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.

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

∫_L^HxP(x)dx
and
∫_L^Hx²P(x)dx
respectively.

Note that the variance of x is given by

E{x²} − (E{x})²

The quantum probability density function is approximately

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

Let the cumulative distribution be defined as

R(x) = ∫_L^xP(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.,

∫_L^HUdV = [UV]_L^H − ∫_L^HVdU

Now consider

∫_L^HP_Sdx = ∫_L^H((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

∫_L^HP_S = [((2cos²(λx))R(x)]_L^H − ∫_L^H R(x)(4cos(λx)sin(λx)λdx

Then

[((2cos²(λx))R(x)]_L^H = 1 − 0 = 1

If ∫_L^HP_S=1 then

∫_L^H R(x)(4cos(λx)sin(λx)λdx = 0
and hence
∫_L^H 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

E_S = ∫_L^HxP_Sdx = ∫_L^Hx((2cos²(λx))P(x)dx

Let

∫_L^xzP_S(z)dz = S(x)

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

Then

E_S(x) = [(2cos²(λx)]_L^H − 4λx)∫_L^H R(x)cos(λx)sin(λx)dx

Since ∫_L^H R(x)cos(λx)sin(λx)dx can be expected to be zero

E_S{x}= E{x}

By a similar argument

E_S{x²}= E{x²)

Thus the variances are equal

Var_S = Var
and likewise for the
standard deviations
σ_Sx = σ_x

The same analysis applied to the momentum of the system yields

σ_Sp = σ_p
and hence
σ_Sxσ_Sp = σ_xσ_p

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

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