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8.1

Chapter 8

The Schrodinger Equation in One Dimension

8.1 Introduction

8.2 Classical Standing Waves

8.3 Standing Waves in Quantum Mechanics

8.4 The Particle in a Rigid Box

8.5 The Time-independent Schrödinger Equation; Stationary States

8.6 The Rigid Box Again

8.7 The Free Particle

8.8 The Nonrigid Box

8.9 The Simple Harmonic Oscillator (optional)

Problems for Chapter 8

8.1 Introduction

In classical mechanics, the state of motion of a particle is specified by giving the particle's position and velocity. In quantum mechanics, the state of motion of a particle isspecified by giving the wave function. In either case, the fundamental question is to predict how the state of motion will evolve as time goes by, and in each the answer is

given by an equation of motion. The classical equation of motion is Newton's secondlaw, m=F a ; if we know the particle's position and velocity at time 0t = , Newton's

second law determines the position and velocity at any other time. In quantummechanics, the equation of motion is called the time-dependent Schrödinger equation. If we know a particle's wave function at 0t = , the time-dependent Schrödinger equationdetermines the wave function at any other time.

The time-dependent Schrödinger equation is a partial differential equation, acomplete understanding of which requires more mathematical preparation than we areassuming here. Fortunately, the majority of interesting problems in quantum mechanicsdo not require use of the equation in its full generality. By far the most interesting statesof any quantum system are those states in which the system has a definite total energy,

and it turns out that for these states the wave function is a standing wave, analogous tothe familiar standing waves on a string. When the time-dependent Schrödinger equationis applied to these standing waves, it reduces to a simpler equation called the time-independent Schrödinger equation. We shall need only this time-independent equation,which will let us find the wave functions of the standing waves and the correspondingallowed energies. Because we shall be using only the time-independent Schrödinger equation we shall often refer to it as just "the Schrödinger equation." Nevertheless, youshould know that there are really two Schrödinger equations (the time-dependent and the

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8.2

time-independent). Unfortunately, it is almost universal to refer to either as "theSchrbdinger equation" and to let the context decide which is being discussed. In this book, however, "the Schrödinger equation" will always mean the simpler, time-independent equation.

In Section 8.2 we review some properties of classical standing waves, using

waves on a uniform, stretched string as our example. In Section 8.3 we discuss quantumstanding waves. Then, in Section 8.4, we show how the familiar properties of classicalstanding waves let one find the allowed energies of one simple quantum system, namelya particle that moves freely inside a perfectly rigid box.

Using our experience with the wave functions of a particle in a rigid box we nextwrite down the time-independent Schrödinger equation, with which one can, in principle,find the allowed energies and wave functions for any system. Then in Sections 8.6 to 8.9we use the Schrödinger equation to find the allowed energies of various simple systems.

Throughout this chapter we treat particles that move nonrelativistically in onedimension. All real systems are, of course, three-dimensional. Nevertheless, just as is thecase in classical mechanics, it is a good idea to start with the simpler problem of a particle confined to move in just one dimension. In the classical case it is easy to findexamples of systems that are at least approximately one-dimensional — a railroad car ona straight track, a bead threaded on a taut string. In quantum mechanics there are fewer examples of one-dimensional systems. However, we can for the moment imagine anelectron moving along a very narrow wire.1 The main importance of one-dimensionalsystems is that they provide a good introduction to three-dimensional systems, and thatseveral one-dimensional solutions find direct application in three-dimensional problems.

8.2 Classical Standing Waves

We start with a review of some properties of classical standing waves. We could discuss

waves on a string, for which the wave function is the string's transverse displacement y( x,t ); or we might consider sound waves, for which the wave function is the pressurevariation, p( x, t ). If we considered electromagnetic waves, then the wave function would be the electric field strength,(x, t ). In this section we choose to discuss waves on astring, but since our considerations apply equally to all waves, we shall use the general

notation ( , ) x t Ψ to represent the wave function.

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

1 ( , ) sin( ) x t B kx t ω Ψ = −

(this is the wave sketched in Fig. 7.8) and the other moving to the left with the sameamplitude,

2 ( , ) sin( ). x t B kx t ω Ψ = +

If we imagine both of these waves traveling simultaneously in the same string, then theresultant wave is their sum2:

1 More realistic examples include the motion of electrons along one axis in certain crystals andin some linear molecules.

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8.3

1 2( , ) ( , ) ( , ) [sin( ) sin( )]. x t x t x t B kx t kx t ω ω Ψ = Ψ + Ψ = − + +

If we recall the important trigonometric identity (Appendix B)

sin sin 2 sin cos ,2 2

a b a ba b

+ −+ =

we can rewrite the wave (8.1) as

( , ) 2 sin cos x t B kx t ω Ψ =

or if we set 2 B A= ,

( , ) sin cos . x t A kx t ω Ψ =

A series of snapshots of the resultant wave (8.2) is sketched in Fig. 8.1. The important point to observe is that the resultant wave is not traveling to the right or left. At certain

fixed points called nodes, where sin kx is zero, ( , ) x t Ψ is always zero and the string is

stationary. At any other point the string simply oscillates up and down in proportion tocos t ω , with amplitude sin A kx . By superposing two traveling waves we have formed a

standing wave.

Because the string never moves at the nodes, we could clamp it at two nodes andremove the string outside the clamps, leaving a standing wave on a finite length of stringas in Fig. 8.2. This is the kind of wave produced on a piano or guitar string when itsounds a pure musical tone.

If we now imagine a string clamped between two fixed points separated by adistance a, we can ask: What are the possible standing waves that can fit on the string?The answer is that a standing wave is possible, provided that it has nodes at the two fixedends of the string. Since the distances between the nodes of a wave are / 2λ , λ , 3 / 2λ ,and so on, we conclude that a standing wave is possible if it has

3or or or ;

2 2a a a

λ λ λ = = = L

that is, if

2, where 1, 2, 3, .

an

nλ = = L

We see that the possible wavelengths of a standing wave on a string of length a arequantized , the allowed values being 2a divided by any positive integer. The first three of these allowed waves are sketched in Fig. 8.3.

It is important to recognize that the quantization of wavelengths arises from therequirement that the wave function must always be zero at the two fixed ends of thestring. We refer to this kind of requirement as a boundary condition, since it relates to

2 We are assuming here that whenever 1Ψ , and 2Ψ are possible waves, the same is true of

1 2Ψ + Ψ . This important property, called the superposition principle, is true of many waves, including

those considered here.

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8.4

the boundaries of the system. We shall find that for quantum waves, just as for classicalwaves, it is the boundary conditions that lead to quantization.

8.3 Standing Waves in Quantum Mechanics; Stationary States

Before we discuss quantum standing waves we need to examine more closely the form of

the classical standing wave (8.2):

( , ) sin cos . x t A kx t ω Ψ =

This function is a product of one function of x (namely sin A kx) and one function of t

(namely cos t ω ). We can emphasize this by rewriting (8.4) as

( , ) ( )cos , x t x t ψ ω Ψ =

where we have used the capital letter Ψ for the full wave function ( , ) x t Ψ and the lower

case letter ψ for its spatial part ( ) xψ . The spatial function ( ) xψ gives the full wave

function ( , ) x t Ψ at time 0t = (since cos 1t ω = when t = 0); more generally, at any time

t the full wave function ( , ) x t Ψ is ( ) xψ times the oscillatory factor cos t ω .

In our particular example (a wave on a uniform string) the spatial function ( ) xψ

was a sine function

( ) sin , x A kxψ =

but in more general problems, such as waves on a nonuniform string, ( ) xψ can be a

more complicated function of x. On the other hand, even in these more complicated problems the time dependence is still sinusoidal; that is, it is given by a sine or cosinefunction of t . The difference between the sine and the cosine is just a difference in thechoice of origin of time. Thus either function is possible, and the general sinusoidalstanding wave is a combination of both:

( , ) ( )( cos sin ). x t x a t b t ψ ω ω Ψ = +

Different choices for the ratio of the coefficients a and b correspond to different choicesof the origin of time. (See Problem 8.11.)

