﻿ The Oscillation of a Particle in a Potential Field, Energy Quantization and the Uncertainty Principle
San José State University

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Thayer Watkins
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The Oscillation of a Particle
in a Potential Field,
Energy Quantization and
the Uncertainty Principle

The purpose of this material is to show that a particle moving in a field with a potential energy function undergoes an oscillation that can be associated with a probability distribution that satisfies the Uncertainty Principle. This means that the satisfaction of the Uncertainty Principle does not mean there is any indeterminacy of the particle itself.

Let V(x) be the potential energy function for a particle of mass m. The total energy of the particle is then

#### E = ½m(dx/dt)² + V(x)

It is presumed that the origin of the coordinate system is chosen such that V(0) is the minimum of V(x). This means that V'(0)=0 and V"(0)≥0.

The dynamics of the particle are found by solving the energy function for (dx/dt); i.e.,

#### (dx/dt) = (2(E−V(x))/m)½

Now consider a solution where x is small and the restoring force on the particle is closely approximated by −kx, with k=V"(0). In other words the potential function is taken to be as a Maclaurin series

#### V(x) = V(0) + V'(0)x + ½V"(0)x² + higher order terms

Since V'(0)=0 and V(0) can be taken to be zero, the potential function for small levels of energy reduces to

#### V(x) = ½kx²

with k=V"(0) and k>0.

Since the restoring force is equal to −kx the dynamics of the particle's trajectory satisfies

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

This equation has a solution of the form

#### x(t) = A·cos(2πω(t−t0)

where A and t0 are constants and ω=(k/m)½.

By the proper choice of the initial time t0 can be made equal to zero. A maximum deviation from equilibrium xM occurs at t=0 so the solution can be expressed as

## The Time Spent in the Allowable Locations

The time spent in an interval dx at x is equal to dx/|v(x)|. The probability of finding the particle in the interval dx at x at a randomly chosen time is then proportional to dx/|v(x)|. Thus the probability density function for x is

#### P(x) = (1/|v(x)|)/T

where T is the total time spent in a cycle. This works out to be

#### P(x) = (1/π)[1/(xM²−x²)½]

The expected value for this probability density function is zero and its variance is

#### σx² = xM²/(2π) and hence σx = xM/(2π)½

For the details of the calculation see Harmonic Oscillator.

This is the shape of the probability distribution. ## 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 the particle under consideration 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. 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 the formulas derived above (which are the same as for a harmonic oscillator) the result is

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

With the quantization of energy the minimum energy of the particle is 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 particle in a potential field. The satisfaction of the Uncertainty Principle thus does not imply any intrinsic indeterminism of the system.