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

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Discussion Overview

The discussion revolves around the behavior of the Schrödinger equation in the context of a potential step when the energy (E) equals the potential (V0). Participants explore the implications of this scenario, contrasting it with cases where E is greater or less than V0.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the outcome of the potential step case when E = V0, noting a lack of available answers for this specific situation.
  • Another participant derives the ordinary differential equation from the Schrödinger equation when E is set to V0, concluding that the solution y = Ax + b does not satisfy boundary conditions, suggesting no valid solution exists for E = V0.
  • It is mentioned that solutions for cases where E ≠ V0 often do not meet boundary conditions either, due to E being a continuous parameter and the actual solution being a Fourier integral over basis states.
  • One participant proposes that outside the well, the solution should be constant, while inside it should be sinusoidal, emphasizing the importance of matching boundaries to find relative constants.
  • Another participant clarifies that the discussion pertains to the potential step barrier rather than a finite square well, noting that a finite square well solution with E = V0 is non-normalizable.
  • A participant argues that a plane wave, Exp(ipx), is not normalizable but is still commonly accepted, suggesting that the case of p = 0 (relevant to E = V0) is similarly manageable through regularization procedures.
  • It is noted that in the case of the step potential, there exists an uncountably infinite number of basis states, contrasting with the countably infinite bound states of a square well, and discussing the implications for normalization of wavefunctions.

Areas of Agreement / Disagreement

Participants express differing views on the existence and nature of solutions when E = V0, with some asserting that no valid solutions exist while others challenge this notion by discussing normalization and the implications of infinite states.

Contextual Notes

Participants highlight limitations regarding normalization conditions and the nature of solutions in different potential scenarios, but these remain unresolved within the discussion.

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

Thanks in advance!
 
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take the Schrödinger 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 Schrödinger 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 Schrödinger 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.
 

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