The standing waves of a quantum system have the same form (8.7), but with one

important difference. For a classical wave, the function ( , ) x t Ψ is, of course, a real

number. (It would make no sense to say that the displacement of a string, or the pressure

of a sound wave, had an imaginary part.) Therefore, the function ( ) xψ and the

coefficients a and b in (8.7) are always real for any classical wave.1 In quantummechanics, on the other hand, the wave function can be a complex number; and for

quantum standing waves it usually is complex. Specifically, the time-dependent part of the wave function (8.7) always occurs in precisely the combination

cos sin ,t i t ω ω −

1 As some readers may know, it is sometimes a mathematical convenience to introduce a certaincomplex wave function. Nonetheless, in classical physics the actual wave function is always the real partof this complex function.

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8.5

where i is the imaginary number 1i = − . That is, the standing waves of a quantum

particle have the form

( , ) ( )(cos sin ). x t x t i t ψ ω ω Ψ = −

In more advanced texts, this specific time dependence is derived from the time-

dependent Schrödinger equation. Here we simply state it as a fact that quantum standingwaves have the sinusoidal time dependence of the particular combination of cos t ω andsin t ω in (8.9).

The form (8.9) can be simplified if we use Euler's formula from the theory of complex numbers (Problem 8.12),

cos sin .ii e θ θ θ + =

This identity can be illustrated in an Argand diagram, as in Fig. 8.4, where the complex

number z x i y= + is represented by a point with coordinates x and y in the complex

plane. Since the number ie θ (with θ any real number) has coordinates cosθ and sinθ ,

we see from Pythagoras's theorem that its absolute value is 1:

2 2| | (cos ) (sin ) 1.ie θ θ θ = + =

Thus the complex number ie θ lies on a circle of radius 1, with polar angle θ as shown.

Notice that since cos( ) cosθ θ − = and sin( ) sinθ θ − = − ,

cos sin .ii e θ θ θ −− =

Returning to (8.9) and using the identity (8.10), we can write for the generalstanding wave of a quantum system

( , ) ( ) . i t xt x e

ω ψ

−Ψ=

Since this function has a definite angular frequency, ω , any quantum system with thiswave function has a definite energy given by the de Broglie relation E ω = h . Conversely(although we shall not prove this), any quantum system that has a definite energy has awave function of the form (8.11).

We saw in Chapter 7 that the probability density associated with a quantum wave

function ( , ) x t Ψ is the absolute value squared,2

( , ) x t Ψ . For the complex standing wave

(8.11) this has a remarkable property:

2 2 2| ( , )| | ( )| | |i t x t x e ω ψ −Ψ =

or, since1i t e ω − =

,2 2| ( , )| | ( )| (for quantum standing waves). x t xψ Ψ =

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

cos sin ,i t e t i t ω ω ω − = −

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8.6

is complex, with two parts that oscillate 90o 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 quantumstanding wave the distribution of matter (of electrons in an atom, or nucleons in anucleus, for example) is time independent or stationary. For this reason a quantumstanding wave is often called a stationary state. The stationary states are the moderncounterpart of Bohr's stationary orbits and are precisely the states of definite energy.Because their charge distribution is static, atoms in stationary states do not radiate.1

An important practical consequence of (8.12) is that in most problems the only

interesting part of the wave function ( , ) x t Ψ is its spatial part ( ) xψ . We shall see that a

large part of quantum mechanics is devoted to finding the possible spatial function ( ) xψ

and their corresponding energies. Our principal tool in finding these will be the time-independent Schrödinger equation.

8.4 The Particle in a Rigid Box

Before we write down the Schrödinger equation, we consider a simple example that we

can solve using just our experience with standing waves on a string. We consider a particle that is confined to some finite interval on the x axis, and moves freely inside thatinterval — a situation we describe as a one-dimensional rigid box. For example, inclassical mechanics we could consider a bead on a frictionless straight thread betweentwo rigid knots; the bead can move freely between the knots, but cannot escape outsidethem, In quantum mechanics we can imagine an electron inside a length of very thinconducting wire; to a fair approximation the electron would move freely back and forthinside the wire, but could not escape from it.

Let us consider, then, a quantum particle of mass m moving nonrelativistically ina one-dimensional rigid box of length a, with no forces acting on it between x = 0 and x= a. The absence of forces means that the potential energy is constant inside the box, and

we are free to choose that constant to be zero. Therefore, its total energy is just its kineticenergy. In quantum mechanics, it is almost always more convenient to think of the

kinetic energy as p2/2m, rather than21

2mv , because of the de Broglie relation, /h pλ = ,

between the momentum and wavelength. Therefore, we write the energy as

2

.2

p E K

m= =

As we have said, the states of definite energy are the standing waves. Therefore,to find the allowed energies we must find the possible standing waves for the particle's

wave function ( , ) x t Ψ . We know that the standing waves have the form

( , ) ( ) .i t

x t x e ω ψ −

Ψ =By analogy with waves on a string, one might guess that the spatial function ( ) xψ will

be a sinusoidal function inside the box; that is, ( ) xψ should have the form sin kx or cos

kx or a combination of both:

1 Of course, atoms do radiate from excited states, but as we discuss in Chapter 15, this is always because some external influence disturbs the stationary state.

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8.7

( ) sin cos x A kx B kxψ = +

for 0 x a≤ ≤ . (We make no claim to have proved this; but it is certainly a reasonableguess, and we shall prove in Section 8.6 that it is correct.)

Since it is impossible for the particle to escape from the box, the wave function

must be zero outside; that is, ( ) xψ = 0 when x < 0 and when x > a. If we make thereasonable (and, again, correct) assumption that ( ) xψ is continuous, then it must also

vanish at x = 0 and x = a:

(0) ( ) 0.aψ ψ = =

These are the boundary conditions that the wave function (8.15) must satisfy. Notice thatthese boundary conditions are identical to those for a classical wave on a string clampedat x = 0 and x = a.

From (8.15) we see that (0) Bψ = . Thus the wave function (8.15) can satisfy the

boundary condition (8.16) only if the coefficient B is zero; that is, the condition (0) 0ψ =

restricts ( ) xψ to have the form( ) sin . x A kxψ =

Next, the boundary condition that ( ) 0aψ = requires that

sin 0, A ka=

which implies that1

, or 2 , or 3 ,ka π π π = L

or

, 1,2,3, .

n

k na

π

= = L

We conclude that the only standing waves that satisfy the boundary conditions

(8.16) have the form ( ) xψ = A sin kx with k given by (8.20). In terms of wavelength, this

condition implies that

2 2, 1,2,3, .

an

k n

π λ = = = L

which is precisely the condition (8.3) for standing waves on a string. This is, of course,not an accident. In both cases, the quantization of wavelengths arose from the boundarycondition that the wave function must be zero at x = 0 and x = a.

For our present discussion the important point is that quantization of wavelengthλ implies quantization of momentum, and hence also of energy. Specifically,

substituting (8.2 1) into the de Broglie relation / p h λ = , we find that

1 Strictly speaking, (8.18) implies either that k satisfies (8.19) or that A = 0; but if A = 0, then ψ =0 for all x, and we get no wave at all. Thus only the solution (8.19) corresponds to a particle in a box.

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8.8

, 1, 2,3, .2

nh n p n

a a

π = = =

hL

Since E = K + U , and U = 0 in this case, we have E = p2/2m. Therefore, (8.22) meansthat the allowed energies for a particle in a one-dimensional rigid box are

2 22

2, 1,2,3, .

2n

E n nma

π = =

h L

The lowest energy for our particle is obtained when n = 1 and is

2 2

1 2.

