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Introduction to Quantum Mechanics (3rd Edition)
Introduction to Quantum Mechanics (3rd Edition)
David J. Griffiths, Darrell F. Schroeter
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Changes and additions to the new edition of this classic textbook include a new chapter on symmetries, new problems and examples, improved explanations, more numerical problems to be worked on a computer, new applications to solid state physics, and consolidated treatment of timedependent potentials.
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2018
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3
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Cambridge University Press
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english
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511
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1107189632
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9781107189638
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6 comments
Zhifeng Zhu
this is a very good version!
14 July 2021 (07:34)
FitzleBomber
10/10 completely transcripted into latex mode
27 July 2021 (19:43)
kazem
This is the best book for introduction to quantum mechanics.
02 August 2021 (23:27)
Sai Misaki
The BEST scan of this book because the formulas are in LaTex rather than images. ヽ(￣▽￣)ﾉ
03 August 2021 (12:15)
Lophycia
This one is the best edition.
09 October 2021 (16:22)
chenze
This is the best version
09 November 2021 (04:09)
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Introduction to Quantum Mechanics 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. INTRODUCTION TO Quantum Mechanics Third edition DAVID J. GRIFFITHS and DARRELL F. SCHROETER Reed College, Oregon 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 Forum, J; asola 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 c David Griffiths 2017 Second edition c Cambridge University Press 2018 Third edition 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. Contents Preface I 1 1.1 1.2 1.3 1.3.1 1.3.2 page xi THEORY 1 The Wave Function 3 The Schrödinger Equation The Statistical Interpretation Probability 3 3 8 Discrete Variables 8 Continuous Variables 11 Normalization Momentum The Uncertainty Principle Further Problems on Chapter 1 14 16 19 20 TimeIndependent Schrödinger Equation 25 Stationary States The Infinite Square Well The Harmonic Oscillator 25 31 39 2.3.1 Algebraic Method 40 2.3.2 Analytic Method 48 The Free Particle The DeltaFunction Potential 55 61 2.5.1 Bound States and Scattering States 61 2.5.2 The DeltaFunction Well 63 The Finite Square Well Further Problems on Chapter 2 70 76 3 Formalism 91 3.1 3.2 Hilbert Space Observables 91 94 3.2.1 Hermitian Operators 94 3.2.2 Determinate States 96 1.4 1.5 1.6 2 2.1 2.2 2.3 2.4 2.5 2.6 Contents vi 3.3 Eigenfunctions of a Hermitian Operator 97 3.3.1 Discrete Spectra 98 3.3.2 Continuous Spectra 99 3.4 3.5 Generalized Statistical Interpretation The Uncertainty Principle 102 105 3.5.1 Proof of the Generalized Uncertainty Principle 105 3.5.2 The MinimumUncertainty Wave Packet 108 3.5.3 The EnergyTime Uncertainty Principle 109 Vectors and Operators 113 3.6.1 Bases in Hilbert Space 113 3.6.2 Dirac Notation 117 3.6.3 Changing Bases in Dirac Notation 121 Further Problems on Chapter 3 124 Quantum Mechanics in Three Dimensions 131 3.6 4 The Schrödinger Equation 131 4.1.1 4.1 Spherical Coordinates 132 4.1.2 The Angular Equation 134 4.1.3 The Radial Equation 138 The Hydrogen Atom 143 4.2.1 The Radial Wave Function 144 4.2.2 The Spectrum of Hydrogen 155 Angular Momentum 157 4.2 4.3 4.3.1 Eigenvalues 157 4.3.2 Eigenfunctions 162 Spin 165 4.4 4.4.1 Spin 1/2 167 4.4.2 Electron in a Magnetic Field 172 4.4.3 Addition of Angular Momenta 176 Electromagnetic Interactions 181 Minimal Coupling 181 4.5 4.5.1 4.5.2 The Aharonov–Bohm Effect 182 Further Problems on Chapter 4 187 Identical Particles 198 TwoParticle Systems 198 5.1.1 Bosons and Fermions 201 5.1.2 Exchange Forces 203 5 5.1 Contents 5.1.3 Spin 206 5.1.4 Generalized Symmetrization Principle 207 5.2 Atoms 209 5.2.1 Helium 210 5.2.2 The Periodic Table 213 Solids 216 The Free Electron Gas 216 5.3 5.3.1 5.3.2 6 6.1 6.1.1 6.2 6.2.1 6.2.2 Band Structure 220 Further Problems on Chapter 5 225 Symmetries & Conservation Laws 232 Introduction 232 Transformations in Space 232 The Translation Operator 235 How Operators Transform 235 Translational Symmetry 238 Conservation Laws Parity 242 243 6.4.1 Parity in One Dimension 243 6.4.2 Parity in Three Dimensions 244 6.4.3 Parity Selection Rules 246 Rotational Symmetry 248 Rotations About the z Axis 248 6.3 6.4 6.5 6.5.1 6.5.2 Rotations in Three Dimensions 249 Degeneracy Rotational Selection Rules 252 255 6.7.1 Selection Rules for Scalar Operators 255 6.7.2 Selection Rules for Vector Operators 258 6.6 6.7 6.8 Translations in Time 262 6.8.1 The Heisenberg Picture 264 6.8.2 TimeTranslation Invariance 266 Further Problems on Chapter 6 268 APPLICATIONS 277 TimeIndependent Perturbation Theory 279 Nondegenerate Perturbation Theory 279 7.1.1 General Formulation 279 7.1.2 FirstOrder Theory 280 II 7 7.1 vii Contents viii 7.1.3 SecondOrder Energies 284 Degenerate Perturbation Theory 286 7.2.1 TwoFold Degeneracy 286 7.2.2 “Good” States 291 7.2.3 HigherOrder Degeneracy 294 The Fine Structure of Hydrogen 295 The Relativistic Correction 296 7.2 7.3 7.3.1 7.3.2 SpinOrbit Coupling 299 7.4 The Zeeman Effect 304 7.4.1 WeakField Zeeman Effect 305 7.4.2 StrongField Zeeman Effect 307 7.4.3 IntermediateField Zeeman Effect 309 7.5 Hyperfine Splitting in Hydrogen Further Problems on Chapter 7 311 313 8 The Variational Principle 327 8.1 8.2 8.3 8.4 Theory The Ground State of Helium The Hydrogen Molecule Ion The Hydrogen Molecule Further Problems on Chapter 8 327 332 337 341 346 9 The WKB Approximation 354 9.1 9.2 9.3 The “Classical” Region Tunneling The Connection Formulas Further Problems on Chapter 9 354 358 362 371 10 Scattering 10.1 Introduction 376 376 10.1.1 Classical Scattering Theory 376 10.1.2 Quantum Scattering Theory 379 Partial Wave Analysis 380 10.2.1 Formalism 380 10.2.2 Strategy 383 Phase Shifts The Born Approximation 385 388 10.2 10.3 10.4 Contents 10.4.1 Integral Form of the Schrödinger Equation 388 10.4.2 The First Born Approximation 391 10.4.3 The Born Series 395 Further Problems on Chapter 10 397 11 Quantum Dynamics 11.1 402 TwoLevel Systems 403 11.1.1 The Perturbed System 403 11.1.2 TimeDependent Perturbation Theory 405 11.1.3 Sinusoidal Perturbations 408 Emission and Absorption of Radiation 411 11.2.1 11.2 Electromagnetic Waves 411 11.2.2 Absorption, Stimulated Emission, and Spontaneous Emission 412 11.2.3 Incoherent Perturbations 413 11.3 Spontaneous Emission 416 11.3.1 Einstein’s A and B Coefficients 416 11.3.2 The Lifetime of an Excited State 418 11.3.3 Selection Rules 420 Fermi’s Golden Rule The Adiabatic Approximation 422 426 11.5.1 Adiabatic Processes 426 11.5.2 The Adiabatic Theorem 428 Further Problems on Chapter 11 433 11.4 11.5 12 Afterword 12.1 12.2 12.3 446 The EPR Paradox Bell’s Theorem Mixed States and the Density Matrix 447 449 455 12.3.1 Pure States 455 12.3.2 Mixed States 456 12.3.3 Subsystems 458 The NoClone Theorem Schrödinger’s Cat 459 461 12.4 12.5 Appendix Linear Algebra A.1 A.2 Vectors Inner Products 464 464 466 ix Contents x A.3 A.4 A.5 A.6 Matrices Changing Bases Eigenvectors and Eigenvalues Hermitian Transformations 468 473 475 482 Index 486 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. 