (2.67)

2.4.1 Addition formula

Using the operator methods developed in previous sections, we will obtain here

the addition formula for the Bessel functions of the first kind of integer order.

First, we will derive the following expression for any ”well behaved” function f ,

f(x+ y) = eiypxf(x)e−iypx1, (2.68)

where px = −id/dx is the operator introduced in Section 1, expression (2.12).

Because e−iypx1 = 1, developing in a Taylor series the f function (we call cn to



the coefficients in the expansion) and using the linearity of the eiypx operator,

eiypxf(x)e−iypx1 = eiypxf(x) = eiypx

∞∑k=0

ckxk =

∞∑k=0

ckeiypxxk.

Now

eiypxxk =

∞∑l=0

(iy)l

l!(−i)l d

l

dxlxk =

k∑l=0

(y)l(k

l

)xk−l = (x+ y)k,

then

eiypxf(x)e−iypx1 =∞∑k=0

ck(x+ y)k = f(x+ y), (2.69)

as we wanted to prove.

Now consider the Bessel function Jn evaluated at x+ y. From expression (2.68)

we have

Jn(x+ y) = eiypxJn(x)e−iypx1,

because e−iypx1 = 1, and developing the first exponential in Taylor series, we

obtain

Jn(x+ y) =∞∑

m=0

ym

m!

dm

dxmJn(x). (2.70)

To calculate the m-derivative of Jn, we use the integral representation (2.65) to

write

dm

dxmJn(x) = im

1

2π

∫ π

−π

sinm τe−i(nτ−x sin τ)dτ,

substituting sin τ = (eiτ − e−iτ )/2i, and using the binomial expansion,

dm

dxmJn(x) =

1

2m1

2π

m∑k=0

(−1)k(m

k

)∫ π

−π

ei(m−k)τe−ikτe−i(nτ−x sin τ)dτ

=1

2m1

2π

m∑k=0

(−1)k(m

k

)∫ π

−π

e−i[(n−m+2k)τ−x sin τ ]dτ,

and therefore, using again the integral representation (2.65), we obtain

dm

dxmJn(x) =

1

2m

m∑k=0

(−1)k(m

k

)Jn−m+2k(x). (2.71)



Substituting this last expression in equation (2.70), we obtain (we have taken

the sum up to infinite as we add only zeros)

Jn(x+ y) =∞∑

m=0

ym

m!

1

2m

∞∑k=0

(−1)k(m

k

)Jn−m+2k(x).

We now change the order of summation and start the second sum at m = k

(because for m < k all the terms are zero)

Jn(x+ y) =∞∑k=0

(−1)k

k!

∞∑m=k

ym

2m(m− k)!Jn−m+2k(x).

We do now j = m− 2k and obtain

Jn(x+ y) =∞∑k=0

(−1)k

k!

∞∑j=−k

yj+2k

2j+2k(j + k)!Jn−j(x),

take the second sum from minus infinite, and exchange the order of the sums

Jn(x+ y) =∞∑

j=−∞Jn−j(x)

∞∑k=0

(−1)k

k!

yj+2k

2j+2k(m+ k)!=

∞∑j=−∞

Jn−j(x)Jj(y).

The final expression

Jn(x+ y) =

∞∑k=−∞

Jn−k(x)Jk(y) (2.72)

is known as the addition formula for the Bessel functions of the first kind of

integer order.

2.4.2 Series of the Bessel functions of the first kind of integer

order

We will derive in this section the solution of some sums of the Bessel functions

of the first kind of integer order that appear in several applications. We will

demonstrate that

∞∑k=1

k2νJ2k (x) =

(−1)ν

4π

π∫−π

B2ν

(g′ (y) , g′′ (y) , ..., g(2ν) (y)

)dy, (2.73)

where ν is a positive integer, g (y) = ix sin y and Bn (x1, x2, ..., xn) is the com-

plete Bell polynomial [Bell 27; Boyadzhiev 09; Comtet 74] given by the following

determinant:



Bn(x1, x2, ..., xn) =

=

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

x1(n−11

)x2

(n−12

)x3

(n−13

)x4

(n−14

)x5 · · · · · · xn

−1 x1(n−21

)x2

(n−22

)x3

(n−23

)x4 · · · · · · xn−1

0 −1 x1(n−31

)x2

(n−32

)x3 · · · · · · xn−2

0 0 −1 x1(n−41

)x2 · · · · · · xn−3

0 0 0 −1 x1 · · · · · · xn−4

0 0 0 0 −1 · · · · · · xn−5

......

......

.... . .

. . ....

0 0 0 0 0 · · · −1 x1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

.

(2.74)

To demonstrate (2.73), we take the Jacobi-Anger expansion, expression (2.66),

to write

dn

dyneix sin y = in

∞∑k=−∞

knJk (x) eiky.

To calculate the n-derivative in the left side of equation above, we use the Faa

di Bruno’s formula ([Gradshtein 99], page 22) for the n-derivative of the com-

position

dn

dxnf (g (x)) =

n∑k=0

f (k) (g (x)) ·Bn,k

(g′ (x) , g′′ (x) , ..., g(n−k+1) (x)

), (2.75)

where Bn,k (x1, x2, ..., xn−k+1) is a Bell polynomial [Bell 27; Boyadzhiev 09;

Comtet 74], given by

Bn,k (x1, x2, ..., xn−k+1) =∑ n!

j1!j2!...jn−k+1!

(x11!

)j1 (x22!

)j2...

(xn−k+1

(n− k + 1)!

)jn−k+1

,(2.76)

the sum extending over all sequences j1, j2, j3, ..., jn−k+1 of non-negative integers

such that j1+j2+...+jn−k+1 = k and j1+2j2+3j3+...+(n− k + 1) jn−k+1 = n.



Using (2.75),

dn

dyneix sin y = eix sin y

n∑k=0

Bn,k

(g′ (x) , g′′ (x) , ..., g(n−k+1) (x)

).

We multiply now for the complex conjugate of (2.67) and obtain

in∞∑

k,l=−∞

knJl (x) Jk (x) ei(k−l)y = Bn

(g′ (y) , g′′ (y) , ..., g(n) (y)

).

Integrating both sides of the above equation from −π to π, and using that∫ π

−πei(k−l)ydy = δkl, we arrive to the formula we wanted

∞∑k=1

k2νJ2k (x) =

(−1)ν

4π

π∫−π

B2ν

(g′ (y) , g′′ (y) , ..., g(2ν) (y)

)dy. (2.77)

In particular, as the complete Bell polynomials for n = 2 and n = 4, are

B2(x1, x2) = x21 + x2 (2.78)

and

B4(x1, x2, x3, x4) = x41 + 6x21x2 + 4x1x3 + 3x22 + x4, (2.79)

it is very easy to show that

∞∑k=1

k2J2k (x) =

1

4x2 (2.80)

and

∞∑k=1

k4J2k (x) =

3

16x4 +

1

4x2. (2.81)

2.4.3 Relation between the Bessel functions of the first kind of

integer order and the Chebyshev polynomials of the second

kind

We finally present here, without demonstration, a useful formula that relates

the Fourier transform of the Bessel functions of the first kind of integer order



with the Chebyshev polynomials of the second kind [Campbell 48]:

FJn(ω)

ω

=

√2

π

i

n(−i)nUn−1(ξ)

√1− ξ2 rect

(ξ2

), (2.82)

where rect is the indicator function of the interval [−1, 1]; i.e.

rect(ξ

2) =

1, if xϵ[−1, 1]

0, otherwise.(2.83)


Chapter 3

Finite systems of differential equations

We will study in this Chapter finite systems of coupled first-order ordinary dif-

ferential equations. For pedagogical reasons, we start with systems 2× 2, after

we study systems 4× 4 that has in some sense the essence of the n× n system

without the cumbersome, and with the obtained experience, we generalize to

arbitrary finite dimension.

3.1 Systems 2 × 2 first

We consider initially a system of two coupled differential equations with constant

coefficients of the form

dx1dt

= a11x1 + a12x2,

dx2dt

= a21x1 + a22x2; (3.1)

that may be re-written in compact form as

dx

dt= Ax, (3.2)

where A is the matrix

A =

(a11 a12a21 a22

)and x is the column vector

x =

(x1x2

).

63


64 Finite systems of differential equations

The formal solution to the differential equations system (3.2) is

x = eAtx(0), (3.3)

with x(0) the initial conditions.

Indeed, by substituting (3.3) into (3.2), we obtain that

dx

dt= AeAtx(0), (3.4)

so that we recover (3.2).

3.1.1 Eigenvalue equations

Suppose now that the 2× 2 matrix A has two different eigenvalues, λ1 and λ2,

with the two column eigenvectors, y1 and y2, respectively.

We already know from Chapter 1, that the matrix T = (y1, y2), whose columns

are the eigenvectors, is the similitude matrix that transforms the matrix A into

its diagonal form D; i.e.,

D = T−1AT. (3.5)

We can invert this expression, and write the matrix A in terms of the similitude

matrix T, and the diagonal matrix D; we get,

A = TDT−1. (3.6)

In the solution (3.3), we have the exponential of the matrix A. As we mentioned

in Chapter 1, this exponential is defined in terms of the Taylor series of the

exponential, i.e.,

eAt =

∞∑n=0

tn

n!An. (3.7)

If we substitute (3.6) in the right side of this definition, we obtain

eAt =∞∑

n=0

tn

n!(TDT−1)n,

but as (TDT−1)n = TDT−1TDT−1...TDT−1 = TDnT−1, it may be rewritten

as

eAt = T

( ∞∑n=0

tn

n!Dn

)T−1,


Systems 2× 2 first 65

or better yet, as

eAt = TeDtT−1. (3.8)

As the D matrix is the diagonal matrix

D =

(λ1 0

0 λ2

),

it is very easy to show that

Dn =

(λn1 0

0 λn2

),

and that

etD =

(eλ1t 0

0 eλ2t

).

So, finally

eAt = T

(eλ1t 0

0 eλ2t

)T−1. (3.9)

At this point it is convenient to explicitly calculate the eigenvalues of the

matrix A. The characteristic equation is

| λI−A |= 0,

where I is the identity matrix. In other words, the characteristic equation is∣∣∣∣( λ− a11 a12a21 λ− a22

)∣∣∣∣ = 0,

that it is reduced to

λ2 − tr(A)λ+ |A| = 0,

where tr(A) means the trace of the A matrix and |A| its determinant.

Then the eigenvalues are given by

λ1,2 =tr(A)±

√tr2(A)− 4|A|2

and then we can find the exponential (3.9), and therefore, the solution to the

system (3.1).



Example 1.

We will solve in this example the following first-order ordinary differential equa-

tions system,

dx1dt

= 4x1 − 5x2,

dx2dt

= 2x1 − 3x2;

with the initial conditions x1(0) = 2 and x2(0) = −4.

The matrix associated with this system is

A =

(4 −5

2 −3

).

The characteristic equation is then, λ2 − λ− 2 = 0, the eigenvalues are −1 and

2, and the corresponding eigenvectors are

(1

1

)and

(5

2

), respectively. The

similarity matrix T is formed with the eigenvectors as columns, so

T =

(1 5

1 2

),

and the diagonal representation is,

D = T−1AT =

=

(−2/3 5/3

1/3 −1/3

)(4 −5

2 −3

)(1 5

1 2

)=

(−1 0

0 2

).

We use now (3.8), to get

eAt = TeDtT−1 =

=

(1 5

1 2

)(e−t 0

0 e2t

)(−2/3 5/3

1/3 −1/3

)=

1

3

(5e2t − 2e−t 5e−t − 5e2t

2e2t − 2e−t 5e−t − 2e2t

).

Finally, we use that the solution is x = eAtx(0) (equation (3.3)), to write the



solution of the systems as,

x1(t) = 10e2t − 7e−t,

x2(t) = 4e2t − 7e−t.

Example 2.

As a second example, we consider now the following first-order ordinary differ-

ential equations system,

dx1dt

= x1 + 2x2,

dx2dt

= 4x1 + 3x2;

with the initial conditions x1(0) = 1 and x2(0) = 2.

The matrix associated with this system is

A =

(1 2

4 3

).

The characteristic equation is then, λ2 − 4λ− 5 = 0, the eigenvalues are −1 and

5, and the corresponding eigenvectors are

(−1

1

)and

(1

2

), respectively.

The similarity matrix T is formed with the eigenvectors as columns, so

T =

(−1 1

1 2

).

and the diagonal representation is,

D = T−1AT =

=

(−2/3 1/3

1/3 1/3

)(1 2

4 3

)(−1 1

1 2

)=

(−1 0

0 5

).



We use now (3.8) to get

eAt = TeDtT−1 =

=

(−1 1

1 2

)(e−t 0

0 e5t

)(−2/3 1/3

1/3 1/3

)=

1

3

(2e−t + e5t e5t − e−t

2e5t − 2e−t e−t + 2e5t

).

Finally, as x = eAtx(0) (equation (3.3)), we apply this matrix to the initial

conditions (x1(0), x2(0)) = (1, 2), and we obtain the final result

x1(t) = e−5t,

x2(t) = 2e−5t.

If the two eigenvalues are equal, we have two possibilities. First, the matrix

can be already diagonal, and then it is a multiple of the identity matrix and

formula (3.9) works fine. Second, if the matrix is not diagonal, then it is not

possible to find a diagonal representation.

In this second case, we can use the Jordan form [Shores 07; Larson 09; Lang 87]

to transform the matrix A into a triangular matrix R by changing the basis.

If the matrix that change the basis is denoted by P, what we said before is

represented mathematically by the equation

R = P−1AP. (3.10)

Our original differential equations system (3.1), can then be written as

dx

dt= PRP−1x.

If we multiply from the left by P−1 the above equation, we obtain

d(P−1x)

dt= R

(P−1x

).

If we define now Y ≡ P−1x, we get the equivalent differential equations system

dY

dt= RY , (3.11)

that is already partially decoupled, and then it is very easily solved.



3.1.2 Cayley-Hamilton theorem method

From the above discussion, it is clear that if it is possible to find the diagonal

representation D of the matrix of the system A, the similitude matrix T, and its

inverse T−1, the solution of the differential equations system is formally found.

However, that is not always the case, and we have to develop other methods to

solve systems like (3.1); in what follows, we will develop one method in the two

dimensional case.

We go back to the characteristic equation. For simplicity, we define b0 = detA

and b1 = tr(A), then the characteristic equation is

λ2 − b1λ− b0 = 0. (3.12)

Remember the Cayley-Hamilton theorem: All square matrices satisfy its char-

acteristic equation; so

A2 − b1A− b0I = 0. (3.13)

Thus, we can write any power of A in terms of I and A. From (3.13),

A2 = b1A+ b0I

A3 = A2A = b1A2 + b0A = (b21 + b0)A+ b0b1I

A4 = A3A =[(b21 + b0)A+ b0b1I

]A = b1(b

21 + 2b0)A+ b0(b

21 + b0)I

and so on.

Applying all these results to the definition of the exponential of a matrix, we get

eAt =∞∑

n=0

tn

n!An

= I+ tA+t2

2!(b1A+ b0I) +

t3

3!

[(b21 + b0)A+ b0b1I

]+ . . .

= I

[1 +

t2

2!b0 +

t3

3!b0b1 +

t4

4!b0(b

21 + b0) + . . .

]+ A

[t+

t2

2!b1 +

t3

3!(b21 + b0) +

t4

4!b1(b

21 + 2b0) + . . .

