Chapter 2

Wave-Particle Duality, Probability, and the Schrodinger Equation

The developments outlined in Chapter 1 are often described as the Old Quantum
Theory. The rules devised were all ad hoc, and the connection between various separate
discoveries, such as the particle nature of radiation, the wave nature of electrons and the
Bohr atom (as well as other rules not discussed in our brief survey) did not rest on any
firm foundation. Quantum mechanics was discovered twice: first, by Werner
Heisenberg in 1925 as matrix mechanics, and then again by Erwin Schrodinger in 1926
as wave mechanics. The two forms were soon found to be identical in content, but wave
mechanics became a more useful tool because the mathematics of waves were familiar
to many physicists. In this chapter we begin our study of quantum mechanics, and we
follow, in spirit, the path laid out by Schrodinger.

2.1. RADIATION AS PARTICLES, ELECTRONS AS WAVES

The fact that radiation and electrons exhibit both particle and wave
properties raises deep conceptual difficulties, as can be seen from the following
considerations: There is no doubt that light consists of individual particles, called
photons, which carry energy and momentum, as was first unequivocally
demonstrated by the Compton effect. The human eye cannot detect individual
photons, but it is fairly close to being a photon counter, since under optimal
conditions it takes only 5-10 photons to activate the darkness- adapted eye. There
are devices, known as photomultipliers, that can easily detect individual photons.
An interesting thought experiment is discussed in Dirac’s wonderful book
on quantum mechanics. When light of a certain polarization is used to produce
electrons (as in the photoelectric effect), the latter are emitted with an angular
distribution that depends on the direction of the polarization of the incident photon
beam. Since in the photoelectric effect single photons eject single electrons, this
implies that individual photons, in addition to carrying energy and momentum,
also have polarization properties. Suppose we now send a beam of polarized light
with an initial intensity I0 through a crystal, which has the property that only the
 component of light polarized along a particular axis can pass through it. Thus if
the polarization of the initial beam is in the direction of the axis, then the
emerg¬ing beam will have intensity I0. If the polarization vector makes an angle 6
with the axis, then the intensity of the emerging beam is I0 cos2 Ø. Let us look at
this result in terms of individual photons. If the beam is totally polarized along the
direction of the axis, then all
of the photons that make up the beam must have been polarized in that direction. For a
beam polarized in a different direction, the intensity is reduced by a factor cos2 8. This
im-plies that only this fraction of the photons passes through the crystal. However,
photons cannot be split into pieces, so that a given photon will either pass through the
crystal, or it will not. We have no way of predicting whether an individual photon will
pass through. All we can say is that for N incident photons, N cos2 8 will get through,
so that the odds, or probability, that a particular photon will get through is cos2 8.
We also know from classical optics that a beam of light consisting of many photons will
exhibit wavelike properties—that is, diffraction and interference. An experiment
car¬ried out by G. I. Taylor in 1909 was the first to show that a beam of light gave rise
to a diffraction pattern around a needle even when the intensity of the light was so low
that only one photon at a time passed by the needle. Since then, many more experiments
showed that the interference and diffraction properties cannot be due to the collective
ef¬fect of the many photons in a beam. This raises new problems. Consider a thought
experi¬ment, which is a variant of the Taylor experiment, in which a very low intensity
beam of light is directed at a screen with two slits in it. The photons are then detected at
a second screen (Fig. 2-1). The intensity is such that at a given time no more than one
photon passes through the two-slit screen. After very many photons have passed by, we
see the classically expected diffraction pattern. Classically this is well understood: If the
electric fields at a particular point r on the detecting screen due to electromagnetic
waves crossing slits 1 and 2 are E,(r, t) and E2(r, /) respectively, then the total field at
the point r at the time t is the sum of the fields. This is a consequence of the
superposition rules for electric fields, which in turn is a consequence of the fact that
Maxwell’s equations for the electro¬magnetic fields arc linear. The intensity at the
screen is proportional to the square of the total electric field, and thus to (E,(r, r) + E2(r.
t))2. The interference pattern is due to the presence of the E,(r, t) • E2(r, t) cross term in
the square of the sum of the fields. If only slit 1 were open, the intensity would be
proportional to E,(r, r)2, and if only slit 2 were open, the intensity would be proportional
to E2(r, t)2. If we now translate intensity into probability, as suggested by our
discussion about polarization, we find that if only slit 1 is open, the probability of
finding a photon at r is /^(r, t), and if only slit 2 is open, the prob¬ability of finding a
photon at r is P2(r, t). However, if both slits are open, the probability is not the sum of
the probabilities associated with each slit.
The only way to resolve these difficulties is to assume that each photon interferes with
itself. This can be handled by assuming that each photon is described by its own
elec¬tric field, e(r, /), and that in the presence of two slits, the photon field at the
detector is the sum of two terms. These are associated with the presence of two slits, so
that
e(r, t) = e,(r, r) + e2(r, t)

Figure 2-1 Interference pattern

resulting from the passage of a beam
of photons through a screen with two
open slits.

