The Quantum Harmonic Oscillator 9 February 2009 Sean Mortara

Introduction A quantum harmonic oscillator is used to describe a quantum particle subject to a symmetric potential

energy source, V . The particle is constrained by the potential energy field about a

reference displacement. On a macro scale this can be illustrated by a mass-spring system for which time

trajectory solutions are readily available.

2/)( xmx 22ω=

2ωmk = m

x

Figure 1: simple spring-mass particle system

On a nano-scale the system can be used to describe the bond interaction between atoms of a molecule

assuming the potential forces in the bond are linear over the distances of interest. For example, by

understanding the dynamics involved in molecular bonds, one can understand the electromagnetic

emission and absorption properties of various molecules. The scale of these systems prohibits the

tracking of a precise time trajectory since the energy (ie. the frequency of the photon striking the

particle) required to resolve accurate states of both position and momentum is sufficient elevate the

particle to a different energy level (the quantization of the energy is the subject of this paper) thereby

destroying the understanding of particle behavior within the lowest of energies. The solution to these

systems necessitates the application of probability distributions, or probability waves, into the

mechanical model.

The following equation identifies the Hamiltonian for the system.

(1) )()(ˆ xExH ψψ =

where

222

ˆ21

2ˆˆ xmm

pH x ω+= (2)

This states that for the Hamiltonian operator, H , acting on )(xψ there exists a scalar value E that

operates on )(x )(xψ identically where ψ is one or a sum of unique basis states for which the quantum

particle exists. The energy scalar E is called the eigenvalue of the system, while )(xψ is known as the

eigenfunction that represents an amplitude – related to probability – for the particle being located at x.

The hats represent an operator and not necessarily a variable. When working in the domain of the

operator the hat can be removed treating the operator as a variable; however, an operator of a different

domain must be converted to the domain being worked. Since the solution, outlined hereafter, is

performed on the spatial domain x , the momentum must be defined in terms of p x , hence the

subscript. In quantum mechanics the momentum operator is shown from its relationship to the

wavelength (see Appendix – Quantum Momentum in Position Space) to be:

dxd

ipx

h=ˆ (3)

Equations (1), (2), and (3) form the basis of the Schrödinger equation for a quantum system influenced

by the described potential field, V.

0)(21))((

222

2

2

=⎟⎠⎞

⎜⎝⎛ −+ xxmE

dxxd

mψωψh (4)

Quantum Solution Equation (4) is a nonlinear differential equation and requires an in-depth solution. The eigenfunction

cannot be deduced through direct manipulation of the differential equation. Instead guesses of the

eigenfunction must be made that will simplify the differential equation to a form that can be solved

directly. It is helpful to simplify equation (4) to a more general form through the following changes of

variables.

h/ωmxy = , and )/( ωε hE= (5)

0)2( 2 =−+ ψεψ yyy (6)

To start let us assume the eigenfunction takes the following form 2/2

)()( yeyy φψ = , with the second

differential 2/2 2y))1(2( yyyyy eyy φφφψ +++= . It is important to note that one of the requirements of

)(yψ is to be square integrable (ie. , βψψ =∫ ∞−dyyy )()(*∞

)(* yψ is the complex conjugate of )(yψ , β

is some finite constant value) over the entire domain of . This is a necessary stipulation to ensure that

the total probability of the particle existing in the entire spatial domain is unity. By using the above

transformation we are also assuming that

y

)(yφ is also square integrable and successfully envelops e

over the entire domain. Noting that can not be zero for all of , equation

2/2y

2/2ye y (6) can be rewritten.

0)21(2 =+++ φεφφ yyy y (7)

At first glance, it does not appear that any progress has been made toward the solution. This is only the

first step, though, to a much more complex solution. Once the total solution is known, the complete

eigenfunction can be inserted into the original differential equation producing an immediate verification.