]= f0(t)I+ f1(t)A, (3.14)

where

f0(t) = 1 +t2

2!b0 +

t3

3!b0b1 +

t4

4!b0(b

21 + b0) + . . . (3.15)



and

f1(t) = t+t2

2!b1 +

t3

3!(b21 + b0) +

t4

4!b1(b

21 + 2b0) + . . . . (3.16)

We have then translated the solution of the exponentiation problem of the matrix

A, to that of finding a closed form for the functions f0 and f1. To find a really

useful form for those functions, we shall derive now a second order linear ordinary

differential equation for each one. From the Cayley-Hamilton theorem (equation

(3.13)), it is evident that

d2eAt

dt2− b1

deAt

dt− b0e

At =(A2 − b1A− b0A

)eAt = 0,

but from (3.14), we also know that eAt = f0(t)I+ f1(t)A, and so

=d2eAt

dt2− b1

deAt

dt− b0e

At =(d2f0dt2

− b1df0dt

− b0f0

)I+

(d2f1dt2

− b1df1dt

− b0f1

)A = 0.

Hence, both functions, f0 and f1, satisfy the equation

d2f

dt2− b1

df

dt− b0f = 0, (3.17)

with distinct initial conditions.

We have to solve now these second-order linear ordinary differential equations

with constant coefficients, and compare these solutions with the expressions

(3.15) and (3.16) that we already have. To solve (3.17), we need the roots of the

auxiliary equation η2 − b1η − b0 = 0 [Zill 97, Chapter 4, section 4.3]. However,

this auxiliary equation is exactly the same as the characteristic equation (3.12)

of the matrix A of our system. Thus, to proceed further, we have to make some

assumptions about the eigenvalues of the matrix A; we assume first that they

are distinct.

3.1.2.1 Case A: λ1 = λ2

If the characteristic equation (3.12) of the matrix A of our system has two dis-

tinct roots, that means also that the auxiliary equation of the differential equa-

tion (3.17) has two different roots, and we have [Zill 97; Agarwal 08; Coddington

89]

f0 = c10eλ1t + c20e

λ2t (3.18)



and

f1 = c11eλ1t + c21e

λ2t, (3.19)

where the coefficients c′s are arbitrary integration constants, to be determined.

Expanding the exponentials in their Taylor series, we obtain

f0 = c10

(1 + λ1t+ λ1

t2

2!+ . . .

)+ c20

(1 + λ2t+ λ2

t2

2!+ . . .

)(3.20)

and

f1 = c11

(1 + λ1t+ λ1

t2

2!+ . . .

)+ c21

(1 + λ2t+ λ2

t2

2!+ . . .

). (3.21)

Comparing the expression (3.20) with expression (3.15), and expression (3.21)

with expression (3.16), we can find the following two systems of equations for

the coefficients,

c10 + c20 = 1

c10λ1 + c20λ2 = 0

and

c11 + c21 = 0

c11λ1 + c21λ2 = 1.

We can write these systems in a more compact form as,(1 1

λ1 λ2

)(c10 c11c20 c21

)=

(1 0

0 1

)(3.22)

or equivalently,

Λc = I, (3.23)

being

Λ =

(1 1

λ1 λ2

), (3.24)

a Vandermonde type matrix ([Fielder 86; Gautschi 62] and also Appendix B),

and

c =

(c10 c11c20 c21

). (3.25)



Multiplying (3.22) from the left by Λ−1, we find the coefficients as

c = Λ−1. (3.26)

Substituting back these coefficients in equations (3.18) and (3.19), we get f0 and

f1, and substituting these functions in (3.14), we finally obtain the exponential

of matrix A, and so, the solution to the system.

3.1.2.2 Case B: λ1 = λ2 = λ

In this case the solutions to equations (3.17) are [Zill 97; Agarwal 08; Coddington

89]

f0 = c10eλ1t + c20te

λ2t (3.27)

and

f1 = c11eλ1t + c21te

λ2t. (3.28)

where the coefficients c’s are arbitrary integration constants, to be determined.

Expanding, as in the previous case, the exponentials in a Taylor series, we get

f0 = c10

(1 + λ1t+ λ1

t2

2!+ . . .

)+ c20t

(1 + λ2t+ λ2

t2

2!+ . . .

)(3.29)

and

f1 = c11

(1 + λ1t+ λ1

t2

2!+ . . .

)+ c21t

(1 + λ2t+ λ2

t2

2!+ . . .

). (3.30)

Comparing (3.29) with (3.15), and (3.30) with (3.16), we obtain the two systems

of equations,

c10 = 1

c10λ1 + c20λ2 = 0 (3.31)

and

c11 = 0

c11λ1 + c21λ2 = 1. (3.32)

These systems of equations can be written in a shorter form as(1 0

λ 1

)(c10 c11c20 c21

)=

(1 0

0 1

). (3.33)



Denoting again

Λ =

(1 0

λ 1

)(3.34)

and

c =

(c10 c11c20 c21

), (3.35)

the coefficients can be found as

c = Λ−1. (3.36)

As in the other case, from these coefficients, it is very easy to find eAt, and the

solution to the system.

Example 3. Consider the first-order ordinary differential equations system

dx

dt= 3x+ 2y,

dy

dt= −2x− y,

with the initial conditions x(0) = 1 and y(0) = 0.

The matrix A associated with this system is(3 2

−2 −1

).

In Example 4, Section 1.4.3, Chapter 1, we showed that this matrix does not

have a diagonal representation. However, the characteristic equation for this

matrix is

λ2 − 2λ+ 1 = (λ− 1)2= 0

and there is just one eigenvalue, 1, with multiplicity 2; i.e., λ1 = λ2 = 1.

Therefore, applying (3.34) and (3.36),

(c10 c11c20 c21

)=

(1 0

1 1

)−1

=

(1 0

−1 1

)



Using now expressions (3.14), (3.27) and (3.28), we obtain successively

eAt = f0I+ f1A

=(c10e

t + c20tet)I+

(c11e

t + c21tet)A

= et (1− t) I+ tetA

Already knowing the exponential of the matrix of the system, we have the solu-

tion to it,(x

y

)= eAt

(x(0)

y(0)

)= et

[(1− t)

(1 0

0 1

)+ t

(3 2

−2 −1

)](1

0

)= et

[(1− t

0

)+

(3t

−2t

)]= et

(1 + 2t

−2t

).

Then, we have finally

x(t) = (1 + 2t)et,

y(t) = −2tet.

3.2 Systems 4 × 4

We study now, coupled first-order ordinary differential equations systems with 4

equations and 4 unknowns. The idea is to explain the complexity of the problem,

without being involved in the complicatedness of the notation of the n×n case.

We consider then the following 4 × 4 system of first-order ordinary differential

equations with constant coefficients,

dx1dt

= a11x1 + a12x2 + a13x3 + a14x4,

dx2dt

= a21x1 + a22x2 + a23x3 + a24x4,

dx3dt

= a31x1 + a32x2 + a33x3 + a34x4,

dx4dt

= a41x1 + a42x2 + a43x3 + a44x4; (3.37)


Systems 4× 4 75

with a given initial condition.

As in the two dimensional case, we can associate the following matrix with this

system,

A =

a11 a12 a13 a14a21 a22 a23 a24a31 a32 a33 a34a41 a42 a43 a44

. (3.38)

The characteristic equation for any 4× 4 matrix is of the form

λ4 − b3λ3 − b2λ

2 − b1λ− b0 = 0, (3.39)

and applying the Cayley-Hamilton theorem, we have

A4 = b3A3 + b2A

2 + b1A+ b0I

A5 =(b23 + b2

)A3 + (b1 + b2b3)A

2 + (b0 + b1b3)A+ b0b3I

A6 =(b33 + 2b2b3 + b1

)A3 +

(b22 + b2b

23 + b1b3 + b0

)A2

+(b1b

23 + b0b3 + b1b2

)A+

(b0b

23 + b0b2

)I

and so on; in such a way, that all powers of the A matrix are given in terms of

I,A,A2 and A3. Using this last fact, and the definition of the exponential of a

matrix in terms of its Taylor expansion, we have

eAt =n∑

i=0

tn

n!An = I+ tA+

t2

2!A2 +

t3

3!A3 +

t4

4!A4 + . . .

=

[1 +

t4

4!b0 +

t5

5!b0b3 +

t6

6!

(b0b

23 + b0b2

)+ ...

]I

+

[t+

t4

4!b1 +

t5

5!(b0 + b1b3) +

t6

6!

(b1b

23 + b0b3 + b1b2

)+ ...

]A

+

[t2

2!+t4

4!b2 +

t5

5!(b1 + b2b3) +

t6

6!

(b22 + b2b

23 + b1b3 + b0

)+ ...

]A2

+

[t3

3!+t4

4!b3 +

t5

5!

(b23 + b2

)+t6

6!

(b33 + 2b2b3 + b1

)+ ...

]A3.

Defining now the functions fi, i = 0, 1, 2, 3 by means of the following expression,

eAt = f0I+ f1A+ f2A2 + f3A

3, (3.40)



we obtain

f0 = 1 +t4

4!b0 +

t5

5!b0b3 +

t6

6!

(b0b

23 + b0b2

)+ ...

f1 = t+t4

4!b1 +

t5

5!(b0 + b1b3) +

t6

6!

(b1b

23 + b0b3 + b1b2

)+ ...

f2 =t2

2!+t4

4!b2 +

t5

5!(b1 + b2b3) +

t6

6!

(b22 + b2b

23 + b1b3 + b0

)+ ...

f3 =t3

3!+t4

4!b3 +

t5

5!

(b23 + b2

)+t6

6!

(b33 + 2b2b3 + b1

)+ ... (3.41)

It is possible to show, in a similar way as in the two dimensional case, that each of

the functions above satisfy a four order linear ordinary differential equation with

constants coefficients, whose auxiliary equation is the same as the characteristic

equation of the matrix of the system. Indeed, from the Cayley-Hamilton theorem

we have that A4 − b3A3 − b2A

2 − b1A− b0I = 0, so

d4eAt

dt4− b3

d3eAt

dt3− b2

d2eAt

dt2− b1

deAt

dt− b0e

At =

=(A4 − b3A

3 − b2A2 − b1A− b0

)eAt = 0.

From (3.40), eAt = f0I+ f1A+ f2A2 + f3A

3, and then

d4eAt

dt4− b3

d3eAt

dt3− b2

d2eAt

dt2− b1

deAt

dt− b0e

At =

=

(d4f0dt4

− b3d3f0dt3

− b2d2f0dt2

− b1df0dt

− b0f0

)I

+

(d4f1dt4

− b3d3f1dt3

− b2d2f1dt2

− b1df1dt

− b0f1

)A

+

(d4f2dt4

− b3d3f2dt3

− b2d2f2dt2

− b1df2dt

− b0f2

)A2

+

(d4f3dt4

− b3d3f3dt3

− b2d2f3dt2

− b1df3dt

− b0f3

)A3,

and as the set of matrices I,A,A2,A3 is linearly independent, each one of the

functions fi, i = 0, 1, 2, 3 satisfy the equation

d4f

dt4− b3

d3f

dt3− b2

d2f

dt2− b1

df

dt− b0f = 0, (3.42)

with distinct initial conditions.

From now on, it is necessary, as in the 2 × 2 case, to distinguish four cases,

depending on the eigenvalues of the A matrix, that are also the roots of the

auxiliary equation of the differential equation (3.42).


Systems 4× 4 77

3.2.1 Case 1. All the eigenvalues are different (λ1 = λ2 = λ3 =λ4)

In this case, the solution to the differential equation (3.42) [Zill 97] give us,

fi =4∑

k=1

ckieλkt, (3.43)

where i = 0, ..., 3 and the c’s are integration constants. In this last expression,

we take i = 0 and expand in Taylor series the exponentials, to obtain

f0 = c10 + c20 + c30 + c40

+ t (c10λ1 + c20λ2 + c30λ3 + c40λ4)

+t2

2!

(c10λ

21 + c20λ

22 + c30λ

23 + c40λ

24

)+

t3

3!

(c10λ

31 + c20λ

32 + c30λ

33 + c40λ

34

)+ . . . . (3.44)

Comparing this last expression with the one obtained from the Cayley-Hamilton

theorem, formula (3.41), we get

4∑k=1

ck0 = 1,

4∑k=1

ck0λk = 0,

4∑k=1

ck0λ2k = 0,

4∑k=1

ck0λ3k = 0.

Proceeding in exactly the same way for the other functions fi, with i = 1, 2, 3,

it is very easy to prove that the 16 constants in equation (3.43) satisfy a set of

algebraic equations that can be written briefly as1 1 1 1

λ1 λ2 λ3 λ4λ21 λ22 λ23 λ24λ31 λ32 λ33 λ34

c10 c11 c12 c13c20 c21 c22 c23c30 c31 c32 c33c40 c41 c42 c43

=

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

, (3.45)

or in a shorter matrix form

Λc = I, (3.46)



with the obvious definitions. For its structure, matrix Λ is called a Vandermonde

matrix [Fielder 86; Shores 07].

The coefficients c’s are given by the inverse of this Vandermonde matrix; i.e., we

have the final result

c = Λ−1. (3.47)

The inverse of a Vandermonde matrix always exists (Appendix B), and there

are elaborated methods and technics to find its inverse [Fielder 86; Gautschi 62;

Gautschi 78].

3.2.2 Case 2. λ1 = λ2 = λ3 = λ4

In this case, the functions fi, as solutions of the differential equation (3.42), are

fi = c1ieλ1t + c2ite

λ1t + c3ieλ3t + c4ite

λ3t, (3.48)

where i = 0, 1, 2, 3 and the c’s are integration constants.

Expanding again the exponentials in Taylor series, we obtain

fi = c1i + c3i

+ t(c1iλ1 + c3iλ3 + c2i + ci4)

+ t2(c1iλ212!

+ c3iλ232!

+ c2iλ1 + c4iλ3

)+ t3

(c1iλ316

+ c3iλ336

+ c2iλ212

+ c4iλ232

)+ . . . ,

i = 0, 1, 2, 3. (3.49)

Comparing equations (3.49) and (3.41) for i = 0, we get

c10 + c30 = 1,

c10λ1 + c30λ3 + c20 + c40 = 0,

c10λ212

+ c30λ232

+ c20λ1 + c40λ3 = 0,

c10λ316

+ c30λ336

+ c20λ212

+ c40λ232

= 0.


Systems 4× 4 79

Proceeding in the same way for i = 1,

c11 + c31 = 0,

c11λ1 + c31λ3 + c21 + c41 = 1,

c11λ212

+ c31λ232

+ c21λ1 + c41λ3 = 0,

c11λ316

+ c31λ336

+ c21λ212

+ c41λ232

= 0.

For i = 2 and for i = 3 is also similar. We obtain then the following 16 algebraic

equations for the 16 constants c’s,

1 0 1 0

λ1 1 λ3 1

λ21 2λ1 λ23 2λ3λ31 3λ21 λ33 3λ23


=

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

.

(3.50)

Then, with the logical and obvious definitions, we can calculated the c’s coeffi-

cients as

c = Λ−1. (3.51)

Matrices of the same structure as Λ are called confluent Vandermonde matrices

([Fielder 86] and Appendix B).

3.2.3 Case 3. λ1 = λ2 = λ3 = λ = λ4

In this case the fi functions take the following form,

fi = c1ieλt + c2ite

λt + c3it2eλt + c4ie

λ4t, (3.52)

where i = 0, 1, 2, 3 and the c’s are constants of the integration.

Expanding the exponentials in Taylor series,

fi = c1i + c4i + t(c1iλ+ c4iλ4)

+ t2(c1iλ2

2!+ c2iλ+ c3i ++

λ42!c4i

)+ t3

(c1iλ3

6+ c2i

λ2

2+ c3iλ+ c4i

λ346

). (3.53)



Comparing (3.53) for i = 0 with (3.41),

c10 + c40 = 1,

t(c10λ+ c40λ4) = 0,

t2(c1iλ2

2!+ c2iλ+ c3i ++

λ42!c4i

)= 0,

t3(c1iλ3

6+ c2i

λ2

2+ c3iλ++c4i

λ346

)= 0.

