quantum harmonic oscillator.pdf

7/27/2019 quantum harmonic oscillator.pdf

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F L E X I B L E L E A R N I N G A P P R O A C H T O P H Y S I C S

FLAP P11.2 The quantum harmonic oscillator

COPYRIGHT 1998 THE OPEN UNIVERSITY S570 V1.1

Module P11.2 The quantum harmonic oscillator

1 Opening items1.1 Module introduction

1.2 Fast track questions

1.3 Ready to study?

2 The harmonic oscillator2.1 Classical description of the problem

2.2 The Schrdinger equation for a simple harmonic

oscillator

2.3 The energy eigenfunctions2.4 The energy eigenvalues

2.5 Probability densities and comparison with classical

predictions

5 Closing items5.1 Module summary

5.2 Achievements

5.3 Exit test


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1 Opening items

1.1 Module introduction

A study of the simple harmonic oscillator is important in classical mechanics and in quantuThe reason is that any particle that is in a position of stable equilibrium will execute simple ha(SHM) if it is displaced by a small amount. A simple example is a mass on the end of a springgravity. The system is stable because the combination of the tension in the spring and the gravitatalways tend to return the mass to its equilibrium position if the mass is displaced. Another examphydrogen in a molecule of hydrogen chloride HCl. The mean separation between the hydrogen aatoms corresponds to a position of stable equilibrium. The electrical forces between the atoms w

to return the atom to its equilibrium position provided the displacements are not too large. Sucmotion about a position of stable equilibrium can be found in all branches of mechanics, molecular and nuclear physics.

The key to understanding both the classical and quantum versions of harmonic motion is the beparticle potential energy as a function of position. The potential energy function of a particle simple harmonic motion has a parabolic graph (see Figure 2), and it may be shown that sufficieposition of stable equilibrium almost all systems have a parabolic potential energy graph andSHM. For oscillations of large amplitude, the potential energy often deviates from the parabolic fmotion is not pure SHM.


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In this module, we will review the main features of the harmonic oscillator in the realm of cla

scale physics, and then go on to study the harmonic oscillator in the quantum or microscopic wsolve the time-independent Schrdinger equation for a particle with the harmonic oscillator potenhence determine the allowed energy levels of the quantum oscillator, the corresponding spatial and the probability density distributions. Comparisons will be made between the predictions oquantum theories, bearing in mind their very different regions of applicability.

Study comment Having read the introduction you may feel that you are already familiar with the materiamodule and that you do not need to study it. If so, try theFast track questionsgiven in Subsection 1.2directly toReady to study?in Subsection 1.3.


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1.2 Fast track questions

Study comment Can you answer the following Fast track questions?. If you answer the questions succeonly glance through the module before looking at the Module summary (Subsection 3.1) and theAchievSubsection 3.2. If you are sure that you can meet each of these achievements, try theExit testin Subsectiondifficulty with only one or two of the questions you should follow the guidance given in the answers and

parts of the module. However, if you have difficulty with more than two of the Exit questions you are strstudy the whole module.

Question F1

What is meant by the term simple harmonic oscillation in classical mechanics? Suggest a criteriwhether classical mechanics or quantum mechanics should be used in a problem involving harmo

Question F2

Write down an expression for the allowed energies of the harmonic oscillator in quantum mechanthe quantum number n, Plancks constant and the frequency of the corresponding classical oscillaSketch the energy eigenfunctions (i.e. spatial wavefunctions) of the n = 0 and n = 1 states.


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Question F3Write down the time-independent Schrdinger equation for a particle of mass m in a one-dimens

oscillator potential centred atx = 0. Show that the spatial wavefunction

(x) = Aexp( 12 x2 )

is an energy eigenfunction with total energy eigenvalue E= 12 , where is the angular frcorresponding classical oscillator. Can the quantum oscillator have a lower energy?

Study comment

Having seen the Fast track questions you may feel that it would be wiser to follow the normal route throughto proceed directly toReady to study? in Subsection 1.3.

Alternatively, you may still be sufficiently comfortable with the material covered by the module to procee

Closing items.


