Mathematical Background QM

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Mathematical Methodsin Quantum Mechanics

With Applicationsto Schrodinger Operators

Gerald Teschl

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Graduate Studies

in Mathematics

Volume 99

American Mathematical Society

Providence, Rhode Island
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Editorial Board

David Cox (Chair) Steven G. KrantsRafe Mazzeo Martin Scharlemann

2000 Mathematics subject classification. 81-01, 81Qxx, 46-01, 34Bxx, 47B25

Abstract. This book provides a self-contained introduction to mathematical methods in quan-

tum mechanics (spectral theory) with applications to Schrodinger operators. The first part cov-ers mathematical foundations of quantum mechanics from self-adjointness, the spectral theorem,quantum dynamics (including Stones and the RAGE theorem) to perturbation theory for self-

adjoint operators.

The second part starts with a detailed study of the free Schrodinger operator respectively

position, momentum and angular momentum operators. Then we develop WeylTitchmarsh the-ory for SturmLiouville operators and apply it to spherically symmetric problems, in particularto the hydrogen atom. Next we investigate self-adjointness of atomic Schrodinger operators and

their essential spectrum, in particular the HVZ theorem. Finally we have a look at scatteringtheory and prove asymptotic completeness in the short range case.

For additional information and updates on this book, visit:

http://www.ams.org/bookpages/gsm-99/

Typeset by LATEXand Makeindex. Version: February 17, 2009.

Library of Congress Cataloging-in-Publication DataTeschl, Gerald, 1970

Mathematical methods in quantum mechanics : with applications to Schrodinger operators

/ Gerald Teschl.

p. cm. (Graduate Studies in Mathematics ; v. 99)

Includes bibliographical references and index.

ISBN 978-0-8218-4660-5 (alk. paper)

1. Schrodinger operators. 2. Quantum theoryMathematics. I. Title.

QC174.17.S3T47 2009 2008045437

515.724dc22

Copying and reprinting. Individual readers of this publication, and nonprofit libraries actingfor them, are permitted to make fair use of the material, such as to copy a chapter for use

in teaching or research. Permission is granted to quote brief passages from this publication inreviews, provided the customary acknowledgement of the source is given.

Republication, systematic copying, or multiple reproduction of any material in this pub-

lication (including abstracts) is permitted only under license from the American MathematicalSociety. Requests for such permissions should be addressed to the Assistant to the Publisher,

American Mathematical Society, P.O. Box 6248, Providence, Rhode Island 02940-6248. Requestscan also be made by e-mail to [email protected].

c 2009 by the American Mathematical Society. All rights reserved.The American Mathematical Society retains all rights

except those granted too the United States Government.
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To Susanne, Simon, and Jakob


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Contents

Preface xi

Part 0. Preliminaries

Chapter 0. A first look at Banach and Hilbert spaces 3

0.1. Warm up: Metric and topological spaces 30.2. The Banach space of continuous functions 120.3. The geometry of Hilbert spaces 160.4. Completeness 22

0.5. Bounded operators 22

0.6. Lebesgue Lp spaces 250.7. Appendix: The uniform boundedness principle 32

Part 1. Mathematical Foundations of Quantum Mechanics

Chapter 1. Hilbert spaces 37

1.1. Hilbert spaces 371.2. Orthonormal bases 391.3. The projection theorem and the Riesz lemma 431.4. Orthogonal sums and tensor products 451.5. The C algebra of bounded linear operators 471.6. Weak and strong convergence 491.7. Appendix: The StoneWeierstra theorem 51

Chapter 2. Self-adjointness and spectrum 55

vii


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viii Contents

2.1. Some quantum mechanics 552.2. Self-adjoint operators 582.3. Quadratic forms and the Friedrichs extension 672.4. Resolvents and spectra 732.5. Orthogonal sums of operators 792.6. Self-adjoint extensions 812.7. Appendix: Absolutely continuous functions 84

Chapter 3. The spectral theorem 87

3.1. The spectral theorem 873.2. More on Borel measures 993.3. Spectral types 1043.4. Appendix: The Herglotz theorem 107

Chapter 4. Applications of the spectral theorem 111

4.1. Integral formulas 1114.2. Commuting operators 1154.3. The min-max theorem 1174.4. Estimating eigenspaces 1194.5. Tensor products of operators 120

Chapter 5. Quantum dynamics 123

5.1. The time evolution and Stones theorem 1235.2. The RAGE theorem 1265.3. The Trotter product formula 131

Chapter 6. Perturbation theory for self-adjoint operators 133

6.1. Relatively bounded operators and the KatoRellich theorem 1336.2. More on compact operators 1366.3. HilbertSchmidt and trace class operators 1396.4. Relatively compact operators and Weyls theorem 145

6.5. Relatively form bounded operators and the KLMN theorem 149

6.6. Strong and norm resolvent convergence 153

Part 2. Schrodinger Operators

Chapter 7. The free Schrodinger operator 161

7.1. The Fourier transform 1617.2. The free Schrodinger operator 167


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Contents ix

7.3. The time evolution in the free case 1697.4. The resolvent and Greens function 171

Chapter 8. Algebraic methods 173

8.1. Position and momentum 1738.2. Angular momentum 1758.3. The harmonic oscillator 1788.4. Abstract commutation 179

Chapter 9. One-dimensional Schrodinger operators 181

9.1. SturmLiouville operators 181

9.2. Weyls limit circle, limit point alternative 187

9.3. Spectral transformations I 1959.4. Inverse spectral theory 2029.5. Absolutely continuous spectrum 2069.6. Spectral transformations II 2099.7. The spectra of one-dimensional Schrodinger operators 214

Chapter 10. One-particle Schrodinger operators 221

10.1. Self-adjointness and spectrum 22110.2. The hydrogen atom 222

10.3. Angular momentum 225

10.4. The eigenvalues of the hydrogen atom 22910.5. Nondegeneracy of the ground state 235

Chapter 11. Atomic Schrodinger operators 239

11.1. Self-adjointness 23911.2. The HVZ theorem 242

Chapter 12. Scattering theory 247

12.1. Abstract theory 24712.2. Incoming and outgoing states 250

12.3. Schrodinger operators with short range potentials 253

Part 3. Appendix

Appendix A. Almost everything about Lebesgue integration 259

A.1. Borel measures in a nut shell 259A.2. Extending a premeasure to a measure 263A.3. Measurable functions 268


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x Contents

A.4. The Lebesgue integral 270A.5. Product measures 275A.6. Vague convergence of measures 278A.7. Decomposition of measures 280A.8. Derivatives of measures 282

Bibliographical notes 289

Bibliography 293

Glossary of notation 297

Index 301


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Preface

Overview

The present text was written for my course Schrodinger Operatorsheldat the University of Vienna in winter 1999, summer 2002, summer 2005,and winter 2007. It gives a brief but rather self-contained introductionto the mathematical methods of quantum mechanics with a view towardsapplications to Schrodinger operators. The applications presented are highlyselective and many important and interesting items are not touched upon.

Part 1 is a stripped down introduction to spectral theory of unboundedoperators where I try to introduce only those topics which are needed for

the applications later on. This has the advantage that you will (hopefully)not get drowned in results which are never used again before you get tothe applications. In particular, I am not trying to present an encyclopedicreference. Nevertheless I still feel that the first part should provide a solidbackground covering many important results which are usually taken forgranted in more advanced books and research papers.

My approach is built around the spectral theorem as the central object.Hence I try to get to it as quickly as possible. Moreover, I do not take thedetour over bounded operators but I go straight for the unbounded case.In addition, existence of spectral measures is established via the Herglotztheorem rather than the Riesz representation theorem since this approach

paves the way for an investigation of spectral types via boundary values ofthe resolvent as the spectral parameter approaches the real line.

xi


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xii Preface

Part 2 starts with the free Schrodinger equation and computes the

free resolvent and time evolution. In addition, I discuss position, momen-tum, and angular momentum operators via algebraic methods. This isusually found in any physics textbook on quantum mechanics, with theonly difference that I include some technical details which are typicallynot found there. Then there is an introduction to one-dimensional mod-els (SturmLiouville operators) including generalized eigenfunction expan-sions (WeylTitchmarsh theory) and subordinacy theory from Gilbert andPearson. These results are applied to compute the spectrum of the hy-drogen atom, where again I try to provide some mathematical details notfound in physics textbooks. Further topics are nondegeneracy of the groundstate, spectra of atoms (the HVZ theorem), and scattering theory (the En

method).

Prerequisites

I assume some previous experience with Hilbert spaces and boundedlinear operators which should be covered in any basic course on functionalanalysis. However, while this assumption is reasonable for mathematicsstudents, it might not always be for physics students. For this reason thereis a preliminary chapter reviewing all necessary results (including proofs).In addition, there is an appendix (again with proofs) providing all necessaryresults from measure theory.

Literature

The present book is highly influenced by the four volumes of Reed andSimon [40][43] (see also [14]) and by the book by Weidmann [60] (anextended version of which has recently appeared in two volumes [62], [63],however, only in German). Other books with a similar scope are for example[14],[15],[21],[23],[39],[48], and[55]. For those who want to know moreabout the physical aspects, I can recommend the classical book by Thirring[58]and the visual guides by Thaller [56],[57]. Further information can befound in the bibliographical notes at the end.

Readers guide

There is some intentional overlap between Chapter 0, Chapter 1, andChapter 2. Hence, provided you have the necessary background, you canstart reading in Chapter 1 or even Chapter 2. Chapters 2 and 3 are key


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Preface xiii

chapters and you should study them in detail (except for Section2.6 which

can be skipped on first reading). Chapter 4should give you an idea of howthe spectral theorem is used. You should have a look at (e.g.) the firstsection and you can come back to the remaining ones as needed. Chapter 5contains two key results from quantum dynamics: Stones theorem and theRAGE theorem. In particular the RAGE theorem shows the connectionsbetween long time behavior and spectral types. Finally, Chapter 6is againof central importance and should be studied in detail.

The chapters in the second part are mostly independent of each otherexcept for Chapter7,which is a prerequisite for all others except for Chap-ter9.