Doing the same for i = 1, 2, 3, it is possible to write the following equation

between matrices,

1 0 0 1

λ 1 0 λ4λ2 2λ 2 λ24λ3 3λ2 6λ λ34


=

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

, (3.54)

so that the c’s coefficients can be evaluated from the expression

c = Λ−1.

Matrix Λ is a confluent Vandermonde type matrix [Fielder 86].

3.2.4 Case 4. λ1 = λ2 = λ3 = λ4 = λ

In this last case, the functions fi take the form,

fi = c1ieλt + c2ite

λt + c3it2eλt + c4it

3eλt, (3.55)

with i = 0, 1, 2, 3 and the c’s are integration constants.

Expanding the exponentials in Taylor series,

fi = c1i + t(c1iλ+ c2i) + t2(c1iλ2

2!+ c2iλ+ c3i

)+ t3

(c1iλ3

6+ c2i

λ2

2+ c3iλ+ c4i

). (3.56)


Systems 4× 4 81

Comparing this last equation for f0 with (3.41),

c10 = 1,

c10λ+ c20 = 0,

c10λ2

2!+ c20λ+ c30 = 0,

c10λ3

6+ c20

λ2

2+ c30λ+ c40 = 0.

Following the same procedure for f1, f2 y f3, we get the following equation for

matrices,1 0 0 0

λ 1 0 0

λ2 2λ 2 0

λ3 3λ2 6λ 6


=

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

. (3.57)

And again,

c = Λ−1.

Matrix Λ is a confluent Vandermonde type matrix.

Example 1. Let us consider the first-order ordinary differential equations sys-

tem

dx1dt

= x1 − x3 + x4,

dx2dt

= x2 + x4,

dx3dt

= −2x1 + 2x3 − 2x4,

dx4dt

= 2x2 + 2x4;

with initial conditions x1(0) = 1, x2(0) = −1, x3(0) = 0, x4(0) = −1. The

associated matrix is 1 0 −1 1

0 1 0 1

−2 0 2 −2

0 2 0 2

.

The eigenvalues are 0, with multiplicity 2, and 3, also with multiplicity 2. How-

ever, for this matrix it does not exist a diagonal representation. This can be



understood looking for the eigenvectors; in the case of the eigenvalue 3, the in-

variant subspace is the straight line in four dimensions, given by the parametric

equation (x1, x2, x3, x4) = (1, 0,−2, 0)t, with t as the parameter. As this sub-

space has dimension 1, it is not possible to find a basis of eigenvectors, and there

is not a diagonal representation.

We proceed with our method. We are in case 2, so


=

1 0 1 0

λ1 1 λ3 1

λ21 2λ1 λ23 2λ3λ31 3λ21 λ33 3λ23

−1

.

With the data of our example,


=

1 0 1 0

0 1 3 1

0 0 9 6

0 0 27 27

−1

=

1 0 −1/3 2/27

0 1 −2/3 1/9

0 0 1/3 −2/27

0 0 −1/3 1/9

.

Substituting the c coefficients in the expressions (3.48) for the functions f , we

get

f0 = 1,

f1 = t,

f2 = −1

3− 2

3t+

1

3e3t − 1

3te3t,

f3 =2

27+

1

9t− 2

27e3t +

1

9te3t.

Now we make

eAt = f0I+ f1A+ f2A2 + f3A

3,


Systems n× n 83

and applying to the initial conditions, we obtain the final solution

x1(t) =4

3(1

3− t)e3t +

5

9,

x2(t) = −2

3e3t − 1

3,

x3(t) = −8

3(1

3− t)e3t +

8

9,

x4(t) = −4

3e3t +

1

3.

3.3 Systems n × n

We shall analyze now a coupled first-order ordinary differential equations system

with n equations and n unknowns. The procedure we will follow is a straight

generalization of the 4× 4 case. The general n× n system is

dx1dt

= a11x1 + a12x2 + · · ·+ a1nxn,

dx2dt

= a21x1 + a22x2 + · · ·+ a2nxn,

dx3dt

= a31x1 + a32x2 + · · ·+ a3nxn,

...,

dxndt

= an1x1 + an2x2 + · · ·+ annxn; (3.58)

with some given initial conditions.

The associated matrix A is

A =

a11 a12 a13 a14 . . . a1na21 a22 a23 a24 . . . a2na31 a32 a32 a34 . . . an3. . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . .

an1 an2 . . . . . . . . . ann

. (3.59)

The characteristic equation for all n×nmatrices, and in particular for the matrix

A of our system, is of the form

λn − bn−1λn−1 − bn−2λ

n−2 − . . .− b1λ1 − b0 = 0. (3.60)



Using the Cayley-Hamilton theorem, it is possible to write

eAt =n∑

i=0

fi(t)Ai, (3.61)

where the functions fi are determined comparing this last expression with the

Taylor expansion eAt =∑∞

i=0ti

i!Ai of the exponential of the matrix tA, obtain-

ing a very complicated series in terms of t and the constants bi.

In a completely analog way, as we did in the 4 dimension case, it is possible

also to show that the functions fi satisfy the following n-order linear differential

equation with constant coefficients,

dnf

dtn− bn−1

dn−1f

dtn−1− . . .− b1

df

dt− b0f = 0. (3.62)

To solve this differential equation, the roots of its auxiliary equation have to be

found. The auxiliary equation is also the characteristic equation of the matrix

A of the system. Thus, we have to look for the eigenvalues of A and compare

them. To fix ideas, let us analyze two cases, the first one when all the eigenvalues

are distinct and the second one when we have some sets of equal eigenvalues, let

us say, λ1 = λ2 = λ3, λ4 = λ5, and the rest are different.

3.3.1 All the eigenvalues are distinct

In this case the solutions to equation (3.62) are [Zill 97]

fi(t) =

n∑j=1

cjieλjt, (i = 0, 1, 2, ..., n− 1), (3.63)

where the n2 coefficients cji are integration constants.

Comparing these solutions with the expressions obtained from the Taylor ex-

pansion of the exponential, we obtain


Systems n× n 85

c = Λ−1 =

1 1 1 · · · 1

λ1 λ2 λ3 · · · λn

λ21 λ22 λ23 · · · λ2n

λ31 λ32 λ33 · · · λ3n

· · · · · · · · · · · · · · ·

· · · · · · · · · · · · · · ·

λn−11 λn−1

2 λn−13 · · · λn−1

n

−1

. (3.64)

The matrix Λ, a Vandermonde type matrix, always has an inverse (Appendix

B), and there are specific technics to find it [Fielder 86; Gautschi 62; Gautschi

78].

3.3.2 The eigenvalues are: λ1 = λ2 = λ3, λ4 = λ5, and the rest

are different

In this case the solutions to the differential equation (3.62) are [Zill 97]

fi(t) =(c1i + c2it+ c3it2)eλ1t + (c4i + c5it)e

λ4t +n∑

j=6

cjieλjt,

(i = 0, 1, ..., n− 1),

(3.65)

where the n2 coefficients cji are integration constants.

Again, we have to compare these solutions with those obtained from the Taylor

expansion of the exponential; after doing that, we get



c = Λ−1 =

1 0 0 1 0 . . . 1

λ1dλ1

dλ1

d2λ1

dλ21

λ4dλ4

dλ4. . . λn

λ21dλ2

1

dλ1

d2λ21

dλ21

λ24dλ2

4

dλ4. . . λ2n

. . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . .

λn−11

dλn−11

dλ1

d2λn−11

dλ21

λn−14

dλn−14

dλ4. . . λn−1

n

−1

. (3.66)

The inverse of the matrix Λ, a confluent Vandermonde type matrix, exists and

there are explicit expressions for it (Appendix B and [Fielder 86; Gautschi 62;

Gautschi 78]).

Example 1.

Let us consider an ordinary differential equations system, whose associated A

matrix is

A =

3 11 −11

1 3 −2

1 5 −4

,

and with the initial conditions

x(0) =

1

0

0

.

The eigenvalues of this matrix are λ1 = 1, λ2 = −2 and λ3 = 3. In effect,

|λI−A| = det

λ− 3 −11 11

−1 λ− 3 2

−1 −5 λ+ 4

,

and this implies that the characteristic equation is

λ3 − 2λ2 − 5λ+ 6 = 0,


Systems n× n 87

whose roots are 1,−2, 3.

As the eigenvalues are all distinct, it exists a diagonal representation for this

matrix and also the method exposed in Section 1 of this Chapter can be utilized.

However, as we already said, this is not always the case and we will use the

method developed in this section. Now we can use formula (3.64) to calculate

the coefficients cij

c10 c11 c12c20 c21 c22c30 c31 c32

=

1 1 1

1 −2 3

1 4 9

−1

=

1 16 −1

615 − 4

15115

−15

110

110

.

With this coefficients the following functions can be obtained

f0 = et +1

5e−2t − 1

5e−3t,

f1 =1

6et − 4

15e−2t +

1

10e−3t,

f2 = −1

6et +

1

15e−2t +

1

10e−3t,

and they can be substituted in the following expression

x(t) = eAtx(0)

= f0Ix(0) + f1Ax(0) + f2A2x(0)

=

f0 + 3f1 + 9f2f1 + 4f2f1 + 4f2

,

to arrive to the final solution to the system

x(t) =

e3t

12e

3t − 12e

t

12e

3t − 12e

t

.

Example 2. Chebyshev system



Consider the system of ordinary differential equations

E0 = αE1,

E1 = α(E0 + E2),

E2 = α(E1 + E3),

... ,

En = αEn−1,

with initial conditions E0(0) = 1 and Ek =0(0) = 0.

From what we have seen in this Chapter, the solution may be written in the

form

E = eαtcE(0),

where

c =

0 1 0 0 . . . 0

1 0 1 0 . . . 0

0 1 0 1 . . .

. . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . .

0 0 . . . . . . 1 0

and the initial condition is

E(0) =

1

0...

0

.

The eigenvalues of the matrix c are then obtained from the equation∣∣∣∣∣∣∣∣∣∣∣∣∣∣

λ −1 0 . . . 0 0

−1 λ −1 0 . . . 0

0 −1 λ −1 . . . 0

. . . . . . . . . . . .... . . .

. . . . . . . . . . . . . . . . . .

0 0 0 . . . −1 λ

∣∣∣∣∣∣∣∣∣∣∣∣∣∣= 0,


Systems n× n 89

and from the formula 2.60 on page 55 of Section 3 of Chapter 2, we can write

the characteristic equation as

Un

(−λ2

)= 0,

where Un is the Chebyshev polynomial of the second kind studied in Chapter

2, Section 3. From the same section, we know also know that a Chebyshev

polynomial of either kind with degree n, has n different simple roots (see also[Arfken 05]); so, from (3.64), we have

c =

1 1 1 · · · 1

λ1 λ2 λ3 · · · λn

λ21 λ22 λ23 · · · λ2n

λ31 λ32 λ33 · · · λ3n

· · · · · · · · · · · · · · ·

· · · · · · · · · · · · · · ·

· · · · · · · · · · · · · · ·

λn−11 λn−1

2 λn−13 · · · λn−1

n

−1

,

where λk are the roots of the Chebyshev polynomials of the second kind that

are given by (see Chapter 2, Section 3, formula 2.59 on page 55),

λk = cos

(k

n+ 1π

).

As we have said, any Vandermonde type matrix always have an inverse, and there

are specific technics to calculate it (Appendix B and [Fielder 86; Gautschi 62;

Gautschi 78]).


Chapter 4

Infinite systems of differential equations

In this Chapter, we will use operational methods to solve infinite systems of first-

order ordinary differential equations. In each case, we shall demonstrate that the

system is equivalent to a Schrodinger-like equation, we will solve this equation

using operational methods, and finally, from this solution, that of the system

will be deduced. To accomplish this goal, the usual notation is inadequate, so

in the rest of this book, we will use with freedom the Dirac notation (Appendix

A) (see also ([Louisell 90; Jackson 62; Alicki 01; Dennery 95])).

All along this Chapter, we will use frequently the linear operators V and V †,

defined by

V ≡∞∑

n=−∞|n⟩⟨n+ 1|, V † ≡

∞∑n=−∞

|n+ 1⟩⟨n|, (4.1)

acting on the vector space generated by the complete and orthonormal set

|n⟩;n = −∞, ...,+∞. It is very easy to see that the action of these opera-

tors on the ”states” |n⟩ is

V |n⟩ = |n− 1⟩ (4.2)

and

V † |n⟩ = |n+ 1⟩ . (4.3)

Thus,

V V † |n⟩ = V |n+ 1⟩ = |n⟩

and

V †V |n⟩ = V † |n− 1⟩ = |n⟩ .

91


92 Infinite systems of differential equations

From these last equations, we deduce two important properties of the linear

operators V and V †. First, that the two operators commute, or in symbolic

terms that

[V, V †] = 0; (4.4)

and second, that they are unitary, which means that

V † = V −1. (4.5)

4.1 Addition formula for the Bessel functions

In Chapter 2, Section 2.4, we proved the addition formula for the Bessel functions

of the first kind

Jn(x+ y) =

∞∑k=−∞

Jn−k(x)Jk(y). (4.6)

In this section, we will use the linear operators V and V † to make another

demonstration. The aims are emphasize the advantages of the operator methods

and obtain skills in the manipulation of operators in general.

To start up, we recall the generating function

exp

[x

2

(t− 1

t

)]=

∞∑n=−∞

tnJn(x) (4.7)

of the Bessel functions of the first kind, that we exhibited in the mentioned

section.

Now, we build the linear operator exp[x2

(V † − V

)], and from (4.5) and (4.7),

exp[x2

(V † − V

)]= exp

[x

2

(V † − 1

V †

)]=

∞∑k=−∞

Jk (x)V†k. (4.8)

Applying this operator to the vacuum state |0⟩, and using that V †k|0⟩ = |k⟩,

exp[x2

(V † − V

)]|0⟩ =

∞∑k=−∞

Jk (x)V†k |0⟩ =

∞∑k=−∞

Jk (x) |k⟩ . (4.9)


Addition formula for the Bessel functions 93

Multiplying from the left by the bra ⟨n|, we obtain

⟨n| exp[x2

(V † − V

)]|0⟩ = ⟨n|

∞∑k=−∞

Jk (x) |k⟩

=∞∑

k=−∞

Jk (x) ⟨n|k⟩

=∞∑

k=−∞

Jk (x) δnk = Jn (x) ,

where we have used the fact that the set |n⟩;n = −∞, ...,+∞ is orthonormal;

i.e., that ⟨η|µ⟩ = δη,µ.

Summarizing,

Jn (x) = ⟨n| exp[x2

(V † − V

)]|0⟩ . (4.10)

We already have all the necessary elements to demonstrate the addition for-

mula (4.6). From (4.10), we write

Jn (x+ y) = ⟨n| exp[x+ y

2

(V † − V

)]|0⟩

= ⟨n| exp[x2

(V † − V

)+y

2

(V † − V

)]|0⟩ .

As trivially the operatorsx

2

(V † − V

)and

y

2

(V † − V

)commute, we have

Jn (x+ y) = ⟨n| exp[x2

(V † − V

)]exp

[y2

(V † − V

)]|0⟩ .

Using now that V = (V †)−1 (expression (4.5)), we obtain

Jn (x+ y) = ⟨n| exp[x

2

(V † − 1

V †

)]exp

[y

2

(V † − 1

V †

)]|0⟩ .

Now we have to recall the generating function of the Bessel functions, expression

(4.8), to get

Jn (x+ y) = ⟨n|∞∑

j=−∞Jj (x)

(V †)j ∞∑

k=−∞

Jk (y)(V †)k |0⟩ .

