Here I compute the state evolution for a single particle in 1D with a real gaussian as an initial state. Initially
ψ = e−(x/(2σ))2/(π½2σ)
where x is the one real coördinate and σ is the standard deviation of the initial position of the particle. This represents a particle with a somewhat unknown position, near 0, and somewhat unknown momentum, near zero. The boundary conditions are periodic but we view it as a circular space.

The equation is ∂ψ/∂t = i∂2ψ/∂x2. The explicit numerical equation is
ψn+1j − ψnj = Δt/Δx2(2 ψnj − (ψnj+1 + ψnj−1))
Where n is the time step and j is the zone number.

The calculation is explicit and the concomitant numerical instability causes a growing square integral. Relative stability requires Δt < Δx2. High spatial frequency noise dominates which manifests as an unreal high energy. I plan to try an implicit numerical method.

Constants to play with:

• dx and dt are what you would guess; zone size and time step.
• sig (σ) is the width (standard deviation) of the initial position distribution.
• span is the number of zones and controls the period of the space.
• dpt is the time between the printings of ψ.

### Implicit

The corresponding equation for the implicit calculation is
ψn+1j − ψnj = Δt/Δx2 1/2((2 ψnj − (ψnj+1 + ψnj−1)) + (2 ψn+1j − (ψn+1j+1 + ψn+1j−1)))

This follows almost exactly the implicit heat equation elaborated here. One wrinkle is that our topology is circular where the heat solution is for an insulated rod. The other wrinkle, of course, is that ϕ is real but ψ is complex. The only mathematical properties of the reals we need for the implicit calculation is that they be a field. The complex numbers are a field.

By elementary zone-wise algebra we can compute for each j, complex Aj, Bj and Cj such that
ψn+1j = Aj + Bjψn+1j+1 + Cjψn+1j−1
This gives us just enough equations to solve for the ψn+1’s. As time goes by the shape of ψ becomes that of a twisted phone cord where the left end is twisted in one direction and the right end in the other direction. (We imagine plotting ψ in x and the two complex dimensions.) This depicts the left part moving to the left, and the right part moving to the right. Thus if we were to catch the particle in the leftwards region, it would likely be moving to the left.

### Solution in Closed Form

From this we learn that if ψ = pe−(px)2 where p = (1+4it)−1/2 then ∂ψ/∂t = i∂2ψ/∂x2. For physics we want |ψ|2dx = 1 whereas our new ψ is bigger: |ψ|2dx = √(π/2). For the rest of this page
ψ = √(2/π) pe−(px)2.

As time goes by the shape of ψ becomes that of a twisted phone cord where the left end is twisted in one direction and the right end in the other direction. (We imagine plotting ψ in x and the two complex dimensions.) This depicts the left part moving to the left, and the right part moving to the right. Thus if we were to catch the particle in the leftwards region, it would likely be moving to the left.

This solution is easily extended to a 3D spherically symmetric function:
ψ = (2/π)3/2 pe−(pr)2.

This closed form seems useless for the relativistic case. See Relativistic diffusion processes and random walk models for problems finding a relativistic replacement for e−x2.