The Quantum Harmonic Oscillator

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 ( x) = mω 2 x 2 / 2 . 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.

k = mω 2

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.

Hˆ ψ ( x) = Eψ ( x)

(1)

where pˆ 2 1 Hˆ = x + mω 2 xˆ 2 2m 2

(2)

This states that for the Hamiltonian operator, Hˆ , acting on ψ (x ) there exists a scalar value E that operates on ψ (x ) identically where ψ (x ) 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 ψ (x ) 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 pˆ must be defined in terms of 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:

pˆ x =

h d i dx

(3)

Equations (1), (2), and (3) form the basis of the Schrödinger equation for a quantum system influenced by the described potential field, V.

1 h d 2 (ψ ( x)) ⎛ ⎞ + ⎜ E − mω 2 x 2 ⎟ψ ( x) = 0 2 2m dx 2 ⎝ ⎠

(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.

y = x mω / h , and ε = E /(hω )

(5)

ψ yy + (2ε − y 2 )ψ = 0

(6)

To start let us assume the eigenfunction takes the following form ψ ( y ) = φ ( y )e y differential ψ yy = (φ yy + 2 yφ y + (1 + y 2 )φ )e y

ψ ( y ) is to be square integrable (ie.

∫

∞

2

/2

2

/2

, with the second

. It is important to note that one of the requirements of

ψ * ( y )ψ ( y )dy = β , ψ * ( y ) is the complex conjugate of ψ ( y ) , β

−∞

is some finite constant value) over the entire domain of y . 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 φ ( y ) is also square integrable and successfully envelops e y over the entire domain. Noting that e y

2

/2

2

/2

can not be zero for all of y , equation (6) can be rewritten.

φ yy + 2 yφ y + (1 + 2ε )φ = 0

(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 φ ( y ) has already been defined to vanish at the limits due to its requirement to be square integrable. ~

φ (k ) =

∞

∫ [φ ( y)e ]dy

(8)

~ ∫ [φ (k )e ]dk

(9)

−iky

−∞

φ ( y) =

∞

iky

−∞

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

φy =

~ ∫ [ikφ e ]dk

∞

iky

−∞

(10)

[

∞

~

]

φ yy = − ∫ k 2φ e iky dk

(11)

−∞

The first differential of equation (9) is precluded by y . By inserting y directly into equation (10) (recognizing that the integral is not over y ) and integrating by parts results in a form that gets rid of y in the term. ~ u = kφ d ~ du = kφ dk dk

[ ]

yφ y =

~ ~ ∫ [ikyφ e ]dk = kφ e

v = e iky dv = iye iky dk

∞

iky

iky

∞ −∞

−∞

∞

−

and ∫ udv = uv − ∫ vdu

(12)

[ ]

(13)

∞

[ ]

⎡ d ~ iky ⎤ ⎡ d ~ iky ⎤ ∫−∞⎢⎣ dk kφ e ⎥⎦ dk = −−∫∞⎢⎣ dk kφ e ⎥⎦ dk

~ This is valid as long as kφ vanishes at the limits. Equations (9), (11) and (13) can be inserted into equation (7). ∞

~

~

~

∫ (−k φ − 2(φ + kφ 2

k

~ ) + (1 + 2ε )φ )e iky dk = 0

(14)

−∞

This can only be true if the term inside the parenthesis is zero. ~ ~ 2kφk + (1 + k 2 − 2ε )φ = 0

(15)

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 φ = Ck z e f (k ) . 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). df ⎞ f ( k ) ⎛ 2Ck ⎜ nk z −1 + k z + (1 + k 2 − 2ε )Ck z e f ( k ) = 0 ⎟e dk ⎠ ⎝

2k z +1

df + k z + 2 + (1 + 2 z − 2ε )k z = 0 dk

(16)

(17)

df (2ε − 2 z − 1)k z − k z + 2 ⎛ 1⎞ 1 = = ⎜ ε − z − ⎟k −1 − k z +1 dk 2⎠ 2 2k ⎝

(18)

Equation (18) can be integrated directly.