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1.3 Ready to study?Study comment In order to study this module, you will need to be familiar with the following physics teprinciples:simple harmonic motioninclassical mechanics; the nature of a conservative force and the impotential energy; the potential energy function for a particle executing SHM (U(x) = ks 0x

2/02); the time-independent Schrdinger equationfor a particle moving in one dimension in a region of constant poteallowedwavefunctionsoreigenfunctions and the corresponding allowed energies or eigenvaluesof energystationary state of definite energy in a one-dimensional box; photons, thePlanckEinsteinde Broglie wavelength, theHeisenberg uncertainty principle; theBorn probability interpretation (of thIf you are uncertain of any of these terms, you can review them now by referring to the Glossary which wi

in FLAP they are developed. You should be familiar with the mathematics of elementary trigonometric fexponentialfunction; the use ofcomplex numbers(includingcomplex conjugates and the role ofarbitrarysolution of second-order differential equations). We will frequently use both integraland differential caelementary functions. The followingReady to study questions will allow you to establish whether you needof the topics before embarking on this module.
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Question R1

A particle of mass m is confinedin an infinitely deep potential well such that V= 0 for a/2V for |1x1| > a/2. Write down the time-independent Schrdinger equation for the particle iConfirm that within the well 0(x) =A1cos(x0/a) is a solution to the Schrdinger equation when t

E= 22/(2ma2). Show that this solution satisfies the boundary condition 0(x) = 0 atx = a/2.

Question R2

Write down the time-independent Schrdinger equation for a particle of mass m moving in thex-the potential energy has a constant value Vthat is greater than the total energyE. Show by substit

01(x) =A1exp(00x) +B1exp(00x)

is a solution withA andB arbitrary constants and real. Find an expression for in terms ofEan

Question R3

If(x) = Aexp( 12 x2 ) , show that d02(x)/dx2 = (x202)(x).


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2 The harmonic oscillator

2.1 Classical description of the problem; classical predictions

We consider a particle of mass m constrained to move in thex-direction. It is subject to a force Fin thex-direction, proportional to the distance from the origin and directed towards the origin:

Fx = ksx (1)

The constant ks is called the force constant, and it plays an important role in our treatment of harA good example of this kind of force is the restoring force on a particle attached to a spring w

expand or contract. Newtons second law is now applied, and we immediately obtain a differrelating the positionx and the time t:

mx = ksx

so x = ksm

x (2)
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Notice that the negative sign in Equation 2

x = ksm

x (Eqn 2)

says that the acceleration is in the negativex-direction whenx is positive and is in the positivex-x is negative. Equation 2 is often regarded as the definition of classical harmonic oscillation:

A particle executessimple harmonic motionabout a fixed point O if the acceleration is propodisplacement from O and directed towards O.

The solutions of Equation 2 have been obtained elsewhere in FLAP:

x =A1cos1(1t) +B 1sin1(1t)4with4= ks m

Question T1

Confirm by direct substitution that x =A 1cos(1t) + B1sin(1t) with = ks m is the geneEquation 2.4t
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x

a

T

Figure 14A g

oscillationx =

showing how t

and the period

The arbitrary constants A and B may be determined from the velocity anddisplacement of the particle at t= 0. For example, we can takex = a and x = 0 att= 0; then:

x = a1cos1(1t) (3)

The amplitude of this simple harmonic oscillation is a, and it is illustrated in

Figure 1. The period T determines the frequency f = 1/T and the angularfrequency = 2/T. Consequently, in this particular case:

=ks

m

4 T = 2m

ks

4and4 f =1

2

ks

m

(4)


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For comparisons with the quantum-mechanical treatment, we need to relate the total energy of tits amplitude and also find an expression for the particle velocity in terms of the displacement. Th

by Equation 1

Fx = ksx (Eqn 1)

is aconservativeforcesince it can be derived from apotential energy functionU(x):

Fx = dU(x)dx

U(x) = Fx dx = (ksx)dxTherefore U(x) = 12 ksx2 + C

It is usual to put the arbitrary constant C= 0, and the potential energy function then becomes:

U(x) =12 ksx

2

(5)

The potential energy function for a one-dimensional simple harmonic oscillator.
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U(x)

a

A graph of the potential energy function has a parabolic form and is shownin Figure 2.

Since the only force acting on the particle is the restoring force, given byEquation 1,

Fx = ksx (Eqn 1)

the sum of the kinetic and potential energies is constant. We call thisconstant the total energyE:

E= 12 mx2 +12 ksx

2 (6)

However when x = 0 , thenx = a, and we have:E= 12 ksa2

so, a =2E

ks

(7)

Figure 24The potential energy function for a simple harmonic oscillator. A possible energy Eis represente

line, and the corresponding amplitude a is indicated. The potential energy function has a parabolic form.


