schrodinger equation in two dimensions

203

C h a p t e r 7The Schrödinger Equationin One Dimension

7.1 Introduction7.2 Classical Standing Waves7.3 Standing Waves in Quantum Mechanics; Stationary States7.4 The Particle in a Rigid Box7.5 The Time-Independent Schrödinger Equation7.6 The Rigid Box Again7.7 The Free Particle7.8 The Nonrigid Box7.9 The Simple Harmonic Oscillator�

7.10 Tunneling�

7.11 The Time-Dependent Schrödinger Equation�

Problems for Chapter 7�Sections marked with a star can be omitted without significant loss of continuity.

7.1 Introduction

In classical mechanics the state of motion of a particle is specified by giving theparticle’s position and velocity. In quantum mechanics the state of motion of aparticle is specified by giving the wave function. In either case the fundamen-tal question is to predict how the state of motion will evolve as time goes by,and in each case the answer is given by an equation of motion. The classicalequation of motion is Newton’s second law, if we know the particle’sposition and velocity at time Newton’s second law determines the posi-tion and velocity at any other time. In quantum mechanics the equation ofmotion is the time-dependent Schrödinger equation. If we know a particle’swave function at the time-dependent Schrödinger equation determinesthe wave function at any other time.

The time-dependent Schrödinger equation is a partial differential equa-tion, a complete understanding of which requires more mathematical prepa-ration than we are assuming here. Fortunately, the majority of interestingproblems in quantum mechanics do not require use of the equation in its fullgenerality. By far the most interesting states of any quantum system arethose states in which the system has a definite total energy, and it turns outthat for these states the wave function is a standing wave, analogous to thefamiliar standing waves on a string. When the time-dependent Schrödingerequation is applied to these standing waves, it reduces to a simpler equationcalled the time-independent Schrödinger equation. We will need only this

t = 0,

t = 0,F = ma;

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204 Chapter 7 • The Schrödinger Equation in One Dimension

*More realistic examples include the motion of electrons along one axis in certaincrystals and in some linear molecules.

time-independent equation, which will let us find the wave functions of thestanding waves and the corresponding allowed energies. Because we will beusing only the time-independent Schrödinger equation we will often refer toit as just “the Schrödinger equation.” Nevertheless, you should know thatthere are really two Schrödinger equations (the time-dependent and thetime-independent). Unfortunately, it is almost universal to refer to either as“the Schrödinger equation” and to let the context decide which is beingdiscussed. In this book, however, “the Schrödinger equation” will alwaysmean the simpler time-independent equation.

In Section 7.2 we review some properties of classical standing waves,using waves on a uniform, stretched string as our example. In Section 7.3 wediscuss quantum standing waves. Then, in Section 7.4, we show how the famil-iar properties of classical standing waves let one find the allowed energies ofone simple quantum system, namely a particle that moves freely inside aperfectly rigid box.

Using our experience with the wave functions of a particle in a rigid boxwe next write down the time-independent Schrödinger equation, with whichone can, in principle, find the allowed energies and wave functions for anysystem.Then in Sections 7.6 to 7.9 we use the Schrödinger equation to find theallowed energies of various simple systems.

Sections 7.10 and 7.11 treat two further topics, quantum tunneling andthe time-dependent Schrödinger equation. While both of these are veryimportant, we will not be using these ideas until later (the former in Chapters14 and 17, and the latter in Chapter 11), so you could skip these sections onyour first reading.

Throughout this chapter we treat particles that move nonrelativisticallyin one dimension.All real systems are, of course, three-dimensional. Neverthe-less, just as is the case in classical mechanics, it is a good idea to start with thesimpler problem of a particle confined to move in just one dimension. In theclassical case it is easy to find examples of systems that are at least approxi-mately one-dimensional — a railroad car on a straight track, a bead threadedon a taut string. In quantum mechanics there are fewer examples of one-dimensional systems. However, we can for the moment imagine an electronmoving along a very narrow wire.* The main importance of one-dimensionalsystems is that they provide a good introduction to three-dimensional systems,and that several one-dimensional solutions find direct application in three-dimensional problems.

7.2 Classical Standing Waves

We start with a review of some properties of classical standing waves in onedimension. We could discuss waves on a string, for which the wave function isthe string’s transverse displacement or we might consider soundwaves, for which the wave function is the pressure variation, If we con-sidered electromagnetic waves, the wave function would be the electric fieldstrength, In this section we choose to discuss waves on a string, butsince our considerations apply equally to all waves, we will use the generalnotation to represent the wave function.°1x, t2

e1x, t2.p1x, t2.y1x, t2;

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a

FIGURE 7.2Three successive snapshots of astanding wave on a finite string, oflength a, clamped at its two ends.The solid curve shows the string atmaximum displacement; the dashedand dotted curves show it aftersuccessive quarter-cycle intervals.

Section 7.2 • Classical Standing Waves 205

Nodes

Tim

e

FIGURE 7.1Five successive snapshots of thestanding wave of Eq. (7.2). Thenodes are points where the stringremains stationary at all times. Thedistance between successive nodesis half a wavelength, l>2.

*The superposition principle asserts that if and are possible waves, the same istrue of for any constants A and B. This important principle is true of anywave whose medium responds linearly to the disturbance. It applies to all the waves wewill be considering.

A°1 + B°2

°2°1

Let us consider first two sinusoidal traveling waves, one moving to the right,

(this is the wave sketched in Fig. 6.8) and the other moving to the left with thesame amplitude,

The superposition principle guarantees that the sum of these two waves isitself a possible wave motion*:

(7.1)

If we recall the important trigonometric identity (Appendix B)

we can rewrite the wave (7.1) as

or if we set

(7.2)

A series of snapshots of the resultant wave (7.2) is sketched in Fig. 7.1. Theimportant point to observe is that the resultant wave is not traveling to theright or left. At certain fixed points called nodes, where is zero,is always zero and the string is stationary. At any other point the string simplyoscillates up and down in proportion to with amplitude Bysuperposing two traveling waves, we have formed a standing wave.

Because the string never moves at the nodes, we could clamp it at twonodes and remove the string outside the clamps, leaving a standing wave on afinite length of string as in Fig. 7.2. This is the kind of wave produced on apiano or guitar string when it sounds a pure musical tone.

If we now imagine a string clamped between two fixed points separatedby a distance a, we can ask: What are the possible standing waves that can fiton the string? The answer is that a standing wave is possible, provided that ithas nodes at the two fixed ends of the string. The distance between two adja-cent nodes is so the distance between any pair of nodes is an integer mul-tiple of this, Therefore a standing wave fits on the string provided

for some integer n; that is, if

(7.3)l =2an

, where n = 1, 2, 3, Á

nl>2 = anl>2.l>2,

A sin kx.cos vt,

°1x, t2sin kx

°1x, t2 = A sin kx cos vt

2B = A,

°1x, t2 = 2B sin kx cos vt

sin a + sin b = 2 sin a + b

2 cos

a - b

2

°1x, t2 = °11x, t2 + °21x, t2 = B3sin1kx - vt2 + sin1kx + vt24

°21x, t2 = B sin1kx + vt2

°11x, t2 = B sin1kx - vt2

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� � 2a

� � a

� � 2a/3

FIGURE 7.3The first three possible standingwaves on a string of length a, fixedat both ends. Each dashed curve isone half-cycle after thecorresponding solid curve.


We see that the possible wavelengths of a standing wave on a string of length aare quantized, the allowed values being divided by any positive integer.Thefirst three of these allowed waves are sketched in Fig. 7.3.

It is important to recognize that the quantization of wavelengths arisesfrom the requirement that the wave function must always be zero at the twofixed ends of the string. We refer to this kind of requirement as a boundarycondition, since it relates to the boundaries of the system. We will find that forquantum waves, just as for classical waves, it is the boundary conditions thatlead to quantization.

7.3 Standing Waves in Quantum Mechanics;Stationary States

Before we discuss quantum standing waves, we need to examine more closelythe form of the classical standing wave (7.2):

(7.4)

This function is a product of one function of x (namely, ) and one func-tion of t (namely, ). We can emphasize this by rewriting (7.4) as

(7.5)

where we have used the capital letter for the full wave function andthe lower case letter for its spatial part The spatial function givesthe full wave function at time (since when );more generally, at any time t the full wave function is times theoscillatory factor

In our particular example (a wave on a uniform string) the spatialfunction was a sine function

(7.6)

but in more general problems, such as waves on a nonuniform string, canbe a more complicated function of x. On the other hand, even in these morecomplicated problems the time dependence is still sinusoidal; that is, it is givenby a sine or cosine function of t.The difference between the sine and the cosineis just a difference in the choice of origin of time. Thus either function is possi-ble, and the general sinusoidal standing wave is a combination of both:

(7.7)

Different choices for the ratio of the coefficients a and b correspond todifferent choices of the origin of time. (See Problem 7.13.)

The standing waves of a quantum system have the same form (7.7), butwith one important difference. For a classical wave, the function is, ofcourse, a real number. (It would make no sense to say that the displacement ofa string, or the pressure of a sound wave, had an imaginary part.) Therefore,the function and the coefficients a and b in (7.7) are always real for anyc1x2

°1x, t2

°1x, t2 = c1x21a cos vt + b sin vt2

c1x2c1x2 = A sin kx

c1x2cos vt.

c1x2°1x, t2 t = 0cos vt = 1t = 0°1x, t2 c1x2c1x2.c

°1x, t2°

°1x, t2 = c1x2 cos vt

cos vtA sin kx

°1x, t2 = A sin kx cos vt

2a

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Section 7.3 • Standing Waves in Quantum Mechanics; Stationary States 207

*As you may know, it is sometimes a mathematical convenience to introduce a certaincomplex wave function. Nonetheless, in classical physics the actual wave function isalways the real part of this complex function.

classical wave.* In quantum mechanics, on the other hand, the wave functioncan be a complex number; and for quantum standing waves it usually is com-plex. Specifically, the time-dependent part of the wave function (7.7) alwaysoccurs in precisely the combination

(7.8)

where i is the imaginary number (often denoted j by engineers).That is, the standing waves of a quantum particle have the form

(7.9)

In Section 7.11 we will prove this from the time-dependent Schrödinger equa-tion. For now, we simply assert that quantum standing waves have the sinusoidaltime dependence of the particular combination of and in (7.9).

The form (7.9) can be simplified if we use Euler’s formula from thetheory of complex numbers (Problem 7.14),

(7.10)

This identity can be illustrated in the complex plane, as in Fig. 7.4, where thecomplex number is represented by a point with coordinates x andy in the complex plane. Since the number (with any real number) hascoordinates and we see from Pythagoras’ theorem that its absolutevalue is 1:

Thus the complex number lies on a circle of radius 1, with polar angle asshown. Notice that since and

Returning to (7.9) and using the identity (7.10), we can write for thegeneral standing wave of a quantum system

(7.11)°1x, t2 = c1x2e-ivt

cos u - i sin u = e-iu

sin1-u2 = -sin u,cos1-u2 = cos uueiu

ƒeiu ƒ = 41cos u22 + 1sin u22 = 1

sin u,cos uueiu

z = x + iy

cos u + i sin u = eiu

sin vtcos vt

°1x, t2 = c1x21cos vt - i sin vt2

i = 1-1

cos vt - i sin vt

�

�cos

�sin 1

Imaginary part y

Real part x

ei� � � cos � i sin �

FIGURE 7.4The complex number

is representedby a point with coordinates

in the complex plane.The absolute value of any complexnumber is defined as

Since it follows

that ƒeiu ƒ = 1.cos2 u + sin2 u = 1,ƒz ƒ = 3x2 + y2 .

z = x + iy

1cos u, sin u2eiu = cos u + i sin u

TAYL07-203-247.I 1/4/03 1:03 PM Page 207


*Of course, atoms do radiate from excited states, but as we discuss in Chapter 11, this isalways because some external influence disturbs the stationary state.

Since this function has a definite angular frequency, any quantum systemwith this wave function has a definite energy given by the de Broglie relation

(6.23). Conversely, any quantum system that has a definite energy hasa wave function of the form (7.11) — a statement we will prove in Section 7.11.

We saw in Chapter 6 that the probability density associated with a quan-tum wave function is the absolute value squared, For thecomplex standing wave (7.11) this has a remarkable property:

or, since

(7.12)

That is, for a quantum standing wave, the probability density is independent oftime. This is possible because the time-dependent part of the wave function,

is complex, with two parts that oscillate 90° out of phase; when one is growing,the other is shrinking in such a way that the sum of their squares is constant.Thus for a quantum standing wave, the distribution of matter (of electrons inan atom, or nucleons in a nucleus, for example) is time independent orstationary. For this reason a quantum standing wave is often called a stationarystate. The stationary states are the modern counterpart of Bohr’s stationaryorbits and are precisely the states of definite energy. Because their chargedistribution is static, atoms in stationary states do not radiate.*

An important practical consequence of (7.12) is that in most problemsthe only interesting part of the wave function is its spatial part We will see that a large part of quantum mechanics is devoted to finding thepossible spatial functions and their corresponding energies. Our principaltool in finding these will be the time-independent Schrödinger equation.

7.4 The Particle in a Rigid Box

Before we write down the Schrödinger equation, we consider a simple examplethat we can solve using just our experience with standing waves on a string. Weconsider a particle that is confined to some finite interval on the x-axis, andmoves freely inside that interval — a situation we describe as a one-dimensionalrigid box and (for reasons we explain later) is often called the infinite squarewell. For example, in classical mechanics we could consider a bead on a friction-less straight thread between two rigid knots; the bead can move freely betweenthe knots, but cannot escape outside them. In quantum mechanics we can imag-ine an electron inside a length of very thin conducting wire; to a fair approxima-tion, the electron would move freely back and forth inside the wire, but could notescape from it.

Let us consider, then, a quantum particle of mass m moving nonrelativis-tically in a one-dimensional rigid box of length a, with no forces acting on itbetween and The absence of forces means that the potentialx = a.x = 0

c1x2c1x2.°1x, t2

e-ivt = cos vt - i sin vt

ƒ °1x, t2 ƒ2 = ƒc1x2 ƒ2 (for quantum standing waves)

ƒe-ivt ƒ = 1,

ƒ °1x, t2 ƒ2 = ƒc1x2 ƒ2 ƒe-ivt ƒ2

ƒ °1x, t2 ƒ2.°1x, t2

E = Uv

v,

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Section 7.4 • The Particle in a Rigid Box 209

*Strictly speaking, (7.18) implies either that k satisfies (7.19) or that but ifthen for all x, and we get no wave at all. Thus, only the solution (7.19)

corresponds to a particle in a box. Notice also that there is no reason to includenegative integer values of n in (7.20) since is just a multiple of sin1kx2.sin1-kx2

c = 0A = 0,A = 0;

energy is constant inside the box, and we are free to choose that constant to bezero.Therefore, its total energy is just its kinetic energy. In quantum mechanicsit is almost always more convenient to think of the kinetic energy as rather than because of the de Broglie relation, (6.1), betweenthe momentum and wavelength. Therefore, we write the energy as

(7.13)

As we have said, the states of definite energy are the standing waves.Therefore, to find the allowed energies, we must find the possible standingwaves for the particle’s wave function We have asserted that thestanding waves have the form

(7.14)

By analogy, with waves on a string, one might guess that the spatial functionwill be a sinusoidal function inside the box; that is, should have the

form or or a combination of both:

(7.15)

for (We make no claim to have proved this; but it is certainly areasonable guess, and we will prove it in Section 7.6.)