2 E

ma

π =

h

This is consistent with the lower bound derived from the Heisenberg uncertainty principle in Chapter 7, where we argued that for a particle confined in a region of lengtha,

2

2 .2 E ma≥h

For our particle in a rigid box the actual minimum energy (8.24) is larger than the lower

bound (8.25) by a factor of 2π .

In terms of the ground-state energy E , the energy of the nth level (8.23) is

2

1 , 1,2,3, .n

E n E n= = L

These energy levels are sketched in Fig. 8.5. Notice that (quite unlike those of thehydrogen atom) the energy levels are farther and farther apart as n increases, and that E nincreases without limit as n → ∞ . The corresponding wave functions ( ) xψ (which look

exactly like the standing waves on a string) have been superimposed on the same picture,the wave function for each level being plotted on the line that represents its energy. Notice how the number of nodes of the wave functions increases steadily with energy;this is what one should expect, since more nodes mean shorter wavelength and hencelarger momentum and kinetic energy.

The complete wave function ( , ) x t Ψ for any of our standing waves has the form

( , ) ( ) sin( ) .i t i t x t x e A kx eω ω ψ − −Ψ = =

We can rewrite this, using the identity (Problem 8.13)

sin ,2

i ie e

i

θ θ

θ −−

=

to give

( ) ( )( , ) ( ).2

i kx t i kx t A x t e e

i

ω ω − − +Ψ = −

We see that our quantum standing wave (just like the classical standing wave of Section8.2) can be expressed as the sum of two traveling waves, one moving to the right and one

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8.9

to the left. The wave moving to the right represents a particle with momentum k h directed to the right , and that moving to the left, a particle with momentum of the samemagnitude k h but directed to the left. Thus a particle in one of our stationary states has a

definite magnitude, k h , for its momentum but is a 50:50 superposition of momenta ineither direction. To some extent this situation is analogous to the result that on average a

classical particle is equally likely to be moving in either direction as it bounces back andforth inside a rigid box.

8.5 The Time-independent Schrödinger Equation

Our discussion of the particle in a rigid box depended on some guessing as to the form of

the spatial wave function ( ) xψ . There are very few problems where this kind of

guesswork is possible, and no problems where it is entirely satisfying. What we need is

the equation that determines ( ) xψ in any problem, and this equation is the time-

independent Schrödinger equation. Like all basic laws of physics, the Schrödinger equation cannot be derived. It is simply a relation, like Newton's second law, thatexperience has shown us is true. Thus a legitimate procedure would be simply to state theequation and to start using it. Nevertheless, it may be helpful to offer some argumentsthat suggest the equation, and this is what we shall try to do.

Almost all laws of physics can be expressed as differential equations, that is, asequations that involve the variable of interest and some of its derivatives. The mostfamiliar example is Newton's second law for a single particle, which we can write as

2

2.

d xm F

dt = ∑

If, for example, the particle in question were immersed in a viscous fluid that exerted a

drag force bv− , and attached to a spring that exerted a restoring force k x− , then (8.29)

would read2

2.

d x dxm b kx

dt dt = − −

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

The equation of motion for classical waves (which is often not discussed in anintroductory physics course) is a differential equation. It is therefore natural to expect theequation that determines the possible standing waves of a quantum system to be adifferential equation. Since we already know the form of the wave functions for a particle in a rigid box, what we shall do is examine these wave functions and try to spot asimple differential equation which they satisfy and which we can generalize to morecomplicated systems.

We saw in Section 8.4 that the spatial wave functions for a particle in a rigid boxhave the form

( ) sin . x A kxψ =

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8.10

To find a differential equation that this function satisfies, we naturally differentiate it, togive

cos .d

kA kxdx

ψ =

There are several ways in which we could relate cos kx in (8.32) to sin kx in (8.3 1) andhence obtain an equation connecting /d dxψ with ψ . However, a simpler course is to

differentiate a second time to give

22

2sin .

d k A kx

dx

ψ = −

Comparing (8.33) and (8.31) we see at once that2 2/d dxψ is proportional to ψ ;

specifically,

22

2.

d k

dx

ψ ψ = −

We can rewrite k 2 in (8.34) in terms of the particle's kinetic energy, K . We know that p =k h . Therefore,

2 2 2

;2 2

p k K

m m= =

h

hence

2

2

2.

mK k =

h

Thus we can write (8.34) as

2

2 2

2,

d mK

dx

ψ ψ = −

h

which gives us a second-order differential equation satisfied by the wave function ( ) xψ

of a particle in a rigid box.

The particle in a rigid box is an especially simple system, with potential energyequal to zero throughout the region where the particle moves. It is not at all obvious howthe equation (8.37) should be generalized to include the possibility of a nonzero potentialenergy, U ( x), which may vary from point to point. However, since the kinetic energy K isthe difference between the total energy E and the potential energy U ( x), it is perhaps

natural to replace K in (8.37) by( ). K E U x= −

This gives us the differential equation1

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

2 2

2( )

2

d U x E

m dx

ψ ψ ψ − + =

h

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8.11

2

2 2

2[ ( ) ] .

d mU x E

dx

ψ ψ = −

h

This differential equation is called the Schrödinger equation (time-independent

Schrödinger equation, in full), in honor of the Austrian physicist who first published itin 1926. Like us, Schrödinger had no way to prove that his equation was correct. All hecould do was argue that the equation seemed reasonable and that its predictions should betested against experiment. In the 60 years since then, it has passed this test repeatedly. In particular, Schrödinger himself showed that it predicts correctly the energy levels of thehydrogen atom, as we describe in Chapter 9. Today, it is generally accepted that theSchrödinger equation is the correct basis of nonrelativistic quantum mechanics, in justthe same way that Newton's second law is accepted as the basis 'of nonrelativisticclassical mechanics.

The Schrödinger equation as written in (8.39) applies to one particle moving inone dimension. We shall need to generalize it later to cover systems of several particles,in two or three dimensions. Nevertheless, the general procedure for using the equation isthe same in a cases. Given a system whose stationary states and energies we wish toknow, we must first find the potential energy function U ( x). For example, a particle held

in equilibrium at x = 0 by a force obeying Hooke's law ( F k x= − ) has potential energy

212

( ) .U x kx=

An electron in a hydrogen atom has

2

( ) .ke

U r r

= −

(We shall return to this three-dimensional example in Chapter 9.) Once we haveidentified U ( x), the Schrödinger equation (8.39) becomes a well-defined equation that we

can try to solve.1 In most cases, it turns out that for many values of the energy E theSchrödinger equation has no solutions (no acceptable solutions, satisfying the particular conditions of the problem, that is). This is exactly what leads to the quantization of energies. Those values of E for which the Schrödinger equation has no solution are notallowed energies of the system. Conversely, those values of E for which there is a

solution are allowed energies, and the corresponding solutions ( ) xψ give the spatial

wave functions of these stationary states.

As we hinted in the last paragraph, there are usually certain conditions that the

wave function ( ) xψ must satisfy to be an acceptable! solution of the Schrödinger

equation. First, there may be boundary conditions on ( ) xψ , for example, the condition

that ( ) xψ must vanish at the walls of a perfectly rigid box. In addition, there are certain

because the differential operator ( )2 2 2/ 2 /m d dx− h is intimately connected with the kinetic energy.

Nevertheless, for the applications in this book the form (8.39) is the most convenient, and we shallalways write it this way.

1 The necessity of identifying U before one can solve the Schrödinger equation corresponds tothe necessity of identifying the total force F on a classical particle before one can solve Newton's secondlaw, F = ma.

