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Schrodinger's Equation and the Infinite Square Well
The Wave Function and Probability
We have examined some of the arguments that lead scientists in the first years of the
1900's to conclude that light was actually composed of individual elements called “photons”,
each with its own unique energy (I œ h / Ñ and momentum (: œ h5 œ 2Î-ÑÞ We have also
argued that the classical electromagnetic wave equation, which successfully describes such
phenomena as interference and diffraction, could be used to describe the particle nature of
light if we associate the absolute magnitude squared of the solution to the wave equation with
the number density of photons. One can demonstrate that the classical wave equation works
for particles like photons which have zero rest mass. However, this equation cannot be
applied to particles which have non-zero rest mass.
It was Erwin Schrödinger who developed the non-relativistic wave equation for
particles with non-zero rest mass. In 1926 he successfully applied this wave equation to the
problem of the hydrogen atom, obtaining the same basic results that Bohr had obtained some
years earlier. Although Schrödinger's equation is based upon some reasonable associations
between classical and quantum ideas, we can take this equation to be a postulate, much like
Newton's second law of motion. The validity of this equation, then, is determined by how
well it's predictions match observations. So far, the success of this equation in predicting
many experimental observations at the atomic level, in particular energy levels, transition
probabilities, etc., has pretty well established its validity.
Schrödinger's one-dimensional, time-dependent equation that governs the development
of the wave function GÐBß >Ñ in space and time is given by
h# ` #
`
G(Bß >) Z (B) G(Bß >) œ 3h
G(Bß >)
#
`>
27 `B
(3.1)
where 7 is the rest mass of the particle which moves under the influence of a potential energy
function Z ÐBÑ. We postulate that the solution to Schrödinger's equation will be associated
with the number density of non-zero rest mass particles, just as the solution to the classical
electromagnetic wave equation is associated with the number density of massless photons.
Thus, we expect that
3Bß > .B œ G* ÐBß >Ñ GÐBß >Ñ .B
(3.2)
where 3Bß > is the density of particles as a function of position and time. For example, if we
had R particles distributed randomly along the B-axis, we would designate R Bß > .B as the
number of particles located between B and B .B.
The number density 3Bß >
.B œ cR Bß >ÎR d .B is just the relative number of particles at a particular location. If we add
up all the particles along the line (from ∞ to ∞Ñ we would just obtain R total particles.
Now if we apply this same thinking to one particle, we realize that the number density is
always less that unity, and that adding up the number density for all possible values of B
would give us unity. Thus, when we speak of a single particle, we talk about the probability
T Bß > .B of finding a single particle at a particular location and time rather than the number
Quantum Mechanics
1.2
density of particles. Thus, when we talk about a single particle, we write
T Bß > .B œ G* ÐBß >Ñ GÐBß >Ñ .B œ ¹GÐBß >ѹ .B
#
(3.3)
If this interpretation of the wave function is correct, it places some restrictions on allowed
solutions to Schrodinger's equation: 1) the wave function must be continuous and singlevalued, and 2) the wave function must be normalizable. The first requirement arises because
we do not want the probability to be multivalued at some point of space and/or time. The
second requirement arises when we think about applying the Schrodinger equation to a single
particle. If our wave function (and, therefore, the probability) is to be applied to a single
particle, we must require that the probability taken over all space must give a certainty of
finding the particle (the particle must be somewhere!). This means that
T Ð>Ñ+8CA2/</ œ T∞ œ (
∞
T ÐBß >Ñ .B œ "
(3.4)
∞
which is our normalization condition. And, if the particle is to behave reasonably, this
normalization condition must not be time dependent! Otherwise, we have particles that
simply appear or dissappear at random. (There are some situations, however, that might
actually be described in just this manner - radioactive decay, being one.) Similarly, if we
consider a particle that is moving through space, we would expect that the probability of
finding that particle within a certain region of space would have to change in time in some
well-defined manner.
The Infinite Square Well
In this section, we want to apply Schrödinger's equation to a simple problem that will
illustrate many of the important concepts of quantum theory, the so-called infinite square well.
This problem has a simple potential energy function chosen to give simple solutions to
Schrodinger's equation. This particular potential energy function is probably not totally
applicable to any real situation, but will serve as an approximation to some types of systems
and is simple enough for us to develop the basic concepts which will apply to more realistic
situations. We will find that the solutions to Schrödinger's equation can be written in terms of
functions associated with a particular value of the energy of the system. This leads us to the
concepts of eigenfunctions and eigenvalues. We will find that the eigenfunction solutions to
Schrödinger's equation have very interesting characteristics, and that the most general solution
to Schrödinger's equation can be formed from a summation of the eigenfunction solutions.
The general solution to Schrödinger's one-dimensional, time-dependent equation
h# ` #
`
G(Bß >) Z (B) G(Bß >) œ 3h
G(Bß >)
#
27 `B
`>
(3.5)
that governs the development of the wave function GÐBß >Ñ in space and time can always be
written in the form
GÐBß >Ñ œ <ÐBÑ/3=>
(3.6)
Quantum Mechanics
1.3
where = œ IÎh , and where <ÐBÑ satisfies the time-independent Schrödinger equation
h# ` 2
<ÐBÑ Z B<ÐBÑ œ I <ÐBÑ
#7 `B#
(3.7)
Our task is to examine the solutions to this last equation for a particle of mass 7 moving in an
infinite potential well (i.e., trapped in a region space) represented schematically in the diagram
below. The potential energy is infinite for region I, where B ! and for region III, where
B " P, but the potential energy is zero in region II, where ! Ÿ B Ÿ PÞ To develop a proper
description of the motion of such a particle, we must determine the solution to Schrodinger's
time-independent equation in each of these three regions, and then fit these solutions together
at the boundaries.