Since it is impossible for the particle to escape from the box, the wavefunction must be zero outside; that is, when and when If we make the plausible (and, again, correct) assumption that is continu-ous, then it must also vanish at and

(7.16)

These are the boundary conditions that the wave function (7.15) must satisfy.Notice that these boundary conditions are identical to those for a classicalwave on a string clamped at and

From (7.15) we see that Thus the wave function (7.15) cansatisfy the boundary condition (7.16) only if the coefficient B is zero; that is,the condition restricts to have the form

(7.17)

Next, the boundary condition that requires that

(7.18)

which implies that*

(7.19)ka = p, or 2p, or 3p, Á

A sin ka = 0

c1a2 = 0

c1x2 = A sin kx

c1x2c102 = 0

c102 = B.x = a.x = 0

c102 = c1a2 = 0

x = a:x = 0c1x2 x 7 a.x 6 0c1x2 = 0

0 … x … a.

c1x2 = A sin kx + B cos kx

cos kxsin kxc1x2c1x2

°1x, t2 = c1x2e-ivt

°1x, t2.

E = K =p2

2m

l = h>p12 mv2,

p2>2m,

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16E1 n � 4

9E1 3

4E1 2

E1

01

E �

0 a x

FIGURE 7.5A composite picture showing thefirst four energy levels and wavefunctions for a particle in a rigidbox. Each horizontal line indicatesan energy level and is also used asthe axis for a plot of thecorresponding wave function.


or

(7.20)

We conclude that the only standing waves that satisfy the boundaryconditions (7.16) have the form with k given by (7.20). Interms of wavelength, this condition implies that

(7.21)

which is precisely the condition (7.3) for standing waves on a string. This is, ofcourse, not an accident. In both cases, the quantization of wavelengths arose fromthe boundary condition that the wave function must be zero at and

For our present discussion, the important point is that quantization ofwavelength implies quantization of momentum, and hence also of energy.Specifically, substituting (7.21) into the de Broglie relation we find that

(7.22)

Since and in this case, we have Therefore,(7.22) means that the allowed energies for a particle in a one-dimensionalrigid box are

(7.23)

The lowest energy for our particle, termed the ground-state energy, isobtained when and is

(7.24)

This is consistent with the lower bound derived from the Heisenberg uncer-tainty principle in Chapter 6, where we argued — see (6.39) — that for aparticle confined in a region of length a,

(7.25)

For our particle in a rigid box, the actual minimum energy (7.24) is larger thanthe lower bound (7.25) by a factor of

In terms of the ground-state energy the energy of the nth level (7.23) is

(7.26)

These energy levels are sketched in Fig. 7.5. Notice that (quite unlike those ofthe hydrogen atom) the energy levels are farther and farther apart as nincreases and that increases without limit as The correspondingwave functions (which look exactly like the standing waves on a string)c1x2

n : q .En

En = n2 E1 n = 1, 2, 3, Á

E1 ,p2 L 10.

E ÚU2

2ma2

E1 =p2

U2

2ma2

n = 1

En = n2 p2

U2

2ma2 n = 1, 2, 3, Á

E = p2>2m.U = 0E = K + U

p =nh

2a=

npUa n = 1, 2, 3, Á

p = h>l,l

x = a.x = 0

l =2pk

=2an n = 1, 2, 3, Á

c1x2 = A sin kx

k =npa

, n = 1, 2, 3, Á

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Section 7.5 • The Time-Independent Schrödinger Equation 211

have been superimposed on the same picture, the wave function for each levelbeing plotted on the line that represents its energy. Notice how the number ofnodes of the wave functions increases steadily with energy; this is what oneshould expect since more nodes mean shorter wavelength (larger curvatureof ) and hence larger momentum and kinetic energy.

The complete wave function for any of our standing waveshas the form

We can rewrite this, using the identity (Problem 7.16)

(7.27)

to give

(7.28)

We see that our quantum standing wave (just like the classical standing waveof Section 7.2) can be expressed as the sum of two traveling waves, one movingto the right and one to the left. The wave moving to the right represents a par-ticle with momentum directed to the right, and that moving to the left, aparticle with momentum of the same magnitude but directed to the left.Thusa particle in one of our stationary states has a definite magnitude, for itsmomentum but is an equal superposition of momenta in either direction. Thiscorresponds to the result that on average a classical particle is equally likely tobe moving in either direction as it bounces back and forth inside a rigid box.

7.5 The Time-Independent Schrödinger Equation

Our discussion of the particle in a rigid box depended on some guessing as tothe form of the spatial wave function There are very few problemswhere this kind of guesswork is possible, and no problems where it is entirelysatisfying. What we need is the equation that determines in any problem,and this equation is the time-independent Schrödinger equation. Like all basiclaws of physics, the Schrödinger equation cannot be derived. It is simply a rela-tion, like Newton’s second law, that experience has shown to be true. Thus alegitimate procedure would be simply to state the equation and to start usingit. Nevertheless, it may be helpful to offer some arguments that suggest theequation, and this is what we will try to do.

Almost all laws of physics can be expressed as differential equations, thatis, as equations that involve the variable of interest and some of its derivatives.The most familiar example is Newton’s second law for a single particle, whichwe can write as

(7.29)m d2

x

dt2 = a

F

c1x2c1x2.

Uk,Uk

Uk

°1x, t2 =A

2i 1ei1kx -vt2 - e-i1kx +vt22

sin u =eiu - e-iu

2i

°1x, t2 = c1x2e-ivt = A sin1kx2e-ivt

°1x, t2c

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If, for example, the particle in question were immersed in a viscous fluid thatexerted a drag force and attached to a spring that exerted a restoringforce then (7.29) would read

(7.30)

This is a differential equation for the particle’s position x as a function of time t,and since the highest derivative involved is the second derivative, the equationis called a second-order differential equation.

The equation of motion for classical waves (which is often not discussedin an introductory physics course) is a differential equation. It is therefore nat-ural to expect the equation that determines the possible standing waves of aquantum system to be a differential equation. Since we already know the formof the wave functions for a particle in a rigid box, what we will do is examinethese wave functions and try to spot a simple differential equation that theysatisfy and that we can generalize to more complicated systems.

We saw in Section 7.4 that the spatial wave functions for a particle in arigid box have the form

(7.31)

To find a differential equation that this function satisfies, we naturally differ-entiate it, to give

(7.32)

There are several ways in which we could relate in (7.32) to in(7.31) and hence obtain an equation connecting with However, asimpler course is to differentiate a second time to give

(7.33)

Comparing (7.33) and (7.31), we see at once that is proportional to specifically,

(7.34)

We can rewrite in (7.34) in terms of the particle’s kinetic energy, K. Weknow that Therefore,

(7.35)

hence

(7.36)k2 =2mK

U2

K =p2

2m=

U2 k2

2m

p = Uk.k2

d2 c

dx2 = -k2 c

c;d2 c>dx2

d2 c

dx2 = -k2 A sin kx

c.dc>dxsin kxcos kx

dc

dx= kA cos kx

c1x2 = A sin kx

m d2

x

dt2 = -b dxdt

- kx

-kx,-bv,

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Erwin Schrödinger(1887–1961, Austrian)

After learning of de Broglie’s mat-ter waves, Schrödinger proposedthe equation — the Schrödingerequation — that governs thewaves’ behavior and earned himthe 1933 Nobel Prize in physics.He left Austria after Hitler’s inva-sion and became a professor inDublin, Ireland. A person with re-markably broad interests, he wasan ardent student of Italian paintingand botany, as well as chemistryand physics. Late in his career, hebecame a pioneer in the new fieldof biophysics and wrote a popularbook entitled What is Life?.

Section 7.5 • The Time-Independent Schrödinger Equation 213

*In more advanced texts this equation is usually written in the form

Because the differential operator is intimately connected with thekinetic energy, this way of writing the Schrödinger equation is perhaps easier toremember because it looks like Nevertheless, for the applications inthis book, the form (7.39) is the most convenient, and we will almost always write itthis way.

K + U = E.

-1U2>2m2d2>dx2

- U2

2m d2

c

dx2 + U1x2c = Ec

Thus, we can write (7.34) as

(7.37)

which gives us a second-order differential equation satisfied by the wavefunction of a particle in a rigid box.

The particle in a rigid box is an especially simple system, with potentialenergy equal to zero throughout the region where the particle moves. It is not atall obvious how the equation (7.37) should be generalized to include the possi-bility of a nonzero potential energy, which may vary from point to point.However, since the kinetic energy K is the difference between the total energyE and the potential energy it is perhaps natural to replace K in (7.37) by

(7.38)

This gives us the differential equation*

(7.39)

This differential equation is called the Schrödinger equation (time-independent Schrödinger equation, in full), to honor the Austrian physicist,Erwin Schrödinger, who first published it in 1926. Like us, Schrödinger had noway to prove that his equation was correct. All he could do was argue that theequation seemed reasonable and that its predictions should be tested againstexperiment. In the 80 years or so since then, it has passed this test repeatedly.In particular, Schrödinger himself showed that it predicts correctly the energylevels of the hydrogen atom, as we describe in Chapter 8. Today, it is generallyaccepted that the Schrödinger equation is the correct basis of nonrelativisticquantum mechanics, in just the same way that Newton’s second law is accept-ed as the basis of nonrelativistic classical mechanics.

The Schrödinger equation as written in (7.39) applies to one particlemoving in one dimension. We will need to generalize it later to cover systemsof several particles, in two or three dimensions. Nevertheless, the general pro-cedure for using the equation is the same in all cases. Given a system whosestationary states and energies we want to know, we must first find the potentialenergy function For example, a particle held in equilibrium at bya force obeying Hooke’s law has potential energy

(7.40)U1x2 = 12 kx2

1F = -kx2 x = 0U1x2.

d2 c

dx2 =2m

U2 3U1x2 - E4c

K = E - U1x2U1x2,

U1x2,

c1x2

d2 c

dx2 = - 2mK

U2 c

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*The necessity of identifying U before one can solve the Schrödinger equationcorresponds to the necessity of identifying the total force F on a classical particlebefore one can solve Newton’s second law, The two are, of course, closelyrelated since † You may reasonably object that it is circular to apply the Schrödinger equation to aparticle in a box, when we used the latter to derive the former. Nevertheless, it is alegitimate consistency check, as well as an instructive exercise, to see how theSchrödinger equation gives back the known energies and wave functions.

F = -dU>dx.F = ma.

An electron in a hydrogen atom has

(7.41)

(We will return to this three-dimensional example in Chapter 8.) Once wehave identified the Schrödinger equation (7.39) becomes a well-definedequation that we can try to solve.* In most cases, it turns out that for manyvalues of the energy E the Schrödinger equation has no solutions (no accept-able solutions, satisfying the particular conditions of the problem, that is). Thisis exactly what leads to the quantization of energies. Those values of E forwhich the Schrödinger equation has no solution are not allowed energies ofthe system. Conversely, those values of E for which there is a solution areallowed energies, and the corresponding solutions give the spatial wavefunctions of these stationary states.

As we hinted in the last paragraph, there are usually certain conditionsthat the wave function must satisfy to be an acceptable solution of theSchrödinger equation. First, there may be boundary conditions on forexample, the condition that must vanish at the walls of a perfectly rigidbox. In addition, there are certain general restrictions on for example, aswe anticipated in Section 7.4, must always be continuous, and in mostproblems its first derivative must also be continuous. When we speak of anacceptable solution of the Schrödinger equation, we mean a solution thatsatisfies all the conditions appropriate to the problem at hand.

In this section you may have noticed that in quantum mechanics it is thepotential energy that appears in the basic equation, whereas in classicalmechanics it is the force F. Of course, U and F are closely related: F being thederivative of U, U being the integral of F. Nevertheless, it is an importantdifference of emphasis that quantum mechanics focuses primarily on potentialenergies, whereas Newtonian mechanics focuses on forces.

7.6 The Rigid Box Again

As a first application of the Schrödinger equation, we use it to rederive theallowed energies of a particle in a rigid box and check that we get the sameanswers as before.† The first step in applying the Schrödinger equation to anysystem is to identify the potential-energy function Inside the box we canchoose the potential energy to be zero, and outside the box it is infinite. This isthe mathematical expression of our idealized perfectly rigid box — no finiteamount of energy can remove the particle from it. Thus

(7.42)U1x2 = b0 for 0 … x … aq for x 6 0 and x 7 a

U1x2.

U1x2

c1x2 c1x2;c1x2 c1x2,c1x2

c1x2

U1x2,

U1r2 = - ke2

r

TAYL07-203-247.I 1/4/03 1:03 PM Page 214

Section 7.6 • The Rigid Box Again 215

That outside the box implies that the particle can never befound there and hence that the wave function must be zero when and when The potential-energy function is described as an infinitelydeep potential well or an infinite square well because of the square (90°)angles at the bottom of the well. The continuity of then requires that

(7.43)

(all of which we had argued in Section 7.4). Inside the box, where the Schrödinger equation (7.39) reduces to

(7.44)

This is the differential equation whose solutions we must investigate. In partic-ular, we want to find those values of E for which it has a solution satisfying theboundary conditions (7.43).

Before solving (7.44), we remark that it is a nuisance, both for the print-er of a book and for the student taking notes, to keep writing the symbols

and For this reason, we introduce the shorthand

From now on we will use this notation whenever convenient. In particular, werewrite (7.44) as

(7.45)

We now consider whether there is an acceptable solution of (7.45) forany particular value of E, starting with the case that E is negative. (We do notexpect any states with since then E would be less than the minimumpotential energy. But we have already encountered several unexpected conse-quences of quantum mechanics, and we should check this possibility.) If Ewere negative, the coefficient on the right of (7.45) would be positiveand we could call it where

(7.46)

With this notation, (7.45) becomes

(7.47)

The simplification of rewriting (7.45) in the form (7.47) has the disadvantageof requiring a new symbol (namely ); but it has the important advantage ofletting us focus on the mathematical structure of the equation.

