﻿ The Asymptotic Equality of a Spatial Average of the Probability Density Function from Schroedinger's Equation for a Particle in a Potential Field and That of the Time-Spent Classical Version of the Same System
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 The Asymptotic Equality of a Spatial Average of the Probability Density Function from Schrödinger's Equation for a Particle in a Potential Field and That of the Time-Spent Classical Version of the Same System

A previous study established that a spatial average of the quantum mechanical probability density function for a harmonic oscillator near zero displacement is asymptotically equal to the classical probability density function of a harmonic oscillator at zero displacement. Asymptotic means as energy increases without bound the two converge. This webpage is to extend in a modified form that result to all values of displacement and to oscillators in any potential field.

Let V(x) be the potential energy function for a particle of mass m, where x is the displacement distance from equilibrium and let V(0)=0.

The kinetic energy K of the particle is

#### K = ½mv²

where v is the velocity of the particle. Therefore the total energy E of the system is

#### E = ½mv² + V(x)

Total energy remains constant over the cycle of the particle; therefore

#### v(x) = [2(E-V(x)/m]½or, equivalently v(x) = (2/m)½(E-V(x))½

The limits of the oscillation, xmin and xmax, are the solutions to the equation

## The Classical Probability Density Function

In classical analysis the trajectory of the particle is deterministic and so there is no probability involved in its determination. But a probability density function can be defined in terms of the probability of the particle being found in an interval at any randomly chosen time. The probability of the particle being found in an interval [x-½Δx, x+½Δx] is proportional to the amount of time the particle spends in the interval. The particle could be at displacement x twice in a cycle, as in the case of a harmonic oscillator. In such a case the amount of time spent in the interval is 2Δx/|v(x)|. The classical probability density for a harmonic oscillator is thus proportional to 2(m/2)½/(E-V(x))½. Pc(x) is then this quantity divided by the cycle time which is just

#### T = 2∫xminxmaxdz/|v(z)|

The factor of 2(m/2)½ appears in each 1/v(x) and in T. It is therefore eliminated.

Therefore

#### Pc(x) = 1/[T1(E-V(x))½] or, equivalently Pc(x) = 1/[T1K(x)½]

where K(x) is just the kinetic energy of the particle as a function of the particle's location and

#### T1 = ∫xminxmax dz/K(z)½

Note that this means the classical probability density function is independent of the mass of the particle. This is just an example of the fact that any constant factor of probability densities is eliminated in the process of normalization. For example, if the probability densities are proportional to αg(x) then T=∫αg(z)dz=α∫g(z)dz and hence

## The Quantum Mechanical Probability Density Function

The Hamiltonian function for the system of a particle of mass m in a potential field of V(x) is

#### H = ½p²/m + V(x)

where p is the momentum of the particle.

In this case H is the same as the total energy.

The time-independent Schrödinger equation for the system is then

#### −(h²/2m)(d²φ/dx²) + V(x)φ = Eφ

where E is total energy.

This equation can be put into the form

where

## The Oscillatory Character of the Wave Functions

The most salient characteristic of solutions to equations of the form d²φ/dx² = −k(x)²φ is their oscillation. When φ is positive then the slope of dφ/dx is decreasing and eventually will become negative and hence φ will ultimately become negative. However when φ is negative the slope of dφ/dx is positive and eventually dφ/dx will become positive and ultimately φ will become positive.

If K(x) were constant then k(x) would be constant and the solution to the equation would be

#### φ(x) = A·cos(kx)

The wavelength λ of such a solution is given by

#### kλ = 2π and thus λ = 2π/k

Thus as E increases without bound so does k and hence λ goes to zero.

Another significant implication of φ being approximated by A·cos(kx) is that the average value of φ² over an interval between two minima of φ², which happen to be zero, is one half of the maximum value of φ². This is because the average value of cos²(θ) from θ=π/2 to θ=3π/2 is (1/2).

The probability density function for the harmonic oscillator reveals how close the classical time-spent probability density function matches ½φ²max. It is obviously close everywhere except possibility near the end points where there are singularities in the classical function. However even near the end points the fit is close. In the above graph there are small red dots marking the points that are one half of the maximum for the lobes of the quantum probability density function. The match is within the accuracy of the graph. According to the proposition being considered the quantum probability density function does not have to match the classical probability density function. It is a matter of the classical values matching the spatial average of the quantum values.

## Derivation of a Property of the Wave Function and the Corresponding Probability Density Function

The equation for the wave function was put into the form

#### d²φ/dx² = −k²(x)φ

Now multiply each side of the equation by φ(x) to put the equation into the form

#### φ(d²φ/dx²) = −k²(x)φ²

.

Note that the probability density function φ²(x) has its extremes where

#### φ'(x)φ(x) = 0

The minima occur where φ(x)=0 and the maxima where φ'(x)=0.

Now note that

Thus

#### d(φdφ/dx)/dx − (dφ/dx)² = −k(x)²φ²

Let the minima and maxima be numbered and denoted as mj and Mj, respectively. The numbering system that is appropriate is from -N to +N, where N is the principal quantum number of the system.