The issue is that the end solution is not immediately apparent, and therefore must be broken into pieces

of the solution that can be ultimately combined to form the final solution. Solutions to differential

equations tend to center around exponential functions due to their unique relationships with their

differentials.

The use of a Fourier transform (specifically the inverse transform) can be used as a substitution for the

new eigenfunction in equation (7). This is an acceptable substitution since )( yφ has already been

defined to vanish at the limits due to its requirement to be square integrable.

(8) [∫∞

∞−

−= dyeyk iky)()(~ φφ ]

]

y

(9) [∫∞

∞−

= dkeky iky)(~)( φφ

Taking the differentials of equation (9) is relatively straight forward since only the one term is a function

of .

(10) [ ]∫∞

∞−

= dkeik ikyy φφ ~

(11) [ ]∫∞

∞−

−= dkek ikyyy φφ ~2

The first differential of equation (9) is precluded by . By inserting directly into equation y y (10)

(recognizing that the integral is not over ) and integrating by parts results in a form that gets rid of

in the term.

y y

[ ] dkiyedvdkkdkddu

evkuiky

iky

==

==

φ

φ~

~

and ∫∫ −= vduuvudv (12)

[ ] [ ] [ ]∫∫∫∞

∞−

∞

∞−

∞

∞−

∞

∞−⎥⎦⎤

⎢⎣⎡−=⎥⎦

⎤⎢⎣⎡−== dkek

dkddkek

dkdekdkeikyy ikyikyikyiky

y φφφφφ ~~~~ (13)

This is valid as long as φ~k vanishes at the limits. Equations (9), (11) and (13) can be inserted into

equation (7).

(14) 0)~)21()~~(2~( 2 =+++−−∫∞

∞−

dkekk ikyk φεφφφ

This can only be true if the term inside the parenthesis is zero.

(15) 0~)21(~2 2 =−++ φεφ kk k

The system has now been reduced to a first order differential (still nonlinear). As in the previous steps

the choice of a function to solve the differential equation (15), is going to involve an exponential

function. Let us choose φ~ to be a function of the generalized form ~ )(kfzeCk=φ . No restrictions are

placed on z, though it will be shown later that other conditions require z to be an integer. Substitute back

into (15).

0)21(2 )(2)(1 =−++⎟⎠⎞

⎜⎝⎛ +− kfzkfzz eCkke

dkdfknkCk ε (16)

0)221(2 21 =−+++ ++ zzz kzkdkdfk ε (17)

kkzk

kkzdkdf

z

zz

21

21

2)122( 1

1

2

−⎟⎠⎞

⎜⎝⎛ −−=

−−−= −

+

+

εε (18)

Equation (18) can be integrated directly.

022

1

41ln)( Ckkkf z +−⎟

⎠⎞⎜

⎝⎛= −−ε (19)

Inserting back into the assumed form of )(kf φ~ and simplifying.

2

41

21~ k

eCk−−= εφ (20)

Notice that equation (20) is still of the form that was previously assumed even though in its final

form does not look like what was solved. The final piece of the function has been determined that ends

the cycle of transforming the problem from one differential equation into another. What remains now is

to properly integrate each solution back into the previous differential equation, until the original function

(that is in the spatial domain) is achieved. The original function

)(kf

)(yψ can be found by inverting the

Fourier transformation and substituting.

∫∞

∞−

−−= dkeeCk ikyk 2

41

21εφ (21)

∫∞

∞−

−−= dkeekCey ikyky2

241

212/)( εψ (22)

One of the properties of a symmetric potential is that it produces symmetric wave functions (even or

odd). This means that )()( yy ψ±=− . ψ

∫∞

∞−

−−−=− dkeekCey ikyky2

241

212/)( εψ (23)

By assigning , effectively transferring the negative sign onto k, a form similar to equation kk −→ (22)

is found.

∫∞

∞−

−−−−=− dkeekCey ikyky2

241

21

212/ )1()( εεψ (24)

Symmetry requires that 1)1( 21

±=− −ε . This can only be true if 21−ε is an integer, n , otherwise the

value would be complex.