just as for a classical light wave. Note that we are still talking about a single photon.
The only real requirements are (1) that the field e(r, t) obeys a linear equation and that
(2) in the classical limit a large collection of photons acts in accordance with Maxwell’s
equa¬tions. The actual formulation of a quantum theory of photons is somewhat
complicated, and we leave the discussion of this to Supplement 18-A.
At this point we turn our attention to electrons. From the time of their discovery,
electrons were described as particles. They appear to travel along trajectories
determined by the electric and magnetic forces acting on them, they have mass, and they
carry energy and momentum. Nevertheless, they have wavelike properties, as first
determined in the diffraction of electron beams by crystals. As demonstrated by the
beautiful two-slit exper-iment of A. Tonomura presented in Chapter 1 (see Fig. 1-9) the
pattern of the hits of elec¬trons on the absorbing screen slowly builds up to a wavelike
interference pattern. This experiment is a realization of the thought experiments we
discussed above in connection with photons. Individual electrons appear to hit the
screen at random. As the number of electrons increases, the expected interference
pattern emerges. Again, the place where a single electron hits cannot be affected by the
fact that other electrons came before it, or will come after it, so that the emergence of
the pattern must lie in the property of each electron. By analogy of our conjectures in
connection with the photon, we may expect that the properties of a single electron are
described by an analog of the one-photon elec¬tric field e(r, r). We thus expect that an
electron will be described by a wave function tra¬ditionally denoted by i//(r, t). In order
to get an electron to interfere with itself we must insist that i//(r, t) obeys a linear
equation. In that case, a sum of two wave functions is also a wave function, so that the
superposition rules apply. Furthermore, we expect that the predictability of where an
electron hits (as if it were a classical particle) will be replaced by a statement, involving
the wave function i//(r, r), about the probability that an electron arrives at r. The rest of
this chapter is devoted to arguments that lead us to the correct form of the linear
equation obeyed by ip(r, t), the Schrodinger equation, the probability for finding an
electron at r at a time t in terms of i//(r, t) and other general properties of the wave
function. Wc shall approach this by constructing waves that might simulate the
properties of particles. The study of wave packets will be helpful in this matter, even
though the idea that there are real waves that act like particles is not correct.
A harmonic wave propagating in the positive x-direction with wave number k has the
form
2-2 PLANE WAVES AND WAVE PACKETS

(2-1)

k = 2tt/\
(2-2)
and the angular frequency to is related to the period T by
to = 2TTIT
and thus to the frequency v = lIT by
In general co will be related to k in some way. For example, in the case of light
propa¬gating in a vacuum, v = c/A where c is the speed of light. Hence 10 = kc. This
relation¬ship is not true for light in a dispersive medium. There v — c/n\, where n is the
refractive index of the medium, and n is generally a function of the wavelength n =
n(A). We shall determine below the relation between OJ and k for the waves of interest
to us.
Since ipk(x, t) does not depend on y or z, it takes on the same value everywhere on the
y-z plane, and is thus called a plane wave, ipk(x, t) is a plane wave for all possible
values of k, and it is therefore possible to take a superposition of plane waves with
different amplitudes A(k) and B(k). There is nothing in what we have done so far to
require A(k) and B(k) to be real, and, in fact, we shall see that in general we must allow
tp(x, t) to be complex. Let us consider the wave A(k)em~w,) and add such waves for a
variety of values of k.' The superposition is called a wave packet, and it takes the form
oo
if/(x,t) = J dkA(k)em (2-4)
— oo
We begin by considering the wave packet at time t = 0,
co
ip(x, 0) = J dkA(k)eilLX (2-5)
— oo
and illustrate it by considering a special form, called the gaussian form
A(k) = e~aik-^'2 (2-6)
This function is centered about ku, and it falls off rapidly away from that center. We
shall see that it is the width of the square of this function that is of interest. The square
falls to 1/3 of its peak value when a(k — k0f = l. This means that the width can be taken
to be Ak = 2/Va. The integral in (2-5) can now be done in steps. We first make a change
of variables to q’ = k — ku, and we end up with
iP(x, 0) = e'k"xe~^l2a J dq’ e~aq‘lp-
_ 2lT it-,, V/2a
a
(2-7)
Aside from the scale factor, which could have been absorbed by slightly modifying
A(k), we end up with a plane-wave factor characterized by the wave number ku and a
modulat¬ing function that acts to localize the packet about x = 0. The width of that
packet, which is also gaussian, is defined in the same way as before: We square the
function and see where it drops off to about 1/3. This provides the width A.v — 2 Va.
We see that there is a reciprocal relation between the width of the function A(k) that
determines the shape of the wave packet, and the width of the wave packet. In fact, the
product of the two widths has the property that Ak Ax = 4 here. The specific value of
the number on the right side is not important. A slightly different definition of the
widths would have given us a different an-
'We do not consider the wave form B(k) exp[—i'(fct — &>/)] because a term
proportional to exp[ioj(] will be seen to be associated with a negative kinetic energy.

