infinite square well

7/27/2019 Infinite Square Well

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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 ( and momentum ( We have alsoI h : h5 2 / -argued that the , which successfully describes suchclassical electromagnetic wave equation

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 Schrdinger 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 Schrdinger's equation is based upon some reasonable associations

between classical and quantum ideas, we can take this equation to be a , much likepostulate

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.

Schrdinger's one-dimensional, equation that governs the developmenttime-dependent

of the wave function in space and time is given byGB >

B > Z B B > 3h B >h ` `

7 `B `>

# #

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

where is the rest mass of the particle which moves under the influence of a potential energy7function . We postulate that the solution to Schrdinger's equation will be associatedZ Bwith 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

3 G G B > . B B > B > .B (3.2)*where is the density of particles as a function of position and time. For example, if we3 B >had particles distributed randomly along the -axis, we would designate as theR B R B > .B number of particles located between and . The number densityB B .B B >3 .B R B> R .B

c d is just the relative number of particles at a particular location. If we add

up all the particles along the line (from to we would just obtain total particles. R 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 Bwould give us unity. Thus, when we speak of a single particle, we talk about the probability

T B > . B of finding a single particle at a particular location and time rather than the number


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Quantum Mechanics 1.2

density of particles. Thus, when we talk about a single particle, we write

T B > .B B > B > . B B > .B G G G* # (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 single-valued, 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 > T T B > .B "+8CA2/ Z B B > 3h B >h ` `

7 `B `>

# #

#2( ) ( ) ( ) ( ) (3.5)G G G

that governs the development of the wave function in space and time can always beGB >written in the form

G B/3 >= (3.6)


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where , and where satisfies the time-independent Schrdinger equation=


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where

5 I Z B#7

h c d

#(3.11)

Now in region I and III where for any energy, is obviously imaginary. ToZ B I 5

make the imaginary nature of more explicit, we will write where is , and is5 5 3 , , realdefined by the equation

, Z B I #7

h c d

#(3.12)

With this substitution, our differential equation becomes

`

`BB B

#

##< , < (3.13)

The general solution to this differential equation is simple it must be of the form

, , =In the special case of the square well, the potential energy function . Weinfinite Z B can treat this by taking the limit as so that givingZ B p p ,

GB > E/ F/ /, (3.18)lim,

, , =

B B 3 >c dFor , the first term goes to zero, but the second terms blows up unless weB ! requireF ! B !For the case where , the second term goes to zero, but the first term blows upunless we Thus, we require the wave function to be zero whereverrequire E !x Z B (i.e., everywhere in regions I and III). This means that the wave function solution go tomust

zero at the boundaries of the well (i.e., at and at ).B ! B P

Inside the well, where and we have . Again the! B P Z B ! 5 #7Ih#general solution to the differential equation must be of the form


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giving

-# B # BE/ 5 E/- - (3.20)

so that . The general solution to this equation must, therefore, be- 35

E/ F/ /, (3.22) 35B 35B 3 >=or

GB > E/ F/35 B > 35B >= = (3.23)

This equation can be expressed in terms of sines and cosines using Euler's relationships,

G = == =

B > E 5B > F 5B > 3 E 5B > F 5B >

cos cos (3.24)sin sinc d

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 Bdirection, while the second term represents a wave moving in the direction. This can beBclearly seen by observing a point on the wavefront where the phase, is equal to9 = 5B >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 and . We can apply the boundary conditions either toB ! P

the exponential form of the equation or to the sinusoidal form. Each will give the sameresults. The math is somewhat simpler using the complex exponential form of the equation,

since we must require that the real the imaginary parts of the wavefunction must goboth and

to zero at the boundary.

Thus, at the boundary where , we haveB !

GB > E/ F/ / !, (3.25) 35! 35! 3 >=which means that

E F ! E F (3.26)

This means that the general solution is of the formGB > E / / / #3E 5B /, sin (3.27) 35B 35B 3 > 3 >= =

We usually absorb the into the arbitrary constant and write this last equation as#3 E

GB > E 5B /, sin (3.28) 3 >=


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where the constant is understood to be complex in general (but may, in some circumstances,Eturn out to be real).

