Introduction to Quantum Mechanics, Summaries of Quantum Mechanics

Download Introduction to Quantum Mechanics and more Summaries Quantum Mechanics in PDF only on Docsity! INTRODUCTION TO QUANTUM MECHANICS Srcel eens Tamers DAVID J. GRIFFITHS Fundamental Equations Schrédinger equation: Time-independent Schrédinger equation: Hw = EW. Ws pe ten Hamiltonian operator: i? _, H=-—ViV 2m Momentum operator: p=-iAV Time dependence of an expectation value: ag) _ 9Q SE. = = ((H, ~ doh ({H, Q)) + Or Generalized uncertainty principle: 1 oA0g = |z (A, e1| Heisenberg uncertainty principle: Oxon > A/2 Canonical commutator: {x, p) = ih Angular momentum: [Lx, Ly] = fA, [Ly. Ly] = iALy. (Lz, Lx] =ihLy a= (0). m=) a=( 8) Pauli matrices: Editor-in-Chief. Science: John Challice Senior Editor: Erik Fahlgren Associate Editor: Christian Botting Editorial Assistant: Andrew Sobel Vice President and Director of Production and Manufacturing, ESM: David W. Riccardi Production Editor: Beth Lew Director of Creative Services: Paul Belfanti Art Director: Jayne Conte Cover Designer: Bruce Kenselaar Managing Editor, AV Management and Production: Patricia Burns Art Editor: Abigail Bass Manufacturing Manager: Trudy Pisciotti Manufacturing Buyer: Lynda Castillo Executive Marketing Manager: Mark Pfaltzgraff © 2005, 1995 Pearson Education, Inc. Pearson Prentice Hall Pearson Education, Inc. Upper Saddle River, NJ 07458 All rights reserved, No part of this book may be reproduced in any form or by any means, without permission in writing from the publisher. Pearson Prentice Hall® is a trademark of Pearson Education, Inc. Printed in the United States of America 109 8 7 ISBN O-13-191175-9 If you purchased this book within the United States or Canada you should be aware that it has been wrongfully imported without the approval of the Publisher or the Author. Pearson Education LTD.. London Pearson Education Australia Pty. Ltd.. Sydney Pearson Education Singapore, Pte. Ltd. Pearson Education North Asia Lid,. Hong Kong Pearson Education Canada, Inc., Toronto Pearson Educacién de Mexico, S.A. de C.V. Pearson Education—Japan. Tokyo Pearson Education Malaysia, Pte, Ltd. Pearson Education. Upper Saddle River, New Jersey  CONTENTS PREFACE vii PARTI THEORY 1 THE WAVE FUNCTION 1 11 1.2 1.3 14 15 1.6 The Schrédinger Equation 1 The Statistical Interpretation 2 Probability 5 Normalization 12 Momentum 15 The Uncertainty Principle 18 2. TIME-INDEPENDENT SCHRODINGER EQUATION 24 21 2.2 2.3 2.4 2.5 2.6 Stationary States 24 The Infinite Square Well 30 The Harmonic Oscillator 40 The Free Particle 59 The Delta-Function Potential 68 The Finite Square Well 78 3 FORMALISM 93 3.1 3.2 3.3 Hilbert Space 93 Observables 96 Eigenfunctions of a Hermitian Operator 100 iti iv Contents 3.4 Generalized Statistical Interpretation 106 3.5 The Uncertainty Principle 110 3.6 Dirac Notation 118 QUANTUM MECHANICS IN THREE DIMENSIONS 131 4.1 Schrédinger Equation in Spherical Coordinates 131 4.2 The Hydrogen Atom 145 43 Angular Momentum 160 44 Spin 171 IDENTICAL PARTICLES 201 5.1 Two-Particle Systems 201 5.2 Atoms 210 5.3 Solids 218 5.4 Quantum Statistical Mechanics 230 PART I APPLICATIONS 6 10 TIME-INDEPENDENT PERTURBATION THEORY 249 6.1 | Nondegenerate Perturbation Theory 249 6.2 Degenerate Perturbation Theory 257 6.3. The Fine Structure of Hydrogen 266 6.4 The Zeeman Effect 277 6.5 Hyperfine Splitting 283 THE VARIATIONAL PRINCIPLE 293 7.1. Theory 293 7.2 The Ground State of Helium 299 7.3. The Hydrogen Molecule Ion 304 THE WKB APPROXIMATION 315 8.1 The “Classical” Region 316 8.2 Tunneling 320 8.3. The Connection Formulas 325 TIME-DEPENDENT PERTURBATION THEORY 340 9.1 Two-Level Systems 341 9.2 Emission and Absorption of Radiation 348 9.3. Spontaneous Emission 355 THE ADIABATIC APPROXIMATION 368 10.1 The Adiabatic Theorem 368 10.2 Berry’s Phase 376 Vili Preface instructors, for example, may wish to treat time-independent perturbation theory immediately after Chapter 2. This book is intended for a one-semester or one-year course at the junior or senior level. A one-semester course will have to concentrate mainly on Part I; a full-year 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 I would like to emphasize that quantum mechanics is not, in my view, something that flows smoothly and naturally from earlier theories. On the contrary, it represents an abrupt and revolutionary departure from classical ideas, calling forth a wholly new and radically counterintuitive way of thinking about the world. That, indeed, is what makes it such a fascinating subject. At first glance, this book may strike you as forbiddingly mathematical. We encounter Legendre, Hermite, and Laguerre polynomials, spherical harmonics, Bessel, Neumann, and Hankel functions, Airy functions, and even the Riemann zeta function—not to mention Fourier transforms, Hilbert spaces, hermitian oper- ators, Clebsch-Gordan coefficients, and Lagrange multipliers. Is all this baggage teally 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 asking the student to dig a foun- dation with a screwdriver. (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. My own instinct is to hand the students shovels and tell them to start dig- ging. They may develop blisters at first, but I still think this is the most efficient and exciting way to learn.) At any rate, I can assure you that there is no deep mathematics in this book, and if you run into something unfamiliar, and you don’t find my explanation adequate, by all means ask someone about it, or look it up. There are many good books on mathematical methods—I particularly recommend Mary Boas, Mathematical Methods in the Physical Sciences, 2nd ed., Wiley, New York (1983), or George Arfken and Hans-Jurgen Weber, Mathematical Methods for Physicists, 5th ed., Academic Press, Orlando (2000). But whatever you do, don’t let the mathematics—which, for us, is only a too/ —interfere with 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. I 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 Preface ix suggested that I 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); I have adopted the following rating scheme: * an essential problem that every reader should study; ** a somewhat more difficult or more 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 one-star problems appear at the end of the relevant section; most of the three-star problems are at the end of the chapter. A solution manual is available (to instructors only) from the publisher. In preparing the second edition ] have tried to retain as much as possible the spiril of the first. The only wholesale change is Chapter 3, which was much too long and diverting; it has been completely rewritten, with the background material on finite-dimensional vector spaces (a subject with which most students at this level are already comfortable) relegated to the Appendix. I have added some examples in Chapter 2 (and fixed the awkward definition of raising and lowering operators for the harmonic oscillator). In later chapters I have made as few changes as I could, even preserving the numbering of problems and equations, where possible. The treatment is streamlined in places (a better introduction to angular momentum in Chapter 4, for instance, a simpler proof of the adiabatic theorem in Chapter 10, and a new section on partial wave phase shifts in Chapter 11). Inevitably, the second edition is a bit longer than the first, which I regret, but I hope it is cleaner and more accessible. I have benefited from the comments and advice of many colleagues, who read the original manuscript, pointed out weaknesses (or errors) in the first edition, suggested improvements in the presentation, and supplied interesting problems. I would like to thank in particular P. K, Aravind (Worcester Polytech), Greg Benesh (Baylor), David Boness (Seattle), 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), Henry Greenside (Duke), Paul Haines (Dartmouth), J. R. Huddle (Navy), Larry Hunter (Amherst), David Kaplan (Washington), Alex Kuzmich (Georgia Tech), Peter Leung (Portland State), Tony Liss (Illinois), Jeffry Mallow (Chicago Loyola), James McTavish (Liverpool), James Nearing (Miami), Johnny Powell (Reed), Krishna Rajagopal (MIT), Brian Raue (Florida Interna- tional), Robert Reynolds (Reed), Keith Riles (Michigan), Mark Semon (Bates), Herschel Snodgrass (Lewis and Clark), John Taylor (Colorado), Stavros Theodor- akis (Cyprus), A. S. Tremsin (Berkeley), Dan Velleman (Amherst), Nicholas Wheeler (Reed), Scott Willenbrock (Illinois), William Wootters (Williams), Sam Wurzel (Brown), and Jens Zorn (Michigan). Introduction to Quantum Mechanics Section 1.2: The Statistical Interpretation 3 lel? FIGURE 1,2: A typical wave function. The shaded area represents the probability of finding the particle between a and b. The particle would be relatively likely to be found near A, and unlikely to be found near B. The statistical interpretation introduces a kind of indeterminacy into quan- tum 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 mechan- ics has to offer is statistical information about the possible results. This inde- terminacy 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.* Question: Where was the particle just before I made the measurement? There are three plausible answers to this question, and they serve to characterize the main schools of thought regarding quantum indeterminacy: 1. The realist position: The particle was at C. This certainly seems like a sen- sible response, and it is the one 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, indeter- minacy 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.”> Evidently W is not the whole story —some additional infor- mation (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 the particle 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 4Of course, no measuring instrument is perfectly precise: what I mean is thal the particle was found in the vicinity of C. to within the tolerance of the equipment. > Berard d° Espagnat, “The Quantum Theory and Reality” (Scientific American, November 1979, p. 165), 4 Chapter 1 The Wave Function particle) to assume a definite position.”