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8.13

From now on we shall use this notation whenever convenient. In particular, we rewrite(8.44) as

2

2( ) ( ).

mE x xψ ψ ′′ = −

h

We now consider whether there is an acceptable solution of (8.45) for any particular value ofE, starting with the case that E is negative. (We do not expect anystates with E < 0, since then E would be less than the minimum potential energy. But wehave already encountered several unexpected consequences of quantum mechanics, and

we should check this possibility.) If E were negative, then the coefficient 22 /mE − h on

the right of (8.45) would be positive and we could call it a 2α , where

2.

mE α

−=

h

With this notation, (8.45) becomes

2

( ) ( ). x xψ α ψ ′′ =The simplification of rewriting (8.45) in the form (8.47) has the disadvantage of requiring a new symbol (namely α ); but it has the important advantage of letting usfocus on the mathematical structure of the equation.

Equation (8.47) is a second-order differential equation, which has the solutions

(Problem 8.19) xeα and xe α − or any combination of these,

( ) x x x Ae Beα α ψ −= +

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

It is important in what follows that (8.48) is the most general solution of (8.47),that is, that every solution of (8.47). has the form (8.48). This follows from a theoremabout second-order differential equations of the same type as the one-dimensionalSchrödinger equation.1 This theorem states three facts: First, these equations always have

two independent solutions. For example, xeα and xe α − are two independent solutions of

(8.47).2 Second, if 1( ) xψ and 2 ( ) xψ denote two such independent solutions, then the

linear combination

1 2( ) ( ) A x B xψ ψ +

is also a solution, for any constants A and B . Third , given two independent solutions

1( ) xψ and 2 ( ) xψ , every solution can be expressed as a linear combination of the form

(8.49). These three properties are illustrated in Problems 8.19 to 8.24.

1 To be precise, ordinary second-order differential equations that are linear and homogeneous.

2 When we say that two functions are independent , we mean that neither function is just a

constant multiple of the other. For example, xeα and xe α − are independent, but xeα and 2 xeα are not;

similarly sin x and cos x are independent, but 5 cos x and cos x are not.

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8.14

That the general solution of a second-order differential equation contains twoarbitrary constants is easy to understand: A second-order differential equation amounts to

a statement of the second derivative ψ ′′ ; to find ψ , one must somehow accomplish two

integrations, which should introduce two constants of integration; and this is what thetwo arbitrary constants A and B in (8.49) are. The theorem above is very useful in

seeking solutions of such differential equations. If, by any means, we can spot twoindependent solutions, we are assured that every solution is a combination of these two.

Since xeα and xe α − are independent solutions of (8.47), it follows from the theorem that

the most general solution is (8.48).

Equation (8.48) gives all solutions of the Schrödinger equation for negativevalues of E. The important question is now whether any of these solutions could satisfy

the required boundary conditions (8.43), and the answer is "no." With ( ) xψ given by

(8.48), the condition that (0)ψ = 0 implies that

0, A B+ =

while the requirement that( )aψ

= 0 implies that0.a a Ae Beα α −+ =

It is easy to check that the only values of A and B that satisfy these two simultaneousequations are A = B = 0. That is, if E < 0, the only solution of the Schrödinger equationthat satisfies the boundary conditions is the zero function. In other words, with E < 0there can be no standing waves, so negative values of E are not allowed.

Let. us next see if the Schrödinger equation (8.45) has any acceptable solutions

for positive energies (as we expect it does). With E > 0, the coefficient 22 /mE − h on the

right of (8.45) is negative and can conveniently be called 2k − where

2 .mE k =h

With this notation, the Schrödinger equation reads

2( ) ( ). x k xψ ψ ′′ = −

This differential equation has the solutions sin kx and cos kx, or any combination of both:

( ) sin cos x A kx B kxψ = +

(see Example 8.1 below). This is exactly the form of the wave function that we assumedat the beginning of Section 8.4. The important difference is that in Section 8.4 we could

only guess the form (8.52), whereas we have now derived it from the Schrödinger equation. From here on the argument follows precisely the argument given before. As we

saw, the boundary condition (0) 0ψ = requires that the coefficient B in (8.52) be zero,

whereas the condition that ( ) 0aψ = can be satisfied without A being zero, provided that

ka is an integer multiple of π (so that sin ka = 0); that is,

nk

a

π =

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8.15

or, from (8.50),

2 2 2 22

22 2

k E n

m ma

π = =

h h

exactly as before.

EXAMPLE 8.1 Verify explicitly that the function (8.52) is a solution of theSchrödinger equation (8.51) for any values of the constants A and B. [This illustrates partof the theorem stated in connection with (8.49).]

To verify that a given function satisfies an equation, one must substitute thefunction into one side of the equation and then manipulate it until one arrives at the other side. Thus, for the proposed solution (8.52),

2

2

2 2

2

2

( ) ( sin cos )

( cos sin )

sin cos

( sin cos )

( )

d x A kx B kx

dxd

kA kx kB kxdx

k A kx k B kx

k A kx B kx

k x

ψ

ψ

′′ = +

= −

= − −

= − +

= −

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 box that we cannow dispose of. We have seen that the stationary states have wave functions

( ) sin ,n x

x Aa

π ψ =

but we have not yet found the constant A. Whatever the value of A, the function (8.53)satisfies the Schrödinger equation and the boundary conditions. Clearly, therefore,neither the Schrödinger equation nor the boundary conditions fix the value of A.

To see what does fix A, recall that2

( ) xψ is the probability density for finding

the particle at x. This means, in the case of a one-dimensional system, that2

( ) x dxψ is

the probability P of finding the particle between x and x + dx.

2(between and ) | ( )| . P x x dx x dxψ + =

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

2| ( )| 1. x dxψ ∞

−∞

=∫

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8.16

This relation is called the normalization condition and a wave function that satisfies itis said to be normalized. It is the condition (8.55) that fixes the value of the constant A,which is therefore called the normalization constant.

In the case of the rigid box, ( ) xψ is zero outside the box; therefore, (8.55) can be

rewritten as

2

0| ( )| 1

a

x dxψ =∫ or, with the explicit form (8.53) for ( ) xψ ,

2 2

0sin 1.

a n x A dx

a

π = ∫

The integral here is easily seen to be a/2 (Problem 8.25). Therefore, (8.57) implies that

2 2

12

A a=

and hence that1

2. A

a=

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

2( ) sin .

n x x

a a

π ψ =

EXAMPLE 8.2 Consider a particle in the ground state of a rigid box of length a. (a)

Find the probability density2

ψ . (b) Where is the particle most likely to be found? (c)

What is the probability of finding the particle in the interval between x = 0.50 a and x =0.51 a? (d) What is it for the interval [0.75 a, 0.76 a]? (e) What would be the averageresult if the position of a particle in the ground state were measured many times?

(a) The probability density is just2

( ) xψ , where ( ) xψ is given by (8.60) with n = 1.

Therefore, it is

2 22| ( )| sin ,

x x

a a

π ψ

=

1 You may have noticed that, strictly speaking, the argument leading from (8.56) to (8.59)

implies only that the absolute value of A is 2 / a . This is because (8.57) and (8.58) should both contain

A rather than A. However, since the probability density depends only on the absolute value of ψ , we

are free to choose any value of A satisfying 2 / A a= (for example, 2 / A a= − or 2 /i a ); the

choice (8.59) is convenient and customary.

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8.17

which is sketched in Fig. 8.6.

(b) The most probable value x, is the value of x for which2

( ) xψ is maximum. From

Fig. 8.6 this is clearly seen to be

mp .2

a

x =(c) The probability of finding the particle in any small interval from x to x x+ ∆ is given by (8.54) as

2(between and ) | ( )| . P x x x x xψ + ∆ ≈ ∆

(This is exact in the limit 0 x∆ → and is therefore a good approximation for any small

interval x∆ .) Thus the two probabilities are

2 22(0.50 0.51 ) | (0.50 )| sin 0.01

2

0.02 2%.

P a x a a x aa

π ψ

≤ ≤ ≈ ∆ = ×

= =(d) Similarly,

22 3(0.75 0.76 ) sin 0.01 0.01 1%.

