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Delta-function potentials, scattering states

  1. Sep 26, 2009 #1
    I'm reading through the Introduction to Quantum Mechanics book by Griffiths (2nd edition)
    and it is describing delta-function potential wells.
    When it describes how to find bound states the energy is E < 0 (negative) in the region x < 0 (negative).

    It says the general solution is:
    [tex]\psi (x) = Ae^{-\kappa x} + Be^{\kappa x}[/tex]
    It explains that as x goes to -∞ the first term blows up. Which makes sense because it will go to infinity, while the second term only goes to zero.

    However, they then look at the situation when E > 0, x < 0
    again the general solution is:
    [tex]\psi (x) = Ae^{-\kappa x} + Be^{\kappa x}[/tex]

    but this time they say that neither term blows up. I don't understand why that is. Shouldn't the first term still go to infinity as x goes to -∞ ?
     
  2. jcsd
  3. Sep 27, 2009 #2
    No, for E > 0, x < 0 the general solution is
    [tex]
    \psi(x) = Ae^{-ikx} + Be^{ikx}
    [/tex]
    which oscillates but does not exponentially grow or decay (for real k).
     
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