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Introduction to Quantum Mechanics
Introduction to Quantum Mechanics
David J. Griffiths, Darrell F. Schroeter
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İl:
2018
Nəşr:
3rd
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Cambridge University Press
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english
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644
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1107189632
ISBN 13:
9781107189638
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equation^{1369}
particle^{459}
spin^{373}
electron^{323}
matrix^{322}
quantum^{321}
operator^{313}
momentum^{287}
probability^{228}
schrödinger equation^{212}
hamiltonian^{199}
vector^{194}
hydrogen^{193}
perturbation^{188}
classical^{176}
linear^{166}
scattering^{166}
hermitian^{159}
energies^{155}
theorem^{149}
particles^{148}
quantum mechanics^{146}
infinite^{146}
operators^{145}
angular momentum^{140}
measurement^{134}
equations^{129}
approximation^{127}
integral^{126}
eigenvalues^{123}
atom^{118}
electrons^{115}
harmonic oscillator^{115}
solutions^{115}
formula^{110}
magnetic^{107}
calculate^{107}
symmetry^{106}
stationary^{106}
uncertainty^{100}
dimensional^{99}
expectation value^{99}
degeneracy^{97}
spherical^{94}
phys^{93}
construct^{92}
perturbation theory^{91}
vectors^{91}
delta^{91}
infinite square^{90}
degenerate^{90}
eigenfunctions^{89}
eigenvalue^{89}
infinite square well^{85}
transition^{82}
finite^{79}
eigenvectors^{77}
linear combination^{77}
eigenstates^{74}
parity^{74}
Əlaqəli Kitab Siyahıları
10 comments
eberthbruce
The quality of the equations is poor.
08 June 2020 (03:49)
Mukhtar Musa Muhammad
Reading culture development
09 September 2020 (11:23)
Forest
Awesome! This book is quite helpful for my study!
13 September 2020 (14:57)
T.Shi
Damn griffiths is a gift from heavens xD A frikin good book and one of the best for beginners
10 November 2020 (18:44)
Saitama
Please give the download link
02 March 2021 (17:51)
Jason
Griffiths often skips steps and does not bother to explain things that are “obvious”. For me this can sometimes be troublesome. Luckily you can find a full QM course (based on this book) on YouTube by Brant Carlson, who explains things with a little bit more details.
23 April 2021 (11:09)
patel lav ishwarbhai
It's very useful app.?????
18 May 2021 (00:52)
ABHISHEK Mishra
I am here to mark the beginning of journey.
23 June 2021 (15:08)
Sai Misaki
Absolutely classical textbook.
03 August 2021 (10:46)
PengZhw
All equations are replaced with pic, hard to read.
29 September 2022 (18:40)
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I NT ROD UCT I ON TO Q UANT UM MECHANI CS Third edition Changes and additions to the new edition of this classic textbook include: A new chapter on Symmetries and Conservation Laws New problems and examples Improved explanations More numerical problems to be worked on a computer New applications to solid state physics Consolidated treatment of timedependent potentials David J. Griffiths received his BA (1964) and PhD (1970) from Harvard University. He taught at Hampshire College, Mount Holyoke College, and Trinity College before joining the faculty at Reed College in 1978. In 2001–2002 he was visiting Professor of Physics at the Five Colleges (UMass, Amherst, Mount Holyoke, Smith, and Hampshire), and in the spring of 2007 he taught Electrodynamics at Stanford. Although his PhD was in elementary particle theory, most of his research is in electrodynamics and quantum mechanics. He is the author of over fifty articles and four books: Introduction to Electrodynamics (4th edition, Cambridge University Press, 2013), Introduction to Elementary Particles (2nd edition, WileyVCH, 2008), Introduction to Quantum Mechanics (2nd edition, Cambridge, 2005), and Revolutions in TwentiethCentury Physics (Cambridge, 2013). Darrell F. Schroeter is a condensed matter theorist. He received his BA (1995) from Reed College and his PhD (2002) from Stanford University where he was a National Science Foundation Graduate Research Fellow. Before joining the Reed College faculty in 2007, Schroeter taught at both Swarthmore College and Occidental College. His record of successful theoretical research with undergraduate students was recognized in 2011 when he was named as a KITPAnacapa scholar. 2 I NT ROD UCT I ON TO Q UANT UM MECHANI CS Third edition DAVID J. GRIFFITHS and DARRELL F. SCHROETER Reed College, Oregon 3 University Printing House, Cambridge CB2 8BS, United Kingdom One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia 314–321, 3rd Floor, Plot 3, Splendor For; um, Jasola District Centre, New Delhi – 110025, India 79 Anson Road, #06–04/06, Singapore 079906 Cambridge University Press is part of the University of Cambridge. It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning, and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.org/9781107189638 DOI: 10.1017/9781316995433 Second edition © David Griffiths 2017 Third edition © Cambridge University Press 2018 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. This book was previously published by Pearson Education, Inc. 2004 Second edition reissued by Cambridge University Press 2017 Third edition 2018 Printed in the United Kingdom by TJ International Ltd. Padstow Cornwall, 2018 A catalogue record for this publication is available from the British Library. Library of Congress CataloginginPublication Data Names: Griffiths, David J.  Schroeter, Darrell F. Title: Introduction to quantum mechanics / David J. Griffiths (Reed College, Oregon), Darrell F. Schroeter (Reed College, Oregon). Description: Third edition.  blah : Cambridge University Press, 2018. Identifiers: LCCN 2018009864  ISBN 9781107189638 Subjects: LCSH: Quantum theory. Classification: LCC QC174.12 .G75 2018  DDC 530.12–dc23 LC record available at https://lccn.loc.gov/2018009864 ISBN 9781107189638 Hardback Additional resources for this publication at www.cambridge.org/IQM3ed Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or thirdparty internet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. 4 5 Contents Preface I Theory 1 The Wave Function 1.1 The Schrödinger Equation 1.2 The Statistical Interpretation 1.3 Probability 1.3.1 Discrete Variables 1.3.2 Continuous Variables 1.4 Normalization 1.5 Momentum 1.6 The Uncertainty Principle Further Problems on Chapter 1 2 TimeIndependent Schrödinger Equation 2.1 Stationary States 2.2 The Infinite Square Well 2.3 The Harmonic Oscillator 2.3.1 Algebraic Method 2.3.2 Analytic Method 2.4 The Free Particle 2.5 The DeltaFunction Potential 2.5.1 Bound States and Scattering States 2.5.2 The DeltaFunction Well 2.6 The Finite Square Well Further Problems on Chapter 2 3 Formalism 3.1 Hilbert Space 3.2 Observables 3.2.1 Hermitian Operators 3.2.2 Determinate States 3.3 Eigenfunctions of a Hermitian Operator 3.3.1 Discrete Spectra 3.3.2 Continuous Spectra 6 3.4 Generalized Statistical Interpretation 3.5 The Uncertainty Principle 3.5.1 Proof of the Generalized Uncertainty Principle 3.5.2 The MinimumUncertainty Wave Packet 3.5.3 The EnergyTime Uncertainty Principle 3.6 Vectors and Operators 3.6.1 Bases in Hilbert Space 3.6.2 Dirac Notation 3.6.3 Changing Bases in Dirac Notation Further Problems on Chapter 3 4 Quantum Mechanics in Three Dimensions 4.1 The Schröger Equation 4.1.1 Spherical Coordinates 4.1.2 The Angular Equation 4.1.3 The Radial Equation 4.2 The Hydrogen Atom 4.2.1 The Radial Wave Function 4.2.2 The Spectrum of Hydrogen 4.3 Angular Momentum 4.3.1 Eigenvalues 4.3.2 Eigenfunctions 4.4 Spin 4.4.1 Spin 1/2 4.4.2 Electron in a Magnetic Field 4.4.3 Addition of Angular Momenta 4.5 Electromagnetic Interactions 4.5.1 Minimal Coupling 4.5.2 The Aharonov–Bohm Effect Further Problems on Chapter 4 5 Identical Particles 5.1 TwoParticle Systems 5.1.1 Bosons and Fermions 5.1.2 Exchange Forces 5.1.3 Spin 5.1.4 Generalized Symmetrization Principle 5.2 Atoms 5.2.1 Helium 5.2.2 The Periodic Table 5.3 Solids 5.3.1 The Free Electron Gas 5.3.2 Band Structure Further Problems on Chapter 5 7 6 Symmetries & Conservation Laws 6.1 Introduction 6.1.1 Transformations in Space 6.2 The Translation Operator 6.2.1 How Operators Transform 6.2.2 Translational Symmetry 6.3 Conservation Laws 6.4 Parity 6.4.1 Parity in One Dimension 6.4.2 Parity in Three Dimensions 6.4.3 Parity Selection Rules 6.5 Rotational Symmetry 6.5.1 Rotations About the z Axis 6.5.2 Rotations in Three Dimensions 6.6 Degeneracy 6.7 Rotational Selection Rules 6.7.1 Selection Rules for Scalar Operators 6.7.2 Selection Rules for Vector Operators 6.8 Translations in Time 6.8.1 The Heisenberg Picture 6.8.2 TimeTranslation Invariance Further Problems on Chapter 6 II Applications 7 TimeIndependent Perturbation Theory 7.1 Nondegenerate Perturbation Theory 7.1.1 General Formulation 7.1.2 FirstOrder Theory 7.1.3 SecondOrder Energies 7.2 Degenerate Perturbation Theory 7.2.1 TwoFold Degeneracy 7.2.2 “Good” States 7.2.3 HigherOrder Degeneracy 7.3 The Fine Structure of Hydrogen 7.3.1 The Relativistic Correction 7.3.2 SpinOrbit Coupling 7.4 The Zeeman Effect 7.4.1 WeakField Zeeman Effect 7.4.2 StrongField Zeeman Effect 7.4.3 IntermediateField Zeeman Effect 7.5 Hyperfine Splitting in Hydrogen 8 Further Problems on Chapter 7 8 The Varitional Principle 8.1 Theory 8.2 The Ground State of Helium 8.3 The Hydrogen Molecule Ion 8.4 The Hydrogen Molecule Further Problems on Chapter 8 9 The WKB Approximation 9.1 The “Classical” Region 9.2 Tunneling 9.3 The Connection Formulas Further Problems on Chapter 9 10 Scattering 10.1 Introduction 10.1.1 Classical Scattering Theory 10.1.2 Quantum Scattering Theory 10.2 Partial Wave Analysis 10.2.1 Formalism 10.2.2 Strategy 10.3 Phase Shifts 10.4 The Born Approximation 10.4.1 Integral Form of the Schrödinger Equation 10.4.2 The First Born Approximation 10.4.3 The Born Series Further Problems on Chapter 10 11 Quantum Dynamics 11.1 TwoLevel Systems 11.1.1 The Perturbed System 11.1.2 TimeDependent Perturbation Theory 11.1.3 Sinusoidal Perturbations 11.2 Emission and Absorption of Radiation 11.2.1 Electromagnetic Waves 11.2.2 Absorption, Stimulated Emission, and Spontaneous Emission 11.2.3 Incoherent Perturbations 11.3 Spontaneous Emission 11.3.1 Einstein’s A and B Coefficients 11.3.2 The Lifetime of an Excited State 11.3.3 Selection Rules 11.4 Fermi’s Golden Rule 11.5 The Adiabatic Approximation 11.5.1 Adiabatic Processes 9 11.5.2 The Adiabatic Theorem Further Problems on Chapter 11 12 Afterword 12.1 The EPR Paradox 12.2 Bell’s Theorem 12.3 Mixed States and the Density Matrix 12.3.1 Pure States 12.3.2 Mixed States 12.3.3 Subsystems 12.4 The NoClone Theorem 12.5 Schrödinger’s Cat Appendix Linear Algebra A.1 Vectors A.2 Inner Products A.3 Matrices A.4 Changing Bases A.5 Eigenvectors and Eigenvalues A.6 Hermitian Transformations Index 10 Preface Unlike Newton’s mechanics, or Maxwell’s electrodynamics, or Einstein’s relativity, quantum theory was not created—or even definitively packaged—by one individual, and it retains to this day some of the scars of its exhilarating but traumatic youth. There is no general consensus as to what its fundamental principles are, how it should be taught, or what it really “means.” Every competent physicist can “do” quantum mechanics, but the stories we tell ourselves about what we are doing are as various as the tales of Scheherazade, and almost as implausible. Niels Bohr said, “If you are not confused by quantum physics then you haven’t really understood it”; Richard Feynman remarked, “I think I can safely say that nobody understands quantum mechanics.” The purpose of this book is to teach you how to do quantum mechanics. Apart from some essential background in Chapter 1, the deeper quasiphilosophical questions are saved for the end. We do not believe one can intelligently discuss what quantum mechanics means until one has a firm sense of what quantum mechanics does. But if you absolutely cannot wait, by all means read the Afterword immediately after finishing Chapter 1. Not only is quantum theory conceptually rich, it is also technically difficult, and exact solutions to all but the most artificial textbook examples are few and far between. It is therefore essential to develop special techniques for attacking more realistic problems. Accordingly, this book is divided into two parts;1 Part I covers the basic theory, and Part II assembles an arsenal of approximation schemes, with illustrative applications. Although it is important to keep the two parts logically separate, it is not necessary to study the material in the order presented here. Some instructors, for example, may wish to treat timeindependent perturbation theory right after Chapter 2. This book is intended for a onesemester or oneyear course at the junior or senior level. A onesemester course will have to concentrate mainly on