`
I
`
II
x=0
III
x=L
Schrodinger's time-independent equation
h# ` #
<ÐBÑ Z ÐBÑ <ÐBÑ œ I <ÐBÑ
#7 `B#
(3.8)
`#
#7
<ÐBÑ œ # cI Z ÐBÑd <ÐBÑ
#
`B
h
(3.9)
`#
<ÐBÑ œ 5 # <ÐBÑ
`B#
(3.10)
can be written in the form
or
Quantum Mechanics
1.4
where
5œÊ
#7
cI Z ÐBÑd
h#
(3.11)
Now in region I and III where Z ÐBÑ " I for any energy, 5 is obviously imaginary. To
make the imaginary nature of 5 more explicit, we will write 5 œ 3,ß where , is real, and is
defined by the equation
,œÊ
#7
cZ ÐBÑ I d
h#
(3.12)
With this substitution, our differential equation becomes
`#
<ÐBÑ œ ,# <ÐBÑ
`B#
(3.13)
The general solution to this differential equation is simple it must be of the form
<ÐBÑ œ E/-B
(3.14)
-# E/-B œ ,# E/-B
(3.15)
so that
This means that - œ „ ,, giving the general solution Ðthe sum of the two possible solutions
with appropriate arbitrary constants)
<ÐBÑ œ E/,B F/,B
(3.16)
The time-dependent solution of the Schrodinger equation for the case where Z ÐBÑ " I is,
therefore,
GÐB,>Ñ œ cE/,B F/,B d/3=>
(3.17)
In the special case of the infinite square well, the potential energy function Z ÐBÑ œ ∞. We
can treat this by taking the limit as Z ÐBÑ p ∞ß so that , p ∞ß giving
GÐB,>Ñ œ lim cE/,B F/,B d/3=>
, Ä∞
(3.18)
For B !, the first term goes to zero, but the second terms blows up unless we require
F œ !Þ For the case where B " !, the second term goes to zero, but the first term blows up
unless we require E œ !x Thus, we require the wave function to be zero wherever Z ÐBÑ œ ∞
(i.e., everywhere in regions I and III). This means that the wave function solution must go to
zero at the boundaries of the well (i.e., at B œ ! and at B œ P).
Inside the well, where ! Ÿ B Ÿ P and Z ÐBÑ œ ! we have 5 œ É #7I
h # . Again the
general solution to the differential equation must be of the form
<ÐBÑ œ E/-B
(3.19)
Quantum Mechanics
1.5
giving
-# E/-B œ 5 # E/-B
(3.20)
so that - œ „ 35 . The general solution to this equation must, therefore, be
<ÐBÑ œ E/35B F/35B
(3.21)
The time-dependent solution is, therefore,
GÐB,>Ñ œ E/35B F/35B ‘/3=>
(3.22)
GÐBß >Ñ œ E/3Ð5B=>Ñ F/3Ð5B=>Ñ
(3.23)
or
This equation can be expressed in terms of sines and cosines using Euler's relationships,
GÐBß >Ñ œ E cosÐ5B =>Ñ F cosÐ5B =>Ñ
3cE sinÐ5B =>Ñ F sinÐ5B =>Ñd
(3.24)
In both the real and imaginary parts of this equation, we have two sinusoidal waves
moving in opposite directions. The first term in each part represents a wave moving in the B
direction, while the second term represents a wave moving in the B direction. This can be
clearly seen by observing a point on the wavefront where the phase, 9 œ 5B „ => is equal to
zero. Thus, the solution to Schrodinger's equation inside the well is the solution of sinusoidal
waves which are traveling in opposite directions, and interfering with each other.
The only wave functions which are allowed, however, are ones which go to zero at the
boundaries of the well, i.e., at B œ ! and P. We can apply the boundary conditions either to
the exponential form of the equation or to the sinusoidal form. Each will give the same
results. The math is somewhat simpler using the complex exponential form of the equation,
since we must require that both the real and the imaginary parts of the wavefunction must go
to zero at the boundary.
Thus, at the boundary where B œ !, we have
GÐB,>Ñ œ E/35! F/35! ‘/3=> œ !
(3.25)
which means that
EF œ!
Ê
E œ F
(3.26)
This means that the general solution is of the form
GÐB,>Ñ œ E/35B /35B ‘/3=> œ #3E sin5B/3=>
(3.27)
We usually absorb the #3 into the arbitrary constant E and write this last equation as
GÐB,>Ñ œ E sin5B/3=>
(3.28)
Quantum Mechanics
1.6
where the constant E is understood to be complex in general (but may, in some circumstances,
turn out to be real).
We also require the wave function to go to zero at the boundary where B œ P, so that
the equation above becomes
GÐP,>Ñ œ E sin5P/3=> œ !
(3.29)
which can only be true if E œ ! (which is the trivial solution consistent with no particle), or if
81
sinÐ5PÑ œ ! Ê 5P œ 81 Ê 5 œ
(3.30)
P
Recall that the constant 5 is related to the energy of the particle. This last equation, which
gives an infinite set of discrete values of 5 means that we will have an infinite set of discrete
solutions, each with a specific energy associated with it. Thus, the allowed energies for a
particle of mass 7 confined to our infinite square well is found using Equations 3.11 and 3.30
I8 œ
h # 5#
h # 81ÎP#
h # 1#
#
œ
œ 8# ”
• œ 8 I"
#7
#7
#7P#
(3.31)
Thus, we have a set of solutions, or eigenfunctions given by
G8 ÐBß >Ñ œ <8 ÐBÑ/3=8 > œ E8 sinŠ
81B 3=8 >
‹/
P
(3.32)
where =8 œ I8 Îh , one for each possible value of the quantum number 8. (The solution
where 8 œ ! gives the null wave function, which we interpret as no solution.) Notice that for
this one-dimensional problem we have only one quantum number, 8. Typically the number of
quantum numbers required to specify the system is related to the number of degrees of
freedom of the system.
At this point our eigenfunctions are completely defined, except for the constant E8 .
The value of this constant is determined by requiring that each of the eigenfunctions be
normalized. It is left as an exercise for the student to show that the normalization constant E8
is the same for all eigenstates and is equal to È#ÎPÞ
___________________________________________________________________________
Problem 3.1 The eigenfunctions of the infinite square well are given by
G8 ÐBß >Ñ œ E8 sinŠ
81B 3=8 >
‹/
P
where E8 are arbitrary constants that we choose to normalize the individual eigenfunctions.
Determine the value of these normalization constants by normalizing the eigenfunctions, and
show that the constant is the same for each eigenfunction and has the form
E œ È#ÎP
where P is the width of the well.
___________________________________________________________________________
Interpretation of the Eigenfunctions of the Infinite Square Well
Quantum Mechanics
1.7
The general form for the eigenfunction solutions to our infinite square well problem,
then, is given by
G8 ÐBß >Ñ œ Ê
#
81B 3=8 >
sinŠ
‹/
P
P
(3.33)
These eigenfunctions to the wave equation are shown in blue in Figure 3.1 for the first four
values of the quantum number 8. These solutions are plotted on a line which represents the
energy of that particular state in terms of I" . Plotted in red are the corresponding probability
densities T8 ÐBß >Ñ œ T8 ÐBÑ given by
T8 ÐBß >Ñ œ G8* ÐBß >ÑG8 ÐBß >Ñ œ ¹G8 ÐBß >ѹ =
#
#
81 B
sin# Š
‹
P
P
(3.34)
(although both plots are arbitrarily scaled to give the same maximum value).