We also require the wave function to go to zero at the boundary where , so thatB Pthe equation above becomes

GP > E 5P / !, sin (3.29)

3 >=

which can only be true if (which is the trivial solution consistent with no particle), or ifE !

sin (3.30)5P ! 5P 8 5 8

P1

1

Recall that the constant is related to the energy of the particle. This last equation, which5gives an infinite set of discrete values of means that we will have an infinite set of discrete5solutions, each with a specific energy associated with it. Thus, the allowed energies for a

particle of mass confined to our infinite square well is found using Equations 3.11 and 3.307

I 8 8 I h 5 h 8 P h

#7 #7 #7P8 "

# # # # ### #

#

1 1(3.31)

Thus, we have a of solutions, or given byset eigenfunctions

G 3 >B > B/ E /

8 B

P88 8= =sin (3.32)

where , one for each possible value of the . (The solution=8 8 I h 8quantum numberwhere gives the null wave function, which we interpret as no solution.) Notice that for8 !this one-dimensional problem we have only one quantum number, . Typically the number of8quantum 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 .E8The 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 E8is the same for all eigenstates and is equal to #P___________________________________________________________________________

Problem 3.1 The eigenfunctions of the infinite square well are given by

G1

8 83 >B > E /

8 B

Psin =8

where are arbitrary constants that we choose to normalize the individual eigenfunctions.E8Determine 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 #Pwhere is the width of the well.P___________________________________________________________________________

Interpretation of the Eigenfunctions of the Infinite Square Well


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The general form for the eigenfunction solutions to our infinite square well problem,

then, is given by

G1

83 >B > /

# 8 B

P P sin (3.33)=8

These eigenfunctions to the wave equation are shown in blue in Figure 3.1 for the first fourvalues of the quantum number . These solutions are plotted on a line which represents the8energy of that particular state in terms of . Plotted in red are the corresponding probabilityI"densities given byT B > T B8 8

T B > B > B > B ># 8 B

P P8 8 88

##G G G

1* = sin (3.34)(although both plots are arbitrarily scaled to give the same maximum value).

0

20

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 , there are in the8 " nodesprobability density function (places where the probability density goes to zero). In fact, we

find in general there are nodes for each energy state desinated by the quantum number8 "8 8 #. This is a bit wierd! For the quantum state , the probability density is a maximum oneach 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.


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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 , meaningstationary states

that the probabilities, and, therefore, .all expectation (or average) values are time-independentThe eigenstate does not correspond to the of a particle. In fact, if we calculate themotion

expectation value (or average value) of for a particle in of the eigenstates of this systemB any(we will discuss how this is done in the following section) we find that

B 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 must also be zero.: .B.>A plot of the momentum probability functions for the first five eigenstates are shown

below where the argument on the plot is equivalent to the in these notes. Each plot has5: 5

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 statewill 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.

40 20 0 20 400

0.05

0.1

0.15pPMax

0

PP 1 kp,( )

PP 2 kp,( )

PP 3 kp,( )

PP 4 kp,( )

PP 5 kp,( )

5050 kp

In addition, each of the stationary states is a state of total energy . You candefinite I8see this by determining the expectation value of the energy of the system in one of itsLeigenstates. But this seems to lead to a contradiction. How can be zero, but not be: L


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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 , so it might be a mistake to compare with Perhaps we had: : : #7 L# #

better compare with .: #7 L #

___________________________________________________________________________

Problem 3.2 Beginning with the position eigenfunction , evaluate the quantity


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R "% " R "& " R "' $ R ## #

R #% # R #& &

where it is understood that , etc. Now, the total number of people in the room mustR$! !be given by

R R4 " " $ # # & "%4!

4

(3.36)

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 that we would get someone of a particular age? We defineprobabilitythis probability as the number of possible choices which give the desired result divided by the

total number of possible choices, or

T 4 R4

R(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 , etc. Now the probability"% #% " # ""% "% (of picking one of age 5 one of age 6 must be or ." " T "& "' T "& T "' or 2

7

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 theages 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 What is the probability that we will"& T "& ""%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 . OneT "% 9< "& 9< "' T "% T"& T "' &"%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 "& +8. "' " $

"% "$

3

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 of the numbers which areany

available: or ? It must be"% "& "' ###% #&

T +8C>2381 T 4 R 4 "R 4 " R

R R R 4! 4! 4!

4 4 4

(3.39)


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This last equation is related to the so-called normalization principle for probabilities Itbasically 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 age of the group (or average number on the bones)? 2) What isaveragethe age (or the most likely number that you will obtain choosing the bonesmost probable

randomly)? 3) What is the age of the people in the room (or median number of themedian

bones)?