1 This structure was inspired by David Park’s classic text Introduction to the Quantum Theory, 3rd edn, McGrawHill, New York (1992). xii Preface 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 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 Preface 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), 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). xiii Part I THEORY 1 THE WAVE FUNCTION 1.1 THE SCHRÖDINGER EQUATION Imagine a particle of mass m, constrained to move along the x axis, subject to some specified force F(x, t) (Figure 1.1). The program of classical mechanics is to determine the position of the particle at any given time: x(t). Once we know that, out the velocity (v = we can figure d x/dt), the momentum ( p = mv), the kinetic energy T = (1/2)mv 2 , or any other dynamical variable of interest. And how do we go about determining x(t)? We apply Newton’s second law: F = ma. (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 F = −∂ V /∂ x, and Newton’s law reads m d 2 x/dt 2 = −∂ V /∂ x.) This, together with appropriate initial conditions (typically the position and velocity at t = 0), determines x(t). Quantum mechanics approaches this same problem quite differently. In this case what we’re looking for is the particle’s wave function, (x, t), and we get it by solving the Schrödinger equation: i 2 ∂ 2 ∂ =− + V . ∂t 2m ∂ x 2 (1.1) Here i is the square root of −1, and is Planck’s constant—or rather, his original constant (h) divided by 2π : h = 1.054573 × 10−34 J s. = (1.2) 2π The Schrödinger equation plays a role logically analogous to Newton’s second law: Given suitable initial conditions (typically, (x, 0)), the Schrödinger equation determines (x, t) for all future time, just as, in classical mechanics, Newton’s law determines x(t) for all future time.2 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 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 (v c). 2 For a delightful firsthand account of the origins of the Schrödinger equation see the article by Felix Bloch in Physics Today, December 1976. 4 CHAPTER 1 The Wave Function m F(x,t) x x(t) Figure 1.1: A “particle” constrained to move in one dimension under the influence of a specified force. ⏐Ψ⏐2 a b A B C x 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. 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 (x, t)2 gives the probability of finding the particle at point x, at time t—or, more precisely,3 b (x, t) d x = 2 a probability of finding the particle between a and b, at time t. (1.3) Probability is the area under the graph of 2 . For the wave function in Figure 1.2, you would be quite likely to find the particle in the vicinity of point A, where 2 is large, and relatively unlikely to find it near point 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 3 The wave function itself is complex, but 2 = ∗ (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. 1.2 The Statistical Interpretation 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. 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 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). 7 Ibid., p. 40. 5 6 CHAPTER 1 The Wave Function ⏐Ψ⏐2 C x Figure 1.3: Collapse of the wave function: graph of 2 immediately after a measurement has found the particle at point C. 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 8 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. 1.2 The Statistical Interpretation 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 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 a b c d 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. 10 Because the wavelength of electrons is typically very small, the slits have to be extremely close together. Histori cally, 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). 7 8 CHAPTER 1 The Wave Function 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 1.3 PROBABILITY 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 N ( j) represent the number of people of age j, then N (14) = 1, N (15) = 1, N (16) = 3, N (22) = 2, N (24) = 2, N (25) = 5, 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. 1.3 Probability N(j) 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 j Figure 1.5: Histogram showing the number of people, N ( j), with age j, for the example in Section 1.3.1. while N (17), for instance, is zero. The total number of people in the room is N= ∞ N ( j). (1.4) j=0 (In the example, of course, N = 14.) Figure 1.5 is a histogram of the data. The following are some questions one might ask about this distribution. Question 1 If you selected one individual at random from this group, what is the probability that this person’s age would be 15? One chance in 14, since there are 14 possible choices, all equally likely, of whom only one has that particular age. If P( j) is the probability of getting age j, then P(14) = 1/14, P(15) = 1/14, P(16) = 3/14, and so on. In general, Answer P( j) = N ( j) . N (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: ∞ P( j) = 1. (1.6) j=0 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 P( j) 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 294 (14) + (15) + 3(16) + 2(22) + 2(24) + 5(25) = = 21. 14 14 9 10 CHAPTER 1 The Wave Function In general, the average value of j (which we shall write thus: j) is ∞ j N ( j) j = j P( j). = N (1.7) j=0 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 142 = 196, with probability 1/14, or 152 = 225, with probability 1/14, or 162 = 256, with probability 3/14, and so on. The average, then, is j2 = ∞ j 2 P( j). (1.8) j=0 In general, the average value of some function of j is given by f ( j) = ∞ f ( j)P( j). (1.9) j=0 (Equations 1.6, 1.7, and 1.8 are, if you like, special cases of this formula.) Beware: The average of the squares, j 2 , is not equal, in general, to the square of the average, j2 . For instance, if the room contains just two babies, aged 1 and 3, then j 2 = 5, but j2 = 4. 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 N(j) N(j) 1 2 3 4 5 6 7 8 9 10 j 1 2 3 4 5 6 7 8 9 10 j Figure 1.6: Two histograms with the same median, same average, and same most probable value, but different standard deviations. 