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

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Main Question or Discussion Point

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.

Thanks in advance!
 

Answers and Replies

  • #2
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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.
 
  • #3
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Thank you, CPL.Luke! I understood.
 
  • #4
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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.
 
  • #5
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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
 
  • #6
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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.
 
  • #7
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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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