Part I; a fullyear course should have room for supplementary material beyond Part II. The reader must be familiar with the rudiments of linear algebra (as summarized in the Appendix), complex numbers, and calculus up through partial derivatives; some acquaintance with Fourier analysis and the Dirac delta function would help. Elementary classical mechanics is essential, of course, and a little electrodynamics would be useful in places. As always, the more physics and math you know the easier it will be, and the more you will get out of your study. But quantum mechanics is not something that flows smoothly and naturally from earlier theories. On the contrary, it represents an abrupt and revolutionary departure from classical ideas, calling forth a wholly new and radically counterintuitive way of thinking about the world. That, indeed, is what makes it such a fascinating subject. At first glance, this book may strike you as forbiddingly mathematical. We encounter Legendre, Hermite, and Laguerre polynomials, spherical harmonics, Bessel, Neumann, and Hankel functions, Airy functions, and even the Riemann zeta function—not to mention Fourier transforms, Hilbert spaces, hermitian operators, and Clebsch–Gordan coefficients. Is all this baggage really necessary? Perhaps not, but physics is like carpentry: Using the right tool makes the job easier, not more difficult, and teaching quantum mechanics without the appropriate mathematical equipment is like having a tooth extracted with a pair of pliers—it’s possible, but painful. (On the other hand, it can be tedious and diverting if the instructor feels obliged to give elaborate lessons on the proper use of each tool. Our instinct is to hand the students shovels and tell them to 11 start digging. They may develop blisters at first, but we still think this is the most efficient and exciting way to learn.) At any rate, we can assure you that there is no deep mathematics in this book, and if you run into something unfamiliar, and you don’t find our explanation adequate, by all means ask someone about it, or look it up. There are many good books on mathematical methods—we particularly recommend Mary Boas, Mathematical Methods in the Physical Sciences, 3rd edn, Wiley, New York (2006), or George Arfken and HansJurgen Weber, Mathematical Methods for Physicists, 7th edn, Academic Press, Orlando (2013). But whatever you do, don’t let the mathematics—which, for us, is only a tool—obscure the physics. Several readers have noted that there are fewer worked examples in this book than is customary, and that some important material is relegated to the problems. This is no accident. We don’t believe you can learn quantum mechanics without doing many exercises for yourself. Instructors should of course go over as many problems in class as time allows, but students should be warned that this is not a subject about which anyone has natural intuitions—you’re developing a whole new set of muscles here, and there is simply no substitute for calisthenics. Mark Semon suggested that we offer a “Michelin Guide” to the problems, with varying numbers of stars to indicate the level of difficulty and importance. This seemed like a good idea (though, like the quality of a restaurant, the significance of a problem is partly a matter of taste); we have adopted the following rating scheme: an essential problem that every reader should study; a somewhat more difficult or peripheral problem; an unusually challenging problem, that may take over an hour. (No stars at all means fast food: OK if you’re hungry, but not very nourishing.) Most of the onestar problems appear at the end of the relevant section; most of the threestar problems are at the end of the chapter. If a computer is required, we put a mouse in the margin. A solution manual is available (to instructors only) from the publisher. In preparing this third edition we have tried to retain as much as possible the spirit of the first and second. Although there are now two authors, we still use the singular (“I”) in addressing the reader—it feels more intimate, and after all only one of us can speak at a time (“we” in the text means you, the reader, and I, the author, working together). Schroeter brings the fresh perspective of a solid state theorist, and he is largely responsible for the new chapter on symmetries. We have added a number of problems, clarified many explanations, and revised the Afterword. But we were determined not to allow the book to grow fat, and for that reason we have eliminated the chapter on the adiabatic approximation (significant insights from that chapter have been incorporated into Chapter 11), and removed material from Chapter 5 on statistical mechanics (which properly belongs in a book on thermal physics). It goes without saying that instructors are welcome to cover such other topics as they see fit, but we want the textbook itself to represent the essential core of the subject. We have benefitted from the comments and advice of many colleagues, who read the original manuscript, pointed out weaknesses (or errors) in the first two editions, suggested improvements in the presentation, and supplied interesting problems. We especially thank P. K. Aravind (Worcester Polytech), Greg Benesh (Baylor), James Bernhard (Puget Sound), Burt Brody (Bard), Ash Carter (Drew), Edward Chang (Massachusetts), Peter Collings (Swarthmore), Richard Crandall (Reed), Jeff Dunham (Middlebury), Greg Elliott (Puget Sound), John Essick (Reed), Gregg Franklin (Carnegie Mellon), Joel Franklin (Reed), 12 Henry Greenside (Duke), Paul Haines (Dartmouth), J. R. Huddle (Navy), Larry Hunter (Amherst), David Kaplan (Washington), Don Koks (Adelaide), Peter Leung (Portland State), Tony Liss (Illinois), Jeffry Mallow (Chicago Loyola), James McTavish (Liverpool), James Nearing (Miami), Dick Palas, Johnny Powell (Reed), Krishna Rajagopal (MIT), Brian Raue (Florida International), Robert Reynolds (Reed), Keith Riles (Michigan), Klaus SchmidtRohr (Brandeis), Kenny Scott (London), Dan Schroeder (Weber State), Mark Semon (Bates), Herschel Snodgrass (Lewis and Clark), John Taylor (Colorado), Stavros Theodorakis (Cyprus), A. S. Tremsin (Berkeley), Dan Velleman (Amherst), Nicholas Wheeler (Reed), Scott Willenbrock (Illinois), William Wootters (Williams), and Jens Zorn (Michigan). 1 This structure was inspired by David Park’s classic text Introduction to the Quantum Theory, 3rd edn, McGrawHill, New York (1992). 13 Part I Theory ◈ 14 1 The Wave Function ◈ 15 1.1 The Schrödinger Equation Imagine a particle of mass m, constrained to move along the x axis, subject to some specified force (Figure 1.1). The program of classical mechanics is to determine the position of the particle at any given time: . Once we know that, we can figure out the velocity kinetic energy determining , the momentum , the , or any other dynamical variable of interest. And how do we go about ? We apply Newton’s second law: . (For conservative systems—the only kind we shall consider, and, fortunately, the only kind that occur at the microscopic level—the force can be expressed as the derivative of a potential energy function,1 , and Newton’s law reads .) This, together with appropriate initial conditions (typically the position and velocity at ), determines . Figure 1.1: A “particle” constrained to move in one dimension under the influence of a specified force. Quantum mechanics approaches this same problem quite differently. In this case what we’re looking for is the particle’s wave function, , and we get it by solving the Schrödinger equation: (1.1) Here i is the square root of , and is Planck’s constant—or rather, his original constant (h) divided by : (1.2) The Schrödinger equation plays a role logically analogous to Newton’s second law: Given suitable initial conditions (typically, ), the Schrödinger equation determines classical mechanics, Newton’s law determines for all future 16 time.2 for all future time, just as, in 1.2 The Statistical Interpretation But what exactly is this “wave function,” and what does it do for you once you’ve got it? After all, a particle, by its nature, is localized at a point, whereas the wave function (as its name suggests) is spread out in space (it’s a function of x, for any given t). How can such an object represent the state of a particle? The answer is provided by Born’s statistical interpretation, which says that point x, at time t—or, more gives the probability of finding the particle at precisely,3 (1.3) Probability is the area under the graph of . For the wave function in Figure 1.2, you would be quite likely to find the particle in the vicinity of point A, where is large, and relatively unlikely to find it near point B. Figure 1.2: A typical wave function. The shaded area represents the probability of finding the particle between a and b. The particle would be relatively likely to be found near A, and unlikely to be found near B. The statistical interpretation introduces a kind of indeterminacy into quantum mechanics, for even if you know everything the theory has to tell you about the particle (to wit: its wave function), still you cannot predict with certainty the outcome of a simple experiment to measure its position—all quantum mechanics has to offer is statistical information about the possible results. This indeterminacy has been profoundly disturbing to physicists and philosophers alike, and it is natural to wonder whether it is a fact of nature, or a defect in the theory. Suppose I do measure the position of the particle, and I find it to be at point C.4 Question: Where was the particle just before I made the measurement? There are three plausible answers to this question, and they serve to characterize the main schools of thought regarding quantum indeterminacy: 1. The realist position: The particle was at C. This certainly seems reasonable, and it is the response Einstein advocated. Note, however, that if this is true then quantum mechanics is an incomplete theory, since the particle really was at C, and yet quantum mechanics was unable to tell us so. To the realist, indeterminacy is not a fact of nature, but a reflection of our ignorance. As d’Espagnat put it, “the position of the particle was never indeterminate, but was merely unknown to the experimenter.”5 Evidently is not the whole story—some additional information (known as a hidden variable) is needed to provide a complete description of the particle. 2. The orthodox position: The particle wasn’t really anywhere. It was the act of measurement that forced it to “take a stand” (though how and why it decided on the point C we dare not ask). Jordan said it most starkly: “Observations not only disturb what is to be measured, they produce it …We compel [the particle] to assume a definite position.”6 This view (the socalled Copenhagen interpretation), is associated with Bohr and his followers. Among physicists it has always been the most widely accepted position. Note, however, that if it is correct there is something very peculiar about the act of measurement—something that almost a century of debate has done precious little to illuminate. 