We define the (or ) of a group by the equationaverage mean

4 4 4 T 4

4R4

R4

R4

R

4!

4

4!

44! 4!

4 4

(3.40)

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 in quantum mechanics. Noteexpectation value

that it is the most probable value! The value is the value that occurs mostnot most probable

frequently, i.e., the value where is a !T4 maximumA less often used quantity is the . The median is defined to be the numbermedian

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

4 4 T4# #

4!

4 (3.41)and in this case is the number 459.57. We define the average of any function by the04equation

04 04 T44!

4

(3.42)


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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 1

0

2

4

6

8

1 2 3 4 5 6 7 8 9

Age

NumberofStudents

Histogram 2

0

0.5

1

1.5

2

1 2 3 4 5 6 7 8 9

Age

NumberofStudents

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 mightrepresent 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 4 4, is small, while in the second histogram the deviations are somewhat larger. Wemight try to calculate the to see if this would help to distinguish the twoaverage deviation

histograms. The average deviation is given by

4 4 T 4 4 4 T 4 4T 4 4 T 4 !? ? 4

4 4 4 4

4! 4! 4!=0

=

(3.43)

In fact, the average value is to be that number which gives zero average deviation.defined

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 of thesquare

deviation, which must be positive, and average that quantity. That is, we want to consider:

5 ?# # 4 (3.44)

which we call the of the distribution. The quantity is called thevariance standard5deviation. A useful theorem concerning the standard deviation is the following:

5# # # 4 4 (3.45)

___________________________________________________________________________

Problem 3.3 Begin with the basic definition of the standard deviation


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5## 4 4

and show that

5# # # 4 4

___________________________________________________________________________

The way we defined the variance and the standard deviation) means that the square of thestandard deviation is alway greater than or equal to zero. This means that , or4 4 !# #

4 4# #: the average of the square of a quantity is always greater than or equal to thesquare 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

4 # $"!

1

#

##

The standard deviation for each case is given by

5

5

"

#

# #& !%%(

$"! #& #%%*

Notice that in the case of the second histogram, if we take the interval j , which is # #5approximately , 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 of the sample lies in a region given by j , while about2 13 3

5

lies outside this region, and that about 95% of the sample lies within 2 of the average value.5Thus, the standard deviation is useful measure of the of the distribution function.width

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 and by the equationB + B ,

T > T B > .B (+,+

,

(3.46)

The quantity is call the the y of finding aT B > probability density differential probabilitparticle at random between the points and is given by and is dependentB B .B T B > .Bupon the size of . As indicated earlier, the probability density in quantum mechanics is.Brelated to the solution to Schrdinger's equation according to the equation

T B > .B l B >l .B G # (3.47)


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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 > T T B > .B "+8CA2/ 0BT B > . B(

(3.49)

or, in particular, we define the of a particle (the ensemble average of theexpectation value

position of a particle) by the equation

B > B T B > .B (

(3.50)

or, in terms of the solution to Schrdinger's equation,

B> B B > B > .B

B> B > B B > .B((

G G

G G

*

*

(3.51)

where we have explicitly indicated that the average or expectation value) may in general be afunction of time, and that the time dependence arises from the time dependence of the solution

to Schrdinger's equation. It is conventional to write the expectation value of in the form ofBthe second equation above, with the variable sandwiched between the wave function and itsBcomplex 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 ?# # # # B B B (3.52)

is identical to the expression we had before, for a discrete distribution function. This

particular quantity gives us an indication of . Thus,the uncertainty in the location of a particle

to determine this uncertainty in the position of a particle, we not only need the expectation


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value of the position, but also the expectation value of the square of the postion, or

B > B > B B > .B# #

( G G* (3.53)___________________________________________________________________________

Problem 3.4 Consider the time-independent Gaussian distribution function

T B > E/ B+- #

where , , and are constants.E + -(a) Determine the value of which will normalize this distribution function.E(b) Find , , and for this distribution function.B B # 5(c) Plot this distribution function and indicate the location of the average value and the

points .B 5___________________________________________________________________________

Problem 3.5 Consider the wave function

GB > E/ / l Bl 3 >- =

where A, , and are positive, real constants.- =

(a) Normalize this wave function.

