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添付資料は量子力学の教科書

”Elementary Quantum Mechanics”Preliminary Edition

David S. SaxonHolden-Day, Inc., 1964

の第４章である。章末の問題に解答せよ。解答は日本語でよい。

Chapter 4

MOTION OF A FREE PARTICLE

4.1 Motion of a Wave Packet; Group Velocity

We have thus far been concerned with the wave function of a system at some fixed instant.

We now begin our discussion of how such states develop in time. In the present chapter we

shall consider the simplest possible problem, namely that of the motion of a particle free

from external influences. As a starting point, we return to equation (3.2) which provides

a general description of a free particle wave packet

ψ(x, t) =1√2πh

∫ ∞

−∞φ(p) exp

[ipxh

− iω(p)t]dp (4.1)

where ω(p) is as yet unknown. We shall assume that ψ is prescribed at t = 0, and we

shall denote its initial value by ψ0(x),

ψ0(x) = ψ(x, t = 0) =1√2πh

∫ ∞

−∞φ(p) exp

[ipxh

]dp. (4.2)

Our problem then is, given an initial wave packet, ψ0(x), how does it develop in time?

This is, what is ψ(x, t)?

As a preliminary step, and as an example, we first consider dispersionless propagation

(as in the propagation of light in free space) where ω is proportional to p/h. The constant

of proportionality has the dimensions of a velocity and we denote it by c. Thus, in this

example

ω =cp

h=

2πc

λand Eq.(4.1) becomes

ψ(x, t) =1√2πh

∫ ∞

−∞φ(p) eip(x−ct)/h dp.

Comparison with Eq.(4.2) then shows that ψ(x, t) is exactly the same function of (x− ct)

as ψ0 is of x. In other words,

ψ(x, t) = ψ0(x− ct)

1

2 CHAPTER 4. MOTION OF A FREE PARTICLE

and the wave packet simply travels to the right with velocity c, without distortion, no

matter its initial shape. (Our treatmemt of this problem in optical propagation is con-

siderably oversimplified. The case where ω and p have opposite sign, ω = −cp/h, must

also be considered. The motion of a wave packet in optics accordingly is not determined

if only ψ0(x) is given. It turns out that in addition ∂ψ/∂t(x, t = 0) must be prescribed.

This is a consequence of the fact that the electromagnetic wave equation is of second

order in the time.)

We now go back to the general case where ω(p) is unknown. Of course we can’t calculate

anything for arbitrary ω(p) and for a completely arbitrary wave packet. However we do

not need to particularize too much in order to exhibit the essential features. The first

assumption we shall make is that φ(p) is a smooth wave packet in momentum space,

centered about p0, say, and of width Δp. To emphasize this behavior we rewrite φ(p) in

the form

φ(p) = g(p− p0)

where g is a smooth function which falls rapidly to zero when its argument exceeds Δp in

magnitude. This means that the main contribution to the integral in Eq.(4.1) come from

the neighborhood of p0.

Secondly we assume that ω(p) is a smooth function of p. If so, we can expand ω in a

Taylor series about p0,

ω(p) = ω(p0) + (p− p0)dω(p0)

dp+

1

2(p− p0)

2 dω2(p0)

dp2+ · · ·

≈ ω0 + (p− p0)vg

h+ (p− p0)

2α + · · ·

where

ω0 = ω(p0), vg = hdω(p0)

dp, α =

1

2

dω2(p0)

dp2.

Introducing also the new variable s = p− p0, we can now rewrite Eq.(4.1) as

ψ(x, t) = f(x, t) exp[ip0x

h− iω0t

]

where f(x, t), which determines the envelope of the wave packet, is given by

f(x, t) =1√2πh

∫ ∞

−∞ds exp

[ish

(x− vgt) − iαs2t+ · · ·]. (4.3)

At the same time, Eq.(4.2) becomes

ψ0(x) = f0 (x) eip0x/h

where f0 (x), the initial envelope function, is given by

f0 (x) = f(x, t = 0) =1√2πh

∫ ∞

−∞ds g(s) eisx/h. (4.4)

4.2. THE CORRESPONDENCE PRINCIPLE REQUIREMENT 3

Now, the main contribution to these integrals comes when s is less than Δp in magni-

tude. Hence, if t is small enough that

α (Δp)2 t� 1

the term in the exponent of Eq.(4.3) quadratic in s can be neglected and the envelope

function becomes

f(x, t) =1√2πh

∫ ∞

−∞ds g(s) eis(x−vg t)/h.

