CHAPTER III The Solution of the Free-Particle Wave Equation

CHAPTER III

The Solution of the Free-Particle Wave Equation

Introduction

In the last two chapters, we have demonstrated that the probability of finding a particle

somewhere in space should be related to the absolute magnitude squared of the wave-function

solution to Schrödinger's equation. We also demonstrated that for this interpretation to be

valid, the wavefunction should be normalizable (square-integrable). We defined the

expectation value (ensemble average) of the location of a particle and the particle's

momentum, and demonstrated that the measurements of position and momentum in classical

physics are related to the corresponding expectation values of the position and momentum, and

we showed that the time-derivative of the expectation value of momentum was consistent with

Newton's second law.

In this chapter we wish to investigate even further the nature of the solution to

Schrödinger's equation. We will develop the most general solution to Schrödinger's equation

which is consistent with the description of a single free particle localized within some region

of space at time , and demonstrate how this general solution varies in time and space. In> œ !so doing, we will discover the very important Heisenberg uncertainty principle, so

fundamental to quantum physics, which relates the measurement of the position of a particle to

the measurement of the momentum of that same particle.

The General Solution to Schrödinger's Equation

We want to consider the one-dimensional Schrödinger equation for atime-dependent

particle under the influence of a potential energy which is a function of the position ofZ ÐBÑthe particle. This equation is

� Bß > Z B Bß > œ 3h Bß >h ` `

7 `B `>

# #

#2 ( ) ( ) ( ) ( ) (3.1)G G G

where ( ) is the time-dependent wavefunction, and is the mass of the particle. As weG Bß > 7have already pointed out, if the potential energy function is explicitly a function of time, itnot

is possible to separate the Schrödinger equation so that we have the dependence on position

separated from its dependence upon time.

To do this we assume that ( can be written as a function of position only, ( ,G <Bß >Ñ BÑtimes a function of time only, , so thatXÐ>Ñ

G <ÐBß >Ñ œ ÐBÑX Ð>Ñ (3.2)

The Schrödinger equation then becomes:

� ÐBÑXÐ>Ñ Z ÐBÑ ÐBÑX Ð>Ñ œ 3h ÐBÑX Ð>Ñh ` `

#7 `B `>

# #

#< < < (3.3)

Now the partial with respect to position does not act upon and the partial with respect toXÐ>Ñtime does not operate on , so that this can be written<ÐBÑ

The Free Particle 3.2

XÐ>Ñ � ÐBÑ Z ÐBÑ ÐBÑX Ð>Ñ œ ÐBÑ 3h X Ð>Ñh ` `

#7 `B `>Œ Œ # #

#< < < (3.4)

Dividing both sides of this equation by we obtain<ÐBÑX Ð>Ñ

1 1(3.5)

<<

ÐBÑ #7 `B XÐ>Ñ `>� ÐBÑ Z ÐBÑ œ 3h XÐ>Ñh ` `Œ Œ # #

#

The left-hand side of this equation is a function of the position alone, while the right-hand-side

of this equation is a function of time alone. For this equation to be valid for all possible

combinations of times and positions, it should be obvious that both sides of this equation must

be set equal to some arbitrary (undetermined) constant! We will let this constant be This(Þleaves us with two differential equations which must be solved:

1(3.6)

<< (

ÐBÑ #7 `B� ÐBÑ Z ÐBÑ œh `Œ # #

#

and

1(3.7)

XÐ>Ñ `>3h X Ð>Ñ œ`Œ (

The solution of the temporal equation is relatively easy. Rewriting Equ.3.7 we have

3h X Ð>Ñ œ XÐ>Ñ`

`>( (3.8)

or

` 3

`> hXÐ>Ñ œ � XÐ>Ñ( (3.9)

the solution of which is

XÐ>Ñ œ G/�3 >Îh( (3.10)

where is some arbitrary constant. Now since is given by the product of andG ÐBß >Ñ ÐBÑG <XÐ>Ñ G, we can set the constant equal to unity and let any necessary adjustments be made by

changing the constants that will occur in the solution of the spatial equation. This means that

the general solution to Schrodinger's time-dependent equation has the form

G <ÐBß >Ñ œ ÐBÑ/�3 >Îh( (3.11)

where is the solution to the time-independent Schrodinger equation<ÐBÑ

� ÐBÑ Z ÐBÑ ÐBÑ œ ÐBÑh `

#7 `B

# #

#< < ( < (3.12)


In the last chapter, we showed that the momentum can be expressed in terms of a

differential operator acting on a position-dependent wave function according to equation

: œ �3hs`

`B(3.13)

The square of this operator is given by

: œ �hs# # `

`B

#

#(3.14)

so that the time-independent Schrödinger equation can be written in the form

:s#

#7 Z ÐBÑ ÐBÑ œ ÐBÑ< ( < (3.15)

We also pointed out that the position operator is just the variable , so we can write this lastBequation as

:B

ss

#

#7 Z Ð Ñ ÐBÑ œ ÐBÑ< ( < (3.16)

where you should remember that is just an arbitrary separation constant. The left side of this(last equation looks like the sum of the kinetic and potential energy of the particle, so we

associate this operator with the Hamiltonian of the system. For conservative systems, the

Hamiltonian is simply the total energy of that system. Thus, this last equation can be written

in operator form as

L ÐBÑ œ ÐBÑs< (< (3.17)

which is in the form of an equation. The expectation value of the Hamiltonianeigenvalue

operator

ØLÙ œ ÐBÑL ÐBÑ .B œ ÐBÑ ÐBÑ .B œs s( (�∞ �∞

∞ ∞

< < ( < < (* * (3.18)

is equal to the separation constant, so we will associate the separation constant with a symbol

representing the total energy of the system, .( œ I Thus the time-independent Schrödinger equation appears to be just a statement of theßconservation of total mechanical energy, provided we write the momentum as a differential

operator and associate the separation constant with the total energy . Because of thisß I(association, we will write the time-dependent solution in the form

G <ÐBß >Ñ œ ÐBÑ/�3 >= (3.19)

where / is consistent with the Einstein relationship between frequency and= (œ h œ IÎhtotal energy for a particle. Everything we have said so far is totally general. We now want to

examine the special case where the particle is in a region where the potential energy is

constant.


Schrödinger's Equation for a Free Particle and Its Solutions

In this section we will examine the general solution to Schrödinger's equation for a free

particle - i.e., one which moves in a region of space where the potential energy is constant (and

the force is zero). For simplicity, we let this constant potential energy be zero. We have

already demonstrated that the most general solution to Schrödinger's equation is of the form

G <ÐBß >Ñ œ ÐBÑ/�3 >= (3.20)

where must satisfy Schrödinger's time-independent equation. For the case of a free<ÐBÑparticle, this time-independent Schrödinger equation is

� ÐBÑ œ I ÐBÑh `

#7 `B

#

#

2

< < (3.21)

where . Multiplying this last equation by givesI œ h �#7Îh= #

` ÐBÑ #7I

`B hœ � ÐBÑ œ �5 ÐBÑ

2<< <

# ## (3.22)

where is a constant. Notice that this constant energy is just equal to the kinetic5 œ #7IÎh#

energy in a region of space where the potential energy is constant. We assume that the

solution to this differential equation is of the form

<ÐBÑ œ E/-B (3.23)

and find that this is indeed a solution to this equation, provided that

- œ „ 35 (3.24)

where . Now is if , and if . In the case were5 œ 5 I � ! I !É #7Ih# real pure imaginary

I !, we write

5 œ œ � œ 3#7I #7lIl

h hÊ Ê

# #, (3.25)

where / is real., œ #7lIl hÈ #

The possible solutions, then, are

<ÐBÑ œ E/ F/ I � !35B �35B for (3.26)

<ÐBÑ œ G/ H/ I ! B � B, , for (3.27)

These solutions to Schrödinger's equation are general in nature, with the coefficients to be

determined by specific boundary conditions, and by the normalization condition.

