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The Wave Function from Schrodinger Equation and the Time-Spent Probability Density Function |
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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 i~~h~~∇, ~~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

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

Thus the time-independent Schrödinger equation is

which is equivalent to

∇²φ(x) = −μ(E − V(x))

where μ(x)=(2m)/~~h²~~

Density Function

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

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

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

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

But for this case

so

v(x) = [(2/m)(E−V(x)]

This can be rewritten as

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 *P*_{TS}(x) is given by

Now let λ(x) be defined by

Now the Schrödinger equation

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

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

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

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

Therefore

Note that

and

∇²(J

Since

and hence

∇²φ = − λ(z)J

Therefore

= − λ(z)J

Multiplying through by J^{¼}(z) gives

+ λ(z)[−(1/4)(J

= − λ(z)J(z)

Note that

and

∇²J(z) = −∇²(μE)

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 E^{4}.
Furthermore H(z) asymptotically approaches
1 as E→∞. Thus λ(z) asymptotically approaches the solution to
the equation

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.

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