1.3 Probability 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, j = j − j , (1.10) and compute the average of j. Trouble is, of course, that you get zero: j = ( j − j)P( j) = j P( j) − j P( j) = j − j = 0. (Note that j 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 j. But absolute values are nasty to work with; instead, we get around the sign problem by squaring before averaging: σ 2 ≡ (j)2 . (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 j. There is a useful little theorem on variances: (j)2 P( j) = ( j − j)2 P( j) σ 2 = (j)2 = j 2 − 2 j j + j2 P( j) = = j 2 P( j) − 2 j P( j) j P( j) + j2 = j 2 − 2 j j + j2 = j 2 − j2 . Taking the square root, the standard deviation itself can be written as σ = j 2 − j2 . (1.12) In practice, this is a much faster way to get σ than by direct application of Equation 1.11: simply calculate j 2 and j2 , subtract, and take the square root. Incidentally, I warned you a moment ago that j 2 is not, in general, equal to j2 . Since σ 2 is plainly nonnegative (from its definition 1.11), Equation 1.12 implies that j 2 ≥ j2 , (1.13) and the two are equal only when σ = 0, which is to say, for distributions with no spread at all (every member having the same value). 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 11 12 CHAPTER 1 The Wave Function 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 probability that an individual (chosen = ρ(x) d x. (1.14) at random) lies between x and (x + d x) The proportionality factor, ρ(x), 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 ρ(x): Pab = b ρ(x) d x, (1.15) a and the rules we deduced for discrete distributions translate in the obvious way: +∞ ρ(x) d x = 1, −∞ x = f (x) = (1.16) +∞ −∞ +∞ −∞ 2 xρ(x) d x, (1.17) f (x) ρ(x) d x, (1.18) σ 2 ≡ (x) = x 2 − x2 . (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 h/2. Ignoring air resistance, the distance x at time t is 1 x(t) = gt 2 . 2 √ The velocity is d x/dt = gt, and the total flight time is T = 2h/g. The probability that a particular photograph was taken between t and t + dt is dt/T , so the probability that it shows a distance in the corresponding range x to x + d x is dt dx g 1 = = √ d x. T gt 2h 2 hx 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. 1.3 Probability Thus the probability density (Equation 1.14) is 1 ρ(x) = √ , 2 hx (0 ≤ x ≤ h) (outside this range, of course, the probability density is zero). We can check this result, using Equation 1.16: h h 1 1 1/2 √ d x = √ 2x = 1. 2 h 0 2 hx 0 The average distance (Equation 1.17) is h h 2 3/2 1 h 1 x = x √ dx = √ = , x 3 3 2 hx 2 h 0 0 which is somewhat less than h/2, as anticipated. Figure 1.7 shows the graph of ρ(x). 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). ρ(x) 1 2h h Figure 1.7: The probability density in Example 1.2: ρ(x) = 1 ∗ x √ 2 hx . Problem 1.1 For the distribution of ages in the example in Section 1.3.1: (a) Compute j 2 and j2 . (b) Determine j 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. 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? 13 14 CHAPTER 1 ∗ The Wave Function Problem 1.3 Consider the gaussian distribution ρ(x) = Ae−λ(x−a) , 2 where A, a, and λ are positive real constants. (The necessary integrals are inside the back cover.) (a) Use Equation 1.16 to determine A. (b) Find x, x 2 , and σ . (c) Sketch the graph of ρ(x). 1.4 NORMALIZATION We return now to the statistical interpretation of the wave function (Equation 1.3), which says that (x, t)2 is the probability density for finding the particle at point x, at time t. It follows (Equation 1.16) that the integral of 2 over all x must be 1 (the particle’s got to be somewhere): +∞ −∞ (x, t)2 d x = 1. (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 without checking that the two are consistent. Well, a glance at Equation 1.1 reveals that if (x, t) is a solution, so too is A(x, t), 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 = 0. 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 t = 0. How do I know that it will stay normalized, as time goes on, and 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, +∞ +∞ ∂ d (x, t)2 d x. (x, t)2 d x = (1.21) dt −∞ −∞ ∂t 14 Evidently (x, t) must go to zero faster than 1/√x, as x → ∞. Incidentally, normalization only fixes the modulus of A; the phase remains undetermined. However, as we shall see, the latter carries no physical significance anyway. 1.4 Normalization (Note that the integral is a function only of t, so I use a total derivative (d/dt) on the left, but the integrand is a function of x as well as t, so it’s a partial derivative (∂/∂t) on the right.) By the product rule, ∂ ∗ ∂ ∂ ∗ ∂ 2 = = ∗ + . (1.22) ∂t ∂t ∂t ∂t Now the Schrödinger equation says that ∂ i ∂ 2 i = − V , ∂t 2m ∂ x 2 and hence also (taking the complex conjugate of Equation 1.23) ∂ ∗ i i ∂ 2 ∗ + V ∗, =− ∂t 2m ∂ x 2 so ∂ i 2 = ∂t 2m ∂ ∗ ∂ 2 ∂ 2 ∗ ∂ i ∗ ∂ ∗ 2 − − . = ∂ x 2m ∂x ∂x ∂x ∂x2 The integral in Equation 1.21 can now be evaluated explicitly: +∞ +∞ d ∂ ∗ i 2 ∗ ∂ (x, t) d x = − . dt −∞ 2m ∂x ∂x −∞ (1.23) (1.24) (1.25) (1.26) But (x, t) must go to zero as x goes to (±) infinity—otherwise the wave function would not be normalizable.15 It follows that +∞ d (x, t)2 d x = 0, (1.27) dt −∞ and hence that the integral is constant (independent of time); if is normalized at t = 0, it stays normalized for all future time. QED Problem 1.4 At time t = 0 a particle is represented by the wave function ⎧ 0 ≤ x ≤ a, ⎪ ⎨ A(x/a), (x, 0) = A(b − x)/(b − a), a ≤ x ≤ b, ⎪ ⎩ 0, otherwise, where A, a, and b are (positive) constants. (a) Normalize (that is, find A, in terms of a and b). (b) Sketch (x, 0), as a function of x. (c) Where is the particle most likely to be found, at t = 0? (d) What is the probability of finding the particle to the left of a? Check your result in the limiting cases b = a and b = 2a. (e) What is the expectation value of x? 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. 15 16 CHAPTER 1 ∗ The Wave Function Problem 1.5 Consider the wave function (x, t) = Ae−λx e−iωt , where A, λ, 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 x 2 . (c) Find the standard deviation of x. Sketch the graph of 2 , as a function of x, and mark the points (x+σ ) and (x−σ ), 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? 1.5 MOMENTUM For a particle in state , the expectation value of x is x = +∞ −∞ x (x, t)2 d x. (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, x 2 d x 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, x 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: x 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 2 , and compute the average, which should agree with x. (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, x 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 ∂ ∂ ∗ dx ∂ i ∂ = x 2 d x = ∗ − d x. (1.29) x dt ∂t 2m ∂x ∂x ∂x 16 To keep things from getting too cluttered, I’ll suppress the limits of integration (±∞). 