1 2 ε − z − 12 ⎞ f (k ) = ln⎛⎜ k ⎟ − k + C0 ⎝ ⎠ 4

(19)

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

φ = Ck

1

2 ε − 12 − 4 k

e

(20)

Notice that equation (20) is still of the form that was previously assumed even though f (k ) 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 ψ ( y ) can be found by inverting the Fourier transformation and substituting. ∞

φ = ∫ Ck

1 2 ε − 1 2 − 4 k iky

e

(21)

e dk

−∞

ψ ( y ) = Ce y

∞ 2

/2

∫k

1 2 ε − 1 2 − 4 k iky

e

e dk

(22)

−∞

One of the properties of a symmetric potential is that it produces symmetric wave functions (even or odd). This means that ψ (− y ) = ±ψ ( y ) .

ψ (− y ) = Ce

∞ y2 / 2

∫k

1 2 ε − 1 2 − 4 k −iky

e

e

dk

(23)

−∞

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

ψ (− y ) = Ce y

∞ 2

/2

∫ (−1)

−∞

ε − 12

k

1 2 ε − 1 2 − 4 k iky

e

e dk

(24)

Symmetry requires that (−1)

ε − 12

= ±1 . This can only be true if ε − 1

2

is an integer, n , otherwise the

value would be complex.

ε =n+

1 2

(25)

The value ε − 1 2 can take on many values producing many eigenfunctions, ψ n , that are valid for equation (6). Consider negative integers. Since equation (22), a function of k n , would lead to functions that cannot be normalized (equation (22) would be indefinite at y = 0 ) these equations cannot be included in the set of valid functions. Therefore, the solution to equation (6) is the set of all functions,

ψ n , for zero and positive integers. ψ n ( y) = Cn e

∞ y2 / 2

∫k

n

e

1 − k2 4

−∞

e iky dk for n = 0, 2, 3K ∞

(26)

Noting that

d n iky e = (ik ) n e iky n dy

ψ n ( y ) = i −n C n e y

2

/2

dn dy n

∞

∫e

1 − k2 4

(27)

e iky dk

(28)

⎡ b 2 − 4ac ⎤ exp ⎢ ⎥ a ⎣ 4a ⎦

(29)

−∞

The integral is of the following definite form: ∞

−( az ∫e

2

+bz + c )

π

dz =

−∞

This gives

4π e − y . Also note that e y 2

2

/2

= e y e− y 2

ψ n ( y ) = (−1) n Cn e − y / 2e y 2

2

2

/2

. Substitute into (28) and redefine the constant.

2 d n − y2 e = Cn e − y / 2 H n ( y ) n dy

(30)

The imaginary term disappears by assuming another i − n can be factored out of the constant Cn when it was redefined. This can be done because any complex number with a magnitude of unity can be added

to the solution without changing the solution. The constant Cn was never assumed to be real. The interpretation of this is an arbitrary assignment of phase to each of the n discrete solutions. The convenience here is that all of the solutions are now real valued. A number of the terms are consolidated into the set of functions identified as the Hermite polynomials,

H n ( y ) . Results of the first handful of polynomials generated by the Hermite function are provided in the appendix (Properties of Hermite Polynomials). d n − y2 H n ( y ) = (−1) e e dy n n

y2

(31)

The final spatial solution is

ψ n ( x) = Cn e − mωx

2

/( 2 h )

H n ( x mω / h )

(32)

with Cn calculated based upon the necessity for the square magnitude of ψ n (x) , or the total probability, to be unity (see Appendix - Normalizing the Quantum Harmonic Oscillator).

Cn =

mω / h

2 n! π n

(33)

The discrete energy states representing the eigenvalues for each eigenfunction are given as 1⎞ ⎛ En = hω ⎜ n + ⎟ 2⎠ ⎝

(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.

(

)

(35)

(

)

(36)

d 2 d ˆ H= pˆ x + m 2ω 2 x 2 = 0 dx dx or

d ˆ d 2 H= pˆ x + m 2ω 2 x 2 = 0 dt dt

In the classical Newtonian system the momentum is represented as the product of the mass of the particle and its velocity, pˆ x = m(dx / dt ) . With this in mind, both (35) and (36) reduce to the same form. d 2x + ω2x = 0 2 dt

(37)

The solution to this linear differential equation is solved in one step x = Ce iωt 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 x = C (1 / 2)(e i (ωt +θ ) + e −i (ωt +θ ) ) , effectively guaranteeing a real solution, but this form adds unnecessary complexity. 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

inversely proportional to the velocity, P ( x) ∝ 1 / v , normalized so that the total probability across all of x is unity.