Manipulating the sums and the states, and using again that V †k|n⟩ = |n + k⟩,



it is easy to arrive to

Jn (x+ y) = ⟨n|∞∑

j=−∞

∞∑k=−∞

Jj (x)Jk (y)(V †)j+k |0⟩

= ⟨n|∞∑

j=−∞

∞∑k=−∞

Jj (x)Jk (y) |j + k⟩ .

Remembering that the states |n⟩ are orthonormal,

Jn (x+ y) =

∞∑j=−∞

∞∑k=−∞

Jj (x)Jk (y) ⟨n|j + k⟩

=∞∑

k=−∞

∞∑j=−∞

Jj (x)Jk (y) δn,j+k,

and finally, we arrive to the addition formula that we wanted to show,

Jn (x+ y) =∞∑

k=−∞

Jn−k (x) Jk (y) .

4.2 First neighbors interaction

Consider the following infinite system of first order ordinary differential equa-

tions

idEn

dt= α(En+1 + En−1), (4.11)

with n = −∞, ...,∞, and subject to some initial condition, that will be specified

later.

We will show that the previous system (4.11) is equivalent to the following

Schrodinger-like equation

id|ψ⟩dt

= H|ψ⟩, (4.12)

with the Hamiltonian given by

H = α(V + V †), (4.13)

and where the linear operators V and V † are those defined in the introduction of

this Chapter, formulas (4.1). Note that the Hamiltonian is an Hermitian linear

operator.


First neighbors interaction 95

As the set |n⟩;n = −∞, ...,+∞ is complete, we can write the solution to the

Schrodinger-like equation as

|ψ(t)⟩ =∞∑

n=−∞En(t)|n⟩, (4.14)

where the coefficients are given by

En(t) = ⟨n|ψ(t)⟩. (4.15)

We insert it into (4.12) to get

i

∞∑n=−∞

dEn

dt|n⟩

= α

∞∑n=−∞

En

∞∑j=−∞

|j⟩⟨j + 1|n⟩+∞∑

n=−∞En

∞∑n=−∞

|j + 1⟩⟨j|n⟩

,

and using the orthonormality condition ⟨η|µ⟩ = δηµ, we obtain

i∞∑

n=−∞

dEn

dt|n⟩ = α

( ∞∑n=−∞

En|n− 1⟩+∞∑

n=−∞En|n+ 1⟩

).

Changing the indices in the sums, we have

i

∞∑n=−∞

dEn

dt|n⟩ = α

( ∞∑n=−∞

En+1|n⟩+∞∑

n=−∞En−1|n⟩

).

As the set |n⟩;n = −∞, ...,+∞ is linearly independent, we can equate the co-

efficients on both sides of the previous equation and recover the original infinite

system of differential equations (4.11). Thus, the infinite original system (4.11),

and the Schrodinger-like equation are equivalent, as we wanted to show.

The initial condition |ψ(0)⟩ may be, in general, an arbitrary superposition of

states; so as the problem is linear, we can consider, without losing generality,

|ψ(0)⟩ = |m⟩. As |ψ⟩ =∑∞

n=−∞En|n⟩, the established initial condition for |ψ⟩means the En(t) is zero for all n, with exception of m, when it values 1; i.e.,

En(t = 0) = δn,m.

Now, we will use operational methods to solve the Schrodinger-like equation,

and then deduce the solution of the infinite system (4.11). Considering the initial

condition |ψ(0)⟩ = |m⟩, it is simple to write the formal solution of (4.12) as

|ψ(t)⟩ = e−iαt(V †+V )|m⟩. (4.16)



To work out the operator e−iαt(V †+V ), we remember that V † = V −1. Hence, we

can write

e−iαt(V †+V ) = e−αt(iV †− 1

iV † ),

that we have already recognized as the generating function of the Bessel functions

of the first kind, so

e−αt(iV †− 1

iV † ) =

∞∑n=−∞

Jn(−2αt)inV †n,

and we can rewrite the solution of the Schrodinger-like equation as

|ψ(t)⟩ =∞∑

n=−∞Jn(−2αt)inV †n|m⟩.

By using that V †n|m⟩ = |m+ n⟩, we finally arrive at

|ψ(t)⟩ =∞∑

n=−∞Jn(−2αt)in|m+ n⟩. (4.17)

Now we can find the amplitudes Ek, k = −∞, ...,∞, that are the solutions to

the system (4.11), simply by

Ek(t) = ⟨k|ψ(t)⟩ = ik−mJk−m(−2αt), (4.18)

where we have used again the orthonormality condition ⟨k|n⟩ = δkn.

In the particular case, that the initial state is the vacuum state |0⟩, we get

Ek(t) = ⟨k|ψ(t)⟩ = ikJk(−2αt). (4.19)

4.3 Second neighbors interaction

Consider now this other infinite system of first order ordinary differential equa-

tions

idEn

dt=A(t)

2(En+1 + En−1) +

B(t)

2(En+2 + En−2), (4.20)

with n = −∞, ...,∞, and subject to some initial condition, that will be specified

later.

As in the previous section, we will demonstrate that the system (4.20) is equiv-

alent to the following Schrodinger-like equation

id|ψ⟩dt

= H|ψ⟩, (4.21)


Second neighbors interaction 97


H =A(t)

2(V + V †) +

B(t)

2(V 2 + V †2), (4.22)

where A(t) and B(t) are arbitrary functions of t, and where the linear operators,

V and V †, are those presented in the introduction of this Chapter, expressions

(4.1). We remark that also here the Hamiltonian is a linear Hermitian operator.

The procedure to show that (4.20) and (4.21) are equivalent, is completely

analog to that of Section 4.2. However, to reinforce the ideas, we will sketch the

main steps again. We write the solution to the Schrodinger-like equation as

|ψ(t)⟩ =∞∑

n=−∞En(t)|n⟩, (4.23)


En(t) = ⟨n|ψ(t)⟩; (4.24)

we insert it into (4.21) and we use the linearity of the Hamiltonian, to get

i∞∑

n=−∞

dEn

dt|n⟩ = H

∞∑n=−∞

En|n⟩ =∞∑

n=−∞EnH|n⟩. (4.25)

The action of the Hamiltonian H in the states |n⟩ is

H|n⟩ =[A(t)

2(V + V †) +

B(t)

2(V 2 + V †2)

]|n⟩

=A(t)

2(|n− 1⟩+ |n+ 1⟩) + B(t)

2(|n− 2⟩+ |n+ 2⟩),

that substituted in (4.25), give us

i

∞∑n=−∞

dEn

dt|n⟩ = A(t)

2(

∞∑n=−∞

En|n− 1⟩+∞∑

n=−∞En|n+ 1⟩)

+B(t)

2(

∞∑n=−∞

En|n− 2⟩+∞∑

n=−∞En|n+ 2⟩).

Changing the indexes of the sums, we get

i

∞∑n=−∞

dEn

dt|n⟩ =

∞∑n=−∞

[A(t)

2(En+1 + En−1) +

B(t)

2(En+2 + En−2)

]|n⟩.



Finally, invoking the linear independence of the set of states |n⟩;n = −∞, ...,∞,we obtain the system of differential equations (4.20), as we wanted.

We have to solve now the Schrodinger-like equation (4.21), subject to some

arbitrary initial condition. As we already explained, we do not loose generality,

if we consider as initial condition one of the states |l⟩. Thus, we can write

|ψ(t)⟩ = e−iα(t)2 (V †+V )e−i

β(t)2 (V †2+V 2)|l⟩,

where by definition

α(t) ≡∫ t

0

A(t′)dt′, β(t) ≡∫ t

0

B(t′)dt′.

Now we use, as in the previous section, that

e−iα(t)2 (V †+V ) = e−

α(t)2 (iV †− 1

iV † ) =

∞∑n=−∞

Jn[−α(t)]inV †n,

and analogously that

e−iβ(t)2 (V †2+V 2) = e−

β(t)2 (iV †2− 1

iV †2 ) =∞∑

m=−∞Jm[−β(t)]imV †2m,

to rewrite the solution as

|ψ(t)⟩ =∞∑

m=−∞

∞∑n=−∞

Jm[−β(t)]Jn[−α(t)]im+nV †(n+2m)|l⟩.

By using that V †k|l⟩ = |l+k⟩, we finally arrive at the solution of the Schrodinger-

like equation

|ψ(t)⟩ =∞∑

m=−∞

∞∑n=−∞

Jm[−β(t)]Jn[−α(t)]im+n|l + n+ 2m⟩. (4.26)

Now we can find the amplitudes Ek; k = −∞, ...,∞, simply writing

Ek(t) = ⟨k|ψ(t)⟩ =∞∑

m=−∞Jm[−β(t)]Jk−2m−l[−α(t)]ik−m−l,

k, l = −∞, ...,∞,

(4.27)

where we have used once more the orthonormality condition ⟨k|n⟩ = δkn.


Second neighbors interaction 99

Finally, let us write the coefficients Ek, k = −∞, ...,∞ in an integral form.

For this, we write one of the Bessel functions in the sum (4.27) as the integral

(equation 2.65 on page 55)

Jn (x) =1

2π

∫ π

−π

e−i(nθ−x sin θ)dθ,

in such a way that

Ek(t) =ik−l

2π

∫ π

−π

e−i(k−l)θe−iα(t) sin θ∞∑

m=−∞Jm[−β(t)](−i)me2imθdθ.

By using once more the generating function of Bessel functions

exp

[x

2

(t− 1

t

)]=

∞∑n=−∞

tnJn(x),

we obtain the final expression

Ek(t) =ik−l

2π

∫ π

−π

e−i(k−l)θe−iα(t) cos θeiβ(t) cos 2θdθ,

k, l = −∞, ...,∞.

(4.28)

In particular, if the initial state is the vacuum state |0⟩, we obtain the fol-

lowing results

Ek(t) =∞∑

m=−∞Jm[−β(t)]Jk−2m[−α(t)]ik−m (4.29)

and

Ek(t) =ik

2π

∫ π

−π

e−ikθe−iα(t) cos θeiβ(t) cos 2θdθ. (4.30)

where k = −∞, ...,∞.

Also note, that as Jn(0) = 0, except for n = 0 when J0(0) = 1, if we set

B(t) = 0 and A(t) = 2α, we recover

Ek(t) = ik−lJk−l(−2αt),

k, l = −∞, ...,∞,

that is the result 4.18 on page 96, for interaction to only first neighbors with the

initial condition |ψ(t)⟩ = |l⟩.



4.4 First neighbors interaction with an extra interaction

Because in this section we will use the number operator, that normally is denoted

by n, and that can be confused with an integer index, we will place ”a hat” in

all the operators.

4.4.1 Interaction ωn

We will analyze now the following infinite system of first order ordinary differ-

ential equations

idEn

dt= ωnEn + α(En+1 + En−1),

n = −∞, ...,∞,(4.31)

with ω and α arbitrary constants. We already explained, that we can take as

initial condition one arbitrary state |m⟩, without loosing generality.

On the other hand, consider the following Schrodinger-like equation

id|ψ⟩dt

= H|ψ⟩, (4.32)


H = ωn+ α(V + V †), (4.33)

where the linear operators V and V † are those defined in the introduction of this

Chapter, formulas 4.1 on page 91, and where n is the number operator, defined

by

n|n⟩ ≡ n|n⟩. (4.34)

On the light of the previous Chapters, it is obvious that the system (4.31)

and the Schrodinger-like equation are equivalent, and we proceed then to solve

the last one to find the solution to the first one. To solve (4.32), we will need

the commutator of the number operator with V and with V †. We take

[V , n]|n⟩ = V n|n⟩ − nV |n⟩ = n|n− 1⟩ − (n− 1)|n− 1⟩ = |n− 1⟩,

so

[V , n] = V . (4.35)


First neighbors interaction with an extra interaction 101

For the operator V †, we have

[V †, n]|n⟩ = V †n|n⟩ − nV †|n⟩ = n|n+ 1⟩ − (n+ 1)|n+ 1⟩ = −|n+ 1⟩,

or in other words,

[V †, n] = −V †. (4.36)

We also have the following commutator,[n, V + V †

]=[n, V

]+[n, V †

]= −V + V †.

Thus, the

Baker-Hausdorff formula: Given two operators A and B that

obey [[A, B

], A]=[[A, B

], B]= 0,

then

eA+B = e−12 [A,B]eAeB .

can not be used to solve the Schrodinger-like equation, as was the case of those

in the former sections, because[[n, V + V †

], n]=[−V + V †, n

]= −V − V † = 0,

and the hypothesis of the theorem is not satisfied.

In order to solve the Schrodinger-like equation, we will use the so-called

interaction representation. We propose for the solution of the equation (4.32)

the following function

|ψ⟩ = e−iωtn|φ⟩. (4.37)

We substitute it in the Schrodinger-like equation (4.32), and we get

ωn|ψ⟩+ ie−iωnt d|φ⟩dt

=[ωn+ α(V + V †)

]|ψ⟩.

Canceling similar terms and multiplying both sides by eiωnt, we have

id|φ⟩dt

= αeiωnt(V + V †)e−iωnt|φ⟩. (4.38)



Thus, we have translated the problem of the equation (4.32) to this other

Schrodinger-like equation. However, this equation, as it is, can not be inte-

grated because the operators in the right hand side do not commute. We have

to paraphrase it; for that we analyze by parts the operator in this new equation;

first,

eiωntV e−iωnt =

∞∑j=0

(iωnt)j

j!V

∞∑k=0

(−iωnt)k

k!

=∞∑

j,k=0

(−1)k(iωt)j+k

j!k!nj V nk,

but

nj V nk|n⟩ = nknj V |n⟩ = nknj |n− 1⟩ = nk(n− 1)j |n− 1⟩ = nk(n− 1)j V |n⟩

then

eiωntV e−iωnt =

∞∑j,k=0

(−1)k(iωt)j+k

j!k!nk(n− 1)j V

=∞∑j=0

[iω(n− 1)t]j

j!V

∞∑k=0

(−iωnt)k

k!

= eiω(n−1)tV e−iωnt = e−iωtV ;

(4.39)

second,

eiωntV †e−iωnt =∞∑j=0

(iωnt)j

j!V †

∞∑k=0

(−iωnt)k

k!

=

∞∑j,k=0

(−1)k(iωt)j+k

j!k!nj V †nk,

but

nj V †nk|n⟩ = nknj V †|n⟩ = nknj |n+ 1⟩ = nk(n+ 1)j |n+ 1⟩ = nk(n+ 1)j V †|n⟩



hence

eiωntV †e−iωnt =∞∑

j,k=0

(−1)k(iωt)j+k

j!k!nk(n+ 1)j V †

=

∞∑j=0

[iω(n+ 1)t]j

j!V †

∞∑k=0

(−iωnt)k

k!

= eiω(n+1)tV †e−iωnt = eiωtV †

(4.40)

The formulas (4.39) and (4.40) can be obtained directly using the expression

eABe−A = B + [A, B] +1

2![A, [A, B]] +

1

3![A, [A, [A, B]]] + · · · , (4.41)

that is valid for any two linear operators, and that we enunciated as the Hadamard

lemma, formula 2.15 on page 39, in Chapter 2.

Using the commutator [V , n] = V (equation 4.35 on page 100), we obtain

eiωtnV e−iωtn = V + (iωt)[n, V ] +(iωt)2

2![n, [n, V ]] +

(iωt)3

3![n, [n, [n, V ]]] + · · ·

= V − (iωt)V − (iωt)2

2![n, V ]− (iωt)3

3![n, [n, V ]] + · · ·

= V − (iωt)V +(iωt)2

2!V +

(iωt)3

3![n, V ] + · · ·

= V − (iωt)V +(iωt)2

2!V − (iωt)3

3!V + · · · ,

(4.42)

so

eiωtnV e−iωtn =

∞∑k=0

(−iωt)k

k!V = e−iωtV , (4.43)

that is expression (4.39). The deduction of expression (4.40) is completely sim-

ilar.