1.5 Momentum This expression can be simplified using integrationbyparts:17 i ∂ ∗ dx ∂ =− ∗ − d x. dt 2m ∂x ∂x (1.30) (I used the fact that ∂ x/∂ x = 1, and threw away the boundary term, on the ground that goes to zero at (±) infinity.) Performing another integrationbyparts, on the second term, we conclude: i ∂ dx =− d x. (1.31) ∗ dt m ∂x 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: dx v = . (1.32) dt Equation 1.31 tells us, then, how to calculate v directly from . Actually, it is customary to work with momentum ( p = mv), rather than velocity: p = m dx = −i dt ∂ ∗ d x. ∂x Let me write the expressions for x and p in a more suggestive way: x = ∗ [x] d x, p = ∗ −i (∂/∂ x) d x. (1.33) (1.34) (1.35) We say that the operator18 x “represents” position, and the operator −i (∂/∂ x) “represents” momentum; to calculate expectation values we “sandwich” the appropriate operator between ∗ and , and integrate. 17 The product rule says that dg df d ( f g) = f + g, dx dx dx from which it follows that b b df dg dx = − g d x + f g . f dx a a dx a b 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 −i). 17 18 CHAPTER 1 The Wave Function 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 T = p2 1 2 mv = , 2 2m and angular momentum is L = r × mv = r × p (the latter, of course, does not occur for motion in one dimension). To calculate the expectation value of any such quantity, Q(x, p), we simply replace every p by −i (∂/∂ x), insert the resulting operator between ∗ and , and integrate: Q(x, p) = ∗ Q(x, −i ∂/∂ x) d x. For example, the expectation value of the kinetic energy is 2 ∂ 2 T = − ∗ 2 d x. 2m ∂x (1.36) (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. 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 ∂ x/∂t = 0, and conclude that dx/dt = 0? ∗ Problem 1.7 Calculate d p/dt. Answer: ∂V d p = − . dt ∂x (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 V0 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: exp (−i V0 t/). What effect does this have on the expectation value of a dynamical variable? 19 Some authors limit the term to the pair of equations p = m dx/dt and −∂ V /∂ x = d p/dt. 1.6 The Uncertainty Principle 10 20 30 40 50 x (feet) Figure 1.8: A wave with a (fairly) welldefined wavelength, but an illdefined position. 10 20 30 40 50 x (feet) Figure 1.9: A wave with a (fairly) welldefined position, but an illdefined wavelength. 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. 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 p= 2π h = . λ λ (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, σx σ p ≥ , 2 (1.40) 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 sixtyfourth 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 −i (∂/∂ x). Although this is a conceptually cleaner approach, it involves diverting mathematical complications that I would rather save for later. 19 20 CHAPTER 1 The Wave Function where σx is the standard deviation in x, and σ p 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 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 σx and σ p can be—just make 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 (x, t) = Ae−a mx 2 / +it , where A and a are positive real constants. (a) Find A. (b) For what potential energy function, V (x), is this a solution to the Schrödinger equation? (c) Calculate the expectation values of x, x 2 , p, and p 2 . (d) Find σx and σ p . Is their product consistent with the uncertainty principle? 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 V (x), 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 d x (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 d x: (dt/d x) d x 1 dt = = d x, (1.41) ρ(x) d x = T T v(x) T Further Problems on Chapter 1 V(x) E m a dx x b Figure 1.10: Classical particle in a potential well. where v(x) is the speed, and T = T dt = 0 a b 1 d x. v(x) (1.42) Thus ρ(x) = 1 . v(x)T (1.43) This is perhaps the closest classical analog22 to 2 . (a) Use conservation of energy to express v(x) in terms of E and V (x). (b) As an example, find ρ(x) for the simple harmonic oscillator, V (x) = kx 2 /2. Plot ρ(x), and check that it is correctly normalized. (c) For the classical harmonic oscillator in part (b), find x, x 2 , and σx . ∗∗ Problem 1.12 What if we were interested in the distribution of momenta ( p = mv), for the classical harmonic oscillator (Problem 1.11(b)). √ (a) Find√ the classical probability distribution ρ( p) (note that p ranges from − 2m E to + 2m E). (b) Calculate p, p 2 , and σ p . (c) What’s the classical uncertainty product, σx σ p , for this system? Notice that this product can be as small as you like, classically, simply by sending E → 0. But in quantum mechanics, as we shall see in Chapter 2, the energy of a simple harmonic √ oscillator cannot be less than ω/2, where ω = k/m is the classical frequency. In that case what can you say about the product σx σ p ? 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 x(t) = A cos(ωt). (1.44) 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. 21 22 CHAPTER 1 The Wave Function You might as well take ω = 1 (that sets the scale for time) and A = 1 (that sets the scale for length). Make a plot of x at 10,000 random times, and compare it with ρ(x). Hint: In Mathematica, first define x[t_] := Cos[t] then construct a table of positions: snapshots = Table[x[π RandomReal[j]], {j, 10000}] and finally, make a histogram of the data: Histogram[snapshots, 100, PDF , PlotRange → {0,2}] Meanwhile, make a plot of the density function, ρ(x), and, using Show, superimpose the two. Problem 1.14 Let Pab (t) be the probability of finding the particle in the range (a < x < b), at time t. (a) Show that d Pab = J (a, t) − J (b, t), dt where ∂ ∗ ∂ i − ∗ . J (x, t) ≡ 2m ∂x ∂x What are the units of J (x, t)? Comment: J is called the probability current, because it tells you the rate at which probability is “flowing” past the point x. If Pab (t) 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 d dt ∞ −∞ 1∗ 2 d x = 0 for any two (normalizable) solutions to the Schrödinger equation (with the same V (x)), 1 and 2 . Problem 1.16 A particle is represented (at time t = 0) by the wave function A a 2 − x 2 , −a ≤ x ≤ +a, (x, 0) = 0, otherwise. (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 p = mdx/dt. Why not?) Further Problems on Chapter 1 (d) Find the expectation value of x 2 . (e) Find the expectation value of p 2 . (f) Find the uncertainty in x (σ x ). (g) Find the uncertainty in p σ p . (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: +∞ (x, t)2 d x = e−t/τ . P(t) ≡ −∞ 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: V = V0 − i , where V0 is the true potential energy and is a positive real constant? (a) Show that (in place of Equation 1.27) we now get 2 dP =− P. dt (b) Solve for P(t), and find the lifetime of the particle in terms of . Problem 1.18 Very roughly speaking, quantum mechanics is relevant when the de Broglie wavelength of the particle in question (h/ p) is greater than the characteristic size of the system (d). In thermal equilibrium at (Kelvin) temperature T , the average kinetic energy of a particle is 3 p2 = kB T 2m 2 (where k B is Boltzmann’s constant), so the typical de Broglie wavelength is λ= √ h . 