20
0
0
0.5
1
Fig. 3.1 The first four eigenfunctions (blue) and probability functions (red) for the
infinite square-well potential problem. (The amplitudes of each function are arbitrarily
scaled to give the same maximum amplitude.)
In the diagram above notice that for all eigenstates with 8 " ", there are nodes in the
probability density function (places where the probability density goes to zero). In fact, we
find in general there are 8 " nodes for each energy state desinated by the quantum number
8. This is a bit wierd! For the quantum state 8 œ #, the probability density is a maximum on
each side of the center of the well, but is zero in between. How does that particle get from one
side of the well to the other without going through the middle? It is easy to give an answer to
this question, but not at all easy to really understand the implications.
Quantum Mechanics
1.8
Since we define the probability in terms of the square of the absolute magnitude of the
wavefunction, the time dependent terms of a particular eigenfunction cancel so that the
probability density of an eigenfunction is time independent! This is a direct consequence of
the fact that the solution to Schrodinger's equation can be written as a time-dependent part
times a position-dependent part. These eigenfunctions are called stationary states, meaning
that the probabilities, and, therefore, all expectation (or average) values are time-independent.
The eigenstate does not correspond to the motion of a particle. In fact, if we calculate the
expectation value (or average value) of B for a particle in any of the eigenstates of this system
(we will discuss how this is done in the following section) we find that
ØBÙ œ
P
#
(3.35)
If the expectation value of the position is not a function of the time, then the expectation value
of the momentum in each eigenstate Ø:Ù œ .ØBÙÎ.> must also be zero.
A plot of the momentum probability functions for the first five eigenstates are shown
below where the argument 5: on the plot is equivalent to the 5 in these notes. Each plot has
been offset vertically so they can be more easily seen. Each probability goes to zero as 5 p
„ ∞. You can readily see that for each eigenfunction, the average momentum of that state
will be zero, but for each momentum eigenfunction, other than the first, there is a positive and
a negative component consistent with a particle having an equal probability of moving to the
right or to the left.
pPM ax 0.15
PP( 1 , kp) 0.1
PP( 2 , kp)
PP( 3 , kp)
PP( 4 , kp)
PP( 5 , kp)
0.05
0
0
40
− 50
20
0
kp
20
40
50
In addition, each of the stationary states is a state of definite total energy I8 . You can
see this by determining the expectation value of the energy of the system ØLÙ in one of its
eigenstates. But this seems to lead to a contradiction. How can Ø:Ù be zero, but ØLÙ not be
Quantum Mechanics
1.9
zero? The particle is simply moving back and forth in the well with constant kinetic energy
isn't it? To try and answer this carefully using the ideas we have developed so far, let's notice
that Ø:Ù Á Ø:# Ù, so it might be a mistake to compare Ø:Ù# Î#7 with ØLÙÞ Perhaps we had
better compare Ø:# ÙÎ#7 with ØLÙ.
___________________________________________________________________________
Problem 3.2 Beginning with the position eigenfunction <8 ÐBÑ, evaluate the quantity Ø:# Î#7Ù
using the definition s
: œ 3h `Î`B and compare it with the energy for that particular
eigenstate.
___________________________________________________________________________
But if all the eigenstates of this system are time independent, how can this possibly
describe the motion of a particle as it moves inside the well. This doesn't seem to make sense.
We will show later that the most general solution to our differential equation is the sum of all
possible solutions, consistent with the initial conditions of the problem and that such a sum of
solutions leads to a time-dependent probability and a time-dependent average value for the
position of a particle. You might be tempted to conclude, however, that since there is no time
dependence in ØBÙ in any of the eigenstates, the eigenfunctions are not really acceptable
solutions in and of themselves, but must be combined to obtain suitable solutions to a given
problem. However, the ground state of atomic Hydrogen (the eigenstate with lowest energy)
seems to exist!
Probabilities and Expectation Values
Discrete Probability Functions
To gain a better understanding of the statistical interpretation of the wave function
solution to Schrodinger's equation, we need to look briefly into some of the basic concepts of
probability. Consider a room containing 14 people of different ages. We poll the people
present and find that there is one person of age 14, one of age 15, three of age 16, two or age
22, two of age 24, and five of age 25. This information can be represented in the form of a
histogram, as shown below.
Histogram of Ages
Numbe r of Pe ople
5
4
3
2
1
0
10
12
14
16
18
20
22
24
26
28
30
Age
Figure 1.3 A histogram of the ages of the 14 people in the room.
We can also represent this data symbolically by letting R Ð4Ñ be equal to the number of
persons with age 4, giving, in tabular form,
Quantum Mechanics
1.10
R Ð"%Ñ œ "
R Ð"&Ñ œ "
R Ð"'Ñ œ $
R Ð##Ñ œ #
R Ð#%Ñ œ #
R Ð#&Ñ œ &
where it is understood that R Ð$!Ñ œ !, etc. Now, the total number of people in the room must
be given by
R œ R Ð4Ñ œ " " $ # # & œ "%
4ϰ
(3.36)
4œ!
Now we can ask some questions regarding probabilities and other statistical
information about the people in the room. For example, if we close our eyes and pick a person
at random, what is the probability that we would get someone of a particular age? We define
this probability as the number of possible choices which give the desired result divided by the
total number of possible choices, or
R Ð4Ñ
R
T Ð4Ñ œ
(3.37)
So the probability of picking a person of age 13 is zero, the probability of picking a person of
"
#
age "% is "%
, the probability of picking a person of age #% is "%
œ (" , etc. Now the probability
of picking one of age "5 or one of age "6 must be T Ð"& or "'Ñ œ T Ð"&Ñ T Ð"'Ñ œ 27 .