When the above equation is integrated from mj to Mj the first term is zero. (The following analysis works equally for integrations from a maximum to a minimum, or from a minimum to a minimum or a maximum to a maximum.) The integration from a minimum to a maximum leaves

Now consider

#### dφ²/dx = 2φ(dφ/dx) and hence (dφ/dx) = ½(dφ²/dx)/φ and further (dφ/dx)² = ¼(dφ²/dx)²/φ²

Thus the previous equation reduces to

#### ¼∫mjMj[(dφ²/dx)²/φ²]dx = ∫mjMjk(x)²φ²dx

The quantity k²(x) varies relatively slowly compared to φ(x) so by an extension of the Mean Value Theorem

#### ∫mjMj k(x)²φ²dx reduces to k(x̅j)²∫mjMjφ²(z)dz

where x̅j is somewhere in the interval [mj, Mj].

The integral

#### ∫mjMjφ²dz

is the probability that the particle is in the interval [mj, Mj].

Likewise, by the extension of the Mean Value Theorem used above, the integral on the left of the previous equation can be replaced by

#### [1/φ²(xj)]∫mjMj(dφ²/dz)²dz

where xj is a value of x in the interval [mj, Mj].

Let

#### ∫mjMjφ(z)²dz be represented as φ(xj*)²δj

where δj is the length of the integration interval (Mj-mj) and xj* is a value of x in that interval.

Thus

#### ¼∫mjMj(dφ²/dz)²)dz/φ(xj)² = k(x̅j)²φ(xj*)²δjand hence φ(xj)²φ(xj*)² = ¼∫mjMj(dφ²/dz)²dz/δj]/k(x̅j)²

The values x̅j, xj and xj* are approximately equal; i.e., they are all in the interval [mj, . Mj] and this interval goes to zero as the energy E increases without bound. Also

#### [∫mjMj(dφ²/dz)²dz/δj]is equal to the average value of (dφ²/dz)² in the interval of integration.

Let the average value of (dφ²/dz)² in the interval be denoted as μj².

Thus, taking xj and xj* to be the same value,

#### (φ(xj)²)² = ¼μj²/k(x̅j)² or, upon taking the square root of each side φ(xj)² = ½μj/k(x̅j)

Note that this probability density φ(xj)² is an average over the interval of integration.

The graph below shows the quantum mechanical probability density function for a harmonic oscillator with V(x) equal to ½γx² where γ is a constant called the stiffness coefficient of the oscillator. The energy of the quantum level oscillator is proportional to an integer called the principal quantum number. For the case below the principal quantum number is 60. The principal quantum number determines the number of peaks in the probability density function. Over a wide range of values of x the probability density function is nearly the same and thus μ is nearly the same.

Presume that the μj's are approximately equal for all j, say μ. Then in the normalization of the probability density function the constant factor of ½μ is eliminated.

Since k(x) is proportional to K(x)½ then

#### φ(xj)² = 1/(T1K(xj)½) where T1 = ∫ dz/K(z)½

This is identical to the expression found for the classical probability density function. Thus the spatially averaged probability density function from Schrödinger's equation would be equal to the time-spent classical probability density function.

This has to be considered as an asymptotical result. In the derivation it was presumed that the three values of x, x̅j, xj and xj*, in the interval [mj, Mj] were the same. As the energy of the system increases without bound the length of the interval between the minimum for φ² and its maximum goes to zero and thus the three values of x must converge to the same value. In other words, the spatial average of the quantum mechanical probability density function is asymptotically equal to the classical probability density function for the same system based upon the time averaged location of the particle.

Although the values of μj are approximately constant over a limited range of values of j they are not equal for all values of j. Let μ be the average value of the μj's and let

Then

#### φ²(xj) = 2μ/k(x̅j) + 2Δμj/k(x̅j)

So the spatially average probability density function from the Schrödinger equation can be considered to be made up of the classical time spent probability density distribution plus perturbations which sum to approximately zero since ΣΔμj=0.

This means that the quantum mechanical probability density functions do not represent some pure indeterminacy of the particle as in the Copenhagen Interpretation but instead the proportion of the time a moving particle spends near the various points.

In a sense the quantum mechanical probability density distribution represents a quantization of allowable locations for the particle. The particle still moves but spastically from allowable location to allowable location. The time average of the particle velocity looks like the graph shown below. The unaveraged velocity would vary between infinity and zero.

This is in contrast to the smoothly varying velocity according to classical mechanics. ## Conclusions

The implication of the above is not just that the spatial average of the quantum mechanical probability density function is asymptotically equal to the classical probability density function. The classical probability density function is derived from a particle moving in a cycle. This means that the quantum mechanical probability density function also describes the time averaged behavior of a particle in motion. The quantum mechanical motion of the particle involves near infinitesimal pauses at discrete locations and then near instantaneous shifts to the next location. Even quantum mechanically the particle is not at all of the discrete locations simultaneously.

The proper interpretation of the wave function of Schrödinger's equation is then the time-spent probability density function. The analysis was limited to the one dimensional case, but it is rather implausible that there could be a drastic difference in the nature of reality for the one dimensional case in contrast to cases of more than one dimension.