About Schrodinger eq. and the potential step case with E = V0

In summary, The conversation discusses the Schrodinger Equation and its application in the Potential Step case. When E = V0, there is no solution to the equation that satisfies the boundary conditions. However, for E ≠ V0, there are solutions that may not fully satisfy the boundary conditions due to the continuous nature of E. The conversation also mentions the difference between countably and uncountably infinite states and their relationship to normalizability.
  • #1
I have a simple question but can't find the answer:
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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  • #2
take the schrodinger equation


and set E=v0

you'll get the ordinary differential equation


with solution


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
Thank you, CPL.Luke! I understood.
  • #4
It should be a constant outside the well, and a sinusoid inside. Match boundaries as usual to find the relative constants. Nothing special is going on here.
  • #5
he's talking about the potential step barrier, not the well.

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
A plane wave, Exp(ipx), is also not normalisable by that logic, yet we usually have no problem with it. There is nothing more problematic with p=0, which is really what we're considering. Everything can be made rigorous by using some regularisation procedure.
  • #7
however in the case of the step potential there is an uncountably infinite number of basis states, however the bound states for a square well are countable infinite.

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.

1. What is the Schrodinger equation?

The Schrodinger equation is a fundamental equation in quantum mechanics that describes how the quantum state of a physical system changes with time. It is used to calculate the probability of finding a particle in a specific location at a specific time.

2. How does the Schrodinger equation apply to the potential step case?

In the potential step case, the Schrodinger equation is used to determine the behavior of a particle encountering a sudden change in potential energy. This allows us to understand how the particle's wave function will change and how it will affect the particle's probability of being found in a certain location.

3. What is the value of E in the potential step case with E = V0?

In the potential step case, E represents the energy of the particle. When E is equal to V0, it means that the particle's energy is equal to the potential energy in the region of the potential step. This is known as the particle being in the "classically allowed" region.

4. How does the potential step case with E = V0 differ from other cases?

When E is equal to V0 in the potential step case, the particle is in the classically allowed region and its behavior can be described using classical mechanics. In other cases where E is not equal to V0, the particle's behavior is governed by quantum mechanics and can exhibit wave-like properties.

5. What can we learn from studying the potential step case with E = V0?

Studying the potential step case with E = V0 allows us to better understand how particles behave when encountering sudden changes in potential energy. It also helps us understand the implications of classical and quantum mechanics and how they differ in describing the behavior of particles in different scenarios.

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