This same type of reasoning could be applied to the case where we have a bag
containing 14 bones, each with a number carved on it (which just happens to correspond to the
ages of the people in our room). If we reach into the bag and pull out a bone, the probability
"
of getting a bone with the number "& is just T Ð"&Ñ œ "%
Þ What is the probability that we will
select a bone with the number 14, or 15, or 16 if we reach into the bag and pick a bone at
&
random? This must be given by T Ð"% 9< "& 9< "'Ñ œ T Ð"%Ñ T Ð"&Ñ T Ð"'Ñ œ "%
. One
might also ask what is the probability of picking a bone with the number 15 on it and then
picking a bone with the number 16 on it. This would be given by the equation
T Ð"& +8. "'Ñ œ
"
$
3
‚
œ
"% "$
182
(3.38)
where the second term takes into account the once the 15 bone is chosen, there are only 13
possibilities left, not 14, three of which give the desired result.
And what is the probability that we pull out a bone with any of the numbers which are
available: "%ß "&ß "'ß ##ß #%ß or #&? It must be
T Ð+8C>2381Ñ œ T Ð4Ñ œ 4œ∞
4œ!
R Ð4Ñ
"
R
œ R Ð4Ñ œ
œ"
R
R
R
4œ!
4œ!
4ϰ
4ϰ
(3.39)
Quantum Mechanics
1.11
This last equation is related to the so-called normalization principle for probabilitiesÞ It
basically states that if you reach into the bag and pull out a bone you will certainly select a
bone with one of the allowed numbers on it! The normalization principle is a very important
one and will guide us in the formal development and understanding of quantum mechanics.
Other questions we might ask regarding the people in the room (or bones in the bag)
are: 1) What is the average age of the group (or average number on the bones)? 2) What is
the most probable age (or the most likely number that you will obtain choosing the bones
randomly)? 3) What is the median age of the people in the room (or median number of the
bones)?
We define the average (or mean) of a group by the equation
4 R Ð4Ñ
4ϰ
Ø4Ù œ
4œ!
R Ð4Ñ
4ϰ
R Ð4Ñ
œ 4
œ 4 T Ð4Ñ
R
4œ!
4œ!
4ϰ
4ϰ
(3.40)
4œ!
In this particular case, the agerage age of people in the room turns out to be #". Notice that no
one happens to be this age - that is not necessary. In fact, more often than not, the average age
of a group of people will turn out to be a fraction, not a whole number. From the equation
above, it is obvious that the average value of the age of the people in the room can be
determined from the probabilities for the different ages. In the same way, the average location
of a particle in quantum mechanics can be determined based upon the probability for finding a
particle at a particular spot. We call this the expectation value in quantum mechanics. Note
that it is not the most probable value! The most probable value is the value that occurs most
frequently, i.e., the value where T Ð4Ñ is a maximum!
A less often used quantity is the median. The median is defined to be the number
where the probability of getting a larger number is equal to the probability of getting a smaller
number. In our particular example, the median is 23: there are seven people who are older
and seven who are younger. Again, notice that no individual is the age of 23!
One might also ask the question: What is the average of the square of the ages of the
people in our room? Although you might wonder why anyone would find that a useful
question, it is stall a valid one. Perhaps you would be more comfortable asking about the
average of the square of the numbers on our bones. The answer is given by our basic
definition of the average
Ø4 Ù œ 4# T Ð4Ñ
4ϰ
#
(3.41)
4œ!
and in this case is the number 459.57. We define the average of any function 0 Ð4Ñ by the
equation
Ø0 Ð4ÑÙ œ 0 Ð4Ñ T Ð4Ñ
4ϰ
4œ!
(3.42)
Quantum Mechanics
1.12
Another parameter which is useful in discussing probabilities or distributions
(histograms) is the standard deviation. The standard deviation is helpful in determining how
sharply peaked the distribution is. For example, consider the histograms shown below for two
different groups of 10 people.
Histogram 2
8
Numbe r of S tud ents
Numbe r of Stude nts
Histogram 1
6
4
2
0
1
2
3
4
5
6
7
8
2
1.5
1
0.5
0
1
9
2
3
4
5
6
7
8
9
Age
Age
The first histogram might represent the ages of students in a single class in a large city daycare
facility where all the students are practically the same age. The second histogram might
represent the ages of students in a rural day care facility where the number of students of a
particular age is too small to warrant a single classroom. In both of these cases, the average
age, the median age, and the most probable age are the same (5), but the distribution of ages is
quite different. In the first histogram the deviation of the individual ages from the average
?4 œ Ø4Ù 4, is small, while in the second histogram the deviations are somewhat larger. We
might try to calculate the average deviation to see if this would help to distinguish the two
histograms. The average deviation is given by
Ø?4Ù œ ?4 T Ð4Ñ œ Ø4Ù 4 T Ð4Ñ œ Ø4ÙT Ð4Ñ 4 T Ð4Ñ œ !
4=∞
4ϰ
4ϰ
4ϰ
4=0
4œ!
4œ!
4œ!
(3.43)
In fact, the average value is defined to be that number which gives zero average deviation.
Clearly the average deviation is of little use to distinguish between the two histograms. A
moments thought will reveal that the reason the average distribution is zero is that there are
just as many positive deviations from the average as there are negative deviations. We might
consider, then, using the absolute value of the deviation. Because absolute values are often
quite complicated to deal with mathematically, an easier approach is to use the square of the
deviation, which must be positive, and average that quantity. That is, we want to consider:
5 # œ ØÐ?4Ñ# Ù
(3.44)
which we call the variance of the distribution. The quantity 5 is called the standard
deviation. A useful theorem concerning the standard deviation is the following:
5 # œ Ø4# Ù Ø4Ù#
(3.45)
___________________________________________________________________________
Problem 3.3 Begin with the basic definition of the standard deviation
Quantum Mechanics
1.13
5 # œ ØØ4Ù 4# Ù
and show that
5 # œ Ø4# Ù Ø4Ù#
___________________________________________________________________________
The way we defined the variance Ðand the standard deviation) means that the square of the
standard deviation is alway greater than or equal to zero. This means that Ø4# Ù Ø4Ù# !, or
Ø4# Ù Ø4Ù# : the average of the square of a quantity is always greater than or equal to the
square of the average.
Now let's determine the standard deviation of the two histograms representing our day
care facilities. The average age in both cases is five. The average of the square of the ages in
each case is given by
Ø4#1 Ù œ #&Þ#
Ø4## Ù œ $"Þ!