Comparison with Eq.(4.4) then shows that f(x, t) is the same function of (x− vgt) as is

f0 of x, that is to say

f(x, t) = f0 (x− vgt). (4.5)

We have thus demonstrated that for t� t0, where

t0 =1

α (Δp)2=

2

(Δp)2 d2ω(p0)/dp2

(4.6)

the wave packet travels undistorted with velocity vg, where

vg = hωdω(p0)

dp. (4.7)

The quantity vg is called the group velocity of the waves, since it represents the velocity

of motion of a group of waves, namely, those which make up the wave packet. It should be

contrasted with the phase velocity, which is the velocity with which the phase of a given

pure wave advances and which is given by vp = hω/p0. For dispersionless propagation,

where ω is proportional to p, these two velocities are equal, but in general they are quite

different.

We emphasize that our result holds only for times short compared to t0, defined in

Eq.(4.6). Eventually, when t exceeds t0, the exponential begins to oscillate very rapidly

for s smaller than Δp. When that happens, the effective domain of integration in p

is reduced in size and this produces a corresponding increase in the width of the wave

packet in configuration space, according to the analysis leading to our discussion of the

uncertainty principle. This all means that in general, a wave packet initially travels

undistorted with the group velocity vg, but eventually begins to spread out in space. We

shall give some examples later.

4.2 The Correspondence Principle Requirement

We now use the following argument to deduce the form of ω(p) for quantum mechanical

waves. If we have a well-defined wave packet in configuration space, we have seen that it

4 CHAPTER 4. MOTION OF A FREE PARTICLE

travels with the group velocity vg, at least for short enough times. In the classical limit,

this limitation on the time must become unimportant, that is, t0 must become very large

compared to all relevant times. Assuming this to be so, we accordingly demand that

vg = vclassical =p0

m.

Hence, from Eq.(4.5)

hdω(p0)

dp=

p0

m.

and, dropping the subscript on p0,

hω =p2

2m= E (4.8)

which is the Planck relation. In a sense, we have thus deduced the Planck relation as a

consequence of our formulation and its interpretation.

Note that Eq.(4.8) is arbitrary up to a constant of integration, which we have sim-

ply set equal to zero. This is related to the freedom of choice of the zero of en-

ergy and implies a similar freedom in the choice of quantum mechanical frequency.

Only differences in energy, and also therefore in frequency, are physically significant.

It is of interest to examine the time t0 when a quantum mechanical wave packet begins

to spread out significantly. We have, from Eqs.(4.6) and (4.8),

t0 =2mh

(Δp)2. (4.9)

An instructive alternative way to write this is to use the uncertainty relation to express

Δp as h/Δx, where Δx is the spatial extent of the initial wave packet. Thus,

t0 ≈ 2m(Δx)2

h.

For a macroscopic particle, say of mass one gram, whose position is defined to even 10−4

cm (1 micron), we have

t0 ≈ 1019 sec.

The age of the universe is about 3×1010 years or about 1018 sec. Hence a wave packet for

a macroscopic particle holds together without spreading for a period comparable to the

age of the universe. This extablishes that classical trajectories for macroscopic systems

are not in conflict with quantum mechanics. On the other hand, for an electron whose

position is defined to say 10−8 cm,

t0 ≈ 10−16 sec

and a classical description is meaningless.

4.3. PROPAGATION OF A FREE PARTICLE WAVE PACKET 5

It is possible to give a simple physical interpretation of the time t0 along the following

lines. The group velocity of propagation of de Broglie waves is p/m. In a time t, two

segments of a wave packet differing in momentum by Δp/2, say, will differ in distance

travelled by an amount Δp t/2m. When this distance becomes comparable with the width

Δx of the initial packet, the width of the packet will begin to increase significantly. Hence

the packet begins to spread at a time t0 defined by

Δp

2mt0 ≈ Δx ≈ h

Δp

which is Eq.(4.9). This argument implies that once wave packets start spreading they do

so at a rate linear in the time. We shall shortly verify that this is indeed the case.

4.3 Propagation of a Free Particle Wave Packet

With our identification of ω(p), we can now rewrite Eq.(4.1) as

ψ(x, t) =1√2πh

∫ ∞

−∞φ(p) exp

[ipxh

− ip2t

2mh

]dp (4.10)

which is a general representation of a time-dependent wave function for a free particle. If

ψ0(x) = ψ(x, 0) is prescribed, φ(p) is given, according to Eq.(3.17), by

φ(p) =1√2πh

∫ ∞

−∞ψ0(x)e

−ipx/h dx.