swer. What is quite general is that the product is independent of a. It is actually a

general result of Fourier integrals that
AA Ax > | (2-8)
so that this reciprocal relation is true for wave packets in general. We illustrate this in
the example that follows.
EXAMPLE 2-1
Consider a wave packet for which
A(k) = N -K^k^K
—0 elsewhere
Calculate ij/(x, 0), and use some reasonable definition of the width to show that (2-8) is
satisfied. SOLUTION We have
K
<//(x, 0) = f dk Ne^ = ^ (eiKx - *-**) = 2N
J IX -*
-K
The definition of A(k) easily shows that AA = 2K. A reasonable definition of Ax might
be the dis¬tance between the two points at which ip(x) first vanishes as it gets away
from x — 0. This happens when Kx = ±TT, SO that Ax = 2TT/K. It follows that
AA Ax = 47T
which certainly satisfies (2-8).
The mathematical description of how wave packets move is a little messy, so we set the
ma-terial off in a subsection. The important result is that for a wave packet for which
the spread about a particular value of A—say, A0—is small, so that it looks a lot like a
plane wave mod-ulated by a very wide function in x, then the wave packet moves with
the group velocity
v
s

(2-9)
and its width in x spreads as a function of time. In the following subsection we see this
in detail for our Gaussian wave packet.
*How Wave Packets Move
As indicated in (2-4), the motion of the wave package can be obtained by integration,
pro-
vided we know how w depends on A. Let us assume that A(k) is sharply peaked about
the
value of A = A0. We may make the approximation
“c--+-■*»>(t L+ 2'*■-■‘KSFL
With this,
(Ax - wt) = (kox - w(k0)t) + (A - Ao) Tx - ~ \~ *o)2(^) J
(2-10)

When this is inserted into (2-4) we get, after changing variables to q = k — ICQ, the
following:
i//(x, t) = J**~«*M J dqA(q + ^)e-^-V)g-^C^*V2 (2-11)
If the last factor in the integrand were absent, as is the case when w = kc for example,
then the integral becomes a function of (x — v^t). If this function peaks at x = 0 at time
t — 0, then it will peak at x = vgt at a later time t. The velocity with which the packet
moves is the group velocity. The second term in the integrand, when it does not vanish,
modifies the am¬plitude A(q + k0) and thus the shape of the wave packet; this is best
seen in terms of our gaussian packet. The integral is worked out just like the one that
leads to (2-7), since the extra term is also gaussian, as the exponent is quadratic in q2.
With the notation.

= /3
We see that in addition to changing x to x-vgt, we change a to a + 2ij8t. We therefore
get
ip{x, t) -

27r
a + 2i(3t
ei(kr,x-blJ)e_
2a+4i(3t
(2-12)
This is a rather untransparent expression, involving a complex function of x and t, but as
we shall see soon, it’s absolute magnitude has a physical meaning. We therefore
calculate
| iffx, t) |2 = ip*(x, t)ifi(x, t)
The pre-factor is easily handled; the phase factor has magnitude 1, and in the second ex-
ponential, the exponent has to be added to its complex conjugate. When all of this is
done, we get
|<K*. t)|2
277 c,(x-v„tj‘
Va2 + 4/3V C a2+^¥
(2-13)
Comparison with the value at t — 0 shows that the width, initially given by 2Va, now
be-comes 2Va + 4/32r/a = 2V/aV l + A0li1ta1. The result is that the wave packet
travel¬ing with speed vg spreads as time increases. When a is large, so that the packet is
broad, the spreading is small. Nevertheless it is there, and this stands in the way of
interpreting the wave packet as describing the particle itself.
2-3 THE PROBABILITY INTERPRETATION
OF THE WAVE FUNCTION
At this point we recall that in the case of photons, the intensity, proportional to [e(r, r)]2,
was interpreted as being proportional to the probability of finding a photon in the
vicinity
of r at the time t. Since we were led to the conclusion that i//(r, r) had to be complex, we
assume that it is |i//(r, f)|2 that is related to the corresponding probability of finding an
electron in the vicinity of r at time t. For simplicity we deal with motion in one
dimension
(though the generalization is straightforward) and assert
The probability of finding an electron, described by the wave function ip(x, t), in the
region lying between x and x + dx is given by
P(x, t)dx = | tp(x, t) |2 dx
(2-14)