Total Energy Less than Zero Everywhere Let us first consider the case where the total

energy is less than zero ( ). In the region where , equation 3.27 will blow up as I ! B � ! B


p∞ G œ ! B ! Bp�∞ H œ !, unless ; likewise, if , it will blow up as , unless . Thus, in

order for the wave function solution to be normalizable, equation 3.27 must be written

<ÐBÑ œ G/ I !ß B ! B, for (3.28)

and

<ÐBÑ œ H/ I ! B � !� B, for , (3.29)

Continuity of the Wave Function

There are two additional conditions we must also impose on these wave functions for

them to be physically meaningful. The first is that the wave function must be a continuous,

single-valued function. This requirement is simply an extension of the fact that the wave

function is related to the probability. If the probability of finding a particle at a particular

location is to be single-valued (i.e., if there cannot be two different values for the probabilityBat a single location), then the wave function must be continuous. If there is a boundary

between two regions in which the solutions to Schrödinger's equation differ (this is often

associated with a point where there is a change in potential energy), the wave function must be

continuous across that boundary so that it remains single-valued.

For our free-particle solution, the requirement that the wave function be continuous

means that at , so we can write the last two equations in the formG œ H B œ !

<ÐBÑ œ G/ I !� lBl, for (3.30)

A plot of this function is shown below. The wavefunction is continuous at the point ,B œ !but the derivative of this function is obviously not continuous at the point .B œ !

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-10 -5 0 5 10

Fig. 3.1 Wave function solution for a free particle with .I !

Continuity of the First Derivative of the Wave Function

In addition to the requirement that the wave function be continuous, the Schrödinger

equation imposes a restriction on the of the wave function at any boundary.first derivative

This can be demonstrated by integrating the Schrödinger equation with respect to over aB


small interval which encloses the boundary?%

� .B Z B ÐBÑ .B œ I ÐBÑ .Bh ` ÐBÑ

#7 `B

#

B � B � B �

B B B

#( ( ( 9 9 9

9 9 9

% % %

% % %2<< < (3.31)

The term on the right side of this last equation is a constant times the integral of the

wavefunction over some finite interval. If we take the limit as this integral (the area?% p !under the function between and ) must go to zero, since the function is a finite,B � B 9 9% %continuous, single-valued function. The integral on the far left is an integral of the second

derivative of the wave function, and must be equivalent to first derivative of the wave

function. This gives, in the limit as ?% p !

lim lim?% ?%%

%

%

%

Ä! Ä!B �

B

#B �

B ` ÐBÑ #7

`B hœ Z B ÐBÑ .B

<<º (

9

9

9

9

(3.32)

Now the integral of the potential energy function times the wave function can be represented

by the diagram below. The integral is just the area under the curve defined by the product of

the potential energy function and the wave function. The wave function must be finite,

continuous, and single-valued, but the potential energy function may exhibit discontinuities.

As long as the potential energy function has only a finite number of finite steps (i.e., it is

“piecewise-continuous”) the integral of the product of the potential energy function and theßwave function remains finite. In the limit as this integral (the area under the curve?% p !between and ) will go to zero, requiring that B � B % % the first derivative of the wave

function solution to Schrödinger's equation be continuous at a boundary so long as the

potential energy function is piecewise-continuous.

∆ε

x

V(x) ψ(X)

Fig. 3.2 A plot of the product of the potential energy function and the wavefunction. The integral

of this product is the area under the curve. In the limit as 0, this area goes to zero.?% p

Now, we will carry out the operations defined in the last equation. The solution for the

free particle where is given byI !


<ÐBÑ œ G/ I !ß B ! B, for (3.33)

<ÐBÑ œ G/ I ! B � !� B, for , (3.34)

At the point , the derivative is given byB œ �%

G/, �,% (3.35)

and at the point , the derivative is given byB œ %

� G/, �,% (3.36)

The requirement that the derivative be continuous (at the point ) means that these twoB œ !equations must be equal in the limit as , or% p !

G œ � G, , (3.37)

Now, since is generally not zero (unless ), this last equation requires, œ #7IÎh I œ !È #

that . This is just the null solution and is not normalizable. This simply means that noG œ !physical solution exists for the case where everywhere! This makes sense because theI !total energy for a free particle is just the kinetic energy which, classically, can never be less

than zero. This does not mean, however, that a solution does not exist if there are finite

regions where and other regions where .I ! I � !

Total Energy Greater than Zero Everywhere We next consider the case where the total

energy is greater that zero ( ). The wave function solution to the time-independentI � !Schrödinger equation is

<ÐBÑ œ E/ F/ I � !35B �35B for (3.38)

which is a sinusoidal solution and does not blow up as . The time-dependentB p „∞solution to this equation is

GÐB >Ñ œ E/ F/ / I � !, for (3.39)ˆ ‰35B �35B �3 >=

which can be written in the form

GÐB >Ñ œ E/ F/, (3.40)3 5B� > �3 5B > = =

The first term of this solution is just the equation of a plane wave of well-defined energy (or

wavelength, since ) traveling in the direction. This can be seen by using Euler's5 œ # Î B1 -equation and writing

/ œ 5B � > 3 5B � >3 5B� > = cos sin (3.41) = =

If we look at the cosine function we can move along with the wave by chosing to ride along at

a point of constant phase. Choosing the point where the phase angle is zero, we obtain

5B � > œ ! Ê B œ > œ @ >5

==Š ‹ :2+=/ (3.42)


Likewise, the second term in equation 3.40 represents a plane wave of the same well-defined

energy traveling in the direction. If this wave function is to represent a particle�B single

moving through space, it cannot be moving in directions at the same time, so we have toboth

choose either or to be zero. This means that E F the solution for a free particle can be written

in the form of a traveling plane wave

GÐB >Ñ œ E/, (3.43)3 5B� > =

where

5 ´ „#7I

h

5 � ! Ê5 ! Ê

È œ, with (3.44)traveling to the right

traveling to the left

We can also express this solution in the form of sines and cosines, using Euler's equation

G = =ÐBß >Ñ œ E 5B � > 3 5B � >c d cos sin (3.45)

Now, to see if this is a reasonable solution to Schrödinger's equation, we examine the

probability function and find that

G G G*ÐBß >Ñ ÐBß >Ñ œ l ÐBß >Ñl œ lEl# # (3.46)

which means that the probability density of finding a particle is constant everywhere. But this

means that the wave function ! The reason the wave function is notis not normalizable

normalizable is that the sine and cosine functions have no - i.e., they do not go to zero as end Bp „∞. We must, therefore, truncate these functions somehow in order to produce a wave

function solution to Schrödinger's equation which represents the motion of a single, free

particle in space.

This problem is the same as that encountered in classical wave theory when we wish to

describe sound or light in a more or less sense. As an example, consider awaves localized

clap of thunder which is a relatively localized (in time) acoustical pulse. If we were to analyze

such a clap, we would find that the sound is composed of a large number of different

frequencies (as can be demonstrated by the electrical noise picked up on the radio at all

frequencies). Likewise, a pulse of light emitted from a pulsed laser is somewhat localized in

space and time. The answer to this normalization problem lies in the application of Fourer

series and integrals. As we will demonstrate in the next two sections, periodic or non-any

periodic waveform can be expressed in terms of a complex Fourier series or Fourier integral

(i.e., a combination of harmonic waveforms).

Note: We can sometimes get around the problem of normalization if we examine

some finite region of space and develop our equations in terms of the probability

current density. One can show (problem 2.6) that the probability current density

for a traveling wave is given by

4 œ „ lElh5

7#

The term is just the equivalent of the classical velocity of a particle, and h5Î7 E

is the amplitude of the wave function. Notice that the units of must be 1/E PÈ


where is some length, since This means that the units of mustP .B œ "Þ 4' G G*

be . This means we can let , represent the "ÎP PÎ> œ "Î> lEl œ RÎP#

number of particles per unit length (the number of particles per unit volumn in

one-dimension). Thus, the units of correspond to the number of particles4passing a given point per unit time. Since is not a function of the position orB 4the time, this means that the traveling wave equation represented by a sine of

cosine wave represents a situation where the number of particles per second

passing a particular location is independent of time or position along the -axis.B

Fourier Series

One can show that periodic function, no matter how bizarre, can be expressedany

mathematically in terms of an infinite series of sines and cosines. Thus, for a function ,0Ð Ñ)which is periodic with period , such that#1

0Ð # Ñ œ 0Ð Ñ) 1 ) (3.47)

we can always express in terms of the infinite series:0Ð Ñ)

0Ð Ñ œ + + + # + $ â"

# , , # , $ â

) ) ) )

) ) )

9 " # $

" # $

cos cos cos

sin sin sin

(3.48)

where the coefficients are chosen so that the Fourier series actually reproduces the waveform

of interest in the interval ! # Þ) 1 Now it turns out that the Fourier series can also be expressed in terms of complex notation,

since both the sine and the cosine function can be expressed in terms of complex exponentials.