Substitution of this expression into Eq.(4.10) then gives the highly interesting result,

ψ(x, t) =1√2πh

∫ ∫ ∞

−∞dx′ dp ψ0(x

′) exp[ ip(x− x′)

h− ip2t

2mh

]or

ψ(x, t) =∫ ∞

−∞ψ(x′, 0)K(x′, x; t) dx′ (4.11)

where K is the improper integral

K(x′, x; t) =1

2πh

∫ ∞

−∞dp exp

[ip(x− x′)h

− ip2t

2mh

]. (4.12)

Evaluating this integral, using the methods of Appendix I, we obtain

K(x′, x : t) =

√m

2πihtexp

[i(x− x′)2m

2ht

]. (4.13)

The quantity K, which characterizes the propagation of a free particle wave function,

is called the propagator, in this case, the free particle propagator. Note, from either

Eq.(4.12) or (4.13), that

K(x′, x; 0) = δ(x− x′)

6 CHAPTER 4. MOTION OF A FREE PARTICLE

and that, as t increases from zero, K begins to spread out, so that contribution to the

integral come from an increasing range of values of (x− x′).

We now discuss the time development of a free particle wave packet in momentum

space. For this purpose, we first define time-dependent momentum space wave functions

φ(p, t) by writing the obvious generalization

φ(p, t) =1√2πh

∫ ∞

−∞ψ(x, t) e−ipx/h dx. (4.14)

Of course t here simply plays the role of a parameter, which was taken to be zero for con-

venience in our earlier discussion. Otherwise stated, the qunatity we previously denoted

by φ(p) is merely φ(p, t = 0). By the Fourier theorem,

ψ(x, t) =1√2πh

∫ ∞

−∞φ(p, t)eipx/h dp. (4.15)

Comparing this expression with Eq.(4.10), we see at once that,

φ(p, t) = φ(p) e−ip2t/2mh = φ(p, t = 0) e−ip2t/2mh (4.16)

while

ρ(p, t) = |φ(p, t) |2 = |φ(p) |2 = ρ(p, t = 0).

In other words, only the phase of an arbitrary free particle wave packet in momentum

space changes in time. The probability density is stationary and the expectation value of

any function of momentum is independent of time. This result is no surprise, of course.

We are talking about the states of a free particle and the momentum is a constant of the

motion, classically. In view of our use of the correspondence principle, the momentum of

a free particle must be a constant of the motion quantum mechanically, as well, and so it

has turned out.

We have emphasized that the expectation value of any function of the momentum is

simple for free particles. As a particular and important example, consider the energy E

which is just p2/2m. We have

〈E〉 =∫ ∞

−∞φ∗(p, t)

p2

2mφ(p, t) dp

now, according to Eq.(4.16),

p2

2mφ(p, t) = − h

i

∂φ(p, t)

∂t

and hence this expression can be rewritten as

〈E〉 =∫ ∞

−∞φ∗(p, t)

[− h

i

∂φ(p, t)

∂t

]dp.

4.4. TIME DEVELOPMENT OF A GAUSSIAN WAVE PACKET 7

Indeed, for any function f(E), we have, by an obvious extension of the argument,

〈f(E)〉 =∫ ∞

−∞φ∗(p, t) f

(− h

i

∂

∂t

)φ(p, t) dp. (4.17)

Since φ(p, t) is an arbitrary free particle wave function, we conclude that in momentum

space the energy E is represented by the operation of time differentiation multiplied by

(−h/i), that is,

E = − h

i

∂

∂t. (4.18)

What about E in configuration space? Since t enters only as a parameter with respect

to the transformation between momentum and configuration space, it is clear that E

has exactly the same representation in both spaces. That this is correct follows upon

expressing φ(p, t) and φ∗(p, t) in Eq.(4.17) in terms of ψ and ψ∗, using Eq.(4.18), and

evaluating the integral over p. One obtains at once,

〈f(E)〉 =∫ ∞

−∞ψ∗(x, t) f

(− h

i

∂

∂t

)ψ(x, t) dx (4.19)

as asserted.

4.4 Time Development of a Gaussian Wave Packet

Before continuing with the general discussion, it is helpful to work out a specific example

in some detail. We shall now do so for the case of a wave packet which, initially, is

Gaussian in from. In particular we consider the wave packet of Eq.(III-32),

ψ(x, 0) = ψ0(x) =1√L√π

exp[ip0x

h− x2

2L2

](4.20)

which describes a particle initially localized about the origin within a distance L and

whose momentum has the expectation value p0. Using Eqs.(4.11) and (4.13), we then

obtain, upon evaluating the integral by the methods of Appendix I

ψ(x, t) =

[√π(L+

iht

mL

)]−1/2

exp[L(−x2/2L2 + ip0x/h− ip 2

0 t/2mh)

L+ iht/mL

](4.21)

and the probability density is

ρ(x, t) = |φ |2 =

[π(L2 +

h2t2

m2L2

)]−1/2

exp[− (x− p0t/m)2

L2 + h2t2/m2L2

]. (4.22)

8 CHAPTER 4. MOTION OF A FREE PARTICLE

Comparing Eq.(4.22) to ρ(x, t = 0), which is simply,

ρ(x, t = 0) =1√πL

e−x2/L2

we see that ρ(x, t) involves only two changes. First, the center of the wave packet moves

with the group velocity p0/m. Secondly, the width of the wave packet increases with time.