17 3. The agnostic position: Refuse to answer. This is not quite as silly as it sounds—after all, what sense can there be in making assertions about the status of a particle before a measurement, when the only way of knowing whether you were right is precisely to make a measurement, in which case what you get is no longer “before the measurement”? It is metaphysics (in the pejorative sense of the word) to worry about something that cannot, by its nature, be tested. Pauli said: “One should no more rack one’s brain about the problem of whether something one cannot know anything about exists all the same, than about the ancient question of how many angels are able to sit on the point of a needle.”7 For decades this was the “fallback” position of most physicists: they’d try to sell you the orthodox answer, but if you were persistent they’d retreat to the agnostic response, and terminate the conversation. Until fairly recently, all three positions (realist, orthodox, and agnostic) had their partisans. But in 1964 John Bell astonished the physics community by showing that it makes an observable difference whether the particle had a precise (though unknown) position prior to the measurement, or not. Bell’s discovery effectively eliminated agnosticism as a viable option, and made it an experimental question whether 1 or 2 is the correct choice. I’ll return to this story at the end of the book, when you will be in a better position to appreciate Bell’s argument; for now, suffice it to say that the experiments have decisively confirmed the orthodox interpretation:8 a particle simply does not have a precise position prior to measurement, any more than the ripples on a pond do; it is the measurement process that insists on one particular number, and thereby in a sense creates the specific result, limited only by the statistical weighting imposed by the wave function. What if I made a second measurement, immediately after the first? Would I get C again, or does the act of measurement cough up some completely new number each time? On this question everyone is in agreement: A repeated measurement (on the same particle) must return the same value. Indeed, it would be tough to prove that the particle was really found at C in the first instance, if this could not be confirmed by immediate repetition of the measurement. How does the orthodox interpretation account for the fact that the second measurement is bound to yield the value C? It must be that the first measurement radically alters the wave function, so that it is now sharply peaked about C (Figure 1.3). We say that the wave function collapses, upon measurement, to a spike at the point C (it soon spreads out again, in accordance with the Schrödinger equation, so the second measurement must be made quickly). There are, then, two entirely distinct kinds of physical processes: “ordinary” ones, in which the wave function evolves in a leisurely fashion under the Schrödinger equation, and “measurements,” in which suddenly and discontinuously collapses.9 Figure 1.3: Collapse of the wave function: graph of immediately after a measurement has found the particle at point C. Example 1.1 Electron Interference. I have asserted that particles (electrons, for example) have a wave nature, encoded in . How might we check this, in the laboratory? The classic signature of a wave phenomenon is interference: two waves in phase interfere constructively, and out of phase they interfere destructively. The wave nature of light was confirmed in 18 1801 by Young’s famous doubleslit experiment, showing interference “fringes” on a distant screen when a monochromatic beam passes through two slits. If essentially the same experiment is done with electrons, the same pattern develops,10 confirming the wave nature of electrons. Now suppose we decrease the intensity of the electron beam, until only one electron is present in the apparatus at any particular time. According to the statistical interpretation each electron will produce a spot on the screen. Quantum mechanics cannot predict the precise location of that spot—all it can tell us is the probability of a given electron landing at a particular place. But if we are patient, and wait for a hundred thousand electrons—one at a time—to make the trip, the accumulating spots reveal the classic twoslit interference pattern (Figure 1.4). 11 Figure 1.4: Buildup of the electron interference pattern. (a) Eight electrons, (b) 270 electrons, (c) 2000 electrons, (d) 160,000 electrons. Reprinted courtesy of the Central Research Laboratory, Hitachi, Ltd., Japan. Of course, if you close off one slit, or somehow contrive to detect which slit each electron passes through, the interference pattern disappears; the wave function of the emerging particle is now entirely different (in the first case because the boundary conditions for the Schrödinger equation have been changed, and in the second because of the collapse of the wave function upon measurement). But with both slits open, and no interruption of the electron in flight, each electron interferes with itself; it didn’t pass through one slit or the other, but through both at once, just as a water wave, impinging on a jetty with two openings, interferes with itself. There is nothing mysterious about this, once you have accepted the notion that particles obey a wave equation. The truly astonishing thing is the blipbyblip assembly of the pattern. In any classical wave theory the pattern would develop smoothly and continuously, simply getting more intense as time goes on. The quantum process is more like the pointillist painting of Seurat: The picture emerges from the cumulative contributions of all the individual dots.12 19 20 1.3 Probability 21 1.3.1 Discrete Variables Because of the statistical interpretation, probability plays a central role in quantum mechanics, so I digress now for a brief discussion of probability theory. It is mainly a question of introducing some notation and terminology, and I shall do it in the context of a simple example. Imagine a room containing fourteen people, whose ages are as follows: one person aged 14, one person aged 15, three people aged 16, two people aged 22, two people aged 24, five people aged 25. If we let while represent the number of people of age j, then , for instance, is zero. The total number of people in the room is (1.4) (In the example, of course, .) Figure 1.5 is a histogram of the data. The following are some questions one might ask about this distribution. Figure 1.5: Histogram showing the number of people, Question 1 , with age j, for the example in Section 1.3.1. If you selected one individual at random from this group, what is the probability that this person’s age would be 15? Answer One chance in 14, since there are 14 possible choices, all equally likely, of whom only one has that particular age. If is the probability of , and so on. In general, 22 getting age j, then (1.5) Notice that the probability of getting either 14 or 15 is the sum of the individual probabilities (in this case, 1/7). In particular, the sum of all the probabilities is 1—the person you select must have some age: (1.6) Question 2 What is the most probable age? Answer 25, obviously; five people share this age, whereas at most three have any other age. The most probable j is the j for which is a maximum. Question 3 What is the median age? Answer 23, for 7 people are younger than 23, and 7 are older. (The median is that value of j such that the probability of getting a larger result is the same as the probability of getting a smaller result.) Question 4 What is the average (or mean) age? Answer In general, the average value of j (which we shall write thus: ) is (1.7) Notice that there need not be anyone with the average age or the median age—in this example nobody happens to be 21 or 23. In quantum mechanics the average is usually the quantity of interest; in that context it has come to be called the expectation value. It’s a misleading term, since it suggests that this is the outcome you would be most likely to get if you made a single measurement (that would be the most probable value, not the average value)—but I’m afraid we’re stuck with it. Question 5 What is the average of the squares of the ages? Answer You could get , with probability 1/14, or , with probability 1/14, or , with probability 3/14, and so on. The average, then, is (1.8) In general, the average value of some function of j is given by (1.9) (Equations 1.6, 1.7, and 1.8 are, if you like, special cases of this formula.) Beware: The average of the squares, , is not equal, in general, to the square of the average, babies, aged 1 and 3, then , but . 23 . For instance, if the room contains just two Now, there is a conspicuous difference between the two histograms in Figure 1.6, even though they have the same median, the same average, the same most probable value, and the same number of elements: The first is sharply peaked about the average value, whereas the second is broad and flat. (The first might represent the age profile for students in a bigcity classroom, the second, perhaps, a rural oneroom schoolhouse.) We need a numerical measure of the amount of “spread” in a distribution, with respect to the average. The most obvious way to do this would be to find out how far each individual is from the average, (1.10) and compute the average of (Note that . Trouble is, of course, that you get zero: is constant—it does not change as you go from one member of the sample to another—so it can be taken outside the summation.) To avoid this irritating problem you might decide to average the absolute value of . But absolute values are nasty to work with; instead, we get around the sign problem by squaring before averaging: (1.11) This quantity is known as the variance of the distribution; σ itself (the square root of the average of the square of the deviation from the average—gulp!) is called the standard deviation. The latter is the customary measure of the spread about . Figure 1.6: Two histograms with the same median, same average, and same most probable value, but different standard deviations. There is a useful little theorem on variances: Taking the square root, the standard deviation itself can be written as 24 (1.12) In practice, this is a much faster way to get σ than by direct application of Equation 1.11: simply calculate and , subtract, and take the square root. Incidentally, I warned you a moment ago that general, equal to . Since is not, in is plainly nonnegative (from its definition 1.11), Equation 1.12 implies that (1.13) and the two are equal only when , which is to say, for distributions with no spread at all (every member having the same value). 25 1.3.2 Continuous Variables So far, I have assumed that we are dealing with a discrete variable—that is, one that can take on only certain isolated values (in the example, j had to be an integer, since I gave ages only in years). But it is simple enough to generalize to continuous distributions. If I select a random person off the street, the probability that her age is precisely 16 years, 4 hours, 27 minutes, and 3.333… seconds is zero. The only sensible thing to speak about is the probability that her age lies in some interval—say, between 16 and 17. If the interval is sufficiently short, this probability is proportional to the length of the interval. For example, the chance that her age is between 16 and 16 plus two days is presumably twice the probability that it is between 16 and 16 plus one day. (Unless, I suppose, there was some extraordinary baby boom 16 years ago, on exactly that day—in which case we have simply chosen an interval too long for the rule to apply. If the baby boom lasted six hours, we’ll take intervals of a second or less, to be on the safe side. Technically, we’re talking about infinitesimal intervals.) Thus (1.14) The proportionality factor, , is often loosely called “the probability of getting x,” but this is sloppy language; a better term is probability density. The probability that x lies between a and b (a finite interval) is given by the integral of : (1.15) and the rules we deduced for discrete distributions translate in the obvious way: (1.16) (1.17) (1.18) (1.19) Example 1.2 Suppose someone drops a rock off a cliff of height h. As it falls, I snap a million photographs, at random intervals. On each picture I measure the distance the rock has fallen. Question: What is the average of all these distances? That is to say, what is the time average of the distance traveled?13 Solution: The rock starts out at rest, and picks up speed as it falls; it spends more time near the top, so the average distance will surely be less than The velocity is . Ignoring air resistance, the distance x at time t is , and the total flight time is 26 . The probability that a The velocity is , and the total flight time is particular photograph was taken between t and distance in the corresponding range x to is . The probability that a , so the probability that it shows a is Thus the probability density (Equation 1.14) is (outside this range, of course, the probability density is zero). We can check this result, using Equation 1.16: The average distance (Equation 1.17) is which is somewhat less than , as anticipated. Figure 1.7 shows the graph of . Notice that a probability density can be infinite, though probability itself (the integral of ρ) must of course be finite (indeed, less than or equal to 1). Figure 1.7: The probability density in Example 1.2: ∗ . Problem 1.1 For the distribution of ages in the example in Section 1.3.1: (a) Compute and . (b) Determine for each j, and use Equation 1.11 to compute the standard deviation. (c) Use your results in (a) and (b) to check Equation 1.12. 27 Problem 1.2 (a) Find the standard deviation of the distribution in Example 1.2. (b) What is the probability that a photograph, selected at random, would show a distance x more than one standard deviation away from the average? ∗ Problem 1.3 Consider the gaussian distribution where A, a, and are positive real constants. (The necessary integrals are inside the back cover.) (a) Use Equation 1.16 to determine A. (b) Find , , and σ. (c) Sketch the graph of . 