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QM: Finite square well with V>0

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    When talking about the finite square well with a potential V > 0 for - A < x < A, I have never seen an example of bound states (i.e. E<0). They only treat examles with scattering states (i.e. E>0). Is there any reason for this? My book (Griffith's Intro. to QM) does not talk about this scenario.
  2. jcsd
  3. Sep 21, 2008 #2
    If V>0, then this is not a square well, but a square barrier. Bound states do not exist in this situation.
  4. Sep 21, 2008 #3
    Ahh yes, barrier - not well! Why is that?
  5. Sep 21, 2008 #4
    A well is when V<0, so the particle can "fall in". A barrier is V>0, so there's an obstacle. I'm not sure about a formal proof, but it is hard to imagine a particle bound to a wall. Total energy must be negative somewhere if there is to be a bound state, but kinetic energy is always positive, plus positive potential = no bound state.
  6. Sep 21, 2008 #5
    Can one use the explanation that the energy E always has to be larger than the minimum potential?
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