The standard deviation for each case is given by
5" œ È#&Þ# #& œ !Þ%%(
5# œ È$"Þ! #& œ #Þ%%*
Notice that in the case of the second histogram, if we take the interval Øj# Ù „ 5# , which is
approximately & „ #, we find that six out of the 10 children's ages fall into this interval. A
similar statement can be made about histogram 1. For an “ideal” (Gaussian or standard)
distribution we find that about 23 of the sample lies in a region given by ØjÙ „ 5 , while about 13
lies outside this region, and that about 95% of the sample lies within 25 of the average value.
Thus, the standard deviation is useful measure of the width of the distribution function.
The Continuous Probability Function
What we have said so far about averages and probability we have applied to a finite
number of samples. These same ideas can be carried over to the case of a continuous
probability function. In this case, we might define the probability of finding a particle in the
region between B œ + and B œ , by the equation
T >+, œ ( T ÐBß >Ñ .B
,
(3.46)
+
The quantity T ÐBß >Ñ is call the probability densityà the differential probability of finding a
particle at random between the points B and B .B is given by T ÐBß >Ñ .B and is dependent
upon the size of .B. As indicated earlier, the probability density in quantum mechanics is
related to the solution to Schrödinger's equation according to the equation
T Bß > .B œ lGÐBß >Ñl# .B
(3.47)
Quantum Mechanics
1.14
If our wave function (and, therefore, the probability) is to be applied to a single particle, we
must require that the probability taken over all space must give a certainty of finding the
particle (the particle must be somewhere!). This means that
T Ð>Ñ+8CA2/</ œ T∞ œ (
∞
T ÐBß >Ñ .B œ "
(3.48)
∞
which is our normalization condition. And, if the particle is to behave reasonably, this
normalization condition must not be time dependent! Otherwise, we have particles that
simply appear or dissappear at random. (There are some situations, however, that might
actually be described in just this manner - radioactive decay, being one.) Similarly, if we
consider a particle that is moving through space, we would expect that the probability of
finding that particle within a certain region of space would have to change in time in some
well-defined manner. We will examine these issues in more detail in the following section.
The average value of some function of position follows directly from our definition of
average values for a discrete probablilty function, so that in general
Ø0 Ð>ÑÙ œ (
∞
0 ÐBÑT ÐBß >Ñ .B
(3.49)
∞
or, in particular, we define the expectation value of a particle (the ensemble average of the
position of a particle) by the equation
ØB>Ù œ (
∞
B T ÐBß >Ñ .B
(3.50)
∞
or, in terms of the solution to Schrödinger's equation,
ØBÐ>ÑÙ œ (
ØBÐ>ÑÙ œ (
∞
B G* ÐBß >Ñ GÐBß >Ñ .B
(3.51)
∞
∞
G* ÐBß >Ñ B GÐBß >Ñ .B
∞
where we have explicitly indicated that the average Ðor expectation value) may in general be a
function of time, and that the time dependence arises from the time dependence of the solution
to Schrödinger's equation. It is conventional to write the expectation value of B in the form of
the second equation above, with the variable B sandwiched between the wave function and its
complex conjugate – with the complex conjutgate always on the left. The reason for this will
become more evident later on.
The standard deviation in position,
5 # ´ ØÐ?BÑ# Ù œ ØB# Ù ØBÙ#
(3.52)
is identical to the expression we had before, for a discrete distribution function. This
particular quantity gives us an indication of the uncertainty in the location of a particle. Thus,
to determine this uncertainty in the position of a particle, we not only need the expectation
Quantum Mechanics
1.15
value of the position, but also the expectation value of the square of the postion, or
ØB# Ð>ÑÙ œ (
∞
G* ÐBß >Ñ B# GÐBß >Ñ .B
(3.53)
∞
___________________________________________________________________________
Problem 3.4 Consider the time-independent Gaussian distribution function
T ÐBß >Ñ œ E/-B+
#
where E, +, and - are constants.
(a) Determine the value of E which will normalize this distribution function.
(b) Find ØBÙ, ØB# Ù, and 5 for this distribution function.
(c) Plot this distribution function and indicate the location of the average value and the
points ØBÙ „ 5 .
___________________________________________________________________________
Problem 3.5 Consider the wave function
GÐBß >Ñ œ E/-lBl /3=>
where A, -, and = are positive, real constants.
(a) Normalize this wave function.
(b) Determine the expectation values of B and B# .
(c) Graph the probability density |GÐBß >Ñ|# as a function of B, and indicate on this graph the
location of the points 5 and 5 . What is the probability of finding the particle
described by this wave function outside the range ØBÙ „ 5 ?
___________________________________________________________________________
Problem 3.6 Beginning with the eigenfunctions of the infinite square well
G8 ÐBß >Ñ œ Ê
#
81B 3=8 >
sinŠ
‹/
P
P
show that the expectation value for any of the eigenstates is given by
ØBÙ œ
P
#
___________________________________________________________________________
The Normalization of the Wave Function
The statistical interpretation of the wave function relates the differential probability
T ÐBß >Ñ .B to the wave function GÐBß >Ñ according to Equation 3.47. And the requirement that
the particle must be somewhere requires that the probability density integrated over all space
must be unity:
T∞ œ (
+∞
-∞
¸G(Bß >Ѹ# .B œ "
(3.54)
Quantum Mechanics
1.16
But an addition requirement is also placed upon the wavefunction – it must also be the
solution to Schrodinger's equation. For Schrödinger's equation to be valid, these two
conditions must be compatible. Looking again at Schrodinger's equation (Equ. 3.5) it should
be obvious that the wavefunction can be multiplied by any arbitrary (complex) constant
without changing the equation. Thus, we are free to select this constant in such a way as to
satisfy Equation 3.54.
Some solutions to Schrodinger's equation, however, cannot be normalized (the wave
functions may get very large as Bp „ ∞, for example). If our interpretation of the wave
function is correct, these solutions can not correspond to “real” systems and, therefore, must
be rejected (e.g., notice that the trivial solution GÐBß >Ñ œ ! is not normalizable, nor is the
solution GÐBß >Ñ œ -98=>+8>, so these solutions cannot be associated with a particle). The
only valid solutions to Schrodinger's equation, then, are functions which are square
integrable, i.e., functions where
(
+∞
-∞
¸G(Bß >Ѹ# .B œ Q
Q
∞
(3.55)
And since any function which is a solution to Schrodinger's equation can be multiplied by an
arbitrary, finite constant and still remain a solution to Schrodinger's equation, we generally
normalize these solutions to be consistent with our interpretation of the wave function for a
single particle.