The probability interpretation is due to Max Bom who, shortly after the discovery of the
Schrodinger equation, studied the scattering of a beam of electrons by a target and was
led to the above form.
With this interpretation of \ip(x, t)p the spreading of the wave packet presents no
problems. AH it implies is that an electron, known to be in a certain region with some
probability distribution, will, with increasing time, have an increasing probability of
being found outside that region.
The appearance of probability in quantum mechanics differs from its appearance in
clas¬sical physics. Here it is not a statement of ignorance about what is “really” going
on, as is the case when we speak of the probability of a coin-toss leading to heads or
tails, but it is a basic limitation on what we can know when the wave function is known.
The mathematical impli¬cations of this interpretation of ip(x, 1) will be discussed at
length in the next chapter.
The probability interpretation aUows us to understand electron interference. As a con-
sequence of the linearity of the equation for ip(x, t), a wave function of the form
4>(x, t) = N(th(x, t) + ip2(x, t)) (2-15)
is a solution, if both ip, and ip2 are solutions. Let ip, be the wave function of the
electron that describes the system with slit 2 closed. This wave function is then
definitely associ¬ated with passage through slit 1. Similarly, if ip2 is the wave function
with slit 1 closed, then the wave function with both slits open will then be the sum of
the wave functions i/q and ip2. Consequently, the probability density of finding an
electron at a point x on the photographic plate behind the slits is proportional to t) +
ip2(x, t)p.
We have
|i/q(x, t) + ip2(x, t)p = |!//,(*, Op + |tp2(x, 0|2 + 2 Re(i//,(*, t)tp%(x, t) (2-16)
and the third term clearly exhibits the interference. This effect requires that there be a
sin¬gle electron source, so that the phase difference of the two wave functions i//, and
tp2 does not vary randomly. If the phase difference varied unpredictably, then the
probability would be determined by
P(x, 0= l</'i(*,0|2+ Wi(x,0\2 (2-17)
EXAMPLE 2-2
Consider a two-slit experiment, in which the wave function at slit 1 acquires an arbitrary
random phase that is to be averaged over, so that the total wave function at the screen is
ip(x, t) = e'^ipiix, t) + tp2(x, t). (Such a situation might arise if there were two
incoherent electron sources, one at each slit). Show that under these circumstances the
interference averages out.
SOLUTION We need to calculate
|Mx, Op = (<?"ViU, 0 + fcfr, 0)(e '-fyf(x, 0 + <P*(x, 0)
= |i//,(x, Op + |ih(x, Op + e^tpiix, t)ip*(x, 0 + e t)ip*(x, t)
= \<p,(x, Op + |ip2(x. Op + 2 cos <p Re (i//|(x, t)\p*(x, t)) + 2 sin <p Im (ip\(x, t)ip*(x,
t))
The angle <p varies randomly from electron to electron, so that in the pattern of dots,
the terms in¬volving this angle average out to zero.
There is a potential difficulty with the probability interpretation. Consider a beam of
elec-trons passing through a screen with a double slit. Suppose we could determine in
some

way through which slit each given electron passes. If the electrons are far apart, then as
far as any given electron is concerned, we might as well have closed the other slit. If
such
a detection means were available to us, we could divide all the electrons, arriving at the
screen on which they are detected, into two classes: those that went through slit 1 and
those that went through slit 2. In that case, however, we would get the distribution (2-
17).
We are forced to the conclusion that we get an interference pattern only if the
experiment
does not allow us to determine which slit the electrons go through. If we somehow
arrange to find out what the slit of passage was, then the interference pattern disappears
and the probability is just the sum of the individual probabilities. The rule is simple:
If the paths are not determined, add the wave functions and square; if the paths are
determined, square the wave fimetions and add.
If we look at the discussion in Example 2-2, we see that somehow the acquisition of
knowledge in which way the electron went must introduce a random phase into the
compo-
nent of the wave function that is being “looked at.” For further discussion see Chapter
20.
2-4 THE SCHRODINGER EQUATION
We have constructed a wave function that may be used to describe satisfactorily the
prob-
ability of finding a freely traveling electron at x, at a time t. We make the connection
with
physics by first recalling that according to de Broglie k = p/h, and as suggested by the
Planck relation, to = E/ti. Thus the wave packet may be rewritten in the form
oo
iKx, t) = —L= [ dp (2-18)
V2TThJm
We now take E(p) = p2/2m for a free particle. We see that the group velocity is
= dto _ dE = If
Vg dk dp rn
which confirms our association of hto with the energy. Here <^>(p)/V2trfi plays the role
played by A(/c) in eq. (2-5).
Now suppose that the particle under consideration is not free, so that instead of
E = p2/2m. we have
E = p2/2m + V(x) (2-19)
If we were to mindlessly insert this into the exponent in eq. (2-18), the wave function
would be changed rather trivially: the new wave function ip(x, t) is just the product of
the
free particle wave function, and a factor e'V{x)llh. Now this factor is a pure phase
factor,
whose absolute square is 1. This would mean that the addition of a potential to the
energy
in the exponent does not change the uniform motion of the wave packet. This is patently
wrong since in a potential, the velocity of a particle changes from place to place What
we
need to do is to find the equation that ip(x, t) is a solution of, and then modify that equa-
tion to take into account the presence of a potential V(x).
We proceed as follows: If we differentiate with respect to time, we find that
(2-20)
dP e‘^-E,n

On the other hand,

and hence

1
V27Th
J dp fi(p)p2e^
(2-21)
We may combine these two results to get an equation for i/r(jc, t) that is solved by (2-
18). This is
dij/(x, t) _ h1 d2ip{x, t)
1 at ~ 2m
(2-22)
This is the Schrodinger equation for a free particle. Although we started from a solution
of (2-22), the equation takes precedence over the solution. It is easy to see that starting
from the equation, a solution of the form would correspond to a negative kinetic
energy and that A cos(fct — cot) + B s\n(kx — cot) will only be a solution if B — iA.
To the extent that (2-22) is a translation of E = p2/2m, with E being replaced by iti - and
p by
A
—ih —, we can generalize the energy equation in the presence of a potential V(x),
E=
P_
2m
+ V(x)
to the general Schrodinger equation
di[>(x, t)
lh~d,— =
h2 d2lp(X, t)
2m dx2
+ V(x)</<x, t)
(2-23)
This is the basic equation that we will be working with in much of this book.
Let us return to the free particle case. The most general solution of that equation de-
pends on the form of cj)(p), and we shall now show that this is determined by the initial
condition—that is, by ip(x, 0). In fact, if we set t = 0 in the solution (2-18), we get
00