* This view (the so-called 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 over half a century of debate has done precious little to illuminate. 3. The agnostic position: Refitse 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 conduct 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.””? For decades this was the “fall-back” position of most physicists: They’d try to sell you the orthodox answer, but if you were persistent they'd retreat to the agnostic response, and terminate the conversation, Until fairly recently, all three positions (realist, orthodox, and agnostic) had their partisans. But in 1964 John Bell astonished the physics community by showing that it makes an observable difference whether the particle had a precise (though unknown) position prior to the measurement, or not. Bell’s discovery effectively eliminated agnosticism as a viable option, and made it an experimental question whether | or 2 is the correct choice. Ill 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 interpreta~ tion:® 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 1 made a second measurement, immediately after the first? Would | get C again, or does the act of measurement cough up some completely new num- ber 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 SQuoted in a lovely article by N. David Mermin, “Is the moon there when nobody looks?” (Physics Today. April 1985. p. 38). 7 Quoted by Mermin (footnote 6). p. 40. 8This statement is a little too strong: There remain a few theoretical and experimental loopholes, some of which T shall discuss in the Afterword, There exist viahle nonlocal hidden variable theories (notably David Bohms's), and other formulations (such as the many worlds interpretation) that do not fit cleanly inte any of my three categories, But | think it is wise. ut least fram a pedagogical point of view, to adopt a clear and coherent platform at this stage. and worry about the alternatives later, Section 1.3: Probability 5 lw? FIGURE 1.3: Collapse of the wave function: graph of [W|? immediately after a measurement has found the particle at point C. interpretation account for the fact that the second measurement is bound to yield the value C? Evidently 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 func- tion collapses, upon measurement, to a spike at the point C (it soon spreads out again, in accordance with the Schrédinger equation, so the second measurement must be made quickly). There are, then, two entirely distinct kinds of physical pro- cesses: “ordinary” ones, in which the wave function evolves in a leisurely fashion under the Schrédinger equation, and “measurements,” in which Y suddenly and discontinuously collapses. 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, *The role of measurement in quantum mechanics is so. critical and so bizarre that you may well be wondering what precisely constitutes a measurement. Does it have to do with the interaction between a microscopic (quantum) system and a macroscopic (classical) measuring apparatus (as Bohr insisted), or is it characterized by the leaving of a permanent “record” (as Heisenberg claimed), or does it involve the intervention of a conscious “observer™ (as Wigner proposed)? 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 does in the laboratory. with rulers, stopwatches, Geiger counters, and so on. 8 Chapter1 The Wave Function NU) NC) 1 123 4656 67 89 10 j; 123 45 6 7 8 910 jf FIGURE 1.5: Two histograms with the same median, same average, and same most probable value, but different standard deviations. (Equations 1.6, 1.7, and 1.8 are, if you like, special cases of this formula.) Beware: The average of the squares, (j*), is not equal, in general, to the square of the average, ay. For instance, if the room contains just two babies, aged 1 and 3, then (x?) = 5, but (xj? =4. Now, there is aconspicuous difference between the two histograms in Figure 1.5. 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 big-city classroom, the second, perhaps, a rural one-room school- house.) We need a numerical measure of the amount of “spread” in a distribution, with respect to the average. The most obvious way to do this would be to find out how far each individual deviates from the average, Aj=i-(). [1.10] and compute the average of Aj. Trouble is, of course, that you get zero, since, by the nature of the average, Aj is as often negative as positive: (Af) = S20 — PW = iPM WI PW) = (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 Aj. But absolute values are nasty to work with; instead, we get around the sign problem by squaring before averaging: o* = ((Ajy’). [1.11] Section 1.3: Probability 9 This quantity is known as the variance of the distribution; o 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 (/). There is a useful little theorem on variances: o? = (As) = SAPP) = OU - YP PU) =u? - 24 + YP = PPA YL P+ UP YP) =P) - 2G) + UP = 7) - GY. Taking the square root, the standard deviation itself can be written as o = (i?) -(j)*- [1.12] In practice, this is a much faster way to get o: Simply calculate (j?) and (j)?, subtract, and take the square root. Incidentally, I warned you a moment ago that (j?) is not, in general, equal to (j)?. Since o? is plainly nonnegative (from its definition in Equation 1.11), Equation 1.12 implies that (P) = UY, [1.13] and the two are equal only when o = 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 1 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 nwo days is presumably twice the probability that it is between 16 and 16 plus one day. (Unless, I suppose, there was some extraordinary baby boom 16 years ago, on exactly that day—in which case we have simply chosen an interval too long for the rule to apply. If the baby boom 10 Chapter 1 The Wave Function 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 al random) lies between x and (x + dx) | = pide, [1.14] The proportionality factor, o(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 6 (a finite interval) is given by the integral of p(x): b Pa = [ p(x) dx, [1.15] a and the rules we deduced for discrete distributions translate in the obvious way: HOO 1 -/ p(x) dx, [1.16] mood +00 (x) -/ xp(x) dx, [1.17] 20 oo (f@)) = F)e@) dx, {1.18] OO 2 = (Ax?) = (7) — Gy, [1.19] Example 1.1 Suppose I drop a rock off a cliff of height 7. As it falls. I snap a wt million photographs, at random intervals. On each picture [ measure the d ” the rock has fallen. Question: What is the average of all these distances” to say, what is the time average of the distance traveled’?! 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 must be less than 4/2. Ignoring air resistance, the distance x at time f is i, x(t) = xgt?. x(t) 378 The velocity is dv/dr = gt, and the total flight time is T = ./2//g. The probability that the camera flashes in the interval dr is dt/T, so the probability that a given 104 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 tor the experimentalist, especially when the sample size is small, but here | am only concerned, of course, with the true average. to which the sample average is presumably a good approximation. Section 1.4: Normalization 13 glance at Equation 1.1 reveals that if Y(x. r) is a solution, so too is AY(x. 1), 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 Schrédinger equation the integral is infinite; in that case no multiplicative factor is going to make it 1. The same goes for the trivial solution VY = 0. Such non-normalizable solutions cannot represent particles, and must be rejected. Physically realizable states correspond to the square-integrable solutions to Schrédinger’s equation.!! 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 ft, and you no longer have a solution to the Schrédinger equation.) Fortunately, the Schrédinger equation has the remarkable property that it automatically preserves the normalization of the wave function—without this crucial feature the Schrédinger 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 O90 2 HOC a > af. Wo. oP as -/ 5 Wen DP dx. (1.21) 00 (Note that the integral is a function only of t, so I use a total derivative (d/drt) in the first expression, but the integrand is a function of x as well as 1, so it’s a partial derivative (0/8f) in the second one.) By the product rule, aw* a. 8. , aw —|¥r = —(*) = ww, 1,22 ar! | ar ) ar + ar {1.22] Now the Schrédinger equation says that aw ik ePw i Est 1.23 at am ax? hk 1.23] and hence also (taking the complex conjugate of Equation 1.23) aw* ihaews Ff =-_ + ivy, [1.24] at 2m ax? hh so a if aap a2 a [ia Yaw “wea? (v5 -Sav)- 245 (ws 2 ¥)]. [1.25] ar ~ 2m ax? ax? ax |2m Ox ax ' Evidently W(x. 1) must go to zero faster than 1/,/|x], as |x| + oo. Incidentally. normalization only fixes the modulus of A: the phase remains undetermined. However. as we shall see, the latter carries no physical significance anyway. 14 Chapter 1 The Wave Function The integral in Equation 1.21 can now be evaluated explicitly: te° i vw awe \i+ =| Wo.nPdr = ot (ws 2 ¥) ~ [1.26] t Joo 2m oo x ax But W(x, ) must go to zero as x goes to (+) infinity—otherwise the wave function would not be normalizable.!? It follows that +00 — Von dx =0, [1.27] dt Juoo and hence that the integral is constant (independent of time); if Y is normalized at ¢ = 0, it srays normalized for all future time. QED Problem 1.4 At time t = 0 a particle is represented by the wave function x : A-, if0<x <a, a Wx O=] gl") gcx<s, (b—a) =*= 0, otherwise, where A, a, and 6 are constants. (a) Normalize © (that is, find A, in terms of a and 5). (b) Sketch ¥(x, 0), as a function of x. (c) Where is the particle most likely to be found, at ¢ = 0? (d) What is the probability of finding the particle to the left of @? Check your result in the limiting cases b = a and b = 2a. (e) What is the expectation value of x? *Problem 1.5 Consider the wave function Wx. t) = Ae Mle ter where A, A, and w are positive real constants. (We'll see in Chapter 2 what potential (V) actually produces such a wave function.) (a) Normalize wv. (b) Determine the expectation values of x and x?. Ra good mathematician can supply you with pathological counterexamples, but they do not arise in physics: for us the wave function afways goes to Zero at infinity. Section 1.5: Momentum 15 (c) Find the standard deviation of x. Sketch the graph of |(?, as a function of x, and mark the points ({x) +o) and ({x) — o), to illustrate the sense in which o represents the “spread” in x, What is the probability that the particle would be found outside this range? 