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U(x)

a

The amplitude of the oscillation increases as the square root of the totalenergy. The relation between the total energy Eand the amplitude a is also

illustrated in Figure 2.

We can rearrange Equation 6

E= 12 mx2 + 12 ksx2 (Eqn 6)

to obtain the desired expression for the particle velocity:

x2 =2E

m

ks

m

x

2

Substituting Equation 7:

a =2E

ks(Eqn 7)

x2 =k

sm (a

2 x2 )




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It is convenient to replace the ratio ks/m by 02 (Equation 4),

=ks

m 4 T = 2m

ks4and4 f =

1

2ks

m (Eqn 4)

so that:

x = a2 x2 (8)

The speed | x | is maximum atx = 0 and is zero at the extremesx = 0a.

Now, imagine making observations on the position of the particle as it oscillates, and assume thrandom times. Obviously, you are more likely to find the particle in regions where it is moviconversely less likely to find it where it is moving quickly. We can quantify this argument as follo

Let the probability of finding the particle in a narrow region of length x at position x be P(x)1the time required for the particle to cross x . Since the particle crosses x twice during oscillation, we have:

P(x)1x = 2t/T

where Tis the period. In the limit as x and ttend to zero, x/ttends to x and

P(x) =2

| x |T
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a

0.6

0.4

0.2

0

P(x)

Figure 34

The classicdensity for a particle exharmonic motion. The oscillation is a. The coclassical probability of

particle in any finite re

x =x01 andx =x02 is giv

Using Equations 4 and 8,

=ks

m 4 T = 2m

ks4and4 f =

1

2ks

m (Eqn 4)

x = a2 x2 (Eqn 8)

we find:

P(x) =

1

a2 x2

Hence P(x)

=

1

a2 x2(9)

The function P(x) is the classical probability density, and we willcompare it with the corresponding quantum probability density in duecourse. Figure 3 shows the graph ofP(x). You can see the probability

density increasing as the displacement increases and the speed decreases;eventually, the probability density goes asymptotically to infinity whenx 0a. However, the probability of finding the particle in any finite

region remains finite.
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2.2 The Schrdinger equation for a simple harmonic oscillator

Before you embark on any mechanics problem, it is important to decide whether to useapproximation or quantum mechanics. One test is to compare a typical de Broglie wavelimportant linear dimension in the problem. If the de Broglie wavelength is negligibly small,mechanics may safely be used. In the harmonic oscillator problem, we can compare the de Brogwith the amplitude of the oscillation. This leads to the following conditions:

IfEis the total energy, h is Plancks constant andfis the classical oscillator frequency:

Use classical mechanics ifE>> hf. Otherwise use quantum mechanics!

In the quantum-mechanical description of particle motion, the concept of a particle trajectory is cWe cannot know the particle position and momentum simultaneously, and this fundamentaformalized in theHeisenberg uncertainty relation x1px. In the case of the quantum sim

motion, you must stop visualizing a particle oscillating about a mean position and concwavefunction! The wavefunction corresponding to a particular state tells you all that can be knbehaviour of the particle in that particular state.
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Study comment The one-dimensional wavefunction 1(x, t) is time-dependent and satisfies the tSchrdinger equation. For a stationary stateof definite energyEthe wavefunction takes the form

(x, t) = (x)exp iE

t

Since this module is entirely concerned with such states we will concentrate on determining the spatial wa

which satisfy thetime-independent Schrdinger equation, and the corresponding values ofE. Because of thmay conveniently refer to 1(x) as the wavefunction since 1(x, t) follows immediately from 1(x) andE.4t

The time-independent Schrdinger equation for particle motion in one dimension is:

2

2md

2

(x)dx2

+U(x)(x) = E(x) (10)

Here, U(x) is the potential energy function, and we have to solve the equation wiboundary conditionsto obtain the allowed values of the total energyEand the corresponding

Solutions of the time-independentSchrdinger equation for a particle trapped in a one-dimdiscussed elsewhere in FLAP, show that confinement leads toquantized energy levelslabellequantum number n and that each energy level has a corresponding wavefunction n(x). Much from this example and the lessons applied to the harmonic oscillator.
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First we substitute Equation 5 into Equation 10

U(x) = 12 ksx2 (Eqn 5)2

2m

d2(x)

dx2+U(x)(x) = E(x) (Eqn 10)

to produce the Schrdinger equation for the quantum harmonic oscillator:2

2m

d2(x)

dx2+ 12 ksx2 (x) = E(x)

and after rearrangement:

2

2m

d2(x)

dx2= 12 ksx2 E( )(x) (11)

The time-independentSchrdinger equation for a quantum harmonic oscillator.