Equation (7.47) is a second-order differential equation, which has thesolutions (Problem 7.22) and or any combination of these,

(7.48)

where A and B are any constants, real or complex.

c1x2 = Aeax + Be-ax

e-axeax

a

c–1x2 = a2 c1x2

a =2-2mE

U

a2,-2mE>U2

E 6 0,

c–1x2 = - 2mE

U2 c1x2

c¿ Kdc

dx and c– K

d2 c

dx2

d2 c>dx2.dc>dx

d2 c

dx2 = - 2mE

U2 c for 0 … x … a

U1x2 = 0,

c102 = c1a2 = 0

c1x2x 7 a.

x 6 0c1x2U1x2 = q

TAYL07-203-247.I 1/4/03 1:03 PM Page 215


*To be precise, ordinary second-order differential equations that are linear andhomogeneous.† When we say that two functions are independent, we mean that neither function is justa constant multiple of the other. For example, and are independent, but and

are not; similarly and are independent, but and are not.cos x5 cos xcos xsin x2eaxeaxe-axeax

It is important in what follows that (7.48) is the most general solution of(7.47), that is, that every solution of (7.47) has the form (7.48). This followsfrom a theorem about second-order differential equations of the same type asthe one-dimensional Schrödinger equation.* This theorem states three facts:First, these equations always have two independent solutions. For example,and are two independent solutions of (7.47).† Second, if and denote two such independent solutions, then the linear combination

(7.49)

is also a solution, for any constants A and B. (This is the superposition princi-ple.) Third, given two independent solutions and every solutioncan be expressed as a linear combination of the form (7.49). These three prop-erties are illustrated in Problems 7.22 to 7.28.

That the general solution of a second-order differential equation con-tains two arbitrary constants is easy to understand:A second-order differentialequation amounts to a statement about the second derivative to find one must somehow accomplish two integrations, which should introduce twoconstants of integration; and this is what the two arbitrary constants A and Bin (7.49) are. The theorem above is very useful in seeking solutions of suchdifferential equations. If, by any means, we can spot two independent solu-tions, we are assured that every solution is a combination of these two. Since

and are independent solutions of (7.47), it follows from the theoremthat the most general solution is (7.48).

Equation (7.48) gives all solutions of the Schrödinger equation for nega-tive values of E. The important question now is whether any of these solutionscould satisfy the required boundary conditions (7.43), and the answer is “no.”With given by (7.48), the condition that implies that

while the requirement that implies that

One can verify (Problem 7.23) that the only values of A and B that satisfythese two simultaneous equations are That is, if the onlysolution of the Schrödinger equation that satisfies the boundary conditions isthe zero function. In other words, with there can be no standing waves,so negative values of E are not allowed. A similar argument gives the sameconclusion for

Let us next see if the Schrödinger equation (7.45) has any acceptablesolutions for positive energies (as we expect it does). With the coeffi-cient on the right of (7.45) is negative and can conveniently becalled where

(7.50)k =22mE

U

-k2-2mE>U2

E 7 0,

E = 0.

E 6 0,

E 6 0,A = B = 0.

Aeaa + Be-aa = 0

c1a2 = 0

A + B = 0

c102 = 0c1x2

e-axeax

c,c–;

c21x2,c11x2

Ac11x2 + Bc21x2

c21x2c11x2e-axeax

TAYL07-203-247.I 1/4/03 1:03 PM Page 216

With this notation, the Schrödinger equation reads

(7.51)

This differential equation has the solutions and or any combina-tion of both:

(7.52)

(see Example 7.1 below). This is exactly the form of the wave function that weassumed at the beginning of Section 7.4. The important difference is that inSection 7.4 we could only guess the form (7.52), whereas we have now derivedit from the Schrödinger equation. From here on, the argument follows precise-ly the argument given before. As we saw, the boundary condition requires that the coefficient B in (7.52) be zero, whereas the condition that

can be satisfied without A being zero, provided that is an integermultiple of (so that ); that is,

or, from (7.50),

exactly as before.

Example 7.1

Verify explicitly that the function (7.52) is a solution of the Schrödingerequation (7.51) for any values of the constants A and B. [This illustrates partof the theorem stated in connection with (7.49).]

To verify that a given function satisfies an equation, one must substitutethe function into one side of the equation and then manipulate it until onearrives at the other side. Thus, for the proposed solution (7.52),

and we conclude that the proposed solution does satisfy the desired equation.

There is one loose end in our discussion of the particle in a rigid boxthat we can now dispose of. We have seen that the stationary states havewave functions

(7.53)c1x2 = A sin npx

a

= -k2 c1x2

= -k21A sin kx + B cos kx2 = -k2

A sin kx - k2 B cos kx

=d

dx 1kA cos kx - kB sin kx2

c–1x2 =d2

dx2 1A sin kx + B cos kx2

E =U2

k2

2m= n2

p2

U2

2ma2

k =npa

sin ka = 0p

kac1a2 = 0

c102 = 0


cos kx,sin kx

c–1x2 = -k2 c1x2


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*You may have noticed that, strictly speaking, the argument leading from (7.55) to(7.59) implies only that the absolute value of A is However, since the probabilitydensity depends only on the absolute value of we are free to choose any value of Asatisfying (for example, or ); the choice (7.59) is con-venient and customary.

i12>aA = -12>aƒA ƒ = 12>ac,12>a .

but we have not yet found the constant A. Whatever the value of A, the func-tion (7.53) satisfies the Schrödinger equation and the boundary conditions.Clearly, therefore, neither the Schrödinger equation nor the boundary condi-tions fix the value of A.

To see what does fix A, recall that is the probability density forfinding the particle at x. This means, in the case of a one-dimensional system,that is the probability P of finding the particle between x and

(7.54)

Since the total probability of finding the particle anywhere must be 1, it followsthat

(7.55)

This relation is called the normalization condition and a wave function thatsatisfies it is said to be normalized. It is the condition (7.55) that fixes the valueof the constant A, which is therefore called the normalization constant.

In the case of the rigid box, is zero outside the box; therefore, (7.55)can be rewritten as

(7.56)

or, with the explicit form (7.53) for

(7.57)

The integral here turns out to be (Problem 7.29).Therefore, (7.57) implies that

(7.58)

and hence that*

(7.59)

We conclude that the normalized wave functions for the particle in a rigid boxare given by

(7.60)c1x2 = A 2a

sin npx

a

A = A 2a

A2 a

2= 1

a>2

A2 L

a

0 sin2anpx

ab dx = 1

c1x2,L

a

0 ƒc1x2 ƒ2 dx = 1

c1x2

Lq

-q ƒc1x2 ƒ2 dx = 1

P1between x and x + dx2 = ƒc1x2 ƒ2 dx

x + dx.ƒc1x2 ƒ2 dx

ƒc1x2 ƒ2

TAYL07-203-247.I 1/4/03 1:03 PM Page 218

Example 7.2

Consider a particle in the ground state of a rigid box of length a. (a) Find theprobability density (b) Where is the particle most likely to be found?(c) What is the probability of finding the particle in the interval between

and (d) What is it for the interval (e) What would be the average result if the position of a particle in theground state were measured many times?

(a) The probability density is just where is given by (7.60) withTherefore, it is

(7.61)

which is sketched in Fig. 7.6.

(b) The most probable value x, is the value of x for which ismaximum. From Fig. 7.6 this is clearly seen to be

(7.62)

(c) The probability of finding the particle in any small interval from x tois given by (7.54) as

(7.63)

(This is exact in the limit and is therefore a good approximationfor any small interval ) Thus, the two probabilities are

(d) and, similarly,

(e) The average result if we measure the position many times (always withthe particle in the same state) is the integral, over all possible positions,of x times the probability of finding the particle at x:

(7.64)

(If you are not familiar with this argument, see the following paragraphs.)This average value (also denoted or ) is often called theexpectation value of x. (But note that it is not the value we expect in anyone measurement; it is rather the average value expected after manymeasurements.) In the present case

(7.65)8x9 =2a

La

0 x sin2apx

ab dx

xavx8x9

8x9 = La

0 x ƒc1x2 ƒ2 dx

P10.75a … x … 0.76a2 L2a

sin2a3p4b * 0.01a = 0.01 = 1%

= 0.02 = 2%

P10.50a … x … 0.51a2 L ƒc10.50a2 ƒ2 ¢x =2a

sin2ap2b * 0.01a

¢x.¢x : 0

P1between x and x + ¢x2 L ƒc1x2 ƒ2 ¢x

x + ¢x

xmp = a>2

ƒc1x2 ƒ2

ƒc1x2 ƒ2 =2a

sin2apxab

n = 1.c1x2ƒc1x2 ƒ2,

30.75a, 0.76a4?x = 0.51a?x = 0.50a

ƒc ƒ2.


0 x

� �� 2

a

FIGURE 7.6The probability density fora particle in the ground state of arigid box. Inside the box, isgiven by (7.61); outside, it is zero.

ƒc ƒ2ƒc1x2 ƒ2

TAYL07-203-247.I 1/4/03 1:03 PM Page 219


This integral can be evaluated to give (Problem 7.33)

(7.66)

an answer that is easily understood from Fig. 7.6: Since is symmet-ric about the middle position the average value must be Wesee from (7.62) and (7.66) that for the ground state of a rigid box, themost probable position and the mean position are the same. Wewill see in the next example that and are not always equal.

Expectation Values

In Example 7.2 we introduced the notion of the expectation value of x.This is not the value of x expected in any one measurement; rather it is the av-erage value expected if we repeat the measurement many times (always withthe system in the same state). This kind of average comes up in many otherbranches of physics, especially in statistical mechanics, and is worth discussingin a little more detail. In particular, we want to justify the expression (7.64).

Suppose that we are interested in a quantity x that can take on variousvalues with definite probabilities. The quantity x could be a continuous vari-able, such as the position of a quantum particle, or a discrete variable, such asthe number of offspring of a female fruit fly chosen at random in a largecolony of fruit flies. Let us consider first the discrete case: Suppose that thepossible results of the measurement are and that these re-sults occur with probabilities This statement means that if alarge number N of statistically independent measurements are made, the num-ber of measurements resulting in value will be in other words,

is the fraction of the measurements that yield the value The av-erage value of x is the sum of all the results of all the measurements divided bythe total number N. Since of the measurements produce the value thissum of all the measurement results is and the average value is

This expression can be rewritten in terms of the probabilities as

(7.67)

If x is a continuous variable, the probability is replaced with a proba-bility increment where is the probability density. [For example,in the case of interest to us now, x is the position of a quantum particle and theprobability density is ] The sum (7.67) becomes an integral,

(7.68)

In particular, for a quantum particle and we have the expres-sion (7.64) for the expectation value of the position. More generally, if we

p1x2 = ƒc1x2 ƒ2,

8x9 = a

i x

i Pi : 8x9 = L

xp1x2 dx.

p1x2 = ƒc1x2 ƒ2.p1x2p1x2 dx,

Pi

8x9 =1N

a

i n

i xi = a

i

ni

N xi = a

i P

i xi

Pi

8x9 =1N

a

i n

i xi

a

i n i xi ,

xi ,ni

xi .Pi = ni>Nni = Pi

# N;xi

P1 , P2 , Á , Pi , Á .x1 , x2 , Á , xi , Á

8x9

8x9xmp

8x9xmp

a>2.x = a>2,ƒc1x2 ƒ2

8x9 =a

2

TAYL07-203-247.I 1/4/03 1:03 PM Page 220

measure or or any function we can repeat the same argument,simply replacing x with and conclude that

(7.69)

Example 7.3

Answer the same questions as in Example 7.2 but for the first excited state ofthe rigid box.

The wave function is given by (7.60) with so

This is plotted in Fig. 7.7, where it is clear that has two equal maxima at

The expectation value is easily found without actually doing any integra-tion. Since is symmetric about contributions to the integral(7.64) from either side of exactly balance one another, and we findthe same answer as for the ground state

The probabilities of finding the particle in any small intervals are given by(7.63) as

(7.70)

since* and

In particular, notice that although is the average value of x, the prob-ability of finding the particle in the immediate neighborhood of iszero. This result, although a little surprising at first, is easily understood byreference to Fig. 7.7.

x = a>2x = a>2

P10.75a … x … 0.76a2 L ƒc10.75a2 ƒ2 ¢x =2a

sin2a3p2b * 0.01a = 0.02

c10.50a2 = 0;

P10.50a … x … 0.51a2 L ƒc10.50a2 ƒ2 ¢x = 0

8x9 =a

2

x = a>2 x = a>2,ƒc1x2 ƒ28x9

xmp =a

4 and

3a4

ƒc1x2 ƒ2

ƒc1x2 ƒ2 =2a

sin2a2pxab

n = 2,

8f1x29 = L

f1x2p1x2 dx

f1x2, f1x2,x3x2


0 x

� �� 2

a

FIGURE 7.7The probability density fora particle in the first excited state

of a rigid box.1n = 22ƒc1x2 ƒ2

*Note that the probability for the interval is not exactly zero since theprobability density is zero only at the one point The significance of (7.70)is really that the probability for this interval is very small compared to the probabilityfor intervals of the same width elsewhere.

0.50a.ƒc1x2 ƒ230.50a, 0.51a4

TAYL07-203-247.I 1/4/03 1:03 PM Page 221


7.7 The Free Particle

As a second application of the Schrödinger equation, we investigate the possi-ble energies of a free particle; that is, a particle subject to no forces and com-pletely unconfined (still in one dimension, of course). The potential energy ofa free particle is constant and can be chosen to be zero. With this choice, wewill show that the energy of the particle can have any positive value,That is, the energy of a free particle is not quantized, and its allowed values arethe same as those of a classical free particle.

To prove these assertions, we must write down the Schrödinger equationand find those E for which it has acceptable solutions. With theSchrödinger equation is

(7.71)

This is the same equation that we solved for a particle in a rigid box.However, there is an important difference since the free particle can beanywhere in the range

Thus we must look for solutions of (7.71) for all x rather than just those xbetween 0 and a.

If we consider first the possibility of states with the coefficientin front of in (7.71) is positive and we can write (7.71) as

where Just as with the rigid box, this equation has the solu-tions and or any combination of both:

(7.72)

But in the present case we can immediately see that none of these solutionscan possibly be physically acceptable. The point is that (7.72) is the solution inthe whole range Now, as the exponential growswithout limit or “blows up,” and it is not physically reasonable to have a wavefunction that grows without limit as we move farther from the origin.Such a cannot be normalized. The only way out of this difficulty is tohave the coefficient A of in (7.72) equal to zero. Similarly, as theexponential blows up; thus by the same argument the coefficient B mustalso be zero, and we are left with just the zero solution That is, thereare no acceptable states with just as we expected.

The argument just given crops up surprisingly often in solving theSchrödinger equation. If a solution of the equation blows up as or as

that solution is obviously not acceptable.Thus we can add to our listof conditions that must be satisfied by an acceptable wave function therequirement that must not blow up as We speak of a functionthat satisfies this requirement as being “well behaved” as Thisrequirement is actually another example of a boundary condition, since the“points” are the boundaries of our system.x = ; q

x : ; q .x : ; q .c1x2 c1x2x : - q ,

x : q

E 6 0,c1x2 K 0.

e-axx : - q ,eax

c1x2c1x2eaxx : q ,- q 6 x 6 q .

c1x2 = Aeax + Be-ax

e-axeaxa = 1-2mE>U.

c–1x2 = a2 c1x2

c-2mE>U2E 6 0,

- q 6 x 6 q

c–1x2 = - ¢2mE

U2 ≤c1x2

U1x2 = 0,

E Ú 0.