If you are interested in one-dimensional models (SturmLiouville equa-

tions), Chapter9 is all you need.If you are interested in atoms, read Chapter 7, Chapter10, and Chap-

ter11. In particular, you can skip the separation of variables (Sections 10.3and10.4, which require Chapter9) method for computing the eigenvalues ofthe hydrogen atom, if you are happy with the fact that there are countablymany which accumulate at the bottom of the continuous spectrum.

If you are interested in scattering theory, read Chapter 7, the first twosections of Chapter10,and Chapter12. Chapter5 is one of the key prereq-uisites in this case.

Updates

The AMS is hosting a web page for this book at

http://www.ams.org/bookpages/gsm-99/

where updates, corrections, and other material may be found, including alink to material on my own web site:

http://www.mat.univie.ac.at/~gerald/ftp/book-schroe/

Acknowledgments

I would like to thank Volker En for making his lecture notes [18] avail-able to me. Many colleagues and students have made useful suggestions andpointed out mistakes in earlier drafts of this book, in particular: KerstinAmmann, Jorg Arnberger, Chris Davis, Fritz Gesztesy, Maria Hoffmann-Ostenhof, Zhenyou Huang, Helge Kruger, Katrin Grunert, Wang Lanning,Daniel Lenz, Christine Pfeuffer, Roland Mows, Arnold L. Neidhardt, Harald
http://www.ams.org/bookpages/gsm-99/http://www.mat.univie.ac.at/~gerald/ftp/book-schroe/http://www.mat.univie.ac.at/~gerald/ftp/book-schroe/http://www.mat.univie.ac.at/~gerald/ftp/book-schroe/http://www.mat.univie.ac.at/~gerald/ftp/book-schroe/http://www.ams.org/bookpages/gsm-99/


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xiv Preface

Rindler, Johannes Temme, Karl Unterkofler, Joachim Weidmann, and Rudi

Weikard.If you also find an error or if you have comments or suggestions

(no matter how small), please let me know.

I have been supported by the Austrian Science Fund (FWF) during muchof this writing, most recently under grant Y330.

Gerald Teschl

Vienna, AustriaJanuary 2009

Gerald TeschlFakultat fur MathematikNordbergstrae 15

Universitat Wien1090 Wien, Austria

E-mail: [email protected]: http://www.mat.univie.ac.at/~gerald/
mailto:[email protected]://www.mat.univie.ac.at/~gerald/http://www.mat.univie.ac.at/~gerald/http://www.mat.univie.ac.at/~gerald/http://www.mat.univie.ac.at/~gerald/mailto:[email protected]


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Part 0

Preliminaries


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Chapter 0

A first look at Banach

and Hilbert spaces

I assume that the reader has some basic familiarity with measure theory and func-

tional analysis. For convenience, some facts needed from Banach andLp spaces are

reviewed in this chapter. A crash course in measure theory can be found in the

Appendix A.If you feel comfortable with terms like Lebesgue Lp spaces, Banach

space, or bounded linear operator, you can skip this entire chapter. However, you

might want to at least browse through it to refresh your memory.

0.1. Warm up: Metric and topological spaces

Before we begin, I want to recall some basic facts from metric and topologicalspaces. I presume that you are familiar with these topics from your calculuscourse. As a general reference I can warmly recommend Kellys classicalbook[26].

A metric space is a space X together with a distance function d :X X R such that

(i) d(x, y)0,(ii) d(x, y) = 0 if and only ifx = y,

(iii) d(x, y) =d(y, x),

(iv) d(x, z)d(x, y) + d(y, z) (triangle inequality).If (ii) does not hold,d is called a semi-metric. Moreover, it is straight-

forward to see the inverse triangle inequality(Problem0.1)

|d(x, y) d(z, y)| d(x, z). (0.1)

3


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4 0. A first look at Banach and Hilbert spaces

Example. Euclidean spaceRn together withd(x, y) = (nk=1(xkyk)2)1/2

is a metric space and so is Cn together withd(x, y) = (nk=1 |xkyk|2)1/2. The set

Br(x) ={yX|d(x, y)< r} (0.2)is called an open ballaround x with radius r >0. A point x of some setUis called an interior pointofU ifUcontains some ball around x. Ifx isan interior point ofU, thenU is also called a neighborhoodofx. A pointxis called a limit pointofU if (Br(x)\{x}) U=for every ball aroundx. Note that a limit pointx need not lie in U, but Umust contain pointsarbitrarily close tox.

Example. Consider R with the usual metric and let U = (

1, 1). Then

every point xU is an interior point ofU. The points1 are limit pointsofU.

A set consisting only of interior points is called open. The family ofopen setsO satisfies the properties

(i), X O,(ii) O1, O2 O implies O1 O2 O,

(iii){O} O implies O O.

That is,O is closed under finite intersections and arbitrary unions.In general, a space Xtogether with a family of sets

O, the open sets,

satisfying (i)(iii) is called a topological space. The notions of interiorpoint, limit point, and neighborhood carry over to topological spaces if wereplace open ball by open set.

There are usually different choices for the topology. Two usually notvery interesting examples are the trivial topologyO ={, X} and thediscrete topologyO = P(X) (the powerset ofX). Given two topologiesO1 andO2 on X,O1 is called weaker (or coarser) thanO2 if and only ifO1 O2.Example. Note that different metrics can give rise to the same topology.For example, we can equip Rn (or Cn) with the Euclidean distance d(x, y)as before or we could also use

d(x, y) =nk=1

|xk yk|. (0.3)

Then

1n

nk=1

|xk| nk=1

|xk|2 nk=1

|xk| (0.4)


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0.1. Warm up: Metric and topological spaces 5

shows Br/n(x) Br(x)Br(x), where B, B are balls computed using d,

d, respectively. Example. We can always replace a metric d by the bounded metric

d(x, y) = d(x, y)

1 + d(x, y) (0.5)

without changing the topology. Every subspaceYof a topological space Xbecomes a topological space

of its own if we call OYopen if there is some open set OX such thatO= O Y (induced topology).Example. The set (0, 1]R is not open in the topology ofX=R, but it isopen in the induced topology when considered as a subset ofY = [1, 1].

A family of open setsB O is called abase for the topology if for eachxand each neighborhood U(x), there is some setO B withxOU(x).Since an open set O is a neighborhood of every one of its points, it can bewritten as O =

OOB O and we have

Lemma 0.1. IfB O is a base for the topology, then every open set canbe written as a union of elements fromB.

If there exists a countable base, then X is called second countable.

Example. By construction the open balls B1/n(x) are a base for the topol-ogy in a metric space. In the case ofRn (or Cn) it even suffices to take ballswith rational center and hence Rn (and Cn) is second countable.

A topological space is called a Hausdorff space if for two differentpoints there are always two disjoint neighborhoods.

Example. Any metric space is a Hausdorff space: Given two differentpoints x and y, the balls Bd/2(x) and Bd/2(y), where d = d(x, y) > 0, aredisjoint neighborhoods (a semi-metric space will not be Hausdorff).

The complement of an open set is called a closed set. It follows fromde Morgans rules that the family of closed setsC satisfies

(i), X C,(ii) C1, C2 C implies C1 C2 C,

(iii){C} C impliesC C.

That is, closed sets are closed under finite unions and arbitrary intersections.

The smallest closed set containing a given set U is called the closure

U=

CC,UCC, (0.6)


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and the largest open set contained in a given set U is called the interior

U= OO,OU

O. (0.7)

We can define interior and limit points as before by replacing the wordball by open set. Then it is straightforward to check

Lemma 0.2. Let X be a topological space. Then the interior of U is theset of all interior points ofUand the closure ofU is the union ofU withall limit points ofU.

A sequence (xn)n=1 X is said to converge to some point x X ifd(x, xn) 0. We write limn xn = x as usual in this case. Clearly thelimit is unique if it exists (this is not true for a semi-metric).

Every convergent sequence is a Cauchy sequence; that is, for every >0 there is some N N such that

d(xn, xm), n, mN. (0.8)If the converse is also true, that is, if every Cauchy sequence has a limit,then X is called complete.

Example. Both Rn and Cn are complete metric spaces. A pointxis clearly a limit point ofUif and only if there is some sequence

xnU\{x}converging to x. HenceLemma 0.3.

A closed subset of a complete metric space is again a completemetric space.

Note that convergence can also be equivalently formulated in terms oftopological terms: A sequence xn converges to x if and only if for everyneighborhood U ofx there is some N N such that xnU for nN. Ina Hausdorff space the limit is unique.

A set U is called dense if its closure is all ofX, that is, ifU = X. Ametric space is called separable if it contains a countable dense set. Notethat X is separable if and only if it is second countable as a topologicalspace.

Lemma 0.4. LetXbe a separable metric space. Every subset ofXis againseparable.

Proof. Let A ={xn}nN be a dense set in X. The only problem is thatA Ymight contain no elements at all. However, some elements ofA mustbe at least arbitrarily close: LetJ N2 be the set of all pairs (n, m) forwhich B1/m(xn)Y = and choose some yn,m B1/m(xn)Y for all(n, m) J. ThenB ={yn,m}(n,m)J Y is countable. To see that B is


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dense, chooseyY. Then there is some sequence xnk withd(xnk , y)< 1/k.Hence (nk, k)J and d(ynk,k, y)d(ynk,k, xnk) + d(xnk , y)2/k0.

A function between metric spaces X and Y is called continuous at apointxXif for every >0 we can find a >0 such that

dY(f(x), f(y)) if dX(x, y)< . (0.9)Iffis continuous at every point, it is called continuous.

Lemma 0.5. LetX, Ybe metric spaces andf :XY. The following areequivalent:

(i) fis continuous atx (i.e, (0.9) holds).

(ii) f(xn)f(x) wheneverxnx.(iii) For every neighborhoodV off(x), f1(V) is a neighborhood ofx.

Proof. (i) (ii) is obvious. (ii) (iii): If (iii) does not hold, there isa neighborhood V of f(x) such that B(x) f1(V) for every . Hencewe can choose a sequence xn B1/n(x) such that f(xn) f1(V). Thusxn x but f(xn) f(x). (iii) (i): Choose V = B(f(x)) and observethat by (iii), B(x)f1(V) for some .

The last item implies that fis continuous if and only if the inverse imageof every open (closed) set is again open (closed).