28 1.4 Normalization We return now to the statistical interpretation of the wave function (Equation 1.3), which says that is the probability density for finding the particle at point x, at time t. It follows (Equation 1.16) that the integral of over all x must be 1 (the particle’s got to be somewhere): (1.20) Without this, the statistical interpretation would be nonsense. However, this requirement should disturb you: After all, the wave function is supposed to be determined by the Schrödinger equation—we can’t go imposing an extraneous condition on two are consistent. Well, a glance at Equation 1.1 reveals that if without checking that the is a solution, so too is , where A is any (complex) constant. What we must do, then, is pick this undetermined multiplicative factor so as to ensure that Equation 1.20 is satisfied. This process is called normalizing the wave function. For some solutions to the Schrödinger equation the integral is infinite; in that case no multiplicative factor is going to make it 1. The same goes for the trivial solution . Such nonnormalizable solutions cannot represent particles, and must be rejected. Physically realizable states correspond to the squareintegrable solutions to Schrödinger’s equation.14 But wait a minute! Suppose I have normalized the wave function at time will stay normalized, as time goes on, and . How do I know that it evolves? (You can’t keep renormalizing the wave function, for then A becomes a function of t, and you no longer have a solution to the Schrödinger equation.) Fortunately, the Schrödinger equation has the remarkable property that it automatically preserves the normalization of the wave function—without this crucial feature the Schrödinger equation would be incompatible with the statistical interpretation, and the whole theory would crumble. This is important, so we’d better pause for a careful proof. To begin with, (1.21) (Note that the integral is a function only of t, so I use a total derivative a function of x as well as t, so it’s a partial derivative on the left, but the integrand is on the right.) By the product rule, (1.22) Now the Schrödinger equation says that (1.23) and hence also (taking the complex conjugate of Equation 1.23) (1.24) 29 so (1.25) The integral in Equation 1.21 can now be evaluated explicitly: (1.26) But must go to zero as x goes to normalizable.15 infinity—otherwise the wave function would not be It follows that (1.27) and hence that the integral is constant (independent of time); if is normalized at , it stays normalized for all future time. QED Problem 1.4 At time a particle is represented by the wave function where A, a, and b are (positive) constants. (a) Normalize (that is, find A, in terms of a and b). (b) Sketch , as a function of x. (c) Where is the particle most likely to be found, at ? (d) What is the probability of finding the particle to the left of a? Check your result in the limiting cases and . (e) What is the expectation value of x? ∗ Problem 1.5 Consider the wave function where A, , and ω are positive real constants. (We’ll see in Chapter 2 for what potential (V) this wave function satisfies the Schrödinger equation.) (a) Normalize . (b) Determine the expectation values of x and . (c) Find the standard deviation of x. Sketch the graph of of x, and mark the points and , as a function , to illustrate the sense in which σ represents the “spread” in x. What is the probability that the particle would be found outside this range? 30 31 1.5 For a particle in state Momentum , the expectation value of x is (1.28) What exactly does this mean? It emphatically does not mean that if you measure the position of one particle over and over again, is the average of the results you’ll get. On the contrary: The first measurement (whose outcome is indeterminate) will collapse the wave function to a spike at the value actually obtained, and the subsequent measurements (if they’re performed quickly) will simply repeat that same result. Rather, is the average of measurements performed on particles all in the state , which means that either you must find some way of returning the particle to its original state after each measurement, or else you have to prepare a whole ensemble of particles, each in the same state , and measure the positions of all of them: is the average of these results. I like to picture a row of bottles on a shelf, each containing a particle in the state (relative to the center of the bottle). A graduate student with a ruler is assigned to each bottle, and at a signal they all measure the positions of their respective particles. We then construct a histogram of the results, which should match , and compute the average, which should agree with . (Of course, since we’re only using a finite sample, we can’t expect perfect agreement, but the more bottles we use, the closer we ought to come.) In short, the expectation value is the average of measurements on an ensemble of identicallyprepared systems, not the average of repeated measurements on one and the same system. Now, as time goes on, will change (because of the time dependence of ), and we might be interested in knowing how fast it moves. Referring to Equations 1.25 and 1.28, we see that16 (1.29) This expression can be simplified using integrationbyparts:17 (1.30) (I used the fact that , and threw away the boundary term, on the ground that goes to zero at infinity.) Performing another integrationbyparts, on the second term, we conclude: (1.31) What are we to make of this result? Note that we’re talking about the “velocity” of the expectation value of x, which is not the same thing as the velocity of the particle. Nothing we have seen so far would enable us to calculate the velocity of a particle. It’s not even clear what velocity means in quantum mechanics: If the particle doesn’t have a determinate position (prior to measurement), neither does it have a welldefined velocity. All we could reasonably ask for is the probability of getting a particular value. We’ll see in Chapter 3 how to construct the probability density for velocity, given ; for the moment it will suffice to postulate that the expectation value of the velocity is equal to the time derivative of the expectation value of position: 32 (1.32) Equation 1.31 tells us, then, how to calculate directly from Actually, it is customary to work with momentum . , rather than velocity: (1.33) Let me write the expressions for and in a more suggestive way: (1.34) (1.35) We say that the operator18 x “represents” position, and the operator calculate expectation values we “sandwich” the appropriate operator between “represents” momentum; to and , and integrate. That’s cute, but what about other quantities? The fact is, all classical dynamical variables can be expressed in terms of position and momentum. Kinetic energy, for example, is and angular momentum is (the latter, of course, does not occur for motion in one dimension). To calculate the expectation value of any such quantity, and , we simply replace every p by , insert the resulting operator between , and integrate: (1.36) For example, the expectation value of the kinetic energy is (1.37) Equation 1.36 is a recipe for computing the expectation value of any dynamical quantity, for a particle in state ; it subsumes Equations 1.34 and 1.35 as special cases. I have tried to make Equation 1.36 seem plausible, given Born’s statistical interpretation, but in truth this represents such a radically new way of doing business (as compared with classical mechanics) that it’s a good idea to get some practice using it before we come back (in Chapter 3) and put it on a firmer theoretical foundation. In the mean time, if you prefer to think of it as an axiom, that’s fine with me. 33 Problem 1.6 Why can’t you do integrationbyparts directly on the middle expression in Equation 1.29—pull the time derivative over onto x, note that , and conclude that ∗ Problem 1.7 Calculate ? . Answer: (1.38) This is an instance of Ehrenfest’s theorem, which asserts that expectation values obey the classical laws.19 Problem 1.8 Suppose you add a constant to the potential energy (by “constant” I mean independent of x as well as t). In classical mechanics this doesn’t change anything, but what about quantum mechanics? Show that the wave function picks up a timedependent phase factor: the expectation value of a dynamical variable? 34 . What effect does this have on 1.6 The Uncertainty Principle Imagine that you’re holding one end of a very long rope, and you generate a wave by shaking it up and down rhythmically (Figure 1.8). If someone asked you “Precisely where is that wave?” you’d probably think he was a little bit nutty: The wave isn’t precisely anywhere—it’s spread out over 50 feet or so. On the other hand, if he asked you what its wavelength is, you could give him a reasonable answer: it looks like about 6 feet. By contrast, if you gave the rope a sudden jerk (Figure 1.9), you’d get a relatively narrow bump traveling down the line. This time the first question (Where precisely is the wave?) is a sensible one, and the second (What is its wavelength?) seems nutty—it isn’t even vaguely periodic, so how can you assign a wavelength to it? Of course, you can draw intermediate cases, in which the wave is fairly well localized and the wavelength is fairly well defined, but there is an inescapable tradeoff here: the more precise a wave’s position is, the less precise is its wavelength, and vice versa.20 A theorem in Fourier analysis makes all this rigorous, but for the moment I am only concerned with the qualitative argument. Figure 1.8: A wave with a (fairly) welldefined wavelength, but an illdefined position. Figure 1.9: A wave with a (fairly) welldefined position, but an illdefined wavelength. This applies, of course, to any wave phenomenon, and hence in particular to the quantum mechanical wave function. But the wavelength of is related to the momentum of the particle by the de Broglie formula:21 (1.39) Thus a spread in wavelength corresponds to a spread in momentum, and our general observation now says that the more precisely determined a particle’s position is, the less precisely is its momentum. Quantitatively, (1.40) where is the standard deviation in x, and is the standard deviation in p. This is Heisenberg’s famous uncertainty principle. (We’ll prove it in Chapter 3, but I wanted to mention it right away, so you can test it out on the examples in Chapter 2.) Please understand what the uncertainty principle means: Like position measurements, momentum measurements yield precise answers—the “spread” here refers to the fact that measurements made on identically prepared systems do not yield identical results. You can, if you want, construct a state such that 35 position measurements will be very close together (by making a localized “spike”), but you will pay a price: Momentum measurements on this state will be widely scattered. Or you can prepare a state with a definite momentum (by making a long sinusoidal wave), but in that case position measurements will be widely scattered. And, of course, if you’re in a really bad mood you can create a state for which neither position nor momentum is well defined: Equation 1.40 is an inequality, and there’s no limit on how big just make and can be— some long wiggly line with lots of bumps and potholes and no periodic structure. ∗ Problem 1.9 A particle of mass m has the wave function where A and a are positive real constants. (a) Find A. (b) For what potential energy function, , is this a solution to the Schrödinger equation? (c) Calculate the expectation values of (d) Find and , and . . Is their product consistent with the uncertainty principle? 