The requirement the the wavefunction be normalizable, i.e., that
(
+∞
-∞
¸G(Bß >Ѹ# .B œ "
(3.56)
means that the wavefunction itself, must tend toward zero in the limit as B p ∞. Let's look a
little closer at what this means. Let's assume, for example, that GÐBß >Ñ has the form
GÐBß >Ñ œ E B8 . Our normalization condition then requires that
(
+∞
E# B#8 .B œ "
(3.57)
-∞
or that
∞
B#8"
E# ”
•
#8 " »
œ"
(3.58)
∞
For this last equation to be valid, the exponential of the variable B must be less than unity, or
the function will blow up as B p ∞. Thus, we require that
#8 " Ÿ !
(3.59)
8 Ÿ "Î#
(3.60)
or
Quantum Mechanics
1.17
This means thatß in general, the wavefunction GÐBß >Ñ must go to zero at least a fast as "ÎÈB.
This also implies that the first derivative ` GÐBß >Ñ/`B of the wavefunction must go to zero at
$
least as fast as "ΈÈB ‰ !
The General Time-Dependent Solution to Schrodinger's Equation
for the Infinite Square Well
Since any combination of solutions to a linear differential equation is itself a solution,
the most general solution to the time-dependent Schrodinger equation is given by
GÐBß >Ñ œ -8 /3=8 > <8 ÐBÑ
(3.61)
8
where the -8 's are arbitrary coefficients. Sometimes the time dependence of the wavefunction
is incorporated into these coefficients, giving
GÐBß >Ñ œ -8 Ð>Ñ <8 ÐBÑ
(3.62)
8
where -8 Ð>Ñ œ -8 /3=8 > .
If we know the form of the solution at time > œ !, expressed as GÐBß !Ñ, then we
should be able to determine the coefficients for that time and use them in determining the
time-dependent solution to the equation. We can do this by exploiting one of the fundamental
consequences of the solution to Schrodinger's equation: as long as the energies of the different
states are unique (non-degenerate), the eignenfunctions are orthogonal to one another. Since
we have chosen to normalize each of the eigenfunction of our system, we say that each of our
eigenfunctions are orthonormal — i.e., they satisfy the condition
*
( <7 ÐBÑ<8 ÐBÑ .B œ $78
P
(3.63)
!
where
!
$78 œ œ
"
7Á8
7œ8
(3.64)
is known as the Kronecker delta.
___________________________________________________________________________
Problem 3.7 Prove that the eigenfunctions of the infinite square well as we have defined
them are orthonormal by showing that the integral
# P
71 B
81 B
‹ sinŠ
‹ .B œ $78
( sinŠ
P !
P
P
___________________________________________________________________________
We can make use of this orthonormality condition to obtain the expansion coefficients
for our general wavefunction, just as we do for a Fourier series (if fact, that is just what we are
dealing with here, since our eigenfunctions are sine waves). If we multiply equation 3.62 by
Quantum Mechanics
1.18
the complex conjugate of an arbitrary eigenfunction and integrate over the region accessible to
the particle, we obtain
*
3= >
*
( <7 ÐBÑGÐBß >Ñ .B œ -8 / 8 ( <7 ÐBÑ<8 ÐBÑ .B
ðóóóóóóñóóóóóóò
8
$78
(3.65)
Because of the Kronecker delta, each term in the summation gives zero except the one where
7 œ 8, giving, for time > œ !
*
3= Ð!Ñ
( <7 ÐBÑGÐBß !Ñ .B œ -7 / 7 œ -7
(3.66)
Once we know the value for each -8 , obtained from the initial wavefunction at time > œ ! for a
particle in the square well, we can determine the general solution for this particle at all 6+>/<
times using the equation
GÐBß >Ñ œ -8 /3=8 > <8 ÐBÑ
(3.67)
8
We should point out that although we have chosen to normalize each of the individual
eigenfunctions, the normalization condition is a requirement of our general solution if it is to
have physical meaning. This requirement imposes a restriction on the expansion coefficients,
-8 . It requires that the sum of the squares of the expansion coefficents must be equal to unity,
as shown below:
*
( G ÐBß >ÑGÐBß >Ñ .B œ "
(3.68)
* 3= > *
-3= >
( -7 / 7 <7 ÐBÑ-8 / 8 <8 ÐBÑ.B œ "
7
8
*
*
-7
-8 /3=7 =8 > ( <7
ÐBÑ<8 ÐBÑ .B œ "
*
-7
-8 /3=7 =8 > $78 œ "
78
l-7 l# œ "
78
7
This last result gives us some insight into the physical significance of the expansion
coefficients. The fact that the sum of the square of the expansion coefficients must always be
unity, would seem to indicate that the square of the expansion coefficient must be related to
the probability of finding the particle in a particular eigenstate of the system. This would
mean, for example, that the expectation value for the energy a particle would be given by the
ensemble average of the energies for each of the available eigenstates according to the
Quantum Mechanics
1.19
equation
ØLÙ œ I7 l-7 l#
(3.69)
7
where we take ØLÙ to be the ensemble average energy, I7 is the energy for the eigenstate
designated by the quantum number 7 and l-7 l# is the probability of finding the particle in that
particular eigenstate.
___________________________________________________________________________
Problem 3.8 Beginning with the definition of the expectation value for the energy of an
arbitrary wave function
s GÐBß >Ñ .B
ØLÙ œ ( GÐBß >Ñ L
expand these wave functions in terms of a sum over eigenfunctions and show that the
expectation value for the energy can be written as
ØLÙ œ l-7 l# I7
7
___________________________________________________________________________
Expectation Values for the General Time-Dependent Solution
Let's determine the expectation value of B for a general time-dependent solution for the
infinite square well. We begin with our basic defining relation
ØBÙ œ ( G* ÐBß >Ñ B GÐBß >Ñ .B
P
(3.70)
!
The general wave function can be written as an infinite sum of eigenfunctions in the form
GÐBß >Ñ œ -8 /3=8 > <8 ÐBÑ
(3.71)
8
giving
* 3=7 > *
ØBÙ œ ( -7
/
<7 ÐBÑ B -8 /3=8 > <8 ÐBÑ .B
P
!
7
(3.72)
8
Factoring out everything that is not a function of the position, we obtain
*
*
ØBÙ œ -7
-8 /3=8 =7 > ( <7
ÐBÑ B <8 ÐBÑ .B
P
7
(3.73)
!
8
In order to evaluate the expectation value, we must evaluate the integrals
*
Ø7lBl8Ù œ ( <7
ÐBÑ B <8 ÐBÑ .B
P
!
