﻿ A Property of Generalized Helmholtz Equations of Dimension One or More
San José State University

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A Property of Generalized Helmholtz
Equations of Dimension One or More

## The Helmholtz Equation and Its Generalization

The Helmholtz equation is of the form

#### ∇²φ = −kφ

Where k is a real positive constant. For the one dimensional case this reduces to

#### (d²φ/dx²) = −kφ(x)

The nature of the solution can be simply described. 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 constant coefficient k 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 k is, the smaller is the interval between the values of x at which φ(x) is equal to zero. This is illustrated below.

The above Helmholtz equation has a special property. The LHS is the derivative of dφ/dx so it has a value of zero where φ(x) is equal to zero. Thus dφ/dx has an extreme value where φ(x) is equal to zero. But φ(x) has an extreme value where dφ/dx is equal to zero.

A generalized Helmholtz equation is one in which the coefficient, f(z), on the RHS is not necessarily constant but is a positive function of the vector of spatial dimensions, z; i.e.,

#### ∇²φ = −f(z)φ(z)

The solutions to this equation also has the property that φ(z) has an extreme value where the derivatives of φ are all equal to zero (∇φ is equal to the zero vector) and the magnitude of the derivative of φ has an extreme value where φ(z) is equal to zero.

The Laplacian ∇²φ on the LHS is explicitly the divergence of the gradient of the scalar field φ(z), ∇·(∇φ).

Consider the product of the generalized Helmholtz equation with the gradient of φ; i.e.,

#### (∇²φ)∇φ = −f(z)φ∇φ

The RHS can be expressed as

#### −½f(z)∇φ²

What is needed is a repesentation of the LHS of the previous equation as the gradient of some function. The Laplacian ∇² is in the nature of a second derivative. The first derivative is the gradient of a scalar function, in this case ∇φ.

Now consider ∇(∇φ·∇φ). The vector calculus formula for this expression is

#### ∇(∇φ·∇φ) = 2(∇φ·∇)∇φ

In this formula the operator which can operate on the vector ∇φ is the divergence so the above formula reduces to

#### ∇(∇φ·∇φ) = 2(∇φ·∇·)∇φ which is the same as ½∇(∇φ·∇φ) = (∇φ)∇²φ

Therefore the product of the generalized Helmholtz equation with the gradient of φ; is equivalent to

#### ½∇(∇φ·∇φ) = −½f(z)∇φ² or, eliminating the ½ ∇(∇φ·∇φ) = −f(z)∇φ²

The integration of this equation from any point z0 where φ=0 to the point Z where ∇φ=0 gives

#### 0 − ∇φ·∇φ = −∫z0Z[f(z)(φ²(Z)]dz

This integration is along the trajectory of the particle of the system. Let the position of the particle along the path be denoted by s.

Since f(z) is positive the Extended Mean Value Theorem may be used to express the integration as

#### 0 − ∇φ·∇φ = −[f(z)(φ²(Z) −0]

where z is somewhere in the interval of the path of the particle from z0 to Z.

The quantity ∇φ·∇φ is the square of the magnitude of ∇φ, |∇φ|², so:

#### 0 − |∇φ(z0)|² = −[f(z)(φ²(Z) −0]

The end points of the integration are from z0 where dφ/ds is an maximum to Z where φ, hence also φ², is a maximum.

This means that

#### (φ²(Z)max = |∇φ(z0)|²/[f(z)] and hence upontaking the square roots of both sides φ(Z)max = |∇φ|max/[f(z)]½

This is the property of the solutions to a generalized Helmholtz equation; i.e., that value of the solution along a path are inversely proportional to the square root of the coefficient in the equation.

## The Interval between a Point Where φ Equals Zero and Where φ is a Maximum

If the coefficient in the Helmholtz equation is constant then the solution is sinusoidal; i.e., φ(s)=A·sin(f½s). The wavelength L of such is solution is given by

#### f½L = 2π and hence L = 2π/f½

The distance from a point where φ equals zero to a maximum for φ is one fourth of the wavelength L.

## Application to the Time Independent SchrödingerEquation for a Particle in a Potential Field

The relevant Schrödinger equation is

#### ∇²φ = −βK(z)φ(z)

where z represents the vector of coordinates, β is a constant and K(r) is the kinetic energy of the particle.

It was found above that the maximum of φ²(z) in an interval between values of z such that φ²(z) is equal to zero is given by

#### (φ²(Z)max = |∇φ(z0)|²/[βK(z)]

The probability of finding the particle in that interval is the integral of φ²(z) from z0 to Z.

The average probability density is approximately one half of the value of the maximum probability density. A closer approximation can be achieved by allowing

#### φ(s) = A·sin((βK(s))½s) and consequently φ²(s) = A²·sin²((βK(s))½s)

The probability of finding the particle in the interval from z0 to Z is then

#### ∫z0ZA²·sin²((βK(z))½s)dz which can be reduced to A²∫z0Zsin²((βK(z))½z)dz

The integral of sin²(x) from x=0 to π/2 is equal to 1/4. Thus the probability of finding the particle in the interval z0 to Z is A²/4 and the average probability density is then

#### [A²/4]/(L/4)

(To be continued.)