﻿ The Evolution of the Wave Function of an Electron according to the Time-Dependent Schroedinger Equation
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 The Evolution of the Wave Function of an Electron according to the Time-Dependent Schroedinger Equation

## Background

The conventional quantum theoretic analysis of an electron in a hydrogen atom makes use of the separation-of-variables assumption; i.e.; that the wave function Φ in spherical coordinates (r, ψ, θ) for the electron is of the form

#### Φ(r, ψ, θ) = R(r)Ψ(ψ)Θ(θ)

This results in mathematically correct solutions to the time-independent Schroedinger equation which are physically irrelevant because the separation-of-variables assumption is incompatible with particleness. The only other alternative to the quantum analysis of an electron is to numerically solve the time-dependent Schroeding equation.

Consider first the case of polar coordinates (r, θ). Let H(pr, pθ, r, θ) be the Hamiltonian function for the system under consideration where pr and (pθ are the radial and angular momenta, respectively. Let ih be the square root of negative one times Planck's constant divided by 2π. With the powers of the momenta converted to ih times derivatives with respect to r and θ H becomes the Hamiltonian operator, H^.

The time-dependent Schroedinger equation for the wave function Φ(r, θ) is

#### ih(∂Φ/∂t) = H^Φ

It is notablle that this equation involves only the first derivative with respect to time and thus only one initial condition is involved rather than two such as intitial position and initial velocity.

The Hamiltonian function for the electron is of the form

#### H = p²/(2m) + V(r)

where p is momentum and V(r) is the potential function.

Thus the Hamiltonian operator for the system is

#### H^ = −(h²/(2m))∇² + V(r)

where ∇² is the Laplacian for the coordinate system.

## The Free Electron

The free electron is the case in which V(r) is identically zero. Let us consider this case with a simply two dimensional coordinate system of (x, y) in which

#### ∇² = (∂²/∂x²) + (∂²/∂y²)

With V(r) equal to zero the time-dependent Schroedinger equation reduces to

#### (∂Φ/∂t) = i(h/(2m))∇²Φ

Since Φ is complex this equation is two equations. Let Φ equal φ+iγ. Then

#### (∂φ/∂t) = −(h/(2m))[(∂²γ/∂x²) + (∂²γ/∂x²)] (∂γ/∂t) = (h/(2m))[(∂²φ/∂x²) + (∂²φ/∂x²)]

The big question is what are appropriate initial values for φ and γ? Anything can be tried so the initial choice is taken to be simple; the initial values of γ are all zero and the initial values of φ are for a distribution proportional to cos²(kx).

The value of (h/(2m)) is 5.788×10-5 m²/s.

The discrete approximation of the first derivative with respect to time is

#### (∂φ/∂t) ≅ [φ(t+τ)−φ(t)]/τ

where τ is the time step. It is likewise for (∂γ/∂t).

The discrete approximation of the second derivative with respect to x is

#### (∂²φ/∂x²) ≅ [φ(x+δ, y) −2φ(x, y) + φ(x−δ, y)]/δ²

where δ is the grid spacing for x. It is likewise for y

The coefficient in the equations is therefore (τh/(2mδ²)). If δ is equal to 10-2 m; i.e., centimeters, and τ is equal to 2 seconds then the coefficient is equal to 1.1577. Let this coefficient be denoted as α.

The interation equations are then

#### φ(x, y, t+τ) = φ(x, y, t) − α[γ(x+δ, y, t) − 2γ(x, y, t) + γ(x−δ, y, t) + γ(x, y+δ, t) − 2γ(x, y, t) + γ(x, y−δ, t)] γ(x, y, t+τ) = γ(x, y, t) + α[φ(x+δ, y, t) − 2φ(x, y, t) + φ(x−δ, y, t) + φ(x, y+δ, t) − 2φ(x, y, t) + φ(x, y−δ, t)]

These may be simplified slightly to:

## The Numerical Results

The first case will be two dimensional with x and y in the interval [-2, 2]. The initial condition is for a PDF which is zero except for y=1. For y=1 the PDF is proportional to cos²(5x). Its graph is shown below. Relative probability density just means the values have not been normalize. Normalization in the case of the infinite range would result in all values being zero.

After one time step with no central force field the PDF is distributed over y=0.8 to y=1.2, as is shown below. The curves for y=0.8 and y=1.2 coincide.

For the second time step the probability density is distributed over y=0.6 to y=1.4. The curves for y=0.6, 0.8, 1.2 and 1.4 coincide. Only the probability for y=1 stands out. The peaks for all of the curves are at the same values of x.

(To be continued.)