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Bound State Condition

  1. May 4, 2009 #1

    OB1

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    What is the precise definition of the bound state condition? Thanks in advance.
     
  2. jcsd
  3. May 5, 2009 #2

    malawi_glenn

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    E < 0 and that the wavefunction "decays" to 0 as r-> infty
     
  4. May 5, 2009 #3
    Disagree with you, though i'm not insisting. Consider a harmonic oscillator: all energies > 0, still bound states. Wave function should "decay" at infinity, but something must be said about when it decays: e.g. a moving Gaussian wave packet decays faster than an exponent, but does not correspond to a bound state.
     
  5. May 5, 2009 #4

    alxm

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    Well, a harmonic oscillator is bound by definition though. It's in an infinite potential well.

    If it wasn't an infinite potential well, the particle could tunnel out sooner or later. So it wouldn't be bound then.
     
  6. May 5, 2009 #5
    Yes, "If it wasn't an infinite potential well, the particle could tunnel out sooner or later. So it wouldn't be bound then".
     
  7. May 5, 2009 #6

    OB1

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    Doesn't any allowed wave function decay to zero at infinity as part of the L2 condition? But not all wave functions are bound states, and we do have bound states in finite wells or attractive delta functions, so the bound state condition can't be just that.
    Or am I missing the point, maybe?
     
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