36 Further Problems on Chapter 1 Problem 1.10 Consider the first 25 digits in the decimal expansion of π (3, 1, 4, 1, 5, 9, … ). (a) If you selected one number at random, from this set, what are the probabilities of getting each of the 10 digits? (b) What is the most probable digit? What is the median digit? What is the average value? (c) Find the standard deviation for this distribution. Problem 1.11 [This problem generalizes Example 1.2.] Imagine a particle of mass m and energy E in a potential well , sliding frictionlessly back and forth between the classical turning points (a and b in Figure 1.10). Classically, the probability of finding the particle in the range dx (if, for example, you took a snapshot at a random time t) is equal to the fraction of the time T it takes to get from a to b that it spends in the interval dx: (1.41) where is the speed, and (1.42) Thus (1.43) This is perhaps the closest classical analog22 to (a) Use conservation of energy to express (b) As an example, find . Plot . in terms of E and . for the simple harmonic oscillator, , and check that it is correctly normalized. (c) For the classical harmonic oscillator in part (b), find 37 , , and . Figure 1.10: Classical particle in a potential well. ∗∗ Problem 1.12 What if we were interested in the distribution of momenta , for the classical harmonic oscillator (Problem 1.11(b)). (a) Find the classical probability distribution to (b) Calculate (c) (note that p ranges from ). , , and . What’s the classical uncertainty product, , for this system? Notice that this product can be as small as you like, classically, simply by sending . But in quantum mechanics, as we shall see in Chapter 2, the energy of a simple harmonic oscillator cannot be less than , where is the classical frequency. In that case what can you say about the product ? Problem 1.13 Check your results in Problem 1.11(b) with the following “numerical experiment.” The position of the oscillator at time t is (1.44) You might as well take (that sets the scale for time) and (that sets the scale for length). Make a plot of x at 10,000 random times, and compare it with . Hint: In Mathematica, first define then construct a table of positions: and finally, make a histogram of the data: Meanwhile, make a plot of the density function, , and, using Show, superimpose the two. Problem 1.14 Let be the probability of finding the particle in the range , at time t. (a) Show that where What are the units of 38 ? Comment: J is called the probability What are the units of ? Comment: J is called the probability current, because it tells you the rate at which probability is “flowing” past the point x. If is increasing, then more probability is flowing into the region at one end than flows out at the other. (b) Find the probability current for the wave function in Problem 1.9. (This is not a very pithy example, I’m afraid; we’ll encounter more substantial ones in due course.) Problem 1.15 Show that for any two (normalizable) solutions to the Schrödinger equation (with the same ), and . Problem 1.16 A particle is represented (at time ) by the wave function (a) Determine the normalization constant A. (b) What is the expectation value of x? (c) What is the expectation value of p? (Note that you cannot get it from . Why not?) (d) Find the expectation value of . (e) Find the expectation value of . (f) Find the uncertainty in . (g) Find the uncertainty in . (h) Check that your results are consistent with the uncertainty principle. ∗∗ Problem 1.17 Suppose you wanted to describe an unstable particle, that spontaneously disintegrates with a “lifetime” τ. In that case the total probability of finding the particle somewhere should not be constant, but should decrease at (say) an exponential rate: A crude way of achieving this result is as follows. In Equation 1.24 we tacitly assumed that V (the potential energy) is real. That is certainly reasonable, but it leads to the “conservation of probability” enshrined in Equation 1.27. What if we assign to V an imaginary part: where is the true potential energy and Γ is a positive real constant? (a) Show that (in place of Equation 1.27) we now get 39 (b) Solve for , and find the lifetime of the particle in terms of Γ. Problem 1.18 Very roughly speaking, quantum mechanics is relevant when the de Broglie wavelength of the particle in question characteristic size of the system is greater than the . In thermal equilibrium at (Kelvin) temperature T, the average kinetic energy of a particle is (where is Boltzmann’s constant), so the typical de Broglie wavelength is (1.45) The purpose of this problem is to determine which systems will have to be treated quantum mechanically, and which can safely be described classically. (a) Solids. The lattice spacing in a typical solid is around the temperature below which the unbound23 nm. Find electrons in a solid are quantum mechanical. Below what temperature are the nuclei in a solid quantum mechanical? (Use silicon as an example.) Moral: The free electrons in a solid are always quantum mechanical; the nuclei are generally not quantum mechanical. The same goes for liquids (for which the interatomic spacing is roughly the same), with the exception of helium below 4 K. (b) Gases. For what temperatures are the atoms in an ideal gas at pressure P quantum mechanical? Hint: Use the ideal gas law to deduce the interatomic spacing. Answer: . Obviously (for the gas to show quantum behavior) we want m to be as small as possible, and P as large as possible. Put in the numbers for helium at atmospheric pressure. Is hydrogen in outer space (where the interatomic spacing is about 1 cm and the temperature is 3 K) quantum mechanical? (Assume it’s monatomic hydrogen, not H .) 1 Magnetic forces are an exception, but let’s not worry about them just yet. By the way, we shall assume throughout this book that the motion is nonrelativistic 2 . For a delightful firsthand account of the origins of the Schrödinger equation see the article by Felix Bloch in Physics Today, December 1976. 3 The wave function itself is complex, but (where is the complex conjugate of ) is real and nonnegative—as a probability, of course, must be. 4 Of course, no measuring instrument is perfectly precise; what I mean is that the particle was found in the vicinity of C, as defined by the precision of the equipment. 5 Bernard d’Espagnat, “The Quantum Theory and Reality” (Scientific American, November 1979, p. 165). 6 Quoted in a lovely article by N. David Mermin, “Is the moon there when nobody looks?” (Physics Today, April 1985, p. 38). 40 7 8 Ibid., p. 40. This statement is a little too strong: there exist viable nonlocal hidden variable theories (notably David Bohm’s), and other formulations (such as the many worlds interpretation) that do not fit cleanly into any of my three categories. But I think it is wise, at least from a pedagogical point of view, to adopt a clear and coherent platform at this stage, and worry about the alternatives later. 9 The role of measurement in quantum mechanics is so critical and so bizarre that you may well be wondering what precisely constitutes a measurement. I’ll return to this thorny issue in the Afterword; for the moment let’s take the naive view: a measurement is the kind of thing that a scientist in a white coat does in the laboratory, with rulers, stopwatches, Geiger counters, and so on. 10 Because the wavelength of electrons is typically very small, the slits have to be extremely close together. Historically, this was first achieved by Davisson and Germer, in 1925, using the atomic layers in a crystal as “slits.” For an interesting account, see R. K. Gehrenbeck, Physics Today, January 1978, page 34. 11 See Tonomura et al., American Journal of Physics, Volume 57, Issue 2, pp. 117–120 (1989), and the amazing associated video at www.hitachi.com/rd/portal/highlight/quantum/doubleslit/. This experiment can now be done with much more massive particles, including “Buckyballs”; see M. Arndt, et al., Nature 40, 680 (1999). Incidentally, the same thing can be done with light: turn the intensity so low that only one “photon” is present at a time and you get an identical pointbypoint assembly of the interference pattern. See R. S. Aspden, M. J. Padgett, and G. C. Spalding, Am. J. Phys. 84, 671 (2016). 12 I think it is important to distinguish things like interference and diffraction that would hold for any wave theory from the uniquely quantum mechanical features of the measurement process, which derive from the statistical interpretation. 13 A statistician will complain that I am confusing the average of a finite sample (a million, in this case) with the “true” average (over the whole continuum). This can be an awkward problem for the experimentalist, especially when the sample size is small, but here I am only concerned with the true average, to which the sample average is presumably a good approximation. 14 Evidently must go to zero faster than , as . Incidentally, normalization only fixes the modulus of A; the phase remains undetermined. However, as we shall see, the latter carries no physical significance anyway. 15 A competent mathematician can supply you with pathological counterexamples, but they do not arise in physics; for us the wave function and all its derivatives go to zero at infinity. 16 To keep things from getting too cluttered, I’ll suppress the limits of integration 17 The product rule says that . from which it follows that Under the integral sign, then, you can peel a derivative off one factor in a product, and slap it onto the other one—it’ll cost you a minus sign, and you’ll pick up a boundary term. 18 An “operator” is an instruction to do something to the function that follows; it takes in one function, and spits out some other function. The position operator tells you to multiply by x; the momentum operator tells you to differentiate with respect to x (and multiply the result by ). 19 Some authors limit the term to the pair of equations 20 That’s why a piccolo player must be right on pitch, whereas a doublebass player can afford to wear garden gloves. For the piccolo, a sixty and fourth note contains many full cycles, and the frequency (we’re working in the time domain now, instead of space) is well defined, whereas for the bass, at a much lower register, the sixtyfourth note contains only a few cycles, and all you hear is a general sort of “oomph,” with no very clear pitch. 21 I’ll explain this in due course. Many authors take the de Broglie formula as an axiom, from which they then deduce the association of momentum with the operator . Although this is a conceptually cleaner approach, it involves diverting mathematical complications that I would rather save for later. 22 If you like, instead of photos of one system at random times, picture an ensemble of such systems, all with the same energy but with random starting positions, and photograph them all at the same time. The analysis is identical, but this interpretation is closer to the quantum notion of indeterminacy. 23 In a solid the inner electrons are attached to a particular nucleus, and for them the relevant size would be the radius of the atom. But the outermost electrons are not attached, and for them the relevant distance is the lattice spacing. This problem pertains to the outer electrons. 41 2 TimeIndependent Schrödinger Equation ◈ 42 2.1 Stationary States In Chapter 1 we talked a lot about the wave function, and how you use it to calculate various quantities of interest. The time has come to stop procrastinating, and confront what is, logically, the prior question: How do you get in the first place? We need to solve the Schrödinger equation, (2.1) for a specified potential1 . In this chapter (and most of this book) I shall assume that V is independent of t. In that case the Schrödinger equation can be solved by the method of separation of variables (the physicist’s first line of attack on any partial differential equation): We look for solutions that are products, (2.2) where (lowercase) is a function of x alone, and is a function of t alone. On its face, this is an absurd restriction, and we cannot hope to obtain more than a tiny subset of all solutions in this way. But hang on, because the solutions we do get turn out to be of great interest. Moreover (as is typically the case with separation of variables) we will be able at the end to patch together the separable solutions in such a way as to construct the most general solution. For separable solutions we have (ordinary derivatives, now), and the Schrödinger equation reads Or, dividing through by : (2.3) Now, the left side is a function of t alone, and the right side is a function of x alone.2 The only way this can possibly be true is if both sides are in fact constant—otherwise, by varying t, I could change the left side without touching the right side, and the two would no longer be equal. (That’s a subtle but crucial argument, so if it’s new to you, be sure to pause and think it through.) For reasons that will appear in a moment, we shall call the separation constant E. Then or (2.4) 43 and or (2.5) Separation of variables has turned a partial differential equation into two ordinary differential equations (Equations 2.4 and 2.5). The first of these is easy to solve (just multiply through by dt and integrate); the general solution is , but we might as well absorb the constant C into interest is the product . (since the quantity of Then3 (2.6) The second (Equation 2.5) is called the timeindependent Schrödinger equation; we can go no further with it until the potential is specified. The rest of this chapter will be devoted to solving the timeindependent Schrödinger equation, for a variety of simple potentials. But before I get to that you have every right to ask: What’s so great about separable solutions? After all, most solutions to the (time dependent) Schrödinger equation do not take the form . I offer three answers—two of them physical, and one mathematical: 1. They are stationary states. Although the wave function itself, (2.7) does(obviously) depend on t, the probability density, (2.8) does not—the timedependence cancels out.4 The same thing happens in calculating the expectation value of any dynamical variable; Equation 1.36 reduces to (2.9) Every expectation value is constant in time; we might as well drop the factor use in place of . (Indeed, it is common to refer to altogether, and simply as “the wave function,” but this is sloppy language that can be dangerous, and it is important to remember that the true wave function always carries that timedependent wiggle factor.) In particular, is constant, and hence (Equation 1.33) . Nothing ever happens in a stationary state. 