Using Euler's relations, we have

sin and cos (3.49)8 œ 8 œ/ � / / /

#3 #) )

38 �38 38 �38) ) ) )

If we substitute these expressions into the equation above, we can write the Fourier series

expansion of a periodic function ( ) in the form:0 )

0Ð Ñ œ - /) �8œ�∞

∞

838) (3.50)

Notice here that the sum is from to , and that the constant termnegative infinity positive infinity

arises when you set . This method of representing the Fourier series is often a bit8 œ !simpler to use. Again, we must determine the correct values of the constants, , to represent-8the desired function. These coefficients can be found by using the orthogonality relationships

for the exponentials:

(�

3Ð7�8Ñ

781

1)/ . œ #) 1 $ (3.51)

where the integral is over a 2 .period 1


____________________________________________________________________________

Problem 3.1 Prove the orthogonality relationship

(�

3Ð7�8Ñ

781

1)/ . œ #) 1 $

____________________________________________________________________________

If we multiply both sides of equation (3.50) by and integrate from to we obtain:/ � -37) 1 1

( (��

� �

�37 3Ð8�7Ñ

8œ�∞

∞

8

8œ�∞

∞

8 78 7

1 1

1 1) )/ 0Ð Ñ . œ - / .

œ - # œ # -

) ) )

1 $ 1

(3.52)

The coefficients, are, therefore, given by:- ß8

- œ 0Ð Ñ / ."

#8

�

�38

1) )(

1

1) (3.53)

This whole proceedure can also be written in terms of some spatial displacement, , byBexpressing the angle in terms of . Thus, we let , where is the spacial period.) ) 1B œ # BÎT TIt is, however, even more convenient to write this as , where , or one half) 1 0 0œ BÎ œ TÎ#the spatial period. In this case, we can write ( ) ( , where0 BÎ œ 0 BÑ1 0

0ÐB # Ñ œ 0ÐBÑ0 (3.54)

We can therefore, express in terms of the infinite series:ß 0ÐBÑ

0ÐBÑ œ + + BÎ + # BÎ + $ BÎ â"

# , BÎ , # BÎ , $ BÎ â

9 " # $

" # $

cos cos cos (3.55)

sin sin sin

1 0 1 0 1 0

1 0 1 0 1 0

or in complex notationß ß

0ÐBÑ œ - /�8œ�∞

∞

838 BÎ1 0 (3.56)

Again we must determine the constants which will give us our desired function! The-8orthogonality relationships for the exponentials can be expressed in terms of the spatial

displacement, , by the equation:B

(-

+

0

01 0/ .B œ #3Ð7�8Ñ BÎ

780 $ (3.57)

Multiplying both sides of equation (3.56) by and integrating from to we/ � �37 BÎ1 0 0 0obtain:


( (� – —�

-

+

0 0

0 01 0 1 0/ 0ÐBÑ .B œ - / .B

œ - # œ # -

�37 BÎ 3Ð8�7Ñ BÎ

8œ�∞

∞

8�

8œ�∞

∞

8 78 70 $ 0

(3.58)

Thus the coefficients are given by:-8

- œ / 0ÐBÑ .B"

#8

�

�38 BÎ

0(

0

01 0 (3.59)

where 2 is the period!0

DIRICHLET CONDITIONS

We have demonstrated that a Fourier series can be found for a periodic function , but0ÐBÑwe must ask under what conditions this series actually converges to that function. Dirichlet's

theorem states that:

If f(x) is periodic in 2 , and if between and it is single-valued, has a1 1 1�finite number of maximum and minimum values, and a finite number of

discontinuities, and if

(-

+

1

1

l0ÐBÑl .B

is finite, then the Fourier series converges to f(x) at all the points where f(x) is

continuous; at the Fourier series converges to the of the jump.jumps midpoint

An example of a function which satisfies the Dirichlet condition is shown below.

x

f(x)

−π π

The figure below illustrates how the Fourier series converges to a square-wave function.

Notice that the Fourier series is equal to the midpoint of a discontinuity, and that there is an

overshoot in the Fourier series at a discontinuity which does not go away as we add more and

more terms! This is called the Gibbs phenomena.


-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

____________________________________________________________________________

Problem 3.2 Expand in a complex Fourier series the periodic function which is defined0ÐBÑin the interval by the relation:! B #P

0ÐBÑ œ! ! B P" P B #Pœ

ÒIf this were a periodic function in time, it would correspond to a square wave of one volt. The

terms of the Fourier series would then correspond to the different a-c frequencies which are

combined in this "square wave" voltage, and the magnitude of the Fourier coefficients would

indicate the relative importance of the various frequencies. Show that the solution can beÓwritten as:

0ÐBÑ œ � â" # B " $ B " & B

# P $ P & P1

1 1 1” •sin sin sin

____________________________________________________________________________

Problem 3.3 When a guitar string is plucked, it generally does not oscillate with a single

frequency, unless it is plucked in a very particular way. If we assume that the string is plucked

in the center (displaced from equilibrium a distance ), but remains fixed on each end, then we$can determine the frequency components by performing a Fourier series expansion to describe

the initial waveform of the string! Assume the length of the string is , and express thePfunctional form of this plucked string by expanding the waveform in terms of a complex

Fourier expansion [the waveform will look something like a roof top]. In addition, write the


expansion in terms of sines and/or cosines. The result you get will depend upon the particular

coordinate system you use, and . Choose aupon your choice of the periodic function

coordinate system so that one end is located at 0, while the other end is at , andB œ B œ Pshow that the expansion can be written in terms of or , depending upon the form=38/= -9=38/=of the wave function chosen. [In one case, you will have a series of roof tops, all above the

C œ ! line; in the other, you will have a function that looks like a triangular sine wave with

both positive and negative values of , but the negative values are outside the region ofCinterest].

____________________________________________________________________________

Fourier Integrals and the Dirac Delta Function

The question now arises: What do we do about wave functions which are not periodic, and

which cannot be made to look periodic? In addition, we have noticed that periodic functions

can be formed from a discrete set of frequencies, i.e., not all frequencies are needed to give a

certain periodic function. An expansion of the concept of the Fourier series allows us to treat

any wave function whatsoever, whether it is periodic or not, and also to include all possible

frequencies.

The complex Fourier expansion of a function which is periodic with period 2 is given by0

0ÐBÑ œ - /�8œ�∞

∞

838 BÎ1 0 (3.60)

where can be determined from:-8

- œ / 0ÐB Ñ .B"

#8

�

�38 B Î w w

0(

0

01 0w (3.61)

Now if we substitute this last equation into the previous one, we have

0ÐBÑ œ / 0ÐB Ñ .B /"

#� – —(8œ�∞

∞

�

�38 B Î w w 38 BÎ

0 0

01 0 1 0w

(3.62)

If we now let , and / , so that / , we have5 œ 8 Î 5 œ 5 � 5 œ œ 58 8 " 81 0 ? 1 0 0 1 ?+

0ÐBÑ œ / 0ÐB Ñ .B /5

#� Ô ×Ö ÙÕ Ø(∞

8œ�∞

5

� 5

�35B w w 35B

/

/

?

1 (3.63)

1 ?