1.55 MOMENTUM For a particle in state Y, the expectation value of x is +00 (x) = | x|Wr. np? dx. (1.28] 08 What exactly does this mean? It emphatically does not mean that if you measure the position of one particle over and over again, f x|%[?dx is the average of the results you’! get. On the contrary: The first measurement (whose outcome is inde- terminate) 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 VY, 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 YW, and measure the positions of all of them: (x) is the average of these results. (1 like to picture a row of bottles on a shelf, each containing a particle in the state W (relative to the center of the bottle). A graduate student with a ruler is assigned to each bottle, and at a signal they all measure the positions of their respective particles. We then construct a histogram of the results, which should match |/[*, and compute the average, which should agree with (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 repeated measurements on an ensem- ble of identically prepared 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 Y), and we might be interested in knowing how fast it moves. Referring to Equations 1.25 and 1.28, we see that!? d{x) | a ih a ~ov aur —- = fx x= — | x—(V*F¥— —- WY) dx. 1.29 dt <9, ax 2m * 9x ax ax * [1.29] BT keep things from getting too cluttered. I'll suppress the limits of integration. 18 Chapter 1 The Wave Function Problem 1.6 Why can’t you do integration-by-parts directly on the middle expres- sion in Equation 1.29—pull the time derivative over onto x, note that 6x/dt = 0, and conclude that d(x)/dt = 0? «Problem 1.7 Calculate d(p)/dt. Answer: dip) -( ay APF (5 13 dt ax (1-38) Equations 1.32 (or the first part of 1.33) and 1.38 are instances of Ehrenfest’s theorem, which tells us that expectation values obey classical laws. Problem 1.8 Suppose you add a constant Vp to the potential energy (by “constant™ I mean independent of x as well as /). In classical mechanics this doesn’t change anything, but what about guantum mechanics? Show that the wave function picks up a time-dependent phase factor: exp(—i Vor/). What effect does this have on the expectation value of a dynamical variable? 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 1.7). If someone asked you “Precisely where is that wave?” you’d probably think he was a little bit nutty: The wave isn’t precisely any where—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 1.8), 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 trade-off here: The more precise a wave’s position is, the less precise is its wavelength, and vice versa.!® A theorem in Fourier analysis makes all this rigorous, but for the moment I am only concerned with the qualitative argument. '6That’s why a piccolo player must be right on pitch. whereas a double-bass player can alford to wear garden gloves. For the piccolo, a sixty-fourth note contains many full cycles. and the frequency (we're working in the time domain now, instead of space) is well defined, whereas for the bass. al a much lower register. the sixty-fourth note contains only a few cycles, and all you hear is a general sort of “oomph,” with no very clear pitch. Section 1.6: The Uncertainty Principle 19 — | 10 20 30 40 50 x (feet) FIGURE 1.7: A wave with a (fairly) well-defined wavelength, but an ill-defined position. A N 10 20 30 40 50 x (feat) FIGURE 1.8: A wave with a (fairly) well-defined position, but an ill-defined wave- length, This applies, of course, to any wave phenomenon, and hence in particular to the quantum mechanical wave function. Now the wavelength of is related to the momentum of the particle by the de Broglie formula:!7 A Qnh = oa, {1.39] a A 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, = Ox0p >= 5: [1.40] N where oy is the standard deviation in x, and op 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 mea- surements, Momentum measurements yield precise answers—the “spread” here refers to the fact that measurements on identically prepared systems do not yield identical results. You can, if you want, construct a state such that repeated posi- tion 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 slate with a reproducible momentum (by making "P11 prove 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 (#/7)(8/ax). Although this is a conceptually cleaner approach, it involves diverting mathematical complications that I would rather save for later. 20 Chapter 1 The Wave Function W 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 inequal- ity, and there’s no limit on how big oy and op can be—just make V some long wiggly line with lots of bumps and potholes and no periodic structure. «Problem 1.9 A particle of mass m is in the state Va.n= AoW alma? Jatin] where A and @ are positive real constants. (a) Find A. (b) For what potential energy function V(x) does W satisfy the Schrédinger equation? {c) Calculate the expectation values of x, x, p, and p?. (d) Find o, and op. Is their product consistent with the uncertainty principle? FURTHER PROBLEMS FOR CHAPTER i Problem 1.10 Consider the first 25 digits in the decimal expansion of m (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 The needle on a broken car speedometer is free to swing, and bounces perfectly off the pins at either end, so that if you give it a flick it is equally likely to come to rest at any angle between 0 and z. {a) What is the probability density, o(@)? Hint: p(@)d@ is the probability that the needle will come to rest between 6 and (9 +40). Graph o(@) as a function of 6, from —x/2 to 32/2. (Of course, part of this interval is excluded, so p is zero there.) Make sure that the total probability is 1. Further Problems for Chapter 1 23 (g) Find the uncertainty in p (cp). ({h) Check that your results are consistent with the uncertainty principle. Problem 1.18 In general, quantum mechanics is relevant when the de Broglie wavelength of the particle in question (#/p) is greater than the characteristic size of the system (d). In thermal equilibrium at (Kelvin) temperature 7, the average kinetic energy of a particle is 2 pe 3 — = 5kaT am 28 (where kg is Boltzmann’s constant), so the typical de Broglie wavelength is a-— [141] ~ 3mkgT , The purpose of this problem is to anticipate 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 free!® electrons in a solid are quantum mechan- ical. Below what temperature are the nuc/ei in a solid quantum mechanical? (Use sodium as a typical case.) Moral: The free electrons in a solid are always quantum mechanical; the nuclei are almost never quantum mechani- ca]. 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 (PV = NkgT) to deduce the interatomic spacing. Answer: T < (1/kg)(h7/3m)9/ P25. Obviously (for the gas to show quantum behavior) we want #7 to be as small as possible. and P as /arge 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? 'Sty a solid the inner electrons are allached 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 latice spacing. This problem pertains to the auer electrons. CHAPTER 2 TIME-INDEPENDENT SCHRODINGER 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 V(x. f) in the first place? We need to solve the Schrédinger equation, aw row -.—-—st VY, 2.1 ar 2m ax? * 2.1 for a specified potential! V(x. 1). In this chapter (and most of this book) I shall assume that V is independent of t. In that case the Schrédinger 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 simple products, YO. =O) en. [2.2] where y (/ower-case) is a function of x alone, and ¢ is a function of tf alone, On its face, this is an absurd restriction, and we cannot hope to get more than a tiny ‘it is tiresome to keep saying “potential energy function.” so most people just call V the “potential.” even though this invites occasional confusion with elecric potential. which is actually potential energy per unit charge. 24 Section 2.1: Stationary States 25 subset of all solutions in this way. But hang on, because the solutions we do obtain 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 av dg Fw & or dt Gxt det? (ordinary derivatives, now), and the Schrédinger equation reads dy dy hy— =-— thy dt 2m dx? gt Vy. Or, dividing through by yg: 2 2 nite? FP idv iy [2.3] gat 2m w dx? Now, the left side is a function of t alone, and the right side is a function of x alone.” 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 ldg in-— = E, i o dt E or I iE dg i oP _ ls 24 dt 7? [2.4] and Ridy "3a dx? +V=E, or ay -— Vw =E. . 2m dx +¥y ¥ [2.5] Separation of variables has turned a partial differential equation into two ordi- nary differential equations (Equations 2.4 and 2.5). The first of these (Equation 2.4) 2Note that this would nor be true if V were a function of 1 as well as x. 28 Chapter 2 Time-Independent Schrédinger Equation is itself a solution. Once we have found the separable solutions, then, we can immediately construct a much more general solution, of the form ~ Wat) = DP envn de ntl, [2.15] n=l It so happens that every solution to the (time-dependent) Schrodinger equation can be written in this form—it is simply a matter of finding the right constants (c}. cz, ...) 