Calling this width L(t), we have

L(t) =

√L2 +

h2t2

m2L2

which is entirely in agreement with our predictions. In particular, it is seen that the wave

packet is initially undistorted and begins to spread appreciably only when ht/mL2 is of

the order of unity, in agreement with Eq.(4.9). Further, when t becomes very large, the

width of the packet grows linearly, at the rate

h

mL≈ Δp

m

also as predicted.

4.5 The Free Particle Schrodinger Equation

We return now to a discussion of the time development of an arbitrary free particle wave

function. We have derived several equivalent integral representations for the wave function

ψ(x, t) and we now seek a differential characterization. This is most easily obtained from

Eq.(4.10). Note that, differentiating under the integral sign,

− h

i

∂ψ(x, t)

∂t=

1√2πh

∫ ∞

−∞φ(p)

p2

2mexp

[ipxh

− ip2t

2mh

]dp

and that exactly the same result is obtained upon evaluating

− h2

2m

∂2ψ(x, t)

∂x2,

again by differentiating uder the integral sign. We thus conclude that any function ψ(x, t)

defined by Eq.(4.10), and therefore any free particle wave function, satisfies the partial

differential equation

− h2

2m

∂2ψ(x, t)

∂x2= − h

i

∂ψ(x, t)

∂t. (4.23)

Otherwise stated, Eq.(4.10) is simply a way of writing the general solution to Eq.(4.23),

which is Schrodinger equation in configuration space for a free particle in one dimension.

4.5. THE FREE PARTICLE SCHRODINGER EQUATION 9

Note that it is explicitly complex and that it is of the first order in the time, and not of

second order as in optics.

The interpretation of this equation is quite simple and direct if we recall that in con-

figuration space, the momentum operator is

p =h

i

∂

∂x,

according to Eq.(III-20) while the energy operator is

E = − h

i

∂

∂t

according to Eq.(4.18). Hence, Schrodinger equation is the operator equation

p2

2mψ(x, t) = Eψ(x, t). (4.24)

Classically, of course, for a free particle

E =p2

2m

and we see that the corresponding quantum mechanical equation requires that the wave

function for a free particle be such that it yields the same result when operated upon

by the total energy operator E as when operated upon by the kinetic energy operator

p2/2m. This condition ensures that the expectation value of any function of total energy

is exactly the same as the expectation value of the same function of the kinetic energy,

〈f(E)〉 = 〈f(p2/2m)〉,

which, in turn, ensures that the requirement of the correspondence principle are satisfied.

Of course Schrodinger equation can also be written in momentum space

p2

2mφ(p, t) = Eφ(p, t) = − h

i

∂φ

∂t(4.25)

which is actually much simpler than Eq.(4.24) because p is now a purely numerical op-

erator. Since Eqs.(4.24) and (4.25) have exactly the same form, it is now necessary to

explicitly identify the representation and we can write symbolically

p2

2mΨ = EΨ (4.26)

which is intended to convey the idea that the representation is unspecified. In configura-

tion space Ψ = ψ(x, t), in momentum space Ψ = φ(p, t).

10 CHAPTER 4. MOTION OF A FREE PARTICLE

Because of its simple form, the solution to Eq.(4.25) is trivial and immediate. We can

rewrite Eq.(4.25) in the form

∂ lnφ(p, t)

∂t= − i

h

p2

2m

whence

φ(p, t) = φ(p, t0) exp[− i

h

p2(t− t0)

2m

](4.27)

where φ(p, t0) is arbitrary. This is recognized as an obvious generalization of our earlier

result to arbitrary initial times, t0. The solution of Schrodinger equation in configuration

space, where it is a partial differential equation, is more complicated and we defer this

question for the moment.