1 ?

w

and, in the limit as 0, and / , and giving,? ? 1 ?5 Ä 5 Ä .5 5 Ä ∞ Ä� 'ß

0ÐBÑ œ / 0ÐB Ñ .B / .5"

#( (” •�∞ �∞

∞ ∞�35B w w 35B

1

w

(3.64)

Notice that the term in brackets is a function of only, which we will call . Thus, we can5 1Ð5Ñwrite as0ÐBÑ


0ÐBÑ œ 1Ð5Ñ / .5 ß(�∞

∞35B (3.65)

where

1Ð5Ñ œ 0ÐBÑ / .B Þ"

#1(�∞

∞�35B (3.66)

These equations define the Fourier transform and its inverse. This transform is valid for any

type of function , periodic or not, and also allows for all possible values of the frequency0ÐBÑ(since ). Since these last two equations are nearly symmetric except for5 œ # Î œ # Î-1 - 1/the factor , it is a common practice to slightly alter our definitions in order to make these"Î#1two equations more nearly alike. (It is important to check how a text defines the transform and

it's inverse because this is not always done in the same way!) Thus, we will the Fourierdefine

transform and its inverse by the equations:

0ÐBÑ œ 1Ð5Ñ / .5 ß"

#È (1 �∞

∞35B (3.67)

and

1Ð5Ñ œ 0ÐBÑ / .B Þ"

#È (1 �∞

∞�35B (3.68)

Now in order for these two equations to actually be useful we require consistency in our

definition. That is, if we substitute the equation for back into the first equation1Ð5Ñ

0ÐBÑ œ 0ÐB Ñ / .B / .5" "

# #È È( (Ô ×Õ Ø1 1

(3.69)

�∞ �∞

∞ ∞

w �35B w 35Bw

we must still get an equality. Now if we change the order of integration, we get

0ÐBÑ œ 0ÐB Ñ / .5 .B"

# (3.70)

1( (Ô ×

Õ Ø∞ ∞

�∞ �∞

w 35ÐB�B Ñ ww

and, in order for the right-hand-side of this equation to be , we require that the integral in0ÐBÑthe braces be a very special integral indeed. We require that the integral over all be equal to52 at the point and zero everywhere else! We write this in a special notation:1 only B œ Bw

2 (3.71)(∞

�∞

35ÐB�B Ñ w/ .5 œ ÐB � B Ñw

1 $

where is called the This function can be thought of as a$ÐB � B Ñw Dirac delta function.

function which is zero everywhere except where the argument is zero, and at this single point,


it is equal to , but in such a way that the area under the curve is unity! This definition∞enables us to write:

0ÐBÑ œ 0ÐB Ñ ÐB � B Ñ .B"

# 2 (3.72)

11 $( c d

∞

�∞

w w w

or

0ÐBÑ œ 0ÐB Ñ ÐB � B Ñ .B (3.73)(∞

�∞

w w w$

As we stated above, the Dirac delta function can be thought of as a special function that forces

the integrand to take on the value where the argument of the delta function equals to zero.

This is equivalent to the equation

0Ð Ñ œ 0ÐBÑ ÐBÑ .B0 (3.74)(∞

�∞

$

These last two equations can be considered the of the Dirac delta function. Fromdefinition

this basic definition, one can derive (by using a change of variables in most cases) the

following relationships:

B ÐB � +Ñ œ + ÐB � +Ñ

Ð � BÑ œ ÐBÑ

Ð+BÑ œ ÐBÑ + Á !"

± + ±

0ÐBÑ ÐBÑ .B œ � 0 Ð!Ñ

$ $

$ $

$ $

$

for

(3.75)

( w w

As an example of the utility of the Dirac delta function we will derive an extremely

important relationship, known as :Parseval's theorem

( (�∞ �∞

∞ ∞

0 ÐBÑ0ÐBÑ.B œ 1 Ð5Ñ1Ð5Ñ.5* * (3.76)

which relates the absolute magnitude squared of the wave function to the absolute magnitude

squared of the Fourier transform of the wave function. We can verify this last equation by

using the Dirac delta function, as follows:

( ( ( (Ô ×Ô ×Õ ØÕ ØÈ È

�∞ �∞ �∞ �∞

∞ ∞ ∞ ∞

35B w �35 B w0 ÐBÑ0ÐBÑ.B 1 Ð5Ñ/ .5 1Ð5 Ñ/ .5 .B" "

# #

* * = (3.77)1 1

w


( ( ( (�∞ �∞ �∞ �∞

∞ ∞ ∞ ∞

w 3Ð5�5 ÑB w0 ÐBÑ0ÐBÑ.B 1 Ð5Ñ1Ð5 Ñ/ .5 .5 .B"

#* * = (3.78)

1

w

( ( (�∞

∞

w w w0 ÐBÑ0ÐBÑ.B 1 Ð5Ñ1Ð5 Ñ Ð5 � 5 Ñ .5 .5

�∞

∞* * = (3.79)$

( (�∞ �∞

∞ ∞

0 ÐBÑ0ÐBÑ.B œ 1 Ð5Ñ1Ð5Ñ.5* * (3.80)

This equation illustrates the fact that normalizing the function normalizes0ÐBÑ simultaneously

the function .1Ð5Ñ

Fourier Transforms and the Free-Particle Wave Equation

Earlier in this chapter we pointed out that the localized nature of a pulse might serve as a

wave-like representation of a particle, and that such a representation required the use of

Fourier analysis. We postulate, therefore, that a possible solution to Schrödinger's equation

which could represent a single, localized particle might be expressed in the form

< 91

ÐBÑ œ Ð5Ñ / .5"

#È ( (3.81)

�∞

∞

35B

where is the Fourier integral expansion of the wavefunction, and where<ÐBÑ

9 <1

Ð5Ñ œ ÐBÑ / .B"

#È (�∞

∞�35B (3.82)

is the inverse transform of . Now the function can also be expressed as since< 9 9ÐBÑ Ð5Ñ : : œ h5 Ð:Ñ, and we will discover in what follows, that the function is the probability9amplitude for measuring a particular value of the momentum. Because of ,Parseval's theorem

we know that if we normalize the position wave function we will automatically<ÐBÑnormalize the momemtum wave function ( In addition, all the consequences arising from9 :ÑÞthe normalization of the position wave function also apply to the momentum wave function.

Although the idea of forming a wave-packet from the sinusoidal solutions to

Schrödinger's equation may seem reasonable, we must now determine if a wave function of

this formwill in fact satisfy the Schrödinger equation for a free particle, and thus be an

acceptable solution to our problem. To do this, we assume, even though the equations above

are not explicitly time dependent, that these equations are valid at any particular instant of

time, or, more generally,


G 91

ÐBß >Ñ œ Ð5 >Ñ / .5"

#È ( , . (3.83)

�∞

∞

35B

We now insert this wave function into the Schrödinger equation for a free particle and

determine if this equation is indeed a valid solution. This give us

� Ð5 >Ñ / .5 œ 3h Ð5 >Ñ / .5h ` " ` "

#7 `B `># #

# #

#

�∞ �∞

∞ ∞

35B 35B , , (3.84)Ô × Ô ×Õ Ø Õ ØÈ È( (

1 19 9

The partial derivatives can be moved into the intergrals to obtain

� � 5 Ð5 >Ñ / .5 œ 3h Ð5 >Ñ / .5h `

#7 `>

#

�∞ �∞

∞ ∞

# 35B 35B , , (3.85)Ô × Ô ×Õ Ø Õ Ø( (9 9

where you should notice that is a function of . Moving the constants into the9 5ß > Bnot

integrals gives

, , (3.86)( (�∞ �∞

∞ ∞# #

35B 35Bh 5 `

#7 `>Ð5 >Ñ / .5 œ 3h Ð5 >Ñ / .59 9

which will always be valid if the integrands are equal, or if

h 5 `

#7 `>Ð5ß >Ñ œ 3h Ð5ß >Ñ

# #

9 9 (3.87)

We can solve this equation to obtain

9 9 9Ð5ß >Ñ œ Ð5ß !Ñ/ œ Ð5ß !Ñ/�3h 5 >Î#7h �3 ># # = (3.88)

where , and where .= œ IÎh I œ h 5 Î#7 œ : Î#7# # #

This last equation means that the solution to Schrödinger's equation for a free particle can

always be written in the form

G 91

ÐBß >Ñ œ Ð5 !Ñ / .5"

#È ( , . (3.89)

�∞

∞

35B�3 >=

where is some arbitrary function that depends upon the initial conditions of the9Ð5ß !Ñproblem and where . Now at time , we get the Fourier transform= œ h 5 Î#7 > œ !# # back

G 91

ÐBß !Ñ œ Ð5ß !Ñ / .5"

#È ( (3.90)

�∞

∞

35B

We can, therefore, determine the function from the Fourier transform of 9 GÐ5ß !Ñ ÐBß !Ñ


9 G1

Ð5 !Ñ œ ÐBß !Ñ / .B"

#, . (3.91)È (

�∞

∞

�35B

Thus, we have the tools to determine the wave function at any time if we know theGÐBß >Ñ >wave function ( at time . From Parseval's theorem, we also know that theG Bß !Ñ > œ !required normalization condition on the wave function imposes a similarGÐBß !Ñnormalization condition on the function , and once the wavefunction is normalized, it9Ð5ß !Ñremains normalized for all time.