3mk B T (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 d = 0.3 nm. Find the temperature below which the unbound23 electrons in a solid are quantum mechanical. Below what temperature are the nuclei in a solid quantum mechanical? (Use silicon as an example.) 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. 23 24 CHAPTER 1 The Wave Function 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 (P V = N k B T ) to deduce the interatomic spacing. 3/5 2/5 P . Obviously (for the gas to show quantum Answer: T < (1/k B ) h 2 /3m 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 H2 .) 2 TIMEINDEPENDENT SCHRÖDINGER EQUATION 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 (x, t) in the first place? We need to solve the Schrödinger equation, i 2 ∂ 2 ∂ =− + V , ∂t 2m ∂ x 2 (2.1) for a specified potential1 V (x, t). 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, (x, t) = ψ(x) ϕ(t), (2.2) where ψ (lowercase) is a function of x alone, and ϕ 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 dϕ ∂ =ψ , ∂t dt ∂ 2 d 2ψ = ϕ ∂x2 dx2 (ordinary derivatives, now), and the Schrödinger equation reads iψ 2 d 2 ψ dϕ =− ϕ + V ψϕ. dt 2m d x 2 1 It is tiresome to keep saying “potential energy function,” so most people just call V the “potential,” even though this invites occasional confusion with electric potential, which is actually potential energy per unit charge. 26 CHAPTER 2 TimeIndependent Schrödinger Equation Or, dividing through by ψϕ: i 1 dϕ 2 1 d 2 ψ + V. =− ϕ dt 2m ψ d x 2 (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 1 dϕ = E, i ϕ dt or iE dϕ = − ϕ, dt (2.4) and − 2 1 d 2 ψ + V = E, 2m ψ d x 2 or − 2 d 2 ψ + V ψ = Eψ. 2m d x 2 (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 C exp (−i Et/), but we might as well absorb the constant C into ψ (since the quantity of interest is the product ψϕ). Then3 ϕ(t) = e−i Et/ . (2.6) The second (Equation 2.5) is called the timeindependent Schrödinger equation; we can go no further with it until the potential V (x) 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 ψ(x) ϕ(t). I offer three answers—two of them physical, and one mathematical: 2 Note that this would not be true if V were a function of t as well as x. 3 Using Euler’s formula, eiθ = cos θ + i sin θ, you could equivalently write ϕ(t) = cos(Et/) + i sin(Et/); the real and imaginary parts oscillate sinusoidally. Mike Casper (of Carleton College) dubbed ϕ the “wiggle factor”—it’s the characteristic time dependence in quantum mechanics. 2.1 Stationary States 1. They are stationary states. Although the wave function itself, (x, t) = ψ(x)e−i Et/ , (2.7) does (obviously) depend on t, the probability density, (x, t)2 = ∗ = ψ ∗ e+i Et/ ψe−i Et/ = ψ(x)2 , (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 d Q(x, p) = ψ ∗ Q x, −i ψ d x. (2.9) dx Every expectation value is constant in time; we might as well drop the factor ϕ(t) altogether, and simply use ψ in place of . (Indeed, it is common to refer to ψ 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, x is constant, and hence (Equation 1.33) p = 0. 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: p2 + V (x). (2.10) 2m The corresponding Hamiltonian operator, obtained by the canonical substitution p → −i(∂/∂ x), is therefore5 H (x, p) = Ĥ = − 2 ∂ 2 + V (x). 2m ∂ x 2 (2.11) Thus the timeindependent Schrödinger equation (Equation 2.5) can be written Ĥ ψ = Eψ, (2.12) and the expectation value of the total energy is ∗ 2 H = ψ Ĥ ψ d x = E ψ d x = E 2 d x = E. (2.13) (Notice that the normalization of entails the normalization of ψ.) Moreover, Ĥ 2 ψ = Ĥ Ĥ ψ = Ĥ Eψ = E Ĥ ψ = E 2 ψ, and hence H2 = ψ ∗ Ĥ 2 ψ d x = E 2 ψ2 d x = E 2 . So the variance of H is σ H2 = H 2 − H 2 = E 2 − E 2 = 0. (2.14) 4 For normalizable solutions, E must be real (see Problem 2.1(a)). 5 Whenever confusion might arise, I’ll put a “hat” (ˆ) on the operator, to distinguish it from the dynamical variable it represents. 27 28 CHAPTER 2 TimeIndependent Schrödinger Equation But remember, if σ = 0, 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 (ψ1 (x), ψ2 (x), ψ3 (x), . . . , which we write as {ψn (x)}), each with its associated separation constant (E 1 , E 2 , E 3 , . . . = {E n }); thus there is a different wave function for each allowed energy: 1 (x, t) = ψ1 (x)e−i E 1 t/ , 2 (x, t) = ψ2 (x)e−i E 2 t/ , . . . . 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 (x, t) = ∞ cn ψn (x)e−i E n t/ . (2.15) n=1 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 (c1 , c2 , . . .) 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 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 V (x), and the starting wave function (x, 0); your job is to find the wave function, (x, t), 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, {ψn (x)}, each with its own associated energy, {E n }. To fit (x, 0) you write down the general linear combination of these solutions: (x, 0) = ∞ cn ψn (x); n=1 6 A linear combination of the functions f (z), f (z), . . . is an expression of the form 1 2 f (z) = c1 f 1 (z) + c2 f 2 (z) + · · · , where c1 , c2 , . . . are (possibly complex) constants. (2.16) 2.1 Stationary States the miracle is that you can always match the specified initial state7 by appropriate choice of the constants {cn }. To construct (x, t) you simply tack onto each term its characteristic time dependence (its “wiggle factor”), exp (−i E n t/):8 (x, t) = ∞ cn ψn (x)e−i E n t/ = n=1 ∞ cn n (x, t). (2.17) n=1 The separable solutions themselves, n (x, t) = ψn (x)e−i E n t/ , (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 2 . Example 2.1 Suppose a particle starts out in a linear combination of just two stationary states: (x, 0) = c1 ψ1 (x) + c2 ψ2 (x). (To keep things simple I’ll assume that the constants cn and the states ψn(x) are real.) What is the wave function (x, t) at subsequent times? Find the probability density, and describe its motion. Solution: The first part is easy: (x, t) = c1 ψ1 (x)e−i E 1 t/ + c2 ψ2 (x) e−i E 2 t/ , where E 1 and E 2 are the energies associated with ψ1 and ψ2 . It follows that c1 ψ1 e−i E 1 t/ + c2 ψ2 e−i E 2 t/ = c12 ψ12 + c22 ψ22 + 2c1 c2 ψ1 ψ2 cos (E 2 − E 1 ) t/ . (x, t)2 = c1 ψ1 ei E 1 t/ + c2 ψ2 ei E 2 t/ The probability density oscillates sinusoidally, at an angular frequency ω = (E 2 − E 1 ) /; this is certainly not a stationary state. But notice that it took a linear combination of stationary states (with different energies) to produce motion.9 7 In principle, any normalized function (x, 0) is fair game—it need not even be continuous. How you might actually get a particle into that state is a different question, and one (curiously) we seldom have occasion to ask. 8 If this is your first encounter with the method of separation of variables, you may be disappointed that the solution takes the form of an infinite series. Occasionally it is possible to sum the series, or to solve the timedependent Schrödinger equation without recourse to separation of variables—see, for instance, Problems 2.49, 2.50, and 2.51. But such cases are extremely rare. 9 This is nicely illustrated in an applet by Paul Falstad, at www.falstad.com/qm1d/. 