4.6 Conservation of Probability

In our development of the quantum mechanical laws, we have idenfied Ψ ∗Ψ as a probability

density and have assumed that Ψ is normalizable. In particular, we assumed that, at any

arbitrarily chosen instant, t, we can choose Ψ in such a way that

∫ ∞

−∞ψ∗(x, t)ψ(x, t) dx = 1

and similarly in momentum space. Since the time development of Ψ is prescribed by

Schrodinger equation, we must verify that this condition, if imposed at one instant, will

continue to be satisfied as time goes on. Given our interpretation of ψ as the probability

amplitude, the normalization condition is simply the statement that the probability of

finding the particle somewhere is unity. We are thus seeking to verify that this probability

remains unity as time goes on, that is, that probability is conserved.

The proof that this is so is straightforward. Writing

P (t) =∫ ∞

−∞ψ∗(x, t)ψ(x, t) dx,

we want to show that dP/dt is zero for any arbitrary solution of Schrodinger equation.

We havedP

dt=∫ ∞

−∞∂

∂t

[ψ∗(x, t)ψ(x, t)

]dx =

∫ ∞

−∞

[∂ψ∗

∂tψ + ψ∗∂ψ

∂t

]dx.

Now according to Schrodinger equation in configuration space, Eq.(4.23),

∂ψ

∂t=

ih

2m

∂2ψ

∂x2,

∂ψ∗

∂t= − ih

2m

∂2ψ∗

∂x2.

4.7. DIRAC NOTATION 11

Thus,

dP

dt=

ih

2m

∫ ∞

−∞

[ψ∗∂

2ψ

∂x2− ψ

∂2ψ∗

∂x2

]dx =

ih

2m

∫ ∞

−∞∂

∂x

[ψ∗∂ψ∂x

− ψ∂ψ

∂x

]dx

=ih

2m

[ψ∗∂ψ∂x

− ψ∂ψ

∂x

]∣∣∣∣∣∞

−∞.

Since the normalization of ψ requires that it vanishes at ±∞, we see that the right side

vanishes and dP/dt is indeed zero.

This is a crucial point that for our whole interpretation and we want to emphasize that

the proof would have failed had Schrodinger equation been other than first order in the

time derivative. This is even more apparent in momentum space, where we have

P (t) =∫ ∞

−∞φ∗(p, t)φ(p, t) dp

dP

dt=∫ ∞

−∞

[∂φ∗

∂tφ+ φ∗∂φ

∂t

]dp =

∫ ∞

−∞

[ ih2m

p2φ∗φ− ih

2mp2φ∗φ

]dp = 0.

If, instead of (4.25), Schrodinger equation were, for example,

( p2

2m

)2φ = E2φ = −h2 ∂

2φ

∂t2

the proof fails, as is readily verified.

4.7 Dirac Notation

We have consistently developed our formulation in both momentum and configuration

space and have attempted to present the basic laws in a form independent of the particular

representation under consideration. We now introduce a notation, originated by Dirac,

which will permit us to write our equations in a representation independent way. Dirac’s

notation is actually far more general than we require at present, or are prepared to

understand. We shall merely introduce one or two definitions and will generalize these,

and add others, as the need arises.

First, we introduce a representation-independent way of writing the expectation value.

Let Ψ denote some arbitrary, but definite wave function, in whatever representation. Let

A be some arbitrary operator. The expectation value of A in the state Ψ is written

〈A〉 = 〈Ψ |A|Ψ 〉.

In the position representation, say,

〈Ψ |A|Ψ 〉 =∫ψ∗(x, t)Aψ(x, t) dx,

12 CHAPTER 4. MOTION OF A FREE PARTICLE

while in the momentum representation

〈Ψ |A|Ψ 〉 =∫φ∗(p, t)Aφ(p, t) dp.

Suppose A is the numerical operator unity. We then introduce, as our second definition

〈Ψ |Ψ 〉 = 〈Ψ |1|Ψ 〉.

In configuration space, we have

〈Ψ |Ψ 〉 =∫ψ∗(x, t)ψ(x, t) dx

and similarly in momentum space. In this notation, the normalization condition is simply

〈Ψ|Ψ〉 = 1.

If it is desirable, or necessary, to specify the representation in this notation, this is

easily accomplished by explicitly giving the argument of the wave function. Thus, in the

position representation, we can write 〈ψ(x, t)|A|ψ(x, t)〉 and so on, which is rather more

compact than writing out the integral.

4.8 Stationary States

We now return to the question of solving Schrodinger equation (4.23) in configuration

space. For this purpose we use the conventional method of separation of variables. Thus

we seek a special solution in the form of a product

ψ(x, t) = ψ(x)T (t) (4.28)

substitution into Eq.(4.23) yields

− h2

2m

d2ψ

dx2T = − h

iψdT

dt

or, rearranging,

− h2

2m

1

ψ

d2ψ

dx2= − h

i

1

T

dT

dt.