(3.74)
Quantum Mechanics
1.20
where we have introduced as useful shorthand notation Ø7lBl8Ù. This symbol represent the
integral of B over the product of two different eigenfunctions, with the one on the left-handside being the one that is a complex conjugate.
Similarly, if we want to evaluate the uncertainty in the position measurement we will
also need to determine ØB# Ù, which means we will need to evaluate the integrals
*
Ø7lB# l8Ù œ ( <7
ÐBÑ B# <8 ÐBÑ .B
P
(3.75)
!
This same process can be followed to evaluate the expectation value of the momentum (in
position space) and its square. This will gives us
*
*
Ø:Ù œ -7
-8 /3=8 =7 > ( <7
ÐBÑ Œ3h
P
7
!
8
*
*
Ø: Ù œ -7
-8 /3=8 =7 > ( <7
ÐBÑ Œh #
P
#
7
!
8
`
 <8 ÐBÑ .B
`B
(3.76)
`#
 <8 ÐBÑ .B
`B#
(3.77)
which means we will need to evaluate the intergrals
*
Ø7l:l8Ù œ 3h ( <7
ÐBÑ
P
!
`
`
<8 ÐBÑ .B œ 3hØ7l l8Ù
`B
`B
(3.78)
`#
`#
#
<
ÐBÑ
.B
œ
h
Ø7l
l8Ù
8
`B#
`B#
(3.79)
and
*
Ø7l:# l8Ù œ h # ( <7
ÐBÑ
P
!
Using our shorthand notation, you should note that
*
Ø7l8Ù œ ( <7
ÐBÑ <8 ÐBÑ .B œ $7ß8
P
!
The values of these integrals are given in the table on the next page.
(3.80)
Quantum Mechanics
1.21
Energy Eigenfunctions and Eigenvalues for the Infinite Square Well
G8 ÐBß >Ñ œ Ê
#
81B 3=8 >
sinŠ
‹/
P
P
h # 1#
#
I8 œ 8 ”
• œ 8 I"
#
#7P
#
Orthonormalization Conditions of these Wavefunctions
*
( <7 ÐBÑ<8 ÐBÑ .B œ $78
P
!
Useful Matrix Elements of the Infinite Square Well
Ú
1Î#
Ý
Ý
!
Ø7lBl8Ù œ P † Û
)78
Ý
Ý
Ü
1 # c 7 # 8# d #
if 7 œ 8
if 7 Á 8à 7 „ 8 even
if 7 Á 8à 7 „ 8 odd
Ú
"
"
Ý
Ý
Ý
$ #7# 1#
Ø7lB# l8Ù œ P# † Û
)78
7„8
Ý
Ý
Ý "
 # #
Ÿ
1 c 7 8# d #
Ü
Ú
!
Ý
Ý
`
"
!
Ø7l l8Ù œ † Û 478
`B
P Ý
Ý
Ü 7 # 8# Ø7l
`#
"
8# 1 #
l8Ù
œ
†
œ
!
`B#
P#
if 7 œ 8
if 7 Á 8
if 7 œ 8
if 7 Á 8ß 7 „ 8 even
if 7 Á 8ß 7 „ 8 odd
if 7 œ 8
if 7 Á 8
Quantum Mechanics
1.22
As an example of how all this works, let's consider the special case where the general
form of the wave function at time > œ ! is given by
GÐBß !Ñ œ Ec<" ÐBÑ <# ÐBÑd
(3.81)
This is the case where the general wave function is formed from equal amounts of the first two
eigenfunctions. The constant E is require for normalization purposes. First we need to
determine the normalization constant E. Remember that once we normalize the function, it
will always remain normalized. The normalization condition is given by
*
*
( G ÐBß !Ñ GÐBß !Ñ .B œ lEl ( <" ÐBÑ <# ÐBÑ‘c<" ÐBÑ <# ÐBÑd .B œ "
P
P
#
*
!
(3.82)
!
" œ lEl# ( <"* ÐBÑ <#* ÐBÑ‘c<" ÐBÑ <# ÐBÑd .B
P
(3.83)
!
œ lEl# ( <"* ÐBÑ<" ÐBÑ .B lEl# ( <"* ÐBÑ<# ÐBÑ .B â
P
!
P
lEl (
!
P
#
!
<#* ÐBÑ<" ÐBÑ .B
lEl ( <#* ÐBÑ<# ÐBÑ .B
P
#
!
which can be written in our shorthand notation as
" œ lEl# eØ"l"Ù Ø"l#Ù Ø#l"Ù Ø#l#Ùf
(3.84)
*
Ø7l8Ù œ ( <7
ÐBÑ <8 ÐBÑ .B œ $7ß8
(3.85)
Since
P
!
we see that the normalization condition gives
" œ #lEl# Ê E œ
"
È#
(3.86)
assuming that E is real. This means that the general wave function at any time > can be
expressed by the equation
GÐBß >Ñ œ
"
< ÐBÑ/3=" > <# ÐBÑ/3=# > ‘
È# "
(3.87)
and that the probability density for finding a particle at some position B will be timedependent, giving a time-dependence to the expectation value of the position ØBÙ, the
momentum Ø:Ù, and perhaps the energy ØLÙ.
To find the expectation value of B we write
Quantum Mechanics
1.23
ØBÙ œ ( G* ÐBß >Ñ B GÐBß >Ñ .B
P
œ(
!
P
!
(3.88)
"
"
<"* ÐBÑ/3=" > <#* ÐBÑ/3=# > ‘ B
< ÐBÑ/3=" > <# ÐBÑ/3=# > ‘ .B
È#
È# "
"
œ Ø"lBl"Ù Ø"lBl#Ù/3=" =# > Ø#lBl"Ù/3=" =# > Ø#lBl#Ù‘
#
where we have made use of our shorthand notation on the last line above. Looking back at
our table for the matrix elements, you should convince yourself that Ø#lBl"Ù œ Ø"lBl#Ù so that
we can express the expectation value of B in the form
"
Ø"lBl"Ù Ø"lBl#Ù/3=" =# > /3=" =# > ‘ Ø#lBl#Ù‘
#
"
œ cØ"lBl"Ù Ø#lBl#Ù #Ø"lBl#Ù-9=c=" =# >dd
#
ØBÙ œ
(3.89)
Plugging in the specific numberical values for the matrix elements, we obtain
ØBÙ œ
P
$#
”" # -9=c=" =# >d•
#
*1
(3.90)
Notice that $#Î*1# œ !Þ$' so the particle doesn't move outside the well as it oscillates back
and forth. At time > œ ! the particle has an expectation value of ØBÙ œ !Þ$# P. The
expectation value then changes to become PÎ# after going through one-quarter period and
becomes ØBÙ œ !Þ') P after half a period, then moves back toward the center of the well. The
overall diplacement (on average) from the center of the well is 0.18 P. In addition, we know
the values of =" and =# , since we know the eigenenegies associated with the eigenfunctions.