29 30 CHAPTER 2 TimeIndependent Schrödinger Equation You may be wondering what the coefficients {cn } represent physically. I’ll tell you the answer, though the explanation will have to await Chapter 3: cn 2 is the probability that a measurement of the energy would return the value E n . (2.19) A competent measurement will always yield one of the “allowed” values (hence the name), and cn 2 is the probability of getting the particular value E n .10 Of course, the sum of these probabilities should be 1: ∞ cn 2 = 1, (2.20) n=1 and the expectation value of the energy must be H = ∞ cn 2 E n . (2.21) n=1 We’ll soon see how this works out in some concrete examples. Notice, finally, that because the constants {cn } 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 E 0 + i (with E 0 and real), and show that if Equation 1.20 is to hold for all t, must be zero. (b) The timeindependent wave function ψ(x) can always be taken to be real (unlike (x, t), 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 ψs that are real. Hint: If ψ(x) satisfies Equation 2.5, for a given E, so too does its complex conjugate, and hence also the real linear combinations (ψ + ψ ∗ ) and i (ψ − ψ ∗ ). (c) If V (x) is an even function (that is, V (−x) = V (x)) then ψ(x) can always be taken to be either even or odd. Hint: If ψ(x) satisfies Equation 2.5, for a given E, so too does ψ(−x), and hence also the even and odd linear combinations ψ(x) ± ψ(−x). 10 Some people will tell you that c 2 is “the probability that the particle is in the nth stationary state,” but this is bad n language: the particle is in the state , not n , and anyhow, in the laboratory you don’t “find the particle to be in a particular state,” you measure some observable, and what you get is a number, not a wave function. 2.2 The Infinite Square Well V(x) a x Figure 2.1: The infinite square well potential (Equation 2.22). ∗ Problem 2.2 Show that E must exceed the minimum value of V (x), 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 2m d 2ψ = 2 [V (x) − E] ψ; 2 dx if E < Vmin , then ψ and its second derivative always have the same sign—argue that such a function cannot be normalized. 2.2 THE INFINITE SQUARE WELL Suppose V (x) = 0, ∞, 0 ≤ x ≤ a, otherwise (2.22) (Figure 2.1). A particle in this potential is completely free, except at the two ends (x = 0 and x = a), 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.) Outside the well, ψ(x) = 0 (the probability of finding the particle there is zero). Inside the well, where V = 0, the timeindependent Schrödinger equation (Equation 2.5) reads − or 2 d 2 ψ = Eψ, 2m d x 2 (2.23) √ d 2ψ 2m E 2 . (2.24) = −k ψ, where k ≡ 2 dx (By writing it in this way, I have tacitly assumed that E ≥ 0; we know from Problem 2.2 that E < 0 won’t work.) Equation 2.24 is the classical simple harmonic oscillator equation; the general solution is 31 32 CHAPTER 2 TimeIndependent Schrödinger Equation ψ(x) = A sin kx + B cos kx, (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 ψ(x)? Ordinarily, both ψ and dψ/d x are continuous,11 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 V = ∞, in Section 2.5; for now I hope you will trust me.) Continuity of ψ(x) requires that ψ(0) = ψ(a) = 0, (2.26) so as to join onto the solution outside the well. What does this tell us about A and B? Well, ψ(0) = A sin 0 + B cos 0 = B, so B = 0, and hence ψ(x) = A sin kx. (2.27) Then ψ(a) = A sin ka, so either A = 0 (in which case we’re left with the trivial—nonnormalizable—solution ψ(x) = 0), or else sin ka = 0, which means that ka = 0, ±π, ±2π, ±3π, . . . . (2.28) But k = 0 is no good (again, that would imply ψ(x) = 0), and the negative solutions give nothing new, since sin(−θ ) = − sin(θ ) and we can absorb the minus sign into A. So the distinct solutions are nπ , with n = 1, 2, 3, . . . . (2.29) kn = a Curiously, the boundary condition at x = a does not determine the constant A, but rather the constant k, and hence the possible values of E: En = n 2 π 2 2 2 kn2 = . 2m 2ma 2 (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 ψ:13 a a 2 A2 sin2 (kx) d x = A2 = 1, so A2 = . 2 a 0 This only determines the magnitude of A, but it is simplest to pick the positive real root: √ A = 2/a (the phase of A carries no physical significance anyway). Inside the well, then, the solutions are nπ 2 sin x . ψn (x) = (2.31) a a 11 That’s right: ψ(x) is a continuous function of x, even though (x, t) need not be. 12 Notice that the quantization of energy emerges as a rather technical consequence of the boundary conditions on solutions to the timeindependent Schrödinger equation. 13 Actually, it’s (x, t) that must be normalized, but in view of Equation 2.7 this entails the normalization of ψ(x). 2.2 The Infinite Square Well Ψ1(x) Ψ2(x) a Ψ3(x) x a x a x Figure 2.2: The first three stationary states of the infinite square well (Equation 2.31). As promised, the timeindependent Schrödinger equation has delivered an infinite set of solutions (one for each positive integer n). The first few of these are plotted in Figure 2.2. They look just like the standing waves on a string of length a; ψ1 , which carries the lowest energy, is called the ground state, the others, whose energies increase in proportion to n 2 , are called excited states. As a collection, the functions ψn (x) have some interesting and important properties: 1. They are alternately even and odd, with respect to the center of the well: ψ1 is even, ψ2 is odd, ψ3 is even, and so on.14 2. As you go up in energy, each successive state has one more node (zerocrossing): ψ1 has none (the end points don’t count), ψ2 has one, ψ3 has two, and so on. 3. They are mutually orthogonal, in the sense that15 ψm (x)∗ ψn (x) d x = 0, (m = n) . (2.32) Proof: nπ mπ 2 a x sin x dx sin ψm (x)∗ ψn (x) d x = a 0 a a 1 a m−n m+n = cos π x − cos πx dx a 0 a a a 1 m−n m+n 1 πx − π x sin sin = a a (m − n) π (m + n) π 0 1 sin [(m − n) π ] sin [(m + n) π ] − = 0. = π (m − n) (m + n) Note that this argument does not work if m = n. (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: ψm (x)∗ ψn (x) d x = δmn , (2.33) 14 To make this symmetry more apparent, some authors center the well at the origin (running it now from −a to +a). The even functions are then cosines, and the odd ones are sines. See Problem 2.36. 