﻿ Means and Variances of Probability Distributions, Quantum and Classical
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Means and Variances of
Probability Distributions,
Quantum and Classical

This material is to introduce a topic in simplified form that will later be analyzed rigorously. Consider the probability density distribution for a harmonic oscillator as derived from time independent Schrödinger equation. 2

The heavy line is for the time-spent probability density function derived from the classical analysis of a harmonic oscillator. As can be seen the spatial average of the quantum mechanical probability density function is the classical distribution function.

Let P(x) be the time-spent distribution function ranging from L up to H. Let E{x} and E{x²} denote

#### ∫LHxP(x)dx and ∫LHx²P(x)dx respectively.

Note that the variance of x is given by

#### E{x²} − (E{x})²

The quantum probability density function is approximately

#### PS(x) = (2cos²(λx))P(x) where λ is such that 2cos²(λH) = 1

Let the cumulative distribution be defined as

#### R(x) = ∫LxP(z)dz

Note that R(L)=0 and R(H)=1.

Use will be made of the formula for integration by parts for a definite integral; i.e.,

Now consider

#### ∫LHPSdx = ∫LH((2cos²(λx))P(x)dx

In the integration by parts formula let U=(2cos²(λx)) and dV=P(x)dx. Thus V=R(x). Then

Then

If ∫LHPS=1 then

#### ∫LH R(x)(4cos(λx)sin(λx)λdx = 0 and hence ∫LH R(x)cos(λx)sin(λx)dx = 0

This is eminently reasonable and established rigorously in Asymptotic limit of a sine integral. The quantity cos(λx)sin(λx) is equivalent to sin(2λx) and the rapid fluctuation of this quantity between positive and negative values eliminates any accumulation of nonzero values.

Now consider

Let

#### ∫LxzPS(z)dz = S(x)

Note that S(L)=0 and S(H)=E(x).

Then

#### ES(x) = [(2cos²(λx)]LH − 4λx)∫LH R(x)cos(λx)sin(λx)dx

Since ∫LH R(x)cos(λx)sin(λx)dx can be expected to be zero

#### ES{x}= E{x}

By a similar argument

#### ES{x²}= E{x²)

Thus the variances are equal

#### VarS = Var and likewise for the standard deviations σSx = σx

The same analysis applied to the momentum of the system yields

#### σSp = σpand hence σSxσSp = σxσp

Thus the classical distributions either satisfy the uncerainty principle together or fail it together.