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2.3 The energy eigenfunctions

There are very many solutions of Equation 11,

2

2m

d2(x)

dx2= 12 ksx2 E( )(x) (Eqn 11)

and we have to select those which satisfy the appropriate boundary conditions. In the cadimensional box, or infinite square well, the allowed wavefunctions are constrained to zero at ththe potential energy goes to infinity. However, the harmonic oscillator potential energy functrigid boundary but it does go to infinity at infinite distance from the origin. Our boundary condi

allowed wavefunctions approach zero as x approaches + or . You may think that this condarrange with any value of the total energyEsince the solution to any second-order differential eqtwo arbitrary constants. This is not so! One of the constants fixes the overallnormalization of theand the remaining constant and the value ofEare used to satisfy the two boundary conditions. out that there are an infinite number ofdiscrete values ofEwhich we labelE1, E2, E3, , Enthese there is a corresponding allowed wavefunction 1, 2, 3, , n. IfEis varied, even infinany one of the allowed values, then the wavefunction will diverge to infinity as x approachThe allowed values ofE are called eigenvalues of total energy and the corresponding waveigenfunctionsof total energy.
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U(x)

a

Physical intuition can give us some idea of the form of the allowedwavefunctions (eigenfunctions). For any given energy E, there will be a

region of space (see Figure 2) where the potential energy is less than thetotal energy; this is the so-called classically allowed region. Here, weexpect the wavefunction to have properties similarto the standing wavesinside a one-dimensional box. In particular, we expect the number of pointsat which

n

0(x) = 0, the number ofnodes of the wavefunction, to increasewith increasing energy. In the region oflargex where the particle energy ismuch less than the potential energy, the classically forbidden region, wemight expect a solution of the Schrdinger equation of an exponentialform. It would be incorrect to anticipate n 0(x) exp(0x) here since this

would not fall asymptotically to zero for negativex. We require instead asymmetricfunction ofx such as n 0(x) exp(0x0

2), which tends to zero asx orx, as required physically.


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In fact, each wavefunction will include a term of the form

(x) = Aexp( 12 x2 ) (12)whereA is a constant and = 0m/ .

Study comment

You can omit the following question at first reading if you wish.

3 Show by substitution that Equation 12

(x) = Aexp( 12 x2 ) (Eqn 12)

is a solution of Equation 11

2

2m

d2(x)

dx2= 12 ksx2 E( )(x) (Eqn 11)

whenx is large.
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Solutions to Equation 11

2

2md

2

(x)dx2 =

12 ksx2 E( )(x) (Eqn 11)

for all values ofx can now be found by multiplying Equation 12

(x) = Aexp( 12 x2 ) (Eqn 12)

by suitable polynomial functionsfn(x) of degree n. This results in wavefunctions similar to standiclassically allowed region joining smoothly to the falling exponential shape in the classically forb

n (x) = An fn (x)exp(1

2 x2

) , where n = 0, 1, 2, 3 (13)

These are the energy eigenfunctions, and to each one there is a corresponding energy eigenvalue(seeSubsection 2.4).


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0(x)

543211 02345

1.5

1.5n = 0

1(x)

211 0345

2

2

2

(b)(a)

2(x)

543211 02345

3

3n = 2

(c)

x

3(x)

211 02345

10

10

(d)

x

Figure 44The first four energy eigenfunctions for the quantum simple harmon

normalized). The limits of motion for a classical oscillator with the same energ

by the vertical dashed lines. The horizontal axis is marked in the dimensionless

The first four of therelevant polynomial

functions fn(x) are listedbe low, and t heeigenfunct ions areillustrated in Figure 4:

f0 = 1f1 = 2 x (14)

f2 = 2 40x 02

f3 = 12 x 8 x3

The exponentialcombined with thepolynomials produces

functions that have nnodes in the classicallyallowed regions.


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0(x)

543211 02345

1.5

1.5n = 0

1(x)

211 0345

2

2

2

(b)(a)

2(x)

543211 02345

3

3n = 2

(c)

x

3(x)

211 02345

10

10

(d)

x




As the number of nodesincreases, so does the

corresponding energyas expected.

The boundaries of theclassically allowed

regions are marked inFigure 4, so you canclearly see the transitionfrom the standing waveforms to the falling

exponential.

The falling exponentialw i t h a r g u m e n tproportional to x02

ensures that thewavefunct ions gosmoothly to zero asx0.


