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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?

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# Solutions to the TISE for unbound states

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