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The Asymptotic Limits of the Spatially
Averaged Probability Distributions from
the Solution to the Time-Independent
Schrödinger Equation are Equal to the
Time-Spent Probability Distribution
from Classical Analysis

To prove the proposition involved in the title it will be shown that that wave equation which is the solution to the time-independent Schrödinger equation for a particle moving in a potential field is the product of two functions, one of which corresponds to the Classical time-spent probability distribution for the particle and the other is asymptotically equal to a solution to the Helmholtz equation. The solution to the Helmholtz equation consists of sinusoidal fluctuations the spatial averages of which are constant.

Consider a particle of mass m moving in space subject to a potential function V(z), such that V(0)=0, V(−z)=V(z) and V"(z)≥0 where z is the point location coordinates of the particle. The time-independent Schrödinger equation for the wave function φ(z) for this physical system is

(−h²/2m)∇²φ(z) + V(z)φ(z) = Eφ(z)

where h is Planck's constant divided by 2π and E is the energy of the system. It can be reduced to

∇²φ = −μK(z)φ(z)

where μ=(2m/h²) and K(z)=E−V(z), the kinetic energy of the system as a function of particle location. This is an example of what the K(r) might look like.

Note that the above equation may also be expressed as

∇²φ = −μE(1−V(z)/E)φ(z)

This suggests that the variation in the energy E relative to the potential V(z) is important in determining the profile of φ(z) and hence of the probability density φ(z)². Let V(z)/E be denoted as U(z). Then instead of thinking of the issue being what happens to φ(z) as E increases without bound, it is what happens to φ(z) as U(z)→0 for all z. But first it is necessary to find a way to deal with the rapid oscillations in φ(z). Here is an example of φ²(z) for 1D space. It is for a harmonic oscillator, where V(z)=½kz².

What happens when E increases is not so much that the level of φ(z) increases but instead the density of the fluctuations increases. The range over which φ(z)² is nonzero may also increase.

By ignoring the constant factors the equation for the wave function can be reduced to

∇²φ = −(1−U(z))φ(z)

where φ²(z) must be normalized.

The Classical Model

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

E = ½mv² + V(z)


v = (2/m)½(E−V(z))½

The Time-Spent
Probability Density Distribution

For a particle executing a periodic trajectory the time spent in an interval ds of the trajectory is ds/|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(z) = 1/(T|v(z)|)

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

P(z) = [(m/2)½/T]/(E½(1−U(z))½)

The constant factor of (m/2)½/ E½ is irrelevant in determining P(z) because it is also a factor of T and thus cancels out.

The time-spent probability distribution is thus inversely proportional to (1−U(z))½.

It is convenient for typographic reasons to represent (1−U(z)) as H(z). H(x) is proportional to kinetic energy and particle velocity is proportional to (H(x))½, as is also momentum p. Therefore the time-spent probability density function is inversely proportional to (H(z))½.

The Asymptotic Limit of the
Quantum Theoretic Solution

The quantum theoretic equation for the wave function is:

∇²φ = −μE(1 − U(x))φ
or, equivalently
= − μEH(x)φ(x)

Now let λ(z) be defined by

φ(z) = λ(z)μE(H(x))−¼

Again for typographic convenience let μE(H(x))−¼=J(x)−¼ and thus J(x)=(μE)4H(x). Therefore

φ(z) = λ(z)(J(x))−¼

A Mathematical Property
of the Laplacian ∇²

The Laplacian ∇² of the product of two functions f·g is given by

∇²(f·g) = (∇²f)g + 2(∇f)·(∇g) + f(∇²g)



∇²φ =(∇²λ)(J−¼) + 2(∇λ)·∇(J−¼) + λ(∇²(J−¼))

Note that

∇(J−¼) = −(1/4)(J−5/4)∇J(z)
∇²(J−¼) = −(1/4)(J−5/4)∇²J(z) + (5/16)(J−9/4)(∇J(z))² − (1/4)(J−5/4)∇²(J(z))


∇²φ = − J(z)φ(z) = − J(z)λ(z)J−¼
= − λ(z)J¾(z)


(∇²λ)(J−¼) − 2(1/4)J−5/4)(∇λ)·(∇J) + λ(z)[−(1/4)(J−5/4)∇²J(z) + (5/16)(J−9/4)(∇J(z))² − (1/4)(J−5/4)∇²(J(z))]
= − λ(z)J¾(z)

Multiplying through by J¼(z) gives

(∇²λ) − (1/2)J−1)(∇λ)·(∇J)
+ λ(z)[−(1/4)(J−1)∇²J(z) + (5/16)(J−2)(∇J(z))² − (1/4)(J−1)∇²(J(z))]
= − λ(z)J(z)

Note that

∇J(z) = −(μE)4∇V(z)/E = −μ4E3∇V(z)
∇²J(z) = −∇²(μE)4V(z)/E = −μ4E3∇²V(z)

and ∇V(z) and ∇²V(z) are fixed as E→∞. Therefore all of the terms on the LHS of the previous equation above except (∇²λ) go to zero as E increases without bound. They approach zero because they have a derivative of J in their numerators involving E³ and a power of J in their denominators involving a power of E4. Furthermore H(z) asymptotically approaches 1 as E→∞. Thus λ(z) asymptotically approaches the solution to the equation

∇²λ(z) = −(μE)4λ(z)

This is a Helmholtz equation. Its solutions in Cartesian coordinate systems are sinusoidal. The values of λ² then oscillate between a maximum value and zero. λ(x) could called the flutter function for the system. Its spatial average is a constant Helmholtz equation averages.

The fact that the coefficient of the Helmholtz equation is proportional to the fourth power of energy means that as E increases the solution very quickly goes to ultradense fluctuations of probability density such that any degree of spatial averaging results is a close approximation of the Classical time-spent probability density distribution.

So the probability density λ(z)² generally consists of a function which oscillates between relative maxima and zero values. The spatial average of that function is a constant. Therefore the probability densities are inversely proportional to J(x)½=(1-V(x)/E)½ just as the classical time-spent probabilities are.

Here is an illustration of J(x), J(x)½, and 1/J(x)½ for the one dimensional case of a harmonic oscillator.

What was shown above is that the wave function that is the solution to the equation

∇²φ(x) = −J(x)φ(x)

can be factored as follows

φ(x) = λ(x)(J(x))¼

where λ(x) is a purely oscillatory function which is asymptotically equal to the solution to a Helmoltz equation. The other factor, (J(x))¼, is the wave function associated with the time-spent probability distribution for a particle moving in a potential field.

The time-spent probability distributions are for particles that maintain their physical existence. Thus there is no justification for the assertion by the Copenhagen Interpretation of quantum theory that particles do not have a physical existence until their characteristics are measured. In effect, the time-independent Schrödinger equation give the dynamic appearance of a physical system rather than its static appearance. The Copenhagen Interpretation treats the solurion to Schrödinger equation as if it were the static condition of the system. This is like treating the appearance of a rapidly rotating fan as if it is a unchanging translucent disk which is a single particle. It is simply wrong.


For the fundamental case of a particle moving in a potential field the spatial average of the probability densities coming from the solution of time-independent Schrödinger equation are equal to a product of two functions. One of those functions is the time-spent probability density function from a Classical analysis of the system. The other function is a purely oscillatory function derived from a solution to a Helmholtz equation. Thus the spatial averages of the probability densities derived from the time-independent Schrödinger equation are asymptotically equal to the probability densities of the time-spent distribution from Classical analysis.

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