κ∫

dx κ = v ω

∫

dx x

2 max

− x2

=1

(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: P( x) =

ω 2 − x2 π xmax

(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 ( 2π radians/second). 2.00E+02

1.60E+02

P(x)

1.20E+02

8.00E+01

4.00E+01

0.00E+00

-1.00E-02

1.00E-02

x (m)

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 E=

2 mω 2 xmax ≈ 1.97392 × 10 -6 joules 2

(40)

A quantum number can be calculated from (34) using the correctly dimensional value of the reduced Planck constant, h = 1.054571628(53) × 10 −34 joules-seconds.

n=

E 1 − ≈ 2.98 × 10 27 hω 2

(41)

It is computationally inefficient to prepare a plot of the probability distribution function using the quantum solution, ψ n ( x)*ψ n ( x) , 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 n = 50 and n = 100 respectively. 2.00E+15

1.60E+15

P(x)

1.20E+15

8.00E+14

4.00E+14

0.00E+00 -1.43E-15

1.43E-15

x (m)

Figure 2: Comparison of quantum and classic oscillator when n = 50. 1.60E+15

P(x)

1.20E+15

8.00E+14

4.00E+14

0.00E+00 -2.02E-15

2.02E-15

x (m)

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. 6.00E+15

4.80E+15

P(x)

3.60E+15

2.40E+15

1.20E+15

0.00E+00 -2.59E-16

2.59E-16

x (m)

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, k = 1 / D , can be measured quantities with uncertainty defined by the probability densities: P ( x) = ψ ( x) *ψ ( x)

P( p) = φ ( p)*φ ( p)

P (k ) = ϕ (k )* ϕ (k )

The functions ψ (x) , φ ( p ) , and ϕ (k ) 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: ∞

∫ ϕ (k ) ϕ (k )dk = 1 *

−∞

In quantum mechanics the momentum and wave propagation are related by the expression p = hk . 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 ϕ (k ) = hφ ( p) . The constant h 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. The mean value of a space (position, momentum, or any other function) with a given probability distribution is by definition: x=

p=

∞

∞

−∞

−∞

∞

∞

−∞

−∞

* ∫ xP( x)dx = ∫ψ xψ dx

* ∫ pP( p)dp = ∫ φ pφdp

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. ∞

p = ∫ψ * pˆ ψ dx −∞

The momentum operator, pˆ , used here needs to be identified as a function of x . Consider the Fourier transform integral and its inverse function ∞

1

F (k ) =

2π

−ikx

f ( x )dx

∞

1

f ( x) =

∫e

−∞

2π

∫e −∞

ikx

F (k )dx

The momentum wave function can be transformed into position space through the following transformation: φ ( p) =

ϕ (k ) h

=

∞

1

∫e 2πh −∞

ψ ( x)dx

−ikx

Making this substitution in the average momentum equation, and noting that p = hk , gives the following expression: h p= 2π

p=

*

⎡ ∞ −ikx′ ⎤ ⎡ ∞ −ikx ⎤ ′ ′ ( ) e ψ x d x ⎢ ⎥ k ⎢ ∫ e ψ ( x)dx ⎥dk ∫−∞ ⎢−∫∞ ⎥⎦ ⎢⎣−∞ ⎥⎦ ⎣ ∞

h 2π

∞ ∞ ∞

∫ ∫ ∫ψ ( x′) e −∞ −∞ −∞

* ikx′

ke −ikxψ ( x)dx′dxdk

The primes, x′ , are designated to ensure that when the integrals are rearranged the independence of the transformation integrals are maintained. We can observe that since: ke −ikx = i

p=

h 2π

( )

∂ e − ikx ∂x

∞ ∞ ∞

* ikx′ ∫ ψ ( x′) e i

∫∫ − ∞ −∞ − ∞

( )

∂ e −ikx ψ ( x)dx′dxdk ∂x

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

∫i

−∞

( )

∂ e − ikx ψ ( x)dx ∂x

This integral can be rearranged by utilizing the rule for integrating by parts ( ∫ udv = uv − ∫ vdu ). If we choose u ( x) = ψ ( x) and dv = i(∂ (e −ikx ) ∂x )dx so that du = (∂ψ ( x) ∂x )dx and v( x) = ie −ikx then: ∞