Thus, we can rewrite the interaction picture Schrodinger-like equation, ex-

pression (4.38), as

id|φ⟩dt

= α(e−iωtV + eiωtV †)|φ⟩. (4.44)

We should stress that because V and V † commute, we can integrate this equation



with respect to t, getting

|φ⟩ = exp[− iα

∫ t

0

(e−iωτ V + eiωτ V †)dτ]|φ(0)⟩

= exp[− iαV

∫ t

0

e−iωτdτ − iαV †∫ t

0

eiωτdτ]|φ(0)⟩

= expαω

[(e−iωt − 1)V − (eiωt − 1)V †

]|φ(0)⟩.

(4.45)

As the operators in the exponential of the last expression commute, we can

factorize it and write

|φ⟩ = exp[η(t)V ] exp[−η∗(t)V †]|φ(0)⟩,

where we have defined, just for simplicity,

η(t) ≡ (α/ω)[exp(−iωt)− 1]. (4.46)

Using the definition of the exponential of an operator as a Taylor series, we have

|φ⟩ =∞∑k=0

ηk

k!V k

∞∑l=0

(−1)lη∗l

l!V †

l=

∞∑k,l=0

(−1)lηkη∗l

k!l!V kV †

l|φ(0)⟩.

We have to apply this operator to the initial condition |φ(0)⟩ as is indicated.

However, we have specified the initial condition on the |ψ⟩ function and not

in the |φ⟩ function, but as |ψ(t)⟩ = e−itωn|φ(t)⟩, it is obvious that |φ(0)⟩ =

|ψ(0)⟩ = |m⟩. Thus, with the initial condition, we have

|φ⟩ =∞∑

k,l=0

(−1)lηkη∗l

k!l!V kV †

l|m⟩.

Using that V kV †l|m⟩ = |m+ l − k⟩, we obtain

|φ⟩ =∞∑

k,l=0

(−1)lηkη∗l

k!l!|m+ l − k⟩.

To finally obtain the solution of the Schrodinger-like equation (4.32), we remem-

ber that |ψ⟩ = e−iωtn|φ⟩, so

|ψ⟩ = e−iωtn∞∑

k,l=0

(−1)lηkη∗l

k!l!|m+ l − k⟩

=

∞∑k,l=0

(−1)lηkη∗l

k!l!e−iωtn|m+ l − k⟩,



as the operator e−iωtn is linear. It is an easy exercise to show that e−iωtn|ν⟩ =e−iωtν |ν⟩, and then

|ψ⟩ =∞∑

k,l=0

(−1)lηkη∗l

k!l!e−iω(m+l−k)t|m+ l − k⟩.

Noting that eiωtη = −η∗ and that e−iωtη∗ = −η, we finally get the solution to

the Schrodinger-like equation (4.32) with the specified initial condition

|ψ⟩ = e−imωt∞∑

k,l=0

(−1)kηlη∗k

k!l!|m+ l − k⟩. (4.47)

To obtain now the solution to the original infinite system of first order dif-

ferential equations, expression (4.31), we recall that by definition En = ⟨n|ψ⟩,so

En =e−imωt∞∑

k,l=0

(−1)kηlη∗k

k!l!⟨n|m+ l − k⟩

=e−imωt∞∑

k,l=0

(−1)kηlη∗k

k!l!δm+l−k,n

=e−imωt∞∑k=0

(−1)kηn−m+kη∗k

k!(n−m+ k)!.

After some trivial algebra, this can be rewrite as

En = e−imωt

(η

|η|

)n−m ∞∑k=0

(−1)k|η|2k+n−m

k!(n−m+ k)!.

Using the series 2.64 on page 55 for the Bessel functions of the first kind, we get

finally the wanted solution to the system (4.31),

En(t) = e−imωt

(η

|η|

)n−m

Jn−m (2|η|) , (4.48)

where we remember that η(t) ≡ (α/ω)[exp(−iωt)− 1] and subject to the initial

condition En(t = 0) = δn,m; i.e.,

En(t = 0) = 0 if n = m

1 if n = m.(4.49)



The factorization that we choose in equation (4.45) took us to the solution

(4.48), that in fact is the simplest one. However, there are other possibilities that

give us complicated solutions, but that establish a generalized addition formula

for the Bessel functions; one of those possibilities is to write the exponential

operator in the equation (4.45) as

expαω

[(e−iωt − 1)

ˆV − (eiωt − 1)V †

]= exp

[αω(e−iωtV − eiωtV †)− α

ω(V − V †)

]We can factorize the exponential operator in the last equation to obtain

expαω

[(e−iωt − 1)V − (eiωt − 1)V †

]= exp

[αω(e−iωtV − eiωtV †)

]exp

[−αω(V − V †)

]or using the fact that V † = V −1,

expαω

[(e−iωt − 1)V − (eiωt − 1)V †

]= exp

[α

ω(e−iωtV − 1

e−iωtV)

]exp

[−αω(V − 1

V)

].

The two exponential operators that we have obtained are equal to those studied

in Section 1, then using again the generating function of the Bessel function, we

get

expαω

[(e−iωt − 1)V − (eiωt − 1)V †

]=

∞∑j,k=−∞

e−iωjtJj(2α/ω)Jk(−2α/ω)V j+k.

From now on, we follow the same procedure. We apply the operator to the initial

condition |m⟩, we let the operators act in that state, we do all the algebra and

finally we evaluate En as ⟨n|ψ⟩, to get

En(t) =∞∑

j=−∞e−iω(n+j)tJj(2α/ω)Jm−n−j(−2α/ω). (4.50)

So we obtain, as a by-product of this section, the following relation for the Bessel

functions of the first kind,(η

|η|

)n−m

Jn−m (2|η|) =∞∑

j=−∞eiω(m−n−j)tJj(2α/ω)Jm−n−j(−2α/ω), (4.51)



where η(t) ≡ (α/ω)[exp(−iωt)− 1].

4.4.2 Interaction ω(−1)n

Consider the Schrodinger-like equation

id|ψ⟩dt

= H|ψ⟩, (4.52)


H = ω(−1)n +α

2(V + V †), (4.53)

where α and ω are arbitrary constants, the linear operators V and V † are those

defined in the introduction of this Chapter, formulas 4.1 on page 91. Equation

(4.52) is also subject to the initial condition |ψ(t = 0)⟩ = |m⟩.The action of the linear operator (−1)n on the basis states |m⟩, can be deduced

writing −1 = eiπ; we have

(−1)n|m⟩ =eiπn|m⟩ =∞∑k=0

(iπ)k

k!(n)k|m⟩

=

∞∑k=0

(iπ)k

k!mk|m⟩ = eiπm|m⟩ = (−1)m|m⟩.

(4.54)

In fact, any function of the number operator, when acts on the base states |m⟩,is the same function substituting n by m; this is f(n)|m⟩ = f(m)|m⟩.As the set |n⟩; n = −∞, ...,∞ is complete and orthonormal, it is logical to

propose that the solution of the Schrodinger-like equation be of the form

|ψ(t)⟩ =∞∑

k=−∞

En(t)|n⟩, (4.55)


En(t) = ⟨n|ψ(t)⟩. (4.56)

If we substitute this proposition in the Schrodinger-like equation (4.52), we ob-

tain the following infinite system of first order ordinary differential equations

idEn

dt= ω(−1)nEn +

α

2(En+1 + En−1),

n = −∞, ...,∞.(4.57)



The initial condition |ψ(t = 0)⟩ = |m⟩ is translated to the initial condition

En(t = 0) = δn,m for the system (4.57). Thus, solve the Schrodinger-like equa-

tion (4.52) means to solve the system (4.57).

The formal solution to the Schrodinger-like equation is

|ψ⟩ = e−itH |m⟩,

where the Hamiltonian H is given by (4.53) and where we have already used the

fact that |ψ(0)⟩ = |m⟩.The action of the Hamiltonian in the states |m⟩ is very complicated and it is

not possible to solve directly the equation. The terms in the Hamiltonian do

not satisfy the conditions of the Baker-Hausdorff formula, then it is impossible

to factorize the exponential operator. In this case, we have to follow another

approach; we will make a change of basis, we will go from the discrete basis

|n⟩; n = −∞, ...,∞ to a continuous basis defined by

|ϕ⟩ =∞∑

k=−∞

einϕ|n⟩. (4.58)

The procedure to find the inverse transformation is outlined below

|ϕ⟩ =∞∑

k=−∞

einϕ|n⟩ ⇒

e−ikϕ|ϕ⟩ =∞∑

k=−∞

ei(n−k)ϕ|n⟩ ⇒

∫ π

−π

dϕe−ikϕ|ϕ⟩ =

∞∑k=−∞

∫ π

−π

dϕei(n−k)ϕ|n⟩,

as∫ π

−πdϕei(n−k)ϕ = 2πδn,k,we obtain

|n⟩ = 1

2π

∫ π

−π

dϕe−inϕ|ϕ⟩. (4.59)

Therefore, the solution to the Schrodinger-like equation can be written as follows

|ψ⟩ = 1

2π

∫ π

−π

dϕe−imϕe−itH |ϕ⟩. (4.60)

Thus, we have to analyze only the last part of the above equation, the action of

the operator e−itH on |ϕ⟩. We write the definition of the exponential operator,



Then we have,

H|ϕ⟩ =[ω(−1)n +

α

2(V + V †)

]|ϕ⟩

=ω|ϕ+ π⟩+ α

2(eiϕ + e−iϕ)|ϕ⟩ = ω|ϕ+ π⟩+ α cosϕ|ϕ⟩.

We analyze now the linear operator H2,

H2 =[ω(−1)n +

α

2(V + V †)

]2=ω2 +

α2

4(V + V †)2 +

αω

2

[(−1)n(V + V †) + (V + V †)(−1)n

].

(4.63)

To go on, we need the commutator of (−1)n with V and V †. In the case of V ,

we have

[(−1)n, V ]|k⟩ = (−1)nV |k⟩ − V (−1)n|k⟩

= (−1)n|k − 1⟩ − V (−1)k|k⟩

= (−1)k−1|k − 1⟩ − (−1)kV |k − 1⟩=[(−1)k−1 − (−1)k

]|k − 1⟩

= (−1)k[(−1)−1 − 1

]|k − 1⟩

= −2(−1)k|k − 1⟩

= −2(−1)kV |k⟩,

so

[(−1)n, V ] = 2(−1)nV . (4.64)

For V †,

[(−1)n, V †]|k⟩ = (−1)nV †|k⟩ − V †(−1)n|k⟩

= (−1)n|k + 1⟩ − V †(−1)k|k⟩= (−1)k+1|k + 1⟩ − (−1)k|k + 1⟩= (−1)k [−1− 1] |k + 1⟩

= −2(−1)kV †|k⟩,

so

[(−1)n, V †] = 2(−1)nV †. (4.65)



With these commutators, we get back to analyze the cross terms in H2, expres-

sion (4.63). We have

(−1)n(V + V †) + (V + V †)(−1)n = (−1)nV + V (−1)n + (−1)nV † + V †(−1)n,

but

[(−1)n, V ] = 2(−1)nV

(−1)nV = 2(−1)nV + V (−1)n

(−1)nV = −V (−1)n

and

[(−1)n, V †] = 2(−1)nV †

(−1)nV † = 2(−1)nV † + V †(−1)n

(−1)nV † = −V †(−1)n,

so

(−1)n(V + V †) + (V + V †)(−1)n =

= (−1)nV + V (−1)n + (−1)nV † + V †(−1)n

= −V (−1)n + V (−1)n − V †(−1)n + V †(−1)n

= 0.

We conclude that the crossed terms in the Hamiltonian are zero, and then square

of the Hamiltonian is simply

H2 = ω2 +α2

4(V + V †)2. (4.66)

then the action on the functions |ϕ⟩ is

H2|ϕ⟩ =[ω2 +

α2

4(V + V †)2

]|ϕ⟩ =

[ω2 + α2 cos2 ϕ

]|ϕ⟩. (4.67)

We also need the action of H2 on the function |ϕ + π⟩. The action of V on

|ϕ+ π⟩ is

V |ϕ+ π⟩ =V∞∑

k=−∞

eik(ϕ+π)|k⟩ =∞∑

k=−∞

eik(ϕ+π)V |k⟩ =∞∑

k=−∞

eik(ϕ+π)|k − 1⟩

=∞∑

k′=−∞

ei(k′+1)(ϕ+π)|k′⟩ = −eiϕ

∞∑k′=−∞

eik′(ϕ+π)|k′⟩ = −eiϕ|ϕ+ π⟩,



With all these results, we get back to equation (4.61), that we reproduce in the

first line, to obtain

e−itH |ϕ⟩ =∞∑k=0

(−it)2k

(2k)!H2

k|ϕ⟩+

∞∑k=0

(−it)2k+1

(2k + 1)!H2

kH|ϕ⟩

=

∞∑k=0

(−it)2k

(2k)!Ω2k|ϕ⟩ − α cosϕ

∞∑k=0

(−it)2k+1

(2k + 1)!Ω2k|ϕ⟩

+ ω∞∑k=0

(−it)2k+1

(2k + 1)!Ω2k|ϕ+ π⟩

=∞∑k=0

(−1)kt2k

(2k)!Ω2k|ϕ⟩+ i

α cosϕ

Ω

∞∑k=0

(−1)kt2k+1

(2k + 1)!Ω2k+1|ϕ⟩

− iω

Ω

∞∑k=0

(−1)kt2k+1

(2k + 1)!Ω2k+1|ϕ+ π⟩

=cos(Ωt)|ϕ⟩+ iα cosϕsin(Ωt)

Ω|ϕ⟩ − iω

sin(Ωt)

Ω|ϕ+ π⟩.

Using now expression (4.60), we get

|ψ⟩ = 1

2π

∫ π

−π

dϕe−imϕe−itH |ϕ⟩

=1

2π

∫ π

−π

dϕe−imϕ

[cos(Ωt)|ϕ⟩+ iα cosϕ

sin(Ωt)

Ω|ϕ⟩ − iω

sin(Ωt)

Ω|ϕ+ π⟩

].

Changing variables in the integral corresponding to the last term, we obtain the

final solution to the Schrodinger-like equation (4.53),

|ψ(t)⟩ = 1

2π

∫ π

−π

dϕ

e−imϕ

[cos(Ωt) + iα cosϕ

sin(Ωt)

Ω

]− iωeimϕ sin(Ωt)

Ω

|ϕ⟩.

(4.73)

To obtain the solution to the infinite system of differential equations, we

recall that by definition En = ⟨n|ψ⟩, so

En(t) =⟨n|ψ(t)⟩

=1

2π

∫ π

−π

dϕ

e−imϕ


sin(Ωt)

Ω

]− iωeimϕ sin(Ωt)

Ω

⟨n|ϕ⟩.

It is very easy to prove that ⟨n|ϕ⟩ = einϕ, and substituting this in the above



expression, we get the final answer to our problem,

En(t) =

=1

2π

∫ π

−π

dϕ

ei(n−m)ϕ


sin(Ωt)

Ω

]− iωei(n+m)ϕ sin(Ωt)

Ω

.

(4.74)

In particular, if the initial state is the vacuum state |m⟩ = |0⟩, the result is

reduced to

En(t) =1

2π

∫ π

−π

dϕeinϕ[

cos(Ωt) + iα cosϕsin(Ωt)

Ω

]− iω

sin(Ωt)

Ω

. (4.75)


Chapter 5

Semi-infinite systems of differentialequations

5.1 First a semi-infinite system

Consider a semi-infinite system of coupled first-order differential equations de-

scribed by

idE0

dt= αE1,

idEn

dt= α

(√n+ 1 En+1 +

√n En−1

), n = 1, 2, 3, ..., (5.1)

where α is an arbitrary constant.

