Consider a particle of mass m moving in a one dimensional space subject to a potential function V(x), such that V(0)=0 and V(−x)=V(x). The time-independent Schrödinger equation for the wave function φ(x) for this physical system reduces to

(d²φ/dx²) = −μK(x)φ(x)

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

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

(d²φ/dx²) = −μE(1−V(x)/E)φ(x)

This indicates that it is the variation in the energy E relative to the potential V(x) that is important. Let V(x)/E be denoted as U(x). Then instead of thinking of the issue being what happens to φ(x) as E increases without bound, it is what happens to φ(x) as U(x)→0 for all x. But first it is necessary to find a way to deal the rapid oscillations in φ(x). Here is an example of φ²(x). It is for a harmonic oscillator, where V(x)=½kx².

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

The equation for the wave function can be reduced to

(d²φ/dx²) = − K(x)φ(x) = −(1−U(x))φ(x)

where φ²(x) must be normalized.

The Classical Model

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

E = ½mv² + V(x)

Thus

v = (2/m)^½(E−V(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−V(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 time-spent probability distribution is thus inversely proportional to (K(x))^½. It may also be represented as being inversely proportional to (1−V(x)/E)^½, or equivalently (1−U(x))^½, where U(x)=V(x)/E.

The Asymptotic Limit of the
Quantum Theoretic Solution

Define λ(x) by

φ(x) = λ(x)(K(x))^−¼
and hence
(dφ/dx) = (dλ/dx)(K(x))^−¼ − λ(x)[¼(K(x))^−5/4(dK/dx)]
and furthermore
(d²φ/dx²) = (d²λ/dx²)(K(x))^−¼ − 2(dλ/dx)[¼(K(x))^−5/4(dK/dx)]
+ (5/16)λ(x)(K(x))^−9/4(dK/dx)² − ¼λ(x)(K(x))^−5/4(d²K/dx²)
which reduces to
(d²φ/dx²) = (d²λ/dx²)(K(x))^−¼ − ½(dλ/dx)K(x))^−5/4(dK/dx)
+ (5/16)λ(x)(K(x))^−9/4(dK/dx)² − ¼λ(x)K(x))^−5/4(d²K/dx²)

Since

(d²φ/dx²) = −K(x)φ(x) = −K(x)[λ(x)(K(x))^−¼]

then

(d²λ/dx²)(K(x))^−¼ − ½(dλ/dx)K(x))^−5/4(dK/dx)
+ (5/16)λ(x)(K(x))^−9/4(dK/dx)² − ¼λ(x)K(x))^−5/4(d²K/dx²)
= (d²λ/dx²)(K(x))^−¼ − ½(dλ/dx)K(x))^−5/4(dK/dx)
+ (5/16)λ(x)(K(x))^−9/4(dK/dx)² − ¼λ(x)K(x))^−5/4(d²K/dx²) = − λ(K(x))^¾

Multiplying through by (K(x))^¼ gives

(d²λ/dx²) − ½(dλ/dx)K(x))⁻¹(dK/dx)
+ (5/16)λ(x)(K(x))⁻²(dK/dx)² − ¼λ(x)K(x))^{− 1}(d²K/dx²)
= − K(x)λ

Since

(dK/dx) = − (dV/dx)
and
(d²K/dx²) = − (d²V/dx²)

for fixed V(x) as E→∞, (dK/dx) and (d²K/dx²) are finite. Thus all of the terms on the LHS of the above equation except for (d²λ/dx²) go to zero as E→∞, because of of the derivative of K in their numerator is finite and there is a power of K in their denomerator. Therefore (d²λ/dx²)→ − K(x)λ. But as E→∞, K(x)→E. Therefore asymptotically λ(x) goes to the solution of the equation

(d²λ/dx²) = − Eλ(x)
which is
λ(x) = A·cos(E^½x−b)

where A and b are constants.

This means that the probability density (φ(x))² asymptotically approaches

A²cos²(E^½x−b)/(K(x))^½

Eliminating the cos² functions is equivalent to taking the spatial average of the probability density functions. Thus the spatial average of the quantum theoretic probability density function for a particle moving in a potential field is asymptotically inversely proportional to (K(x))^½=(1−U(x))^½, the same as the time-spent probability distribution from classical analysis.

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. The Danish dragon that ate up quantum reality and spit out entanglements is dead.