21

+= nε (25)

The value 21−ε can take on many values producing many eigenfunctions, nψ , that are valid for

equation (6). Consider negative integers. Since equation (22), a function of , would lead to functions

that cannot be normalized (equation

n

0

k

(22) would be indefinite at =y ) these equations cannot be

included in the set of valid functions. Therefore, the solution to equation (6) is the set of all functions,

nψ , for zero and positive integers.

∫∞

∞−

−= dkeekeCy ikykny

nn

22

41

2/)(ψ for ∞= K3,2,0n (26)

Noting that

ikynikyn

n

eikedyd )(= (27)

∫∞

∞−

−−= dkeedydeCiy ikyk

n

ny

nn

n

22

41

2/)(ψ (28)

The integral is of the following definite form:

⎥⎦

⎤⎢⎣

⎡ −=∫

∞

∞−

++−

aacb

adze cbzaz

44exp

2)( 2 π (29)

This gives 2

4 ye−π . Also note that . Substitute into 2/2/ 222 yyy eee −= (28) and redefine the constant.

)()1()( 2/2/ 2222

yHeCedydeeCy n

yn

yn

nyy

nn

n−−− =−=ψ (30)

The imaginary term disappears by assuming another can be factored out of the constant when it

was redefined. This can be done because any complex number with a magnitude of unity can be added

ni− nC

to the solution without changing the solution. The constant C was never assumed to be real. The

interpretation of this is an arbitrary assignment of phase to each of the discrete solutions. The

convenience here is that all of the solutions are now real valued.

n

n

)(yH

A number of the terms are consolidated into the set of functions identified as the Hermite polynomials,

. Results of the first handful of polynomials generated by the Hermite function are provided in

the appendix (Properties of Hermite Polynomials).

n

22

)1()( yn

nyn

n edydeyH −−= (31)

The final spatial solution is

)/()( )2/(2

hh ωψ ω mxHeCx nxm

nn−= (32)

with calculated based upon the necessity for the square magnitude of nC )(xnψ , or the total probability,

to be unity (see Appendix - Normalizing the Quantum Harmonic Oscillator).

π

ω!2

/n

mCnn

h= (33)

The discrete energy states representing the eigenvalues for each eigenfunction are given as

⎟⎠⎞

⎜⎝⎛ +=

21nEn ωh (34)

Appendix

Comparison to the Classic Harmonic System

When considering the classic harmonic oscillator, it is assumed that observing the system does not

destroy the state of the system so that both the position and momentum can be simultaneously known

and projected with arbitrary precision. If the momentum and position can be precisely known, it

becomes difficult to understand the meaning of assigning an amplitude wave relating to the probability

density of the particle being located at a certain position. Therefore the classical solution avoids solving

the Hamiltonian system outright, but rather uses the fact that the energy in the system (the scalar

eigenvalue) is invariant to either space or time.

( ) 0ˆˆ 2222 =+= xmpdxdH

dxd

x ω (35)

or ( ) 0ˆˆ 2222 =+= xmpdtdH

dtd

x ω (36)

In the classical Newtonian system the momentum is represented as the product of the mass of the

particle and its velocity, . With this in mind, both )/(ˆ dtdxmp =x (35) and (36) reduce to the same form.