Note: In a topological space, (iii) is used as the definition for continuity.However, in general (ii) and (iii) will no longer be equivalent unless one usesgeneralized sequences, so-called nets, where the index set N is replaced byarbitrary directed sets.

The supportof a functionf :X Cn is the closure of all points x forwhichf(x) does not vanish; that is,

supp(f) ={xX|f(x)= 0}. (0.10)IfXand Yare metric spaces, then X Ytogether with

d((x1, y1), (x2, y2)) =dX(x1, x2) + dY(y1, y2) (0.11)

is a metric space. A sequence (xn, yn) converges to (x, y) if and only ifxnx and yny. In particular, the projections onto the first (x, y)x,respectively, onto the second (x, y)y, coordinate are continuous.

In particular, by the inverse triangle inequality (0.1),

|d(xn, yn) d(x, y)| d(xn, x) + d(yn, y), (0.12)we see that d : X X R is continuous.


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Example. If we consider R R, we do not get the Euclidean distance ofR2 unless we modify (0.11) as follows:

d((x1, y1), (x2, y2)) =

dX(x1, x2)2 + dY(y1, y2)2. (0.13)

As noted in our previous example, the topology (and thus also conver-gence/continuity) is independent of this choice.

IfXandYare just topological spaces, the product topologyis definedby calling O XY open if for every point (x, y) O there are openneighborhoods U of x and V of y such that U V O. In the case ofmetric spaces this clearly agrees with the topology defined via the productmetric (0.11).

A cover of a set Y

Xis a family of sets{

U}

such that Y U.A cover is called open if all U are open. Any subset of{U} which still

coversY is called a subcover.

Lemma 0.6 (Lindelof). If X is second countable, then every open coverhas a countable subcover.

Proof. Let{U} be an open cover for Y and letB be a countable base.Since every U can be written as a union of elements fromB, the set of allB B which satisfy BU for some form a countable open cover for Y .Moreover, for every Bn in this set we can find an n such that Bn Un .By construction{Un} is a countable subcover.

A subset K X is called compact if every open cover has a finitesubcover.

Lemma 0.7. A topological space is compact if and only if it has thefiniteintersection property: The intersection of a family of closed sets is emptyif and only if the intersection of some finite subfamily is empty.

Proof. By taking complements, to every family of open sets there is a cor-responding family of closed sets and vice versa. Moreover, the open setsare a cover if and only if the corresponding closed sets have empty intersec-tion.

A subsetKXis calledsequentially compactif every sequence hasa convergent subsequence.

Lemma 0.8. LetXbe a topological space.

(i) The continuous image of a compact set is compact.

(ii) Every closed subset of a compact set is compact.

(iii) IfXis Hausdorff, any compact set is closed.


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(iv) The product of finitely many compact sets is compact.

(v) A compact set is also sequentially compact.

Proof. (i) Observe that if{O}is an open cover for f(Y), then{f1(O)}is one for Y .

(ii) Let{O} be an open cover for the closed subset Y. Then{O} {X\Y}is an open cover for X.

(iii) LetYXbe compact. We show that X\Y is open. FixxX\Y(if Y = X, there is nothing to do). By the definition of Hausdorff, forevery yYthere are disjoint neighborhoods V(y) ofy and Uy(x) ofx. Bycompactness ofY, there are y1, . . . , yn such that the V(yj) cover Y. Butthen U(x) =

nj=1 Uyj(x) is a neighborhood ofx which does not intersect

Y.(iv) Let{O} be an open cover for X Y. For every (x, y) X Y

there is some (x, y) such that (x, y)O(x,y). By definition of the producttopology there is some open rectangleU(x, y)V(x, y)O(x,y). Hence forfixedx,{V(x, y)}yY is an open cover ofY. Hence there are finitely manypoints yk(x) such that the V(x, yk(x)) coverY . SetU(x) =

kU(x, yk(x)).

Since finite intersections of open sets are open,{U(x)}xXis an open coverand there are finitely many points xj such that the U(xj) cover X. Byconstruction, the U(xj) V(xj, yk(xj))O(xj ,yk(xj)) cover X Y.

(v) Let xn be a sequence which has no convergent subsequence. ThenK =

{xn

} has no limit points and is hence compact by (ii). For every n

there is a ball Bn(xn) which contains only finitely many elements of K.However, finitely many suffice to cover K, a contradiction.

In a metric space compact and sequentially compact are equivalent.

Lemma 0.9. LetXbe a metric space. Then a subset is compact if and onlyif it is sequentially compact.

Proof. By item (v) of the previous lemma it suffices to show that X iscompact if it is sequentially compact.

First of all note that every cover of open balls with fixed radius > 0has a finite subcover since if this were false we could construct a sequence

xnX\n1m=1 B(xm) such that d(xn, xm)> for m < n.In particular, we are done if we can show that for every open cover

{O} there is some > 0 such that for every x we have B(x) O forsome = (x). Indeed, choosing{xk}nk=1 such that B(xk) is a cover, wehave that O(xk) is a cover as well.

So it remains to show that there is such an . If there were none, forevery >0 there must be an x such that B(x)O for every . Choose


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= 1n and pick a correspondingxn. SinceXis sequentially compact, it is no

restriction to assumexn converges (after maybe passing to a subsequence).Letx= lim xn. Thenxlies in someOand henceB(x)O. But choosingn so large that 1n 0 we can find an N such thatam an1 for m, nN. This implies in particular|amj anj | forany fixed j. Thus anj is a Cauchy sequence for fixed j and by completenessofC has a limit: limn anj =aj. Now consider

kj=1

|amj anj | (0.23)

and take m :kj=1

|aj anj | . (0.24)

Since this holds for any finitek, we even have aan1. Hence (aan)1(N) and sincean1(N), we finally conclude a = an+ (aan)1(N). Example. The space (N) of all bounded sequences a = (aj)j=1 togetherwith the norm

a = supjN |aj| (0.25)is a Banach space (Problem0.10).

Now what about convergence in the space C(I)? A sequence of functionsfn(x) converges to fif and only if

limnf fn= limn supxI |fn(x) f(x)|= 0. (0.26)


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That is, in the language of real analysis, fn converges uniformly to f. Now

let us look at the case where fn is only a Cauchy sequence. Thenfn(x) isclearly a Cauchy sequence of real numbers for any fixed xI. In particular,by completeness ofC, there is a limitf(x) for eachx. Thus we get a limitingfunctionf(x). Moreover, letting m in

|fm(x) fn(x)| m,n > N, xI , (0.27)we see

|f(x) fn(x)| n > N, xI; (0.28)that is, fn(x) converges uniformly to f(x). However, up to this point wedo not know whether it is in our vector space C(I) or not, that is, whetherit is continuous or not. Fortunately, there is a well-known result from realanalysis which tells us that the uniform limit of continuous functions is againcontinuous. Hence f(x) C(I) and thus every Cauchy sequence in C(I)converges. Or, in other words

Theorem 0.14. C(I) with the maximum norm is a Banach space.

Next we want to know if there is a countable basisfor C(I). We will calla set of vectors{un} Xlinearly independent if every finite subset is andwe will call a countable set of linearly independent vectors{un}Nn=1 Xa Schauder basis if every element f X can be uniquely written as acountable linear combination of the basis elements:

f=N

n=1 cnun, cn = cn(f)C, (0.29)

where the sum has to be understood as a limit ifN =. In this case thespanspan{un}(the set of all finite linear combinations) of{un}is dense inX. A set whose span is dense is calledtotaland if we have a countable totalset, we also have a countable dense set (consider only linear combinationswith rational coefficients show this). A normed linear space containing acountable dense set is called separable.

Example. The Banach space 1(N) is separable. In fact, the set of vectorsn, with nn = 1 and

nm = 0, n= m, is total: Let a = (aj)j=1 1(N) be

given and setan =

nj=1 aj

j. Then

a an1=

j=n+1|aj| 0 (0.30)

sinceanj =aj for 1jn and anj = 0 for j > n. Luckily this is also the case for C(I):

Theorem 0.15 (Weierstra). LetIbe a compact interval. Then the set ofpolynomials is dense inC(I).


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0.2. The Banach space of continuous functions 15

Proof. Let f(x) C(I) be given. By considering f(x)f(a) + (f(b)f(a))(x b) it is no loss to assume that fvanishes at the boundary points.Moreover, without restriction we only consider I= [12 ,

12 ] (why?).

Now the claim follows from the lemma below using

un(x) = 1

In(1 x2)n,

where

In=

11

(1 x2)ndx= nn + 1

11

(1 x)n1(1 + x)n+1dx

= = n!(n + 1)

(2n + 1)

22n+1 = n!

12 (

12+ 1)

( 12 + n)

=

(1 + n)

( 32 + n)=

n(1 + O(

1

n)).

In the last step we have used ( 12 ) =

[1, (6.1.8)] and the asymptoticsfollow from Stirlings formula [1,(6.1.37)].

Lemma 0.16 (Smoothing). Letun(x)be a sequence of nonnegative contin-uous functions on [1, 1] such that

|x|1un(x)dx= 1 and

|x|1

un(x)dx0, >0. (0.31)

(In other words, un has mass one and concentrates nearx= 0 asn .)Then for everyf C[12 , 12 ] which vanishes at the endpoints, f(12 ) =

f( 12 ) = 0, we have that

fn(x) =

1/21/2

un(x y)f(y)dy (0.32)

converges uniformly to f(x).

Proof. Sincefis uniformly continuous, for given we can find a(indepen-dent ofx) such that |f(x)f(y)| whenever |xy| . Moreover, we canchoose n such that |y|1 un(y)dy. Now abbreviate M= max{1, |f|}and note

|f(x) 1/21/2

un(x y)f(x)dy|=|f(x)| |1 1/21/2

un(x y)dy| M .

In fact, either the distance ofx to one of the boundary points12 is smallerthanand hence|f(x)| or otherwise the difference between one and theintegral is smaller than .


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Using this, we have

|fn(x) f(x)| 1/21/2

un(x y)|f(y) f(x)|dy+ M

|y|1/2,|xy|

un(x y)|f(y) f(x)|dy

+

|y|1/2,|xy|

un(x y)|f(y) f(x)|dy+ M

= + 2M + M= (1 + 3M), (0.33)

which proves the claim.