∫i

−∞

( )

∞ ∂ (ψ ( x) ) ∂ e − ikx dx ψ ( x)dx = ie −ikxψ ( x) − ∫ ie −ikx −∞ ∂x ∂x −∞ ∞

Furthermore, since ψ (±∞ ) = 0 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. 1 2π

p=

∞ ∞ ∞

∫ ∫ ∫ ψ ( x′) e − ∞ − ∞ −∞

* ikx′

h −ikx ∂ (ψ ( x) ) dx′dxdk e ∂x i

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

p=

1

π

1 2π

∞ ∞

∫

* ∫ ψ ( x′)

−∞ −∞

∞ ∞

∫ ∫ ψ ( x′) −∞ − ∞

*

K h ∂ (ψ ( x) ) ⎡ ik ( x′− x ) ⎤ dk ⎥dx′dx ⎢ e i ∂x ⎣⎢−∫K ⎦⎥

h ∂ (ψ ( x) ) ⎡ sin K ( x′ − x) ⎤ ⎢ ⎥dx′dx ( x′ − x ) ⎦ i ∂x ⎣ Lim K →∞

The limit function is only significant in the vicinity very near x′ − x = 0 . Since ψ (x′)* is a continuous function, it varies little over this region and therefore can be evaluated at x (removing it from the integral over dx′ ). p=

1

∞

ψ ( x) π −∫∞

*

∞ h ∂ (ψ ( x) ) ⎡ sin K ( x′ − x) ⎤ ⎢ ⎥ dx′dx i ∂x −∫∞ ⎣ Lim ( x′ − x ) ⎦ K →∞

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

p=

∫ ψ ( x) −∞

*

h ∂ ψ ( x)dx i ∂x

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

h ∂ i ∂x

Properties of Hermite Polynomials The physics based Hermite function set is classically defined as: H n ( x) = (−1) n e x

2

d n − x2 e dx n

(42)

The first eleven polynomials generated are: H 0 ( x) = 1 H1 ( x) = 2 x H 2 ( x) = 4 x 2 − 2 H 3 ( x) = 8 x 3 − 12 x H 4 ( x) = 16 x 4 − 48 x 2 + 12 H 5 ( x) = 32 x 5 − 160 x 3 + 120 x H 6 ( x) = 64 x 6 − 480 x 4 + 720 x 2 − 120 H 7 ( x) = 128 x 7 − 1344 x 5 + 3360 x 3 − 1680 x H 8 ( x) = 256 x8 − 3584 x 6 + 13440 x 4 − 13440 x 2 + 1680 H 9 ( x) = 512 x 9 − 9216 x 7 + 48384 x 5 − 80640 x 3 + 30240 x H10 ( x) = 1024 x10 − 23040 x8 + 161280 x 6 − 403200 x 4 + 302400 x 2 − 30240

Through observation each Hermite polynomial can also be described as a summation series: 2 n−2 k (−1) k n! n−2 k x k =0 k!( n − 2k )!

n/2

H n ( x) = ∑

(43)

Recursion From the classic definition the derivative of the Hermite polynomial is H n′ ( x) =

n d ⎡ n x2 d − x2 ⎤ (−1) e e ⎢ ⎥ dx n dx ⎣ ⎦

n n+1 2 d 2 ⎤ ⎡ x2 d − x2 e−x ⎥ = (−1) n ⎢2 xe x + e e n n +1 dx dx ⎣ ⎦

H n′ ( x) = 2 xH n ( x) − H n+1 ( x)

(44)

Appell Sequence From the series summation the following two relationships can be observed H n′ ( x) =

H n−1 ( x) =

( n−1) / 2

∑ k =0

( n−1) / 2

∑ k =0

2 n−2 k (−1) k n! n−2 k −1 x k!(n − 2k − 1)!

2 n−2 k −1 (−1) k (n − 1)! n−2 k −1 x k!(n − 2k − 1)!