15 In this case the ψs are real, so the complex conjugation (*) of ψ is unnecessary, but for future purposes it’s a m good idea to get in the habit of putting it there. 33 34 CHAPTER 2 TimeIndependent Schrödinger Equation where δmn (the socalled Kronecker delta) is defined by 0, m = n, δmn = 1, m = n. (2.34) We say that the ψs are orthonormal. 4. They are complete, in the sense that any other function, f (x), can be expressed as a linear combination of them: ∞ ∞ 2 nπ x . (2.35) cn ψn (x) = cn sin f (x) = a a n=1 n=1 √ I’m not about to prove the completeness of the functions 2/a sin(nπ x/a), but if you’ve studied advanced calculus you will recognize that Equation 2.35 is nothing but the Fourier series for f (x), and the fact that “any” function can be expanded in this way is sometimes called Dirichlet’s theorem.16 The coefficients cn can be evaluated—for a given f (x)—by a method I call Fourier’s trick, which beautifully exploits the orthonormality of {ψn }: Multiply both sides of Equation 2.35 by ψm (x)∗ , and integrate. ψm (x)∗ f (x) d x = ∞ cn ψm (x)∗ ψn (x) d x = n=1 ∞ cn δmn = cm . (2.36) n=1 (Notice how the Kronecker delta kills every term in the sum except the one for which n = m.) Thus the nth coefficient in the expansion of f (x) is17 cn = ψn (x)∗ f (x) d x. (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 nπ 2 2 2 2 sin x e−i(n π /2ma )t . (2.38) n (x, t) = a a I claimed (Equation 2.17) that the most general solution to the (timedependent) Schrödinger equation is a linear combination of stationary states: 16 See, for example, Mary Boas, Mathematical Methods in the Physical Sciences, 3rd edn (New York: John Wiley, 2006), p. 356; f (x) can even have a finite number of finite discontinuities. 17 It doesn’t matter whether you use m or n as the “dummy index” here (as long as you are consistent on the two sides of the equation, of course); whatever letter you use, it just stands for “any positive integer.” 18 Problem 2.45 explores this property. For further discussion, see John L. Powell and Bernd Crasemann, Quantum Mechanics (AddisonWesley, Reading, MA, 1961), Section 57. 2.2 The Infinite Square Well (x, t) = ∞ cn n=1 2 nπ 2 2 2 sin x e−i(n π /2ma )t . a a (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, (x, 0) by appropriate choice of the coefficients cn : (x, 0) = ∞ cn ψn (x) . n=1 The completeness of the ψs (confirmed in this case by Dirichlet’s theorem) guarantees that I can always express (x, 0) in this way, and their orthonormality licenses the use of Fourier’s trick to determine the actual coefficients: 2 a nπ x (x, 0) d x. (2.40) cn = sin a 0 a That does it: Given the initial wave function, (x, 0), we first compute the expansion coefficients cn , using Equation 2.40, and then plug these into Equation 2.39 to obtain (x, t). 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 ψs and the equation for the allowed energies. Example 2.2 A particle in the infinite square well has the initial wave function (x, 0) = Ax (a − x) , (0 ≤ x ≤ a) , for some constant A (see Figure 2.3). Outside the well, of course, = 0. Find (x, t). Ψ(x, 0) Aa2 4 a x Figure 2.3: The starting wave function in Example 2.2. Solution: First we need to determine A, by normalizing (x, 0): a a a5 2 2 (x, 0) d x = A 1= x 2 (a − x)2 d x = A2 , 30 0 0 so 30 A= . a5 The nth coefficient is (Equation 2.40) 35 36 CHAPTER 2 TimeIndependent Schrödinger Equation 2 a 30 nπ cn = sin x (a − x) d x x a 0 a a5 √ a a 2 15 nπ nπ 2 a x dx − x dx = x sin x sin a a a3 0 0 √ a a 2 ax nπ 2 15 nπ a x − cos x = sin 3 nπ a nπ a a 0 a 2 a 2 nπ nπ (nπ x/a) − 2 x − x − 2 x sin cos 3 nπ a a (nπ/a) 0 √ 2 2 15 a3 (nπ )2 − 2 cos (nπ ) + a 3 = cos (nπ ) + a 3 cos (0) − 3 nπ a (nπ )3 (nπ )3 √ 4 15 [cos (0) − cos (nπ )] = (nπ )3 0, n even, = √ 8 15/ (nπ )3 , n odd. Thus (Equation 2.39): (x, t) = 30 a 3 2 π n=1,3,5... 1 nπ 2 2 2 x e−in π t/2ma . sin a n3 Example 2.3 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 ψ1 (Figure 2.2). This suggests that c1 2 should dominate,19 and in fact ! √ "2 8 15 c1 2 = = 0.998555 . . . . π3 The rest of the coefficients make up the difference:20 19 Loosely speaking, c tells you the “amount of ψ that is contained in .” n n 20 You can look up the series 1 1 1 π6 + 6 + 6 + ··· = 6 960 1 3 5 and 1 1 π4 1 + + + · · · = 96 14 34 54 in math tables, under “Sums of Reciprocal Powers” or “Riemann Zeta Function.” 2.2 The Infinite Square Well ∞ n=1 ! √ "2 8 15 cn 2 = π3 ∞ n=1,3,5,... 1 = 1. n6 The most likely outcome of an energy measurement is E 1 = π 2 2 /2ma 2 —more than 99.8% of all measurements will yield this value. The expectation value of the energy (Equation 2.21) is ! √ "2 ∞ ∞ 8 15 n 2 π 2 2 1 4802 52 H = = = . n3π 3 2ma 2 π 4 ma 2 n4 ma 2 n=1,3,5,... n=1,3,5,... As one would expect, it is very close to E 1 (5 in place of π 2 /2 ≈ 4.935)—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 cn s are independent of time, so I’m going to do the proof for t = 0; if this bothers you, you can easily generalize the argument to arbitrary t). "∗ ! ∞ " ! ∞ 2 cm ψm (x) cn ψn (x) d x 1 = (x, 0) d x = = = ∞ ∞ m=1 n=1 ∞ ∞ ∗ cm cn m=1 ψm (x)∗ ψn (x) d x ∗ cm cn δmn = n=1 m=1 n=1 ∞ cn 2 . n=1 (Again, the Kronecker delta picks out the term m = n 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 Ĥ ψn = E n ψn , so (2.41) ∗ Ĥ d x = cm ψm Ĥ cn ψn d x ∗ cn 2 E n . = cm cn E n ψm∗ ψn d x = H = ∗ Problem 2.3 Show that there is no acceptable solution to the (timeindependent) Schrödinger equation for the infinite square well with E = 0 or E < 0. (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 x, x 2 , p, p 2 , σx , and σ p , 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? 37 38 CHAPTER 2 ∗ TimeIndependent Schrödinger Equation 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: (x, 0) = A [ψ1 (x) + ψ2 (x)] . (a) Normalize (x, 0). (That is, find A. This is very easy, if you exploit the orthonormality of ψ1 and ψ2 . Recall that, having normalized at t = 0, you can rest assured that it stays normalized—if you doubt this, check it explicitly after doing part (b).) (b) Find (x, t) and (x, t)2 . Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let ω ≡ π 2 /2ma 2 . (c) Compute x. 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 a/2, go directly to jail.) (d) Compute p. (As Peter Lorre would say, “Do it ze kveek vay, Johnny!”) (e) 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 E 1 and E 2 ? 