The left side depends only upon x, the right side only upon t, but the two must be equal

to each other for all x and t. Hence each must equal the same constant, say α, which is

called the separation constant,

− h

i

1

T

dT

dt= α (4.29)

4.8. STATIONARY STATES 13

and, also,

− h2

2m

1

ψ

d2ψ

dx2= α. (4.30)

The solution to Eq.(4.29) is immediate:

T (t) = e−iαt/h.

Hence, from Eq.(4.28)

ψα(x, t) = ψα(x) e−iαt/h (4.31)

where ψα(x) must satisfy Eq.(4.30) and where the subscript denotes that we are discussing

the solution corresponding to the particular separation constant α.

The interpretation of α is the following. Consider the wave function Eq.(4.31) and

suppose it to be normalized. Then no matter the particulars of ψα(x), we have

〈E〉 = 〈ψα(x, t)|E|ψα(x, t)〉 =∫ψ ∗

α

[− h

i

∂ψα

∂t

]dx = α

and similarly, for any function of energy, f(E),

〈ψα|f(E)|ψα〉 = f(α).

Since all fluctuations of E about α accordingly vanish, this means that ψα is a state in

which the energy E has the precise and definite numerical value α. To remind ourselves of

this fact, we shall replace α by the number E, which is simply the numerical value of the

energy for the state in question. Compare this notation with that for the states φp defined

earlier. It will be recalled that these were states in which the linear momentum had the

precise numerical value p. In either case this notation is somewhat unfortunate since

the symbol which is being used to denote a numerical value is the same as that used to

denote an operator in other contexts. However, it is standard practice to do this, and the

literature can hardly be read without keeping this ambiguity in mind. The reader cannot

proceed blindly, but must distinguish which is meant from the context. Fortunately, this

is relatively easy to do.

In any case, we thus write

ψE(x, t) = ψE(x) e−iEt/h (4.32)

while Eq.(4.30) can be rewritten in the form

− h2

2m

d2ψE(x)

dx2= E ψE(x). (4.33)

14 CHAPTER 4. MOTION OF A FREE PARTICLE

Eq.(4.33), which no longer contains the time, is called the time independent Schrodinger

equation, in this case, for a free particle.

The states ψE(x, t) have one very important property. The expectation value of any

operator A is independent of time provided A does not itself explicitly depend on t. For

example, if A = f(p, x), then 〈ψE|A|ψE〉 does not change with time. For this reason,

the states ψE are called stationary states. Stationary states are solutions of the time-

dependent Schrodinger equation; they are states of definite energy. Accordingly, they are

particularly simple states quantum mechanically.

We now complete our discussion by exhibiting the solutions to Eq.(4.31). We have at

once

ψE(x) = e±i√

2mEx/h

as is easily verified. Hence

ψE(x, t) = exp[± i

√2mEx

h− iEt

h

].

Note that E must be positive if ψE is not to increase exponentially in one direction or

the other. Now, since

p =√

2mE

where p is the numerical value of the momentum corresponding to the energy E, we can

equally well write

ψE(x, t) = exp[± ipx

h− ip2t

2mh

], E =

p2

2m

where the two signs in the exponent make explicit the two possible values of linear momen-

tum corresponding to a given energy E. This is then a solution of Schrodinger equation

(4.23) for any positive value of E, or, equally, for any value of p, positive or negative. The

general solution of Eq.(4.23) is a superposition of these stationary states with arbitrary

amplitudes. Our original representation, Eq.(4.10), is recognized as just such a super-

position. We have thus demonstrated that Eq.(4.10) is a representation of the general

solution of Eq.(4.23), as claimed.

4.9 A Particle in a Box

As an example of the use of Schrodinger equation, we now examine the stationary states

of a particle which is constrained to move in the interior of a (one-dimensional) box, but

which is otherwise free. This constraint is intended to mean that outside of the box the

probability of finding the particle is zero, and hence that ψ must vanish. If ψ is not to

be discontinuous, then its behavior in the interior must be such that it is zero at the

4.9. A PARTICLE IN A BOX 15

constraining walls. (We are not in a position to prove that ψ must indeed be continuous,

as assumed. The problem under consideration is not a free particle problem in actuality

since the walls of the container exert impulsive forces on the particle. In the next two

chapters we discuss the general question of motion under the influence of external forces.

The case of motion in a box can be considered as the limiting case of motion in a square

well potential as the potential becomes infinitely deep. When so considered, the stated

continuity and boundary conditions can be verified.) Taking the walls of the box to be at

x = 0 and x = L, we thus seek those solutions of the free particle Schrodinger equation

(4.33), which are such that

ψE(x = 0) = ψE(x = L) = 0. (4.34)

We saw earlier, that for a fixed energy E,

ψE = e±i√

2mEx/h.