We can express the equations above in terms of the momentum and the energy by using

the fact that , and . (Remember that the deBroglie relationship between wave-: œ h5 I œ h=like and particle-like phenomena is .) The general solution to Schrödinger's equation: œ 2Î-can thus be written as

G 91

ÐBß >Ñ œ Ð: !Ñ / .:"

h #È ( , , (3.92)

�∞

∞

3 :Îh B�3 IÎh >

where

9 G1

Ð: !Ñ œ ÐBß !Ñ / .B"

#, . (3.93)È (

�∞

∞

�3:B h/

and where we must always remember that the energy is in general a function of theI œ h=momentum. Again, to make these two equations more symmetric, we will divide the hbetween these two equations (using the flexibility of normalization to allow us to absorb

constants into the wave functions) so that we have the two equations:

G 91

9 G1

ÐBß >Ñ œ Ð: !Ñ / .:"

# h

Ð: !Ñ œ ÐBß !Ñ / .B"

# h

È (

È (

, (3.94)

,

�∞

∞

3 :Îh B�3 IÎh >

�∞

∞

�3:B h

/

The Interpretation of 9Ð:ß >Ñ Now, let's look at a few examples. First we will consider the case where the initial

momentum of a particle is well defined, i.e., the case where the particle has a definite value of

the momentum, . Mathematically, we can express this by saying that the function is:o 9Ð:ß !Ñgiven by

9 T$Ð:ß !Ñ œ Ð: � :o) (3.95)


where is some constant. The wave function which is a solution to Schrödinger's equation isTtherefore given by

< T$1

ÐB >Ñ œ Ð: � : Ñ / .:"

# h, . (3.96)È (

�∞

∞

3 :Îh B�3 IÎh >o

or

<T

1ÐBß >Ñ œ /

# hÈ (3.97)3 : Îh B�3 I Îh > o o

where /2 for a free particle. But this is just the equation of a plane wave, which weI œ : 7o o#

already know is not normalizable and . The only way weimplies a constant density everywhere

can represent a distribution of particles is to make use of the equationlocalized

< 91

ÐB >Ñ œ Ð: !Ñ / .:"

# h, , . (3.98)È (

�∞

∞

3 :Îh B�3 IÎh >

where , cannot be associated with one single, well-defined momentum, but must be9Ð: !Ñassociated with a in the value of the momentum of the particle. [One might thinkdistribution

of this by considering a group of particles, and assuming each particle might have a slightly

different momentum - however, this must be applicable to a particle as well, so oursingle

interpretation must be consistent with a description of a single particle.] We might, therefore,

interpret the function | | as the 9Ð:ß >Ñ # probability for measuring a particular value ofdensity

momentum at a particular time momentum: >. If this is true, then must be the 9Ð:ß >Ñprobability amplitude for measuring a particular value of momentum, in direct

correspondence with the interpretation of as the probability amplitude for measuring a<ÐBß >Ñparticular value of position, , as some time .B > As further justification for this association, let's look again at the expectation value of

the momentum. This is given by

Ø:Ù œ ÐBß >Ñ h ÐBß >Ñ .B`

`B( Œ �∞

∞

G G* �3 (3.99)

Each of the wave functions of position and time can be written as

G 91

ÐBß >Ñ œ Ð:ß >Ñ/ .:"

# hÈ (�∞

∞3:BÎh (3.100)

However, in order to keep the various parameters identifiable, we will use for the dummy:variable in the second integral and for the dummy variable in the first integral. The partial:w

derivative acts only on the integrand, so that

�3h ÐBß >Ñ œ Ð:ß >Ñ / .:` �3h 3:

`B h# hG 9

1È (�∞

∞3:BÎh (3.101)


Now writing the expectation value of using the momentum representation we obtain:

Ø:Ù œ .B .: Ð: ß >Ñ/ .: Ð:ß >Ñ:/"

# h19 9( ( (

�∞ �∞ �∞

∞ ∞ ∞w w 3: BÎh 3:BÎh* w

(3.102)

Rearranging terms and integrating first over position, we obtain

Ø:Ù œ .:.: Ð: ß >Ñ : Ð:ß >Ñ / .B"

# h19 9( ( (

�∞ �∞ �∞

∞ ∞ ∞w w 3 :�: BÎh* w (3.103)

Recall that we defined the Dirac delta function as

$1

Ð5 � 5 Ñ œ / .B"

#w 3 5�5 B

�∞

∞( w (3.104)

which is equivalent to

$1

(: � : œ / .B"

# hw 3 :�: BÎh

�∞

∞ w (3.105)

This means that the expectation value of the momentum can be written as

Ø:Ù œ .:.: Ð: ß >Ñ : Ð:ß >Ñ Ð: � : Ñ( (�∞ �∞

∞ ∞w w w9 9 $* (3.106)

Integrating over , the delta function gives zero everywhere except where , so we are: : œ :w w

left with

Ø:Ù œ .: Ð:ß >Ñ : Ð:ß >Ñ(�∞

∞

9 9* (3.107)

Notice that this equation for the expectation value of momentum has the same form as the

expression

ØBÙ œ .B ÐBß >Ñ B ÐBß >Ñ(�∞

∞

< <* (3.108)

for the expectation value of the position. This would seem to indicate that the wave function

solutions are wave functions which represent the probability amplitude in position,GÐBß >Ñwhile are wave functions which represent the probability amplitude in momentum.9Ð:ß >ÑThe wave functions are often described as being the wave functions in position space,GÐBß >Ñwhile the wave functions are described as being the wave functions in momentum9Ð:ß >Ñspace. Thus, the momentum operator in momentum space is just the variable just like the:s :ßposition operator in position space is just the variable Bs B. Now let's find an expression for the position operator Bs in momentum space. To do

that we start with the expectation value of position and write

ØBÙ œ .B ÐBß >Ñ B ÐBß >Ñ(�∞

∞

G G* (3.109)


and then use the general expression for in terms of the momentumGÐBß >Ñ

ØBÙ œ .B .: Ð: >Ñ / B .: Ð: >Ñ /"

# h19 9( ( (

�∞

∞

�∞ �∞

∞ ∞

�3 :Îh B w w 3 : Îh B , , (3.110)* w

The variable can be carried into the last integral, since it is independent of . And thisB :w

variable can be written as a parital with respect to momentum, as in the next equation

ØBÙ œ .B .: Ð: >Ñ / .: Ð: >Ñ /" h `

# h 3 `:19 9( ( ( ” •ˆ ‰

�∞

∞

�∞ �∞

∞ ∞

�3 :Îh B w w 3 : Îh B , , (3.111)'

* w

The last integral in the equation above can be integrated by parts to give

( (” • ” •ˆ ‰ ˆ ‰ ˆ ‰»

�∞ �∞

∞ ∞

w 3 : Îh B w w 3 : Îh B w

w 3 : Îh B

∞

�∞

9 9

9

Ð: >Ñ / .: œ � Ð: >Ñ / .:h ` h `

3 `: 3 `:

Ð: >Ñ /

, , (3.112)' '

,

w w

w

Because the momentum wave function must be normalizable, this function must go to9 : ß >w

zero faster than , so that the last term in the expression above must be zero. Thus, we are"Î:w

left with an expression for the expectation value of given byB

ØBÙ œ .B .: Ð: >Ñ / .: 3h Ð: >Ñ /" `

# h `:19 9( ( ( ” • ˆ ‰

�∞

∞

�∞ �∞

∞ ∞

�3 :Îh B w w 3 : Îh B , , (3.113)'

* w

Now, if we reorder the way in which we take the integral, we obtain

ØBÙ œ .: .: Ð: >Ñ 3h Ð: >Ñ .B /" `

# h `:19 9( (Œ

�∞

∞w w �3 :�: BÎh

w

�∞

∞

* , , (3.114) w

The last integral in the equation above is equal to , giving# h Ð: � : Ñ1 $ w

ØBÙ œ .: .: Ð: >Ñ 3h Ð: >Ñ : � :`

`:( Œ �∞

∞w w w

w9 9 $* , , (3.115)

or, integrating over first, we obtain:w

ØBÙ œ .: Ð: >Ñ 3h Ð: >Ñ`

`:( Œ �∞

∞

9 9* , , (3.116)


This last equation can be expressed in the form

ØBÙ œ .: Ð: >Ñ Ð: >Ñ(�∞

∞

9 9* , , (3.117)Bs

where is the position operator in momentum representation.B œ 3s h `Î`: Commutation of Operators

Since both the momentum and postition operators may need to be expressed in terms of

partial derivatives, we will need to take special care when we deal with the products of

operators. For example, let's consider the product of the position and the mometum operators

operating on a position dependent wavefunction

B: œ B �3ss < <<

ÐBÑ h ÐBÑ œ �3hB` ` ÐBÑ

`B `BŒ (3.118)

The product of these same two operators operating in reverse order gives

:B œ �3ss < < <<

ÐBÑ h B ÐBÑ œ �3h ÐBÑ B` ` ÐBÑ

`B `BŒ ” • (3.119)

These two operations are clearly not the same. If we take the difference between these two

operations we obtain

B: � :Bss ss < <ÐBÑ œ 3h ÐBÑ (3.120)

We define the for and by the equationcommutation operator B :

c dBß : B: � :Bs s ss ssœ (3.121)

so that equation (3.120) can be written as

c dBß :s s œ 3h

Two operators are said to commute when the commutation operator is equal to zero, i.e., when

the order of operation of the two operators is insignificant. They are said to be non-

commuting when the commutator is non-zero, i.e., when the order of operation is significant.