Now consider the following Schrodinger-like equation

idψ(x, t)

dt= α(A+A†) ψ(x, t). (5.2)

The linear operators A and A† are the down and up operators defined in the

Section 2.1 of the Hermite polynomials in Chapter 2; for our purposes, only its

action

A|n⟩ =√n|n− 1⟩ (5.3)

and

A†|n⟩ =√n+ 1|n+ 1⟩ (5.4)

on the basis |n⟩, n = 0, 1, 2, ... is needed.

We shall show first, that the Schrodinger-like equation (5.2) is equivalent to

the system (5.1). After, we will solve the Schrodinger-like equation, and translate

115


116 Semi-infinite systems of differential equations

that solution to the solution of the original system (5.1).

We have to remember that the set of functions |n⟩;n = 0, 1, 2, ... constitutes a

complete orthonormal set for the space of square integrable functions; then, we

can expand any function in that space as

|f⟩ =∞∑

n=0

cn|n⟩, (5.5)

where

cn = ⟨n|f⟩. (5.6)

From (5.5), it is clear that we can write the solution of the Schrodinger-like

equation (5.2), in terms of |n⟩, as

|ψ(x, t)⟩ =∞∑

n=0

En(t)|n⟩, (5.7)

and the coefficients En(t) in this expansion are given by equation (5.6),

En(t) = ⟨n|ψ(x, t)⟩. (5.8)

We substitute (5.7) in the Schrodinger-like equation (5.2), and obtain

i∞∑

n=0

dEn

dt|n⟩ = α(A+A†)

∞∑n=0

En|n⟩ = α∞∑

n=0

En(A+A†)|n⟩

Using now the action of the operators A and A†, equations (5.3) and (5.4), we

get (note that the first sum in the right hand starts in 1),

i∞∑

n=0

dEn

dt|n⟩ = α

∞∑n=1

En

√n|n− 1⟩+ α

∞∑n=0

En

√n+ 1|n+ 1⟩,

Changing the indexes of the sums, we find

i∞∑

n=0

dEn

dt|n⟩ = α

∞∑n=0

En+1

√n+ 1|n⟩+ α

∞∑n=1

En−1

√n|n⟩,

and as the set |n⟩;n = 0, 1, 2, ... is linearly independent, we arrive to our

original semi-infinite system

idE0

dt= αE1,

idEn

dt= α

(√n+ 1En+1 +

√nEn−1

), n = 1, 2, 3, ...,


First a semi-infinite system 117

as we wanted to show.

We proceed now to solve the Schrodinger-like equation (5.2). The formal

solution to that equation is

|ψ(x, t)⟩ = e−iαt(A+A†)|ψ(0)⟩, (5.9)

where |ψ(0)⟩ represents the initial condition.

As the commutator of A and A† is 1; i.e.,

[A,A†]ψn(x) ≡ (AA† −A†A)ψn(x) = ψn(x), (5.10)

we can use the

Baker-Hausdorff formula: Given two operators A and B,

that obey

[[A,B] , A] = [[A,B] , B] = 0,

then

eA+B = e−12 [A,B]eAeB ,

to factorize the solution (5.9) as

|ψ(x, t)⟩ = e−α2 t2

2 e−iαtA†e−iαtA|ψ(0)⟩,

and then, from (5.8), we obtain

En = ⟨n|ψ(x, t)⟩ = e−α2t2

2 ⟨n|e−iαtA†e−iαtA|ψ(0)⟩.

The initial condition |ψ(0)⟩ may be, in general, an arbitrary superposition of

states; so, we can consider, without losing generality, |ψ(0)⟩ = |m⟩. In this last

case

En = ⟨n|ψ(x, t)⟩ = e−α2t2

2 ⟨n|e−iαtA†e−iαtA|m⟩.

We can expand the exponentials in Taylor series, obtaining

En = e−α2t2

2

∞∑j,k=0

(−iαt)j+k

j!k!⟨n|A†jAk|m⟩.



From equations (5.3) and (5.4), we have

Ak|m⟩ =

√m!

(m− k)!|m− k⟩

and

A†j |m− k⟩ =

√(m− k + j)!

(m− k)!|m− k + j⟩,

so

En = e−α2t2

2

∞∑j,k=0

(−iαt)j+k

√m!(m− k + j)!

j!k!(m− k)!⟨n|m− k + j⟩.

As the set |n⟩;n = 0, 1, 2, ... is orthonormal, ⟨n|m − k + j⟩ = δn,m−k+j , and

the last expression reduces to

En = e−α2t2

2

∞∑j=0

(−iαt)m−n+2j

√m!n!

j!(m− n+ j)!(n− j)!.

We first consider the case when n = m+ s with s = 0, 1, 2, ..., getting

Em+s = e−α2t2

2

√m!

(m+ s)!

∞∑j=0

(−iαt)2j−s (m+ s)!

j!(j − s)!(m+ s− j)!. (5.11)

We use now the fact that limz→n1

Γ(−z) = 1(−n−1)! = 0 with n = 0, 1, 2, ...

[Abramowitz 72] , to rewrite (5.11) as the following finite sum,

Em+s = e−α2t2

2

√m!

(m+ s)!

m+s∑j=s

(−iαt)2j−s (m+ s)!

j!(j − s)!(m+ s− j)!,

and changing the index of the sum, we get

Em+s = e−α2t2

2

√m!

(m+ s)!

m∑j=0

(−iαt)2j+s (m+ s)!

j!(j + s)!(m− j)!. (5.12)

Getting back again to Chapter 2, this time to the associated Laguerre polyno-

mials section (Section 2.2), we have that

Lηr(x) =

r∑k=0

(r + η

r − k

)(−1)k

xk

k!,


Semi-infinite system with first neighbors interaction 119

and using it in equation (5.12) allow us to write

Em+s = exp

(−α

2t2

2

)(−iαt)s

√m!

(m+ s)!Lsm(α2t2). (5.13)

When n = m− s with s = 0, 1, 2..., we obtain, by mean of an identical process,

Em−s = exp

(−α

2t2

2

)(−iαt)s

√(m− s)!

m!Lsm−s(α

2t2). (5.14)

In case the initial state is |0⟩, what means that m = 0, we find

En = exp

(−α

2t2

2

)(−iαt)n√

n!. (5.15)

In several applications, En represents an amplitude, so its module squared is

interesting. In this case we get

P = EnE∗n = exp(−α2t2

) (α2t2)n

n!. (5.16)

i.e. a Poisson distribution with mean α2t2.

5.2 Semi-infinite system with first neighbors interaction

We define the Suskind-Glogower (SG) linear operators

V ≡∞∑

n=0

|n⟩ ⟨n+ 1| (5.17)

and

V † ≡∞∑

n=0

|n+ 1⟩ ⟨n| (5.18)

where |n⟩ are vectors of the orthonormal basis |n⟩, n = 0, 1, 2, ....The action of these two operators on the vectors |n⟩, is

V |n⟩ = |n− 1⟩ (5.19)

and

V † |n⟩ = |n+ 1⟩ . (5.20)

It is very important to make explicit the result

V |0⟩ = 0,



that comes naturally from (5.17). The SG operators possess a non-commuting

and non-unitary nature, that resides in the expressions

V V † = 1,

and

V †V = 1− |0⟩ ⟨0| .

Indeed, we have

[V, V †] = |0⟩⟨0|,

and

V † = V −1.

Using the SG operators, we build the Hamiltonian

H = η(V + V †) , (5.21)

where η is the coupling coefficient, and we consider the Schrodinger-like equation

id|ψ(t)⟩dt

= H|ψ(t)⟩ (5.22)

associated with the Hamiltonian and with the initial condition |ψ(0)⟩ = |m⟩.To find the system of first order ordinary differential equations generated by this

Schrodinger-like equation, we propose that

|ψ(t)⟩ =∞∑

n=0

En(t)|n⟩. (5.23)

Substituting this proposition in the Schrodinger-like equation, and introducing

the explicit form of the Hamiltonian and its action on the basis functions |n⟩,



given by expressions (5.19) and (5.20), we obtain

i∞∑

n=0

dEn(t)

dt|n⟩ =

[η(V + V †)] ∞∑

n=0

En(t)|n⟩

=∞∑

n=0

[η(V + V †)]En(t)|n⟩

=η∞∑

n=1

En(t)|n− 1⟩+ η∞∑

n=0

En(t)|n+ 1⟩

=η

∞∑n=0

En+1(t)|n⟩+ η

∞∑n=1

En−1(t)|n⟩.

As the set |n⟩ n = 0, 1, 2, ... is linearly independent, we arrive to the following

coupled system of differential equations

idE0

dt= ηE1,

idEn

dt= η (En+1 + En−1) , n = 1, 2, 3, ..., (5.24)

with the initial condition |n⟩ = δn,m.

We proceed then to solve the Schrodinger-like equation (5.22) with the initial

condition |ψ(0)⟩ = |m⟩. We will use in this case a new basis, the basis formed

by the eigenfunctions of the Hamiltonian. That means, that we have to solve

the eigenvalue problem

H|φ⟩ = E|φ⟩. (5.25)

We begin writing the eigenfunctions in terms of the basis |n⟩; n = 0, 1, 2, ...,as

|φ⟩ =∞∑k=0

ck|k⟩, (5.26)



then

H|φ⟩ =∞∑k=0

ckH|k⟩ = η

∞∑k=0

ck(V + V †)|k⟩

=η∞∑k=1

ck|k − 1⟩+ η∞∑k=0

ck|k + 1⟩

=η∞∑k=0

ck+1|k⟩+ η∞∑k=1

ck−1|k⟩

=ηc1|0⟩+ η

∞∑k=1

(ck+1 + ck−1)|k⟩

,

and substituting in the stationary Schrodinger-like equation (5.25), we arrive to

ηc1|0⟩+ η∞∑k=1

(ck+1 + ck−1)|k⟩ = E∞∑k=0

ck|k⟩.

Therefore, the coefficients ck, k = 0, 1, 2, ... satisfy the following recurrence

equations,

c1 =E

ηc0,

η(ck+1 + ck−1) = Eck, k = 1, 2, 3, ...

These are the recurrence relations of the Chebyshev polynomials of the second

kind that we studied in Section 2.3 of Chapter 2 (see also [Arfken 05; Gradshtein

99; Abramowitz 72]), and we can write

ck = Uk(ξ), k = 0, 1, 2, ..., (5.27)

where

ξ ≡ E

2η. (5.28)

Hence, the eigenfunctions of the Hamiltonian (5.21) are

|φξ⟩ =∞∑k=0

Uk(ξ)|k⟩. (5.29)

With this knowledge, we can go back to the Schrodinger-like equation (5.22),

and write the formal solution as

|ψ⟩ = e−itH |m⟩. (5.30)



From this, it is clear that we need to write now the initial condition in terms

of the basis constituted by the eigenfunctions of the Hamiltonian. For that,

we have to invert the relation |φξ⟩ =∑∞

k=0 Uk(ξ)|k⟩, with the following direct

procedure

|φξ⟩ =∞∑k=0

Uk(ξ)|k⟩ ⇒

Un(ξ)|φξ⟩ =

∞∑k=0

Un(ξ)Uk(ξ)|k⟩ ⇒

√1− ξ2Un(ξ)|φξ⟩ =

∞∑k=0

√1− ξ2Un(ξ)Uk(ξ)|k⟩ ⇒∫ 1

−1

dξ√1− ξ2Un(ξ)|φξ⟩ =

∞∑k=0

∫ 1

−1

dξ√1− ξ2Un(ξ)Uk(ξ)|k⟩,

but [Arfken 05; Gradshtein 99]∫ 1

−1

dξ√

1− ξ2Un(ξ)Uk(ξ) =π

2δn,k,

and then

|n⟩ = 2

π

∫ 1

−1

dξ√1− ξ2Un(ξ)|φξ⟩, (5.31)

that express the elements of the basis |n⟩; n = 0, 1, 2, ... in terms of the basis

of eigenfunctions of the Hamiltonian.

With all these results, we can write the solution to the Schrodinger-like equation,

equation (5.30), as

|ψ⟩ = 2

πe−itH

∫ 1

−1

dξ√1− ξ2Um(ξ)|φξ⟩, (5.32)

and using the linearity property of the e−itH operator,

|ψ⟩ = 2

π

∫ 1

−1

dξ√1− ξ2Um(ξ)e−itH |φξ⟩.

We analyze now the action of e−itH on the eigenfunctions of the Hamiltonian

e−itH |φξ⟩ =∞∑k=0

(−it)k

k!Hk|φξ⟩,



but as H|ϕ⟩ = E|ϕ⟩

e−itH |φξ⟩ =∞∑k=0

(−it)k

k!Ek|φξ⟩ = e−itE |φξ⟩,

so

|ψ⟩ = 2

π

∫ 1

−1

dξe−iτξ√1− ξ2Um(ξ)|φξ⟩,

where

τ = 2ηt.

Rewriting |φξ⟩ in terms of the basis |n⟩; n = 0, 1, 2, ..., we obtain

|ψ(t)⟩ = 2

π

∫ 1

−1

dξe−iτξ√1− ξ2Um(ξ)

∞∑k=0

Uk(ξ)|k⟩,

or finally the solution to the Schrodinger-like equation satisfying the initial con-

ditions

|ψ(t)⟩ = 2

π

∞∑k=0

∫ 1

−1

dξe−iτξ√1− ξ2Um(ξ)Uk(ξ)|k⟩. (5.33)

In the particular case, when the initial state is the vacuum state, or in other

words m = 0, the above expression reduces to

|ψ(t)⟩ = 2

π

∞∑k=0

∫ 1

−1

dξe−iτξ√1− ξ2U0(ξ)Uk(ξ)|k⟩,

but U0(x) = 1 for all x, so

|ψ(t)⟩ = 2

π

∞∑k=0

∫ 1

−1

dξe−iτξ√1− ξ2Uk(ξ)|k⟩. (5.34)

The last integral can be done using the formula that we exposed in Chapter 2,

Section 2.4, [Campbell 48]

FJn (ω)

ω

=

√2

π

i

n(−i)n Un−1 (ξ)

√1− ξ2rect

(ξ

2

), (5.35)

where F is the Fourier transform and

rect

(ξ

2

)=

1 , −1 ≤ ξ ≤ 1

0 , otherwise.(5.36)



From (5.35), we write

Jn(ω)

ω=

(−i)n−1

nπ

∫ 1

−1

dξeiωξ√1− ξ2Un−1(ξ).

and substituting it in (5.34), we get

|ψ(t)⟩ = 2∞∑k=0

ik(k + 1)Jk+1(−τ)

−τ|k⟩.

Writing explicitly τ ,

|ψ(t)⟩ = −1

η

∞∑k=0

ik(k + 1)Jk+1(−2ηt)|k⟩,

and using the parity properties of the Bessel functions [Arfken 05; Abramowitz

72]

|ψ(t)⟩ = 1

ηt

∞∑k=0

ik(k + 1)Jk+1(2ηt)|k⟩. (5.37)

To finish this section, we have to calculate the amplitudes En, n = 0, 1, 2, ...,

that satisfy the system (5.24). For that, we remember that En = ⟨n|ψ⟩, thus

En(t) = ⟨n|ψ(t)⟩ =2

π

∞∑k=0

∫ 1

−1

dξe−iτξ√1− ξ2Um(ξ)Uk(ξ)⟨n|k⟩

=2

π

∞∑k=0

∫ 1

−1

dξe−iτξ√1− ξ2Um(ξ)Uk(ξ)δn,k,

or

En(t) =2

π

∫ 1

−1

dξe−iτξ√1− ξ2Um(ξ)Un(ξ). (5.38)

When the initial state is the vacuum state, we obtain

En =1

ηtin(n+ 1)Jn+1(2ηt). (5.39)



5.3 Nonlinear system with first neighbors interaction

We will study in this last section of this Chapter the following coupled system

of first order differential equations

idE0

dt+ f(1)E1 = 0,

idEn

dt+ f(n+ 1)En+1 + f(n)En−1 = 0, n = 1, 2, 3, ..., (5.40)

where by definition

f(n) ≡√n+ xn2

x, (5.41)

and with the initial condition En = δn,0.