2. They are states of definite total energy. In classical mechanics, the total energy (kinetic plus potential) is called the Hamiltonian: (2.10) The corresponding Hamiltonian operator, obtained by the canonical substitution 44 , is The corresponding Hamiltonian operator, obtained by the canonical substitution , is therefore5 (2.11) Thus the timeindependent Schrödinger equation (Equation 2.5) can be written (2.12) and the expectation value of the total energy is (2.13) (Notice that the normalization of entails the normalization of .) Moreover, and hence So the variance of H is (2.14) But remember, if , then every member of the sample must share the same value (the distribution has zero spread). Conclusion: A separable solution has the property that every measurement of the total energy is certain to return the value E. (That’s why I chose that letter for the separation constant.) 3. The general solution is a linear combination of separable solutions. As we’re about to discover, the timeindependent Schrödinger equation (Equation 2.5) yields an infinite collection of solutions , , which we write as , each with its associated separation constant ; thus there is a different wave function for each allowed energy: Now (as you can easily check for yourself) the (timedependent) Schrödinger equation (Equation 2.1) has the property that any linear combination6 of solutions is itself a solution. Once we have found the separable solutions, then, we can immediately construct a much more general solution, of the form (2.15) It so happens that every solution to the (timedependent) Schrödinger equation can be written in this form—it is simply a matter of finding the right constants so as to fit the initial conditions for the problem at hand. You’ll see in the following sections how all this works out in practice, and in Chapter 3 we’ll put it into more elegant language, but the main point is this: Once you’ve solved the timeindependent Schrödinger equation, you’re essentially done; getting from there 45 to the general solution of the timedependent Schrödinger equation is, in principle, simple and straightforward. A lot has happened in the past four pages, so let me recapitulate, from a somewhat different perspective. Here’s the generic problem: You’re given a (timeindependent) potential ; your job is to find the wave function, , and the starting wave function , for any subsequent time t. To do this you must solve the (timedependent) Schrödinger equation (Equation 2.1). The strategy is first to solve the timeindependent Schrödinger equation (Equation 2.5); this yields, in general, an infinite set of solutions, each with its own associated energy, . To fit , you write down the general linear combination of these solutions: (2.16) the miracle is that you can always match the specified initial state7 by appropriate choice of the constants To construct . you simply tack onto each term its characteristic time dependence (its “wiggle factor”), :8 (2.17) The separable solutions themselves, (2.18) are stationary states, in the sense that all probabilities and expectation values are independent of time, but this property is emphatically not shared by the general solution (Equation 2.17): the energies are different, for different stationary states, and the exponentials do not cancel, when you construct . Example 2.1 Suppose a particle starts out in a linear combination of just two stationary states: (To keep things simple I’ll assume that the constants wave function and the states are real.) What is the at subsequent times? Find the probability density, and describe its motion. Solution: The first part is easy: where and are the energies associated with and . It follows that The probability density oscillates sinusoidally, at an angular frequency ; this is certainly not a stationary state. But notice that it took a linear combination of stationary states (with 46 different energies) to produce motion.9 You may be wondering what the coefficients represent physically. I’ll tell you the answer, though the explanation will have to await Chapter 3: (2.19) A competent measurement will always yield one of the “allowed” values (hence the name), and probability of getting the particular value .10 is the Of course, the sum of these probabilities should be 1: (2.20) and the expectation value of the energy must be (2.21) We’ll soon see how this works out in some concrete examples. Notice, finally, that becausethe constants are independent of time, so too is the probability of getting a particular energy, and, a fortiori, the expectation value of H. These are manifestations of energy conservation in quantum mechanics. ∗ Problem 2.1 Prove the following three theorems: (a) For normalizable solutions, the separation constant E must be real. Hint: Write E (in Equation 2.7) as (with and Γ real), and show that if Equation 1.20 is to hold for all t, Γ must be zero. (b) The timeindependent wave function (unlike can always be taken to be real , which is necessarily complex). This doesn’t mean that every solution to the timeindependent Schrödinger equation is real; what it says is that if you’ve got one that is not, it can always be expressed as a linear combination of solutions (with the same energy) that are. So you might as well stick to s that are real. Hint: If satisfies Equation 2.5, for a given E, so too does its complex conjugate, and hence also the real linear combinations (c) If and is an even function (that is, then always be taken to be either even or odd. Hint: If 2.5, for a given E, so too does linear combinations can satisfies Equation , and hence also the even and odd . 47 . ∗ Problem 2.2 Show that E must exceed the minimum value of , for every normalizable solution to the timeindependent Schrödinger equation. What is the classical analog to this statement? Hint: Rewrite Equation 2.5 in the form if , then and its second derivative always have the same sign—argue that such a function cannot be normalized. 48 2.2 The Infinite Square Well Suppose (2.22) (Figure 2.1). A particle in this potential is completely free, except at the two ends and , where an infinite force prevents it from escaping. A classical model would be a cart on a frictionless horizontal air track, with perfectly elastic bumpers—it just keeps bouncing back and forth forever. (This potential is artificial, of course, but I urge you to treat it with respect. Despite its simplicity—or rather, precisely because of its simplicity—it serves as a wonderfully accessible test case for all the fancy machinery that comes later. We’ll refer back to it frequently.) Figure 2.1: The infinite square well potential (Equation 2.22). Outside the well, (the probability of finding the particle there is zero). Inside the well, where , the timeindependent Schrödinger equation (Equation 2.5) reads (2.23) or (2.24) (By writing it in this way, I have tacitly assumed that ; we know from Problem 2.2 that won’t work.) Equation 2.24 is the classical simple harmonic oscillator equation; the general solution is (2.25) where A and B are arbitrary constants. Typically, these constants are fixed by the boundary conditions of the problem. What are the appropriate boundary conditions for continuous,11 ? Ordinarily, both and are but where the potential goes to infinity only the first of these applies. (I’ll justify these boundary conditions, and account for the exception when Continuity of , in Section 2.5; for now I hope you will trust me.) requires that (2.26) 49 so as to join onto the solution outside the well. What does this tell us about A and B? Well, so , and hence (2.27) Then , so either solution (in which case we’re left with the trivial—nonnormalizable— , or else , which means that (2.28) But is no good (again, that would imply since , and the negative solutions give nothing new, and we can absorb the minus sign into A. So the distinct solutions are (2.29) Curiously, the boundary condition at does not determine the constant A, but rather the constant k, and hence the possible values of E: (2.30) In radical contrast to the classical case, a quantum particle in the infinite square well cannot have just any old energy—it has to be one of these special (“allowed”) values.12 To find A, we normalize :13 This only determines the magnitude of A, but it is simplest to pick the positive real root: (the phase of A carries no physical significance anyway). Inside the well, then, the solutions are (2.31) As promised, the timeindependent Schrödinger equation has delivered an infinite set of solutions (one for each positive integer . The first few of these are plotted in Figure 2.2. They look just like the standing waves on a string of length a; , which carries the lowest energy, is called the ground state, the others, whose energies increase in proportion to , are called excited states. As a collection, the functions have some interesting and important properties: 1. They are alternately even and odd, with respect to the center of the well: even, and so is even, is odd, is on.14 2. As you go up in energy, each successive state has one more node (zerocrossing): points don’t count), has one, has none (the end has two, and so on. 3. They are mutually orthogonal, in the sense that15 (2.32) 50 Figure 2.2: The first three stationary states of the infinite square well (Equation 2.31). Proof: Note that this argument does not work if . (Can you spot the point at which it fails?) In that case normalization tells us that the integral is 1. In fact, we can combine orthogonality and normalization into a single statement: (2.33) where (the socalled Kronecker delta) is defined by (2.34) We say that the 4. s are orthonormal. They are complete, in the sense that any other function, , can be expressed as a linear combination of them: (2.35) I’m not about to prove the completeness of the functions , but if you’ve studied advanced calculus you will recognize that Equation 2.35 is nothing but the Fourier series for the fact that “any” function can be expanded in this way is sometimes called Dirichlet’s The coefficients can be evaluated—for a given beautifully exploits the orthonormality of , and theorem.16 —by a method I call Fourier’s trick, which : Multiply both sides of Equation 2.35 by , and integrate. (2.36) (Notice how the Kronecker delta kills every term in the sum except the one for which 51 .) Thus the (Notice how the Kronecker delta kills every term in the sum except the one for which nth coefficient in the expansion of .) Thus the is17 (2.37) These four properties are extremely powerful, and they are not peculiar to the infinite square well. The first is true whenever the potential itself is a symmetric function; the second is universal, regardless of the shape of the potential.18 Orthogonality is also quite general—I’ll show you the proof in Chapter 3. Completeness holds for all the potentials you are likely to encounter, but the proofs tend to be nasty and laborious; I’m afraid most physicists simply assume completeness, and hope for the best. The stationary states (Equation 2.18) of the infinite square well are (2.38) I claimed (Equation 2.17) that the most general solution to the (timedependent) Schrödinger equation is a linear combination of stationary states: (2.39) (If you doubt that this is a solution, by all means check it!) It remains only for me to demonstrate that I can fit any prescribed initial wave function, The completeness of the express by appropriate choice of the coefficients : s (confirmed in this case by Dirichlet’s theorem) guarantees that I can always in this way, and their orthonormality licenses the use of Fourier’s trick to determine the actual coefficients: (2.40) That does it: Given the initial wave function, , we first compute the expansion coefficients using Equation 2.40, and then plug these into Equation 2.39 to obtain , . Armed with the wave function, we are in a position to compute any dynamical quantities of interest, using the procedures in Chapter 1. And this same ritual applies to any potential—the only things that change are the functional form of the s and the equation for the allowed energies. Example 2.2 A particle in the infinite square well has the initial wave function for some constant A (see Figure 2.3). Outside the well, of course, 52 . Find . Figure 2.3: The starting wave function in Example 2.2. Solution: First we need to determine A, by normalizing : so The nth coefficient is (Equation 2.40) Thus (Equation 2.39): Example 2.3 Check that Equation 2.20 is