(2-24)
and this equation determines <p(p). The solution of (2-23) is also determined by tjr(x,
0). This is in contrast to the familiar wave equation
c)2fix, t) _ 2 rffix, t)
si2 “ dx2
in which both f{x, 0) and (dflx, t)ldt),=0 have to be specified. The difference is a
conse¬quence of the fact that the Schrodinger equation is of first order in t. We shall see
that this is closely related to the probability interpretation of <p(x, t).
*The Relation between c})(p) and fi(x, 0)
The relation between ip(x, 0) and 4>(p) is obtained by noting that (2-23) is a Fourier
inte¬gral and thus can be inverted. Here we make use of the properties of Fourier
integrals out¬lined in Supplement 2-A. [www.wiley.com/college/gasiorowicz]

We multiply 0) by e 'p xJfl and integrate over all x. This leads to

00
/
dx IJJ(X, 0)e
00 oc
= 1 f dx [ dp 4>(p)eiip ,nM (2-25)
VlTTh J J
— 00 —00
We now use dx e‘(p p )xlh — 2irh8{p — /;') to get V27rft<^(p') on the right-hand side,
so that
00

(2-26)
2-5 THE HEISENBERG UNCERTAINTY RELATIONS
Let us return to the reciprocal relation obtained in eq. (2-8),
A k Ax > 1/2
Recalling the identification of hk with the momentum, this relation takes the form
Ax Ap a: tU2 (2-27)
This is called the Heisenberg uncertainty relation, or the Heisenberg indeterminacy rela-
tion. It arose in the context of our discussion of the wave packets, but as we now see, it
is a statement about the wave function. We saw that ip(x) cannot describe a particle that
is both well-localized in space and has a sharp momentum. This is in great contrast to
classi¬cal mechanics. What the relation states is that there is a quantitative limitation on
the ac¬curacy with which we can describe a system using our familiar, classical notions
of position and momentum. Position and momentum are said to be complementary
variables. We can illustrate this limitation by a couple of examples.2
Diffraction of a Photon Beam
Consider a beam of photons, passing through a slit of width a (Fig. 2-2). When the beam
is treated as an electromagnetic wave, one can show that the beam is diffracted with an
angular spread of magnitude

where A is the wavelength of the light. The particle-wave duality of quantum

mechan¬ics allows us to describe the beam as a sequence of photons passing through
the slit. Under those circumstances, one would not expect a spreading of the beam. The
uncer¬tainty relation rescues us, so to speak, from a paradox. The photon momentum in
the x-di recti on is given by p, = h/\. The y-coordinate of the photon is indeterminate to
the extent that
Ay < a
’Many examples are discussed in W. Heisenberg, The Physical Principles of the
Quantum Theory, Dover Publications, New York, 1930, and in most books of quantum
mechanics listed in the bibliography. See also J. Bernstein, P. M. Fishbane, and S.
Gasiorowicz, loc.cit.

e
I
Figure 2-2 Diffraction of a beam of photons that pass
through a slit in a screen.
This implies that the momentum of the photon in the y-direction is uncertain to the
extent that
A h_h
Ty>5
This, however corresponds to a spread of the beam of the order of
«~ h/a _ A
Px hiA a
in agreement with the “wave description” of the incoming beam.3
Inability to Localize Bohr Orbits
Quantum mechanics does not allow us to talk about classical orbits of an electron in the
Coulomb field of a proton. Suppose we want to conduct an experiment to study the
location of the electron in an orbit. Let us limit ourselves to circular orbits. We will
want to distin¬guish between an orbit of radius characterized by the quantum number n
and a neighboring orbit. If we use a photon beam to do this, we need a beam of
wavelength such that
A « r,
n+J
r„ =
mfiot
[(« + l)2 - »2]
meca ‘
(where we have again neglected factors of the order of 27T). Such a photon, by particle-
wave duality, will transfer momentum of the order of
Py~
h meca
A«
to the electron. This means that the energy transfer to the electron is
A C ~ P ^ ~ PP? m^Ca)2
AE ~ me ~ me >;> n2
which is large enough to kick the electron out of its “orbit.” Again, the attempt to
localize the electron must be accompanied by an uncontrollable momentum transfer to
it.
One more comment: The heading of this section included Relations. We shall see in
Chapter 5 that the uncertainty relation we used above follows formally from the
3In this example, we have worked with orders of magnitude, so that we have ignored
factors of 2-JT in the difference between h and ft.