022

2

=+ xdt

xd ω (37)

The solution to this linear differential equation is solved in one step where the amplitude

coefficient is determined by the initial conditions. Representing the solution as a complex exponential is

not the only form that works. The solution could have just as easily been identified through the use of

sine and cosine. The fact that the exponential is complex should not discourage anyone. The use of the

complex number helps to preserve a phase relationship of the solution to a set of initial conditions. The

solution could have been presented as , effectively guaranteeing a real

solution, but this form adds unnecessary complexity.

tiCex ω=

))(2/1( eeCx )()( θωθω +−+ += titi

This is a typical solution to the classic harmonic; however, we have gone a step further than what has

been accomplished in the quantum solution by introducing time. This is really not an equivalent

comparison. In the quantum model we were concerned with understanding the degree to which we can

resolve both the position and momentum of the particle at an instant of time as a probability of

observation. In the classical model it is assumed that the position and momentum of the particle can be

measured at an instant in time with arbitrary precision. The probability density then is a dirac delta

centered at the measurement. Once a single position and momentum measurement is made on the

classical particle we know precisely everything about the system for all time (assuming we are

protecting the oscillating particle from all external disturbances other than the potential field and our

observation, the latter of which is assumed to not impact the particle’s behavior).

A better comparison would be to take a series of randomly timed measurements of the position to see

how the classic particle is distributed generally. We can theorize what this distribution would look like

based on our classic solution by noting that in a specific small measurable distance, dx, the particle will

be less likely to be detected at an instant in time if it is moving fast. The probability distribution is

vxP /1)(inversely proportional to the velocity, ∝ , normalized so that the total probability across all of x

is unity.

122

max

=−

= ∫∫ xxdx

vdx

ωκκ (38)

The integral evaluates to π over the valid range of x (which is not infinite in the classical system). The

unknown proportionality coefficient can now be determined and the overall probability distribution is

given as:

22

max

)(xx

xP−

=π

ω (39)

Let us consider a 0.001 kilograms mass oscillating about a reference with amplitude of 0.01 meters and

a frequency of 1 hertz ( π2 radians/second).

0.00E+00

4.00E+01

8.00E+01

1.20E+02

1.60E+02

2.00E+02

-1.00E-02 1.00E-02

x (m)

P(x)

Figure 1: Probability distribution of locating an oscillating particle (1gram,+/-1cm,1Hz) without previous knowledge of position or velocity to compute trajectory.

The energy for this system is

6-2max

2

101.973922

×≈=xmE ω joules (40)

A quantum number can be calculated from (34) using the correctly dimensional value of the reduced

Planck constant, joules-seconds. 34−10)53(054571628.1 ×=h

271098.221

×≈−=ωhEn (41)

It is computationally inefficient to prepare a plot of the probability distribution function using the

quantum solution, )()( xx nn ψ*ψ , of this order. However, it may be useful to look at an energy level that

is manageable and generalize the trend as energy increases. Figure 2 and Figure 3 illustrate the classic

and quantum solution for an energy level calculated by 50=n and 100=n respectively.

0.00E+00

4.00E+14

8.00E+14

1.20E+15

1.60E+15

2.00E+15

-1.43E-15 1.43E-15

x (m)

P(x)

Figure 2: Comparison of quantum and classic oscillator when n = 50.

0.00E+00

4.00E+14

8.00E+14

1.20E+15

1.60E+15

-2.02E-15 2.02E-15

x (m)

P(x)

Figure 3: Comparison of quantum and classic oscillator when n = 100.

The classic solution appears to run along the mean of the peaks and valleys of the quantum solution.

The outermost peaks of the quantum solution rise along with the classic solution giving ever increasing

likelihood to observe the particle closer to the classical limits. The quantum solution does have some

probability to exceed the classical limits; however, this probability decreases with increasing energy

levels. As n approaches infinity the quantum solution becomes exactly the classical solution with an

infinite number of peaks, spanning the classic limits, with each representing a dirac delta function whose

magnitude is equal to the classic solution at that position.

On a large enough scale solving the harmonic oscillator by way of quantum mechanics becomes

impractical, and the classic solution is with insignificant error considered sufficient. On small scales,

however, the two solutions are far from similar, and the classic solution can no longer be considered

representative.

0.00E+00

1.20E+15

2.40E+15

3.60E+15

4.80E+15

6.00E+15

-2.59E-16 2.59E-16

x (m)

P(x)

Figure 4: Comparison of quantum and classic oscillator when n = 0.