4 P a x a a

a

π ≤ ≤ ≈ × = =

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

2

av0 | ( )| .

a

x x x dxψ = ∫ This average value xav is also denoted x or x and is often called the expectation value

of x. (But note that it is not the value we expect in any one measurement; it is rather theaverage value expected after many measurements.) In the present case

2

av0

2sin .

a x x x dx

a a

π = ∫

This integral can be evaluated to give (Problem 8,29)

av

,2

a x

=an answer that is easily understood from Fig. 8.6; since

2( ) xψ is symmetric about x =

a/2, any two points an equal distance on either side of a/2 are equally likely, and theaverage value must be a/2. We see from (8.62) and (8.66) that for the ground state of arigid box, the most probable position xmp and the mean position xav are the same. Weshall see in the next example that xmp and xav are not always equal.

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8.18

EXAMPLE 8.3 Answer the same questions as in Example 8.2 but for the first excitedstate of the rigid box.

The wave function is given by (8.60) with n = 2, so

2 22 2| ( )| sin .

x x

a a

π ψ

=

This is plotted in Fig. 8.7, where it is clear that2

( ) xψ has two equal maxima at

mp

3aand .

4 4

a x =

Since2

( ) xψ is symmetric about x = a/2, the average value xav, is the same as for the

ground state,

av .2

a x =

The probabilities of finding the particle in any small intervals are given by (8.63) as

2(0.50 0.51 ) | (0.50 )| 0 P a x a a xψ ≤ ≤ ≈ ∆ =

since1 (0.50 ) 0aψ = ; and

2 22 3(0.75 0.76 ) | (0.75 )| sin 0.01 0.02.

2 P a x a a x a

a

π ψ

≤ ≤ ≈ ∆ = × =

In particular notice that although x = a/2 is the average value of x, the probability of finding the particle in the immediate neighborhood of x = a/2 is zero. This result,although a little surprising at first, is easily understood by reference to Fig. 8.7.

8.7 The Free Particle

As a second application of the Schrödinger equation, we investigate the possible energiesof a free particle; that is, a particle subject to no forces and completely unconfined (stillin one dimension, of course). The potential energy of a free particle is constant and can be chosen to be zero. With this choice, we shall show that the energy of the particle canhave any positive value, 0 E ≥ . That is, the energy of a free particle is not quantized, andits allowed values are the same as those of a classical free particle,

1 Note that the probability for the interval [0.50a, 0.51a] is not exactly zero, since the probability

density2

( ) xψ is zero only at the one point 0.50a. The significance of (8,67) is really that the

probability for this interval is very small compared to the probability for intervals of the same widthelsewhere.

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8.19

To prove these assertions we must write down the Schrödinger equation and findthose E for which it has acceptable solutions. With U ( x) = 0, the Schrödinger equation is

2

2( ) ( ).

mE x xψ ψ

′′ = − h

This is the same equation that we solved for a particle in a rigid box, However, there isan important difference, since the free particle can be anywhere in the range

. x−∞ < < ∞

Thus we must look for solutions of (8.68) for all x rather than just those x between 0 anda.

If we consider first the possibility of states with E < 0, the coefficient 22 /mE − h

in front of ψ in (8.68) is positive and we can write (8.68) as

2( ) ( ), x xψ α ψ ′′ =

where 2 /mE α = − h. Just as with the rigid box, this equation has the solutions xeα and xe

α − , or any combination of both:

( ) . x x x Ae Beα α ψ −= +

But in the present case we can immediately see that none of these solutions can possibly be physically acceptable. The point is that (8,69) is the solution in the whole range

x−∞ < < ∞ . Now, as x → ∞ , the exponential xeα grows without limit or "blows up,"

and it obviously makes no sense to have a wave function ( ) xψ that gets bigger and

bigger without limit as we move farther and farther away from the origin. The only way

out of this difficulty is to have the coefficient A of xe α − in (8.69) equal to zero. Similarly,

as x → −∞ , the exponential xe α − blows up; thus by the same argument the coefficient B

must also be zero, and we are left with just the zero solution ( ) 0 xψ ≡ . That is, there are

no acceptable states with E < 0, just as we expected.

The argument just given crops up surprisingly often in solving the Schrödinger equation. If a solution of the equation blows up as x → ∞ , or as x → −∞ , that solution isobviously not acceptable. Thus we can add to our list of conditions that must be satisfied

by an acceptable wave function ( ) xψ the requirement that ( ) xψ must not blow up as

x → ±∞ . We speak of a function that satisfies this requirement as being "well behaved"as x → ±∞ . This requirement is actually another example of a boundary condition, sincethe "points" x = ±∞ are the boundaries of our system.

Let us next examine the possibility of states of our free particle with 0 E ≥ . In

this case the Schrödinger equation can be written as

2

2

2( ) ( ) ( )

mE x x k xψ ψ ψ

′′ = − = − h

where

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8.20

2.

mE k =

h

As before, the general solution of this equation is

( ) sin cos . x A kx B kxψ = +

The important point about this solution is that neither sin kx nor cos kx blows up as x → ±∞ . Thus neither function suffers the difficulty that we encountered with negativeenergies. Thus for any value of k the function (8.72) is an acceptable solution, for anyconstants A and B. According to (8.71), this means that all energies in the continuousrange 0 E ≤ < ∞ are allowed. In particular, the energy of a free particle is not quantized.Evidently, it is only when a particle is confined in some way that its energy is quantized.

To understand what the positive-energy wave functions (8.72) represent, it ishelpful to recall the identities (Problem 8.13)

sin and cos .

2 2

ikx ikx ikx ikxe e e ekx kx

i

− −− += =

Substituting these expansions into the wave function (8.72), we can write

( ) ,ikx ikx x Ce Deψ −= +

where you can easily find C and D in terms of the original coefficients A and B. It isimportant to note that since A and B were arbitrary, the same is true of C and D; that is,(8.74) is an acceptable solution for any values of C and D.

The full, time-dependent wave function ( , ) x t Ψ for the spatial function (8.74) is

( ) ( )( , ) ( ) .i t i kx t i kx t x t x e Ce Deω ω ω ψ − − − +Ψ = = +

This is a superposition of two traveling waves, one moving to the right (with coefficientC ) and the other moving to the left (with coefficient D). If we choose the coefficient D =

0, then (8.75) represents a particle with definite momentum k h to the right; if we choose

C = 0, then (8.75) represents a particle with momentum of the same magnitude k h but

directed to the left. If both C and D are nonzero, then (8.75) represents a superposition of both momenta.

8.8 The Nonrigid Box

So far our only example of a particle that is confined, or "bound," is the rather unrealisticcase of a particle in a perfectly rigid box. In this section we apply the Schrödinger

equation to a particle in the more realistic nonrigid box. This is a rather long section; butthe ideas it Contains are all fairly simple and are central to an understanding of manyquantum systems.

The first step in applying the Schrödinger equation to any system is to determinethe potential energy function. Therefore, we must first decide what is the potentialenergy, U ( x), of a particle in a nonrigid box. For a rigid box we know that

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8.21

, 0 and .

0, 0( )

x a

x x aU x

≤ ≤

∞ < >

=

This is infinite outside the box because no finite amount of energy can remove the particle from a perfectly rigid box.1 For most systems a more realistic assumption would

be that there is a finite minimum energy needed to remove a stationary particle from the box. If we call this minimum energy U 0, the potential energy function would be

0 , 0 and .

0, 0( )

x a

U x x aU x

≤ ≤

< >

=

In Fig. 8.8(a) and (b) we plot the potential energy functions (8.76) and (8.77). For obvious reasons these functions are often called potential wells, and we speak of a particle "moving in a well" when its potential energy is given by one of these functions.The rigid box is an infinitely deep well and the nonrigid box a finite well.