satisfied, for the wave function in Example 2.2. If you measured the 53 Check that Equation 2.20 is satisfied, for the wave function in Example 2.2. If you measured the energy of a particle in this state, what is the most probable result? What is the expectation value of the energy? Solution: The starting wave function (Figure 2.3) closely resembles the ground state This suggests that should dominate,19 (Figure 2.2). and in fact The rest of the coefficients make up the difference:20 The most likely outcome of an energy measurement is —more than 99.8% of all measurements will yield this value. The expectation value of the energy (Equation 2.21) is As one would expect, it is very close to (5 in place of —slightly larger, because of the admixture of excited states. Of course, it’s no accident that Equation 2.20 came out right in Example 2.3. Indeed, this follows from the normalization of (the s are independent of time, so I’m going to do the proof for bothers you, you can easily generalize the argument to arbitrary (Again, the Kronecker delta picks out the term ; if this . in the summation over m.) Similarly, the expectation value of the energy (Equation 2.21) can be checked explicitly: The timeindependent Schrödinger equation (Equation 2.12) says (2.41) so 54 Problem 2.3 Show that there is no acceptable solution to the (timeindependent) Schrödinger equation for the infinite square well with or . (This is a special case of the general theorem in Problem 2.2, but this time do it by explicitly solving the Schrödinger equation, and showing that you cannot satisfy the boundary conditions.) ∗ Problem 2.4 Calculate , and , for the nth stationary state of the infinite square well. Check that the uncertainty principle is satisfied. Which state comes closest to the uncertainty limit? ∗ Problem 2.5 A particle in the infinite square well has as its initial wave function an even mixture of the first two stationary states: (a) Normalize . (That is, find A. This is very easy, if you exploit the orthonormality of and . Recall that, having normalized at , you can rest assured that it stays normalized—if you doubt this, check it explicitly after doing part (b).) (b) Find and . Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let (c) Compute . . Notice that it oscillates in time. What is the angular frequency of the oscillation? What is the amplitude of the oscillation? (If your amplitude is greater than (d) Compute (e) , go directly to jail.) . (As Peter Lorre would say, “Do it ze kveek vay, Johnny!”) If you measured the energy of this particle, what values might you get, and what is the probability of getting each of them? Find the expectation value of H. How does it compare with and ? Problem 2.6 Although the overall phase constant of the wave function is of no physical significance (it cancels out whenever you calculate a measurable quantity), the relative phase of the coefficients in Equation 2.17 does matter. For example, suppose we change the relative phase of where ϕ is some constant. Find 55 and in Problem 2.5: , and , and compare your where ϕ is some constant. Find , and , and compare your results with what you got before. Study the special cases and . (For a graphical exploration of this problem see the applet in footnote 9 of this chapter.) ∗ Problem 2.7 A particle in the infinite square well has the initial wave function (a) Sketch , and determine the constant A. (b) Find . (c) What is the probability that a measurement of the energy would yield the value ? (d) Find the expectation value of the energy, using Equation 2.21.21 Problem 2.8 A particle of mass m in the infinite square well (of width starts out in the state for some constant A, so it is (at equally likely to be found at any point in the left half of the well. What is the probability that a measurement of the energy (at some later time would yield the value ? Problem 2.9 For the wave function in Example 2.2, find the expectation value of H, at time , the “old fashioned” way: Compare the result we got in Example 2.3. Note: Because time, there is no loss of generality in using 56 . is independent of 2.3 The Harmonic Oscillator The paradigm for a classical harmonic oscillator is a mass m attached to a spring of force constant k. The motion is governed by Hooke’s law, (ignoring friction), and the solution is where (2.42) is the (angular) frequency of oscillation. The potential energy is (2.43) its graph is a parabola. Of course, there’s no such thing as a perfect harmonic oscillator—if you stretch it too far the spring is going to break, and typically Hooke’s law fails long before that point is reached. But practically any potential is approximately parabolic, in the neighborhood of a local minimum (Figure 2.4). Formally, if we expand in a Taylor series about the minimum: subtract (you can add a constant to recognize that long as (since with impunity, since that doesn’t change the force), is a minimum), and drop the higherorder terms (which are negligible as stays small), we get which describes simple harmonic oscillation (about the point , with an effective spring constant . That’s why the simple harmonic oscillator is so important: Virtually any oscillatory motion is approximately simple harmonic, as long as the amplitude is small.22 57 Figure 2.4: Parabolic approximation (dashed curve) to an arbitrary potential, in the neighborhood of a local minimum. The quantum problem is to solve the Schrödinger equation for the potential (2.44) (it is customary to eliminate the spring constant in favor of the classical frequency, using Equation 2.42). As we have seen, it suffices to solve the timeindependent Schrödinger equation: (2.45) In the literature you will find two entirely different approaches to this problem. The first is a straightforward “brute force” solution to the differential equation, using the power series method; it has the virtue that the same strategy can be applied to many other potentials (in fact, we’ll use it in Chapter 4 to treat the hydrogen atom). The second is a diabolically clever algebraic technique, using socalled ladder operators. I’ll show you the algebraic method first, because it is quicker and simpler (and a lot more fun);23 if you want to skip the power series method for now, that’s fine, but you should certainly plan to study it at some stage. 58 2.3.1 Algebraic Method To begin with, let’s rewrite Equation 2.45 in a more suggestive form: (2.46) is the momentum operator.24 The basic idea is to factor the Hamiltonian, where (2.47) If these were numbers, it would be easy: Here, however, it’s not quite so simple, because commute is not the same as and x are operators, and operators do not, in general, , as we’ll see in a moment—though you might want to stop right now and think it through for yourself). Still, this does motivate us to examine the quantities (2.48) (the factor in front is just there to make the final results look nicer). Well, what is the product ? As anticipated, there’s an extra term, involving . We call this the commutator of x and ; it is a measure of how badly they fail to commute. In general, the commutator of operators and (written with square brackets) is (2.49) In this notation, (2.50) We need to figure out the commutator of x and . Warning: Operators are notoriously slippery to work with in the abstract, and you are bound to make mistakes unless you give them a “test function,” , to act on. At the end you can throw away the test function, and you’ll be left with an equation involving the operators alone. In the present case we have: (2.51) 59 Dropping the test function, which has served its purpose, (2.52) This lovely and ubiquitous formula is known as the canonical commutation relation.25 With this, Equation 2.50 becomes (2.53) or (2.54) Evidently the Hamiltonian does not factor perfectly—there’s that extra ordering of and is important here; the same argument, with on the right. Notice that the on the left, yields (2.55) In particular, (2.56) Meanwhile, the Hamiltonian can equally well be written (2.57) In terms of , then, the Schrödinger equation26 for the harmonic oscillator takes the form (2.58) (in equations like this you read the upper signs all the way across, or else the lower signs). Now, here comes the crucial step: I claim that: If satisfies the Schrödinger equation with energy E (that is: Schrödinger equation with energy : , then satisfies the . Proof: (I used Equation 2.56 to replace by 60 in the second line. Notice that whereas the (I used Equation 2.56 to replace ordering of and by in the second line. Notice that whereas the does matter, the ordering of and any constants—such as , and E—does not; an operator commutes with any constant.) By the same token, is a solution with energy : Here, then, is a wonderful machine for generating new solutions, with higher and lower energies—if we could just find one solution, to get started! We call down in energy; is the raising operator, and ladder operators, because they allow us to climb up and the lowering operator. The “ladder” of states is illustrated in Figure 2.5. Figure 2.5: The “ladder” of states for the harmonic oscillator. But wait! What if I apply the lowering operator repeatedly? Eventually I’m going to reach a state with energy less than zero, which (according to the general theorem in Problem 2.3) does not exist! At some point the machine must fail. How can that happen? We know that 61 is a new solution to the Schrödinger equation, but there is no guarantee that it will be normalizable—it might be zero, or its squareintegral might be infinite. In practice it is the former: There occurs a “lowest rung” (call it such that (2.59) We can use this to determine : or This differential equation is easy to solve: so We might as well normalize it right away: so , and hence (2.60) To determine the energy of this state we plug it into the Schrödinger equation (in the form of Equation 2.58), , and exploit the fact that : (2.61) With our foot now securely planted on the bottom rung (the ground state of the quantum oscillator), we simply apply the raising operator (repeatedly) to generate the excited states,27 increasing the energy by with each step: (2.62) where is the normalization constant. By applying the raising operator (repeatedly) to principle) construct all28 , then, we can (in the stationary states of the harmonic oscillator. Meanwhile, without ever doing that explicitly, we have determined the allowed energies! 62 Example 2.4 Find the first excited state of the harmonic oscillator. Solution: Using Equation 2.62, (2.63) We can normalize it “by hand”: so, as it happens, . I wouldn’t want to calculate this way (applying the raising operator fifty times!), but never mind: In principle Equation 2.62 does the job—except for the normalization. You can even get the normalization algebraically, but it takes some fancy footwork, so watch closely. We know that is proportional to , (2.64) but what are the proportionality factors, and ? First note that for “any”29 functions and , (2.65) In the language of linear algebra, is the hermitian conjugate (or adjoint) of . Proof: and integration by parts takes to the reason indicated in footnote 29), so In particular, But (invoking Equations 2.58 and 2.62) 63 (the boundary terms vanish, for (2.66) so But since and are normalized, it follows that , and hence30 and (2.67) Thus and so on. Clearly (2.68) which is to say that the normalization factor in Equation 2.62 is (in particular, , confirming our result in Example 2.4). As in the case of the infinite square well, the stationary states of the harmonic oscillator are orthogonal: (2.69) This can be proved using Equation 2.66, and Equation 2.65 twice—first moving Unless , then, value : must be zero. Orthonormality means that we can again use Fourier’s trick (Equation 2.37) to evaluate the coefficients states (Equation 2.16). As always, and then moving , when we expand as a linear combination of stationary is the probability that a measurement of the energy would yield the . 