Quantum Momentum in Position Space

Let P be a probability density function to represent the uncertainty in the measurement of a particular

state or observable. A Gaussian function is an example, but the density function can take on the form of

any number of functions as long as it conforms to fundamental rules regarding the probability of a

particular event. The actual form of this function is heavily dependant on the problem that is being

studied. One requirement of the function is that it has a finite integral over the span of the observable

that can be normalized to unity representing the fact that the total probability that the observable is

within its space should be one. In the study of particle dynamics, position, x , momentum, p mv= , and

wave propagation, , can be measured quantities with uncertainty defined by the probability

densities:

D/1=k

)()()( * xxxP ψψ=

)()()( * pppP φφ=

)()()( * kkkP ϕϕ=

The functions , , and are known as wave functions defined by the square root of the

probability function. These functions can be (and usually are) complex requiring the use of the complex

conjugate in the square to produce the real valued probability. As stated above, the probability function

needs to be integrable to a finite value that can be scaled to unity. More formally, the wave functions

should be of a scaled form such that:

)(xψ )( pφ )(kϕ

1)()( * =∫∞

∞−

dkkk ϕϕ

In quantum mechanics the momentum and wave propagation are related by the expression . This

relationship was established by Planck in his work on black body radiation, and also confirmed by

Einstein in describing the photoelectric effect. As a result, the wave functions are identical except for

scaling, thus

kp h=

)()( pk φϕ h= . The constant relates the momentum to a characteristic wavelength (or

distance in position space). If the momentum of a particle changes so does the characteristic

wavelength. Additionally, the wave propagation is a vector not an absolute scalar, thus it can also be

negative.

h

The mean value of a space (position, momentum, or any other function) with a given probability

distribution is by definition:

dxxdxxxPx ∫∫∞

∞−

∞

∞−

== ψψ *)(

dppdpppPp ∫∫∞

∞−

∞

∞−

== φφ *)(

It would be nice to express the averages of both position and momentum as the function of one

particular space and its associated wave function. For example the average momentum could also be

given as a function of the spatial integral.

dxpp ∫∞

∞−

= ψψ ˆ*

The momentum operator, , used here needs to be identified as a function of p x . Consider the Fourier

transform integral and its inverse function

dxxfekF ikx∫∞

∞−

−= )(21)(π

dxkFexf ikx∫∞

∞−

= )(21)(π

The momentum wave function can be transformed into position space through the following

transformation:

dxxekp ikx∫∞

∞−

−== )(21)()( ψπ

ϕφhh

Making this substitution in the average momentum equation, and noting that , gives the following

expression:

kp h=

dkdxxekxdxep ikxxik∫ ∫∫∞

∞−

∞

∞−

−∞

∞−

′−

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡′′= )()(

2

*

ψψπh

dkdxxdxkeexp ikxxik∫ ∫ ∫∞

∞−

∞

∞−

∞

∞−

−′ ′′= )()(2

* ψψπh

The primes, , are designated to ensure that when the integrals are rearranged the independence of the

transformation integrals are maintained. We can observe that since:

x′

( )x

eikeikx

ikx

∂∂

=−

−

( ) dkdxxdxx

eiexpikx

xik∫ ∫ ∫∞

∞−

∞

∞−

∞

∞−

−′ ′

∂∂′= )()(

2* ψψ

πh

Consider just the integral over , excluding all terms that are not associated with dx x to be outside the

integral.

( )∫∞

∞−

−

∂∂ dxx

xei

ikx

)(ψ

This integral can be rearranged by utilizing the rule for integrating by parts ( ). If we

choose

∫ ∫−= vduuvudv

)()( xxu ψ= ( ) and ( )ikx ∂∂= − dxxeidv so that ( )dxxxdu ∂∂= )(ψ and then: ikx−= iexv )(

( ) ( )dxxxeixiedxx

xei ikxikx

ikx

∂∂

−=∂

∂ −∞

∞−

∞

∞−

−∞

∞−

−

∫∫)()()( ψψψ

Furthermore, since 0)( =±∞ψ because of the requirement for square integrability mentioned above the

first term vanishes. The second term can then be substituted into the average momentum equation.

