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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.]