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0(x)

543211 02345

1.5

1.5n = 0

1(x)

211 0345

2

2

2

(b)(a)

2(x)

543211 02345

3

3n = 2

(c)

x

3(x)

211 02345

10

10

(d)

x




Notice that eachallowed wavefunction

h a s a d e fi n i t esymmetry, it is either anodd or aneven functionofx. A wavefunction isodd or even dependingon whether or not 0(x)changes sign under thetransformationxx:

If0(x) = +0(x), then

0(x) is even.If0(x) = 0(x), then0(x) is odd.

Inspection of the first

four eigenfunctions(Figure 4) shows thatn(x) is even or oddwhen n is even or odd.
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This property of the wavefunction is extremely important and follows from the nature of the poU(x)

=1

2kx2 . Clearly, U(x) = U(x), i.e. the potential is symmetric aboutx = 0. This means th

observable must also be symmetric aboutx = 0, including the stationary state probability dePn (x) = n* (x)n (x) = |n (x) |2 . We must have Pn(x) = Pn(x), and therefore:

|n (x) |2 = |n (x) |2

In this case the wavefunctions are real, so the condition becomes

n2 (x) = n2 (x)

and taking the square root we obtain

n(x) = n(x)

The eigenfunctions are therefore necessarily either odd or even when the potential function is sythe origin.

Question T2Within a one-dimensional box between x = a/2 and x = a/2 the eigenfunctions of a confinn(x) =A1cos(nx0/a) for n = 2, 4, 6 and n(x) =A1sin(nx0/a) for n = 1, 3, 5 . Confirm thaand odd functions, as required by symmetry.4t


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2.4 The energy eigenvalues

The allowed total energies or eigenvalues of total energy corresponding to the first few eigenfunEquations 13 and 14

n (x) = An fn (x)exp( 12 x2 ) , where n = 0, 1, 2, 3 (Eqn 13)

f0 = 1

f1 = 2 x (Eqn 14)

f2 = 2 40x 02

f3 = 12 x 8 x3

may be found by direct substitution into the Schrdinger equation. We will do the first one, and t

the second!


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Starting with 0 (x) = A0 exp( 12 x2 ) we get by successive differentiation:

= (02x2)

Substitute this into Equation 11:

2

2m

d2(x)

dx2 =12 ksx

2

E( )(x) (Eqn 11)

2

2m(2x2 )= ( 12 ksx2 E)

Collecting terms: x2 222m

12 ks + E 2

2m

= 0

This is not an equation to be solved for a particular value ofx11rather it is an identity that is truofx. It follows that the coefficient ofx2 and of the constant term must each be equal to zero. From

22

2m=ks

24i.e.42 =

ksm

2
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Using = ks m , we confirm that =m

x222

2m 12 ks

+ E

2

2m

= 0

Now, from the constant term E

=

2

2m

and substituting = 0m/ gives: E= 12

or in terms of the frequencyf= 0/(2) E= 12 hf

We have confirmed that 0 (x) = A0 exp( 12 x2 ) is an energy eigenfunction of the sycorresponding eigenvalue is E0 = 12 hf . This is in fact the lowest possible value of the energy oharmonic oscillator. There is a zero point energy of the harmonic oscillator (0just as there iconfined in a one-dimensional box). In Question T5, you can show that this is a consequence of

uncertainty principle px1x.


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In exactly the same way, it can be shown that the eigenfunctions 1(x), 2(x) and 3(x) have eig5

2

hf and 72

hf, respectively. This suggests a general rule which, although true, we will not attemp

The allowed energy eigenvalues of the quantum harmonic oscillator are:

En = n + 12( )hf with n = 0, 1, 2, 3 (15)

where thequantum numbern characterizes the allowed energies and wavefunctions.

Question T3

Show that 1(x) = A12 x exp( 12 x2 ) is an eigenfunction of the quantum harmonic oscillatcorresponding total energy eigenvalue is E1 = 32 hf . (A1 is an arbitrary constant.)4t
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U(x)

0

Figure 54The quantum

oscillator energy levelson the potential energy

The energy eigenvalues given by Equation 15

En

= n + 12( )hf with n = 0, 1, 2, 3 (Eqn 15)

are often referred to as the harmonic oscillator energy levels. The levelsare spaced equally by an amount hf. They are shown schematically inFigure 5 superimposed on the potential energy function.