It is interesting to consider that the non-commutation of the position and momentum operators

arises because is finite! In classical physics (where energy levels, etc., can be considered ashcontinuous), we would expect these two operators to commute.

The commutation operation we have just introduced is significant for two reasons: 1)

This commutation operation is related to the uncertainty principle! If the commutator of two

operators is not zero, then there exists an uncertainty relationship between the two observables

associated with these operators. 2) The commutation relationship is independent of

representation! One obtains the same result whether you work in position representation or

momentum representation!


Introduction to The Heisenberg Uncertainty Relation

To gain a better picture of how and are related to one another, we will< 9ÐBß !Ñ Ð:ß !Ñconsider a function which would represent a particle localized in space (with some uncertainty

? <B ÐBß !Ñ. We will choose the wave function so that

< ?

?

?ÐBß Ñ œ

" B

BlBl

#

! lBl �B

#

0 (3.122)

if

if

ÚÝÝÛÝÝÜÈ

The Fourier transform of this function

9 <1

Ð: !Ñ œ ÐB !Ñ / .B"

# h, , . (3.123)È (

�∞

∞

�3:B h/

gives

9 ?1 ?

Ð:ß !Ñ œ Ð: BÎ#hÑ#ÐhÎ:Ñ

# h BÈ sin (3.124)

Notice that this last equation is zero when / , or where so that as: B #h œ 8 : œ 8# hÎ B? 1 1 ?? 9B : Ð:ß !Ñ œ : , the value of where 0 . If we define the uncertainty in asdecreases increases

the of the function (i.e., the distance between the first zeros of the function) thenwidth 9Ð:ß !Ñ? 1 ? ? ? 1 <: œ % hÎ B : B œ h œ 2Ó ÐBß !Ñ [or 4 2 . Thus, as becomes more localized (i.e., as

? 9B Ð:ß !Ñ gets smaller) then becomes more spread out. This is the basic nature of the

Heisenberg uncertainty relation (although at this point we have not been very precise in our

definition of uncertainty). A more exact statement of the Heisenberg uncertainty principle will

be developed later.

This position-momentum uncertainty is a natural consequence of using Fourier

transforms to produce a localized wave packet (a necessity if we want to describe the motion

of a . particle) Our mathematical model seems to imply that the more precisely we know , atBtime , the less precisely we know at time , and vise versa> œ ! : > œ ! !

____________________________________________________________________________

Problem 3.4 Consider the wave function

GÐBß >Ñ œ E/ /� lBl �3 >- =

Determine the Fourier transform in momentum space of this wave function at time [i.e.,> œ !the momentum distribution function ] and the normalization constant . Comment on9Ð:ß !Ñ Ehow the momentum distribution function varys with very large values of , and with very-small values of ? How does this relate to the Heisenberg uncertainty principle?-____________________________________________________________________________


Traveling Waves and the Spreading of a Gaussian Wave Packet

We now want to examine how our mathematical description of the wave function for a

free particle changes in space and time (as it should for a free particle moving through space).

Our general solution for the free-particle Schrödinger's equation is

< 91

ÐB >Ñ œ Ð5 !Ñ / .5"

#, , . (3.125)È (

�∞

∞

35B�3 >=

where

9 <1

Ð5 !Ñ œ ÐBß !Ñ / .B"

#, (3.126)È (

�∞

∞

�35B

Since , and , and since and are generally related in some fashion, we= œ IÎh 5 œ :Îh I :note that, in general, . This is just the simple relationship between the frequency of= =œ Ð5Ñthe wave and the wavelength. For this relationship (derived from the fact thatphotons

I œ :-) is very simple

= = -/œ 5- Ê @ œ - œ Î5 œ (3.127):2

where is the speed of light in a vacuum, and is a constant. This gives-

< 91

ÐB >Ñ œ Ð5 !Ñ / .5"

#, , . (3.128)È (

�∞

∞

35ÐB�@ >Ñ:2

which is just the equation for the propagation of plane waves whose phase velocities remain

constant as the waves propagate.

For particles, however, the relationship is not so simple. Relativistically, the equation for

the energy gives the expressionI œ : - 7 -# # # # %o

= œ 5 - 7 -

hÊ # #

# %

#o (3.129)

Classically, the energy reduces to , and, since this reducesI œ 7 - : Î#7 â I œ h ßo# # =

to

= œ 7 - h5

h #7o# #( )

(3.130)

In neither case do we obtain the simple relationship / constant. However, we know that= 5 œthe general form of the wave function which must describe is given byparticles

< 91

ÐB >Ñ œ Ð5 !Ñ / .5"

#, , . (3.131)È (

�∞

∞

35B�3 Ð5Ñ>=


where (k) is an appropriate function of k.= Unlike the equation for photons, the plane-wave

components of this equation do not travel at the same speed. This causes them to interfere in a

complicated manner which gives rise to an interference . We now want to examineenvelope

this last equation in greater detail to see if it does, indeed, give a reasonable description of

what is observed in nature.

To make this example as concrete as possible, we will assume that ( ) is a simple9 5ß !function of the form

9 #Ð5ß !Ñ œ / (3.132)� Ð5�5 Ñα o#

This is a Gaussian function which is peaked (with a maximum value of ) about the value#5 œ 5o and which has a associated with the constant parameter . (The size of width α αdetermines how faster ( ) falls off as moves away from .) Since the function is9 95 5 5 Ð5ß !Ño

the probability amplitude for finding a particular value of (and thus of ) we see that is a5 : αmeasure of the uncertainty in a measurement of the momentum. We will give a more precise

definition of this uncertainty later.

From Parseval's theorem, we know that if the function satisfies the normalization9Ð5ß !Ñcondition, then the wave function will automatically be normalized. The normalization<ÐBß !Ñcondition,

(�∞

∞

9 9*Ð5ß !Ñ Ð5ß !Ñ.5 œ " (3.133)

fixes the area of the function , and demands that and are related by the equation9 # αÐ5ß !Ñ

#α

1œ

#” •"Î% (3.134)

Since we may not generally know the functional form of (or since it may be very=complicated), we will expand ( ) about the point using a Taylor series:= 5 5 œ 5o

= == =

Ð5Ñ œ Ð5 Ñ Ð5 � 5 Ñ Ð5 � 5 Ñ â` " `

`5 #x `5o o oº º

5 5

#

##

o o

(3.135)

This equation will give us good results as long as is not much different from . Thus, the5 5oequation for our wave function having a Gaussian momentum distribution becomes:

< #1

( ) (3.136)Bß > œ / / .5"

#È (∞

�∞

� 5�5 3Ð5B� Ð5Ñ>Ñα =( )o#

where the exponential term is given by


/ œ /B: 3 Ò 5B � Ð5 Ñ> � > Ð5 � 5 Ñ â`

`5

� > Ð5 � 5 Ñ â" `

#x `5

3Ð5B� >Ñ

5

#

#5

#

= (3.137)

]