Note thatfn will be as smooth as un, hence the title smoothing lemma.The same idea is used to approximate noncontinuous functions by smoothones (of course the convergence will no longer be uniform in this case).

Corollary 0.17. C(I) is separable.

However,(N) is not separable (Problem0.11)!

Problem 0.7. Show that|f g| f g.Problem 0.8. Let Xbe a Banach space. Show that the norm, vector ad-dition, and multiplication by scalars are continuous. That is, if fn f,gng, andn, thenfn f, fn+ gnf+ g, andngng.Problem 0.9. LetXbe a Banach space. Show that

j=1 fj


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0.3. The geometry of Hilbert spaces 17

Suppose H is a vector space. A map.,.. : H H C is called asesquilinear formif it is conjugate linear in the first argument and linearin the second; that is,

1f1+ 2f2, g = 1f1, g + 2f2, g,f, 1g1+ 2g2 = 1f, g1 + 2f, g2, 1, 2C, (0.34)

where denotes complex conjugation. A sesquilinear form satisfying therequirements

(i)f, f> 0 forf= 0 (positive definite),(ii)f, g=g, f (symmetry)

is called an inner product or scalar product. Associated with everyscalar product is a norm

f= f, f. (0.35)The pair (H, .,..) is called an inner product space. IfH is complete, itis called aHilbert space.

Example. Clearly Cn with the usual scalar product

a, b=nj=1

ajbj (0.36)

is a (finite dimensional) Hilbert space. Example. A somewhat more interesting example is the Hilbert space 2(N),that is, the set of all sequences

(aj)j=1

j=1

|aj|2


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Suppose u is a unit vector. Then the projection off in the direction of

uis given byf=u, fu (0.40)

and f defined viaf = f u, fu (0.41)

is perpendicular tou sinceu, f=u, fu, fu=u, fu, fu, u=0.

f

f

f

u

Taking any other vector parallel to u, it is easy to see

f u2 =f+ (f u)2 =f2 + |u, f |2 (0.42)and hence f=u, fu is the unique vector parallel to u which is closest tof.

As a first consequence we obtain the CauchySchwarzBunjakowskiinequality:

Theorem 0.18 (CauchySchwarzBunjakowski). LetH0 be an inner prod-

uct space. Then for everyf, g H0 we have|f, g| f g (0.43)

with equality if and only iff andg are parallel.

Proof. It suffices to prove the caseg = 1. But then the claim followsfromf2 =|g, f|2 + f2.

Note that the CauchySchwarz inequality entails that the scalar productis continuous in both variables; that is, if fn f and gn g, we havefn, gn f, g.

As another consequence we infer that the map

.

is indeed a norm. In

fact,

f+ g2 =f2 + f, g + g, f + g2 (f + g)2. (0.44)But let us return to C(I). Can we find a scalar product which has the

maximum norm as associated norm? Unfortunately the answer is no! Thereason is that the maximum norm does not satisfy the parallelogram law(Problem0.17).


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Theorem 0.19(Jordanvon Neumann). A norm is associated with a scalar

product if and only if the parallelogram law

f+ g2 + f g2 = 2f2 + 2g2 (0.45)holds.

In this case the scalar product can be recovered from its norm by virtueof the polarization identity

f, g= 14

f+ g2 f g2 + if ig2 if+ ig2 . (0.46)Proof. If an inner product space is given, verification of the parallelogramlaw and the polarization identity is straightforward (Problem0.14).

To show the converse, we define

s(f, g) =1

4

f+ g2 f g2 + if ig2 if+ ig2 .Then s(f, f) =f2 and s(f, g) = s(g, f) are straightforward to check.Moreover, another straightforward computation using the parallelogram lawshows

s(f, g) + s(f, h) = 2s(f,g + h

2 ).

Now choosing h = 0 (and using s(f, 0) = 0) shows s(f, g) = 2s(f, g2 ) andthus s(f, g) +s(f, h) = s(f, g +h). Furthermore, by induction we inferm2ns(f, g) =s(f,

m2ng); that is, s(f, g) =s(f,g) for every positive rational

. By continuity (check this!) this holds for all > 0 and s(f, g) =s(f, g), respectively, s(f, ig) = i s(f, g), finishes the proof.

Note that the parallelogram law and the polarization identity even holdfor sesquilinear forms (Problem0.14).

But how do we define a scalar product on C(I)? One possibility is

f, g= ba

f(x)g(x)dx. (0.47)

The corresponding inner product space is denoted byL2cont(I). Note thatwe have

f |b a|f (0.48)and hence the maximum norm is stronger than theL2cont norm.

Suppose we have two norms.1 and.2 on a space X. Then.2 issaid to be strongerthan.1 if there is a constant m >0 such that

f1mf2. (0.49)It is straightforward to check the following.


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Lemma 0.20. If.2 is stronger than.1, then any.2 Cauchy sequenceis also a.1 Cauchy sequence.

Hence if a function F : X Y is continuous in (X, .1), it is alsocontinuous in (X, .2) and if a set is dense in (X, .2), it is also dense in(X, .1).

In particular,L2cont is separable. But is it also complete? Unfortunatelythe answer is no:

Example. Take I= [0, 2] and define

fn(x) =

0, 0x1 1n ,1 + n(x 1), 1 1nx1,

1, 1x2.

(0.50)

Then fn(x) is a Cauchy sequence inL2cont, but there is no limit inL2cont!Clearly the limit should be the step function which is 0 for 0x 0 on the unit sphere (which is compact by the Heine-Boreltheorem). Now choose m1= 1/m.

Problem 0.12. Show that the norm in a Hilbert space satisfiesf+g=f + g if and only iff=g, 0, org= 0.


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Problem 0.13. Show that2(N) is a separable Hilbert space.

Problem 0.14. SupposeQ is a vector space. Lets(f, g) be a sesquilinearform onQ and q(f) = s(f, f) the associated quadratic form. Prove theparallelogram law

q(f+ g) + q(f g) = 2q(f) + 2q(g) (0.52)and the polarization identity

s(f, g) =1

4(q(f+ g) q(f g) + i q(f ig) i q(f+ ig)) . (0.53)

Conversely, show that any quadratic form q(f) : Q R satisfyingq(f) =||

2

q(f)and the parallelogram law gives rise to a sesquilinear formvia the polarization identity.

Problem 0.15. A sesquilinear form is calledbounded if

s= supf=g=1

|s(f, g)|

is finite. Similarly, the associated quadratic formq isbounded if

q= supf=1

|q(f)|

is finite. Show

q s 2q.(Hint: Use the parallelogram law and the polarization identity from the pre-vious problem.)

Problem 0.16. SupposeQ is a vector space. Lets(f, g) be a sesquilinearform onQ andq(f) =s(f, f) the associated quadratic form. Show that theCauchySchwarz inequality

|s(f, g)| q(f)1/2q(g)1/2 (0.54)holds ifq(f)0.

(Hint: Consider0 q(f+g) = q(f) + 2Re( s(f, g)) + ||2

q(g) andchoose= t s(f, g)/|s(f, g)| with tR.)Problem 0.17. Show that the maximum norm (onC[0, 1]) does not satisfythe parallelogram law.

Problem 0.18. Prove the claims made aboutfn, defined in (0.50), in thelast example.


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0.4. Completeness

SinceL2cont is not complete, how can we obtain a Hilbert space from it?Well, the answer is simple: take the completion.

If X is an (incomplete) normed space, consider the set of all Cauchysequences X. Call two Cauchy sequences equivalent if their difference con-verges to zero and denote by Xthe set of all equivalence classes. It is easyto see that X(and X) inherit the vector space structure from X. Moreover,

Lemma 0.22. Ifxn is a Cauchy sequence, thenxn converges.Consequently the norm of a Cauchy sequence (xn)n=1 can be defined by

(xn)n=1= limn xn and is independent of the equivalence class (showthis!). Thus X is a normed space (X is not! Why?).Theorem 0.23. Xis a Banach space containingXas a dense subspace ifwe identifyxXwith the equivalence class of all sequences converging tox.

Proof. (Outline) It remains to show that Xis complete. Let n = [(xn,j)j=1]

be a Cauchy sequence in X. Then it is not hard to see that= [(xj,j)j=1]is its limit.

Let me remark that the completion X is unique. More precisely anyother complete space which contains X as a dense subset is isomorphic to

X. This can for example be seen by showing that the identity map on Xhas a unique extension to X (compare Theorem0.26below).

In particular it is no restriction to assume that a normed linear spaceor an inner product space is complete. However, in the important caseofL2cont it is somewhat inconvenient to work with equivalence classes ofCauchy sequences and hence we will give a different characterization usingthe Lebesgue integral later.

0.5. Bounded operators

A linear map A between two normed spaces Xand Ywill be called a(lin-ear) operator

A: D(A)XY. (0.55)The linear subspace D(A) on which A is defined is called the domainofAand is usually required to be dense. The kernel

Ker(A) ={f D(A)|Af= 0} (0.56)and range

Ran(A) ={Af|f D(A)}= AD(A) (0.57)


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0.5. Bounded operators 23

are defined as usual. The operator A is called bounded if the operator

norm A= supfX=1

AfY (0.58)

is finite.

The set of all bounded linear operators from X to Y is denoted byL(X, Y). IfX=Y, we write L(X, X) = L(X).

Theorem 0.24. The spaceL(X, Y) together with the operator norm (0.58)is a normed space. It is a Banach space ifY is.

Proof. That (0.58) is indeed a norm is straightforward. IfYis complete andAn is a Cauchy sequence of operators, then Anf converges to an element

g for every f. Define a new operator A via Af = g. By continuity ofthe vector operations, A is linear and by continuity of the normAf =limn Anf (limn An)f, it is bounded. Furthermore, given > 0 there is some N such thatAn Am for n, m N and thusAnfAmf f. Taking the limitm , we see AnfAf f;that is, AnA.

By construction, a bounded operator is Lipschitz continuous,

AfY AfX, (0.59)and hence continuous. The converse is also true

Theorem 0.25. An operatorA is bounded if and only if it is continuous.Proof. SupposeA is continuous but not bounded. Then there is a sequenceof unit vectorsun such thatAun n. Thenfn = 1nun converges to 0 butAfn 1 does not converge to 0.