TAYL07-203-247.I 1/4/03 1:03 PM Page 222

Section 7.7 • The Free Particle 223

*Although the function (7.75) doesn’t blow up as it does still suffer a lesserdifficulty, that it cannot be normalized since is infinite. However, thisdifficulty can be circumvented since we can build normalizable functions out of (7.75)using the Fourier integral.

1q-q ƒc1x2 ƒ2 dx

x : ; q ,

Let us next examine the possibility of states of our free particle withIn this case the Schrödinger equation can be written as

(7.73)

where

(7.74)

As before, the general solution of this equation is

(7.75)

The important point about this solution is that neither nor blowsup as Thus neither function suffers the difficulty that we encoun-tered with negative energies,* and, for any value of k, the function (7.75) is anacceptable solution, for any constants A and B.According to (7.74), this meansthat all energies in the continuous range are allowed. In particu-lar, the energy of a free particle is not quantized. Evidently, it is only when aparticle is confined in some way, that its energy is quantized.

To understand what the positive-energy wave functions (7.75) represent,it is helpful to recall the identities (Problem 7.16)

(7.76)

Substituting these expansions into the wave function (7.75), we can write

(7.77)

where you can easily find C and D in terms of the original coefficients A and B.It is important to note that since A and B were arbitrary, the same is true of Cand D; that is, (7.77) is an acceptable solution for any values of C and D.

The full, time-dependent wave function for the spatial func-tion (7.77) is

(7.78)

This is a superposition of two traveling waves, one moving to the right (withcoefficient C) and the other moving to the left (with coefficient D). If wechoose the coefficient then (7.78) represents a particle with definitemomentum to the right; if we choose then (7.78) represents a parti-cle with momentum of the same magnitude but directed to the left. If bothC and D are nonzero, then (7.78) represents a superposition of both momenta.

UkC = 0,Uk

D = 0,

°1x, t2 = c1x2e-ivt = Cei1kx -vt2 + De-i1kx +vt2

°1x, t2

c1x2 = Ceikx + De-ikx

sin kx =eikx - e-ikx

2i and cos kx =

eikx + e-ikx

2

0 … E 6 q

x : ; q .cos kxsin kx


k =22mE

U

c–1x2 = - ¢2mE

U2 ≤c1x2 = -k2 c1x2

E Ú 0.

TAYL07-203-247.I 1/4/03 1:03 PM Page 223

U

0

� �

a

(a)

0

(b)

(c)

U0

a

0

U0

a

FIGURE 7.8Three potential wells: (a) theinfinite well (7.79); (b) the finitesquare well (7.80); (c) a finiterounded well.


*Until about 30 years ago, one would have said that in this respect the perfectly rigidbox is totally unrealistic — a real bound system might require a large energy to pull itapart, but surely not an infinite amount.As we discuss in Chapter 18, we now know thatsubatomic particles like neutrons and protons are made up of sub-subatomic particlescalled quarks, and that an infinite energy is needed to pull them apart (that is, theycannot be pulled apart). Thus a potential energy function like (7.79) may be morerealistic than we had formerly appreciated.

7.8 The Nonrigid Box

So far, our only example of a particle that is confined, or bound, is the ratherunrealistic case of a particle in a perfectly rigid box, the infinite square well. Inthis section we apply the Schrödinger equation to a particle in the more realis-tic nonrigid box, a potential well of finite depth. This is a rather long section,but the ideas it contains are all fairly simple and are central to an understand-ing of many quantum systems.

The first step in applying the Schrödinger equation to any system is todetermine the potential-energy function. Therefore, we must first decidewhat is the potential energy, of a particle in a nonrigid box. For a rigidbox we know that

(7.79)

No finite amount of energy can remove the particle from a perfectly rigidbox.* For most systems, a more realistic assumption would be that there is afinite minimum energy needed to remove a stationary particle from the box. Ifwe call this minimum energy the potential-energy function would be

(7.80)

This potential, which we call the nonrigid box, is often called a finite square well.In Fig. 7.8(a) and (b) we plot the potential-energy functions (7.79) and (7.80).

Even the finite square well of Fig. 7.8(b) is somewhat unrealistic in thatthe potential energy jumps abruptly from 0 to at and For areal particle in a box (for example, an electron in a conductor) the potentialenergy changes continuously near the walls, more like the well shown inFig. 7.8(c).This well is sometimes called a rounded well.To simplify our discus-sion, we will suppose that the rounded well has exactly constant,

for and as shown in Fig. 7.8(c).In this section we want to investigate the energy levels of a particle

confined in a nonrigid box such as either Fig. 7.8(b) or (c). As one mightexpect, the properties of both wells are qualitatively similar.

Like the infinite well, the finite wells allow no states with if wedefine the zero of U at the bottom of the well. (See Problem 7.38.) An impor-tant difference between the infinite and finite wells is that in the finite well theparticle can escape from the well if This means that the wave func-tions for are quite similar to those of a free particle. In particular, thepossible energies for are not quantized, but we will not pursue thispoint here since our main interest is in the bound states, whose energies lie inthe interval 0 6 E 6 U0 .

E 7 U0

E 7 U0

E 7 U0 .

E 6 0

x 7 a,x 6 0U1x2 = U0 ,U1x2

x = a.x = 0U0

U1x2 = b0 0 … x … a

U0 x 6 0 and x 7 a

U0 ,

U1x2 = b0 0 … x … aq x 6 0 and x 7 a

U1x2,

TAYL07-203-247.I 1/4/03 1:03 PM 4

Section 7.8 • The Nonrigid Box 225

Turning points

Ene

rgy

0 b c ax

E

U0

U(x) FIGURE 7.9The classical turning points. Aclassical particle of energy Etrapped in the potential welloscillates back and forth, turningaround at the points and

where the kinetic energy is zero and hence E = U1x2.x = c,

x = b

Ene

rgy

0 b c ax

E

U0

U(x)

U�E positive U�E negative U�E positive

FIGURE 7.10The factor whichappears on the right side of theSchrödinger equation, is positive for

and and is negativefor b 6 x 6 c.

x 7 cx 6 b

3U1x2 - E4,

A classical particle moving in a finite well with energy in the intervalwould simply bounce back and forth indefinitely. In the square

well of Fig. 7.8(b) it would bounce between the points and Forthe rounded well, the points at which the particle turns around are deter-mined by the condition (since the kinetic energy must be zero atthe turning point where the particle comes instantaneously to rest). Thesepoints can be found graphically as in Fig. 7.9 by drawing a horizontal line atthe height representing the energy E. The points and at whichthis line meets the potential-energy curve, are the two classical turning points,and a classical particle with energy E simply bounces back and forth betweenthese points.

We now consider the Schrödinger equation,

for a quantum particle in a finite potential well. We seek values of E in therange which possess physically acceptable solutions. To under-stand when we should expect to find allowed energies, it is useful to examinethe general behavior of solutions of the Schrödinger equation.

Focusing attention on a particular value of E (with ), we candistinguish two important ranges of x: those x where the factor ispositive and those x where it is negative. The dividing points between these re-gions are the classical turning points and where (These were defined in Fig. 7.9 and are shown again in Fig. 7.10.) The regionswhere is positive are outside these turning points ( and

) and are often called the classically forbidden regions since a classicalparticle with energy E cannot penetrate there.The region where isnegative is the interval and is called the classically allowed region.The behavior of the wave function is quite different in these two regions.c1x2b 6 x 6 c

3U1x2 - E4x 7 cx 6 b3U1x2 - E4U1x2 = E.x = c,x = b

3U1x2 - E40 6 E 6 U0

0 6 E 6 U0 ,

c–1x2 =2m

U2 3U1x2 - E4c

x = c,x = b

E = U1x2x = a.x = 0

0 6 E 6 U0

TAYL07-203-247.I 1/4/03 1:03 PM Page 225


x

(a) (b)

(c) (d)

FIGURE 7.11If satisfies an equation of theform (7.81), it is concave away fromthe axis.

c1x2

Wave Functions Outside the Well

In the region where is positive, the Schrödinger equation has theform

(7.81)

where the “positive function” is In an interval whereis positive, this implies that is also positive and hence that is

increasing and is concave upward, as in Fig. 7.11(a) and (b). If isnegative, then (7.81) implies that is negative and hence that is con-cave downward, as in Fig. 7.11(c) and (d). In either case we see that isconcave away from the x-axis.

The behavior shown in Fig. 7.11 can be seen explicitly if we look in eitherof the regions or where The explicitform of is readily found by solving the Schrödinger equation. Since

we can define a number by the equation

(7.82)

and the Schrödinger equation becomes

(7.83)

As we have seen before, the general solution of this differential equationhas the form

(7.84)

with A and B arbitrary. As you can easily check, any function of this form isconcave away from the x-axis as in Fig. 7.11.

If we look on the left of the well then, as the expo-nential blows up and is physically unacceptable.Thus, in the region the solution (7.84) is physically acceptable only if

Similarly, in the region the general solution has the same form

(7.85)

and this is acceptable only if Thus, in these classically forbidden regionsthe physically acceptable wave functions die away exponentially as x goes to

Wave Functions within the Well

In the region where the Schrödinger equation has theform

(7.86)

we can argue that is concave toward the axis and tends to oscillate, as fol-lows: If is positive, then is negative and curves downward, asin Fig. 7.12(a); if is negative, the argument reverses and bendsc1x2c1x2 c1x2c–1x2c1x2 c1x2

c–1x2 = 1negative function2 * c1x2

U1x2 6 E,b 6 x 6 c,

; q .C = 0.

c1x2 = Ceax + De-ax x 7 a

x 7 a,B = 0.

x 6 0,e-axx : - q ,1x 6 02,

c1x2 = Aeax + Be-ax

c–1x2 = +a2 c1x2

2m

U2 3U0 - E4 = +a2

a1U0 - E2 7 0,c1x2 U1x2 = U0 = constant.x 7 a,x 6 0

c1x2c1x2c–1x2 c1x2c1x2 c¿1x2c–1x2c1x2 12m>U223U1x2 - E4.c–1x2 = 1positive function2 * c1x2

3U1x2 - E4

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Section 7.8 • The Nonrigid Box 227

x

(b)(a) (c)

FIGURE 7.12If satisfies an equation of theform (7.86), it curves toward theaxis and tends to oscillate.

c1x2

upward, as in Fig. 7.12(b). In either case, curves toward the axis. If theinterval is sufficiently wide, a function bending toward the axis willcross the axis and immediately start bending the other way, as in Fig. 7.12(c).Thus we can say that in the region the wave function tends tooscillate about the axis.

If the negative function in (7.86) has a large magnitude, then tendsto be large and curves and oscillates rapidly. Conversely, when the nega-tive function is small, bends gradually and oscillates slowly. This is allphysically reasonable: The negative function in (7.86) is proportional to

which is just the negative of the kinetic energy; according to deBroglie, large kinetic energy means short wavelength and hence rapid oscilla-tion, and vice versa.

For the case of a finite square well, where within thewell, the negative function of (7.86) is We can define a constant k bythe equation

(7.87)

and the Schrödinger equation becomes

(7.88)

The most general solution of this equation is

(7.89)

where F and G are arbitrary constants.Solving for the allowed energies of a particle in a finite square well is

rather messy. It involves starting with the general forms (7.84), (7.85), and(7.89), and then using the conditions that and must be continuousat the edges of the well to solve for the coefficients A, D, F, and G (B and C arealready known to be zero). It turns out there is no simple analytic solution tothis problem and a numerical solution is needed, as explored in Problem 7.67.However, as we show in the next subsection, one can determine the detailedqualitative behavior of the solutions without doing any calculations at all.

Searching for Allowed Energies

Now that we understand the qualitative behavior of solutions of theSchrödinger equation, let us return to our hunt for acceptable solutions. Weconsider the most general case of a rounded well of finite depth, and we startin the region where is constant (Fig. 7.10). There, the known,acceptable form is

c1x2 = Aeax 1x 6 02

U1x2x 6 0,

c¿1x2c1x2

c1x2 = F sin kx + G cos kx

c–1x2 = -k2 c

- 2m

U2 E = -k2

-2mE>U2.U0 = constant = 0

3U1x2 - E4,c1x2c1x2 c–1x2

b 6 x 6 c

b 6 x 6 cc1x2

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b c

Ene

rgy

E

U0

�

Curving away Oscillating Curving awayFIGURE 7.13The wave function oscillatesbetween the two turning points

and and curves awayfrom the axis outside them. Theexample shown is well behaved as

but blows up asx : + q .x : - q ,

x = cx = b

0 b c a

FIGURE 7.14When E is very small, the wavefunction bends too slowly inside thewell. The function that has the form

when blows up asx : + q .

x 6 0Aeax

1

2

3

4

0 a

FIGURE 7.15Solutions of the Schrödingerequation for four successively largerenergies. All four solutions have thewell-behaved form for but only number 3 is also wellbehaved as x : + q .

x 6 0,Aeax

(a) (b)

FIGURE 7.16Wave functions for the second andthird energy levels of a nonrigidbox.

When we move to the right, our solution will cease to have this explicit formonce starts to vary, but it will continue to bend away from the axis until xreaches the point b. At it will start oscillating and continue to do sountil where it will start curving away from the axis again. Thus, the gen-eral appearance of will be as shown in Fig. 7.13, with two regions where

bends away from the axis, separated by one region where oscillates.Figure 7.13 shows a solution that blows up on the right. We must now find outif there are any values of E for which the solution is well behaved both on theleft and right.

Let us begin a systematic search for allowed energies, starting with Eclose to zero. We will show first that with E sufficiently close to zero, an ac-ceptable wave function is impossible. We start with the acceptable wave func-tion in the region and follow to the right.When we reach theclassical turning point has a positive slope and starts to bendtoward the axis. But with E very small, bends very slowly. Thus, when wereach the second turning point the slope is still positive. With and both positive, continues to increase without limit, as shown in Fig. 7.14.Therefore, the wave function which is well behaved as blows up as

and we conclude that there can be no acceptable wave function withE too close to zero.

Suppose now that we slowly increase E, continuing to hunt for anallowed energy. For larger values of E, the kinetic energy is larger and, as wehave seen, bends more rapidly inside the well. Thus it can bend enoughthat its slope becomes negative, as shown in Fig. 7.15, curve 2. Eventually, itsvalue and slope at the right of the well will be just right to join onto the solu-tion that is well behaved as and we have an acceptable wave function(curve 3 in Fig. 7.15). If we increase E any further, will bend over too farinside the well and will now approach as like curve 4 in Fig. 7.15.Evidently, there is exactly one allowed energy in the range explored so far.

If E is increased still further, the wave function may bend over and backjust enough to fit onto the function that is well behaved as as inFig. 7.16(a). If this happens, we have a second allowed energy. Beyond this, wemay find a third acceptable wave function, like that in Fig. 7.16(b), and so on.