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 time-independent Schrédinger equation, you're essentially done; getting from there to the general solution of the time-dependent Schrédinger equation is, in principle, simple and straightforward. A lot has happened in the last four pages, so let me recapitulate, from a somewhat different perspective. Here’s the generic problem: You're given a (time- independent) 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 (time-dependent) Schrédinger equation (Equation 2.1). The strategy® is first to solve the time-independent Schrédinger equation (Equation 2.5); this yields, in general, an infinite set of solutions (W(x), W2(x). W3(*). ...), each with its own associated energy (£, £2, £3,...). To fit Y(x,0) you write down the general linear combination of these solutions: ox VO, 0) = Den Ynlx): [2.16] n=] the miracle is that you can always match the specified initial state by appropriate choice of the constants cj, cz, ¢3, ... . To construct ¥(x,7) you simply tack onto each term its characteristic time dependence, exp(—iE,1/f): oo w~ Bx) = cyte Ee = VP ey Wa (x, 1)- [2.17] n=l t=] The separable solutions themselves, Watt) = Yn lae fBat/h 2.18] SOccasionally you can solve the time-dependent Schrddinger equation without recourse to sep- aration of variables—-see. for instance. Problems 2.49 and 2.50. But such cases are extremely rare. Section 2.1: Stationary States 29 are stationary states, in the sense that all probabilities and expectation values are independent of time, but this property is emphatically mot shared by the general solution (Equation 2.17); the energies are different, for different stationary states, and the exponentials do not cancel, when you calculate |W|*. Example 2.1 Suppose a particle starts out in a linear combination of just fro stationary states: Wx. 0) = 1 v1 (x) + coax). (To keep things simple I'll assume that the constants c, and the states w,,(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: BOL Scie TE + coo re (EH, where E, and E> are the energies associated with yr, and wo. It follows that or NP = Cyne El + ern Yer ne FU" + exyne TP") = fw} + hW2 + 2ereoi Wr cos[( Ez — Ey)t/A). (L used Euler’s formula, exp/@ = cos +i sin, to simplify the result.) Evidently the probability density oscillates sinusoidally, at an angular frequency (E2— £))/fis this is certainly not a stationary state. But notice that it took a linear combination of states (with different energies) to produce motion.’ «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 Eo + iF (with Eq and [ real), and show that if Equation 1.20 is to hold for all t, F must be zero. (b) The time-independent wave function (x) can always be taken to be real (unlike W(x. t), which is necessarily complex), This doesn’t mean that every solution to the time-independent Schrédinger equation is real: what it says is that if you've got one that is nor, it can always be expressed as a linear combination of solutions (with the same energy) that are. So you might as well stick to y’s that are real. Hint: If r(x) satisfies Equation 2.5, for a given E, so too does its complex conjugate, and hence also the real linear combinations (yr + w*) and f(y — ¥*). 7This is nicely illustrated by an applet at the Web site http:/thorin.adne.com/~topquark/ quantum/deepweillmain-html. 30 Chapter 2 Time-Independent Schridinger Equation (c) If V(x) is an even function (that is, V(—x) = V(x)) then (x) can always be taken to be either even or odd. Hint: If (x) satisfies Equation 2.5. for a given E, so too does y(—x), and hence also the even and odd linear combinations w(x) + w(—x). «Problem 2.2 Show that E must exceed the minimum value of V(x), for every normalizable solution to the time-independent Schrédinger equation. What is the classical analog to this statement? Hint: Rewrite Equation 2.5 in the form ay 2m qe = pvo — Ely: if E < Vmin. then y and its second derivative always have the same sign —argue that such a function cannot be normalized. 2.2: THE INFINITE SQUARE WELL Suppose 0 ifO0<x<a, oo, otherwise [2.19] V(x) =| (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 artifi- cial, 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 accessi- ble test case for all the fancy machinery that comes later. We’ll refer back to it frequently.) V(x) FIGURE 2.1; The infinite square well poten- a x tial (Equation 2.19). Section 2.2: The Infinite Square Well 33 2. As you go up in energy, each successive state has one more node (zero- crossing): yf, has none (the end points don’t count), w2 has one, #3 has two, and so on. 3. They are mutually orthogonal, in the sense that | Vin)" Wn (x) dx = 0, [2.29] whenever m 4 n. Proof: 2 a | Vn 2) Wale) dx = = [ sin(™ x) sin(= 2) ax a Jo a a : {| (“— ) Ce )] =- cos mx | — cos mx )| dx a Jo a a 1 _ f{m—n 1 _ [min = sin ax | — sin WX (m —n)x a (m+n) a i {ee —n)r] _ sin((m aa =0 (m —n) (m+n) —" a 0 at 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:!° | Von Yen (x) dx = bins [2.30] where 5, (the so-called Kronecker delta) is defined in the usual way, 0, if : om = { § nme (2.31) We say that the y’s are orthonormal. 4. They are complete, in the sense that any ofher function, f(x), can be expressed as a linear combination of them: ~ 2— fae f@= » CrWn(X) = a » Cy SID (=) : [2.32] n=l n=l '0Tn this case the sare real, so the * on ys, is unnecessary. but for future. purposes it’s a good idea to get in the habit of putting it there. 34 Chapter 2 Time-Independent Schrodinger Equation I’m not about to prove the completeness of the functions sin (n7.x/a), but if you’ve studied advanced calculus you will recognize that Equation 2.32 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. '! The coefficients c, can be evaluated—for a given f(+)—by a method I call Fourier’s trick, which beautifully exploits the orthonormality of {y,}: Multiply both sides of Equation 2.32 by p»(x)*, and integrate. oo oo J mca 200 dx = Yen f Yin)" Wn(0d dx = enBm = ome (233) n=1 n=l (Notice how the Kronecker delta kills every term in the sum except the one for which 2 = mm.) Thus the nth coefficient in the expansion of f(x) is!” Ch = | Vin (x)* f(x) dx. {2.34] 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.!> Orthog- onality is also quite general—I'Il 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 evidently wep) = [2sin (=) eer h/2na? yt [2.35] a a I claimed (Equation 2.17) that the most general solution to the (time-dependent) Schrédinger equation is a linear combination of stationary states: OS bon = on pz sin (“2 x) elon name, (2.36) n=l "See, for example, Mary Boas, Mathematical Methods in the Physical Sciences, 2d ed. (New York: John Wiley, 1983), p. 313: (+) can even have a finite number of finite discontinuities. '21) doesn't matter whether you use m or # as the “dummy index” here (as long as you are consistent on the two sides of the equation. of course); whatever letter you usc. it just stands for “any positive integer." see. for example, John L, Powell and Bernd Crasemann, Quantin Mechanics (Addison- Wesley, Reading, MA, 1961), p. 126. Section 2.2: The Infinite Square Well 35 (If you doubt that this is a solution, by all means check it!) It remains only for me to demonstrate that J can fit any prescribed initial wave function, U(x. 0), by appropriate choice of the coefficients c,: x W(x, 0) = 0 cuba). n=l The completeness of the y’s (confirmed in this case by Dirichlet’s theorem) guar- antees that I can always express V(x.0) in this way, and their orthonormality licenses the use of Fourier’s trick to determine the actual coefficients: 2 a Cn =\f— [ sin (=:) W(x, 0) dx. (2.37] ad Jo a That does it: Given the initial wave function, ¥(x.0), we first compute the expansion coefficients c,, using Equation 2.37, and then plug these into Equation 2.36 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 y's and the equation for the allowed energies. Example 2.2 A particle in the infinite square well has the initial wave function W(x.0) = Ax(a—x). (052 <a). for some constant A (see Figure 2.3). Outside the well, of course, ¥ = 0. Find Wox.t). W(x, 0) a x FIGURE 2.3: The starting wave function in Example 2.2. 38 Chapter 2 Time-Independent Schridinger Equation Example 2.3. In Example 2.2 the starting wave function (Figure 2.3) closely re- sembles the ground state y (Figure 2.2). This suggests that |c, |? should dominate, and in fact - la? = (2) = 0.998555... The rest of the coefficients make up the difference:!4 Sia? = (xe) y Jen. n=l 4=1.35.... The expectation value of the energy, in this example, is 2 co (33) wm 480n2> SSH? (H) = aero = . » 2ma? x4ma2 » n4 ma? n=1.3.5..., n=),3.5... n3z3 As one might expect, it is very close to Ey = 12h? /2ma?—slightly larger, because of the admixture of excited states. Problem 2.3 Show that there is no acceptable solution to the (time-independent) Schrédinger 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 Schrédinger equation, and showing that you cannot meet the boundary conditions.) «Problem FA catcutate (x), Or), (p), (p?), og, and @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? «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: Wx, 0) = Aly) + Wax)]. '4You can look up the series Iii _ 7 16 + 36 + 56 t= agg and 14d we pita tat 96 in math tables. under “Sums of Reciprocal Powers” or “Riemann Zeta Function.” Section 2.2: The Infinite Square Well 39 (a) Normalize (x, 0). (That is, find A. This is very easy, if you exploit the orthonormality of 4 and yz. Recall that, having normalized Y at ¢ = 0, you can rest assured that it stays normalized—if you doubt this, check it explicitly after doing part (b).) (b) Find (x, 1) and [W(x, nl. Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let # = 77A/2ma?. (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 £) and E2? 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 | and wW2 in Problem 2.5: W(x. 0) = Alyn (x) + ead], where @ is some constant. Find W(x, t), oP, and (x), and compare your results with what you got before. Study the special cases @ = w/2 and @ = 7. (For a graphical exploration of this problem see the applet in footnote 7.) *Problem 2.7 A particle in the infinite square well has the initial wave function!> Ax, O<x <a/2, A(a—x). a/2 5x <a. W(x,0) = | (a) Sketch (x, 0), and determine the constant A. (b) Find W(x. 1). 'SThere is no restriction in principle on the shape of the starting wave function. as long as it is normalizable. In particular, W(x. 0) need not have a continuous derivalive—in fact, it doesn’t even have to be a continuous function. However. if you wy lo calculate (H) using f ¥v.0)* AWC, 0) ax 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 2.48 you'll see how to manage cases like Problem 2.7. 