Moreover, if A is bounded and densely defined, it is no restriction toassume that it is defined on all ofX.

Theorem 0.26(B.L.T. theorem). LetA L(X, Y) and letYbe a Banachspace. IfD(A) is dense, there is a unique (continuous) extension ofA toXwhich has the same norm.

Proof. Since a bounded operator maps Cauchy sequences to Cauchy se-quences, this extension can only be given by

Af = limn Afn, fn D(A), fX.

To show that this definition is independent of the sequence fn f, letgnfbe a second sequence and observe

Afn Agn=A(fn gn) Afn gn 0.


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From continuity of vector addition and scalar multiplication it follows that

our extension is linear. Finally, from continuity of the norm we concludethat the norm does not increase.

An operator in L(X, C) is called abounded linear functionaland thespaceX= L(X, C) is called the dual space ofX. A sequence fn is said toconverge weakly, fn f, if(fn)(f) for every X.

The Banach space of bounded linear operators L(X) even has a multi-plication given by composition. Clearly this multiplication satisfies

(A + B)C=AC+ BC, A(B + C) =AB + BC, A, B , C L(X) (0.60)and

(AB)C=A(BC), (AB) = (A)B=A (B), C. (0.61)Moreover, it is easy to see that we have

AB AB. (0.62)However, note that our multiplication is not commutative (unless X is one-dimensional). We even have an identity, the identity operator I satisfyingI= 1.

A Banach space together with a multiplication satisfying the above re-quirements is called a Banach algebra. In particular, note that (0.62)ensures that multiplication is continuous (Problem0.22).

Problem 0.19. Show that the integral operator

(Kf)(x) =

10

K(x, y)f(y)dy,

where K(x, y) C([0, 1] [0, 1]), defined onD(K) = C[0, 1] is a boundedoperator both inX=C[0, 1] (max norm) andX=L2cont(0, 1).Problem 0.20. Show that the differential operator A = ddx defined on

D(A) =C1[0, 1]C[0, 1] is an unbounded operator.Problem 0.21. Show that

AB

A

B

for everyA, B

L(X).

Problem 0.22. Show that the multiplication in a Banach algebraXis con-tinuous: xnx andyny implyxnynxy.Problem 0.23. Let

f(z) =j=0

fjzj , |z|< R,


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0.6. LebesgueLp spaces 25

be a convergent power series with convergence radiusR > 0. SupposeA is

a bounded operator withA< R. Show thatf(A) =

j=0

fjAj

exists and defines a bounded linear operator (cf. Problem0.9).

0.6. Lebesgue Lp spaces

For this section some basic facts about the Lebesgue integral are required.The necessary background can be found in Appendix A. To begin with,SectionsA.1,A.3,and A.4 will be sufficient.

We fix some -finite measure space (X, , ) and denote byLp(X,d),

1p, the set of all complex-valued measurable functions for which

fp=X

|f|p d1/p

(0.63)

is finite. First of all note thatLp(X,d) is a linear space, since|f+g|p 2p max(|f|, |g|)p 2p max(|f|p, |g|p) 2p(|f|p + |g|p). Of course our hopeis thatLp(X,d) is a Banach space. However, there is a small technicalproblem (recall that a property is said to hold almost everywhere if the setwhere it fails to hold is contained in a set of measure zero):

Lemma 0.27. Letf be measurable. Then

X

|f|p d= 0 (0.64)if and only iff(x) = 0 almost everywhere with respect to .

Proof. Observe that we have A ={x|f(x)= 0} =nAn, where An ={x| |f(x)| > 1n}. If

|f|pd = 0, we must have (An) = 0 for every n andhence(A) = limn (An) = 0. The converse is obvious.

Note that the proof also shows that if f is not 0 almost everywhere,there is an >0 such that ({x| |f(x)| })> 0.Example. Let be the Lebesgue measure on R. Then the characteristicfunction of the rationals Q is zero a.e. (with respect to ).

Let be the Dirac measure centered at 0. Then f(x) = 0 a.e. (withrespect to ) if and only iff(0) = 0.

Thusfp = 0 only implies f(x) = 0 for almost everyx, but not for all!Hence.p is not a norm onLp(X,d). The way out of this misery is toidentify functions which are equal almost everywhere: Let

N(X,d) ={f|f(x) = 0-almost everywhere}. (0.65)


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ThenN(X,d) is a linear subspace ofLp(X,d) and we can consider thequotient space

Lp(X,d) =Lp(X,d)/N(X,d). (0.66)Ifd is the Lebesgue measure on XRn, we simply writeLp(X). Observethatfp is well-defined on Lp(X,d).

Even though the elements of Lp(X,d) are, strictly speaking, equiva-lence classes of functions, we will still call them functions for notationalconvenience. However, note that for f Lp(X,d) the value f(x) is notwell-defined (unless there is a continuous representative and different con-tinuous functions are in different equivalence classes, e.g., in the case ofLebesgue measure).

With this modification we are back in business since Lp

(X,d) turnsout to be a Banach space. We will show this in the following sections.

But before that let us also define L(X,d). It should be the set ofbounded measurable functions B(X) together with the sup norm. The onlyproblem is that if we want to identify functions equal almost everywhere, thesupremum is no longer independent of the equivalence class. The solutionis the essential supremum

f= inf{C| ({x| |f(x)|> C}) = 0}. (0.67)That is, C is an essential bound if|f(x)| Calmost everywhere and theessential supremum is the infimum over all essential bounds.

Example. If is the Lebesgue measure, then the essential sup ofQ withrespect to is 0. If is the Dirac measure centered at 0, then the essentialsup ofQ with respect to is 1 (since Q(0) = 1, and x = 0 is the onlypoint which counts for ).

As before we set

L(X,d) =B(X)/N(X,d) (0.68)and observe thatf is independent of the equivalence class.

If you wonder where thecomes from, have a look at Problem 0.24.As a preparation for proving that Lp is a Banach space, we will need

Holders inequality, which plays a central role in the theory of Lp spaces.In particular, it will imply Minkowskis inequality, which is just the triangleinequality for Lp.

Theorem 0.28 (Holders inequality). Letp andqbedual indices; that is,

1

p+

1

q = 1 (0.69)


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0.6. LebesgueLp spaces 27

with1p . Iff Lp(X,d) andgLq(X,d), thenf gL1(X,d)and f g1 fpgq. (0.70)Proof. The case p = 1, q=(respectively p =, q= 1) follows directlyfrom the properties of the integral and hence it remains to consider 1 g and equal tog whenevern


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Problem 0.28. Find a sequencefnwhich converges to0 inLp([0, 1], dx)but

for whichfn(x)0 for a.e.x[0, 1]does not hold. (Hint: EverynN canbe uniquely written asn = 2m+kwith0m and0k


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0.7. Appendix: The uniform boundedness principle 33

In other words, ifXn X is a sequence of closed subsets which coverX, at least one Xn contains a ball of radius >0.

Now we come to the first important consequence, the uniform bound-edness principle.

Theorem 0.39(BanachSteinhaus). LetXbe a Banach space andY somenormed linear space. Let{A} L(X, Y)be a family of bounded operators.SupposeAx C(x) is bounded for fixed x X. ThenA C isuniformly bounded.

Proof. Let

Xn={x| Ax n for all }=

{x| Ax n}.

ThennXn = Xby assumption. Moreover, by continuity ofA and thenorm, eachXn is an intersection of closed sets and hence closed. By Bairestheorem at least one contains a ball of positive radius: B(x0)Xn. Nowobserve

Ay A(y+ x0) + Ax0 n + Ax0fory< . Setting y = xx , we obtain

Ax n + C(x0)

xfor anyx.


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Part 1

Mathematical

Foundations of

Quantum Mechanics


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Chapter 1

Hilbert spaces

The phase space in classical mechanics is the Euclidean spaceR2n (for thenposition and n momentum coordinates). In quantum mechanics the phasespace is always a Hilbert space H. Hence the geometry of Hilbert spacesstands at the outset of our investigations.

1.1. Hilbert spaces

Suppose H is a vector space. A map .,..: HH C is called a sesquilinearform if it is conjugate linear in the first argument and linear in the second.A positive definite sesquilinear form is called an inner productor scalar

product. Associated with every scalar product is a norm

=

, . (1.1)

The triangle inequality follows from the CauchySchwarzBunjakowskiinequality:

|, | (1.2)with equality if and only if and are parallel.

IfH is complete with respect to the above norm, it is called a Hilbertspace. It is no restriction to assume that H is complete since one can easily

replace it by its completion.Example. The spaceL2(M,d) is a Hilbert space with scalar product givenby

f, g=M

f(x)g(x)d(x). (1.3)

37


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38 1. Hilbert spaces

Similarly, the set of all square summable sequences 2(N) is a Hilbert space

with scalar productf, g=

jN

fj gj. (1.4)

(Note that the second example is a special case of the first one; take M=Rand a sum of Dirac measures.)

A vector H is called normalized or a unit vector if = 1.Two vectors , H are called orthogonalor p erpendicular() if, = 0 and parallel if one is a multiple of the other.

If and are orthogonal, we have the Pythagorean theorem:

+ 2

=2

+ 2

, , (1.5)which is one line of computation.

Suppose is a unit vector. Then the projection of in the direction of is given by

=, (1.6)and defined via

= , (1.7)is perpendicular to.

These results can also be generalized to more than one vector. A set of

vectors{j}is called an orthonormal set(ONS) ifj, k= 0 for j=kandj, j= 1.Lemma 1.1. Suppose{j}nj=0 is an orthonormal set. Then every Hcan be written as

= + , =nj=0

j, j , (1.8)

where and are orthogonal. Moreover,j , = 0 for all1jn.In particular,

2 =

n

j=0 |j, |2 +

2. (1.9)

Moreover, every in the span of{j}nj=0 satisfies (1.10)

with equality holding if and only if =. In other words, is uniquelycharacterized as the vector in the span of{j}nj=0 closest to .


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1.2. Orthonormal bases 39

Proof. A straightforward calculation showsj, = 0 and hence and = are orthogonal. The formula for the norm follows byapplying (1.5) iteratively.

Now, fix a vector

=nj=0

cjj

in the span of{j}nj=0. Then one computes 2 =+ 2 =2 + 2

=2 +n

j=0|cj j, |2

from which the last claim follows.

