﻿ The Solutions to Generalized Helmholtz Equations
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The Solutions to Generalized Helmholtz Equations

## The Helmholtz Equation

The standard form of the Helmholtz equation is

#### ∇²φ + k²φ = 0

where k is a real-valued constant. What is investigated here are characteristics of the solution to an equation like the Helmholtz equation but where the coefficient labeled k above is not a constant; i.e.,

#### ∇²φ + f²(X)φ = 0

where X is the vector of the coordinates.

## The One Dimensional Case

The relevance of this problem is where φ² is probability density and the item of interest is the how the spatial average of φ² is related to the value of f(x). The solution for φ² oscillates very rapidly between a minimum of zero and a maximum. Thus the spatial average of φ² is roughly one half of the maximum of φ². This will be shown below to be inversely related to the magnitude of f(x).

For this case the equation is

#### (d²φ/dx²) = (d/dx)(dφ/dx) = −f²(x)φ

If (dφ/dx) is positive and φ is positive then an increase in x results in a decrease in (dφ/dx). The slope (dφ/dx) continues to decrease as x increases until it reaches 0 and thereafter φ also decreases. The decreases in (dφ/dx) and φ continue until φ reaches zero and then becomes negative. Thereafter (dφ/dx) increases with increasing x.

The larger that the function f²(x) is, the more rapidly (dφ/dx) goes to zero and the smaller are the magnitudes of the maxima and minima of φ. Also the larger that f²(x) is, the smaller is the interval between the values of x at which φ(x) is equal to zero. This is illustrated below. Suppose f(x) is roughly constant over some interval Δx. Let this constant value be denote as f(x).

If both sides of the above equation are multiplied by (dφ/dx) the result is

#### (dφ/dx)(d²φ/dx²) = − f²(x)φ(dφ/dx) which is equivalent to ½(d/dx((dφ/dx)²) = − ½f²(x)(d(φ²)/dx

The factor of ½ can be eliminated and the result integrated from a value of x such that φ is equal to zero and (dφ/dx) is a maximum to a value of x where (dφ/dx) is equal to zero and φ is a maximum. This means that

#### 0 − (dφ/dx)max² = − f²(x)[φmax² − 0] which reduces to (dφ/dx)max² = f²(x)φmax² and hence (dφ/dx)max = f(x)φmaxand further that φmax = (dφ/dx)max/f(x)

This is a rigorous relationship.

The interval Δx over which (dφ/dx) goes from a maximum to zero is the same interval for which φ goes from zero to a maximum.

If f(x) is constant over an interval the solution to the equation is a sinusoidal function. The wavelength L is such that

#### f(x)L = 2π and hence Δx = L/4 = π/(2f(x))

The integral of φ² over the interval Δx is approximately one half of the maximum of φ² times Δx; i.e.,.

#### ∫φ²dx ≅ ½[(dφ/dx)max/f(x)][ π/(2f(x)] which can be expressed as ∫φ²dx ≅ [π(dφ/dx)max/4]/f²(x)

When x corresponds to time (dφ/dx)max corresponds to the velocity of a particle at zero potential energy and hence is constant if the total energy of the system is constant.

Any constant factor, such as the one in the above equation, is irrelevant for probability densities because it is also a factor of the sum of the probabilities and thus cancels out when the term for one interval is divided by the sum for all intervals.

Thus the probability for an interval (state) is inversely proportional to the coefficient in the Helmholtz equation f²(x). The probability density is inversely proportional to f(x), the square root of the coefficient. The spatial average of the probability densities, which is ∫φ²dx/Δx, is then inversely proportional to f(x). Let the spatial average of a probability density function be denoted by an overscore, ̅P.

The problem now is to make rigorous the argument sketched above. This can be achieved by starting with a system that is simple and tractible such as a harmonic oscillator.

## The Harmonic Oscillator

The dynamics of a harmonic oscillator are given by

#### m(d²x/dx²) = −k²x

Its energy E is the sum of its kinetic energy K and potential energy V; i.e.,

#### E = K + V = ½mv² + ½k½x²

where v is equal to (dx/dt). Thus

#### v² = (2/m)(E − V(x)) and hence v = (2/m)½(E − V(x))½

The particle of a harmonic oscillator travels from a minimum value of x, xmin, to a maximum value of x, xmax, and back to xmin. The probability of finding the particle in an interval dx is [(2/|v(x)|)dx]/T where T is the time required to execute a complete cycle. Hence the probability density at a point x of the path of the particle is [2/|v(x)|]/T. This probability density function can be called the time-spent probability density function because it is the proportion of the time spent by the particle in a particular interval. Thus the classical probability density function, PC, for a harmonic oscillator is

#### PC = 2/(T|v(x)|) = (2/m)½)/[E − V(x)]½/T

A useful bit of notation is to let K(x) represent (E−V(x)). K(x) is the kinetic energy of the particle as a function of its displacement. Thus

#### v = [(2/m)K(x)]½/T and hence PC = (m/2)½T)/[K(x)]½

The probability density function according to quantum physics, PQM, is given as the square of the wave function φ where φ satisfies the time independent Schrödinger equation

#### −(h²/2m)(d²φ/dx²) + V(x)φ = Eφ which reduces to (d²φ/dx²) = − (2m/h²)(E − V(x))φ

where h² is Planck's constant divided by 2π. Thus the quantum mechanical wave function satisfies the equation

#### (d²φ/dx²) = − (2m/h²)K(x)φ

This is a generalized Helmholtz equation.

From the previous analysis the average value of φ² is inversely proportional to the square root of the coefficient of φ in the above equation. The constant terms in the coefficient are irrelevant in determining ̅PQM. Therefore

#### ̅PQM is proportional to 1/[K(x)]½ .

It was found previously that

#### PC = (2/T)/[K(x)]½

Thus since both ̅PQM and PC are inversely proportional to [K(x)]½ they are proportional to each other. However the integral of both ̅PQM and PC must be unity. Therefore they must be equal to each other; i.e.,

#### ̅PQM(x) = PC(x)

Here is a diagram that shows the values of PQM and PC for a harmonic oscillator in which the quantum number is 30. Visually one can see that ̅PQM is equal to PC.

## Conclusion

For the significant case of harmonic oscillators the spatial average of the quantum mechanical probability density function is equal to the time-spent probability density function derived from classical analysis.

(To be continued.)