The previous webpage established that the ratio of the wave-function from Schrödinger's time-independent equation to the square root of the time-spent probability density function from classical analysis is asymptotically equal to a function which is purely oscillatory. This material show s that the product of a function satisfying a generalized Helmholtz equation and the square root of the time-spent probability density function satisfies the time-independent Schrödinger equation for the physical system.

For a physical system consider a particle of mass m moving in space subject to a potential function V(z), such that V(0)=0 and V(−z)=V(z) where z is the location coordinates of a point. 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 φ is the wave function for the particle; i.e., |φ(z)|² is the probabability density at z. 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= (m/(2h²) 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.

However, in the determination of probability distributions constant factors are irrelevant because in the normalization process they cancel out. Thus note that the above equation may also be expressed as

∇²φ = −μE(1−V(z)/E)φ(z)

This indicates that it is the variation in the energy E relative to the potential V(z) that is important. 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 also increases.

By eliminating the irrelevant 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 and V(−z)=V(z). Let v be the velocity of the particle, p its momentum and E its total energy. Then

E = ½mv² + V(z)

Thus

v = (2/m)^½(E−V(z))^½

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=∫dx/|v|. It can be called the normalization constant, the constant required to make the probability densities to sum to unity. This is the time-spent probability distribution for the particle. Thus

P(z) = [(m/2)^½/T]/(E^½(1−U(z))^½)

The constant factor of (m/2)^½/TE^½ 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 to (1−V(z)/E)^½, or equivalently (1−U(z))^½.

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

The Asymptotic Limit of the
Quantum Theoretic Solution

By eliminating the irrelevant constant factors the equation determining the quantum theoretic wave function can be reduced to

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

with, as before, J(x)=(1−U(x))

Now let λ(z) be defined by

∇²λ(z) = −Ω(z)λ(z)

Ω(z) is a positive valued real function.

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

∇²(fg) = (∇²f)g + 2(∇f)·(∇g) + f(∇²g)

Therefore

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

Since

∇²λ(z) = −Ω(z)λ(z)

it follows that

∇²φ =(−Ωλ)(J^−¼) + 2(∇λ)·∇(J^−¼) + λ(∇²(J^−¼))
and hence
∇²φ = −Ω(z)φ(z) + 2(∇λ)·∇(J^−¼) + λ(∇²(J^−¼))

Note that

∇(J^−¼) = −(1/4)(J^−5/4)∇J(z)
and
∇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))

Note that

∇J(z) = −∇V(z)/E
and
∇²J(z) = −∇²V(z)/E

and ∇V(z) and ∇²V(z) are fixed as E→∞. Therefore all of the terms on the RHS of the above equation for ∇² except −Ω(z)φ(z) go to zero as E increases without bound. Furthermore J(z) asymptotically approaches 1 as E→∞. Thus φ(z) asymptotically approaches the solution to the equation

(∇²φ) = −Ωφ(z)

This is the equation that the solution to the time-independent Schrödinger equation satisfies.

The time-spent probability distributions are for a particle that maintains its 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.

Conclusions

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 asymptotically equal to the probability densities of the time-spent distribution from classical analysis.

There is no justification for the assertion in the Copenhagen Interpretation that particles generally do not exist materially. Effectively, except for its true believers, the Copenhagen Interpretation of quantum theory is demonstratively invalid.