Solutions to the TISE for unbound states

[trelek2]

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:
<br /> \psi _{I} (x)=Ae ^{ik _{1}x }+ Be ^{-ik _{1}x }<br />
and
<br /> \psi _{II} (x)=De ^{-k _{2}x }<br />
Now these are the eigenfunctions, to get the wavefunctions I balieve that we need to:
<br /> \psi(x,t)=\psi(x)e ^{(-iEt2 \pi)/h} <br />
But is it true for both \psi _{I} and \psi _{II}?
I saw somewhere that the wavefunctions for this particular case are given by
<br /> \psi _{I} (x,t)=\psi _{I} (x)e ^{(-iEt2 \pi)/h} <br />
and
<br /> \psi _{II} (x,t)=\psi _{I} (x)e ^{(-iEt2 \pi)/h} <br />
This is really confusing, I guess it should have been
<br /> \psi _{II} (x,t)=\psi _{II} (x)e ^{(-iEt2 \pi)/h} <br />
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
<br /> e ^{(-iEt2 \pi)/h} <br />
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?

 

[DrClaude]

This is really confusing, I guess it should have been
<br /> \psi _{II} (x,t)=\psi _{II} (x)e ^{(-iEt2 \pi)/h}<br />

This is correct.

But why Is the second sunction a wavefunction anyway.
It is just a decaying exponential , not a wavelike function?

The wave function is called like that because it is a solution to a wave equation, namely the Schrödinger equation. It does not have to be "wavelike" in the classical sense.

In a classically forbidden region, it will be a decaying exponential.

Also I was wondering if you could think of any physical real life situation to which this idealized problem might correspond?

Quantum wells can be fabricated in semiconductors.