40 Chapter 2. Time-Independent Schrédinger Equation (c) What is the probability that a measurement of the energy would yield the value E)? (d) Find the expectation value of the energy. Problem 2.8 A particle of mass m in the infinite square well (of width a) starts out in the left half of the well, and is (at ¢ = 0) equally likely to be found at any point in that region. (a) What is its initial wave function, (x, 0)? (Assume it is real. Don’t forget to normalize it.) (b) What is the probability that a measurement of the energy would yield the value wh? /2ma?? Problem 2.9 For the wave function in Example 2.2, find the expectation value of Hi, at time r = 0, the “old fashioned” way: (H) = f ve. 0" AVC, 0) de. Compare the result obtained in Example 2.3. Note: Because (#) 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 #1 attached to a spring of force constant k. The motion is governed by Hooke’s law, 2 Fo=-kx =m—, m ape (ignoring friction), and the solution is x(t) = A sin(wr) + Bcos(wrt), where w= /* [2.41] AL is the (angular) frequency of oscillation. The potential energy is Vix) = Sexy [2.42] ils graph is a parabola. Section 2.3: The Harmonic Oscillator 43 As anticipated, there’s an extra term, involving (xp — pv). We call this the com- mutator of x and p; it is a measure of how badly they fai! to commute. In general, the commutator of operators A and B (written with square brackets) is [A, B] = AB — BA. (2.48) In this notation, 1 ~ 2hinw We need to figure out the commutator of x and p. 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: d_ay [p> + (mox)] - ity, P). [2.49] 2h . yf Ad hd, .)_hf df df _; ; [x. p/f@) = [7 a Eon =F (Z x as f) = ih f(x). [2.50] Dropping the test function, which has served its purpose, [x. p]) = ih. [2.51] This lovely and ubiquitous result is known as the canonical commutation rela- tion.'8 With this, Equation 2.49 becomes 1 1 a4 = —H+-. 2.52 aa = FH +5 [2.52] or I H=hw (oa, - *) : [2.53] Evidently the Hamiltonian does not factor perfectly—there’s that extra —1/2 on the tight. Notice that the ordering of a; and a_ is important here; the same argument. with a4 on the left, yields aQya_ = —H--. [2.54] In particular, [a-, a4] = 1. [2.55] '81n 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 p = (h/i)d/dx. 44 Chapter 2 Time-Independent Schrédinger Equation So the Hamiltonian can equally well be written 1 H=hw (c0- + 3) . [2.56] In terms of a4, then, the Schrédinger equation’? for the harmonic oscillator takes the form hw (2a £ >) w=Ey [2.57] (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 w satisfies the Schrodinger equation with energy E, (that is: Hw = Ew), then ayw satisfies the Schrodinger equation with energy (E + hw): Hazy) = (E + hw)(azy). Proof: Hap) = he (ca. + 4) (ath) =hw (cxaay + sas v =hoay («0 + ;) waa, [re (ea. +14 ;) | = a4(H + how) = as(E + holy = (E + ho)(asw). (I used Equation 2.55 to replace a_a, by a,a_ + 1, in the second line. Notice that whereas the ordering of a; and a_ does matter, the ordering of az and any constants—such as fh, w, and E—does not, an operator commutes with any constant.) By the same token, a_y is a solution with energy (E — hw): Hla_y) = hw (ca = 5) (ad_y) = hiwa_ (coe = s) y =a_ [ne («0 -1- :) ¥] =a_(H —hw)w =a_(E — hoy = (E —ho)(a_y). Here, then, is a wonderful machine for generating new solutions, with higher and lower energies—if we could just find one solution, to get started! We call a4 ladder operators, because they allow us to climb up and down in energy; a+ is the raising operator, and a_ the lowering operator. The “ladder” of states is illustrated in Figure 2.5. '9i'm getting tired of writing “time-independent Schriidinger equation,” so when it’s clear from the context which one I mean, 1'll just call it the “Schrddinger equation.” Section 2.3: The Harmonic Oscillator 45 FIGURE 2.5: The “ladder” of states for the harmonic oscillator. But wait! What if I apply the lowering operator repeatedly? Eventually I’m going to reach a state with energy less than zero, which (according to the general theorem in Problem 2.2) does not exist! At some point the machine must fail. How can that happen? We know that a_y is a new solution to the Schrédinger equation, but there is no guarantee that it will be normalizable —it might be zero, or its square-integral might be infinite. In practice it is the former: There occurs a “lowest rung” (call it Wo) such that a_Wo = 0. [2.58] We can use this to determine W(x): 1 d ——. [ih + mowx =0, J2hima ( dx ) Yo 48 Chapter 2. Time-Independent Schrédinger Equation and integration by parts takes f f*(dg/dx)dx to — f(df/dx)*g dx (the boundary terms vanish, for the reason indicated in footnote 22), so oO 1 OS d * a * = tA (= *odx [os (atg)dx mal. lenz + mo) f] gdx [en gdx, QED In particular, 20 OG | (az Wn) * (ag Wn) dx = | (apa 4 Yn) Un dx, 90 oc But (invoking Equations 2.57 and 2.61) a4d-Wn =n, d-d4 Yn = (H+ Dn. [2.65] so x ~ ox [ (44-00) (a4 bn) dx = lon f (nai dx = (a + vf Wal? dx. oO —90 ied ow > ow 3 ow [ (Vu) "(a Yn) dx = Mey? | Wn-Pax =n | Ini? ax. oO oO 00 But since wy, and w, +1 are normalized, it follows that len? = n+ and |d,{? =n, and hence abn = Vat las, avn = Va Wnt {2.66] Thus Vis avo. We= ean = any 1 = 440. 2S TM = get 0 a 3 = tay, = —L La w= wow = wa vo. Wa= yar = yaaa” Wo. and so on. Clearly 1 n Vn = ya Yo. (2.67] which is to say that the normalization factor in Equation 2.61 is Ay = 1//n! (in particular, A; = 1, confirming our result in Example 2.4). Section 2.3: The Harmonic Oscillator 49 As in the case of the infinite square well, the stationary states of the harmonic oscillator are orthogonal: x / Vn Wn dX = Sina [2.68] 9 This can be proved using Equation 2.65, and Equation 2.64 twice—first moving a, and then moving a_: ~ fo. $] J vitesartnde =n f virtnas —oX —C 2% 00 = | (a_Vin)"(a_-vn) dx = | (ay.a_ Vin Wn dx - oC oD w~ =m | Wein dx. oO Unless m = n, then, f YA, dx must be zero. Orthonormality means that we can again use Fourier’s trick (Equation 2.34) to evaluate the coefficients, when we expand W(x, 0) as a linear combination of stationary states (Equation 2.16), and \cn|? is again the probability that a measurement of the energy would yield the value E,. Example 2.5 Find the expectation value of the potential energy in the nth state of the harmonic oscillator. Solution: 1 2.2 ] » [* «2 (V)= gna’ = yer Wy xo Wn dx. —x There’s a beautiful device for evaluating integrals of this kind (involving powers of x or p): Use the definition (Equation 2.47) to express x and p in terms of the raising and lowering operators: ih ih x= f——(apta_): p=i re ag =a). [2.69] 2inw 2 2; In this example we are interested in x h ea [tox + (a@pa_) + (aay) + a)? . 2mw So (V)= “ | Vi [Cay + aya) + (aay) + (-Y'] Yn de. 50 Chapter 2 Time-Independent Schrédinger Equation But (a+)?¥, is (apart from normalization) ¥n42, which is orthogonal to y,, and the same goes for (a_)?W,_, which is proportional to W,—2. So those terms drop out, and we can use Equation 2.65 to evaluate the remaining two: 1 1 V)= ein tath)= ghe (> + 7) : As it happens, the expectation value of the potential energy is exactly falf the total (the other half, of course, is kinetic). This is a peculiarity of the harmonic oscillator, as we'll see later on. «Problem 2.10 (a) Construct yoCr). (b) Sketch Wo. wi. and yo. (c) Check the orthogonality of wo, yf), and wo, by explicit integration. Hint: If you exploit the even-ness and odd-ness of the functions, there is really only one integral left to do. «Problem 2.11 (a) Compute (x), (p). (x7), and (p*), for the states Wo (Equation 2.59) and vy (Equation 2.62), by explicit integration. Comment: In this and other problems involving the harmonic oscillator it simplifies matters if you introduce the variable £ = /m@/hx and the constant @ = (mw/mh)!/4. (b) Check the uncertainty principle for these states. (c) Compute (7) (the average kinetic energy) and (V) (the average potential energy) for these states. (No new integration allowed!) Is their sum what you would expect? xProblem 2.12 Find (x), (p), (x2), (p?), and (T), for the ath stationary state of the harmonic oscillator, using the method of Example 2.5. Check that the uncertainty principle is satisfied. Problem 2.13 A particle in the harmonic oscillator potential starts out in the state W(x, 0) = ALB ox) + 4410). (a) Find A. (b) Construct W(x, 1) and |W.) /?. Section 2.3: The Harmonic Oscillator 53 Putting these into Equation 2.78, we find oO DOU + DU + Dajss — 2jaj + (K — Daj]éf =O. [2.80} J=0 It follows (from the uniqueness of power series expansions”>) that the coefficient of each power of — must vanish, G+ DG +2)aj42 —2jaj + (K — Daj =0, and hence that | @j+1-K) G+)G +27 This recursion formula is entirely equivalent to the Schrédinger equation. Starting with ag, it generates all the even-numbered coefficients: aja = [2.81] _d-—&) _~ 6-8), _ G-K)d—«) aQ= 2 a0. a= 12 I= 24 dg. and starting with «, it generates the odd coefficients: a= G-K), a =U), -7- 67K), eG 9g 120 ' We write the complete solution as h(E) = Reven(E) + hoaaE), [2.82] where Reven(E) = ay + ark? + ag +++ is an even function of &, built on ao, and Noga) = arg + aa8? + a58° + --- is an odd function. built on a;. Thus Equation 2.81 determines A(&) in terms of two arbitrary constants (aig and a,;)—which is just what we would expect, for a second-order differential equation. However. not all the solutions so obtained are normalizable. For at very large J, the recursion formula becomes (approximately) ajq2 yu 23 See, for example. Arfken (footnote 24). Section 5.7. 54 Chapter 2 Time-Independent Schrédinger Equation with the (approximate) solution a,x 1 G2 for some constant C, and this yields (at large §, where the higher powers dominate) 1 ; 1 >; 2 new cy’ el wc Yo eY » cel Gi/ Now, if # goes like exp(£?), then y (remember y/?—that’s what we're trying to calculate) goes like exp(£?