(b) Determine the expectation values of and .B B#

(c) Graph the probability density | | as a function of , and indicate on this graph theGB > B#

location of the points and . What is the probability of finding the particle 5 5described by this wave function outside the range ?B 5

___________________________________________________________________________

Problem 3.6 Beginning with the eigenfunctions of the infinite square well

G1

83 >B > /

# 8 BP P sin =8

show that the expectation value for any of the eigenstates is given by

B P

#

___________________________________________________________________________

The Normalization of the Wave Function

The statistical interpretation of the wave function relates the differential probability

T B > .B B >to the wave function according to Equation 3.47. And the requirement thatGthe particle must be somewhere requires that the probability density integrated over all space

must be unity:

T B > .B "

#(

-

+

G( (3.54)


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But an addition requirement is also placed upon the wavefunction it must also be the

solution to Schrodinger's equation. For Schrdinger'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 , for example). If our interpretation of the waveBp function is correct, these solutions can not correspond to real systems and, therefore, must

be rejected (e.g., notice that the trivial solution is not normalizable, nor is theGB > !solution , so these solutions cannot be associated with a particle). TheGB > -98=>+8>only valid solutions to Schrodinger's equation, then, are functions which are square

integrable, i.e., functions where

( -

+

#

G( (3.55)B > .B Q Q

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 functionfor a

single particle.

The requirement the the wavefunction be normalizable, i.e., that

( -

+

#

G( (3.56)B > .B "

means that the wavefunction itself, must tend toward zero in the limit as . Let's look aB p little closer at what this means. Let's assume, for example, that has the formGB >

GB > E B8. Our normalization condition then requires that

(-

+

# #8E B .B " (3.57)

or that

E "B

#8 "#

#8"

(3.58)For this last equation to be valid, the exponential of the variable must be less than unity, orB

the function will blow up as . Thus, we require thatB p #8 " ! (3.59)

or

8 "# (3.60)


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This means that in general, the wavefunction must go to zero at least a fast as . B > " BG This also implies that the first derivative / of the wavefunction must go to zero at` B > `BG

least as fast as !" B $The General Time-Dependent Solution to Schrodinger's Equation

for the Infinite Square WellSince 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

G - / B8

8 83 >=8 (3.61)

where the 's are arbitrary coefficients. Sometimes the time dependence of the wavefunction-8is incorporated into these coefficients, giving

G - > B8

8 8 (3.62)

where .- > - /8 83 >=8

If we know the form of the solution at time , expressed as , then we> ! B !Gshould 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

(!P

7 8 78< < $

*

B B .B (3.63)

where

$78 ! 7 8" 7 8 (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

# 7 B 8 BP P P

.B ( !P

78sin sin1 1

$

___________________________________________________________________________

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


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the complex conjugate of an arbitrary eigenfunction and integrate over the region accessible to

the particle, we obtain

( ( < G <

78

B B > .B - / B B .B=8 (3.65)

Because of the Kronecker delta, each term in the summation gives zero except the one where

7 8 > !, giving, for time

(< G*7 7 73 !B B ! .B - / -=7 (3.66)Once we know the value for each , obtained from the initial wavefunction at time for a- > !8particle in the square well, we can determine the general solution for this particle at all6+>/=8

(3.67)

We should point out that although we have to normalize each of the individualchosen

eigenfunctions, the normalization condition is a of our general solution if it is torequirement

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 G

< .B "

- / B - / B .B "

- - / B B .B "

- - / "

l- l "

7 87 73 > 3 >8 8

787 78 8

3 >

787 8 78

3 >

77

#

= =

= =

= =

7 8

7 8

7 8

(3.68)

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


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equation

L I l- l7

7 7# (3.69)

where we take to be the ensemble average energy, is the energy for the eigenstateL I 7

designated by the quantum number and is the probability of finding the particle in that7 l- l7 #particular eigenstate.

___________________________________________________________________________

Problem 3.8 Beginning with the definition of the expectation value for the energy of an

arbitrary wave function

L B > L B > .B(G Gsexpand these wave functions in terms of a sum over eigenfunctions and show that the

expectation value for the energy can be written as

L l- l I 7 7# 7

___________________________________________________________________________

Expectation Values for the General Time-Dependent SolutionLet's determine the expectation value of for a general time-dependent solution for theB

infinite square well. We begin with our basic defining relation

B B > B B > .B(!

P

G G* (3.70)

The general wave function can be written as an infinite sum of eigenfunctions in the form

G - / B8

8 83 >=8 (3.71)

giving

B - / B B - / B .B( !

