﻿ Time Spent Probability Distributions, their Variances and their Satisfaction of the Uncertainty Principle
San José State University

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Thayer Watkins
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Time Spent Probability Distributions,
their Variances and their Satisfaction
of the Uncertainty Principle

Consider a particle traversing a periodic trajectory. Let s(t) and v(t) denote its path position and velocity, respectively, as functions of time t with

#### v(t) = (ds/dt)

The times dt and dτ spent in path and velocity intervals ds and dv are given by

#### dt = dx/v dτ = dv/(dv/dt)

The proportions of time spent probability distributions are then

#### PS(s) = 1/(TS|v(s)|) andPV(v) = 1/(TV|dv(s)/dt|)

where TS is the time period for executing the path; i.e.,

#### TS = ∫ ds/|v(s)|

The time period for the velocity cycle TV is defined as

Note that

Thus

#### TV = ∫dv/|(dv(s)/ds)v|

Consider a change in the variable of integration from v to s based on the relation dv=(dv/ds)ds

From this it follows that

#### TV = ∫dv/|(dv(s)/ds)v| = ∫(dv/ds)ds/|(dv(s)/ds)v| = ∫ ds/|v| hence TV = TS

which is what intuition would have told us.

## Formulation of a Model

Consider a particle of mass m moving in one dimensional space whose position is denoted as x. The potential field given by U(x) where U(0)=0 and U(−x)=U(x). Let v be the velocity of the particle, p its momentum E its total energy. Then

Thus

#### v = (2/m)½(E−U(x))½

For a particle executing a periodic trajectory the time spent in an interval dx of the trajectory is dx/|v|, where |v| is the absolute value of the particle's velocity. Thus the probability density of finding the particle in that interval at a random time is

#### P(x) = 1/(Tv(x))

where T is the total time spent in executing a cycle of the trajectory; i.e., T=∫dx/|v|a. It can be called the normalization constant, the constant required to make the probability densities to sum to unity. Thus

#### P(x) = [(m/2)½/T]/(E−U(x))½

It is convenient to represent (E−U(x)) as K(x), the kinetic energy expressed as a function of the location. Therefore

#### P(x) = [(m/2)½/T]/(K(x))½

The constant factor (m/2)½ is irrelevant in determining P(x) because it is also a factor of T and thus cancels out.

The variance of P(x) is defined as

#### VarP(x) = ∫(x−x)²P(x)dx

where x is the mean value of x, ∫xP(x)dx. For the case being considered x is zero, so

#### VarP(x) = ∫(x²/(T(K(x)½)dx

The Uncertainty Principle also involves the variance of momentum and that is determined by the variance of velocity. As noted previously, the time the particle spends in a velocity interval dv is the interval length divided by the acceleration of the particle; i.e.,

#### dt = dv/(dv/dt) = dv/[(dv/dx)(dx/dt)] = dv/[(dv/dx)v]

Since K(x) is equal to E−U(x) and v(x) = (2/m)½(K(x))½) it follows that

#### (dv/dx) = −½(−dU/dx)/[T(K(x))½] which reduces to (dv/dx) = ½(dU/dx)/(TK(x)½)

The mean value of the velocity is zero. The variance of velocity is therefore

#### VarP(v) = ∫ [v²/(T((dv/dx)v))|dv which reduces to VarP(v) = ∫ |v/(T(dv/dx))|dv and hence to VarP(v) = ∫ [K(x)½/(T(dU/dx))]dv

So the variance of velocity is directly dependent on (K(x))½ whereas the variance of location is inversely dependent on (K(x))½. Anything that increaes K(x) decreases the variance of x but increases the variance of v. Thus the product of the variances is subject to the contrary influences due to any changes in K(x). Likewise the product of the variance of x and the variance of momentum is subject to those contrary influences.

## Harmonic Oscillators

The levels of K(x) depends upon the total energy and the potential energy function U(x). The potential energy function can be considered by its Taylor's series

#### U(x) = U(0) + (dU/dx)0x + ½(d²U/dx²)0x² + …

It is assumed that U(0)=0. As the level of energy E goes to zero the system under consideration converges to a harmonic oscillator with a stiffness coefficient k equal to (dU/dx) at x equal to zero. From the previous analysis it was found that σx=xmax/(2π)½ where xmax is the maximum deviation of the oscillator from equilibrium. That maximum deviation is where all of the energy is potential and none of it is kinetic; i.e., xmax=(2E/k)½. Likewise the standard deviation of velocity is

#### σv=vmax/(2π)½

The maximum velocity vmax occurs where all of the particle's energy is kinetic and none of it is potential; i.e., at x=0 so vmax=(2E/m)½. Thus

But

#### σp = mσv = (2mE)½/(2π)½

Thus the product of the standard deviations of displacement and momentum for a harmonic oscillator is equal to

#### σpσx = [ (2mE)½/(2π)½][ (2E/k)½/(2π)½ = 2E(m/k)½/(2π) which can be expressed as σpσx = (E/(k/m)½)/π

It has already been established that the harmonic oscillator has a minimum quantum of energy equal to hω, where ω is the frequency of the oscillator and h is Planck's constant. The frequency of a harmonic oscillator is equal to (k/m)½ and hence the minimum E divided by the frequency (k/m)½ is equal to h, Planck's constant.

Thus, from the above

#### σpσx = h/π

In order to sastisfy the uncertainty relation that product has to be greater than or equal to h/(4π). Thus the product of the standard deviations of displacement and momentum for a harmonic oscillator exceeds the required amount for a factor of 4.