Even the finite well of Fig. 8.8(b) is unrealistic in that the potential energy jumpsabruptly from 0 to U 0 at x = 0 and x = a. For a real particle in a box (for example, anelectron in a conductor) the potential energy changes continuously near the walls, morelike the well shown in Fig. 8.8(c). This well is sometimes called a rounded well , whilethose of Fig. 8.8(a) and (b) are called square wells. To simplify our discussion we shallsuppose that the rounded well has U ( x) exactly constant, U ( x) = U 0, for x < 0 and x > a,as shown in Fig. 8.8(c).

In this section we wish to investigate the energy levels of a particle confined in anonrigid box such as either Fig. 8.8(b) or (c). As one might expect, the properties of bothwells are qualitatively similar. To be definite we shall consider mostly the rounded wellof Fig. 8.8(c).

Like the infinite well, the finite wells allow no states with E < 0, if we define the

zero of U at the bottom of the well. (See Problem 8.34.) An important difference is thatfor E > U 0 a particle in the finite well is not confined; that is, it has enough energy toescape to x = ±∞ . This means that the wave functions for E > U 0 are quite similar tothose of a free particle. In particular, the possible energies for E > U 0 are not quantized; but we shall not pursue this point here since our main interest is in the bound states,whose energies lie in the interval 0 < E < U 0.

A classical particle moving in a finite well with energy in the interval 0 < E < U 0.would simply bounce back and forth indefinitely. In the square well of Fig. 8.8(b) itwould bounce between the points x = 0 and x = a. For the rounded well, the points atwhich the particle turns around are determined by the condition E = U ( x) (since thekinetic energy must be zero at the turning point where the particle comes instantaneously

to rest). These points can be found graphically as in Fig. 8.9, by drawing a horizontal line

1 Until a few years ago one would have said that in this respect the perfectly rigid box is totallyunrealistic — a real bound system might require a large energy to pull it apart, but surely not an infiniteamount. As we discuss in Chapter 14, it now appears that subatomic particles like neutrons and protonsare made up of sub-subatomic particles called quarks, and that an infinite energy is needed to pull themapart (that is, they cannot be pulled apart). Thus a potential energy function like (8.76) may be morerealistic than we had formerly appreciated.

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8.22

at the height representing the energy E . The points, x = b and x = c, at which this linemeets the potential-energy curve are the two classical turning points, and a classical particle with energy E simply bounces back and forth between these points.

Let us now consider the Schrödinger equation,

2

2( ) [ ( ) ] ,

m x U x E ψ ψ ′′ = −h

for a quantum particle with the rounded potential well for its potential energy. We focusattention on a particular energy E in the range 0 < E < U 0 and consider whether or not theSchrödinger equation has an acceptable solution for this energy.

WAVE FUNCTIONS OUTSIDE THE WELL

We first note that in the regions x < 0 and x > a, the potential energy is constant,U ( x) = U 0 , and the Schrödinger equation is easily solved. If we set

2

02

2

[ ] ,

m

U E α − =h

then in the region x < 0 the general solution has the form

( ) , 0, x x x Ae Be xα α ψ −= + <

with A and B arbitrary. Now, as x → −∞ , the exponential xeα − blows up and is

physically unacceptable, Thus the form (8.79) is physically acceptable only if thecoefficient B = 0.

Similarly, in the region x > a, the general solution has the form

( ) , , x x x Ce De x aα α ψ −= + >

and this is acceptable only if the coefficient C = 0.

NUMERICAL SOLUTION OF THE SCHRÖDINGER EQUATION

Equations (8.79) and (8.80) give the form of the general solution of theSchrödinger equation in the particular regions x < 0 and x > a. We must now find a singlesolution for the entire interval −∞ < x < ∞ . For most potential energy functions U ( x) the

solution cannot be expressed in terms of known elementary functions, such as xe α − or sin

kx, in the interval 0 < x < a. When this is possible we say that the equation can be solved analytically, but there are very few real problems where there are analytic solutions of the Schrödinger equation (or Newton's second law, or any other differential equation).

Usually, one has to find the solution numerically using a calculator or computer. It iseasy to understand the general principles of such numerical calculations, and once youdo, we can discuss and understand the way in which acceptable solutions of theSchrödinger equation exist for some energies and not for others.

To start solving the Schrödinger equation, or any other second-order differential

equation, one needs to know the values of ψ and its derivative ψ ′ at one point x = x0.

(This is exactly analogous to the familiar statement that to solve Newton's second law

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8.23

one has to know the particle's position and velocity at one time t 0.) If the potential energyfunction is known, the Schrödinger equation,

2

2( ) [ ( ) ] ( ),

m x U x E xψ ψ ′′ = −

h

tells us 0( ) xψ ′′ in terms of the known 0( ) xψ . We therefore know ψ and its first twoderivatives at x = x0.

Suppose now that we want to find ψ at some other point x. Our first step is todivide the interval between x0 and x into n small intervals at points

0 1 2 1 ,n n

x x x x x x−

< < < < < =L

a distance x∆ apart. Knowing ψ , ψ ′ , and ψ ′′ at x0, we can now set up a procedure for

finding the same three functions at x1, then at x2, and so on, all the way to xn = x.

To find 1( ) xψ we use the approximation

( ) ( ) ( ) , f x x f x f x x′+ ∆ = + ∆which gives the value of f at x + x∆ , by approximating the curve of f between x and x +

x∆ , with a straight line of slope ( ) f x′ , as shown in Fig. 8.10. Applying this

approximation to the function ( ) xψ , we find that

1 0 0( ) ( ) ( ) , x x x xψ ψ ψ ′= + ∆

which gives 1( ) xψ in terms of known quantities. This is, of course, only an

approximation, but by making x∆ smaller we can achieve any desired accuracy — at theexpense of a longer overall computation. Applying the same approximation (8.82) to

( ) xψ ′ , we find similarly that

1 0 0( ) ( ) ( ) , x x x xψ ψ ψ ′ ′ ′′= + ∆

which gives 1( ) xψ ′ in terms of known quantities. Knowing 1( ) xψ and 1( ) xψ ′ , we can use

the Schrödinger equation to find 1( ) xψ ′′ , and we are now ready to start again and move

from x1 on to x2, and so on. After n steps this procedure gives us the wave function at thedesired point xn = x, as illustrated in Fig. 8.11.

In the problem of the particle in a nonrigid box we know that for x < 0 the wave

function must have the form ( ) xψ = x Aeα if it is to be physically acceptable. We can now

imagine using the numerical procedure just outlined to follow ( ) xψ across the well from

x = 0 to x = a and beyond. In the region x > a we already know that any solution has theform

. x xCe Deα α −−

By comparing our numerical solution with this form we can identify coefficients C and D for our solution. If we find that 0C ≠ , our solution blows up as x → ∞ and is not physically acceptable. In this case the energy E is not an allowed energy of the particle,and we must try a different value of E. If we find that C is zero, our solution is well behaved both as x → −∞ and as x → ∞ and E is an allowed energy.

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8.24

SOME GENERAL PROPERTIES OF WAVE FUNCTIONS

To understand when we should expect to find allowed energies, it is useful toexplore a little farther the general behavior of solutions of the Schrödinger equation

2

2( ) [ ( ) ] ( ).

m x U x E xψ ψ ′′ = −

hIf, as usual, we focus attention on a particular value of E (with 0 < E < U 0) we candistinguish two important ranges of x: those x where the factor [U ( x) – E ] in (8.85) is positive, and those x where it is negative. The dividing points between these regions arethe classical turning points x = b and x = c, where U ( x) = E. (These were defined in Fig.8.9 and are shown again in Fig. 8.12.) The region where [U ( x) – E ] is positive is outsidethese turning points (x < b and x > c) and is often called the classically forbidden region,since a classical particle with energy E cannot penetrate there. The region where [U ( x) – E ] is negative is the interval b < x < c and is called the classically allowed region. We

shall see that the behavior of the wave function ( ) xψ is quite different in these two

regions.