/2) (Equation 2.77), which is precisely the asymptotic behavior we didn't want.2° There is only one way to wiggle out of this: For nornalizable solutions the power series must terminate. There must occur some “highest” j (call it 7), such that the recursion formula spits out a,42 = 0 (this will truncate either the series heyen or the series toga; the other one must be zero from the start: a} = 0 if is even, and ag = 0 if » is odd). For physically acceptable solutions, then, Equation 2.81 requires that K=2n+1, for some non-negative integer 7, which is to say (referring to Equation 2.73) that the energy must be E, = (: + ) hw. form =0.1,2..... [2.83] Thus we recover, by a completely different method, the fundamental quantization condition we found algebraically in Equation 2.61. It seems at first rather surprising that the quantization of energy should emerge from a technical detail in the power series solution to the Schrédinger equation, but let’s look at it from a different perspective. Equation 2.70 has solutions, of course, for any value of E (in fact, it has nvo linearly independent solutions for every £). But almost all of these solutions blow up exponentially at large x, and hence are not normalizable. Imagine, for example, using an E that is slightly /ess than one of the allowed values (say, 0.49/w), and plotting the solution (Figure 2.6(a)): the “tails” fly off to infinity. Now try an E slightly larger (say, 0.51fiw); the “tails” now blow up in the orher direction (Figure 2.6(b)). As you tweak the parameter in tiny increments from 0.49 to 0.51, the tails flip over when you pass through 0.5—only at precisely 0.5 do the tails go to zero, leaving a normalizable solution.?? 2611's no surprise that the ill-behaved solutions are still contained in Equation 2.81: this recursion relation is equivalent to the Schrédinger equation, so it’s gor to include both the asymptotic forms we found in Equation 2.75. °71 is possible to set this up on a computer. and discover the allowed cnergies “experimentally.” You might call it the wag the dog method: When the tail wags. you know you've just passed over an allowed value. See Problems 2.54-2.56. Section 2.3: The Harmonic Oscillator 55 4 wn T {b) FIGURE 2.6: Solutions to the Schrédinger equation for (a) E = 0.49 ha, and (b) E= 0.51 fa. For the allowed values of K, the recursion formula reads —2@n- SF) G+ng+D (2.84) aj42= If n = 0, there is only one term in the series (we must pick a) = 0 to kill Aoag. and j = 0 in Equation 2.84 yields az = 0): ho(&) = a0. and hence Wolk) = age F? 58 Chapter 2 Time-Independent Schrédinger Equation Iviog(nll? 0.24 : @) ef “a el FIGURE 2.7: (a) The first four stationary states of the harmonic oscillator. This material is used by permission of John Wiley & Sons, Inc.; Stephen Gasiorowicz, Quantum Physics, John Wiley & Sons, Inc., 1974. (b) Graph of IFio0l?, with the classical distribution (dashed curve) superimposed. Section 2.4: The Free Particle 59 (c) If you differentiate an nth-order polynomial, you get a polynomial of order (n — 1). For the Hermite polynomials, in fact, dH, _ = 2n Hy —1 (8). [2.88] Check this, by differentiating Hs and Hg. (d) H,,(&) is the nth z-derivative, at z = 0, of the generating function exp(—z* + 2zé); or, to put it another way, it is the coefficient of z”/m! in the Taylor series expansion for this function: co -2 on ott SE He, (2.89) n=0 Use this to rederive Ho, A, and Hp. 2.4 THE FREE PARTICLE We turn next to what should have been the simplest case of all: the free particle (V(x) = 0 everywhere). Classically this would just mean motion at constant veloc- ity, but in quantum mechanics the problem is surprisingly subtle and tricky. The time-independent Schrédinger equation reads wd? oS = Ey. [2.90] or 2 ¢ v =—ky. where k= ~ ~ [2.91] x7 Yt So far, it's the same as inside the infinite square well (Equation 2.21), where the potential is also zero; this time, however, I prefer to write the general solution in exponential form (instead of sines and cosines), for reasons that will appear in due course: w(x) = Ae™ 4+ Be, [2.92] Unlike the infinite square well, there are no boundary conditions to restrict the possible values of k (and hence of E); the free particle can carry any (positive) energy. Tacking on the standard time dependence, exp(—i Er/h), W(x. 2) = Agiks— FE) + Begikst Hn | [2.93] Now, any function of x and 7 that depends on these variables in the special combination (x + ur) (for some constant v) represents a wave of fixed profile, traveling in the +x-direction, at speed v. A fixed point on the waveform (for 60 Chapter 2 Time-Independent Schrodinger Equation example, a maximum or a minimum) corresponds to a fixed value of the argument, and hence to x and ¢ such that x tut =constant. or x = -vt + constant. Since every point on the waveform is moving along with the same velocity, its shape doesn’t change as it propagates. Thus the first term in Equation 2.93 repre- sents a wave traveling to the right, and the second represents a wave (of the same energy) going to the /eft. By the way, since they only differ by the sign in front of k, we might as well write Wee.) = Ae ete), [2.94] and let & run negative to cover the case of waves traveling to the left: ret VamE with k>0O=- traveling to the right, 2.95] ~R k <0 traveling to the left. , Evidently the “stationary states” of the free particle are propagating waves; their wavelength is A = 27/|k|, and, according to the de Broglie formula (Equation 1.39), they carry momentum pahk. {2.96] The speed of these waves (the coefficient of t over the coefficient of x) is Alk| [LE Uquantum = om Va’ [2.97] On the other hand, the classical speed of a free particle with energy E is given by E= (1/2)mv? (pure kinetic, since V = 0), so 2E Uclassical = hn = 2vquantum: [2.98] Apparently the quantum mechanical wave function travels at half the speed of the particle it is supposed to represent! We'll return to this paradox in a moment—there is an even more serious problern we need to confront first: This wave function is not normalizable. For +00 +00 | WED, dx = lar [ dx = |A[*(oo). [2.99] oO -—o In the case of the free particle, then, the separable solutions do not represent physically realizable states. A free particle cannot exist in a stationary state; or, to put it another way, there is no such thing as a free particle with a definite energy. Section 2.4: The Free Particle 63 hex, tl? 05 0.4 0.3 0. 1 _/ ~. -6 -4 -2 0 2 4 6 [>< FIGURE 2.8: Graph of | (x, £)|? (Equation 2.104) at ¢ = 0 (the rectangle) and at t = ma? /h (the curve). W(x, 0) ok) Vea alr ~ala x k (a) (b) FIGURE 2.9: Example 2.6, for small a. (a) Graph of &(x, 0). (b) Graph of #{4). it’s flat, since the k’s cancelled out (Figure 2.9(b)). This is an example of the uncertainty principle: If the spread in position is small, the spread in momentum (and hence in k—see Equation 2.96) must be large. At the other extreme (large a) the spread in position is broad (Figure 2.10(a)) and _ fa sin(ka) Now, sinz/z has its maximum at z = 0, and drops to zero at z = + m (which, in this context, means k = + 2/a). So for large a, #(k) is a sharp spike about k = 0 (Figure 2.10(b)). This time it’s got a well-defined momentum but an ill-defined position. 64 Chapter 2 Time-Independent Schrédinger Equation P(x, 0) o(k) Vain 1 Vea x _ x ® k a a {a) (b) FIGURE 2.10: Example 2.6, for large a. {a} Graph of (x, 0). (b) Graph of (k). T return now to the paradox noted earlier: the fact that the separable solution W(x, t) in Equation 2.94 travels at the “wrong” speed for the particle it osten- sibly represents. Strictly speaking, the problem evaporated when we discovered that % is not a physically realizable state. Nevertheless, it is of interest to dis- cover how information about velocity is contained in the free particle wave function (Equation 2,100). The essential idea is this: A wave packet is a superposition of sinusoidal functions whose amplitude is modulated by @ (Figure 2.11); it consists of “ripples” contained within an “envelope.” What corresponds to the particle velocity is not the speed of the individual ripples (the so-called phase velocity), but rather the speed of the envelope (the group velocity)—which, depending on the nature of the waves, can be greater than, less than, or equal to, the velocity of the ripples that go to make it up. For waves on a string, the group velocity is the same as the phase velocity. For water waves it is one-half the phase velocity, as you may have noticed when you toss a rock into a pond (if you concentrate on a particular ripple, you will see it build up from the rear, move forward through the group, and fade away at the front, while the group as a whole propagates out at half the speed). What I need to show is that for the wave function of a free particle in quantum mechanics FIGURE 2.11: A wave packet. The “enve- lope” travels at the group velocity; the “rip- ples” travel at the phase velocity. Section 2.4: The Free Particle 65 the group velocity is Avice the phase velocity—just right to represent the classical particle speed. The problem, then, is to determine the group velocity of a wave packet with the general form i te0 i(kx wt) W(x, = sa! GeO ag, Fo dx (In our case w = (Ak?/2m), but what I have to say now applies to any kind of wave packet, regardless of its dispersion relation—the formula for w as a function of k.) Let us assume that ¢(k) is narrowly peaked about some particular value ko. (There is nothing il/egal about a broad spread in k, but such wave packets change shape rapidly—since different components travel at different speeds—so the whole notion of a “group,” with a well-defined velocity, loses its meaning.) Since the integrand is negligible except in the vicinity of ko, we may as well Taylor-expand the function «(k) about that point, and keep only the leading terms: wk) = wo + wo (k — ko). where wis the derivative of w with respect to k, at the point kg. Changing variables from k to s = k — kp (to center the integral at ko), we have 1 ote (lobar : Wor.) = Te fe olka + spell ota —tonreostl ge Atr=0, 1 ote kote W(x. 0) = Te fe (ko + shot ds, and at later times 1. 