From (1.9) we obtain Bessels inequality

nj=0

|j , |2 2 (1.11)

with equality holding if and only if lies in the span of{j}nj=0.Recall that a scalar product can be recovered from its norm by virtue of

the polarization identity

, =1

4 + 2 2 + i i2 i + i2 . (1.12)

A bijective linear operatorU L(H1,H2) is called unitary ifUpreservesscalar products:

U ,U2=, 1, , H1. (1.13)By the polarization identity this is the case if and only ifUpreserves norms:U 2 =1 for all H1. The two Hilbert space H1 and H2 are calledunitarily equivalentin this case.

Problem 1.1. The operator

S :2

(N)2

(N), (a1, a2, a3, . . . )(0, a1, a2, . . . )satisfiesSa=a. Is it unitary?

1.2. Orthonormal bases

Of course, since we cannot assume H to be a finite dimensional vector space,we need to generalize Lemma 1.1 to arbitrary orthonormal sets{j}jJ.


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We start by assuming that J is countable. Then Bessels inequality (1.11)

shows that jJ

|j, |2 (1.14)

converges absolutely. Moreover, for any finite subset KJwe havejK

j, j2 =jK

|j, |2 (1.15)

by the Pythagorean theorem and thusjJj, j is Cauchy if and only

if

jJ|j, |2 is. Now let J be arbitrary. Again, Bessels inequalityshows that for any given >0 there are at most finitely many j for which|j, | . Hence there are at most countably many jfor which |j , |>0. Thus it follows that

jJ|j, |2 (1.16)

is well-defined and so is jJ

j , j . (1.17)

In particular, by continuity of the scalar product we see that Lemma 1.1holds for arbitrary orthonormal sets without modifications.

Theorem 1.2. Suppose{j}jJ is an orthonormal set. Then every Hcan be written as

=

+

,

= jJ

j, j , (1.18)

where and are orthogonal. Moreover,j , = 0 for allj J. Inparticular,

2 =jJ

|j, |2 + 2. (1.19)

Moreover, every in the span of{j}jJ satisfies (1.20)

with equality holding if and only if =. In other words, is uniquelycharacterized as the vector in the span of{j}jJclosest to .

Note that from Bessels inequality (which of course still holds) it followsthat the map is continuous.

An orthonormal set which is not a proper subset of any other orthonor-mal set is called an orthonormal basis(ONB) due to the following result:

Theorem 1.3. For an orthonormal set{j}jJthe following conditions areequivalent:


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1.2. Orthonormal bases 41

(i){j}jJ is a maximal orthonormal set.(ii) For every vector H we have

=jJ

j, j. (1.21)

(iii) For every vector H we have2 =

jJ

|j, |2. (1.22)

(iv)j, = 0 for alljJ implies= 0.Proof. We will use the notation from Theorem1.2.(i)

(ii): If

= 0, then we can normalize

to obtain a unit vector

which is orthogonal to all vectors j. But then{j}jJ {} would be alarger orthonormal set, contradicting the maximality of{j}jJ.(ii)(iii): This follows since (ii) implies = 0.(iii) (iv): If, j = 0 for all j J, we conclude2 = 0 and hence= 0.(iv) (i): If{j}jJwere not maximal, there would be a unit vector such that{j}jJ {} is a larger orthonormal set. Butj, = 0 for alljJ implies = 0 by (iv), a contradiction.

Since is continuous, it suffices to check conditions (ii) and (iii)on a dense set.

Example. The set of functions

n(x) = 1

2einx, nZ, (1.23)

forms an orthonormal basis for H =L2(0, 2). The corresponding orthogo-nal expansion is just the ordinary Fourier series (Problem 1.20).

A Hilbert space isseparableif and only if there is a countable orthonor-mal basis. In fact, ifH is separable, then there exists a countable total set{j}Nj=0. HereN N ifH is finite dimensional andN=otherwise. Afterthrowing away some vectors, we can assume that n+1 cannot be expressedas a linear combinations of the vectors 0, . . . , n. Now we can construct

an orthonormal basis as follows: We begin by normalizing 0,

0= 00 . (1.24)

Next we take 1 and remove the component parallel to 0 and normalizeagain:

1= 1 0, 101 0, 10 . (1.25)


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Proceeding like this, we define recursively

n =n n1j=0 j, nj

n n1j=0 j, nj

. (1.26)

This procedure is known as GramSchmidt orthogonalization. Hencewe obtain an orthonormal set {j}Nj=0such that span{j}nj=0= span{j}nj=0for any finite n and thus also for N (ifN =). Since{j}Nj=0 is total, sois{j}Nj=0. Now suppose there is some = + H for which = 0.Since {j}Nj=1is total, we can find a in its span, such that


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1.3. The projection theorem and the Riesz lemma 43

Let me remark that ifHis not separable, there still exists an orthonor-

mal basis. However, the proof requires Zorns lemma: The collection ofall orthonormal sets in H can be partially ordered by inclusion. Moreover,any linearly ordered chain has an upper bound (the union of all sets in thechain). Hence Zorns lemma implies the existence of a maximal element,that is, an orthonormal basis.

Problem 1.2. Let{j} be some orthonormal basis. Show that a boundedlinear operator A is uniquely determined by its matrix elements Ajk =j, Ak with respect to this basis.Problem 1.3. Show thatL(H) is not separable ifH is infinite dimensional.

1.3. The projection theorem and the Riesz lemmaLet M H be a subset. Then M ={|, = 0, M} is calledthe orthogonal complement of M. By continuity of the scalar prod-uct it follows that M is a closed linear subspace and by linearity that(span(M)) = M. For example we have H ={0}since any vector in Hmust be in particular orthogonal to all vectors in some orthonormal basis.

Theorem 1.7(Projection theorem). LetMbe a closed linear subspace of aHilbert spaceH. Then every H can be uniquely written as= + withM andM. One writes

M

M= H (1.28)

in this situation.

Proof. SinceM is closed, it is a Hilbert space and has an orthonormal basis{j}jJ. Hence the result follows from Theorem 1.2.

In other words, to every H we can assign a unique vector whichis the vector in Mclosest to . The rest, , lies in M. The operatorPM= is called theorthogonal projectioncorresponding toM. Notethat we have

P2M =PM and PM, =, PM (1.29)

sincePM, =, =, PM. Clearly we have PM = PM = . Furthermore, (1.29) uniquely characterizes orthogonal projec-tions (Problem1.6).

Moreover, we see that the vectors in a closed subspace M are preciselythose which are orthogonal to all vectors in M; that is, M =M. IfMis an arbitrary subset, we have at least

M = span(M). (1.30)


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Note that by H={0}we see that M ={0}if and only ifM is total.Finally we turn to linear functionals, that is, to operators : H

C. By the CauchySchwarz inequality we know that : , is abounded linear functional (with norm). It turns out that in a Hilbertspace every bounded linear functional can be written in this way.

Theorem 1.8(Riesz lemma). Suppose is a bounded linear functional on aHilbert spaceH. Then there is a unique vector H such that() =, for all H. In other words, a Hilbert space is equivalent to its own dualspaceH = H.

Proof. If 0, we can choose = 0. Otherwise Ker() ={|() = 0}is a proper subspace and we can find a unit vector

Ker(). For every

H we have () ()Ker() and hence0 =, () ()= () (), .

In other words, we can choose = (). To see uniqueness, let 1,2 betwo such vectors. Then1 2, =1, 2, = () () = 0for any H, which shows 1 2 H={0}.

The following easy consequence is left as an exercise.

Corollary 1.9. Supposes is a bounded sesquilinear form; that is,

|s(, )| C . (1.31)

Then there is a unique bounded operatorA such thats(, ) =A,. (1.32)

Moreover,A C.Note that by the polarization identity (Problem 0.14), A is already

uniquely determined by its quadratic form qA() =,A.Problem 1.4. Suppose U : H H is unitary and M H. Show thatU M = (U M).

Problem 1.5. Show that an orthogonal projectionPM= 0 has norm one.Problem 1.6. SupposeP

L satisfies

P2 =P and P , = , P and setM= Ran(P). Show

P = forM andM is closed, M impliesP M and thusP = 0,

and concludeP =PM.


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1.4. Orthogonal sums and tensor products 45

Problem 1.7. LetP1,P2 be two orthogonal projections. Show thatP1P2(that is,, P1 , P2) if and only ifRan(P1) Ran(P2). Show inthis case that the two projections commute (that is, P1P2 = P2P1) and thatP2 P1 is also a projection. (Hints:Pj= if and only ifPj =andRan(P1)Ran(P2) if and only ifP2P1= P1.)Problem 1.8. Show P : L2(R) L2(R), f(x) 12 (f(x) +f(x)) is aprojection. Compute its range and kernel.

Problem 1.9. Prove Corollary1.9.

Problem 1.10. Consider the sesquilinear form

B(f, g) = 1

0 x

0 f(t)dt x

0 g(t)dt dx

inL2(0, 1). Show that it is bounded and find the corresponding operatorA.(Hint: Partial integration.)

1.4. Orthogonal sums and tensor products

Given two Hilbert spaces H1 and H2, we define their orthogonal sumH1 H2to be the set of all pairs (1, 2)H1 H2 together with the scalarproduct

(1, 2), (1, 2)

=

1, 1

1+

2, 2

2. (1.33)

It is left as an exercise to verify that H1 H2 is again a Hilbert space.Moreover, H1 can be identified with{(1, 0)|1 H1} and we can regardH1 as a subspace ofH1H2, and similarly for H2. It is also customary towrite 1+ 2 instead of (1, 2).

More generally, let Hj,jN, be a countable collection of Hilbert spacesand define

j=1

Hj ={j=1

j | j Hj,j=1

j2j


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and H. Hence we start with the set of all finite linear combinations ofelements ofH H:

F(H,H) ={nj=1

j(j, j)|(j, j) HH, jC}. (1.36)

Since we want (1+2)= 1+2,(1 +2) =1 +2,and () = (), we considerF(H,H)/N(H,H), where

N(H,H) = span{n

j,k=1

jk(j, k) (nj=1

jj ,

nk=1

kk)} (1.37)

and write for the equivalence class of (, ).Next we define

, =, , (1.38)which extends to a sesquilinear form onF(H,H)/N(H,H). To show that weobtain a scalar product, we need to ensure positivity. Let =

i iii=

0 and pick orthonormal bases j, k for span{i}, span{i}, respectively.Then

=j,k

jkjk, jk =i

ij , ik, i (1.39)

and we compute

, =j,k

|jk |2 >0. (1.40)

The completion ofF(H,H)/N(H,

H) with respect to the induced norm iscalled the tensor productH H ofHand H.

