﻿ The Wave Function from Schrodinger Equation and the Time-Spent Probability Density Function
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The Wave Function from
Schrodinger Equation and
the Time-Spent Probability
Density Function

Let H(p, x) be the Hamiltonian function for a physical system. If H does not explictly involve time then it is the same as total energy E, kinetic and potential. Then let H^(∇, x) be its Hamiltonian operator, where momentum p is replaced by ih∇, h is Planck's constant divided by 2π and i is the square root of negative one.

The time-independent Schrödinger equation for the system is the

#### H^(∇, x)φ(x) = Eφ(x)

For a particle of mass m in a potential field V(x), H^((∇x, x) is of the form

#### −h²(∇²/2m) + V(x)

Thus the time-independent Schrödinger equation is

where μ(x)=(2m)/

## The Time-Spent Probability Density Function

The probability density for a particle at point s of its trajectory when moving at a velocity v is given by

#### PTS(x) = 1/T|v(x)|

where T is the total time taken to execute the periodi trajectory.

By Hamiltonian analysis the velocity v(x) is given by

#### v(x) = (∂H/∂p)

For the Hamiltonian function of a particle is a potential field H=p²/2m + V(x)

#### (∂H/∂p) = 2p/(2m) = p/m = v

But for this case

#### E = ½mv² + V(x) so v(x) = [(2/m)(E−V(x)]½

This can be rewritten as

#### v(x) = (2E/m)½(1−V(x)/E)½

In general, velocity may be expressed as a function of kinetic energy K(x)=E−V(x)=E(1−V(x)/E).

Therefore the wavefunction σ(x) associated with the time-spent probability density function PTS(x) is given by

#### σ(x) = 1/(Tv(x))½ = (1/ET)½(1−V(x)/E)−½

Now let λ(x) be defined by

#### φ(x) = λ(x)σ(x) = λ(x)(1/ET)(1−V(x)/E)−¼

Now the Schrödinger equation

#### ∇²φ(x) = −μE(1 − V(x)/E)

must be evaluated and that starts with the Laplacian ∇².

The Laplacian is the divergence of the gradient of a scalar field; i.e.,

#### ∇²φ = ∇·(∇φ)

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

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

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

Therefore

Note that

Since

Therefore

#### (∇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

Note that

#### ∇J(z) = −(μE)4∇V(z)/E = −μ4E3∇V(z) and ∇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. This function λ(x) could called the flutter function for the system. Its spatial average is a constant. See 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.