Schrodinger equation, provided we use a suitable definition of the meaning of A* and

Ap. There is another relation that is not so directly derivable, but it is nevertheless
applicable. This is a relation that is familiar in classical wave theory. A wave train that
is limited in time, so that it lasts a time At, must have a spread in frequencies given by
Av a: which
translates into
AE At > h (2-28)
We shall see an example of this when we discuss the lifetime of an electron in an
excited state and the associated width of the spectral line in Chapter 17.
INTERIM SUMMARY
The particle nature of radiation and electrons (and other particles such as neutrons,
helium nuclei, and even complex molecules) is incompatible with the observed wave
properties that these particles manifest when suitable “wave-detection” experiments are
performed. The suggested solution is that the particles are described by wave functions
that obey a su-perposition principle. These must obey a linear equation (not necessarily
the familiar “wave equation”), they must be complex in general, and the wave functions
must be inter¬preted as yielding only probability statements about the behavior of
individual particles. Some good guesses lead to the linear Schrodinger equation (2-23)
for the wave function ip(x, t) that describes a particle in a potential V(x). We also stated
the content of the Bom interpretation—namely, that the wave function i//(x, t) is a
probability amplitude, and its absolute square yields a probability.
2-6 MORE ON THE PROBABILITY INTERPRETATION
For the interpretation given in (2-15) to hold, we must require that
OC 00
J dx P(x, t) = J dx\ip(x, t)\2 = 1 (2-29)
— OC — 00
since the particle must be somewhere in the range — °o < x < oo. Suppose we find a
solu¬tion of the Schrodinger equation for which the integral in (2-29) is not equal to
unity. As a consequence of the linearity of the Schrodinger equation, if ij/(x, t) is a
solution, so is Aip(x, t), and with a proper choice of A one can normalize the wave
function <p(x, t) so it does indeed satisfy (2-29). We shall see below that all we need is
that
00
J dx\ip(x, 0)12 < oo (2-30)
— 00
that is, the initial state wave functions must be square integrable. Since we may need to
deal with integrals of the type
00
J dx ip*(x, t)xf'i[/(x, t)

— 00
and

we will require that the wave functions tp(x, 0) go to zero rapidly as x -» ±°°, often
faster than any power of*. We shall also require that the wave functions <p(x, t) be
continuous in*.
The Importance of Phases
The emphasis on \tp(x, t)|2 as the physically relevant quantity might lead to the
impres¬sion that the phase of the wave function is of no importance. If we write <// =
Re‘n, then indeed \i]/\2 = R2 independent of 6. However, the linearity of the equation
allows us to add solutions, as in our discussion of the electron interference pattern with
two slits. We see that
\R\ei0' + R^f = R\ + R\ + 2RXR2 COS(0, - 02)
depends on the relative phase. An overall phase in the total wave function can be
ignored, or chosen arbitrarily for convenience.
The Probability Current
The ability to normalize the wave function is only possible if the constant A mentioned
above is a constant, independent of time. We now show that this is the case. We will use
(2-23) and its complex conjugate, with the explicit assumption that the potential energy
V(x) is real. Under these circumstances we have
—ih
d<P*(x, t) _ h2 t)
dt
We may use this to calculate
d
dt
P(x, t)
2m dx2
dtp* dtp
+ V(x)tp*(x, t)
i (h2\(^r .
dx \ lim y dx dx y
If we now define the flux or, equivalently, the probability current by we get
(2-31)
(2-32)

(2-33)
When this is integrated over all space, we find that
00 00
ft j dx P(x, t) = -1 dx j-xj{x, 0 = 0 (2-34)
-00 —00
The last step follows from the fact that for square integrable functions j(x, t) vanishes at
± oo. Incidentally, had we allowed for discontinuities in <//(*, t) we would have been
led to delta functions in the flux, and hence in the probability density, which is
unacceptable in a physically observed quantity. [Eq. (2-34) then implies that A is
constant.]

Equation (2-33) is a conservation law analogous to the charge conservation equation in

classical electrodynamics. It expresses the fact that a change in the density in a region—
say, o < x < b—is compensated by a net change in flux into that region
b b
jtjdx P(x, t) = ~J dx j^-fix, t) = j(ci, t) - j(b, t) (2-35)
EXAMPLE 2-3
What is the probability current for a wave function of the form (a) Ae“-' + B ikx, (b) Ae
(c) R(x)elS(xVf,r>
SOLUTION
(a) With ip(x) = Ad* + we have dip/dx = ik(A,kx - Be-**) and ip* = A*e~ikj +
B*e,h, so that
2ib (^* * “ = l(A*e+ ~ Be~ikx) - c.c]
=§(W2-WVSM’-H2)
(b) Here ip(x) is real, which means that j(x, 0) is manifestly zero. Note that this
result is true for any tp of the form A (real function of x) where A may be a complex
number.
(c) Here
h
2 im

e‘m

\_
m

These comments are sufficient to proceed with out discussion of the consequences of
the probability interpretation in quantum mechanics.
2-7 EXPECTATION VALUES AND THE MOMENTUM IN WAVE MECHANICS
Given the probability density P(x, /), we can calculate the expectation value of x, x2, or,
for that matter, any function of x. (See Supplement 2-B for a discussion of expectation
values) [www.wiley.com/college/gasiorowicz]. We have
00 CC
(fix)) = J dxfix)P(x, t)= J dx ip*(x, t)fix)ip(x, t) (2-36)
— 00 —00
Given our assumptions about the behavior of tp(x, t) as x —» ±°°, there is no problem
with the convergence of the integral. Note that we inserted fix) between ip*(x, t) and
ip(x, t). Clearly the order in which we put the three functions does not matter when we
integrate. We will soon find that sometimes fix) involves derivatives with respect to x,
and then the order matters. Note that (fix)) depends on time, even if it does not have an
explicit time depen¬dence, because in its definition the wave function tj> may have a
nontrivial time dependence.
The above expression does not tell us how to calculate the average of the momentum, or
a function of the momentum, because we don’t know what to insert between ip* and ip
when calculating (p), for example.