Consider now the Schrodinger-like equation

id|ψ(t)⟩dt

= −(W + W †)|ψ(t)⟩, (5.42)

with the initial condition |ψ(t)⟩ = |0⟩. The nonlinear down and up operators

are respectively

W ≡√

1 + x

xf(n+ 1)A (5.43)

and

W † ≡√

1 + x

xf(n)A†, (5.44)

where the linear operators A and A† are those introduced in the Section 1 of

Chapter 2, and whose action on the states of our basis |n⟩; n = 0, 1, 2, .. is

given by

A|n⟩ =√n|n− 1⟩ (5.45)

and

A†|n⟩ =√n+ 1|n+ 1⟩. (5.46)

We recall that the operator n is the number operator, given by n = A†A.

As the number operator will appear frequently, in all this section we will add a

”hat” to all operators to avoid confusion.


Nonlinear system with first neighbors interaction 127

The solution of the Schrodinger-like equation can be expanded in terms of

the complete orthogonal set |n⟩, n = 0, 1, 2, ... as

|ψ(x, t)⟩ =∞∑k=0

En(t)|n⟩, (5.47)

and substituting it in (5.42), we get

i

∞∑n=0

dEn(t)

dt|n⟩ = −

∞∑n=0

[f(n)En(t)|n− 1⟩+ f(n+ 1)En(t)|n+ 1⟩] ,

from which follows the system (5.40).

The formal solution of the Schrodinger-like equation (5.42) is

|ψ(t)⟩ = exp[it(W + W †

)]|ψ(0)⟩. (5.48)

Unlike the operators A and A†, operators W and W † obey a more complicated

commutation relation. We have

[W , W †] =

(1 + x

x2

)[(n+ 1) + x(n+ 1)3 − n2 − xn3

], (5.49)

thus direct use of the Baker–Hausdorff formula is not allowed in factorizing the

exponential in expression (5.48). Instead another way should be pursued. To do

that, we define a new operator W0 as

W0 ≡ n+1

2x+

1

2, (5.50)

such that, jointly these three operators,W , W †, W0

, provide an operator

algebra where the commutation relations between them are closed, and are given

by

[W0, W†] = W †, [W0, W ] = −W , [W , W †] = 2W0. (5.51)

Notice that these operators form an SU(1, 1) group and therefore the exponential

in equation (5.48) may be factorized as follows [Wei 63]

exp[it(W + W †

)]= exp

[iF (t) W †

]exp

[G (t) W0

]exp

[iH (t) W

]. (5.52)



To find the functions F (t), G (t) and H (t) , we take the derivative with respect

to t in the above equation, obtaining

i(W + W †

)exp

[it(W + W †

)]=idF

dtW † exp

(iFW †

)exp

(GW0

)exp

(iHW

)+dG

dtexp

(iFW †

)W0 exp

(GW0

)exp

(iHW

)+ i

dH

dtexp

(iFW †

)exp

(GW0

)W exp

(iHW

).

Multiplying both sides of this equation by

exp(−iHW

)exp

(−GW0

)exp

(−iFW †

)and canceling similar terms, we get

i(W + W †

)= i

dF

dtW † +

dG

dtexp

(iFW †

)W0 exp

(−iFW †

)+ i

dH

dtexp

(iFW †

)exp

(GW0

)W exp

(−GW0

)exp

(−iFW †

).

(5.53)

We work now the term exp(iFW †

)W0 exp

(−iFW †

). For that we use the

Hadamard lemma formula, that establishes that

eT Re−T = R+[T , R

]+

1

2!

[T ,[T , R

]]+

1

3!

[T ,[T ,[T , R

]]]+ ... (5.54)

Thus,

exp(iFW †

)W0 exp

(−iFW †

)== W0 +

[iFW †, W0

]+

1

2!

[iFW †,

[iFW †, W0

]]+

1

3!

[iFW †,

[iFW †,

[iFW †, W0

]]]+ ...,

but we already know that[W †, W0

]= −W †, therefore

exp(iFW †

)W0 exp

(−iFW †

)= W0 − iFW †. (5.55)

Now we analyze the term exp(GW0

)W exp

(−GW0

), using again the Hadamard

formula. We have

exp(GW0

)W exp

(−GW0

)=

= W +[GW0, W

]+

1

2!

[GW0,

[GW0, W

]]+

1

3!

[GW0,

[GW0,

[GW0, W

]]]+ ...,



but[W0, W

]= −W , then

exp(GW0

)W exp

(−GW0

)=

= W −GW −G1

2!

[GW0, W

]−G

1

3!

[GW0,

[GW0, W

]]+ ...

= W −GW +G2 1

2!W +G2 1

3!

[GW0, W

]+ ...

= W −GW +G2 1

2!W −G3 1

3!W + ...

= W∞∑k=0

(−1)kGk

k!= W exp (−G) ,

and finally

exp(GW0

)W exp

(−GW0

)= exp (−G) W . (5.56)

The third term in equation (5.53) is then

exp(iFW †

)exp

(GW0

)W exp

(−GW0

)exp

(−iFW †

)= exp (−G) exp

(iFW †

)W exp

(−iFW †

),

therefore we have to reduce the term

exp(iFW †

)W exp

(−iFW †

).

For that we use once more the Hadamard lemma formula (5.54), to get

exp(iFW †

)W exp

(−iFW †

)=

= W +[iFW †, W

]+

1

2!

[iFW †,

[iFW †, W

]]+

1

3!

[iFW †,

[iFW †,

[iFW †, W

]]]+ ...,

but[W †, W

]= −2W0, hence

exp(iFW †

)W exp

(−iFW †

)=

= W − 2iFW0 − 2[iF ]

2

2!

[W †, W0

]+ 2

[iF ]3

3!

[W †,

[W †, W0

]]+ ...,



and we use use now that[W †, W0

]= −W †, so finally

exp(iFW †

)W exp

(−iFW †

)= W − 2iFW0 − F 2W †. (5.57)

Substituting (5.57), (5.56) and (5.55) in expression (5.53), we get

i(W + W †

)=

= idF

dtW † +

dG

dt

(W0 − iFW †

)+ i

dH

dtexp (−G)

(W − 2iFW0 − F 2W †

).

Therefore, we obtain the following first order ordinary differential equations for

the functions F (t), G (t) and H (t),

exp (−G) dHdt

= 1,

dF

dt− F

dG

dt− exp (−G)F 2 dH

dt= 1,

dG

dt+ 2 exp (−G)F dH

dt= 0.

A word has to be said about initial conditions of this system. From formula

(5.52) it is clear that for t = 0 the exponential operator reduces to the identity,

so we need F (0) = 0, G (0) = 0 and H (0) = 0. With these initial conditions

the system is easily solved obtaining

F (t) = H (t) = tanh (t) and G (t) = −2 ln (cosh t) . (5.58)

In this case the solution of the Schrodinger-like equation (5.48) takes the form

|ψ (t)⟩ = exp[i tanh (t) W †

]exp

[−2 ln (cosh t) W0

]exp

(i tanh tW

)|ψ (0)⟩ .

(5.59)

We have to impose now the initial condition |ψ (0)⟩ = |0⟩. We analyze the action

of the exponential operators in the above equation on the base state. First, as

W |n⟩ = 0, we have

exp(i tanh tW

)|0⟩ =

∞∑k=0

(i tanh t)k

k!W k |0⟩ = |0⟩ .

Second, we get

exp[−2 ln (cosh t) W0

]|0⟩ =

∞∑k=0

[−2 ln (cosh t)]k

k!W k

0 |0⟩ ,



Summarizing

exp(i tanh tW †

)|0⟩ =

∞∑k=0

αk

√k!

[f (k)]! |k⟩ ,

where

α ≡ i

√1 + x

xtanh t,

and by definition

[f (k)]! ≡ f (n) f (n− 1) f (n− 2) ...f (1) .

The final solution of the Schrodinger-like equation, with the specified initial

condition, is

|ψ (t)⟩ = (cosh t)−(1+x)/x

∞∑k=0

αk

√k!

[f (k)]! |k⟩ (5.60)

To find the solution to the original system (5.40), we have to calculate ⟨n|ψ⟩,obtaining

En (t) = (cosh t)−(1+x)/x αn

√n!

[f (n)]! |n⟩ . (5.61)


Chapter 6

Partial differential equations

In this Chapter we will study partial differential equations by applying operator

techniques.

6.1 A simple partial differential equation

In order to introduce the operational method, let us first look a very simple

equation

i∂ψ(x, z)

∂z= −∂

2ψ(x, z)

∂x2. (6.1)

Like in Chapter 2, we use definition ( 2.12 on page 39) p ≡ −i ddx , such that we

rewrite the above equation as

i∂ψ(x, z)

∂z= p2ψ(x, z),

that has the simple solution

ψ(x, z) = exp(−izp2

)ψ(x, 0), (6.2)

where ψ(x, 0) is the boundary condition.

6.1.1 A Gaussian function as boundary condition

Consider the Gaussian boundary condition

ψ(x, 0) = Ne−x2

, (6.3)

133


134 Partial differential equations

where N is a normalization constant. From expression (6.2), we have

ψ(x, z) = N exp(−izp2

)e−x2

= N

∞∑n=0

(iz)n

n!

d2n

dx2ne−x2

,

that by using Rodrigues formula for the Hermite polynomials, ( 2.2 on page 37),

may be rewritten as

ψ(x, z) = ψ(x, 0)

∞∑n=0

(iz)n

n!H2n(x),

and by using the series of even Hermite polynomials ( 2.34 on page 48)

ψ(x, z) =N√

1 + 4izexp

(− x2

1 + 4iz

). (6.4)

The reader can verify easily that this function is in fact a solution of equation

(6.1) that satisfies the boundary condition (6.3).

6.1.2 An arbitrary function as boundary condition

Consider now as boundary condition an arbitrary function that may be written

as the Fourier transform

ψ(x, 0) =

∫ ∞

−∞eixuf(u)du.

From formula (6.2) this implies that

ψ(x, z) = exp(−izp2

) ∫ ∞

−∞eixuf(u)du,

and again developing in Taylor series

ψ(x, z) =∞∑

m=0

1

m!(iz)

m∫ ∞

−∞

d2m

dx2meixuf(u)du.

Note that

d2m

dx2meixu =

(−u2

)meixu,

so we have the final solution

ψ(x, z) =

∫ ∞

−∞

∞∑m=0

(−iz)m

m!u2meixuf(u)du

=

∫ ∞

−∞e−izu2

eixuf(u)du, (6.5)


Airy system 135

that it is known as the Fresnel transform of the function f(u).

6.2 Airy system

We now complicate a little the partial differential equation by adding a linear

term

i∂ψ

∂z= −∂

2ψ

∂x2+ kxψ, (6.6)

where k is an arbitrary constant.

By using operator forms again, we rewrite it as

i∂ψ

∂z= p2ψ + kxψ, (6.7)

with the formal solution

ψ(x, z) = e−i(p2+kx)zψ(x, 0).

We identify the operators

A = −ip2z, B = −ikxz,

with the commutator given by

[A,B] = −z2k[p2, x

]= 2iz2kp. (6.8)

Because the operator given in equation (6.8) commutes with A but not with

B, we can not apply the Baker-Hausdorff formula to obtain a factorized form.

However, we can propose the anszats

ψ(x, z) = ef(z)p2

eg(z)xeh(z)pem(z)ψ(x, 0). (6.9)

By deriving with respect to z, a system of first order of linear differential equa-

tions may be obtained for f , g, h and m.

Here we would like to offer a different way to obtain the solution: sometimes it

is better to transform the differential equation to obtain an easier one. We do

ψ(x, z) = eiαp3

3 ϕ(x, z), (6.10)

where α is a parameter that we will fix later according to our convenience.

The boundary condition is transformed as

ϕ(x, 0) = e−iα p3

3 ψ(x, 0). (6.11)



By plugging (6.10) in (6.7), we have

ieiαp3

3∂ϕ(x, z)

∂z=(p2 + kx

)eiα

p3

3 ϕ(x, z),

and multiplying on the left by e−iα p3

3 we obtain

i∂ϕ(x, z)

∂z= e−iα p3

3

(p2 + kx

)eiα

p3

3 ϕ(x, z)

=

(p2 + ke−iα p3

3 xeiαp3

3

)ϕ(x, z) (6.12)

To simplify the term e−iα p3

3 xeiαp3

3 we use the Hadamard lemma ( 2.15 on

page 39), that establish that given two linear operators T and S then

eξTSe−ξT = S + ξ [T, S] +ξ2

2![T, [T, S]] +

ξ3

3![T, [T, [T, S]]] + · · · , (6.13)

where [T, S] ≡ TS − ST is the commutator of operators T and S.

As we already know [p, x] = −i, and it is an easy exercise to calculate that

[p3, x] = −3ip2, in such a way that all other commutators are zero, and we get

e−iαp3/3xeiαp3/3 = x− αp2.

We can rewrite equation (6.12) as

i∂ϕ(x, z)

∂z=[p2 + k

(x− αp2

)]ϕ(x, z),

and by choosing α = 1k , we obtain a very simple partial differential equation

i∂ϕ(x, z)

∂z= kxϕ(x, z)

with solution

ϕ (x, z) = e−ikxzϕ(x, 0).

Substituting the boundary condition (6.11), we have

ϕ (x, z) = e−ikxze−iα p3

3 ψ(x, 0) (6.14)

and finally from (6.14) and (6.10), we obtain the solution

ψ (x, z) = eip3

3k e−ikxze−i p3

3kψ(x, 0).


Airy system 137

To proceed further, we insert 1, written as eikxze−ikxz, in the previous equation

to arrive to

ψ (x, z) = eip3

3k e−ikxze−i p3

3k eikxze−ikxzψ(x, 0).

The term e−ikxze−i p3

3k eikxz can be reduced to e−i(p+kz)3

3k , writing the p exponen-

tial in Taylor series, writing one as eikxze−ikxz between the powers of p, and using

the fact, derived again from the Hadamard lemma, that e−ikxzpeikxz = p+ kz.

Hence,

ψ (x, z) = eip3

3k e−i(p+kz)3

3k e−ikxzψ(x, 0).

As the operators in the two first exponentials commute, we obtain

ψ (x, z) = e−i3k (3p

2kz+3pk2z2+k3z3)e−ikxzψ(x, 0),

and as all the operators in the first exponential commute, we can factorize them

and get the final solution of partial differential equation (6.6) as

ψ (x, z) = e−i k2z3

3 e−ip2ze−ikz2pe−ikxzψ(x, 0). (6.15)

By comparing equations (6.15) and (6.9), we see that we have found the forms

of f , g, h and m.