In the region where [U ( x) – E ] is positive, the Schrödinger equation has the form

( ) (positive function) ( ). x xψ ψ ′′ = ×

In an interval where ( ) xψ is positive, this implies that ( ) xψ ′′ is also positive and hence

that ( ) xψ curves upward , as in Fig. 8.13(a) and (b). If ( ) xψ is negative then (8.86)

implies that ( ) xψ ′′ is negative and hence that ( ) xψ curves downward , as in Fig. 8.13(c)

and (d). In either case, we see that ( ) xψ curves away from the axis. In particular, if the

Schrödinger equation has the form (8.86) as x → ∞ , as it does in our case, then either ( ) xψ will blow up as x → ∞ [as when ( ) xψ = xCe

α — see Fig. 8.13(a) and (c)], or

( ) xψ will approach zero as x → ∞ [as when ( ) xψ = x De α − — see Fig. 8.13(b) and (d)].

In the region b < x < c, the Schrödinger equation has the form

( ) (negative function) ( ), x xψ ψ ′′ = ×

and we can argue that ( ) xψ curves toward the axis and tends to oscillate, as follows: If

( ) xψ is positive, then ( ) xψ ′′ is negative and ( ) xψ curves downward , as in Fig. 8.14(a);

if ( ) xψ is negative, the argument reverses and ( ) xψ bends upward, as in Fig. 8.14(b). In

either case, ( ) xψ curves toward the axis. If the interval b < x < c is sufficiently wide, a

function bending toward the axis will cross the axis and immediately start bending theother way, as in Fig. 8.14(c). Thus we can say that in the region b < x < c the wavefunction tends to oscillate.

If the "negative function" in (8.87) has a large magnitude, then ( ) xψ ′′ tends to be

large and ( ) xψ curves and oscillates rapidly. Conversely, when the "negative function" is

small, ( ) xψ bends gradually and oscillates slowly. This is all physically reasonable: The

"negative function" in (8.87) is proportional to[U ( x) – E ], which is just minus the kineticenergy; according to de Broglie, large kinetic energy means short wavelength and hencerapid oscillation, and vice versa.

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8.25

Now that we understand the qualitative behavior of solutions of the Schrödinger equation for all x, let us return to our hunt for an acceptable solution. We can start in theregion x < 0, where U ( x) is constant (Fig. 8.12), and begin with the known, acceptableform

( ) , ( 0). x x Ae xα ψ = <

When we move to the right, our solution will cease to have this explicit form once U ( x)starts to vary, but it will continue to bend away from the axis until x reaches the point b.At x = b, it will start oscillating and continue to do so until x = c, where it will start

curving away from the axis again. Thus the general appearance of ( ) xψ will be as shown

in Fig. 8.15, with two regions where ( ) xψ bends away from the axis, separated by one

region where ( ) xψ oscillates. Figure 8.15 shows a solution that blows up on the right.

We must now find out if there are any values of E for which the solution is well behaved both on the left and right.

SEARCHING FOR ALLOWED ENERGIES

Let us begin a systematic search for allowed energies, starting with E close tozero. We shall show first that with E sufficiently close to zero, an acceptable wave

function is impossible. We start with the wave function x Aeα in the region x < 0, and

follow ( ) xψ to the right. When we reach the classical turning point x = b, ( ) xψ has a

positive slope and starts to bend toward the axis. But with E very small, ( ) xψ bends very

slowly. Thus when we reach the second turning point x = c, the slope is still positive.

With ψ and ψ ′ both positive, ( ) xψ continues to increase without limit, as shown in Fig.

8.16. Therefore, the wave function which is well behaved as x → −∞ blows up as x → +∞ , and we conclude that there can be no acceptable wave function with E veryclose to zero.

Suppose now that we slowly increase E , continuing to hunt for an allowed

energy. For larger values of E the kinetic energy is larger and, as we have seen, ( ) xψ

bends more rapidly inside the well. Thus it can bend enough that its slope becomesnegative, as shown in Fig. 8.17, curve 2. Eventually, its value and slope at the right of the well will be just right to join onto the solution, which is well behaved as x and wehave an acceptable wave function (curve 3 in Fig. 8.17). If we increase E any further,

( ) xψ will bend over too far inside the well and will now approach −∞ as x → ∞ , like

curve 4 in Fig. 8.17. Evidently, there is exactly one allowed energy in the range exploredso far.

If E is increased still further, the wave function may bend over and back just

enough to fit onto the function that is well behaved as x → ∞ as in Fig. 8.18(a). If thishappens, we have a second allowed energy. Beyond this, we may find a third acceptablewave function, like that in Fig. 8.18(b), and so on.

Figure 8.19 shows the first three wave functions for a finite square well, besidethe corresponding wave functions for an infinitely deep square well of the same width a. Notice the marked similarity of corresponding functions: The first wave function of theinfinite well fits exactly half an oscillation into the well, while that of the finite well fitssomewhat less than half an oscillation into the well, since it doesn't actually vanish until

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8.26

x = ±∞ . Similarly, the second function of the infinite well makes one completeoscillation, whereas that of the finite well makes just less than one oscillation. For thisreason the energy of each level in the finite well is slightly lower than that of thecorresponding level in the infinite well.

It is useful to note that the ground-state wave function for any finite well has no

nodes, while that for the second level has one node, and that for the nth level has 1n − nodes. This general trend (more nodes for higher energies) could have been anticipated,since higher energy corresponds to a wave function that oscillates more quickly andhence has more nodes.

The wave functions in Fig. 8.19 illustrate two more important points. First, thewave functions of the finite well are nonzero outside the well, in the classically forbiddenregion. (Remember that a classical particle with energy 0 < E < U 0 cannot escape outsidethe turning points.) However, since the wave function is largest inside the well, the

particle is most likely to be found inside the well; and since ( ) xψ approaches zero

rapidly as one moves away from the well, we can say that our particle is bound inside, or close to, the potential well. Nevertheless, there is a definite, nonzero probability of

finding the particle in the classically forbidden regions. This difference between classicaland quantum mechanics is due to the wave nature of quantum particles. The ability of thequantum wave function to penetrate classically forbidden regions has importantconsequences, as we discuss in Section 13.9.

A second important point concerns the number of bound states of the finite well.With the infinite well one can increase E indefinitely and always encounter more boundstates. With the finite well, however, the particle is no longer confined when E reachesU 0, and there are no more bound states. The number of bound states depends on the welldepth U 0 and width a, but it is always finite.

8.9 The Simple Harmonic Oscillator (optional1

)As another example of a one-dimensional bound particle we consider the simpleharmonic oscillator. The name simple harmonic oscillator (or SHO) is used, inclassical and quantum mechanics, for a system that oscillates about a stable equilibrium point to which it is bound by a force obeying Hooke's law.

Familiar classical examples of harmonic oscillators are a mass suspended from anideal spring and a pendulum oscillating with small amplitude. Before we give anyquantum examples, it may be worth recalling why the harmonic oscillator is such animportant system. If a particle is in equilibrium at a point x0, the total force on the particle is zero at x0; that is, F ( xo) = 0. If the particle is displaced to a neighboring point x,the total force will be approximately

0 0 0( ) ( ) ( )( ), F x F x F x x x′= + −

with F ( xo) = 0 in this ease. [This is just the approximation (8.82) and should be good for any x sufficiently close to x0.] If x0 is a point of stable equilibrium, the force is a restoring

1 If you choose to omit this section, be aware that we shall use its results in the optional Sections16.6 and 16.7 on excited states of molecules, and in Section 17.7 on vibrations in solids.

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8.27

force; that is, F ( x) is negative when x – x0 is positive, and vice versa. Therefore, 0( ) F x′

must be negative and we denote it by – k , where k is called the force constant. With

0( ) F x′ = – k and F ( x0) = 0, Eq. (8.88) becomes

0( ) ( ), F x k x x= − −

which is Hooke's law. Thus any particle oscillating about a stable equilibrium point willoscillate harmonically, at least for small displacements ( x - x0).