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0(x)

543211 02345

1.5

1.5n = 0

1(x)

211 0345

2

2

2

(b)(a)

2(x)

543211 02345

3

3n = 2

(c)

x

3(x)

211 02345

10

10

(d)

x




You should now referback to Figure 4 and

a s s o c i a t e e a c heigenfunction shapewith the correspondingenergy eigenvalue.Notice, in particular, the

following:

o The energy levelsare equally spaced.

o The number ofn o d e s i n t h eeigenfunctions, givenby n, increases withenergy.

o The eigenfunctionsspread out in space asthe energy increases.


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For each of the energy eigenvaluesEn, we can work out, from Equation 7,

E= 12 ksa24or4 a = 2Eks

(Eqn 7)

the region of space in which a classical simple harmonic oscillator with that energy would be confiLet the region for energyEn be bounded byx = an then;

since En = 12 ksan2 an =2En

ks

but En = (n + 12 )hf = (n +12 )

However = 0m/ = ks/(), so that4En = (n + 12 )ks 4and4an =2n + 1

making the equation dimensionless:

an = 2n + 1 (16)

( ) ( )


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0(x)

543211 02345

1.5

1.5n = 0

1(x)

211 0345

2

2

2

(b)(a)

2(x)

543211 02345

3

3n = 2

(c)

x

3(x)

211 02345

10

10

(d)

x




These boundaries aremarked in Figure 4

which showed theeigenfunctions.

Notice that for thequantum oscillator the

amplitude ceases tohave direct physicalmeaning, since there isno definite limit to theregion of space in which

the particle may befound. Also theeigenfunct ions arestationary states and

imply no particleoscillation, as in theclassical model.


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If we associate energy emission with a transition between two stationary states then we need time dependence of the full wavefunction. This complication is beyond the scope of this developed in the FLAP module dealing with theone-dimensionalbox.

Question T4

In classical theory, a charged particle executing SHM of frequency femits electromagnetic ra

frequencyf. Write down an expression for the allowed energies of the equivalent quantum oscillat

What is the energy of a photon emitted when the quantum oscillator jumps from level n1 to levShow that it is only when n2 = n1 1 that the frequency of the radiation associated with such phothe classical frequencyf.4t
http://glossary.pdf/http://glossary.pdf/http://glossary.pdf/http://glossary.pdf/http://glossary.pdf/http://glossary.pdf/http://glossary.pdf/http://glossary.pdf/http://glossary.pdf/


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2.5 Probability densities and comparison with classical predictions

We must be particularly careful when comparing the predictions of quantum mechanics and classThe two theories were developed for very different systems. Classical mechanics is the approgeneral, for the large-scale world of objects we sense directly, and it was developed on the basis oobservations on such objects. Quantum mechanics applies, in general, to the world of atoms and

the relevant objects11electrons, nucleons etc.11cannot be sensed directly. It is hardly surppredictions of quantum mechanics seem strange and often defy common sense when compredictions of classical mechanics which accord with everyday experience.

Although it is beyond the scope ofFLAP, one can show that the laws of quantum mechanics do

laws of classical mechanics in the limit of large distances or high energies or large quanWe can therefore regard quantum mechanics as the more fundamental theory and classical mapproximation that becomes more exact as we move from the microscopic to the macroscopic wothat classical physics can be obtained from some limiting case of quantum phyics is correspondence principle.


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The most important result we derived from the quantum mechanics of the harmonic oscillator waof quantized energy levels. If the potential energy function of a particle of mass m moving alo

U(x) = 12 ksx2 , then the total energy can be one of the discrete set En = (n +12 )hf with f =

1

2

quantum number n = 0, 1, 2, 3 . No other value ofE is possible. The difference between a

levels E=En En111 = hfand in the limit of large quantum numbers this becomes negligible

En:

EEn

=hf

n + 12( )hf 0 as n

For very large quantum numbers the energy levels become so close to each other that the essentiis not noticed. This corresponds to classical mechanics, which predicts a continuum of values of tIt is also in agreement with our condition for the applicability of classical mechanics E>> hf, andof the correspondence principle.

Another important prediction of quantum mechanics is the so-called zero point energy. Whenumber n = 0, the total energy is E0 = 12 hf , and this is the smallest possible value of the totalharmonic oscillator. The zero point energy is a purely quantum effect and it has no parallmechanics, which allows energies arbitrarily close to zero.
http://glossary.pdf/http://glossary.pdf/http://glossary.pdf/http://glossary.pdf/http://glossary.pdf/


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Quantum mechanics predicts:

discrete energy levels En = (n + 12 )hf;

the zero point energy E0 = 12 hf .

