solutions to the TISE for "unbound states"

trelek2

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?