The Momentum in Wave Mechanics

We approach the calculation of (p) by noting that classically
dx
p = mv = m
dt
so we try
oo
(p) = (x) = m ~ J dx iJJ*(X, t)x\p{x, t)
— 00
f , (dlfl*(x, t) dl[l(xyt)\
— ml dx I—jt—xil/(x, t) + tfi*(x, t)x———I
— oo
(2-37)
Note that there is no dx/dt under the integral sign. The only quantity that varies with
time is ip(x, /), and it is this variation that gives rise to a change in (x) with time. We
again use the Schrodinger equation and its complex conjugate to evaluate the above. We
end up with
co
«-§/
dx
d2ip*
dx?
d2lp
xip — IJ/*x —- dx2
(2-38)
Now
d2ip*
dx?
XIp
dip* dip* d^
dx ^ dx X dx
d r w* ,i
d ,i , ,*# d r,* w~\ + ^
This means that the integrand in (2-38) has the form
-9 / dip* dip \ dip
Tx \i£x* -rx^-r^) + wTx
Because the wave functions vanish at infinity, the first term does not contribute, and the
integral gives
(p) = J dx ip*{x, t)^y 0
(2-39)
This suggests that the momentum be represented by the differential operator
= hd_
P°P i dx
Once we accept this, we can easily calculate
OO . oo
(p2) = J dx ip*(x, O^y <A(*, t) = -h2 J dx ip*(x, t)8
(2-40)

and more generally

</(/')> = J dx if/*(x, t)f(^rt) (2-41)
The form of the momentum operator in (2-40) raises the question whether the
expectation value of the momentum in some states could be imaginary. We can, in fact,
show that the expectation value of p is always real. We write
(P) ~ (P)* = / *[**(7) dfx ~ *(-f)
co 00
(2-42>
— CC -* —00
The last step follows from the square integrability of the wave functions. Sometimes
one has occasion to use functions that are not square integrable but that obey periodic
bound¬ary conditions, such as
1/4*) = ip{x + L) (2-43)
Also, under these circumstances
L
(p) ~ <P>* = 7 j dx (0*0) = j (|0(T)|2 - |0(O)|2) = 0
0
An operator whose expectation value for all admissible wave functions is real is called a
hermitian operator. We see that the momentum operator is hermitian.
The form of the momentum operator allows us to write the Schrodinger equation in the
form
. rUMx, t) ,
ih t) (2-44)
where the operator//, called the Hamiltonian operator, is a transcription of the energy
2
H = ^t + m (2’45)
into operator form, using the operator form of p given in (2-38). The Hamiltonian
opera¬tor is particularly significant in quantum mechanics, and we shall analyze it in
detail in many examples.
Wave Function in Momentum Space
Now that we have obtained a way of representing the momentum, we can discuss the
physical meaning of 0(p), which was found to have the form
cc
— 00
(2-46)

We can calculate
OQ OO CO
J dp 4>*{p)<p(p) = j dp <p*(p) *— j dx Ip(x)e
— CO — OO — 00
00 00
= —J dx Ip(x) J dp 4>*ip)e~ipxlh
— 00 —OQ
oo
= J dx i[i(x)ip*ix) — 1
(2-47)
This result is known as Parseval's theorem in the mathematical literature. It states that if
a
function is normalized to 1, so is its Fourier transform.
Next consider
oo oo
/_**•«*
dx t!>*(x)eipx,h
= J dp <f>*(p)p<t>(p) (2-48)
oo
This result, together with (2-47), strongly suggests that d>(p) should be interpreted as
the wave function in momentum space, with |</>(p)|2 yielding the probability density
for find¬ing the particle with momentum p.
Lest the reader think that in spite of this symmetry between x- and p-space, p = —ih
(dlfJx) is an operator and x is not, we note that x is, in fact, an operator too. It hap¬pens
to have a simple (multiplicative) form in x-space, but if we want to calculate (f(x)) in
momentum space, then we can show by methods very similar to the ones used earlier
that
(p)= \ dxrW f
h dipjx)
dx
OO CO
/ f
(x)= j dp ^j^(p) (2-49)
— co '
and more generally
(f(x))= f dpd>*(p)f(m^U(p) (2-50)
— co '
In other words, the operator x has the representation
x=ih~ (2-51)
in momentum space.
The following example illustrates some computations for a specific wave function
</<*)•
(a) For what value of x does P(x) = |i/'W|2 peak?
(b) Calculate (x) and (x2).
(c) What is the probability that the particle is found between x = 0 and x = 1/a?
(d) Calculate <f>( p) and use this to calculate ( p) and (p2).
SOLUTION
(a) The peak in P(x) occurs when dP(x)!dx = 0—that is, when
EXAMPLE 2-4
Consider a particle whose normalized wave function is
ip(x) — laX^ce xe ax
=0
x> 0 x <0
-j- (x2e~2or) = 2x(l - ark’2” = 0
which is at x = 1 la.
(b)

