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

  1. May 30, 2008 #1
    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!
     
  2. jcsd
  3. May 30, 2008 #2
    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.
     
  4. May 30, 2008 #3
    Thank you, CPL.Luke! I understood.
     
  5. May 31, 2008 #4
    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.
     
  6. May 31, 2008 #5
    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
     
  7. May 31, 2008 #6
    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.
     
  8. May 31, 2008 #7
    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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