1 +90 se , W(x. t) = ciency f oo + sjelkotsir—eyt) ds. V2n 00 Except for the shift from x to (x — wot), the integral is the same as the one in W(x, 0). Thus W(x. 1) & Hero (x — ashe. 0). [2.105] Apart from the phase factor in front (which won’t affect (|? in any event) the wave packet evidently moves along at a speed wp: dw Usgroup = dk [2.106] (evaluated at k = ko). This is to be contrasted with the ordinary phase velocity wo Uphase = ke [2.107] 68 Chapter 2 Time-Independent Schrédinger Equation 2.5 THE DELTA-FUNCTION POTENTIAL 2.5.1 Bound States and Scattering States We have encountered two very different kinds of solutions to the time-independent Schrédinger equation: For the infinite square well and the harmonic oscillator they are normalizable, and labeled by a discrete index n; for the free particle they are non-normalizable, and labeled by a continuous variable k. The former represent physically realizable states in their own right, the latter do not; but in both cases the general solution to the time-dependent Schrédinger equation is a linear combination of stationary states—for the first type this combination takes the form of a sum (over »), whereas for the second it is an integral (over k). What is the physical significance of this distinction? In classical mechanics a one-dimensional time-independent potential can give tise to two rather different kinds of motion. If V(x) rises higher than the particle’s total energy (£) on either side (Figure 2.]2(a)), then the particle is “stuck” in the potential well—it rocks back and forth between the turning points, but it cannot escape (unless, of course, you provide it with a source of extra energy, such as a motor, but we’re not talking about that). We call this a bound state. If, on the other hand, E exceeds V(x) on one side (or both), then the particle comes in from “infinity,” slows down or speeds up under the influence of the potential, and returns to infinity (Figure 2.12(b)). (It can’t get trapped in the potential unless there is some mechanism, such as friction, to dissipate energy, but again, we're not talking about that.) We call this a scattering state. Some potentials admit only bound states (for instance, the harmonic oscillator); some allow only scattering states (a potential hill with no dips in it, for example); some permit both kinds, depending on the energy of the particle. The two kinds of solutions to the Schrédinger equation correspond precisely to bound and scattering states. The distinction is even cleaner in the quantum domain, because the phenomenon of tunneling (which we'll come to shortly) allows the particle to “leak” through any finite potential barrier, so the only thing that matters is the potential at infinity (Figure 2.12(c)): E>([V(-oo) or V(+oo)}=> © scattering state. [2.109] | E <[V(—oo) and V(+00)] > bound state, In “real life” most potentials go to zero at infinity, in which case the criterion simplifies even further: E<0Q= bound state, | E>0O= scattering state. 2.110] Because the infinite square well and harmonic oscillator potentials go to infinity as x — oo, they admit bound states only; because the free particle potential is zero Section 2.5: The Delta-Function Potential 69 Classical turning points (a) Vix) E 4-H ee | ~ x x Classical turning point (b) (c) FIGURE 2.12: (a) A bound state. (b) Scattering states. (c) A classical bound state, but a quantum scattering state. everywhere, it only allows scattering states.*4 In this section (and the following one) we shall explore potentials that give rise to both kinds of states. Mig you are irtitalingly observant. you may have noticed that the general theorem requiring E > Vmin (Problem 2.2) does not really apply to scattering states. since they are not normalizable anyway. If this bothers you, wy solving the Schrédinger equation with E < 0. for the free particle, and 70 Chapter 2. Time-Independent Schrédinger Equation 8{x) FIGURE 2.13: The Dirac delta function x (Equation 2.111). 2.5.2 The Delta-Function Well The Dirac delta function is an infinitely high, infinitesimally narrow spike at the origin, whose area is | (Figure 2.13): ,_f 0 ifx 40 . too —_ w= | oo. ify £0 \. with [. 8(x) dx =1. [2.111] Technically, it isn’t a function at all, since it is not finite at x = 0 (mathematicians call it a generalized function, or distribution).*> Nevertheless, it is an extremely useful construct in theoretical physics. (For example, in electrodynamics the charge density of a point charge is a delta function.) Notice that (x —a) would be a spike of area | at the point a. If you multiply 8(x — a) by an ordinary function f(x), it’s the same as multiplying by f(a), FQRIMA— a) = f(MS — a), [2.112] because the product is zero anyway except at the point a. In particular, +00 +90 / f(x)bx — a) dx = f(a) | b(x —a)dx = f(a). (2.113] oo That’s the most important property of the delta function: Under the integral sign it serves to “pick out” the value of f(x) at the point a. (Of course, the integral need not go from —oo to +00; all that matters is that the domain of integration include the point a, soa —e€ toa +e would do, for any € > 0.) Let’s consider a potential of the form V(x) = —a@8(x). [2.114] note that even linear combinations of these solutions cannot be normalized. The positive energy solutions by themselves constitute a complete set. 33The delta function can be thought of as the init of a sequence of functions, such as reclangles (or triangles) of ever-increasing height and ever-decreasing width. Section 2.5: The Delta-Function Potential 73 and the allowed energy (Equation 2.117) is a 2 2 pa _ _ me [2.127] 2m 2h ° Finally, we normalize y: +90 5 ot |B? / I(x)? dx = aie f eX dy = =], —90 0 K so (choosing, for convenience, the positive real root): fma fh B=Je= . [2.128] Evidently the delta-function well, regardless of its “strength” a, has exactly one bound state: [2.129] Wo) = ER ermal, B= ma’ , on What about scattering states, with E > 0? For x < 0 the Schrédinger equation reads 2 ss =- ~ v= ky. where k= Vine [2.130] is real and positive. The general solution is v(x) = Ae 4 Bee [2.131] and this time we cannot rule out either term, since neither of them blows up. Similarly, for x > 0, ; ; w(x) = Fe + Ge7*, [2.132] The continuity of w(x) at x = 0 requires that F+G=A+B. [2.133] The derivatives are dw/dx =ik(Fe'™ — Ge). for (x > 0), so dy/dx|, =ik(F - G). diy/dx =ik (Ae — Be) for (x <0). so dyy/dx|_ =ik(A— B). 74 Chapter 2 Time-Independent Schridinger Equation and hence A(dys/dx) = ik(F —-G—A+ B). Meanwhile, ¥(0) = (A + B), so the second boundary condition (Equation 2,125) says 2ma ik(F -G-A+B)=-—, RP (A+B), [2.134] or, more compactly, F-—-G=A(i+2i8) — BU —2ip). where 6 = x [2.135] i Having imposed both boundary conditions, we are left with two equations (Equations 2,133 and 2.135) in four unknowns (A, B, F, and G)—five, if you count k. Normalization won’t help—this isn’t a normalizable state. Perhaps we’d better pause, then, and examine the physical significance of these various con- stants. Recall that exp(/kx) gives rise (when coupled with the time-dependent factor exp(—/Er/h)) to a wave function propagating to the right, and exp(—ikx) leads to a wave propagating to the /eft. It follows that A (in Equation 2.131) is the amplitude of a wave coming in from the left, B is the amplitude of a wave return- ing to the left, F (Equation 2.132) is the amplitude of a wave traveling off to the right, and G is the amplitude of a wave coming in from the right (see Figure 2.15), In a typical scattering experiment particles are fired in from one direction—let’s say, from the left. In that case the amplitude of the wave coming in from the right will be zero: G=0. (for scattering from the left); (2.136] A is the amplitude of the incident wave, B is the amplitude of the reflected wave, and F is the amplitude of the transmitted wave. Solving Equations 2.133 and 2.135 for B and F, we find ip i = . F= . i—ip4 1-ip* (If you want to study scattering from the right, set A = 0; then G is the incident amplitude, F is the refiected amplitude, and B is the transmitted amplitude.) B [2.137] FIGURE 2.15: Scattering from a delta func- tion well. Section 2.5: The Delta-Function Potential 75 Now, the probability of finding the particle at a specified location is given by |'W|?, so the relative?” probability that an incident particle will be reflected back is _ BPP “JAP 1+ 87" [2.138] R is called the reflection coefficient. (If you have a beam of particles, it tells you the fraction of the incoming number that will bounce back.) Meanwhile, the probability of transmission is given by the transmission coefficient \FP 1 T= =—. 2.139 Az 1+ 8? C ] Of course, the sum of these probabilities should be 1—and it is: RiT=1. (2.140] Notice that R and T are functions of 8, and hence (Equations 2.130 and 2.135) of E: 1 1 = —____, T= —_____. 2.141 14+ Qh E/me?) 1+ Gna? /2h7E) [ ] The higher the energy, the greater the probability of transmission (which certainly seems reasonable). This is all very tidy, but there is a sticky matter of principle that we cannot altogether ignore: These scattering wave functions are not normalizable, so they don’t actually represent possible particle states. But we know what the resolution to this problem is: We must form normalizable linear combinations of the stationary states, just as we did for the free particle—true physical particles are represented by the resulting wave packets. Though straightforward in principle, this is a messy business in practice, and at this point it is best to turn the problem over to a computer.*® Meanwhile, since it is impossible to create a normalizable free-particle wave function without involving a range of energies, R and T should be interpreted as the approximate reflection and transmission probabilities for particles in the vicinity of E. Incidentally, it might strike you as peculiar that we were able to analyze a quintessentially time-dependent problem (particle comes in, scatters off a potential, 37 This is not a normalizable wave function. so the absolute probability of finding the particle al a particular location is not well defined: nevertheless. the ratio of probabilities for the incident and reflected waves is meaningful. More on this in the next paragraph. 38Numerical studies of wave packets scattering off welis and barriers reveal extraordinarily rich structure. The classic analysis is A. Goldberg, H. M. Schey. and J. L. Schwartz, Amn. J. Phys. 35, 177 (1967); more recent work can be found on the Web, 78 Chapter 2. Time-Independent Schrédinger Equation (a) Sketch this potential. (b) How many bound states does it possess? Find the allowed energies, for @ = h? /ma and for a = fh? /4ma, and sketch the wave functions. «Problem 2.28 Find the transmission coefficient for the potential in Problem 2.27. 