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What happens in the Potential Step (the Schrodinger Equation aplication) case when E = V0 ?

I only seem to find the cases when E>V0 or E<V0.

Thanks in advance!

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- Thread starter Tavo
- Start date

- #1

- 2

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What happens in the Potential Step (the Schrodinger Equation aplication) case when E = V0 ?

I only seem to find the cases when E>V0 or E<V0.

Thanks in advance!

- #2

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p^2/2m(y)+v0(y)=E(y)

and set E=v0

you'll get the ordinary differential equation

d^2/dx^2(y)=0

with solution

y=Ax+b

this solution is invalid however as it won't match your boundary conditions, therefore there is no solution with E=v0

one thing that may trouble you is that the solutions you see for E=/ V0 usually don't match your boundary conditions either, and this has to do with the fact that E is a continuous parameter, with the actual solution to the full schrodinger equation being a fourier integral over the basis states, and there being a zero probability of E being any particular number.

- #3

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Thank you, CPL.Luke! I understood.

- #4

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- #5

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however it should be noted that there isn't a finite square well solution with E=v0, (as the solution would be non-normalizable

also there is never a constant solution to the schrodinger equation that satisfies normalisation

- #6

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- #7

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countably infinite states have to be normalised, as they represent possible wavefunctions. whereas the uncountably infinite states don't acually represent wavefunctions in and of themselves.

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