An important example of a quantum harmonic oscillator is the motion of any oneatom inside a solid crystal; each atom has a stable equilibrium position relative to itsneighboring atoms and can oscillate harmonically about that position. Another importantexample 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 the potentialenergy U . This is easily found by integrating (8.89) to give

0

2102( ) ( ) ( )

x

xU x F x dx k x x= − = −∫ if we take U to be zero at x0. Thus in quantum mechanics the SHO can be characterizedas a system whose potential energy has the form (8.90). This function is a parabola, withits minimum at x = x0, as shown in Fig. 8.20(a).

It is important to remember that (8.89) and (8.90) are approximations that areusually valid only for small displacements from x0. This point is illustrated in Fig.8.20(b), which shows the potential energy of a typical diatomic molecule, as a functionof the distance r between the two atoms. The molecule is in equilibrium at the separationr 0. For r close to r 0, the potential energy is well approximated by a parabola of the form

U(r) =21

02( )k r r − , but when r is far from r 0, U(r) is quite different. Thus for small

displacements the molecule will behave like an SHO, but for large displacements it willnot. This same statement can be made about almost any oscillating system. (For example,the simple pendulum is well known to oscillate harmonically for small amplitudes, butnot when the amplitude is large.) It is because small displacements from equilibrium arevery 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 our origin at the

equilibrium position, then x0 = 0 and the force is F k x= − . If the particle has mass m,

then Newton's second law reads

.ma kx= −

If we define the classical angular frequency

c

k

mω =

this becomes

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8.28

22

c2,

d x x

dt ω = −

which has the general solution

c csin cos . x a t b t ω ω = +

Thus the position of the classical SHO varies sinusoidally in time with angular frequency

cω . If we choose our origin of time, t = 0, at the moment when the particle is at x = 0 and

moving to the right, then (8.92) takes the form

csin . x a t ω =

The positive number a is the amplitude of the oscillations, and the particle oscillates between x = a and x = – a. In other words, the points x = ± a are the classical turning

points. When the particle is at x = a, all of its energy is potential energy; thus E =21

2ka

and hence

2 . E ak

=

As one would expect, the classical amplitude a increases with increasing energy. To findthe allowed energies of a quantum harmonic oscillator we must solve the Schrödinger equation with U ( x) given by

212

( ) ,U x kx=

where we have again chosen x0 = 0. Qualitatively, the analysis is very similar to that for the finite well discussed in the preceding section. For any choice of E the solutions will bend away from the axis outside the classical turning points x = a and x = a− , and will

oscillate between these points. As before, there is just one solution that is well behaved as x → −∞ , and for most values of E this solution blows up as x → ∞ . Thus the allowedenergies are quantized. The potential energy (8.95) increases indefinitely as x movesaway from the origin [see Fig. 8.20(a)] and the particle is therefore confined for allenergies. 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 wells is thatthe Schrödinger equation for the SHO can be solved analytically. The solution is quitecomplicated and will not be given here. On the other hand, the answer for the energylevels is remarkably simple: The allowed energies turn out to be

1 3 5, , , .

2 2 2

k k k E

m m m= h h h L

The quantity /k m is the frequency cω , defined in (8.91), of a classical oscillator with

the same force constant and mass. It is traditional to rewrite the allowed energies in terms

of cω , as

( )1c2

, 0,1,2, .n

E n nω =+ =h L

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8.29

As anticipated, the allowed energies are all quantized, and there are levels with arbitrarilylarge energies. A remarkable feature of (8.96) is that the energy levels of the harmonicoscillator are all equally spaced. This is a property that we could not have anticipated inour qualitative discussion, but which emerges from the exact solution.

The allowed energies of the harmonic oscillator are shown in Fig. 8.2 1, which

also shows the corresponding wave functions (each drawn on the line representing itsenergy). Notice the similarity of these wave functions to those of the finite well shown inFig. 8.19. In particular, just as with the finite well, the lowest wave function has nonodes, the next has one node, and so on. Notice also that the wave functions with higher energy spread out further from x = 0, just as the classical turning points x = ± a movefarther out when E increases. Finally, note that the ground state (n = 0) has a "zero-point"

energy (equal to 12 cω h ) as required by the uncertainty principle.

The wave functions shown in Fig. 8.21 were plotted using the known analyticsolutions, which we list in Table 8.1, where, for convenience, we have introduced the parameter

c

.bmω

= h

(This parameter has the dimensions of a length and is, in fact, the half-width of the SHOwell at the ground-state energy — see Problem 8.44.) The factors A0, A1, and A2 in thetable are normalization constants1.

TABLE 8.1 The energies and wave functions of the first three levels of a quantum

harmonic oscillator. The length b is defined as / cmω h .

n E n ( ) xψ

0 12 cω h

2 2/ 2

0

x b A e−

1 32 cω h 2 2/ 2

1

x b x A e

b

−

2 52 cω h 2 2

2/ 2

2 21 2 x b x

A eb

− −

1 The three functions in Table 8.1 illustrate, what is true for all n, that the wave function of the

nth level has the form2 2

( ) exp( / 2 )n P x x b− , where ( )n P x is a certain polynomial of degree n, called a

Hermite polynomial.

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8.30

EXAMPLE 8.4 Verify that the n = 1 wave function given in Table 8.1 is a solution of

the Schrödinger equation with E = 32 cω h . (For the cases n = 0 and n = 2, see Problems

8.45 and 8.48.)

The potential energy is U =21

2 k x and the Schrödinger equation is therefore

( )2122

2.

mkx E ψ ψ ′′ = −

h

Differentiating the wave function 1ψ , of Table 8.1, we find that

2 22

/ 2

1 1 3

1 x b x A e

b bψ −

′ = −

and

2 2 2 23 2

/ 2 / 21 1 13 5 4 23 3 , x b x b x x x x A e A e

b b b b bψ − − ′′ = − + = −

where the last three factors together are just 1ψ . Using (8.97) to replace b, we find

2 2 2

c c1 12

3.

m x mω ω ψ ψ

′′ = − h h

Replacing2

cω by k /m in the first term, we get

21 c 12

2 1 3 ,2 2

m kxψ ω ψ ′′ = − h

h

which is precisely the Schrödinger equation (8.98), with E = 32 cω h as required.

The properties of the quantum SHO are nicely illustrated by the example of adiatomic molecule, such as HCl A diatomic molecule is, of course, a complicated, three-dimensional system. However, the energy of the molecule can be expressed as the sum of three terms: an electronic term, corresponding to the motions of the individual electrons;a rotational term, corresponding to rotation of the whole molecule; and a vibrationalterm, corresponding to the in-and-out, radial vibrations of the two atoms. Carefulanalysis of molecular spectra lets one disentangle the possible values of these three terms.

In particular, the vibrational motion of the two atoms is one-dimensional (along the line joining the atoms) and the potential energy approximates the SHO potential. Therefore,the allowed values of the vibrational energy should approximate those of an SHO, andthis is amply borne out by observation, as illustrated in Fig. 8.22.

The molecular potential energy U (r ) in Fig. 8.22 deviates from the parabolicshape of the SHO at higher energies. Therefore, we would expect the higher energylevels to depart from the uniform spacing of the SHO. This, too, is what is observed.

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8.31

The observation of photons emitted when molecules make transitions betweendifferent vibrational levels is an important source of information on interatomic forces,as we discuss further in Chapter 16. These same photons can also be used to identify themolecule that emitted them. For example, the H2 molecule emits infrared photons of

frequency 141.2 10× Hz when it drops from one vibrational level to the next. This

radiation is used by astronomers to locate clouds of H2 molecules in our galaxy.

old chapter 8

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