6.2.1 Airy function as boundary condition

Let us take as boundary condition an Airy function [Beals 10]; i.e.,

ψ(x, 0) = Ai(ϵx) =1

2π

∫ ∞

−∞e−i( t3

3 +tϵx)dt, (6.16)

where ϵ is an arbitrary constant. Therefore, we obtain using (6.15)

ψ (x, z) = e−i k2z3

3 e−ip2ze−ikz2pe−ikxz

∫ ∞

−∞e−i( t3

3 +tϵx)dt,

that, as we know how to evaluate derivatives of exponentials that depend lin-

early on x, and by manipulating the integrand, we can recognize again the Airy

function but with a different argument

ψ(x, z) = e−iη(x,z)Ai[ϵx+ ϵz2(k − ϵ3)], (6.17)

where η(x, z) is the following real function of x and z,

η(x, z) =(k − ϵ3

) [xz +

z3

3

(k − 2ϵ3

)]. (6.18)



6.3 Harmonic oscillator system

We continue introducing difficulties in the partial differential equation (6.1), now

by adding a parabolic function

i∂ψ(x, z)

∂z= −1

2

∂2ψ(x, z)

∂x2+ω2x2

2ψ(x, z)

=

(p2

2+ω2x2

2

)ψ(x, z). (6.19)

This equation may be associated to the quantum harmonic oscillator. We can

solve it via a procedure similar to the one used to obtain the series of even

Hermite polynomials, because the operators involved are the same, or we can

introduce ”ladder” operators

Aω =

√ω

2x+ i

p√2ω, A†

ω =

√ω

2x− i

p√2ω, (6.20)

with commutation relations

[Aω, A†ω] = 1, (6.21)

such that we can rewrite (6.19) as

i∂ψ(x, z)

∂z= ω

(A†

ωAω +1

2

)ψ(x, z). (6.22)

The above equation has as solution

ψ(x, z) = e−izω(A†ωAω+ 1

2 )ψ(x, 0), (6.23)

and, as seen in equation ( 2.9 on page 39), we can expand any function of x in

terms of Hermite-Gauss functions, so

ψ(x, 0) =∞∑

n=0

cnψ(ω)n (x), (6.24)

where we have indexed the function ψ(ω)n (x) to note that it has to be properly

written in terms of ω; i.e.,

ψ(ω)n (x) =

ω1/4

π1/4√2nn!

e−ωx2/2Hn(√ωx). (6.25)

The constants cn in expansion (6.24) have to be determined once the explicit

boundary condition ψ(x, 0) is given.


z-dependent harmonic oscillator 139

By plugging (6.24) into (6.23), and remembering that (A†ωAω + 1/2)ψ

(ω)n (x) =

(n+ 1/2)ψ(ω)n (x), we finally obtain the solution to equation (6.19) as

ψ(x, z) =∞∑

n=0

cne−iωz(n+1/2)ψ(ω)

n (x). (6.26)

Remark that the boundary condition is trivially satisfied.

6.4 z-dependent harmonic oscillator

We now consider the frequency of the harmonic oscillator to vary in z. In other

words, we propose the second order partial differential equation,

i∂ψ(x, z)

∂z=

[−1

2

∂2

∂x2+ω2(z)x2

2

]ψ(x, z)

=

[p2

2+ω2(z)x2

2

]ψ(x, z)

(6.27)

where ω(z) is an arbitrary function of z.

We apply the method outlined in Section 2.1.2, and write the ansatz

ψ(x, z) = exp[f (z)x2

]exp [g (z) (xp+ px)] exp

[h (z) p2

]ψ(x, 0). (6.28)

Deriving with respect to z, inserting the adequate exponential operators, and

reducing conveniently the terms, we obtain after a relatively long process

∂ψ(x, z)

∂z=

df

dzx2 +

dg

dz

(xp+ px+ 4ifx2

)+exp (4ig)

dh

dz

[p2 + 2if (xp+ px)− 4f2x2

]ψ(x, z),

that comparing with (6.27) give us the following system of ordinary differential

equations

df

dz− 2if2 = −iω

2(z)

2,

dg

dz+ f = 0, (6.29)

dh

dz+ i

e−4ig

2= 0.

A brief analysis of (6.28) is enough to see that we must have the initial conditions

f(0) = g(0) = h(0) = 0.



Given a function ω(z), we can solve the first equation in the system (6.29) for

f , and then substitute f in the second equation to find g and finally substitute

g in the third equation to find h.


Appendix A

Dirac notation

In Dirac notation, we denote functions ”f” by means of ”kets” |f⟩. For instancean eigenfunction of the harmonic oscillator (see Chapter 2)

ψn(x) =π−1/4

√2nn!

e−x2/2Hn(x). (A.1)

Is represented by the ket |n⟩, with n = 0, 1, 2, ... In quantum mechanics, these

states are called number or Fock states. Any function can be expanded in terms

of eigenfunctions of the harmonic oscillator

f(x) =∞∑

n=0

cnψn(x), (A.2)

where

cn =

∫ ∞

−∞dxf(x)ψn(x), (A.3)

and in the same way any ket may be expanded in terms of |n⟩’s

|f⟩ =∞∑

n=0

cn|n⟩, (A.4)

where the orthonormalization relation

⟨m|n⟩ =∫ ∞

−∞dxψm(x)ψn(x) = δnm, (A.5)

has been used. The quantity ⟨m| is a so-called ”bra”.

141


142 Dirac notation


Appendix B

Inverse of the Vandermonde andVandermonde confluent matrices

A matrix N ×N of the form

V =

1 1 1 . . . 1 1

λ1 λ2 λ3 . . . λN−1 λN

λ21 λ22 λ23 . . . λ2N−1 λ2N

λ31 λ32 λ33 . . . λ3N−1 λ3N

. . . . . . . .

. . . . . . . .

. . . . . . . .

λN−11 λN−1

2 λN−13 . . . λN−1

N−1 λN−1N

, (B.1)

or

Vi,j = λi−1j i = 1, 2, 3, ..., N ; j = 1, 2, 3, ..., N, (B.2)

is said to be a Vandermonde matrix [Fielder 86; Gautschi 62]. Vandermonde

systems arise in many approximation and interpolation problems [Gautschi 62].

143


144 Inverse of the Vandermonde and Vandermonde confluent matrices

The determinant of the Vandermonde matrix can be expressed as

det (V ) =∏

1≤i≤j≤N

(λj − λi) .

Therefore, if the numbers λ1, λ2, ..., λN are distinct, V is a nonsingular matrix[Fielder 86].

When two or more λi are equal, the corresponding matrix is singular. In

that case, one may use a generalization called confluent Vandermonde matrix[Gautschi 62], which makes the matrix non-singular, while retaining most prop-

erties. If λi = λi+1 = ... = λi+k and λi = λi−1, then the (i+ k)th column is

given by

Ci+k,j =

0 j ≤ k,

(j − 1)!

(j − k − 1)!xj−k−1i j > k.

The confluent Vandermonde matrix looks as

1 ... 1 0 0 .. 1 1

λ1 ... λi 1 0 .. λm−1 λm

λ21 ... λ2i 2λi 0 .. λ2m−1 λ2m

λ31 ... λ3i 3λ2i . .. λ3m−1 λ3m

... ... ... ...(i− 1)!

(i− k − 1)!λi−k−1i .. ... ...

... ... ... ... ... .. ... ...

... ... ... ... ... .. ... ...

λn−11 ... λn−1

i (n− 1)λn−2i

(n− 1)!

(n− k − 1)!λn−k−1i .. λn−1

m−1 λn−1m

.


The inverse of the Vandermonde matrix 145

Another way to write the (i+ k) column is using the derivative, as follows

Ci,j+k =dCi,j+k−1

dλj. (B.3)

B.1 The inverse of the Vandermonde matrix

In applications, a key role is played by the inverse of the Vandermonde and

confluent Vandermonde matrices [Gautschi 62; Gautschi 78]. Both matrices,

Vandermonde and confluent Vandermonde, can be factored into a lower trian-

gular matrix L′ and an upper triangular matrix U ′ where V or C is equal to

L′U ′. The factorization is unique if no row or column interchanges are made

and if it is specified that the diagonal elements of U ′ are unity.

Then, we can write V −1 = (U ′)−1

(L′)−1

. Denoting (U ′)−1

as U , we have found

that U is an upper triangular matrix whose elements are

Ui,j =

0 i > j,

j∏k=1, k =i

1

λi − λkotherwise.

The matrix U can be decomposed as the product of a diagonal matrix D and

other upper triangular matrix W. It is very easy to find that

Di,j =

N∏

k=1, k =i

1

λi − λki = j

0 i = j

and

Wi,j =

0 i > j

N∏k=j+1, k =i

(λi − λk) otherwise.

The matrix L = (L′)−1

is a lower triangular matrix, whose elements are

Li,j =

0 i < j

1 i = j

Li−1,j−1 − Li−1,jλi−1 i = 2, 3, ..., N ; j = 2, 3, ..., i− 1

.

Summarizing, the inverse of the Vandermonde matrix can be written as V −1 =

DWL.



B.2 The inverse of the confluent Vandermonde matrix

We will treat now the case of the confluent Vandermonde matrix. We suppose

that just one of the values λi is repeated, and it is repeated m times. We

make the usual LU decomposition, getting C = L′cU

′c, where L′

c is a lower

triangular matrix and U ′c an upper triangular matrix U ′. Then, we can write

C−1 = (U ′c)

−1(L′

c)−1

. Denoting (U ′c)

−1as Uc, we have found that Uc is an

upper triangular matrix whose elements are

(Uc)i,j = 0 i > j,

(Uc)i,j =δij

(i− 1)!

i = 1, 2, 3, ...,m; j = 1, 2, 3, ...,m,

(Uc)i,j = − 1

(i− 1)!

j∑α=m+1

j∏β=i,β =α

1

(λα − λβ),

i = 1, 2, 3, ...,m, j = m+ 1,m+ 2, ..., N,

(Uc)i,j =

j∏β=1,β =α

1

(λi − λβ)

i = m+ 1,m+ 2, ..., N ; j = i, ..., N

(B.4)

where it is understood that λm = λm−1 = ... = λ2 = λ1, and where the numbers

λm, λm−1, ..., λ2 appear, they must be substituted by λ1.

The matrix Lc = (L′c)

−1is a lower triangular matrix, whose elements are given

by the following recurrence relation,

(Lc)i,j =

0 i < j,

1 i = j,

(Lc)i,ji−1,j−1 − (Lc)i,ji−1,j λi−1 i = 2, 3, ..., N ; j = 2, 3, ..., i− 1,

also here it is understood that λm = λm−1 = ... = λ2 = λ1, and where the

numbers λm, λm−1, ..., λ2 appear, they must be substituted by λ1.


The inverse of the confluent Vandermonde matrix 147

When more than one value is repeated, the inverse has blocks with the same

structure that we have already found.


Appendix C

Tridiagonal matrices

A tridiagonal matrix

M =

α1 β1 0 0 0 · · · · · · 0

β1 α2 β2 0 0 · · · · · · 0

0 β2 α3 β3 0 · · · · · · 0

0 0 β3 α4 β4 0 · · · 0

0 0 0 β4 α5 β5 · · · 0

· · · · · · · · · · · · · · · · · ·. . . · · ·

· · · · · · · · · · · · · · ·. . . αn−1 βn−1

0 0 0 0 0 · · · βn−1 αn

, (C.1)

may be associated with a set of n polynomials given by

p0(λ) = 1, p1(λ) = λ− α1, (C.2)

pk(λ) = (λ− αk)pk−1(λ)− β2k−1pk−2(λ), k = 1, 2, ...n. (C.3)

Then, pn(λ) is the characteristic polynomial. It is not difficult to show that

149


150 Tridiagonal matrices

yj =1√Nj

1

p1(λj)β1

p2(λj)β1β2

p3(λj)β1β2β3

...

...

pn−1(λj)β1β2...βn−1

≡ 1√Nj

p0(λj)

p1(λj)

p2(λj)

p3(λj)

...

...

pn−1(λj)

, (C.4)

is the normalized eigenvector associated to the eigenvalue λj , with

Nj =n−1∑m=0

p2m(λj), (C.5)

such that the eigenvectors matrix is simply

rij =pi−1(λj)√

Nj

, (C.6)

and its inverse is given by the transposition operation.

C.1 Fibonacci system

In Chapter 2 we introduced some tridiagonal matrices related to specific poly-

nomials: Hermite, Laguerre and Chebyshev. An n × n tridiagonal matrix that

may be related to the Fibonacci numbers is similar to the one we related to the

Chebyshev polynomials, except from its diagonal part


Fibonacci system 151

M =

1 1 0 0 0 · · · · · · 0

1 −1 1 0 0 · · · · · · 0

0 1 1 1 0 · · · · · · 0

0 0 1 −1 1 0 · · · 0

0 0 0 1 1 1 · · · 0

· · · · · · · · · · · · · · · · · ·. . . · · ·

· · · · · · · · · · · · · · ·. . . −q 1

0 0 0 0 0 · · · 1 q

, (C.7)

where q = 1 (q = −1) for odd (even) n. It is not difficult to show that, if we call

γn = det(M), |γn| = Fn+2, where Fn is a Fibonacci number.


152 Tridiagonal matrices


Bibliography

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Fifth edition. Springer, 2007.G. Campbell and R. Foster. Fourier Integral for Practical Applications. NY D Van

Nostrand Company, 1948.Earl A. Coddington. An Introduction to Ordinary Differential Equations. Dover, 1989.Comtet L. Advanced Combinatorics: The Art of Finite and Infinite Expansions.

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Index

adjoint operator, 32Airy function, 137angle between vectors, 8ansatz, 44, 139anti-Hermitian operator, 33anti-symmetric operator, 33associated Laguerre polynomials, 49

differential equation, 50generating function, 49recurrence relations, 49Rodrigues’ formula, 49

Baker-Hausdorff formula, 40Bell polynomials, 59Bessel functions of the first kind, 55

addition formula, 56, 92differential equation, 55generating function, 55Jacobi-Anger expansions, 56recurrence relations, 55series of integer order, 58

Cauchy-Schwarz inequality, 8Cayley-Hamilton theorem, 30characteristic polynomial, 25Chebyshev polynomials, 52Chebyshev polynomials of the first kind,

52generating function, 52recurrence relations, 53

Chebyshev polynomials of the secondkind, 53

determinant, 55generating function, 53recurrence relations, 54roots, 54, 55

complete Bell polynomials, 58confluent Vandermonde matrix, 79–81,

86, 143inverse of, 143

diagonal matrices, 28

eigenvaluefinite dimension case, 24

eigenvalues, 20Hermitian operator, 34

eigenvectorfinite dimension case, 24

eigenvectors, 20Hermitian operator, 34

Euclidian space, 8exponential operator, 19

Faa di Bruno’s formula, 59Fibonacci numbers, 150Fibonacci system, 150finite systems of differential equations,

63Cayley-Hamilton theorem method,

69, 74, 83systems 2 by 2, 63systems 4 by 4, 74systems n by n, 83

155


156 Index

Fresnel transform, 135

Hadamard lemma, 39

Hermite polynomials, 37

addition formula, 48

differential equation, 39

generating function, 37

recurrence relations, 37

Rodrigues’ formula, 37

Hermitian operator, 33

infinite systems of differential equations,91

first neighbors interaction, 94

first neighbors interaction with anextra interaction, 100

second neighbors interaction, 96

isomorphism, 14

Laguerre polynomials, 51

matrix, 51

linear operator, 31

linear transformation, 11

associated matrix, 14

codomain, 11

commutator, 17

composition, 16

image, 13

injective, 14

inverse, 19

kernel, 12

product, 16

surjective, 14

norm, 8

orthogonality, 9

Parseval formula, 10

partial differential equations, 133

Airy system, 135

harmonic oscillator, 138

variable frequency harmonicoscillator, 139

scalar product, 7

self-adjoint operator, 33semi-infinite systems of differential

equations, 115first neighbors interaction, 119

similar matrices, 27Suskind-Glogower linear operators, 119symmetric operator, 33

tridiagonal matrices, 149

Vandermonde matrix, 78, 85, 143inverse of, 143

vector space, 1bases, 6dimension, 6finite dimension, 6infinite dimension, 6linear independence, 4span of a set of vectors, 3subspace, 3

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