2.6 THE FINITE SQUARE WELL As a last example, consider the finite square well potential -VW, for -a<x <a, 0, for |x| > a, [2.145] VaX)= | where Vo is a (positive) constant (Figure 2.17). Like the delta-function well, this potential admits both bound states (with E < 0) and scattering states (with E > 0). We'll look first at the bound states. In the region x < —a the potential is zero, so the Schrédinger equation reads Pey ay 4 —~—-—_= Ey, HK. Im dx PY OT Gy He where k= vine [2.146] is real and positive. The general solution is w(x) = A exp(—kx) + Bexp(kx), but the first term blows up (as x — —oo), so the physically admissible solution (as before—see Equation 2.119) is w(x) = Be**, forx <a. [2.147] Vix) —a a x -V, FIGURE 2.17: The finite square well {Equation 2.145). Section 2.6: The Finite Square Well 79 In the region —a@ < x < a, V(x) = —Vp, and the Schrédinger equation reads he Py ey 5 Sa da? —Vow=Ey, or de = ly. where Vamn(E + Vi l= ae [2.148] ft Although E is negative, for bound states, it must be greater than —Vo, by the old theorem E > Vmin (Problem 2.2); so / is also real and positive. The general solution is*? w(x) =Csin(lx) + Deos(ix), for -a<x <a, [2.149] where C and D are arbitrary constants. Finally, in the region « > a the potential is again zero; the general solution is (x) = F exp(—«x) + Gexp(«x), but the second term blows up (as + > oo), so we are left with wx) = Fe**. for x > a. [2.150] The next step is to impose boundary conditions: y and dy/dx continuous at —a and +a. But we can save a little time by noting that this potential is an even function, so we can assume with no loss of generality that the solutions are either even or odd (Problem 2.1(c)). The advantage of this is that we need only impose the boundary conditions on one side (say, at +a); the other side is then automatic, since ¥(—x) = +y(x). I'll work out the even solutions; you get to do the odd ones in Problem 2.29. The cosine is even (and the sine is odd), so I’m looking for solutions of the form Fe, forx >a, w(x) = 4% Deos(ix), forO<x <a, [2.151] w(—x). for + <0. The continuity of y(x), al x =a, says Fe~“" = Deos(la). [2.152] and the continuity of dw/dx, says —«Fe*" = -1Dsin(la). (2.153] Dividing Equation 2.153 by Equation 2.152, we find that « =I tan(la). [2.154] ¥You can, if you like, write the general solution in exponential form (Ce + D'e). This leads to the same final result. but since the potential is symmetric we know the solutions will be either even or odd, and the sine/cosine notation allows us to exploit this directly. 80 Chapter 2 Time-Independent Schridinger Equation tanz FIGURE 2.18: Graphical solution to Equation 2.156, for zo = 8 (even states). This is a formula for the allowed energies, since « and / are both functions of £. To solve for E, we first adopt some nicer notation: Let zsla, and w= 5 2m Vo. [2.155] 7? According to Equations 2.146 and 2.148, («?+ 1?) = 2m Vo/h*, so xa = ,/22 — 22, ° and Equation 2.154 reads tanz = \/ (zo/2)? — 1. [2.156] This is a transcendental equation for z (and hence for E) as a function of zo (which is a measure of the “size” of the well). It can be solved numerically, using a computer, or graphically, by plotting tanz and ./(zo/z)? — 1 on the same grid, and looking for points of intersection (see Figure 2.18). Two limiting cases are of special interest: 1. Wide, deep well. If zo is very large, the intersections occur just slightly below z, = 27/2, with » odd; it follows that 2_322 were E,t Vy =o. nt Mo = 7 Gay [2.157] But E + Vo is the energy above the bottom of the well, and on the right side we have precisely the infinite square well energies, for a well of width 2a (see Equation 2.27)—or rather, half of them, since this x is odd. (The other ones, of course, come from the odd wave functions, as you’ll discover in Problem 2.29.) So the finite square well goes over to the infinite square well, as Vo > oo; however, for any finite Vo there are only a finite number of bound states, 2. Shallow, narrow well. As zo decreases, there are fewer and fewer bound states, until finally (for zy < 2/2, where the lowest odd state disappears) only one remains. It is interesting to note, however, that there is always one bound state, no matter ow “weak” the well becomes. Section 2,6: The Finite Square Well 83 Problem 2.30 Normalize y(x) in Equation 2.151, to determine the constants D and F. Problem 2.31 The Dirac delta function can be thought of as the limiting case of a rectangle of area 1, as the height goes to infinity and the width goes to zero. Show that the delta-function well (Equation 2.114) is a “weak” potential (even though it is infinitely deep), in the sense that za > 0. Determine the bound state energy for the delta-function potential, by treating it as the limit of a finite square well. Check that your answer is consistent with Equation 2,129. Also show that Equation 2.169 reduces to Equation 2.141 in the appropriate limit. Problem 2.32 Derive Equations 2.167 and 2.168. Hint: Use Equations 2.165 and 2.166 to solve for C and D in terms of F: k ik k ik C= [sna +iF costa] Lc ee [cost iF sincay| elf F, Plug these back into Equations 2.163 and 2.164. Obtain the transmission coefficient, and confirm Equation 2.169. «*Problem 2.33 Determine the transmission coefficient for a rectangular barrier (same as Equation 2.145, only with V(x) = +Vp > 0 in the region -a <x <a). Treat separately the three cases E < Vo, E = Vo, and E > Vo (note that the wave function inside the barrier is different in the three cases). Partial answer: For E < Vo," ve 2 T=14 — oO — sinh? (Fvae = B)). 4E(Vo — E) h «Problem 2.34 Consider the “step” potential: 0 ifx <0. V@)= | Vo. ifx>O. (a) Calculate the reflection coefficient, for the case E < Vo, and comment on the answer. (b) Calculate the reflection coefficient for the case E > Vo. (c) For a potential such as this, which does not go back to zero to the right of the barrier, the transmission coefficient is nor simply |F|?/|A|? (with A the “This is a good example of tunneling—classically the particle would bounce back. 84 Chapter 2 Time-Independent Schridinger Equation Vix) Vo FIGURE 2.20: Scattering from a “cliff” (Problem 2.35). (4) incident amplitude and F the transmitted amplitude), because the transmitted wave travels at a different speed. Show that E—Vo |F[? T= E~VolFY E |A/P for E > Vo, Hint: You can figure it out using Equation 2.98, or—more ele- gantly, but less informatively —from the probability current (Problem 2.19). What is T, for E < Vo? [2.172] For E > Vo, calculate the transmission coefficient for the step potential, and check that T+ R=1. Problem 2.35 A particle of mass #7 and kinetic energy E > 0 approaches an abrupt potential drop Vo (Figure 2.20). (a) (b) (c) What is the probability that it will “reflect” back, if E = Vo/3? Hint: This is just like Problem 2.34, except that the step now goes down, instead of up. I drew the figure so as to make you think of a car approaching a cliff, but obviously the probability of “bouncing back” from the edge of a cliff is far smaller than what you got in (a)—-unless you’re Bugs Bunny, Explain why this potential does nor correctly represent a cliff. Hint: In Figure 2.20 the potential energy of the car drops discontinuously to — Vo, as it passes x = 0, would this be true for a falling car? When a free neutron enters a nucleus, it experiences a sudden drop in poten- tial energy, from V = 0 outside to around —12 MeV (million electron volts) inside. Suppose a neutron, emitted with kinetic energy 4 MeV by a fission event, strikes such a nucleus. What is the probability it will be absorbed, thereby initiating another fission? Hint: You calculated the probability of reflection in part (a); use T = 1 — R to get the probability of transmission through the surface. Further Problems for Chapter 2 85 FURTHER PROBLEMS FOR CHAPTER 2 Problem 2.36 Solve the time-independent Schrédinger equation with appropri- ate boundary conditions for the “centered” infinite square well: V(x) = 0 (for -—a <x < +a), V(x) = oo (otherwise). Check that your allowed energies are consistent with mine (Equation 2.27), and confirm that your y’s can be obtained from mine (Equation 2.28) by the substitution x — (x + a)/2 (and appropriate renormalization). Sketch your first three solutions, and compare Figure 2.2. Note that the width of the well is now 2a. Problem 2.37 A particle in the infinite square well (Equation 2.19) has the initial wave function WX OHA sin? (rx /a) (O<x <a). Determine A, find W(x. tf), and calculate (v), as a function of time. What is the expectation value of the energy? Hint: sin" @ and cos” 6 can be reduced, by repeated application of the trigonometric sum formulas, to linear combinations of sin(#7@) and cos(#@), with 1 = 0.1.2. ... .7. «Problem 2.38 A particle of mass m is in the ground state of the infinite square well (Equation 2.19). Suddenly the well expands to twice its original size—the right wall moving from a to 2a —leaving the wave function (momentarily) undisturbed. The energy of the particle is now measured. (a) What is the most probable result? What is the probability of getting that result? (b) What is the next most probable result, and what is its probability? (c) What is the expectation yalue of the energy? Hint: If you find yourself confronted with an infinite series, try another method. Problem 2.39 (a) Show that the wave function of a particle in the infinite square well returns to its original form after a quantum revival time T = 4a? /sh. That is: Wx, T) = V(x. 0) for any state (vor just a stationary state). (b) What is the classical revival time, for a particle of energy E bouncing back and forth between the wails? (c) For what energy are the two revival times equal?” 3 The fact that the classical and quantum revival times bear no obvious relation to one another {and the quantum one doesn’t even depend on the energy) is a curious paradox: see Daniel Styer. Am. J. Phys. 69, 56 (2001).