Classical mechanics predicts:a continuum of energies from zero upwards.

Quantum mechanics classical mechanics as n.

The essential lumpy or grainy characteristic of quantum mechanics and the smoothness of classicalso evident in the probability density distributions P(x). We worked out a probability density fharmonic oscillator under the assumption that observations on the particle position were made aduring the oscillation. This was found to be:

Pcl (x) = 1 a2 x2(Eqn 9)


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In quantum mechanics, there is a fundamental uncertainty in the position of the particle before themade, and the probability of finding the particle in the range x to x + x is given

The quantum probability density is then P(x) = |1(x)1|02

. The quantum probabilities are readilysquaring the real eigenfunctions given by Equations 13 and 14:

n (x) = An fn (x)exp( 12 x2 ) with n = 0, 1, 2, 3 (Eqn 13)

f0 = 1

f1 = 2 x (Eqn 14)

f2 = 2 40x 02

f3 = 12 x 8 x3

Pn (x) = An2fn2 (x)exp(x2 ) with n = 0, 1, 2, (17)

Quantum probability densities for the harmonic oscillator.

The graphs of the probability functions given by Equation 17

f h h l


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P0(x)

543211 02345

n = 0

P1(x)

211 0345

0.5

2

(b)(a)

P2(x)

543211 02345

0.5

n = 2

(c)

x

P6(x)

211 02345

0.5

(d)

x

1.01.0

1.0

Figure 64The probability

density functions for theharmonic oscillator. Fourcases are shown forparticles in energy levelsgiven by the quantumnumber n = 0, 1, 2, and 6.The dotted curves showthe corresponding classicalcalculat ion of theprobability density, withthe dashed lines theclassical limits of theoscillation.

for the three lowestenergy levels and for

the n = 6 energy levelare shown in Figure 6,plotted as a function ofthe dimensionlessvariable x .

P ( )Al h i Fi 6


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P0(x)

543211 02345

n = 0

P1(x)

211 0345

0.5

2

(b)(a)

P2(x)

543211 02345

0.5

n = 2

(c)

x

P6(x)

211 02345

0.5

(d)

x

1.01.0

1.0

Figure 64The probability

density functions for theharmonic oscillator. Fourcases are shown forparticles in energy levelsgiven by the quantumnumber n = 0, 1, 2, and 6.The dotted curves showthe corresponding classicalcalculat ion of theprobability density, withthe dashed lines theclassical limits of theoscillation.

Also shown in Figure 6are the corresponding

classical probabilitydistributions given byEquation 9.

Pcl (x) =1

a2

x2


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These are obtained by finding the amplitude of the classical oscillator from Equation 16:Of course, it is really absurd to suggest that the classical calculation is relevant in the region of

numbers, and the classical and quantum probability distributions bear little resemblance to each 2 or 3. But for n = 6, you can see that the quantum distribution is beginning to approach theIt does not stretch the imagination too much to see that the two forms become indistinguishaThis is another instance of the correspondence principle.

Question T5

Write down the probability density function P(x) for the n = 0 state of the quantum oscillator.

Show that at a distance x = 1 from the origin, P(x) falls to 1/e of its maximum. x is runcertainty in the position of the particle.

Now estimate the momentum of the particle using the relation px 2mE , and henceuncertainty in the momentum px. Find the product x1px, and comment on the result.4t

3 Cl i it


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3 Closing items

3.1 Module summary1 If a particle of mass m is subject to a force Fx = ks0x, then classical (Newtonian) mechanic

will execute simple harmonic motion (SHM) about the origin with frequency

f =1

2ks

m

2 The potential energy function of a particle executing SHM is U(x) = 12 ksx2 . The total enerexecuting SHM is related to the amplitude of the oscillation by the equation

E= 12 ksa24or4 a = 2Eks(Eqn 7)

3 The classicalprobability density for a particle executing SHM is

P(x) =

1

a2 x2 (Eqn 9)4 E >> h0f is a condition for the use of classical mechanics in oscillator problems. Otherwis

mechanics.

5 Th S h di i f h h i ill i


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5 The Schrdinger equation for the quantum harmonic oscillator is:

2

2m

d2(x)

dx2 = ( 12 ksx2 E)(x) (Eqn 11)

6 The energy eigenfunctionsare given by

n (x) = An fn (x)exp( 12 x2 ) with n = 0, 1, 2, 3 (Eqn 13)

where fn(x) is a polynomial function of degree n and has n nodes. The first few polynomiaEquations 14and the eigenfunctions illustrated in Figure 4.

