﻿ The Uncertainty Principle and Harmonic Oscillators
San José State University

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Thayer Watkins
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The Uncertainty Principle and
Harmonic Oscillators

This is an application of the concept of Heisenberg's Uncertainty Principle to a classical system. A classical system is deterministic and does not inherently involve probabilities. However for a system that goes through a cycle the time spent in the allowable states is in the nature of a probability distribution. It represents the probability of finding the system in its various states at a randomly chosen time. The system utilized for this application is a harmonic oscillator.

## A Harmonic Oscillator

A harmonic oscillator is a system with an equilibrium and such that the restoring force for a displacement from the equilibrium is proportional to the displacement. Let x be the displacement. Then

#### F = m(d²x/dt²) = −kx and hence (d²x/dt²) = −(k/m)x

The parameter k is called the stiffness coefficient. The solution to the above equation is

#### x(t) = A·cos(2πωt) + B·sin(2πωt)

where A and B are constants and ω, the frequency, is equal to (k/m)½. For now the important point is the formula for the frequency; i.e., ω=(k/m)½.

## The Time-Spent Probability Distribution

The time spent in an interval dx is dx/|v(x)| where v(x) is the velocity at point x. Thus the probability density is inversely proportional to the velocity v(x). The velocity is found from the total energy function

#### E = ½mv² + ½kx²

where ½mv² is the kinetic energy and ½kx² is the potential energy. The velocity is then

#### v(x) = [(2E−kx²)/m]½

When the term (k/m)½ is factored out of the above expression the result is

#### v(x) = (k/m)½[(2E/k)−x²]½

The extreme displacements, ±xm, are where the entire energy is potential and none kinetic; i.e.,

#### E = ½kxm² and hence xm = (2E/k)½

Thus the expression for velocity takes the form

#### v(x) = (k/m)½[(xm²−x²]½

Now it can be noted that the probability density is inversely proportional to [(xm²−x²]½. The displacement ranges from −xm to +xm. The normalizing factor T for the probability density function is then

#### T = ∫−xmxmdx/[(xm²−x²]½

The variable of integration can be changed to z=x/xm to obtain

#### T = ∫−1+1dz/(1−z²)½

The change z=sin(θ) gives dz=cos(θ)dθ and (1−z²)½=cos(θ) so

#### T = ∫−π/2π/2cos(θ)dθ/cos(θ) = ∫−π/2π/2dθ T = (π/2) − (−π/2) = π

The probability density function P(x) for displacement is then

#### P(x) = (1/π)[1/(xm²−x²]½

Here is its shape. From this distribution the variance of displacement σx² may be computed. Since the average value of x is zero this takes the form of

#### σx² = ∫−xmxmx²dx/[(xm²−x²]½

By the same changes of variable as was used in determining the normalizing factor the formula for variance reduces to

## The Probability Density Function for Velocity

The time the system spends in a velocity interval dv is given by

#### dt = dv/|(dv/dt)| = dv/|a(v)|

where a(v) is acceleration. For a harmonic oscillator the acceleration is given by

#### a = F/m = −kx/m = −(k/m)x

Thus the probability density for velocity is inversely proportional to the magnitude of displacement. There is a perfect symmetry between displacement and velocity for a harmonic oscillator. The displacement as a function of v is given by

#### x(v) = [(2E−mv²)/k]½

The extreme velocities occur where all of the energy is kinetic and none is potential; i.e.,

#### E = ½mvm² and hence vm = (2E/m)½

Velocity goes through a cycle from 0 to vm and then back down again to 0 and decreasing to −vm before increasing back to 0. All of the values of velocity between −vm and +vm are covered. The probability density function Q(v) for velocity is then

#### Q(v) = (1/S)[1/(vm²−v²]½

where S is the normalizing factor. It is given by

#### S = ∫−vmvmdv/[(vm²−v²]½

This is identical to the evaluation of the normalizing factor T for displacements. Thus the value of S is π. Thus

#### Q(s) = (1/π)(1/[(vm²−v²]½) and hence σv² = vm²/(2π) and σv = vm/(2π)½

This is the shape of the probability distribution for particle velocity. Since the momentum p of the particle is equal to mv

## The Uncertainty Relation

The Uncertainty Principle requires that

#### σxσp ≥ h/(4π)

When σx and σp are replaced by their formulas in a harmonic oscillator; i.e.,

the result is

#### σxσp = (2E/(k/m)½)/(2π) and since (k/m)½=ωthis reduces to σxσp = (1/π)(E/ω)

The minimum energy of the oscillator equal to hω and therefore the expression (E/ω) is equal to Planck's constant h and hence

#### σxσp = h/π = 4(h/(4π))

Thus the Uncertainty Principle is satisfied by the time-spent probability distributions for displacement and velocity of a harmonic oscillator. Therefore the satisfaction of the Uncertainty Principle does not imply any intrinsic indeterminism for the particle of the system.