Therefore

H n′ ( x) = 2nH n−1 ( x)

(45)

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

H n+1 ( x) − 2 xH n ( x) + 2nH n−1 ( x) = 0

(46)

Normalizing the Quantum Harmonic Oscillator The coefficient Cn 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. ∞

∫ψ

* m

( x)ψ n ( x)dx = δ m ,n

−∞

This can be simplified by returning to y space. (Note dy / dx = mω / h ) ∞

∫ψ

* m

( y )ψ n ( y )dy =

−∞

∞

mω δ m ,n = βδ m ,n h

Cn2 ∫ e − y H m ( y ) H n ( y )dy = βδ m,n 2

−∞

(47)

What is interpreted here is that the Hermite functions are orthogonal to one another when weighted by the exponential function, e − y . 2

Define ∞

I m ,n = ∫ e − y H m H n dy = 2

−∞

βδ m ,n C n2

(48)

so that ∞

I n−1,n+1 = ∫ e − y H n−1 H n+1dy = 0 2

(49)

−∞

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

H n +1 − 2 yH n ( y ) + 2nH n −1 ( y ) = 0

(50)

∞

I n−1,n+1 = ∫ e − y H n−1 [2 yH n − 2nH n−1 ]dy = 0 2

−∞

∞

∫ 2 ye

− y2

H n−1 H n dy = 2nI n−1,n−1

−∞

∞

n −1 n 2 ⎤ n −1 y 2 d − y2 ⎡ n y2 d − y2 ⎤ ⎡ − ye e − e e 2 ( 1 ) ( 1 ) e − y ⎥ dy = 2nI n−1,n−1 ⎢ ⎥ ⎢ n −1 n ∫−∞ dy dy ⎣ ⎦⎣ ⎦

∞

−

d y 2 ⎡ d n−1 − y 2 ⎤ ⎡ d n − y 2 ⎤ ∫ dy e ⎢⎣ dy n−1 e ⎥⎦ ⎢⎣ dy n e ⎥⎦ dy = 2nI n−1,n−1 −∞

By using the product rule for differentiation d y 2 ⎡ d n−1 − y 2 ⎤ d ⎡ y 2 d n−1 − y 2 ⎤ y 2 ⎡ d n − y 2 ⎤ e e ⎥= e ⎥−e ⎢ n e ⎥ ⎢e n −1 dy ⎢⎣ dy n−1 ⎦ dy ⎣ dy ⎦ ⎣ dy ⎦ ∞

∞ ⎡ d n − y2 ⎤ ⎡ d n − y2 ⎤ d ⎡ y 2 d n−1 − y 2 ⎤ ⎡ d n − y 2 ⎤ ∫ e ⎢⎣ dy n e ⎥⎦ ⎢⎣ dy n e ⎥⎦ dy − −∫∞ dy ⎢⎣e dy n−1 e ⎥⎦ ⎢⎣ dy n e ⎥⎦ dy = 2nI n−1,n−1 −∞ y2

(51)

∞

I n ,n −

d ⎡ y 2 d n−1 − y 2 ⎤ ⎡ d n − y 2 ⎤ ∫−∞ dy ⎢⎣e dy n−1 e ⎥⎦ ⎢⎣ dy n e ⎥⎦ dy = 2nI n−1,n−1

(52)

Using the product rule for integration n −1 2 d n − y2 y2 d u= ne v=e e− y n −1 dy dy and ∫ udv = uv − ∫ vdu Let n +1 d d ⎡ y 2 d n−1 − y 2 ⎤ − y2 du = n+1 e dy dv = e e ⎥ dy dy dy ⎢⎣ dy n−1 ⎦

( )

( )

∞

∞ ⎡ d n − y 2 y 2 d n−1 − y 2 ⎤ ⎡ y 2 d n−1 − y 2 ⎤ ⎡ d n+1 − y 2 ⎤ I n ,n − ⎢ n e e e ⎥ + ∫ ⎢e e ⎥ ⎢ n+1 e ⎥ dy = 2nI n−1,n−1 n −1 dy n−1 ⎣ dy ⎦ −∞ −∞⎣ dy ⎦ ⎣ dy ⎦

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.

I n,n = 2nI n−1,n−1

(53)

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

I n , n = 2 n! I 0,0 = 2 n! ∫ e − y dy = 2 n n! π = n

2

n

−∞

Cn =

β 2 n! π n

=

mω / h 2 n! π n

β C n2

(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.