( ) dkdxxdxxe

iexp ikxxik∫ ∫ ∫

∞

∞−

∞

∞−

∞

∞−

−′ ′∂

∂′=)()(

21 * ψψπ

h

Now the functions involving can be isolated in the integral over dk where the limits k K approach

infinity.

( ) dxxddkexx

ixp

K

K

xxik ′⎥⎥⎦

⎤

⎢⎢⎣

⎡

∂∂′= ∫ ∫∫

∞

∞− −

−′∞

∞−

)(* )()(21 ψψπ

h

( ) dxxdxx

xxKxx

ixp Lim

K

′⎥⎦

⎤⎢⎣

⎡−′−′

∂∂′= ∫ ∫

∞

∞− ∞→

∞

∞− )()(sin)()(1 * ψψ

πh

The limit function is only significant in the vicinity very near 0=−′ xx . Since is a continuous

function, it varies little over this region and therefore can be evaluated at x (removing it from the

integral over ).

*)(x′ψ

xd ′

( ) dxxdxx

xxKxx

ixp Lim

K∫ ∫∞

∞−

∞

∞− ∞→

′⎥⎦

⎤⎢⎣

⎡−′−′

∂∂

=)(

)(sin)()(1 * ψψπ

h

The integral over can be shown to be equal to xd ′ π regardless of the variable x .

dxxxi

xp ∫∞

∞− ∂∂

= )()( * ψψ h

From here it is observed that in x-space the momentum operator, , is a function of the spatial

derivative operator

p

xi

p∂∂

=hˆ

Properties of Hermite Polynomials

The physics based Hermite function set is classically defined as:

22

)1()( xn

nxn

n edxdexH −−= (42)

The first eleven polynomials generated are:

30240302400403200161280230401024)(

3024080640483849216512)(

168013440134403584256)(

168033601344128)(

12072048064)(

12016032)(

124816)(

128)(

24)(

2)(1)(

24681010

35799

24688

3577

2466

355

244

33

22

1

0

−+−+−=

+−+−=

+−+−=

−+−=

−+−=

+−=

+−=

−=

−=

==

xxxxxxH

xxxxxxH

xxxxxH

xxxxxH

xxxxH

xxxxH

xxxH

xxxH

xxH

xxHxH

Through observation each Hermite polynomial can also be described as a summation series:

∑=

−−

−−

=2/

0

22

)!2(!!)1(2)(

n

k

knkkn

n xknk

nxH (43)

Recursion

From the classic definition the derivative of the Hermite polynomial is

⎥⎦

⎤⎢⎣

⎡−=′ − 22

)1()( xn

nxn

n edxde

dxdxH

⎥⎦

⎤⎢⎣

⎡+−= −

+

+− 2222

1

1

2)1( xn

nxx

n

nxn e

dxdee

dxdxe

)()(2)( 1 xHxxHxH nnn +−=′ (44)

Appell Sequence

From the series summation the following two relationships can be observed

∑−

=

−−−

−−−

=′2/)1(

0

122

)!12(!!)1(2)(

n

k

knkkn

n xknk

nxH

∑−

=

−−−−

− −−−−

=2/)1(

0

1212

1 )!12(!)!1()1(2)(

n

k

knkkn

n xknk

nxH

Therefore

)(2)( 1 xnHxH nn −=′ (45)

Recurrence

Using the identities of the (44) and (45) sequence the recurrence identity is expressed as

0)(2)(2)( 11 =+− −+ xnHxxHxH nnn (46)