Lemma 1.10. Ifj, k are orthonormal bases forH, H, respectively, then

jk is an orthonormal basis forH H.Proof. Thatj k is an orthonormal set is immediate from (1.38). More-over, since span{j}, span{k}are dense in H, H, respectively, it is easy tosee that j k is dense inF(H,H)/N(H,H). But the latter is dense inH H. Example. We have H Cn = Hn.

Example. Let (M,d) and ( M, d) be two measure spaces. Then we haveL2(M,d) L2(M, d) =L2(M M,d d).

Clearly we have L2(M,d) L2(M , d) L2(M M,d d). Nowtake an orthonormal basis j k for L2(M,d)L2(M , d) as in ourprevious lemma. Then

M

M

(j(x)k(y))f(x, y)d(x)d(y) = 0 (1.41)


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1.5. TheC algebra of bounded linear operators 47

impliesM

j(x)fk(x)d(x) = 0, fk(x) =

M

k(y)f(x, y)d(y) (1.42)

and hence fk(x) = 0 -a.e. x. But this implies f(x, y) = 0 for -a.e. x and-a.e. y and thus f= 0. Hence jk is a basis for L2(M M,d d)and equality follows.

It is straightforward to extend the tensor product to any finite numberof Hilbert spaces. We even note

(j=1

Hj) H =j=1

(HjH), (1.43)

where equality has to be understood in the sense that both spaces are uni-tarily equivalent by virtue of the identification

(j=1

j) =j=1

j . (1.44)

Problem 1.11. Show that = 0 if and only if= 0 or= 0.Problem 1.12. We have = = 0 if and only if there is someC\{0} such that= and= 1.Problem 1.13. Show (1.43)

1.5. The C algebra of bounded linear operators

We start by introducing a conjugation for operators on a Hilbert space H.LetA L(H). Then the adjoint operatoris defined via

, A=A, (1.45)(compare Corollary1.9).

Example. IfH =Cn and A= (ajk)1j,kn, then A= (akj)1j,kn.

Lemma 1.11. LetA, B L(H). Then(i) (A + B)= A+ B, (A)= A,

(ii) A= A,(iii) (AB)= BA,(iv)A=A andA2 =AA=AA.

Proof. (i) and (ii) are obvious. (iii) follows from , (AB)=A,B=BA, . (iv) follows from

A= sup==1

|, A|= sup==1

|A,|=A


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and

AA= sup==1

|, AA|= sup==1

|A,A|

= sup=1

A2 =A2,

where we have used= sup=1 |, |.

As a consequence ofA =A observe that taking the adjoint iscontinuous.

In general, a Banach algebraAtogether with an involution

(a + b)= a+ b, (a)= a, a= a, (ab)= ba (1.46)

satisfying

a2 =aa (1.47)is called aC algebra. The elementa is called the adjoint ofa. Note thata=afollows from (1.47) andaa aa.

Any subalgebra which is also closed under involution is called a-subalgebra. An ideal is a subspaceI A such that a I, b A implyab Iand ba I. If it is closed under the adjoint map, it is called a -ideal.Note that if there is an identitye, we havee= e and hence (a1)= (a)1

(show this).Example. The continuous functions C(I) together with complex conjuga-tion form a commutative C algebra.

An elementa Ais called normalifaa= aa,self-adjointifa = a,unitary ifaa =aa= I, an (orthogonal) projection ifa= a =a2, andpositive ifa = bb for some b A. Clearly both self-adjoint and unitaryelements are normal.

Problem 1.14. LetA L(H). Show thatA is normal if and only if

A

=

A

,

H.

(Hint: Problem0.14.)

Problem 1.15. Show thatU : H H is unitary if and only ifU1 =U.Problem 1.16. Compute the adjoint of

S :2(N)2(N), (a1, a2, a3, . . . )(0, a1, a2, . . . ).


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Proof. Let k be an orthonormal basis. Then by the usual diagonal se-

quence argument we can find a subsequence nm such thatk, nm con-verges for all k. Since n is bounded,, nm converges for every Hand hence nm is a weak Cauchy sequence.

Finally, let me remark that similar concepts can be introduced for oper-ators. This is of particular importance for the case of unbounded operators,where convergence in the operator norm makes no sense at all.

A sequence of operators An is said to converge stronglyto A,

s-limn An = A : AnA x D(A) D(An). (1.49)

It is said toconverge weaklytoA,

w-limn An=A : An A D(A) D(An). (1.50)

Clearly norm convergence implies strong convergence and strong conver-gence implies weak convergence.

Example. Consider the operator Sn L(2(N)) which shifts a sequence nplaces to the left, that is,

Sn(x1, x2, . . . ) = (xn+1, xn+2, . . . ), (1.51)

and the operatorSn L(2(N)) which shifts a sequence n places to the rightand fills up the first nplaces with zeros, that is,

Sn(x1, x2, . . . ) = (0, . . . , 0 n places

, x1, x2, . . . ). (1.52)

Then Sn converges to zero strongly but not in norm (sinceSn = 1) andSn converges weakly to zero (since, Sn =Sn, ) but not strongly(sinceSn=) .

Note that this example also shows that taking adjoints is not continuous

with respect to strong convergence! IfAnsA, we only have

, An=An, A,=, A (1.53)and hence An A in general. However, ifAn and Aare normal, we have

(An

A)

=

(An

A)

(1.54)

and hence AnsA in this case. Thus at least for normal operators taking

adjoints is continuous with respect to strong convergence.

Lemma 1.14. SupposeAn is a sequence of bounded operators.

(i) s-limn An = A impliesA lim infn An.(ii) Every strong Cauchy sequenceAn is bounded:An C.


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1.7. Appendix: The StoneWeierstra theorem 51

(iii) If An A for in some dense set andAn C, thens-limn An = A.

The same result holds if strong convergence is replaced by weak convergence.

Proof. (i) follows from

A = limn An lim infn An

for every with= 1.(ii) follows as in Lemma1.12(i).(iii) Just use

An A An An + An A + A A

2C

+

An

A

and choose in the dense subspace such that 4C and n largesuch thatAn A 2 .

The case of weak convergence is left as an exercise. (Hint: (2.14).)

Problem 1.17. Supposen andn . Thenn, n , .Problem 1.18. Let{j}j=1 be some orthonormal basis. Show thatn if and only ifn is bounded andj, n j, for everyj. Show thatthis is wrong without the boundedness assumption.

Problem 1.19. A subspace M H is closed if and only if every weakCauchy sequence inMhas a limit inM. (Hint: M=M.)

1.7. Appendix: The StoneWeierstra theorem

In case of a self-adjoint operator, the spectral theorem will show that theclosed-subalgebra generated by this operator is isomorphic to the C al-gebra of continuous functions C(K) over some compact set. Hence it isimportant to be able to identify dense sets:

Theorem 1.15 (StoneWeierstra, real version). SupposeK is a compactset and letC(K, R) be the Banach algebra of continuous functions (with thesup norm).

If F C(K,R

) contains the identity 1 and separates points (i.e., foreveryx1=x2 there is some functionf F such thatf(x1)=f(x2)), thenthe algebra generated byF is dense.

Proof. Denote by A the algebra generated by F. Note that iff A, wehave|f| A: By the Weierstra approximation theorem (Theorem 0.15)there is a polynomial pn(t) such that

|t| pn(t) < 1n for t f(K) andhencepn(f) |f|.


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1.7. Appendix: The StoneWeierstra theorem 53

Proof. There are two possibilities: either all f F vanish at one pointt0 K (there can be at most one such point since F separates points)or there is no such point. If there is no such point, we can proceed as inthe proof of the StoneWeierstra theorem to show that the identity canbe approximated by elements in A (note that to show|f| A if f A,we do not need the identity, since pn can be chosen to contain no constantterm). If there is such a t0, the identity is clearly missing from A. However,adding the identity to A, we get A+ C = C(K) and it is easy to see thatA={f C(K)|f(t0) = 0}. Problem 1.20. Show that the functionsn(x) =

12

einx, nZ, form anorthonormal basis forH =L2(0, 2).

Problem 1.21. LetkN andI R. Show that the-subalgebra generatedby fz0(t) = 1(tz0)k for one z0 C is dense in the C algebra C(I) ofcontinuous functions vanishing at infinity

forI=R ifz0C\R andk= 1, 2, forI= [a, ) ifz0(, a) and anyk, forI= (, a] [b, ) ifz0(a, b) andk odd.

(Hint: Add to R to make it compact.)


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Chapter 2

Self-adjointness and

spectrum

2.1. Some quantum mechanics

In quantum mechanics, a single particle living in R3 is described by acomplex-valued function (the wave function)

(x, t), (x, t)R3 R, (2.1)where x corresponds to a point in space and t corresponds to time. Thequantity t(x) =|(x, t)|2 is interpreted as theprobability densityof theparticle at the time t. In particular, must be normalized according to

R3|(x, t)|2d3x= 1, tR. (2.2)

The location x of the particle is a quantity which can be observed (i.e.,measured) and is hence called observable. Due to our probabilistic inter-pretation, it is also a random variable whose expectationis given by

E(x) =

R3

x|(x, t)|2d3x. (2.3)

In a real life setting, it will not be possible to measure xdirectly and one willonly be able to measure certain functions ofx. For example, it is possible to

check whether the particle is inside a certain area of space (e.g., inside adetector). The corresponding observable is the characteristic function(x)of this set. In particular, the number

E() =

R3

(x)|(x, t)|2d3x=

|(x, t)|2d3x (2.4)

55


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2.1. Some quantum mechanics 57

is a state (i.e.,(t)= 1), we haveU(t)=. (2.8)

Such operators are calledunitary. Next, since we have assumed uniquenessof solutions to the initial value problem, we must have

U(0) =I, U(t + s) =U(t)U(s). (2.9)

A family of unitary operators U(t) having this property is called a one-parameter unitary group. In addition, it is natural to assume that thisgroup is strongly continuous; that is,

limtt0

U(t)= U(t0), H. (2.10)Each such group has an infinitesimal generatordefined by

H= limt0

i

t(U(t) ), D(H) ={ H| lim

t01

t(U(t) ) exists}.