P

7 87 7

3 > 3 >8 8

* *= =7 8< < (3.72)

Factoring out everything that is not a function of the position, we obtain

B - - / B B B .B (7 8

7 78 83 >

!

P* * = =8 7 < < (3.73)

In order to evaluate the expectation value, we must evaluate the integrals

7lBl8 B B B .B(!

P

7 8< #

7 8 7 78 8

!

P #

#

( * * = =8 7 < < (3.77)

which means we will need to evaluate the intergrals

7l:l8 3h B B .B 3h7l l8` `

`B `B(

!

P

7 8< B > /# 8 B

P P sin=8

I 8 8 I h

#7P8 "

# ## #

# 1

Orthonormalization Conditions of these Wavefunctions

(!

P

7 8 78< < $* B B .B

Useful Matrix Elements of the Infinite Square Well

7lBl8 P

# 7 8! 7 8 7 8

7 8 7 8)78

7 8

7lB l8 P

" "

$ #7 7 8

")78

7 8

c d

c d

1 if

if even

if odd

if

1

1

1

# # # #

# ## #

78

# # # #

# #

#

# #

# #

if

if

if even4

if odd

if

if

7 8

7l l8 ` "

`B P

! 7 8! 7 8 7 8

78

7 8

7 8 7 8

7l l8 ` "

`B P 8 7 8

! 7 8

1


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As an example of how all this works, let's consider the special case where the general

form of the wave function at time is given by> !

G < expressed by the equation

G < B/ B/"

# " #3 > 3 >= =" # (3.87)and that the probability density for finding a particle at some position will be time-Bdependent, giving a time-dependence to the expectation value of the position , theBmomentum , and perhaps the energy .: L

To find the expectation value of we writeB


23/26


B B > B B > .B

B/ B/ B B/ B/ .B" "

# #

"lBl" "lBl#/ #lBl"/"#

((

!

P

!

P

" #3 > 3 > 3 > 3 >

" #

3 >

G G

< < < 3 >

" #

= = = =" # " #= =

(3.89)

Plugging in the specific numberical values for the matrix elements, we obtain

B " -9= >P $#

# * c d

1= =

# " #(3.90)

Notice that so the particle doesn't move outside the well as it oscillates back$#* !$'1#

and forth. At time the particle has an expectation value of . The> ! B !$# Pexpectation value then changes to become after going through one-quarter period andP#becomes after half a period, then moves back toward the center of the well. TheB !') Poverall diplacement (on average) from the center of the well is 0.18 . In addition, we knowPthe values of and , since we know the eigenenegies associated with the eigenfunctions.= =" #

The angular frequency is given by

= =8 "8 "

## 8

I 8 I

h h(3.91)

This means we can determine the frequency of oscillation of the particle in this particular

state. We find that

/ = = 1 = 1= =

1

11

% # $ #

#

$I # h $h $2

%7P )7P

" #" " "

" # #

(3.92)

The minus sign here is insignificant since ( .-9= -9= ) )The expectation value of the momentum 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

: =38$ >.B ) h

.> $ P=" (3.93)

So we see that the expectation value of the momentum also varies with time.


24/26


The expectation value of the energy of the system can be determined in a very straight

forward manner, using the equation

L l- l I 7

7#

8 (3.94)

For our particular problem, where the general state of the system is composed of equal parts ofthe first two eigenstates of the system, we have that ,- - " # l- l l- l "#" # " #

# #giving

L I I I %I I " " " & & h

# # # # # #7P" # " " "

# #

#c d 1 (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 obtain an energy value equal to thenotexpectation value. We will, rather, obtain either the value or the value . For theI I" #particular problem we are considering, the probability of measuring the value will be equalI"to the probability of measuring the value because the wavefunction is made up of equalI#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 the wave function must have collapsed toI"the wavefunction because if we measure the energy a moment later we would


25/26


For a position measurement, the eigenvalue equation is given by

Bs<


26/26


where and . Notice that this is equivalent to or5 #7I h I h h 5 #7I # # # #8 88 8 8 8=: #7 I 8

#8 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

: h5 #7I #78 h #7P 82#P 28 8 8 8# # # #

1 - (3.105)

This last equation shows that the possible wavelengths allowed for our infinite square well

system are given by

-8 #P

8(3.106)

infinite square well

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