# About Schrodinger eq. and the potential step case with E = V0

I have a simple question but can't find the answer:
What happens in the Potential Step (the Schrodinger Equation aplication) case when E = V0 ?
I only seem to find the cases when E>V0 or E<V0.

take the schrodinger equation

p^2/2m(y)+v0(y)=E(y)

and set E=v0

you'll get the ordinary differential equation

d^2/dx^2(y)=0

with solution

y=Ax+b

this solution is invalid however as it won't match your boundary conditions, therefore there is no solution with E=v0

one thing that may trouble you is that the solutions you see for E=/ V0 usually don't match your boundary conditions either, and this has to do with the fact that E is a continuous parameter, with the actual solution to the full schrodinger equation being a fourier integral over the basis states, and there being a zero probability of E being any particular number.

Thank you, CPL.Luke! I understood.

It should be a constant outside the well, and a sinusoid inside. Match boundaries as usual to find the relative constants. Nothing special is going on here.

he's talking about the potential step barrier, not the well.

however it should be noted that there isn't a finite square well solution with E=v0, (as the solution would be non-normalizable

also there is never a constant solution to the schrodinger equation that satisfies normalisation

A plane wave, Exp(ipx), is also not normalisable by that logic, yet we usually have no problem with it. There is nothing more problematic with p=0, which is really what we're considering. Everything can be made rigorous by using some regularisation procedure.

however in the case of the step potential there is an uncountably infinite number of basis states, however the bound states for a square well are countable infinite.

countably infinite states have to be normalised, as they represent possible wavefunctions. whereas the uncountably infinite states don't acually represent wavefunctions in and of themselves.