The angular frequency is given by
=8 œ
I8
8# I "
œ
œ 8# ="
h
h
(3.91)
This means we can determine the frequency of oscillation of the particle in this particular
state. We find that
=" =#
/œ
œ =" %=" Î#1 œ $=" Î#1
(3.92)
#1
$h 1
$2
œ $I" Î#1h œ œ #
%7P
)7P#
The minus sign here is insignificant since -9=()Ñ œ -9=Ð)Ñ.
The expectation value of the momentum Ø:Ù can be determined a number of different
ways, but, perhaps the simplest is to take the time derivative of the expectation value of
position. Doing so, we obtain
Ø:Ù œ
.ØBÙ
)h
œ
=38Ð$=" >Ñ
.>
$P
So we see that the expectation value of the momentum also varies with time.
(3.93)
Quantum Mechanics
1.24
The expectation value of the energy of the system can be determined in a very straight
forward manner, using the equation
ØLÙ œ l-7 l# I8
(3.94)
7
For our particular problem, where the general state of the system is composed of equal parts of
the first two eigenstates of the system, we have that -" œ -# œ "ÎÈ# Ê l-" l# œ l-# l# œ "Î#,
giving
ØLÙ œ
"
"
"
&
& h # 1#
I" I# œ cI" %I" d œ I" œ Œ

#
#
#
#
# #7P#
(3.95)
One of the fundamental postulates of quantum physics, however, states that if we attempt to
measure the location, the momentum, or the energy of a system we can only obtain those
values for a measurement of a quantity that are consistent with the eigenvalue equation for
the operator associated with the measurement of that quantity. This means that if we attempt
to measure the energy of our system, we will not obtain an energy value equal to the
expectation value. We will, rather, obtain either the value I" or the value I# . For the
particular problem we are considering, the probability of measuring the value I" will be equal
to the probability of measuring the value I# because the wavefunction is made up of equal
parts of these two eigenfunctions. This would seem to indicate that our measurement process
somehow collapses our general wavefunction to one of the possible eigenfunctions. Thus, if
we measure the energy of the system and obtain I" the wave function must have collapsed to
the wavefunction <" ÐBÑ/3=" > because if we measure the energy a moment later we would
expect to again obtain I" — not some other value. Thus the act of measurement somehow
disturbs our system causing the system to change. Once we make a measurement, the system
is no longer in the state we started with.
Eigenfunctions and Eigenvalues for Position and Momentum
We pointed out above that when we make a measurement of the energy of the system,
we do not obtain the expectation value of the energy, we only have a certain probability of
obtaining one of the eigenvalues of the energy. One can only obtain values of a measurement
which are eigenvalues of the corresponding operator. In this section we want to investigate
how this applies to measurements of position and momentum. To do this, we must examine
the “eigenvalue equations” for position or momentum to see how these concepts apply for
those measurements. The eigenvalue equation for any measurement process has the same
form as the eigenvalue equation for the energy operator. It has the form
s <ÐBÑ œ @+6?/ <ÐBÑ
b
(3.96)
where the operator associated with the act of making some measurement acts on the
eigenfunction of that operator on the left side of the equation to produce a measured value (a
constant) times the eigenfunction of that operator on the right side of the equation. This
equation defines both the eigenfunctions of a given operator associated with a particular
measurement, and the corresponding values that can be obtained for that measurement.
Quantum Mechanics
1.25
For a position measurement, the eigenvalue equation is given by
s
B <ÐBÑ œ B<ÐBÑ
(3.97)
where the operator associated with the act of making a position measurement (in position
space) is simply equal to the variable B. This means that the eigenvalue equation for B is
simply
B<ÐBÑ œ B<ÐBÑ
(3.98)
which is satisfied for any position-dependent wavefunction. This also implies that all possible
values of B can be measured.
For a momentum measurement, the eigenvalue equation is
s
: <ÐBÑ œ :<ÐBÑ
(3.99)
where the momentum operator (in position space) is given by
s
: œ 3h
`
`B
(3.100)
Thus, the eigenvalue equation for the momentum operator is given by
3h
`
<ÐBÑ œ :<ÐBÑ
`B
(3.101)
To determine the allowed eigenfunctions and eigenvalues, we must solve this differential
equation. The solution is straight forward, and we find that the momentum eigenfunctions are
given by
<ÐBÑ œ E/3:BÎh œ E/35B
(3.102)
where we have made use of our familiar relationship : œ h5 . There is no restriction of the
values that 5 can take on – either positive or negative, so again all possible values for the
momentum are possible. The solution to Schrödinger's equation for the infinite square well
gave solutions of the form
GÐB,>Ñ œ E/35B F/35B ‘/3=>
(3.103)
which are obviously just a combination of eigenfunctions of the momentum operator (one
eigenfunction associated with motion in the B direction, and one in the B direction). When
we applied the boundary conditions for this problem, we obtained restrictions on the values of
5 and thus on the possible values of the momentum – meaning that not all possible momentum
values are now allowed, only certain restricted values. This is what gives us the periodic
nature of the solutions for the infinite square well.
The eigenfunctions for the momentum for the infinite square well, then, are the
functions
G8 ÐB,>Ñ œ E/358 B F/358 B ‘/3=8 >
(3.104)
Quantum Mechanics
1.26
where 58# œ #7I8 Îh # and =8 œ I8 Îh . Notice that this is equivalent to h # 58# œ #7I8 or
:8# Î#7 œ I8 Þ Since only certain total energies (kinetic enegies) are allowed, only certain
momentum values are allowed. This will mean that not all values of the momentum can be
measured. Those values which are allowed will be those given by
:8 œ h58 œ „ È#7I8 œ „ È#78# h # 1# Î#7P# œ 82Î#P œ 2Î-8
(3.105)
This last equation shows that the possible wavelengths allowed for our infinite square well
system are given by
-8 œ
#P
8
(3.106)