(2.11)This operator is called the Hamiltonianand corresponds to the energy ofthe system. If (0) D(H), then (t) is a solution of the Schrodingerequation(in suitable units)

id

dt(t) =H (t). (2.12)

This equation will be the main subject of our course.

In summary, we have the following axioms of quantum mechanics.

Axiom 1. The configuration space of a quantum system is a complexseparable Hilbert space H and the possible states of this system are repre-sented by the elements ofHwhich have norm one.

Axiom 2. Each observablea corresponds to a linear operatorA definedmaximally on a dense subset D(A). Moreover, the operator correspond-ing to a polynomial Pn(a) =

nj=0 ja

j, j R, is Pn(A) =nj=0 jA

j ,

D(Pn(A)) = D(An) ={ D(A)|AD(An1)} (A0 =I).

Axiom 3. The expectation value for a measurement of a, when thesystem is in the state D(A), is given by (2.5), which must be real forall D(A).

Axiom 4. The time evolution is given by a strongly continuous one-parameter unitary groupU(t). The generator of this group corresponds tothe energy of the system.

In the following sections we will try to draw some mathematical conse-quences from these assumptions:

First we will see that Axioms 2 and 3 imply that observables corre-spond to self-adjoint operators. Hence these operators play a central role


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58 2. Self-adjointness and spectrum

in quantum mechanics and we will derive some of their basic properties.

Another crucial role is played by the set of all possible expectation valuesfor the measurement ofa, which is connected with the spectrum(A) of thecorresponding operator A.

The problem of defining functions of an observable will lead us to thespectral theorem (in the next chapter), which generalizes the diagonalizationof symmetric matrices.

Axiom 4 will be the topic of Chapter5.

2.2. Self-adjoint operators

Let H be a (complex separable) Hilbert space. Alinear operatoris a linear

mapping

A: D(A)H, (2.13)where D(A) is a linear subspace ofH, called the domainofA. It is calledboundedif the operator norm

A= sup=1

A= sup==1

|,A| (2.14)

is finite. The second equality follows since equality in|,A| Ais attained when A = z for some z C. If A is bounded, it is norestriction to assume D(A) = H and we will always do so. The Banach spaceof all bounded linear operators is denoted by L(H). Products of (unbounded)operators are defined naturally; that is, AB = A(B) for D(AB) ={D(B)|BD(A)}.

The expression,Aencountered in the previous section is called thequadratic form,

qA() =,A, D(A), (2.15)associated to A. An operator can be reconstructed from its quadratic formvia the polarization identity

,A= 14

(qA( + ) qA( ) + iqA( i) iqA( + i)) . (2.16)

A densely defined linear operator A is calledsymmetric(or hermitian) if

,A=A,, , D(A). (2.17)The justification for this definition is provided by the following

Lemma 2.1. A densely defined operatorA is symmetric if and only if thecorresponding quadratic form is real-valued.


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2.2. Self-adjoint operators 59

Proof. Clearly (2.17) implies that Im(qA()) = 0. Conversely, taking the

imaginary part of the identityqA(+ i) =qA() + qA() + i(,A ,A)

shows ReA, = Re,A. Replacing by i in this last equationshows ImA,= Im,Aand finishes the proof.

In other words, a densely defined operatorA is symmetric if and only if

,A=A,, D(A). (2.18)This already narrows the class of admissible operators to the class of

symmetric operators by Axiom 3. Next, let us tackle the issue of the correctdomain.

By Axiom 2, A should be defined maximally; that is, if A is anothersymmetric operator such that A A, then A = A. Here we writeA AifD(A) D(A) and A = A for all D(A). The operator A is calledan extensionofA in this case. In addition, we writeA = A if both AAand A Ahold.

The adjoint operator A of a densely defined linear operator A isdefined by

D(A) = { H| H :,A=, , D(A)},A = .

(2.19)

The requirement that D(A) be dense implies thatA is well-defined. How-ever, note that D(A) might not be dense in general. In fact, it mightcontain no vectors other than 0.

Clearly we have (A) = A for C and (A+ B) A+Bprovided D(A+ B) = D(A) D(B) is dense. However, equality will nothold in general unless one operator is bounded (Problem 2.2).

For later use, note that (Problem 2.4)

Ker(A) = Ran(A). (2.20)

For symmetric operators we clearly have AA. If, in addition,A = Aholds, then A is called self-adjoint. Our goal is to show that observables

correspond to self-adjoint operators. This is for example true in the case ofthe position operatorx, which is a special case of a multiplication operator.

Example. (Multiplication operator) Consider the multiplication operator

(Af)(x) =A(x)f(x), D(A) ={fL2(Rn, d) | Af L2(Rn, d)}(2.21)

given by multiplication with the measurable function A : Rn C. Firstof all note that D(A) is dense. In fact, consider n ={x Rn | |A(x)|


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n} Rn. Then, for everyf L2(Rn, d) the function fn =nf D(A)converges to f as n by dominated convergence.

Next, let us compute the adjoint ofA. Performing a formal computation,we have for h, f D(A) that

h,Af=

h(x)A(x)f(x)d(x) =

(A(x)h(x))f(x)d(x) =Ah,f,(2.22)

where A is multiplication byA(x),

(Af)(x) =A(x)f(x), D(A) ={f L2(Rn, d) | AfL2(Rn, d)}.(2.23)

Note D(A) = D(A). At first sight this seems to show that the adjoint of

A is A. But for our calculation we had to assume hD(A) and there

might be some functions in D(A) which do not satisfy this requirement! Inparticular, our calculation only shows AA. To show that equality holds,we need to work a little harder:

Ifh D(A), there is some gL2(Rn, d) such that h(x)A(x)f(x)d(x) =

g(x)f(x)d(x), f D(A), (2.24)

and thus (h(x)A(x) g(x))f(x)d(x) = 0, f D(A). (2.25)

In particular, n(x)(h(x)A(x)

g(x))f(x)d(x) = 0, fL2(Rn, d), (2.26)

which shows that n(h(x)A(x) g(x)) L2(Rn, d) vanishes. Sincen

is arbitrary, we even have h(x)A(x) = g(x) L2(Rn, d) and thus A ismultiplication by A(x) and D(A) = D(A).

In particular,A is self-adjoint ifA is real-valued. In the general case wehave at leastAf=Af for all f D(A) = D(A). Such operators arecalled normal.

Now note that

A

B

B

A

; (2.27)

that is, increasing the domain of A implies decreasing the domain of A.Thus there is no point in trying to extend the domain of a self-adjointoperator further. In fact, ifA is self-adjoint and B is a symmetric extension,we infer ABBA= A implyingA= B.Corollary 2.2. Self-adjoint operators are maximal; that is, they do not haveany symmetric extensions.


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2.2. Self-adjoint operators 61

Furthermore, ifA is densely defined (which is the case ifA is symmet-

ric), we can consider A. From the definition (2.19) it is clear thatAAand thus A is an extension ofA. This extension is closely related to ex-tending a linear subspace M viaM = M(as we will see a bit later) andthus is called the closureA = A ofA.

IfA is symmetric, we have A A and hence A = A A; that is,A lies between A and A. Moreover,, A=A, for all D(A), D(A) implies that A is symmetric since A= A for D(A).Example. (Differential operator) Take H =L2(0, 2).

(i) Consider the operator

A0f=

id

dx

f, D(A0) =

{f

C1[0, 2]

|f(0) =f(2) = 0

}. (2.28)

That A0 is symmetric can be shown by a simple integration by parts (dothis). Note that the boundary conditions f(0) = f(2) = 0 are chosensuch that the boundary terms occurring from integration by parts vanish.However, this will also follow once we have computed A0. IfgD(A0), wemust have 2

0g(x)(if(x))dx=

20

g(x)f(x)dx (2.29)

for some gL2(0, 2). Integration by parts (cf. (2.116)) shows

2

0

f(x)g(x) i x

0

g(t)dt

dx= 0. (2.30)

In fact, this formula holds for g C[0, 2]. Since the set of continuousfunctions is dense, the general case gL2(0, 2) follows by approximatinggwith continuous functions and taking limits on both sides using dominatedconvergence.

Hence g(x)i x0 g(t)dt {f|f D(A0)}. But{f|f D(A0)} ={hC[0, 2]| 20 h(t)dt= 0}(show this) implying g(x) =g(0)+i x0 g(t)dtsince{f|f D(A0)} ={h H|1, h = 0} ={1} and{1} = span{1}.Thus gAC[0, 2], where

AC[a, b] =

{f

C[a, b]

|f(x) =f(a) +

x

a

g(t)dt, g

L1(a, b)

} (2.31)

denotes the set of all absolutely continuous functions (see Section2.7). Insummary,gD(A0) impliesgAC[0, 2] andA0g= g=ig. Conversely,for every gH1(0, 2) ={f AC[0, 2]|fL2(0, 2)}, (2.29) holds withg=ig and we conclude

A0f=id

dxf, D(A0) =H

1(0, 2). (2.32)


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In particular,A0 is symmetric but not self-adjoint. SinceA0 = A0 A0,

we can use integration by parts to compute

0 =g, A0f A0g, f= i(f(0)g(0) f(2)g(2)) (2.33)and since the boundary values ofg D(A0) can be prescribed arbitrarily,we must have f(0) =f(2) = 0. Thus

A0f=i ddx

f, D(A0) ={f D(A0) | f(0) =f(2) = 0}. (2.34)

(ii) Now let us take

Af=i ddx

f, D(A) ={f C1[0, 2] | f(0) =f(2)}, (2.35)

which is clearly an extension ofA0. Thus AA0 and we compute

0 =g,Af Ag, f= if(0)(g(0) g(2)). (2.36)Since this must hold for all f D(A), we conclude g(0) =g(2) and

Af=i ddx

f, D(

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