PDE is
\[i \frac{\partial \psi}{\partial t} = -\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi.\]
Prove and/or motivate that the 'boundary' values are:
[$]\psi \rightarrow 0[$] as [$]x \rightarrow \pm\infty.[$]
Yes, demanding that [$]\psi[$] tapers off at the infinities is a bit rough-and-ready (I got it from Numerical Recipes (NR)) and it will not work as you say (also not numerically, more later). The mathematical details are missing. If you reason in probabilities then the solution will have compact support but is a bit of hand-waving.Well, generally it's not true. You need to start the particle in a state where it's true and have conditions (at infinity) on the potential V that can keep the particle from escaping to [$]\infty[$].
Assuming all that (i.e., there are only bound state solutions), then for each such solution:
[$]|\psi(t,x)|^2[$] is the probability density for finding the particle at [$]x[$],
[$]\Rightarrow \int_{-\infty}^{\infty} |\psi(t,x)|^2 \, dx = 1[$]
[$]\Rightarrow \psi \rightarrow 0[$] as [$]x \rightarrow \pm\infty[$]
But, if V=0, then [$]\psi(t,x) = e^{\pm i \sqrt{c} x - i c t} [$] for say some positive c and you can see the problem. That would be a scattering state (aka a travelling wave), ignoring the problem that it's not normalizable. Many potentials V will allow similar behavior for [$]\psi[$] far away from the barrier. For example, if V is finite and "too shallow", then regardless of how you prepare the particle at t=0 (the initial condition), it might escape the region and become a scattering state that heads off to [$]\pm \infty[$].
Yeah, but if V=0 everywhere, what is the point of introducing a boundary?? Can't the problem just be solved on the infinite space? Or, maybe I am missing something?The paper is using an ABSORBING boundary condition [(7) for the static domain]. if not, he will have reflexions and noisy pattern in figure 1 for example.
This question can be answered at different levels (which is certainly very relevant for the non-quantum wave equation (Engquist-Majda 1977);Yeah, but if V=0 everywhere, what is the point of introducing a boundary?? Can't the problem just be solved on the infinite space? Or, maybe I am missing something?The paper is using an ABSORBING boundary condition [(7) for the static domain]. if not, he will have reflexions and noisy pattern in figure 1 for example.