0
0

(c) The desired probability is

l/o
2
o
o
(d)
oc

From this we can calculate

With the change of variables p = ha tan 6, we get
1nil

The introduction of operators brings in a new concern. Products of operators need

careful definition, because the order in which they act is important. Consider for
example
xptp(x) = ~ it LX
#&)
dx
(2-52)
whereas
pxip(x) = -ih-j^(xip(x)) — —ihx ^ ^ - itap{x) (2-53)
which is different from (2-52). Note that we can deduce
[p, x]ijAx) = (px — xp)i!>(x) = ifupfx) (2-54)
Since this is true for all ip(x), wc conclude that we have an operator relation, which
reads
[p, x] — —ifi (2-55)
This is a commutation relation, and it is interesting because it is a relation between
opera¬tors, independent of what wave function this acts on. The difference between
classical physics and quantum mechanics lies in that physical variables arc described by
operators and these do not necessarily commute. The commutator (2-55) will be seen to
lie at the heart of the Heisenberg uncertainty relation.
PROBLEMS
1. Given that A(k) — NKjd + a2), calculate tp(x). Plot A(k) and ip(x) and show
that Afc Ax > 1, inde¬pendent of the choice of a.
2. The relation between the wavelength A and the frequency v in a wave guide is
given by
A= =
What is the group of velocity of such waves?
3. For surface tension waves in shallow water, the relation between frequency and
wavelength is given by
I2TTT
" Y pA3
where T is the surface tension and p is the density. What is the group velocity of the
waves?
4. For deep water gravity waves, the relation between frequency and wavelength is
given by

What is the group velocity of such waves?

5. Consider the problem of spreading of a gaussian wave packet describing a free
particle, with
_ Ich W~2m
Calculate the fractional change in the size of the wave packet in one second if
(a) The packet represents an electron of mass 0.9 x 10 30 kg, with the wave packet
having dimen¬sions of 10 ’’in; 10~10m.
(b) The packet represents an object of mass 10-3 kg and size 0.01 m. [It will be
convenient to ex¬press the width in units of h/mc, where m is the mass of the particle
represented by the packet.]

6. A beam of electrons is to be fired over a distance of I04 km. If the size of the
initial packet is 10-3 m, what will be its size upon arrival, if its kinetic energy is (a) 13.6
cV; (b) 100 MeV? [Caution: The relation between kinetic energy (K. E.J and
momentum is not always K. E. = p2/2m\]
7. Consider a wave packet for neutrinos, which are massless to a very good
approximation, so that E — pc. Show that such a wave packet does not spread.
8. Consider a wave function of the form
i/r(x) = Ae ^
Calculate the wave function in momentum space 4>(p).
9. Consider the example in Problem 8. Calculate A so that <//(A) is properly
normalized.
10. Show that the conservation law (2 33) holds when ip(x, t) obeys eq. (2-23), but
only if V(x) is real.
11. Suppose the V(x) is complex. Obtain an expression for
dP{x,t) , d f ——— and — J dx P(x, t)
For absorption of particles the last quantity must be negative (since particles disappear,
the proba¬bility of their being anywhere decreases). What does this tell us about the
imaginary part of V(x)?
12. Consider the distribution of grades in a class of 60 students, given by
Grades 60 55 50 45 40 35 30 25 20 15 10 5
# students 1 279 16 13 36 2 0 10
(a) Plot a histogram of the distribution,
(b) Calculate the class average.
(c) Calculate (Ag)2 = «/> - <g>2).
13. Compare your histogram with a distribution of the form
N(g)= Ce~ls~<g)msf
with C chosen that 2 ?V(g) = 60.
8
14. Show that a grade distribution of the form
Grades 60 50 10
# students 7 34 19
leads to the same average grade, but that the dispersion Ag is different. What is it?
15. Consider the wave function obtained in the first example, 2A7(sin kx)/x. For
what value of N will it be normalized? [Hint: A useful integral is
16. Consider the wave function
ip(x) = {ml 77)1 ^ exp(—ax?/2)
Calculate (x") for n = 1,2. Can you quickly write down the result for (A:* 15 16 17)?
17. Calculate 4>{p) forthe wave function in problem 16. Calculate (//') for n = 1, 2.

18. Use the definitions (A*)2 = (x7) - (x)1 and (Ap)1 = <p2> - (p)7 with the results
of Problems 16 and 17 to show that ApAx > hH.
19. Make an estimate of the strength of the nuclear potential energy given the
following fact: The "size" of the box that roughly describes the nuclear potential is 10
15 m, and it takes 8 McV to eject a par¬ticle from this potential well.
(a) Use the uncertainty principle to estimate (p2) for a nucleon in the box, and given
the fact that the mass of the nucleon is M - 1.67 X 10 27 kg, estimate the kinetic energy
of the nucleon.
(b) Since the potential that gives rise to the binding must more than compensate tor
this, what is the negative potential energy?
20. Monochromatic light passes through a shutter that opens for a time At = 10”1(1
sec. What is the spread in frequencies caused by the shutter?
21. Nuclei typically of size 10 14 m frequently emit electrons with energies in the
range 1-10 McV. In the early days of nuclear physics, people believed that electrons
"lived” inside the nuclei. Use the uncertainty relation to show that electrons of such
energies could not be contained inside the nucleus. 22
22. Show that eq. (2-49) holds.