Figure 7.17 shows the first three wave functions for a finite square well,beside the corresponding wave functions for an infinitely deep square well of

x : q

x : q ,- qc1x2x : q ,

c1x2

x : + q ,x : - q

c1x2 c¿cx = c,c1x2c1x2x = b,c1x2x 6 0,Aeax

c1x2c1x2c1x2x = c,

x = b,U1x2

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Section 7.9 • The Simple Harmonic Oscillator 229

0 0a a

E1

E2

E3

U0

� �

FIGURE 7.17The lowest three energy levels andwave functions for a finite squarewell and for an infinite square wellof the same width. The horizontallines that represent each energylevel have been used as the axes fordrawing the corresponding wavefunctions.

the same width a. Notice the marked similarity of corresponding functions:The first wave function of the infinite well fits exactly half an oscillation intothe well, while that of the finite well fits somewhat less than half an oscillationinto the well since it doesn’t actually vanish until Similarly, the sec-ond function of the infinite well makes one complete oscillation, whereas thatof the finite well makes just less than one oscillation. For this reason, theenergy of each level in the finite well is slightly lower than that of the corre-sponding level in the infinite well.

It is useful to note that the ground-state wave function for any finite wellhas no nodes, while that for the second level has one node, and that for the nthlevel has nodes. This general trend (more nodes for higher energies)could have been anticipated since higher energy corresponds to a wave func-tion that oscillates more quickly and hence has more nodes.

The wave functions in Fig. 7.17 illustrate two more important points.First, the wave functions of the finite well are nonzero outside the well, in theclassically forbidden region. (Remember that a classical particle with energy

cannot escape outside the turning points.) However, since thewave function is largest inside the well, the particle is most likely to be foundinside the well; and since approaches zero rapidly as one moves awayfrom the well, we can say that our particle is bound inside, or close to, the po-tential well. Nevertheless, there is a definite nonzero probability of finding theparticle in the classically forbidden regions. This difference between classicaland quantum mechanics is due to the wave nature of quantum particles. Theability of the quantum wave function to penetrate classically forbiddenregions has important consequences, as we discuss in Section 7.10

A second important point concerns the number of bound states of the fi-nite well. With the infinite well, one can increase E indefinitely and always en-counter more bound states. With the finite well, however, the particle is nolonger confined when E reaches and there are no more bound states. Thenumber of bound states depends on the well depth and width a, but it is al-ways finite. The number of bounds states can be approximately computedusing a simple method described in Problem 7.45.

7.9 The Simple Harmonic Oscillator�

�The simple harmonic oscillator plays an amazingly important role in many areas ofquantum physics. Nevertheless, we won’t be using the results of this section again untilChapter 12 on molecules. Thus you could, if you wish, omit it on a first reading.

U0

U0 ,

c1x20 6 E 6 U0

n - 1

x = ; q .

TAYL07-203-247.I 1/4/03 1:03 PM Page 229


r0

(b)

r

x0

(a)

Pote

ntia

l ene

rgy

x

FIGURE 7.18(a) The potential energy of an idealsimple harmonic oscillator is aparabola. (b) The potential energyof a typical diatomic molecule (solidcurve) is well approximated by thatof an SHO (dashed curve) when r isclose to its equilibrium value r0 .

*Unless, of course, the first derivative happens to vanish at the equilibrium point,This occurs for the quartic potential with but we

will not consider this special case here.F1x2 r -x3,U1x2 r x4,F¿1x02 = 0.

As another example of a one-dimensional bound particle, we consider thesimple harmonic oscillator. The name simple harmonic oscillator (or SHO) isused in classical and quantum mechanics for a system that oscillates about astable equilibrium point to which it is bound by a force obeying Hooke’s law,

Familiar classical examples of harmonic oscillators are a mass suspend-ed from an ideal spring and a pendulum oscillating with small amplitude.Before we give any quantum examples, it may be worth recalling why theharmonic oscillator is such an important system. If a particle is in equilibriumat a point the total force F on the particle is zero at that is,If the particle is displaced slightly to a neighboring point x, the total force willbe approximately

(7.90)

with in this case. If is a point of stable equilibrium, the force is arestoring force; that is, is negative when is positive, and vice versa.Therefore, must be negative and we denote it by where k is calledthe force constant. With and Eq. (7.90) becomes

(7.91)

which is Hooke’s law. Thus, any particle oscillating about a stable equilibriumpoint will oscillate harmonically* for sufficiently small displacements

An important example of a quantum harmonic oscillator is the motion ofany one atom inside a solid crystal; each atom has a stable equilibrium positionrelative to its neighboring atoms and can oscillate harmonically about thatposition. Another important example is a diatomic molecule, such as HCl,whose two atoms can vibrate harmonically, in and out from one another.

In quantum mechanics we work, not with the force F, but with thepotential energy U. This is easily found by integrating (7.91) to give

(7.92)

if we take U to be zero at Thus, in quantum mechanics the SHO can becharacterized as a system whose potential energy has the form (7.92). Thisfunction is a parabola, with its minimum at as shown in Fig. 7.18(a).

It is important to remember that (7.91) and (7.92) are approximationsthat are usually valid only for small displacements from This point is illus-trated in Fig. 7.18(b), which shows the potential energy of a typical diatomicmolecule, as a function of the distance r between the two atoms. The moleculeis in equilibrium at the separation For r close to the potential energy iswell approximated by a parabola of the form but when ris far from is quite different. Thus for small displacements the mole-cule will behave like an SHO, but for large displacements it will not. This samestatement can be made about almost any oscillating system. (For example, the

U1r2r0 ,U1r2 = 1

2 k1r - r022,r0 ,r0 .

x0 .

x = x0 ,

x0 .

U1x2 = -Lx

x0

F1x2 dx = 12 k1x - x022

1x - x02.

F1x2 = -k1x - x02F1x02 = 0,F¿1x02 = -k

-k,F¿1x02x - x0F1x2 x0F1x02 = 0

F1x2 = F1x02 + F¿1x021x - x02

F1x02 = 0.x0 ;x0 ,

F = -kx.

TAYL07-203-247.I 1/4/03 1:03 PM Page 230


simple pendulum is well known to oscillate harmonically for small amplitudes,but not when the amplitude is large.) It is because small displacements fromequilibrium are very common that the harmonic oscillator is so important.

There is a close connection between the classical and quantum harmonicoscillators, and we start with a quick review of the former. If we choose ourorigin at the equilibrium position, then and the force is If theparticle has mass m, Newton’s second law reads

If we define the classical angular frequency

(7.93)

this becomes

which has the general solution

(7.94)

Thus, the position of the classical SHO varies sinusoidally in time with angularfrequency If we choose our origin of time, at the moment when theparticle is at and moving to the right, then (7.94) takes the form

(7.95)

The positive number a is the amplitude of the oscillations, and the particleoscillates between and In other words, the points arethe classical turning points. When the particle is at all of its energy ispotential energy; thus, and hence

(7.96)

As one would expect, the classical amplitude a increases with increasing energy.To find the allowed energies of a quantum harmonic oscillator, we must

solve the Schrödinger equation with given by

(7.97)

where we have again chosen Qualitatively, the analysis is very similarto that for the finite well discussed in the preceding section. For any choice ofE, the solutions will bend away from the axis outside the classical turningpoints and and will oscillate between these points. As before,most values of E do not produce an acceptable solution; only special values ofE produce a solution that satisfies the boundary conditions, that is, decays ex-ponentially as and Thus, the allowed energies are quantized.The potential energy (7.97) increases indefinitely as x moves away from the

x : q .x : - q

x = -ax = a

x0 = 0.

U1x2 = 12 kx2

U1x2

a = A2E

k

E = 12 ka2

x = a,x = ;ax = -a.x = a

x = a sin v c t

x = 0t = 0,vc .

x = a sin v c t + b cos v

c t

d2 x

dt2 = -v c

2x

vc = A km

ma = -kx

F = -kx.x0 = 0

TAYL07-203-247.I 1/4/03 1:03 PM Page 231

n � 0

�

x

n � 1

n � 2

Ene

rgy U(x) � kx2

21

FIGURE 7.19The first three energy levels andwave functions of the simpleharmonic oscillator.


TABLE 7.1The energies and wave functions of the first three levels of a quantumharmonic oscillator. The length b is defined as

n

0

1

2 A2¢1 - 2 x2

b2 ≤e-x2>2b252 Uvc

A1 x

b e-x2>2b23

2 Uvc

A 0 e-x2>2b21

2 Uvc

C1x2En

2U>mvc .

origin [see Fig. 7.18(a)], and the particle is therefore confined for all energies.In this respect, the SHO resembles the infinitely deep potential well, and wewould expect to find infinitely many allowed energies, all of them quantized.

An important difference between the SHO and most other potential wellsis that the Schrödinger equation for the SHO can be solved analytically. Thesolution is quite complicated and will not be given here. However, the answerfor the energy levels is remarkably simple: The allowed energies turn out to be

The quantity is the frequency defined in (7.93), of a classical oscilla-tor with the same force constant and mass. It is traditional to rewrite theallowed energies in terms of as

(7.98)

As anticipated, the allowed energies are all quantized, and there are levelswith arbitrarily large energies. A remarkable feature of (7.98) is that the ener-gy levels of the harmonic oscillator are all equally spaced. This is a propertythat we could not have anticipated in our qualitative discussion, but whichemerges from the exact solution.

The allowed energies of the harmonic oscillator are shown in Fig. 7.19,which also shows the corresponding wave functions (each drawn on the linerepresenting its energy). Notice the similarity of these wave functions to thoseof the finite well shown in Fig. 7.17. In particular, just as with the finite well, thelowest wave function has no nodes, the next has one node, and so on. Noticealso that the wave functions with higher energy spread out farther from just as the classical turning points move farther out when E increases.Finally, note that the ground state has a zero-point energy (equal to

) as required by the uncertainty principle, as discussed in connection withEq. (6.39).

The wave functions shown in Fig. 7.19 were plotted using the knownanalytic solutions, which we list in Table 7.1, where, for convenience, we haveintroduced the parameter

(7.99)b = A Umvc

12 Uvc

1n = 02x = ;a

x = 0,

En = An + 12 BUvc n = 0, 1, 2, Á

vc ,

vc ,1k>m

E =12

U A km

, 32

U A km

, 52

U A km

, Á

TAYL07-203-247.I 1/4/03 1:03 PM Page 232

(This parameter has the dimensions of a length and is, in fact, the half-width ofthe SHO well at the ground-state energy — see Problem 7.48.) The factors

and in the table are normalization constants*

Example 7.4

Verify that the wave function given in Table 7.1 is a solution of theSchrödinger equation with (For the cases and seeProblems 7.49 and 7.52.)

The potential energy is and the Schrödinger equation istherefore

(7.100)

Differentiating the wave function of Table 7.1, we find that

and

where the last three factors together are just Using (7.99) to replace b,we find

Replacing by in the first term, we get

which is precisely the Schrödinger equation (7.100), with as claimed.

The properties of the quantum SHO are nicely illustrated by the exam-ple of a diatomic molecule, such as HCl. A diatomic molecule is, of course, acomplicated, three-dimensional system. However, the energy of the moleculecan be expressed as the sum of three terms: an electronic term, correspondingto the motions of the individual electrons; a rotational term, corresponding torotation of the whole molecule; and a vibrational term, corresponding to thein-and-out, radial vibrations of the two atoms. Careful analysis of molecularspectra lets one disentangle the possible values of these three terms. In

E = 32 Uvc

cfl1 =

2m

U2 a12

kx2 -32

Uvcbc1

k>mvc2

cfl1 = ¢m2

v c

2x2

U2 -3mvc

U≤c1

c1 .

cfl1 = A1¢ -

3x

b3 +x3

b5 ≤e-x2>2b2 = ¢x2

b4 -3

b2 ≤A1 x

b e-x2>2b2

cœ1 = A1¢ 1

b-

x2

b3 ≤e-x2>2b2

c1 ,

c– =2m

U2 A12 kx2 - E Bc

U = 12 kx2,

n = 2,n = 0E = 32 Uvc .

n = 1

A2A0 , A1 ,


*The three functions in Table 7.1 illustrate what is true for all n, that the wave functionof the nth level has the form where is a certain polynomialof degree n, called a Hermite polynomial.

Pn1x2Pn1x2 exp1-x2>2b22,

TAYL07-203-247.I 1/4/03 1:03 PM Page 233


Lx0

U0

E

x1

FIGURE 7.21A rectangular barrier of height and width L.

U0

n � 0

n � 5Ene

rgy

U(r)

n � 10

n � 15

r

U0

FIGURE 7.20Vibrational levels of a typicaldiatomic molecule (right) and theSHO that approximates themolecule for small displacements(left). The first five or so levelscorrespond very closely; the higherlevels of the molecule aresomewhat closer together, and themolecule has no levels above theenergy U0. particular, the vibrational motion of the two atoms is one-dimensional (along

the line joining the atoms) and the potential energy approximates the SHOpotential. Therefore, the allowed values of the vibrational energy shouldapproximate those of an SHO, and this is amply borne out by observation, asillustrated in Fig. 7.20.

The molecular potential energy in Fig. 7.20 deviates from the para-bolic shape of the SHO at higher energies. Therefore, we would expect thehigher energy levels to depart from the uniform spacing of the SHO. This, too,is what is observed.

The observation of photons emitted when molecules make transitionsbetween different vibrational levels is an important source of information oninteratomic forces, as we discuss further in Chapter 12. These same photonscan also be used to identify the molecule that emitted them. For example, the

molecule emits infrared photons of frequency when it dropsfrom one vibrational level to the next.This radiation is used by astronomers tolocate clouds of molecules in our galaxy.

7.10 Tunneling�

�The phenomenon of tunneling is perhaps second only to quantization of energy as auniquely quantum phenomenon. It has many important applications, as we will men-tion here and in Chapters 14 and 17. However, we will not be using this material againuntil Chapter 14, so you could omit this section for now.

So far in this chapter, we have focused mainly on a particle that is confinedinside a potential well, such as the rigid box of Section 7.6, or the harmonicoscillator well of Section 7.9. We have seen that for these confined, or bound,particles, the Schrödinger equation implies quantization of the allowed ener-gies — one of the most dramatic differences between classical and quantummechanics. We have also discussed one example of an unconfined particle, thefree particle of Section 7.7, for which we found that the Schrödinger equationdoes not imply any quantization of energy. In this section we discuss a secondexample of an unconfined particle, for which the Schrödinger equation has im-plications almost as dramatic as the quantization of energy for bound states.Specifically, we will consider a particle whose potential energy has a “barrier”(to be defined in a moment). In classical mechanics such barriers are impene-trable and a particle on either side of the barrier cannot cross over to the otherside. In quantum mechanics we will find that the particle can “tunnel” throughthe barrier and emerge on the other side. This barrier penetration, ortunneling, has dramatic consequences in several natural phenomena, such asradioactive decay (Chapter 17), and is the basis of several modern electronicdevices, such as the scanning tunneling microscope (Chapter 14).

