San José State University |
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The Asymptotic Limit of theProbability Distribution of a Particle Moving in a Potential Field |
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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

where μ=m/(2~~h~~²) 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

Thus the constant factor of μE can be eliminated and the resulting equation will the same normalized probability density function as the original equation.

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.

First however, 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

where φ²(x) must be normalized.

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

Thus

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

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. This is the time-spent
probability density distribution.

Thus

As noted previously, It is convenient to represent (E−V(x)) as K(x), the kinetic energy expressed as a function of the location. Therefore

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.

where S is the normalization constant.

Quantum Theoretic Solution

Define λ(x) by

and hence

(dφ/dx) = (dλ/dx)(K(x))

and furthermore

(d²φ/dx²) = (d²λ/dx²)(K(x))

+ (5/16)λ(x)(K(x))

which reduces to

(d²φ/dx²) = (d²λ/dx²)(K(x))

+ (5/16)λ(x)(K(x))

Since

then

+ (5/16)λ(x)(K(x))

= (d²λ/dx²)(K(x))

+ (5/16)λ(x)(K(x))

Multiplying through by (K(x))^{¼} gives

+ (5/16)λ(x)(K(x))

= − K(x)λ(x)

Since

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 thefinite-valued 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

which is

λ(x) = A·cos(E

where A and b are constants.

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

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}, the same as the
time-spent probability distribution from classical analysis.

Now denote (1−V(x)/E) as J(x) and define λ(x) by

and hence

(dφ/dx) = (dλ/dx)(J(x))

and furthermore

(d²φ/dx²) = (d²λ/dx²)(J(x))

+ (5/16)λ(x)(J(x))

which reduces to

(d²φ/dx²) = (d²λ/dx²)(J(x))

+ (5/16)λ(x)(J(x))

Since

then

+ (5/16)λ(x)(J(x))

= (d²λ/dx²)(J(x))

+ (5/16)λ(x)(J(x))

Multiplying through by (J(x))^{¼} gives

+ (5/16)λ(x)(J(x))

= − J(x)λ

Since

and

(d²J/dx²) = − (d²V/dx²)/E

for fixed V(x) as E→∞, (dJ/dx) and (d²J/dx²) go to zero as E increases without bound. Thus all of the terms on the LHS of the above equation except for (d²λ/dx²) go to zero doubly fast as E→∞, because of of the derivative of J in their numerator and the power of J in their denomerator. Therefore (d²λ/dx²)→ − J(x)λ. But as E→∞, J(x)→1. Therefore asymptotically λ(x) goes to the solution of the equation

which is

λ(x) = A·cos(x−b)

where A and b are constants.

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

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 (J(x))^{½}=(1−U(x))^{½}, the same as the
time-spent probability distribution from classical analysis.

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.

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