64 Example 2.5 Find the expectation value of the potential energy in the nth stationary state of the harmonic oscillator. Solution: There’s a beautiful device for evaluating integrals of this kind (involving powers of x or definition (Equation 2.48) to express x and : Use the in terms of the raising and lowering operators: (2.70) In this example we are interested in : So But is (apart from normalization) , which is proportional to , which is orthogonal to , and the same goes for . So those terms drop out, and we can use Equation 2.66 to evaluate the remaining two: As it happens, the expectation value of the potential energy is exactly half the total (the other half, of course, is kinetic). This is a peculiarity of the harmonic oscillator, as we’ll see later on (Problem 3.37). ∗ Problem 2.10 (a) Construct (b) Sketch . , and . (c) Check the orthogonality of , and , by explicit integration. Hint: If you exploit the evenness and oddness of the functions, there is really only one integral left to do. ∗ Problem 2.11 (a) Compute , and , for the states (Equation 2.60) and (Equation 2.63), by explicit integration. Comment: In this and other problems involving the harmonic oscillator it simplifies matters if you 65 introduce the variable and the constant . (b) Check the uncertainty principle for these states. (c) Compute and for these states. (No new integration allowed!) Is their sum what you would expect? ∗ Problem 2.12 Find , and , for the nth stationary state of the harmonic oscillator, using the method of Example 2.5. Check that the uncertainty principle is satisfied. Problem 2.13 A particle in the harmonic oscillator potential starts out in the state (a) Find A. (b) Construct and . Don’t get too excited if oscillates at exactly the classical frequency; what would it have been had I specified (c) Find , instead of and ?31 . Check that Ehrenfest’s theorem (Equation 1.38) holds, for this wave function. (d) If you measured the energy of this particle, what values might you get, and with what probabilities? 66 2.3.2 Analytic Method We return now to the Schrödinger equation for the harmonic oscillator, (2.71) and solve it directly, by the power series method. Things look a little cleaner if we introduce the dimensionless variable (2.72) in terms of ξ the Schrödinger equation reads (2.73) where K is the energy, in units of : (2.74) Our problem is to solve Equation 2.73, and in the process obtain the “allowed” values of K (and hence of To begin with, note that at very large ξ (which is to say, at very large . completely dominates over the constant K, so in this regime (2.75) which has the approximate solution (check it!) (2.76) The B term is clearly not normalizable (it blows up as ; the physically acceptable solutions, then, have the asymptotic form (2.77) This suggests that we “peel off” the exponential part, (2.78) in hopes that what remains, , has a simpler functional form than 2.78, and 67 itself.32 Differentiating Equation so the Schrödinger equation (Equation 2.73) becomes (2.79) I propose to look for solutions to Equation 2.79 in the form of power series in ξ:33 (2.80) Differentiating the series term by term, and Putting these into Equation 2.80, we find (2.81) It follows (from the uniqueness of power series expansions34 ) that the coefficient of each power of ξ must vanish, and hence that (2.82) This recursion formula is entirely equivalent to the Schrödinger equation. Starting with , it generates all the evennumbered coefficients: and starting with , it generates the odd coefficients: We write the complete solution as (2.83) 68 where is an even function of ξ, built on is an odd function, built on and , and . Thus Equation 2.82 determines in terms of two arbitrary constants —which is just what we would expect, for a secondorder differential equation. However, not all the solutions so obtained are normalizable. For at very large j, the recursion formula becomes (approximately) with the (approximate) solution for some constant C, and this yields (at large ξ, where the higher powers dominate) Now, if h goes like , then (remember ?—that’s what we’re trying to calculate) goes like (Equation 2.78), which is precisely the asymptotic behavior we didn’t want.35 There is only one way to wiggle out of this: For normalizable solutions the power series must terminate. There must occur some “highest” j (call it or the series , such that the recursion formula spits out ; the other one must be zero from the start: (this will truncate either the series if n is even, and if n is odd). For physically acceptable solutions, then, Equation 2.82 requires that for some positive integer n, which is to say (referring to Equation 2.74) that the energy must be (2.84) Thus we recover, by a completely different method, the fundamental quantization condition we found algebraically in Equation 2.62. It seems at first rather surprising that the quantization of energy should emerge from a technical detail in the power series solution to the Schrödinger equation, but let’s look at it from a different perspective. Equation 2.71 has solutions, of course, for any value of E (in fact, it has two linearly independent solutions for every . But almost all of these solutions blow up exponentially at large x, and hence are not normalizable. Imagine, for example, using an E that is slightly less than one of the allowed values (say, plotting the solution: Figure 2.6(a). Now try an E slightly larger (say, , and ; the “tail” now blows up in the other direction (Figure 2.6(b)). As you tweak the parameter in tiny increments from 0.49 to 0.51, the graph 69 “flips over” at precisely the value 0.5—only here does the solution escape the exponential asymptotic growth that renders it physically unacceptable.36 Figure 2.6: Solutions to the Schrödinger equation for (a) , and (b) . For the allowed values of K, the recursion formula reads (2.85) If , there is only one term in the series (we must pick yields to kill , and in Equation 2.85 we take ,37 and Equation : and hence (which, apart from the normalization, reproduces Equation 2.60). For 2.85 with yields , so and hence 70 (confirming Equation 2.63). For yields , and gives , so and and so on. (Compare Problem 2.10, where this last result was obtained by algebraic means). In general, will be a polynomial of degree n in ξ, involving even powers only, if n is an even integer, and odd powers only, if n is an odd integer. Apart from the overall factor called Hermite polynomials, .38 they are the so The first few of them are listed in Table 2.1. By tradition, the arbitrary multiplicative factor is chosen so that the coefficient of the highest power of ξ is normalized39 or . With this convention, the stationary states for the harmonic oscillator are (2.86) They are identical (of course) to the ones we obtained algebraically in Equation 2.68. Table 2.1: The first few Hermite polynomials, In Figure 2.7(a) I have plotted . for the first few ns. The quantum oscillator is strikingly different from its classical counterpart—not only are the energies quantized, but the position distributions have some bizarre features. For instance, the probability of finding the particle outside the classically allowed range (that is, with x greater than the classical amplitude for the energy in question) is not zero (see Problem 2.14), and in all odd states the probability of finding the particle at the center is zero. Only at large n do we begin to see some resemblance to the classical case. In Figure 2.7(b) I have superimposed the classical position distribution (Problem 1.11) on the quantum one (for ; if you smoothed out the bumps, the two would fit pretty well. 71 Figure 2.7: (a) The first four stationary states of the harmonic oscillator. (b) Graph of , with the classical distribution (dashed curve) superimposed. Problem 2.14 In the ground state of the harmonic oscillator, what is the probability (correct to three significant digits) of finding the particle outside the classically allowed region? Hint: Classically, the energy of an oscillator is , where a is the amplitude. So the “classically allowed region” for an oscillator of energy E extends from to . Look in a math table under “Normal Distribution” or “Error Function” for the numerical value of the integral, or evaluate it by computer. 72 Problem 2.15 Use the recursion formula (Equation 2.85) to work out and . Invoke the convention that the coefficient of the highest power of ξ is to fix the overall constant. ∗∗ Problem 2.16 In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials. (a) The Rodrigues formula says that (2.87) Use it to derive (b) and . The following recursion relation gives you in terms of the two preceding Hermite polynomials: (2.88) Use it, together with your answer in (a), to obtain (c) and . If you differentiate an nthorder polynomial, you get a polynomial of order . For the Hermite polynomials, in fact, (2.89) Check this, by differentiating (d) and is the nth zderivative, at . , of the generating function ; or, to put it another way, it is the coefficient of in the Taylor series expansion for this function: (2.90) Use this to obtain , and 73 . 2.4 The Free Particle We turn next to what should have been the simplest case of all: the free particle everywhere). Classically this would just be motion at constant velocity, but in quantum mechanics the problem is surprisingly subtle. The timeindependent Schrödinger equation reads (2.91) or (2.92) So far, it’s the same as inside the infinite square well (Equation 2.24), where the potential is also zero; this time, however, I prefer to write the general solution in exponential form (instead of sines and cosines), for reasons that will appear in due course: (2.93) Unlike the infinite square well, there are no boundary conditions to restrict the possible values of k (and hence of ; the free particle can carry any (positive) energy. Tacking on the standard time dependence, , (2.94) Now, any function of x and t that depends on these variables in the special combination some constant represents a wave of unchanging shape, traveling in the (for direction at speed v: A fixed point on the waveform (for example, a maximum or a minimum) corresponds to a fixed value of the argument, and hence to x and t such that Since every point on the waveform moves with the same velocity, its shape doesn’t change as it propagates. Thus the first term in Equation 2.94 represents a wave traveling to the right, and the second represents a wave (of the same energy) going to the left. By the way, since they only differ by the sign in front of k, we might as well write (2.95) and let k run negative to cover the case of waves traveling to the left: (2.96) Evidently the “stationary states” of the free particle are propagating waves; their wavelength is and, according to the de Broglie formula (Equation 1.39), they carry momentum 74 , (2.97) The speed of these waves (the coefficient of t over the coefficient of is (2.98) On the other hand, the classical speed of a free particle with energy E is given by kinetic, since (pure , so (2.99) Apparently the quantum mechanical wave function travels at half the speed of the particle it is supposed to represent! We’ll return to this paradox in a moment—there is an even more serious problem we need to confront first: This wave function is not normalizable: (2.100) In the case of the free particle, then, the separable solutions do not represent physically realizable states. A free particle cannot exist in a stationary state; or, to put it another way, there is no such thing as a free particle with a definite energy. But that doesn’t mean the separable solutions are of no use to us. For they play a mathematical role that is entirely independent of their physical interpretation: The general solution to the timedependent Schrödinger equation is still a linear combination of separable solutions (only this time it’s an integral over the continuous variable k, instead of a sum over the discrete index : (2.101) (The quantity is factored out for convenience; what plays the role of the coefficient 2.17 is the combination in Equation .) Now this wave function can be normalized (for appropriate . But it necessarily carries a range of ks, and hence a range of energies and speeds. We call it a wave packet.40 In the generic quantum problem, we are given , and we are asked to find particle the solution takes the form of Equation 2.101; the only question is how to determine . For a free so as to match the initial wave function: (2.102) This is a classic problem in Fourier analysis; the answer is provided by Plancherel’s theorem (see Problem 2.19): (2.103) 75 is called the Fourier transform of difference is the sign in the integrals have to exist.42 exponent).41 ; is the inverse Fourier transform of (the only There is, of course, some restriction on the allowable functions: The For our purposes this is guaranteed by the physical requirement that itself be normalized. So the solution to the generic quantum problem, for the free particle, is Equation 2.101, with (2.104) Example 2.6 A free particle, which is initially localized in the range where A and a are positive real constants. Find Solution: First we need to normalize Next we calculate , is released at time : . : , using Equation 2.104: Finally, we plug this back into Equation 2.101: (2.105) Unfortunately, this integral cannot be solved in terms of elementary functions, though it can of course be evaluated numerically (Figure 2.8). (There are, in fact, precious few cases in which the integral for (Equation 2.101) can be carried out explicitly; see Problem 2.21 for a particularly beautiful example.) 76 Figure 2.8: Graph of (Equa