A simple example of a barrier is shown in Fig. 7.21, which is a plot of thepotential energy of an electron moving along an x-axis consisting of two iden-tical conducting wires separated by a small gap from to The gap betweenx1 .x0

H2

1.2 * 1014 HzH2

U1r2

TAYL07-203-247.I 1/4/03 1:03 PM Page 234

Section 7.10 • Tunneling 235

A

U(x) � 0

U(x) � U0

x0 x

FIGURE 7.22A rectangular barrier of infinitelength, with for

The wave function issinusoidal with amplitude A when

and decreasesexponentially when x 7 x0 .x 6 x0 ,

x0 6 x 6 q .U1x2 = U0

the two conductors could be just an air gap, or it could be a thin layer of dirt ina poorly made electrical connection. In either case the gap forms a barrier ofthickness between the two wires. Inside either wire the potentialenergy is a constant, which we can take to be zero, but in the barrier it has ahigher value We will be interested in the case that the particle’s ener-gy is less than the barrier height, as shown in Fig. 7.21. Underthese conditions a classical particle is excluded from the barrier since with

its kinetic energy between and would be negative — an impossi-bility in classical mechanics. Thus, a classical particle could approach thebarrier from the left (for example) but would inevitably rebound straight backon arrival at it could certainly not emerge on the right of unless we gaveit more energy. To see in detail what happens to a quantum particle when ithits this barrier, one must solve the Schrödinger equation with this potential-energy function Although the solution is not especially difficult, we donot need to go through it since we can already understand its main featuresfrom our discussion of the finite square well in Section 7.8.

Consider first a barrier whose length L is infinite. This extreme case isshown in Fig. 7.22. It is sometimes called a potential step and is precisely thesame as the right wall of the finite square well of Section 7.8 (Fig. 7.17). InFig. 7.22 we have also shown the wave function for an energy Tothe left of the kinetic energy is positive and is an oscil-lating sinusoidal wave, some combination of and To the right of

is negative and is a decreasing exponential with the form

(7.101)

where, you may recall [see (7.82) if you don’t],

(7.102)

As we emphasized in Section 7.8, the quantum particle has a nonzero probabil-ity of being found in the classically forbidden region where is negative.

In the infinitely long barrier of Fig. 7.22, goes steadily to zero as xincreases. But if the barrier has finite length, the situation is as shown inFig. 7.23. Just as in Fig. 7.22, is sinusoidal (with amplitude ) to the leftALc1x2

c1x2 E - U

a = A2m1U0 - E2U2

c1x2 = Be-ax

c1x21E - U2x0 ,cos kx.sin kx

c1x2K = 1E - U2x0 ,E 6 U0 .c1x2

U1x2.

x1x0 ;

x1x0E 6 U0

0 6 E 6 U0 ,U0 7 0.

L = x1 - x0

Lx0

AL

AR

x1

e� x�

FIGURE 7.23Wave function for a barrier of finitelength. On the left is sinusoidal, with amplitude inthe barrier it decreasesexponentially; on the right

it is sinusoidal again, withamplitude AR.1x 7 x12,

AL;c1x21x 6 x02,

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*Our discussion here is a little oversimplified. Strictly speaking, we should be usingwave functions that travel in a definite direction, rather than sinusoidal functions.Nevertheless, the result (7.103) is a good enough approximation.† Here, too, we are oversimplifying. In Fig. 7.23 you can see that the value of theexponential wave function at is not quite the same as the amplitude (Similarly,the value at is not quite equal to ) Nevertheless, under the conditions of interest,(7.104) is a satisfactory approximation.

AR.x1

AL.x0

eikx

of and just as in Fig. 7.22, it decreases exponentially within the barrierBut when we reach the barrier stops and, once again,

is positive. Therefore, before has decreased to zero, it starts os-cillating again with amplitude The probability that the particle is to theleft of approaching the barrier, is proportional to the probability thatit is to the right of the barrier is proportional to Therefore, there is anonzero probability P that a particle striking the barrier from the left willescape to the right*:

(7.103)

From Fig. 7.23, we see that is less than because of the exponentialdecrease of within the barrier. Specifically†

(7.104)

Therefore, the probability that a particle which strikes the rectangular barri-er of Fig. 7.23 with energy will tunnel through and emerge on theother side is

(7.105)

where is given by (7.102).The probability (7.105) that a quantum particle will tunnel through the

classically impenetrable barrier depends on the two variables: as given by(7.102), and L, the thickness of the barrier. In many applications the product

is very large and the probability (7.105) is therefore very small. Neverthe-less, if the particle keeps bumping against the barrier, it will eventually passthrough it. This is what happens in the decay of certain radioactive nuclei, aswe will see in Section 17.10.

A modern application of quantum tunneling is the scanning tunnelingmicroscope (STM). Here the surface of a sample is explored by measuring theelectric current between the surface and a conducting probe that is scannedacross the surface, just above it at a fixed height. This current is possible onlybecause of quantum tunneling and the magnitude of the current is proportion-al to the probability (7.105) of tunneling through the barrier. This probabilityis sensitively dependent on L, the barrier thickness, that is, the distance be-tween the probe and the surface. This means that by measuring the current,one can get a sensitive picture of the surface’s topography, as we describe indetail in Chapter 14. (Meanwhile, see the beautiful picture of the surface ofgraphite made with an STM in Fig. 13.6.)

There are many other examples of quantum tunneling, but the two men-tioned here are among the most important: The first is historically important

a

aL

a,

a

P L e-2aL

E 6 U0

AR

ALL

e-ax1

e-ax0= e-aL

c1x2 ALAR

P = ¢AR

AL≤2

AR2 .AL

2 ;x0 ,AR.

c1x21E - U2 x11x0 6 x 6 x12.x0 ;

TAYL07-203-247.I 1/4/03 1:03 PM Page 236

as the first application of quantum mechanics to a subatomic process, and thesecond is the basis for a modern and widely used experimental technique. Cu-riously, barrier penetration will not play an important role in our furtherdevelopment of quantum theory, and our focus for the next several chapterswill be on bound states similar to those described in Sections 7.4 through 7.9.However, we will return to barrier penetration in Chapters 14 and 17.

Example 7.5

Consider two identical conducting wires, lying on the x axis and separated byan air gap of thickness (that is, a few atomic diameters). An elec-tron that is moving inside either conductor has potential energy zero, where-as in the gap its potential energy is Thus the gap is a barrier of thetype illustrated in Fig. 7.21. The electron approaches the barrier from the leftwith energy E such that that is, the electron is 1 eV belowthe top of the barrier.What is the probability that the electron will emerge onthe other side of the barrier? How different would this be if the barrier weretwice as wide?

The required probability is given by Equation (7.105) with

Thus and the transmission probability is or about 0.004%. This probability is not large, but if we send

enough electrons at the barrier, some will certainly get through. If we double L, this will give a transmission probability — a dramatically smaller result. This illustrates the extreme sensitivity of thetransmission probability to the width of the gap — a property that is exploitedin the scanning tunneling microscope.

7.11 The Time-Dependent Schrödinger Equation�

�The time-dependent Schrödinger equation is one of the principal cornerstones ofquantum theory. Nevertheless, it is used surprisingly little in practical calculations, andwe will use it only in Chapter 11. Thus, you could skip this section on a first readingwithout significant loss of continuity.

As we mentioned at the start of this chapter, there are in fact two Schrödingerequations — the time-independent Schrödinger equation, which has been ourmain subject up to now, and the time-dependent Schrödinger equation.The lat-ter is one of the basic equations of quantum mechanics: It determines the timeevolution of any quantum system and is the analog of Newton’s second law,

in classical mechanics. Among all the many solutions of the time-dependent Schrödinger equation, the most important are the stationary states(or states of definite energy); and for these, the time-dependent Schrödingerequation reduces to the time-independent equation, as we will prove in thissection. Because the stationary states are by far the most important states inpractice, it is the time-independent Schrödinger equation that one uses in

F = ma,

P¿ = e-4aL = e-20.4 = 1.4 * 10-9

3.7 * 10-5P = e-2aL = e-10.2 =aL = 5.1

L 42 * 15 * 105 eV2 * 11 eV2197 eV # nm

L 5.1 nm-1

a = 42m1U0 - E2U

= 42mc21U0 - E2Uc

U0 - E = 1 eV;

U0 7 0.

L = 1 nm,

Section 7.11 • The Time-Dependent Schrödinger Equation 237

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almost all practical problems. Nevertheless, you should have some acquain-tance with the time-dependent equation, and that is the purpose of this section.

Before we write down the time-dependent Schrödinger equation, it ishelpful to note that the time-independent Schrödinger equation (7.39) can berewritten (as you should check) in the form

(7.106)

This is the equation that (we have claimed) determines the possible energies Eand the corresponding spatial wave functions for a particle whose poten-tial energy is The full wave function for the same particle, including itstime dependence, is denoted and the time dependence of any suchwave function is determined by the time-dependent Schrödinger equation,

(7.107)

There is no way to prove that this equation is correct. Just like its counterpartin classical mechanics, it is ultimately an axiom, an assumption, that is

justified by the success of its many predictions. Nevertheless, we can motivatethe form of this equation with the following plausibility argument.

We are searching for a wave equation, a differential equation that wavefunctions must obey, and we can use the free-particle wave function as a testcase. We assume that a free particle has the wave function and has a (purely kinetic) energy given by Any candidatewave equation must have this wave function as its solution for the special case

Guided by the energy equation we seek differentialoperators that produce factors of and and we find that the operators and do just that:

And so we can construct an equation,

which can be viewed as a reflection of the relation for a free particle.We have just demonstrated that the equation

is correct for the special case of a free particle (potential ). Schrödingergeneralized this equation by adding a potential energy term, producing hisfamous result:

- iU 00t

° = - U2

2m

02

0x2 ° + U1x2°

U = 0

- iU 00t

° = - U2

2m

02

0x2 °

E = K

- iU 00t

3ei1kx -vt24 = - U2

2m

02

0x2 3ei1kx -vt24

00t

ei1kx -vt2 = - ivei1kx -vt2 and 02

0x2 ei1kx -vt2 = -k2 ei1kx -vt2

02>0x20>0tk2,v

Uv = U2 k2>2m,U1x2 = 0.

E = Uv = U2 k2>2m.

°1x, t2 = ei1kx -vt2

F = ma

iU 00t

°1x, t2 = ¢ - U2

2m

02

0x2 + U1x2≤°1x, t2

°1x, t2U1x2. c1x2

Ec1x2 = ¢ - U2

2m d2

dx2 + U1x2≤c1x2

TAYL07-203-247.I 1/4/03 1:03 PM Page 238


*There is a brief review of partial derivatives in Section 8.2. If your knowledge of thiskind of derivative needs a little refreshing, you might want to read that section beforecoming back to this.

Thus, Schrödinger’s time-dependent equation can be viewed as an energyequation: The total energy is kinetic plus potential,

Notice that the time-dependent Schrödinger equation involves deriva-tives with respect to t and x. Since depends on two variables, these de-rivatives are partial derivatives, and the time-dependent Schrödinger equationis a partial differential equation.* It has a vast array of solutions, some of whichare very complicated, but by far its most important solutions are the stationarystates, and these are relatively simple. Recall that we claimed earlier that theseare standing waves, with the form (7.11)

(7.108)

and that the spatial wave function had to satisfy the time-independentSchrödinger equation. We can now prove these claims.

Let us suppose first that the function (7.108) does satisfy the time-dependent Schrödinger equation (7.107). Substituting into (7.107), we find that

Recognizing that (the de Broglie hypothesis) and canceling thecommon factor of we see that

(7.109)

which is precisely the time-independent Schrödinger equation (7.106). Thus, ifthe standing wave (7.108) satisfies the time-dependent Schrödinger equation,its spatial part must satisfy the time-independent Schrödinger equation. Con-versely, if satisfies the time-independent equation (7.109), we can followthe same steps backward and conclude that the standing wave (7.108) satisfiesthe time-dependent equation.

Although stationary states, given by the standing waves (7.108), aresolutions — and very important solutions — of the time-dependentSchrödinger equation, they are by no means the only solutions. To see this,suppose that we have two stationary states,

(7.110)

where and are the corresponding energies. To be definite,we could consider and to be the lowest two wave functions of the rigidbox as given in Equation (7.60) with and 2. Because each of thesefunctions satisfies the time-dependent Schrödinger equation, it is easy to seethat the same is true of any linear combination of the form

(7.111)

for any two fixed numbers and (That is, the superposition principleapplies to the time-dependent Schrödinger equation.) To prove this, just

b.a

°1x, t2 = a°11x, t2 + b°21x, t2

n = 1c2c1

Uv2 = E2Uv1 = E1

°11x, t2 = c11x2e-iv1 t and °21x, t2 = c21x2e-iv2 t

c1x2

Ec1x2 = ¢ - U2

2m d2

dx2 + U1x2≤c1x2e-ivt,Uv = E

iU 00t

°1x, t2 = Uvc1x2e-ivt = ¢ - U2

2m

02

0x2 + U1x2≤c1x2e-ivt

c1x2°1x, t2 = c1x2e-ivt

°1x, t2E = K + U.

TAYL07-203-247.I 1/4/03 1:03 PM Page 239

substitute the proposed solution into the left side of the time-dependentSchrödinger equation:

The special feature of the separate stationary-state wave functions of(7.110) is that the corresponding probability densities are independent of t(that is, stationary). This is because when we form the time-dependent exponential factor drops out (since ). The two probabili-ty densities (for the case of the lowest two stationary states of the rigid box)are shown in Fig. 7.24, where each picture is frozen and unchanging in time.

It is perhaps a little surprising that when we form a linear combination oftwo stationary-state wave functions, as in (7.111), the resulting state is not astationary state. To see this, we can rewrite (7.111) as

(7.112)

where, in the second line, we have factored out the first exponential andreplaced in the second exponential by which we have abbreviatedas When we form the absolute value squared of thefirst exponential disappears in the now-familiar way, but the second does not.

(7.113)

It is fairly easy to see that since one of the terms on the right is oscillatorywhile the other is not, the right side is definitely not independent of time. Theonly exception to this statement is if one of the two coefficients or is zero,in which case the state is just one of the original stationary states. Rather thanprove these statements in general, we examine one particular case in thefollowing example. (But see also Problem 7.59.)

Example 7.6

For a particle in a rigid box, consider the nonstationary state with wave func-tion (7.112) for the special case that (The particular value

is chosen to guarantee that is normalized — see Problem 7.58.)Evaluate and plot it for several different times.

With the given values of and the wave function is

(7.114)

and from (7.113), we find

(7.115)ƒ °1x, t2 ƒ2 = 12 ƒc11x2 + c21x2e-iv21 t ƒ2

°1x, t2 =112

3c11x2e-iv1 t + c21x2e-iv2 t4

b,aƒ °1x, t2 ƒ2

°1>12a = b = 1>12.

ba

ƒ °1x, t2 ƒ2 = ƒac11x2 + bc21x2e-iv21 t ƒ2

°1x, t2,v21 = 1v2 - v12.v2 - v1 ,v2

e-iv1 t

= e-iv1 t3ac11x2 + bc21x2e-iv21 t4 °1x, t2 = ac11x2e-iv1 t + bc21x2e-iv2 t

ƒe-ivt ƒ = 1ƒ °1x, t2 ƒ2,

= ¢ - U2

2m

02

0x2 + U1x2≤°1x, t2.