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 ψ1 and ψ2 in Problem 2.5: # $ (x, 0) = A ψ1 (x) + eiφ ψ2 (x) , where φ is some constant. Find (x, t), (x, t)2 , and x, and compare your results with what you got before. Study the special cases φ = π/2 and φ = π . (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 Ax, 0 ≤ x ≤ a/2, (x, 0) = A (a − x) , a/2 ≤ x ≤ a. (a) (b) (c) (d) Sketch (x, 0), and determine the constant A. Find (x, t). What is the probability that a measurement of the energy would yield the value E 1 ? Find the expectation value of the energy, using Equation 2.21.21 21 Remember, there is no restriction in principle on the shape of the starting wave function, as long as it is normal izable. In particular, (x, 0) need not have a continuous derivative. However, if you try to calculate H using (x, 0)∗ Ĥ (x, 0) d x in such a case, you may encounter technical difficulties, because the second derivative of (x, 0) is ill defined. It works in Problem 2.9 because the discontinuities occur at the end points, where the wave function is zero anyway. In Problem 2.39 you’ll see how to manage cases like Problem 2.7. 2.3 The Harmonic Oscillator Problem 2.8 A particle of mass m in the infinite square well (of width a) starts out in the state A, 0 ≤ x ≤ a/2, (x, 0) = 0, a/2 ≤ x ≤ a, for some constant A, so it is (at t = 0) 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 t) would yield the value π 2 2 /2ma 2 ? Problem 2.9 For the wave function in Example 2.2, find the expectation value of H , at time t = 0, the “old fashioned” way: H = (x, 0)∗ Ĥ (x, 0) d x. Compare the result we got in Example 2.3. Note: Because H is independent of time, there is no loss of generality in using t = 0. 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, F = −kx = m d2x dt 2 (ignoring friction), and the solution is x(t) = A sin(ωt) + B cos(ωt) , where k m is the (angular) frequency of oscillation. The potential energy is ω≡ V (x) = 1 2 kx ; 2 (2.42) (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 V (x) in a Taylor series about the minimum: 1 V (x) = V (x0 ) + V (x0 ) (x − x0 ) + V (x0 ) (x − x0 )2 + · · · , 2 subtract V (x0 ) (you can add a constant to V (x) with impunity, since that doesn’t change the force), recognize that V (x0 ) = 0 (since x0 is a minimum), and drop the higherorder terms (which are negligible as long as (x − x0 ) stays small), we get 39 40 CHAPTER 2 TimeIndependent Schrödinger Equation V(x) x0 x Figure 2.4: Parabolic approximation (dashed curve) to an arbitrary potential, in the neighborhood of a local minimum. 1 V (x0 ) (x − x0 )2 , 2 which describes simple harmonic oscillation (about the point x0 ), with an effective spring constant k = V (x0 ). 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 The quantum problem is to solve the Schrödinger equation for the potential V (x) ≈ 1 mω2 x 2 (2.44) 2 (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: V (x) = − 1 2 d 2 ψ + mω2 x 2 ψ = Eψ. 2 2m d x 2 (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. 2.3.1 Algebraic Method To begin with, let’s rewrite Equation 2.45 in a more suggestive form: $ 1 # 2 p̂ + (mωx)2 ψ = Eψ, 2m (2.46) 22 Note that V (x ) ≥ 0, since by assumption x is a minimum. Only in the rare case V (x ) = 0 is the oscillation 0 0 0 not even approximately simple harmonic. 23 We’ll encounter some of the same strategies in the theory of angular momentum (Chapter 4), and the technique generalizes to a broad class of potentials in supersymmetric quantum mechanics (Problem 3.47; see also Richard W. Robinett, Quantum Mechanics (Oxford University Press, New York, 1997), Section 14.4). 2.3 The Harmonic Oscillator where p̂ ≡ −i d/d x is the momentum operator.24 The basic idea is to factor the Hamiltonian, $ 1 # 2 p̂ + (mωx)2 . (2.47) Ĥ = 2m If these were numbers, it would be easy: u 2 + v 2 = (iu + v) (−iu + v) . Here, however, it’s not quite so simple, because p̂ and x are operators, and operators do not, in general, commute (x p̂ is not the same as p̂x, 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 1 ∓i p̂ + mωx (2.48) â± ≡ √ 2mω (the factor in front is just there to make the final results look nicer). Well, what is the product â− â+ ? 1 2mω 1 = 2mω â− â+ = i p̂ + mωx −i p̂ + mωx # $ p̂ 2 + (mωx)2 − imω x p̂ − p̂x . As anticipated, there’s an extra term, involving x p̂ − p̂x . We call this the commutator of x and p̂; it is a measure of how badly they fail to commute. In general, the commutator of operators Â and B̂ (written with square brackets) is # $ Â, B̂ ≡ Â B̂ − B̂ Â. (2.49) In this notation, â− â+ = $ i 1 # 2 p̂ + (mωx)2 − x, p̂ . 2mω 2 (2.50) We need to figure out the commutator of x and p̂. 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,” f (x), 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: df df d d −x − f x, p̂ f (x) = x (−i) ( f ) − (−i) (x f ) = −i x dx dx dx dx = i f (x) . (2.51) Dropping the test function, which has served its purpose, x, p̂ = i. (2.52) This lovely and ubiquitous formula is known as the canonical commutation relation.25 24 Put a hat on x, too, if you like, but since x̂ = x we usually leave it off. 25 In a deep sense all of the mysteries of quantum mechanics can be traced to the fact that position and momentum do not commute. Indeed, some authors take the canonical commutation relation as an axiom of the theory, and use it to derive p̂ = −i d/d x. 41 42 CHAPTER 2 TimeIndependent Schrödinger Equation With this, Equation 2.50 becomes â− â+ = or 1 1 Ĥ + , ω 2 1 Ĥ = ω â− â+ − . 2 (2.53) (2.54) Evidently the Hamiltonian does not factor perfectly—there’s that extra −1/2 on the right. Notice that the ordering of â+ and â− is important here; the same argument, with â+ on the left, yields 1 1 (2.55) â+ â− = Ĥ − . ω 2 In particular, â− , â+ = 1. (2.56) Meanwhile, the Hamiltonian can equally well be written 1 . Ĥ = ω â+ â− + 2 (2.57) In terms of â± , then, the Schrödinger equation26 for the harmonic oscillator takes the form 1 ψ = Eψ (2.58) ω â± â∓ ± 2 (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 ψ satisfies the Schrödinger equation with energy E (that is: Ĥ ψ = equation with energy (E + Eψ), then â+ψ satisfies the Schrödinger ω): Ĥ â+ ψ = (E + ω) â+ ψ . Proof: 1 1 â+ ψ = ω â+ â− â+ + â+ ψ Ĥ â+ ψ = ω â+ â− + 2 2 1 1 = ωâ+ â− â+ + ψ = â+ ω â+ â− + 1 + ψ 2 2 = â+ Ĥ + ω ψ = â+ (E + ω)ψ = (E + ω) â+ ψ . QED (I used Equation 2.56 to replace â− â+ by â+ â− + 1 in the second line. Notice that whereas the ordering of â+ and â− does matter, the ordering of â± and any constants— such as , ω, and E—does not; an operator commutes with any constant.) By the same token, â− ψ is a solution with energy (E − ω): 1 1 Ĥ â− ψ = ω â− â+ − â− ψ = ωâ− â+ â− − ψ 2 2 1 = â− ω â− â+ − 1 − ψ = â− Ĥ − ω ψ = â− (E − ω)ψ 2 = (E − ω) â− ψ . 26 I’m getting tired of writing “timeindependent Schrödinger equation,” so when it’s clear from the context which one I mean, I’ll just call it the “Schrödinger equation.” 2.3 The Harmonic Oscillator E E+3h ω â +3ψ E+2h ω â +2ψ â +ψ E+ h ω ψ E E− h ω E−2h ω â+ â_ψ â _2ψ â_ E0 ψ0