š ºº ›

==

=

o o

o

o

o

Where we originally had the term in our integral equation, we now have the combination5Ð5 � 5 Ñ 5B 5 Bo o everywhere in the term . If we add and subtract the term in theexcept

exponential and group the terms appropriately we will have

/ œ /B: 35 B 3 Ò Ð5 � 5 ÑB � > � > Ð5 � 5 Ñ â`

`5

� > Ð5 � 5 Ñ â" `

#x `5

3Ð5B� >Ñ

5

#

#5

#

= [ ] (3.138)

]

š ºº ›

o o o o

o

==

=o

o

We now introduce the symbols and and write this last expression5 œ Ð5 � 5 Ñ œ" `

#x `5w

#

#5

o "= º

o

as

/ œ /B: 35 B � >

‚ /B: 35 B � > � 3 5 > â`

`5

3Ð5B� >Ñ

w w

5

#

= (3.139)

]

š ›š Š ‹ ›º

o o=

="

o

We recognize the term as a term for this waveform in analogy with theÐ` Î`5Ñ= 5o velocity

phase velocity of a plane wave. We call this velocity term the . We will see ingroup velocity

what follows that the term containing determines the “spread” in the wave packet as a"function of time. In those cases where is small, the term with is very smallÒ 5 œ Ð5 � 5 Ñw

o "and quite often negligible.Ó With these substitutions, our integral expression becomes

<#

1( ) (3.140)Bß > œ / / / .5

#È (3Ð5 B� >Ñ � 5 3 5 ÐB�@ > � 5 >â w

∞

�∞

o o= α "w w w# #1[ ) ]

where we have changed the integration variable from [ , since .5 p .5 œ .Ð5 � 5 Ñ œ .5 5wo o

is a constant], and where we have factored out the terms and which# =/B: 35 B � >š ›o o

are functions of .not 5w

We are now in a position to evaluate the integral. The integral can be written in the form

of a generalized Gaussian integral (see Appendix 2.C) of order zero

N œ / .Bo (�∞

∞

�+B �,B (3.141)#


where and are complex constants. The solution of this integral is+ ,

N œ /+

o Ê1 (3.142), Î%+#

Therefore, we rewrite our integral representing ( ) by grouping together powers of to< Bß > 5w

obtain

<#

1( ) (3.143)Bß > œ / / .5

#È (3Ð5 B� >Ñ � 3 > 5 � �3ÐB�@ > 5 w

∞

�∞

o o= α "[ + ] [ )] w w#1

from which we see that and , giving as our solution to the+ œ Ò 3 >Ó , œ Ò � 3ÐB � @ >ÑÓα " 1

wave function

<# 1

1 α "( ) (3.144)Bß > œ / /

# 3 >È ŸÊ3Ð5 B� >Ñ �ÐB�@ >Ñ Î%Ò 3 >Óo o= α "1#

<#

α "( ) (3.145)

)Bß > œ / /

#Ð 3 >35 B� 5 Ñ> �ÐB�@ >Ñ Î%Ò 3 >Óo o o[ ( / ]= α " ŸÈ 1

#

Thus, for the special case where is a Gaussian function, peaked around the value9Ð5Ñ5 œ 5o, the traveling wave function corresponding to this looks like a with a plane wave phase

velocity , and with an amplitude given by the expression within the curly brackets,@ œ Î5:2 =o o

Ö × . We say that this plane wave is . That is, the amplitude of the sineamplitude modulated

waves vary with position and time in such a way that we have a Gaussian wave whichpacket

moves with a speed given by . The width of this packet as well as its amplitude changes@1with time, due to the presence of the constant ." The probability density for finding a particle at a point at time is given by the squareB >of the magnitude of , or | | which gives< <ÐBß >Ñ ÐBß >Ñ #

TÐBß >Ñ œ ‚ /B: � �# %Ð 3 >Ñ %Ð � 3 >Ñ

" "

3 > � 3 >

ÐB � @ >Ñ ÐB � @ >Ñ#

α " α " α " α "

#1 1

# #

(3.146)È È Ÿ

TÐBß >Ñ œ /B: � ÐB � @ >Ñ Î# " " "

> % 3 > � 3 >

#

α " α " α "

#

# # #1

#È Ÿ” • (3.147)


TÐBß >Ñ œ /B:Î#

>

� B � @ >Ñ

#Ò > Ó

#

α "

α

α "

#

# # #

1#

# # #È Ÿ (3.148)(

TÐBß >Ñ œ /B: �Î#

>

B � @ >Ñ

#Ò > Ó

#

α " α " α

#

# # #

1#

# # #È Ÿ (3.149)(

/

or, using the normalization condition requirement for , we have#

TÐBß >Ñ œ /B: �"

# >

B � @ >Ñ

#Ò > ÓÈ Èc d Ÿ1 α " α α " α# # #

1#

# # #/ (3.150)

(

/

The equation shows that the probability density is in the form of a Gaussian

KÐDÑ œ EÐ Ñ /% (3.151)�D Î# %

The maximum of this function [ ] occurs when . Thus the maximum point of ourEÐ Ñ D œ !%function occurs when 0, or when ! This implies that the peak of theB � @ > œ B œ @ >1 1

probability function travels at a speed equal to the group velocity . In addition, since the@1Gaussian is symmetric about the point , the D œ ! average value of is the point whereBB œ @ >1 !

From this equation we see that the probability density of finding a particle at a location x

at a time t changes as a function of time. The probability density is peaked where z x œ �v t 0; and thus we see that the peak moves in the positive-x direction with a speed equal tog œv . For a “free particle" where the group velocity is given byg h œ h 5 Î#7= # #

@ œ œ œ` h5 :

`5 7 71

5º=

o

o o

(3.152)

which is the classical velocity of a free particle!

____________________________________________________________________________

Problem 3.5 For surface tension waves in shallow water, the relation between frequency and

wavelength is given by

/1

3-œ

# XË $

where is the surface tension and is the density. Determine the phase velocity and theX 3group velocity for these surface tension waves.

____________________________________________________________________________

The width of a Gaussian can be determined by noticing that the Gaussian function

decreases to of its maximum value when , or when . Thus, the full"Î/ D Î œ " D œ „# % %Èwidth of the Gaussian is given by the equation


$ % α " αD œ # œ # #Ð > ÑÎ (3.153)È È # # #

which varies as a function of time, getting wider as time progresses in either the positive or

negative direction! We see that the width of the peak depends upon , a function of theαoriginal width of the function , and upon . [We have used the symbol for the width9 " $ 5 Bof the function to differentiate it from the more precise definition of the uncertainty of aBmeasurement , which we will discuss in more detail later.]?B At time , the width is a and is given by> œ ! minimum,

$ αD œ # #oÈ (3.154)

[Notice that this is the width which the 'packet' would maintain if .] Now is a measure" αœ !of the width of the momentum distribution function ( ,9 5Ñ

9 # #( ) (3.155)5 œ / œ /� Ð5�5 Ñ �5 ÎÐ"Î Ñα αo# w#

From what we have just done, we see that the width of that function is given by

$ $ α5 œ 5 œ # "Îw È (3.156)

which gives

$ $ α α5 ‚ D œ # "Î ‚ # # œ % #oÈ È È (3.157)

and we see that if is large, then is small, and vise versa!$ $5 Do A more precise definition of the uncertainty in , which we will designate by , isB B?given by the (rms) , orroot-mean square deviation from the mean

?B œ ØÐB � ØBÙÑ ÙÈ # (3.158)

which can also be written as

?B œ ØÐB � ØBÙÑ Ù œ ØB Ù � ØBÙ# # # # (3.159)

Since we understand to be the probability amplitude, and to be the< <ÐBß >Ñ ± ÐBß >Ñ ± #

probability density, then we know that the of any function is given byaverage 0ÐBÑ

Ø0ÐBÑÙ œ 0ÐBÑ ÐBß >Ñ ÐBß >Ñ .B œ 0ÐBÑ T ÐBß >Ñ .B' '�∞ �∞

∞ ∞

(3.160)< <*

which gives, for the uncertainty in ,B

?B œ ÐB � B Ñ T ÐBß >Ñ.B (3.161)” •'�∞

∞#

"Î#

o


Since the probability density for the location of our particle, , is symmetric about theTÐBß >Ñpeak, the average of is just the peak value, i.e., the value where . Thus, , soB B œ @ > B œ @ >1 1o

that we can write asTÐBß >Ñ

T ÐBß >Ñ œ /B: �"