= a¢- U2

2m

02

0x2 + U1x2≤°11x, t2 + b ¢- U2

2m

02

0x2 + U1x2≤°21x, t2 iU

00t

°1x, t2 = aiU 00t

°11x, t2 + biU 00t

°21x, t2


0 a

x

n � 1

��1(x, t)�2

0 a

x

n � 2

��2(x, t)�2

FIGURE 7.24The probability density for each of the lowest two energylevels of the rigid box. Becausethese are stationary states, theprobability densities are constant in time.

ƒ °1x, t2 ƒ2

TAYL07-203-247.I 1/4/03 1:03 PM Page 240


0 a

t � 0

/8�

/4�

3 /8�

/2�

FIGURE 7.25The probability density for the non-stationary state (7.114)at five equally spaced times.

ƒ °1x, t2 ƒ2

x

n � 1

a

1(x)�

x

n � 2

a

2(x)�

FIGURE 7.26The two lowest stationary-statewave functions and forthe rigid box.

c21x2c11x2

*If you’re not familiar with this result, check it using Euler’s formula (7.10).

where the explicit forms of and can be found in (7.60) andwith the energies given by (7.23). The ab-

solute value on the right can be evaluated in several ways. If you think ofit as then

(7.116)

where in the first line we used the Euler relation for and in the second weused Thus,

(7.117)

The most obvious thing about this result is that, because of the factorthe probability density does vary with time. Notice that in this case

the time dependence is actually periodic, with period We have plotted (7.117) for five equally spaced times, apart, in

Fig. 7.25, where it is quite clear that the probability density is moving aroundinside the box. At the probability is concentrated in the left half of thebox, with almost no probability of finding the particle on the right. As tadvances, the probability shifts across to the right, and by the time theparticle is almost certainly in the right half. If we had shown a few more plots(at the same spacing of t), they would have shown the probability shifting backto the left, and by the density would be the same as at again.

Some features of Fig. 7.25 can be understood if we examine the wavefunctions carefully. At the term inside the absolute value signs in(7.115) is just If you look at the plots in Fig. 7.26, you will seethat and are both positive on the left of the box but have oppo-site signs on the right.Therefore, when we add them, they interfere construc-tively on the left, but destructively on the right. Thus, at theprobability density is large on the left and very small on the right,as in the first of the plots of Fig. 7.25. By the time the exponentialfactor in (7.115) is* and the term inside the absolute value signs is

Referring again to Fig. 7.26, you will see that now the twowaves interfere destructively on the left and constructively on the right, asseen in the last of the plots in Fig. 7.25.

Looking back at (7.114), we can now understand why any linear combi-nation of stationary-state wave functions is itself not stationary: The timedependence of each separate term in (7.114) is contained in a simple expo-nential factor (which disappears when we form or but thesetwo factors have different frequencies and Thus, as time goes by, theyvary at different rates and the interference between the two terms in the sumkeeps changing, as is evident in Fig. 7.25.

The time dependence of nonstationary states, as determined by the time-dependent Schrödinger equation, is fascinating and theoretically very impor-tant. Nevertheless — and perhaps fortunately for us — it plays no role in themajority of elementary quantum problems. For the most part, our main con-cern will be solving the time-independent Schrödinger equation to find theallowed energies and the corresponding stationary-state wave functions for

v2 .v1

ƒ °2 ƒ2 ),ƒ °1 ƒ2e-ivt

c11x2 - c21x2.e-ip = -1

t = t>2,ƒ °1x, 02 ƒ2

t = 0,

c21x2c11x2c11x2 + c21x2.

t = 0,

t = 0t = t

t = t>2,

t = 0,

t>8t = 2p>v21 .cos1v

21 t2,

ƒ °1x, t2 ƒ2 = 12 3c11x22 + c21x22 + 2c11x2c21x2 cos1v

21 t24.cos2 + sin2 = 1.

e-iu

= 1u + v cos u22 + 1v sin u22 = u2 + v2 + 2uv cos u ƒu + ve-iu ƒ2 = ƒu + v cos u - iv sin u ƒ2

ƒu + ve-iu ƒ2,

v21 = 1v2 - v12 = 1E2 - E12>Uc2c1

TAYL07-203-247.I 1/4/03 1:03 PM Page 241


electrons in atoms, molecules, and solids and for nucleons in nuclei. The oneexception to this statement concerns the following problem, which we willaddress in Chapter 11: If the solutions of the time-independent Schrödingerequation were really perfectly stationary states, then they would representstates of an atom (or other system) that never change.This would imply that anatom in one of its energy levels would never make a transition to anotherenergy level — in flagrant contradiction to the observed transitions made bymost atoms. As we will see in Chapter 11, atomic transitions result fromvarious outside influences, such as electromagnetic fields, which mean that thestationary states calculated for the isolated atom are not in fact exactlystationary states for the real system, which must include the outside influences.To calculate how these influences cause the observed transitions, we will needto use the time-dependent Schrödinger equation.

CHECKLIST FOR CHAPTER 7CONCEPT DETAILS

Superposition principle If and are possible waves, so is

Classical standing wave (7.5) (does not move right or left)Nodes Points at which the disturbance remains zero for all times

Time dependence of quantum standing waves (7.11) (fixed spatial part, oscillatorytime part)

Time-independent Schrödinger equation (7.39)

Particle in a one-dimensional rigid box Moves freely inside but cannot escape outside

Boundary conditions (7.16)

Allowed energies (7.23)

Wave functions (7.60)

Normalization condition (7.55)

Behavior of wave functions

In classically forbidden zone Curves away from axis, exponentially as

In classically allowed zone Curves toward the axis and oscillates sinusoidally

Allowed energies and well-behaved functions Section 7.8

Simple harmonic oscillator� (7.97)

Allowed energies (7.98)

Tunneling� Ability of a quantum particle to penetrate into aclassically forbidden zone

Probability of tunneling (7.105)

Time-dependent Schrödinger equation� (7.107)iU 00t

°1x, t2 = ¢ - U2

2m

02

0x2 + U1x2≤°1x, t2P = e-2aL

En = An + 12 BUvc

U = 12 kx2

1E 7 U2x : ; q

c: 01E 6 U2

1

ƒc1x2 ƒ2 dx = 1

c1x2 = 22>a sin1npx>a2En = n21p2

U2>2ma22c102 = c1a2 = 0

0 … x … a

d2 c

dx2 =2m

U2 3U1x2 - E4c

°1x, t2 = c1x2e-ivt

°1x, t2 = c1x2 cos vt

A°1 + B°2°2°1

TAYL07-203-247.I 1/4/03 1:03 PM Page 242

Problems for Chapter 7 243

PROBLEMS FOR CHAPTER 7

SECTION 7.2 (Classical Standing Waves)

7.1 • A string is oscillating with wave function

(7.118)

with and For each of the times 0.05, and 0.07 s, sketchthe string for

7.2 • A standing wave of the form (7.118) has amplitude2 m, wavelength 10 m, and period Writedown the expression for

7.3 • A string is clamped at the points andIt is oscillating sinusoidally with amplitude

2 cm, wavelength 20 cm, and frequency Write an expression for its wave function

7.4 • For the standing wave of Eq. (7.118) calculate thestring’s transverse velocity at any fixed position x.[That is, differentiate (7.118) with respect to t, treat-ing x as a constant.] At certain times the string’s dis-placement is zero for all x, and an instantaneoussnapshot of the string would show no wave at all.Sketch the string at such a moment, and indicate, withseveral small arrows, the velocity of several points onthe string at that moment.

7.5 • In Fig. 7.27 are sketched three successive snapshotsof a standing wave on a string. The first (solid) curveshows the string at maximum displacement. Werethese snapshots taken at equally spaced times? If not,make a sketch in which they were. (Use the same firstand third curves.)

y1x, t2.f = 40 Hz.x = 30 cm.

x = 0

y1x, t2. T = 2 s.

0 … x … 10 cm.t = 0,

v = 10p rad>s.k = 0.2p rad>cm,A = 3 cm,

y1x, t2 = A sin kx cos vt

7.6 • Figure 7.27 shows a standing wave on a string. Inany position other than the dotted one, the string haspotential energy because it is stretched (compared tothe straight, dotted position). Let denote this po-tential energy at the position of the solid curve(which shows the maximum displacement). What isthe potential energy when the string reaches the posi-tion of the dotted line? What is the kinetic energy atthis position?

7.7 • Consider a standing wave on a string, clamped at thepoints and It is oscillating with am-plitude 3 cm and wavelength 80 cm, and its maximumtransverse velocity is Write an expression forits displacement as a function of x and t.

7.8 • Rewrite the expression (7.118) for a standing wave,eliminating the wave number k and angular frequen-cy in favor of the wavelength and period T. Uselv

y1x, t260 cm>s.

x = 40 cm.x = 0

U0

your answer to explain clearly why the wave repeatsitself if we move from any point to (with tfixed). Show similarly what happens if we increase tfrom any to

7.9 •• A string is oscillating with wave function of theform (7.118) but with and

(a) Sketch two complete wavelengthsof the wave at each of the times and

(b) For a fixed value of x, differentiate(7.118) with respect to t to give the string’s transversevelocity at any position x. Graph this velocity as afunction of x for each of the two times in part (a).

SECTION 7.3 (Standing Waves in QuantumMechanics)

7.10 • Prove that any complex number (with xand y real) can be written as (with r and real). Give expressions for x and y in terms of r and and vice versa. [Hint: Use Euler’s formula, (7.10).]

7.11 • (a) Show for any complex number that where is the complexconjugate of z. (b) Hence, prove the useful identity,

(c) Show for any standing wave,with the form (7.11), and hencethat the probability density is independent of time.

7.12 • Consider a complex function of time,where r is a real constant. (a) Write z in terms of itsreal and imaginary parts, x, and y, and show that theyoscillate sinusoidally and 90° out of phase. (b) Show

that is constant.

7.13 •• We claimed in connection with Eq. (7.7) that thegeneral (real) sinusoidal wave has time dependence

(7.119)

Another way to say this is that the general sinusoidaltime dependence is

(7.120)

(a) Show that these two forms are equivalent; that is,prove that the function (7.119) can be expressed inthe form (7.120) and vice versa. Give expressions fora and b in terms of A and and vice versa. (Remem-ber the trig identities in Appendix B.) (b) Show thatby changing the origin of time (that is, rewritingeverything in terms of with a suit-ably chosen constant) you can eliminate the constant

from (7.120) [that is, rewrite (7.120) as ]

7.14 •• The function is defined for any z, real or com-plex, by its power series

Write down this series for the case that z is purelyimaginary, Note that the terms in this seriesare alternately real and imaginary. Group together all

z = iu.

ez = 1 + z +z2

2!+

z3

3!+ Á

ez

A sin vt¿.f

t¿ = t + constant,

f

A sin1vt + f2

a cos vt + b sin vt

ƒz ƒ = 3x2 + y2

z = re-ivt,

ƒ °1x, t2 ƒ = ƒc1x2 ƒ ,ƒzw ƒ = ƒz ƒ # ƒw ƒ .

z* = x - iyƒz ƒ2 = zz*z = x + iy,

u,uz = reiu

z = x + iy

t = 0.1 s.t = 0.05 s

v = 10p rad>s.k = 1 rad>cm,A = 2.5 cm,

t0 + T.t0

x0 + lx0

FIGURE 7.27(Problems 7.5 and 7.6)

TAYL07-203-247.I 1/4/03 1:03 PM Page 243


the real terms and all the imaginary terms, and provethe important identity, called Euler’s formula,

(7.121)

[Hint: You will need to remember the power seriesfor and given in Appendix B. You maywonder whether it is legitimate to regroup the termsof an infinite series as recommended here; for powerseries like those in this problem, it is legitimate.]

7.15 •• Use the identity to prove the trigidentities for and

7.16 •• Use the result (7.121) to prove that

SECTION 7.4 (The Particle in a Rigid Box)

7.17 • Find the lowest three energies, in eV, for an electronin a one-dimensional box of length (about the size of an atom).

7.18 • Find the lowest three energies, in MeV, of a protonin a one-dimensional rigid box of length (atypical nuclear size).

7.19 • What is the spacing, in eV, between the lowest twolevels of an electron confined in a one-dimensionalwire of length 1 cm?

7.20 • Sketch the energy levels and wave functions for thelevels for a particle in a one-dimensionalrigid box. (See Fig. 7.5.)

7.21 • For the ground state of a particle in a rigid box, wehave seen that the momentum has a definite magni-tude but is equally likely to be in either direction.This means that the uncertainty in p is Theuncertainty in position is Verify that theseuncertainties are consistent with the Heisenberguncertainty principle (6.34).

SECTION 7.6 (The Rigid Box Again)

7.22 • Prove that the function satisfiesthe equation for any two constants A and B.

7.23 • We saw that the coefficients A and B in the wavefunction for a negative-energy state in the rigid boxwould have to satisfy and

Show that this is possible only ifand hence, that there are no negative-

energy states.

7.24 • Prove that the function satis-fies the equation for any two constants Aand B.

7.25 •• We have seen that second-order differential equa-tions like the Schrödinger equation have two indepen-dent solutions. Consider the fourth-order equation

where is a positive constant. Provethat each of the four functions and is a solu-tion. (It is a fact — though harder to prove — that anysolution of the equation can be expressed as a linearcombination of these four solutions.)

e;ibxe;bxbd4

c>dx4 = b4 c,

c– = -k2 cc = Aeikx + Be-ikx

A = B = 0Aeaa + Be-aa = 0.

A + B = 0

c– = a2 c

c = Aeax + Be-ax

¢x L a>2.¢p L Uk.

Uk

n = 5, 6, 7

a = 5 fm

a = 0.2 nm

cos u =eiu + e-iu

2 and sin u =

eiu - e-iu

2

sin1u + f2.cos1u + f2ei1u+f2 = eiu eif

sin ucos u

eiu = cos u + i sin u

7.26 •• A second-order differential equation like theSchrödinger equation has two independent solu-tions and These two solutions can bechosen in many ways, but once they are chosen, anysolution can be expressed as a linear combination

(where A and B are constants, realor complex). (a) To illustrate this property, considerthe differential equation where k is aconstant. Prove that each of the three functions

and is a solution. (b) Show thateach can be expressed as a combination of the othertwo.

7.27 •• Many physical problems lead to a differentialequation of the form

where a, b, c are constants. [An example was given in(7.30).] (a) Prove that this equation has two solutionsof the form where is either solution ofthe quadratic equation (b) Provethat any linear combination of these two solutions isitself a solution.