The most general solution for given E is thus the linear combination

ψE = Aei√

2mEx/h +B e−i√

2mEx/h.

Applying the boundary conditions, Eq.(4.34), we obtain,

A+B = 0, A ei√

2mEL/h +B e−i√

2mEL/h = 0,

or, from the first,

B = −Aand, from the second,

sin

√2mEL

h= 0.

This latter condition is not satisfied for arbitrary values of E, but only if E has particular,

discrete values En such that

√2mEn

L

h= nπ, n = 0, 1, 2, · · ·

or

En =n2π2h2

2mL2. (4.35)

The corresponding stationary wave functions are, after normalization,

ψEn =

√2

Lsin

nπx

L, (4.36)

16 CHAPTER 4. MOTION OF A FREE PARTICLE

whence it is seen that solutions are obtained only if a half-integral number of de Broglie

waves can exactly fit into the box. Note, however, that for n = 0, the wave function

vanishes identically. Hence the state of lowest energy is that for n = 1,

ψE1 =

√2

Lsin

πx

L,

E1 =π2h2

2mL2=

10−53

2mL2ergs.

In terms of E1 we can conveniently express En as

En = n2E1.

We have thus found that, in contrast to classical physics, a particle in a box, can exist

only in a discrete set of states. Further, we note that no state of zero energy exists, in

agreement with our understanding of the nature of the uncertainty principle. A particle

in a box cannot simply sit on the floor but must always be in at least some minimal state

of motion.

In spite of the fact that the spacing of these energy levels increase with n, for a macro-

scopic particle this spacing is infinitesimal for all achievable energies. For example, con-

sider a one-gram particle in a box one cm in length. Its ground state energy is about

5 × 10−54 ergs. For the n = 1030 state its energy will be about 107 erg = one joule and

for the n = 1033 state its energy will be about one million joules. The distance between

adjacent states is then about 10−24 ergs in the former case and 10−21 ergs in the lat-

ter. These are enormaously larger than the 10−53 erg spacing in the neighborhood of the

ground state, but both are still undetectably small on the macroscopic energy scale. On

the other hand, for an electron in a box two angstroms in size, E1 = 10−11 ergs = 6 eV,

which is roughly of the order of the spacing actually observed in atoms.

We conclude with a discussion of the properties on an arbitrary wave packet or nonsta-

tionary state for a particle in a box. The most general possible such state is an arbitrary

superposition of the discrete set of stationary states,

ψEn(x, t) = ψEn(x) e−iEnt/h

where En and ψEn are given by Eqs.(4.35) and (4.36). We thus write

ψ(x, t) =∑n

An ψEn(x) e−iEnt/h (4.37)

which evidently describes a state in which the energy of the particle is not precisely fixed

but can take on any of the values En with probability determined by the extent to which

the nth state is represented in the superposition. Specifically, if ψ(x, t) is normalized,

we expect that a measurement of the energy will yield the value En with a likelihood

4.9. A PARTICLE IN A BOX 17

which is just |An|2. This is an important result and we now give a formal proof that our

interpretation is indeed correct.

As a preliminary step, note that the stationary wave functions ψEn , which are simple

sinusoids according to Eq.(4.36), are such that

∫ L

0ψ ∗

Em(x)ψEn

(x) dx =

{1, m = n

0, m �= n

}= δmn. (4.38)

For brevity, we have introduced the Kronecker delta symbol δmn which is defined to be

unity when m is equal to n and to be zero when m and n differ from one another. Any set

of functions satisfying an equation of the type of Eq.(4.38), i.e., such that each function

in the set is normalized and such that different functions in the set are orthogonal to one

another, is called an orthonormal set.

Supposing ψ(x, t) to be normalized,

∫ L

0ψ∗(x, t)ψ(x, t) dx = 1,

we obtain from Eq.(4.37)

∫ L

0

(∑n

A ∗nψ

∗En

(x)eiEnt/h)(∑

m

AmψEm(x)e−iEmt/h

)dx = 1

where we have used n as the summation index in the expression for ψ∗(x, t) and m in the

expression for ψ(x, t). Interchanging the order of the summation and integration, we now

have ∑m,n

A ∗nAm e

i(En−Em)t/h∫ L

0ψ ∗

En(x)ψEm

(x) dx = 1.