# >

B � @ >Ñ

#Ò > Ó

œ EÐ Ñ/

È Èc d Ÿ1 α " α α " α

%

# # #

1#

# # #

�ÐB�B Ñ Î

/ (3.162)

(

/

9# %

where

EÐ Ñ œ"

# >%

1 α " αÈ Èc d# # # / (3.163)

and

% α " αœ #Ò > Ó# # # / (3.164)

Now we are in a position to evaluate the uncertainty in :B

? %B œ EÐ Ñ ÐB � B Ñ / .B (3.165)Ô ×Õ Ø(

�∞

∞

# �ÐB�B Ñ Î

"Î#

oo# %

? %B œ EÐ Ñ D / .D (3.166)Ô ×Õ Ø(

�∞

∞

# � "Î ÑD

"Î#

( % #

which is a Gaussian integral and can easily be evaluated to give

? % 1 %B œ EÐ Ñ Ð"Î Ñ"

# (3.167)” •È �$Î#

"Î#

or, plugging in for and EÐ Ñ% %

?1 α " α

1

α " αB œ

" "

# > # #Ò > Ó (3.168)

/

/– —È Èc dÈ ” •

# # # # # #

�$Î# "Î#

?α " α α " α

B œ# > # #Ò > Ó

" " (3.169)

1

/ / Ÿ” • ” •c d# # # # # #

"Î# �$Î# "Î#

or

?α "

αB œ

>Êc d# # #

(3.170)


This is the expression for any arbitrary time , it reduces> but for the special case where ,> œ !to

? αB œ9 È (3.171)

where the subscript denotes that this is for time .> œ ! If we consider a traveling wavepacket, the width of that packet can be expressed as

? ?? " "

? ?B œ œ B "

B > >

B BËc d Ë9% # # # #

9 9# %9 (3.172)

The relative spread of the wavepacket, then, is given by

? "

? ?

B >

B Bœ "

9

# #

9%Ë (3.173)

For cases where is very small relative to the original width of the wavepacket, the spread of">the wavepacket is negligible.

We determine in the same way beginning with?5

? 95 œ Ð5 � 5 Ñ ± Ð5ß >Ñ ± .5 (3.174)Ô ×Õ Ø(�∞

∞

# #

"Î#

o

where

± Ð5ß >Ñ ± œ / œ # Î /9 # α 1# # �# Ð5�5 Ñ �# Ð5�5 Ñ (3.175)α αo o# #È

Notice that the time dependence in ( cancels, so that the uncertainty in momentum is9 5ß >Ñtime independent! The uncertainty in then becomes5

? α 15 œ # Î Ð5 � 5 Ñ / .5 (3.176)Ô ×Õ ØÈ (

�∞

∞

# �# Ð5�5 Ñ

"Î#

oα o

#

or

? α 15 œ # Î D / .D (3.177)Ô ×Õ ØÈ (

�∞

∞

# �# D

"Î#

α #

which evaluates to give

? α 1 1 α5 œ # Î Ð# Ñ"

# (3.178)” •È È �$Î#

"Î#

or,


?α

5 œ"

#È (3.179)

and, utilizing our previous result of , we have the Heisenberg uncertainty relation? αB œ9 È? ?5 B œ9 9

"# (3.180)

where the subscripts denote the uncertainties at time . If we write this in terms of the> œ !momentum, we have and relation

? ?: B œh

#9 9 (3.181)

At any time other than the position distribution function is wider than at , so that> œ ! > œ !the general Heisenberg uncertainty principle becomes

? ?: B �h

#(3.182)

where the equality holds only for the minimum width of wave packets.Gaussian

____________________________________________________________________________

Problem 3.6 Radioactive materials typically emit electrons in the energy range of 1-10 MeV.

The energies of electrons which orbit the nucleus are typically in the energy range of 10's of

eV or less. In the early days of nuclear physics many scientist thought that these high-energy

electrons were somehow “trapped” in the unstable nuclei associated with radioactive decay.

Show that the Heisenberg uncertainty condition rules out the possibility that electrons can exist

within a nucleus of the size of 10 m. Thus, the electrons are somehow “produced” when a�"%

proton changes to a neutron, or vice versa.

____________________________________________________________________________

The fact that is a function which spreads out in time indicates the necessity ofTÐBß >Ñinterpreting this as a probability function. The particle itself does not get in time, butlarger

rather changes with timethe ability to predict where the particle will be at a later time

such that the uncertainty in location of the particle increases with time!

Since the spreading of the wave packet with time is not a classical phenomenon, this

spreading in the probability function must remain very small for macroscopic objects even for

long periods of time (for example the orbits of the planets are very well known). In order to

get a better understanding of the nature of the spreading we will look again at the source of the

spreading. The spreading is a result of the presence of the constant in the exponential which"arises in the expansion of ( ). If the part of ( ) which contained this term were small= <5 Bß >enough to be ignored, there would be no appreciable spreading. Thus if the term

"=

5 > œ Ð5 � 5 Ñ > ¥ "" `

#x `5w ##

#

#5º (3.183)

o

o


then there will be no appreciable spreading. One way of looking at this is to ask how long the

system can progress before the spreading becomes significant. This would mean that the time

during which the system is being observed must be such that

> ¥ > œ Ð5 � 5 Ñ" `

#x `5o o– —º (3.184)

#

#5

#

�"=

o

where is the characteristic time for spreading of the wave packet. Let's see what this means>ofor classical (non-relativistic) particles in potential-free regions of space. In this case we have

I œ Ê h œ Ê œ: h 5 h5

#7 #7 #7

# # # #

(3.185)= =

from which we can obtain the group velocity

@ œ œ œ œ @. h5 :

.5 7 71

=particle (3.186)

and

"=

œ œ" ` h

#x `5 #7º (3.187)

#

#5o

or, since ,5 œ :Îh

> œ Ð5 � 5 Ñ œ œ" ` h Ð: � : Ñ Ð: � : Ñ

#x `5 #7 h #7ho o

o o– —º ” • ” • (3.188)# # #

# #5

#

�" �" �"=

o

Thus, we have for a free particle

> œ œ#7h #7h

Ð: � : Ñ :o

o# #( )

(3.189)?

where we interpret in an sense and so use the uncertainty in as determined from> :o average

the momentum distribution. This last equation can be written in two revealing forms:

(1) in terms of the energy, since : Î#7 œ I#

> œ œh h

: Î#7 Io

( )(3.190)

? ?#

(2) in terms of the initial uncertainty in position, using the approximate uncertainty

principle /? ?: B � h #

> œ Ê > Ÿ œ œ#7h #7h #7Ð# BÑ )7 B

: ÐhÎ# BÑ h ho o

( ) (3.191)

( )

? ?

? ?# #

# #


Let's see how this works out for a couple of simple examples. If we attempt to determine the

location of a small mg object to an uncertainty of millimeter ( meters), then" " " ‚ "!�3

> œ œ (Þ&)$ ‚ "!) " ‚ "! 51 " ‚ "! 7

"Þ!&& ‚ "! N � =/-o

322 �' � #

�$%seconds

This means that no significant spreading of the wave packet would occur for this situation

within the lifetime of the universe! (One year is seconds, and the age of the$Þ"&' ‚ "!(

universe is estimated to be about 15 billion years old, so that the age of the universe in seconds

is approzimately seconds.) However, if we attempted to locate with%Þ($% ‚ "!"( an electron

a similar precision, then

> œ œ 'Þ*!) ‚ "!) *Þ"" ‚ "! 51 " ‚ "! 7

"Þ!&& ‚ "! N � =/-o

32 �$" � #

�$%� seconds

This means that there would be no significant spreading of the wave packet within about 70

milliseconds. If we try to locate an electron to within the dimensions of an atom, we obtain

> œ œ 'Þ*!) ‚ "!) *Þ"" ‚ "! 51 " ‚ "! 7

"Þ!&& ‚ "! N � =/-o

10 �$" � #

�$%�"' seconds

and significant spreading is seen to occur almost instantaneously.

____________________________________________________________________________

Problem 3.7 A beam of electons is to be fired over a distance of 10 km. If the size of the%

initial packet is 10 m, what will be its size upon arrival, if its kinetic energy is (a) 13.6 eV;�$

(b) 100 MeV? [Remember that the relationship between the kinetic energy and the momentum

is not given by for a relativistic particle.]: Î#7#

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Documents

CHAPTER III The Solution of the Free-Particle Wave Equation