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Solutions to the TISE for unbound states

  1. Feb 28, 2009 #1
    solutions to the TISE for "unbound states"

    Hi!

    Suppose we have a step potential with boundary at x=0
    V=0 for x<0 and V=V for x>0
    Suppose V>E

    I guess I hot pretty far with this problem, I do have one doubt however:
    We obviously have two solutions:
    [tex]
    \psi _{I} (x)=Ae ^{ik _{1}x }+ Be ^{-ik _{1}x }
    [/tex]
    and
    [tex]
    \psi _{II} (x)=De ^{-k _{2}x }
    [/tex]
    Now these are the eigenfunctions, to get the wavefunctions I balieve that we need to:
    [tex]
    \psi(x,t)=\psi(x)e ^{(-iEt2 \pi)/h}
    [/tex]
    But is it true for both [tex] \psi _{I} [/tex] and [tex] \psi _{II} [/tex]?
    I saw somewhere that the wavefunctions for this particular case are given by
    [tex]
    \psi _{I} (x,t)=\psi _{I} (x)e ^{(-iEt2 \pi)/h}
    [/tex]
    and
    [tex]
    \psi _{II} (x,t)=\psi _{I} (x)e ^{(-iEt2 \pi)/h}
    [/tex]
    This is really confusing, I guess it should have been
    [tex]
    \psi _{II} (x,t)=\psi _{II} (x)e ^{(-iEt2 \pi)/h}
    [/tex]
    But why Is the second sunction a wavefunction anyway.
    It is just a decaying exponential , not a wavelike function?
    Or is it that if we multiply it by that
    [tex]
    e ^{(-iEt2 \pi)/h}
    [/tex]
    this somehow changes it into a wave?
    Please explain.
    Also I was wondering if you could think of any physical real life situation to which this idealized problem might correspond?
     
  2. jcsd
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