7 Since the potential function is symmetric about thex = 0 line, the eigenfunctions areoddorThey are odd when n is odd, and even when n is even.

8 The total energy eigenvaluesare given by the formula:

En = n + 12( )hf with n = 0, 1, 2, 3 (Eqn 15)

When n = 0, we have thezero point energy. This prediction of discrete energies, with the low

equal to zero, is in contrast to classical mechanics which allows a continuum of possible eneupwards.

9 Th t b bilit d it f ti i i b


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9 The quantum probability density function is given by:

Pn

(x) = n

2 (x) = An

2fn

2 (x)exp(x2 ) with n = 0, 1, 2, 3 (Eqn 17)

The probability density functions are bounded approximately by the classically allowed regiexponentially outside this region. As n, the quantum probability function becomes cothe classical probability function which is an illustration of thecorrepondence principlefunctions Pn(x) are illustrated inFigure 6.


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3.2 Achievements

Having completed this module, you should be able to:

A1 Define the terms that are emboldened and flagged in the margins of the module.

A2 Define simple harmonic motion (SHM) in classical mechanics in terms of the restoring

potential energy function.A3 Calculate the frequency and period of SHM from the particle mass and the force const

amplitude of the oscillation to the total energy. Sketch the classical probability function.

A4 Decide whether to use classical or quantum mechanics in a particular SHM problem.

A5 Write down the Schrdinger equation for quantum SHM. Verify the first few energy eigeeigenvalues. Recall the general formula En = n + 12( )hf, and use it to calculate the energy eigthe particle mass and the force constant.

A6 Sketch the shape of the first few energy eigenfunctions, and relate the number of nodes

number n. Distinguish the even and odd eigenfunctions.A7 Understand and use the quantum probability density functions Pn (x) = n2 (x) , and sketch t

n = 0, 1, 2, 3. Compare the classical probability density distribution with the quantum distribu

Study comment You may now wish to take the Exit test for this module which tests thes


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Study comment You may now wish to take the Exit test for this module which tests thesIf you prefer to study the module further before taking this test then return to theModule contents to revtopics.


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3.3 Exit test

Study comment Having completed this module, you should be able to answer the following questions eaone or more of the Achievements.

Question E1

(A2, A3 and A4)4A small mass of 0.0021kg is hung on a light spring producing an extensCalculate the spring constant ks. Find the frequency of small vertical oscillations of the equilibrium position.

Show that the conditionE>> h0ffor the validity of the classical approximation is easily satisfied iof the oscillation is 11mm.

Question E2


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Q

(A 5 and A6)4A particle moves in a region where its potential energy is given by

The wavefunction is the eigenfunction

(x) = A(1 2x2 )exp( 12 x2 )with = 5 10101m1 , and A is a normalization constant. Is this an odd or an e

Find the positions where 0(x) = 0, and sketch the shape of the wavefunction (use the dimensix ). What is the result of measuring the total energy if the particle is an electron? Calcula

electronvolts.

Question E3

(A6andA7)4Write down an expression for the probability density P(x) for the n = 1 state of a qharmonic oscillator in one dimension. (Refer to Equations 13and 14 for the eigenfunctions.)

where P(x) is zero, the points where P(x) is a maximum, and the boundaries of the classically aSketch the shape ofP(x). What can you say about the result of a measurement of (a) the position(b) the total energy of the particle?

Question E4


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Q

(A3,A4,A5 andA7)4A simple model of the HCl molecule indicates that the hydrogen ion is h

potential with force constant ks = 4701N1m1 . Calculate the frequency of oscillations inapproximation (neglecting the motion of the chlorine atom). The amplitude of oscillations is10111m, or one-tenth of the interatomic spacing. Is the classical approximation valid? Obtain anthe vibrational energy levels using quantum theory. Where is the boundary of the classically

when n = 0? Calculate the frequency and wavelength of electromagnetic radiation emitted wheHCl jump from one of these energy levels to the one immediately below.

Study comment This is the finalExit testquestion. When you have completed theExit testgo back to Sutry the Fast track questions if you have not already done so.

If you have completed both the Fast track questions and theExit test, then you have finished the module here.
http://p11_2m.pdf/http://p11_2m.pdf/http://p11_2m.pdf/http://p11_2m.pdf/http://p11_2m.pdf/

quantum harmonic oscillator.pdf

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