7.28 •• Consider the second-order differential equation

where g, and h are known functions of x. Prove thatif and are both solutions of this equation,the linear combination is also a so-lution for any two constants A and B — the resultknown as the superposition principle.

7.29 •• Show that the integral which appears in the nor-malization condition for a particle in a rigid boxhas the value

[Hint: Use the identity for in terms of given in Appendix B.]

7.30 •• (a) Write down and sketch the probability distribu-tion for the second excited state of aparticle in a rigid box of length a. (b) What are the mostprobable positions, (c) What are the probabilitiesof finding the particle in the intervals and

7.31 •• Answer the same questions as in Problem 7.30 butfor the third excited state

7.32 •• If a particle has wave function the probabili-ty of finding the particle between any two points band c is

(7.122)

For a particle in the ground state of a rigid box, calcu-late the probability of finding it between and

(where a is the width of the box). Use thehint in Problem 7.29.x = a>3 x = 0

P1b … x … c2 = Lc

b ƒc1x2 ƒ2 dx

c1x2,1n = 42.

30.75a, 0.76a4? 30.50a, 0.51a4xmp ?

1n = 32ƒc1x2 ƒ2

cos 2usin2 u

La

0 sin2anpx

ab dx =

a

2

Ac11x2 + Bc21x2c21x2c11x2

f,

f1x2c–1x2 + b1x2c¿1x2 + h1x2c1x2 = 0

ag2 + bg + c = 0.gc1x2 = egx,

ac–1x2 + bc¿1x2 + cc1x2 = 0

eikxcos kx,sin kx,

c– = -k2 c

Ac11x2 + Bc21x2

c21x2.c11x2

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7.33 •• Evaluate the integral (7.65) to give the average re-sult (the expectation value) found when the posi-tion of a particle in the ground state of a rigid box ismeasured many times. [Hint: Rewrite interms of and use integration by parts.]

7.34 •• Do Problem 7.33 for the case of the second excitedstate of a rigid box.

7.35 ••• Consider a particle in the ground state of a rigidbox of length a. (a) Evaluate the integral (7.122) togive the probability of finding the particle between

and for any (b) What does yourresult give when Explain. (c) What if (d) What if (e) The answer to part (c) is halfthat for part (b), whereas that to part (d) is not halfthat for part (c). Explain.

SECTION 7.8 (The Nonrigid Box)

7.36 • Consider the potential-energy function(a) Sketch U as a function of x. (b) For

a classical particle of energy E, find the turningpoints in terms of E and k. (c) If we double the ener-gy, what happens to the length of the classicallyallowed region?

7.37 •• Consider the potential-energy function (theGaussian well)

where and a are positive constants. (a) Sketch(b) For find the classical turning

points (in terms of and a).

7.38 •• Give an argument that a particle moving in eitherof the finite wells of Figs. 7.8(b) and (c) can have nostates with [Hint: Remember that whenever

is positive, must bend away from theaxis; show that a wave function that is well behaved as

(that is, behaves like ) necessarily blowsup as ]

7.39 •• Make sketches of the probability density for the first four energy levels of a particle in a non-rigid box. Draw next to each graph the correspondinggraph for the rigid box.

7.40 •• Consider the potential well shown in Fig. 7.28(a).Sketch the wave function for the eighth excited levelof this well, assuming that it has energy [Hint: The wave function has eight nodes — notcounting the nodes at the walls of the well.]

E 7 U0 .

ƒc1x2 ƒ2x : q .

eaxx : - q

c1x2U1x2 - EE 6 0.

U0 , E,0 6 E 6 U0 ,U1x2. U0

U1x2 = U011 - e-x2>a22

U1x2 = 12 kx2.

c = a>4?c = a>2?c = a?

c … a.x = cx = 0

1n = 32

cos12px>a2 sin21px>a28x9 7.41 •• Do the same task as in Problem 7.40, but for the

well in Fig. 7.28(b).

7.42 •• Sketch the wave function for the fourth excitedlevel of the well of Fig. 7.28(a), assuming that it hasenergy in the interval

7.43 •• Do Problem 7.42, but for the well of Fig. 7.28(b).

7.44 •• Consider the infinitely deep potential well shownin Fig. 7.29. (a) Argue that this is the potential energyof a particle of mass m above a hard surface in a uni-form gravitational field with x measured verticallyup. (b) Sketch the wave functions for the ground stateand the first two excited states of this well.

0 6 E 6 U0 .

7.45 •• (a) Derive an approximate expression for thenumber of bound states in a finite square well ofdepth To do this, make the assumption that theenergy of the highest state of the finite well is close tothe corresponding energy of the infinite well. Thisstate is the one that has an energy E near the top ofthe well, that is, (b) Compute the approxi-mate number of bound states for the case of an elec-tron in a well of depth and width

These numbers crudely approximatevalence electrons in a solid.

7.46 •• In a region where the potential energy varies,the Schrödinger equation must usually be solved bynumerical methods — generally with the help of acomputer. The simplest such method, though not themost efficient, is called Euler’s method and can be de-scribed as follows: In solving the Schrödinger equa-tion numerically, one needs to know the values of

and at one point For example, for thefinite wells of Fig. 7.8 we know these values at since we know that has the form for

Suppose now we want to find at somepoint We first divide the interval from to xinto n equal intervals each of width

Knowing and at we can use theSchrödinger equation to find approximate values ofthese two functions at and from these, we can findvalues at and so on until we know both functions at

To do this, we use the approximations

andc¿1xi + 12 L c¿1xi2 + c–1xi2 ¢x

c1xi + 12 L c1xi2 + c¿1xi2 ¢x

xn = x.x2

x1 ,

x = x0 ,c¿1x2c1x2x0 6 x1 6 x2 6 Á 6 xn = x

¢x:x0x 7 0.

c1x2x … 0.c1x2 = eaxc1x2 x0 = 0

x0 .c¿1x2c1x2

U1x2a = 0.3 nm.

U0 = 10 eV,

E = U0 .

U0 .

U0

0

U0

(a) (b)

FIGURE 7.28(Problems 7.40 to 7.43)

x

U(x

)

0

FIGURE 7.29(Problem 7.44)

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[The first approximation is just the definition of thederivative, and the second is the same approximationapplied to ] Consider the equation

with the starting values and and do the following (for which youdon’t need a computer): (a) What is the exact solutionof this equation with these initial conditions? What isthe exact value of at (b) Divide the in-terval from to 1 into two equal intervals

and use the Euler’s method to find an ap-proximate value for (c) Repeat with and

Compare your results with the exact answer,and make a plot of the exact and approximate valuesfor the case that Note well how the approxi-mate solution improves as you increase n.

7.47 •• Do Problem 7.46, but for the differential equationwith the starting values

and

SECTION 7.9 (The Simple Harmonic Oscillator)

7.48 • Show that the length parameter b defined for theSHO in Eq. (7.99) is equal to the value of x at theclassical turning point for a particle with the energyof the quantum ground state.

7.49 • Verify that the wave function for the SHO,given in Table 7.1, satisfies the Schrödinger equationwith (The potential-energy function is

)

7.50 • The wave function for the ground state of aharmonic oscillator is given in Table 7.1. Show that itsnormalization constant is

(7.123)

You will need to know the integral which can be found in Appendix B.

7.51 • Show that the normalization constant for thewave function of the first excited state of the SHO is

(The wave function is given inTable 7.1. You will need to know the integral

which is given in Appendix B.)

7.52 •• Verify that the wave function for the SHO,given in Table 7.1, satisfies the Schrödinger equationwith energy

7.53 ••• The wave functions of the harmonic oscillator,like those of a particle in a finite well, are nonzero inthe classically forbidden regions, outside the classicalturning points. In this question you will find the prob-ability that a quantum particle which is in the groundstate of an SHO will be found outside its classicalturning points. The wave function for this state is inTable 7.1, and its normalization constant is givenin Problem 7.50. (a) What are the turning points for aclassical particle with the ground-state energy inan SHO with Relate your answer to theconstant b in Eq. (7.99). (b) For a quantum particle inthe ground state, write down the integral that givesthe total probability for finding the particle betweenthe two classical turning points. The form of the re-quired integral is given in Problem 7.32, Eq. (7.122).

U = 12 kx2

?

12 Uvc

A0

52 Uvc .

n = 21q

-q x2 e-lx2

dx,

A1 = 14>pb221>4.A1

1q-q e-lx2

dx,

A0 = 1pb22-1>4A0

c01x2U = 1

2 kx2.E = 1

2 Uvc .

n = 0

c¿102 = 1.c102 = 0c–1x2 = +c1x2,

n = 4.

n = 4.n = 3c112.1n = 22, x = 0

x = 1?c1x2

c¿102 = 1,c102 = 0c–1x2 = -c1x2,c¿1x2.

To evaluate it, change variables until you get an inte

gral of the form this is a standard integralof mathematical physics (called the error function)with the known value 1.49. What is the probability offinding the particle between the classical turningpoints? (c) What is the probability of finding itoutside the classical turning points?

SECTION 7.10 (Tunneling)

7.54 • Consider two straight wires lying on the x axis, sep-arated by a gap of 4 nanometers. The potential ener-gy in the gap is about 3 eV higher than the energyof a conduction electron in either wire. What is theprobability that a conduction electron in one wire ar-riving at the gap will pass through the gap into theother wire?

7.55 •• The radioactive decay of certain heavy nuclei byemission of an alpha particle is a result of quantumtunneling, as described in detail in Section 17.10.Meanwhile, here is a simplified model: Imagine analpha particle moving around inside a nucleus, suchas thorium 232. When the alpha bounces against thesurface of the nucleus, it meets a barrier caused by theattractive nuclear force. The dimensions of this barri-er vary a lot from one nucleus to another, but as rep-resentative numbers you can assume that thebarrier’s width is (1 fm ) andthe average barrier height is Findthe probability that an alpha hitting the nuclear sur-face will escape. Given that the alpha hits the nuclearsurface about times per second, what is theprobability that it will escape in a day?

SECTION 7.11 (The Time-DependentSchrödinger Equation)

7.56 • Verify that

7.57 • If satisfies the time-independent Schrödingerequation (7.106), verify that the function

satisfies the time-dependent Schrödingerequation (7.107).

7.58 •• Assuming the wave functions and arenormalized, verify that the superposition (7.114) isalso normalized. [Hint: It will help if you can provethat ]

7.59 •• Verify that for the wave function (7.111),

assuming that and are real.

COMPUTER PROBLEMS

7.60 • (Section 7.2) Use appropriate software to drawfive graphs (all on the same plot) of the standingwave (7.2) for and for the five times

0.05, 0.1, 0.15, and 0.2. Take Using your plots, describe clearly the behavior of thestanding wave.

A = l = T = 1.t = 0,0 6 x 6 2

c21x2c11x22Re1a*b2c11x2c21x2 cos1v

21 t2ƒ °1x, t2 ƒ2 = ƒa ƒ2 ƒc11x2 ƒ2 + ƒb ƒ2 ƒc21x2 ƒ2 +

1

c 1 c2 dx = 0.

c2c1

c1x2e-iEt>U °1x, t2 =c1x2

eip = -1.

5 * 1021

U0 - E L 5 MeV.= 10-15 mL L 35 fm

U0

11-1 e-y2

dy;

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7.61 • (Section 7.9) Use appropriate software to plot thelowest two wave functions of the simple harmonicoscillator. The functions are given in Table 7.1, andthe normalization constants in Problems 7.50 and7.51. For the purposes of your plot, you may as wellchoose your unit of length such that

7.62 •• (Section 7.2) If you have access to graphing soft-ware that lets you animate graphics, make 20 separateplots of the standing wave of Problem 7.60 at equallyspaced times 0.05, 0.95 and animatethem to make a movie of the standing wave. Describethe behavior of the wave.

7.63 •• (Section 7.11) Make plots similar to Fig. 7.25, but for eight different times, If possi-ble, animate these pictures to show the movement ofthe distribution in the well.

7.64 ••• (Section 7.8) If you haven’t already done so, doProblem 7.46, then using a programmable calculatoror computer software such as MathCad, extend youranswers to the cases that the number of steps n is 5, 10,and 50. Make a plot showing the exact solution andthe values of your approximate solution for

7.65 ••• (Section 7.8) If you haven’t already done so, doProblem 7.47, then using a programmable calculatoror computer software such as MathCad, extend youranswers to the cases that the number of steps n is 5, 10,and 50. Make a plot showing the exact solution andthe values of your approximate solution for

7.66 ••• (Section 7.8) If you have access to computer soft-ware with preprogrammed numerical solution of dif-ferential equations (for example, the functionNDSolve in Mathematica), do the following: (a) Plotthe Gaussian well of Problem 7.37, using units suchthat and taking (The firstthree choices simply define a convenient system ofunits — for instance, a is the unit of length — and thelast is chosen so that the well supports several boundstates.) (b) We know that to be acceptable, the wave

m = 36.U0 = a = U = 1

n = 50.

n = 50.

78 .3

8 , Á ,28 ,1

8 ,t = 0,

0.01, Á ,t = 0,

b = 1.

function must be proportional to far to the left ofthe well. To ensure this, use the boundary conditions

and and solve theSchrödinger differential equation for energy Plot your solution, and confirm that it does not be-have acceptably as (c) Repeat for What can you conclude about the energy of theground state from the plots of parts (b) and (c)?(d) Repeat for two or three intermediate energiesuntil you know the ground-state energy to two signif-icant figures.

7.67 ••• (Section 7.8) Consider the problem of a particle infinite square well of depth as described in Section7.8.To solve this problem, use a coordinate system cen-tered on the box, with the left edge of the well at

and the right edge at From thesymmetry of the potential, one can argue that the sta-tionary states should have symmetric probability dis-tributions, that is, This implies thatsolutions are either symmetric and satisfy

or are antisymmetric and satisfyThus, with our choice of coordinates,

the solutions within the well are either of the form(for ) or of the form

(for ). (a) For the case of the symmetricsolutions, derive an equation relating k and [Hint:Using the boundary conditions that both and

are continuous at you can producetwo equations that relate k, and the arbitrary coef-ficients A and G in Section 7.8. By dividing theseequations, you can eliminate the coefficients. Thefinal equation you produce is a transcendental equa-tion relating E and it cannot be solved for E usingelementary methods.] (b) Show that the equation de-rived in (a) produces the correct stationary-state ener-gies in the limit (c) Using numericaltechniques, find the ground-state energy of asquare well of depth [Hint: A simple trial-and-error search for the solution works surprisinglywell.]

U0 = 3 E1 .E1

U0 : q .

U0 ;

a,x = -a>2,c¿1x2 c1x2a.

n = 2, 4, 6, Ásin1kx2n = 1, 3, 5, Ácos1kx2

c1x2 = -c1-x2.c1x2 = c1-x2ƒc1x2 ƒ2 = ƒc1-x2 ƒ2.

x = +a>2.x = -a>2

U0 ,

E = 0.2.x : q .

E = 0.1.c¿1-32 = a,c1-32 = 1

eax

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