Normalizing the Quantum Harmonic Oscillator

The coefficient must be determined that will normalize wave function so that the total probability

generated by integrating the absolute square of the wave function is unity. Integrating two different

wave functions produces zero since each function defines a unique space.

nC

(47) nmnm dxxx ,* )()( δψψ =∫

∞

∞−

This can be simplified by returning to space. (Note y h// ωmdxdy = )

nmnmnmmdyyy ,,

* )()( βδδωψψ ==∫∞

∞− h

nmnmy

n dyyHyHeC ,2 )()(

2

βδ=∫∞

∞−

−

What is interpreted here is that the Hermite functions are orthogonal to one another when weighted by

the exponential function, . 2y−e

Define

2,

,2

n

nmnm

ynm C

dyHHeIβδ

== ∫∞

∞−

− (48)

so that

(49) 0111,12

== ∫∞

∞−+−

−+− dyHHeI nn

ynn

Using the recurrence relation (see Appendix - Properties of Hermite Polynomials)

0)(2)(2 11 =+− −+ ynHyyHH nnn (50)

[ ] 022 111,12

=−= ∫∞

∞−−−

−+− dynHyHHeI nnn

ynn

1,11 222

−−

∞

∞−−

− =∫ nnnny nIdyHHye

1,11

11 2)1()1(2

22222

−−

∞

∞−

−−−

−−− =⎥

⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡−∫ nn

yn

nyny

n

nyny nIdye

dydee

dydeye

1,11

1

2222

−−

∞

∞−

−−−

−

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡− ∫ nn

yn

ny

n

ny nIdye

dyde

dyde

dyd (51)

By using the product rule for differentiation

⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡ −−−

−−

−

−222222

1

1

1

1y

n

nyy

n

nyy

n

ny e

dydee

dyde

dyde

dyde

dyd

1,11

1

2222222

−−

∞

∞−

−−−

−∞

∞−

−− =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡∫∫ nn

yn

ny

n

nyy

n

ny

n

ny nIdye

dyde

dyde

dyddye

dyde

dyde

1,11

1

, 2222

−−

∞

∞−

−−−

−

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡− ∫ nn

yn

ny

n

ny

nn nIdyedyde

dyde

dydI (52)

Using the product rule for integration

Let dye

dyde

dyddvdye

dyddu

edydeve

dydu

yn

nyy

n

n

yn

nyy

n

n

⎥⎦

⎤⎢⎣

⎡==

==

−−

−−

+

+

−−

−−

222

222

1

1

1

1

1

1

and ∫∫ −= vduuvudv

( ) ( ) 1,11

1

1

1

1

1

, 2222222

−−

∞

∞−

−+

+−

−

−∞

∞−

−−

−− =⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡− ∫ nn

yn

ny

n

nyy

n

nyy

n

n

nn nIdyedyde

dydee

dydee

dydI

The second term can be shown to vanish at the limits. Additionally it is readily apparent that the third

term is the nothing more than the weighted integral of two Hermite functions of different order and is

therefore zero as well. This produces an obvious recurring relationship between the integrals of each

series.

1,1, 2 −−= nnnn nII (53)

Therefore the only integral that needs to be solved is the trivial one for . 0,0I

20,0, !2!2!22

n

nynnnn C

ndyenInI βπ ==== ∫∞

∞−

−

π

ωπ

β!2

/!2 n

mn

Cnnn

h== (54)

Reference • Bader, Richard F.W., Dr., An Introduction to the Electronic Structure of Atoms and Molecules,

McMaster University, Hamilton, Ontario, http://www.chemistry.mcmaster.ca/esam/intro.html.

• Byron, Frederick W. & Fuller, Robert W., Mathematics of Classical and Quantum Physics,

Dover Publications, 1992.

• Chester, Marvin, Primer of Quantum Mechanics, Dover Publications, 2003.

• Ponomarenko, Sergey A., Quantum Harmonic Oscillator Revisited: A Fourier Transform

Approach, American Association of Physics Teachers, 2004.