Hence, using the orthonormality condition Eq.(4.38), this becomes

∑m,n

A ∗nAm e

i(En−Em)t/h δmn = 1.

and, finally, the normalization condition is seen to require that

∑m

|Am|2 = 1. (4.39)

Next we calculate the expectation value of the energy. We have

〈ψ|E|ψ〉 =∫ L

0ψ∗(x, t)

[− h

i

∂ψ(x, t)

∂t

]dx

whence, using Eq.(4.37)

〈ψ|E|ψ〉 =∫ L

0

(∑n

A ∗nψ

∗En

(x)eiEnt/h)(∑

m

AmEmψEm(x)e−iEmt/h

)dx

18 CHAPTER 4. MOTION OF A FREE PARTICLE

and thus again using the normalization of the ψn, we obtain

〈ψ|E|ψ〉 =∑m

Em |Am|2. (4.40)

Indeed, by the same argument, it follows that for every s

〈ψ|Es|ψ〉 =∑m

E sm |Am|2

and hence that, for any function f(E)

〈ψ|f(E)|ψ〉 =∑m

f(Em) |Am|2. (4.41)

Since f(E) is arbitrary, we thus see that, as asserted, ψ is a state in which a measure-

ment of the energy yields the value Em with probability |Am|2. Otherwise stated, Am

is the probability emplitude, |Am|2 the probability, that the system will be found upon

observation to be in the mth stationary state, with energy Em.

Suppose now that the wave function of a particle in a box is prescribed at some instant,

which we take to be t = 0 for simplicity. Denoting this initial state by ψ0(x) we thus have

ψ(x, t = 0) = ψ0(x)

where ψ0(x) is assumed to be known. The behavior of the system can now be readily

determined. From Eq.(4.37), setting t = 0, we have

ψ0(x) =∑n

AnψEn(x).

Multiplying this expression by ψ ∗Em

(x) and integrating over the box, we obtain

∫ L

0ψ ∗

Em(x)ψ0(x) dx =

∑n

∫ ∫ψ ∗

Em(x)ψEn

(x) dx.

In virtue of the orthogonality of the ψEn, the right side reduces to the single term Am and

hence we find

Am =∫ L

0ψ ∗

Em(x)ψ0(x) dx = 〈ψEm |ψ0 〉, (4.42)

for the given initial state ψ0(x), Eq.(4.42) then gives the probability amplitude for the

system to be in the mth stationary state.

The evolution of the system in time can be studied in the following way. Substitution

of Eq.(4.42) back into Eq.(4.37) yields the expression

ψ(x, t) =∑n

( ∫ L

0ψ ∗

En(x′)ψ0(x

′) dx′)ψEn

(x) e−iEnt/h

4.9. A PARTICLE IN A BOX 19

where we have used x′ to denote the dummy variable of integration in the expression for

the amplitudes An. Rearranging this result, we thus have

ψ(x, t) =∫ L

0ψ0(x

′)K(x′, x; t) dx′ (4.43)

where the propagator K for a particle in a box is given by

K(x′, x; t) =∞∑

n=1

ψ ∗En

(x′)ψEn(x) e−iEnt/h. (4.44)

These results should be compared with those for a free particle, Eq.(4.11) and Eq.(4.12).

The specific form of K is of some interest. From Eq.(4.35) and Eq.(4.36), we have

K(x′, x; t) =2

L

∞∑n=1

sinnπx′

Lsin

nπx

Lexp

(− i

n2π2h

2mL2t).

In the limit in which L→ ∞, the sum can be replaced by an integral and the resemblance

of the result to that for a free particle is readily apparent.

Unfortunately, it is not possible to obtain K in closed from for a particle in a box.

However, in the classical limit, it is not difficult to show that a wave packet propagates

without distortion and bounces successfully off the walls of the container, just as it should.

The algebra involved is quite tedious, however, and we shall not work out the details.

20 CHAPTER 4. MOTION OF A FREE PARTICLE

Problem 1 Consider a wave packet which at t = 0 has the form

ψ(x, 0) = Aeip0x/he−|x|/L.

a. Normalize ψ(x, 0).

b. Calculate φ(p, 0) and φ(p, t), and verify that each is normalized.

c. Calculate 〈p〉 and discuss its time dependence.

d. Calculate 〈x〉 at t = 0.

e. Calculate 〈x〉 at any time t. [Hint: Do this in momentum space.]

Problem 2 A particle is confined in a box of length L.

a. Calculate 〈ψEn|x|ψEn

〉.b. Calculate 〈ψEn

|x2|ψEn〉.

c. Calculate 〈ψEn|p2|ψEn

〉.d. The motion of a classical particle in a box is periodic with period T = 2L/v, where v

is the particle’s speed. The quantum mechanical motion exhibits no such periodicity.

Explain how the classical periodic motion is attained in its appropriate limit. [This

is a non-trivial problem. It requires more thought than algebra for its solution.]