- #1

- 17

- 2

## Summary:

- Even though the potential function is even, the scattering states of the time-independent wave function is not even or odd. (You can't use this theorem in the scattering states.) Why?

## Main Question or Discussion Point

Hello! I have been recently studying Quantum mechanics alone and I've just got this question.

If the potential function V(x) is an even function, then the time-independent wave function can always be taken to be either even or odd. However, I found one case that this theorem is not applied, which is the case of scattering states in finite square well.

If V(x) is -V_0 at -a<x<a and 0 at lxl>a, V(x) is an even function(V_0 is a positive constant), so the time-independent wave function should be even or odd function but in my textbook (Quantum Mechanics by Griffiths), the author doesn't use this theorem even tho he does in case of bound states in finite square well.

I'm assuming because the wave function is not normalizable, so you can't really apply that "even-or-odd-theorem", but I want to know more precise reasons!

If the potential function V(x) is an even function, then the time-independent wave function can always be taken to be either even or odd. However, I found one case that this theorem is not applied, which is the case of scattering states in finite square well.

If V(x) is -V_0 at -a<x<a and 0 at lxl>a, V(x) is an even function(V_0 is a positive constant), so the time-independent wave function should be even or odd function but in my textbook (Quantum Mechanics by Griffiths), the author doesn't use this theorem even tho he does in case of bound states in finite square well.

I'm assuming because the wave function is not normalizable, so you can't really apply that "even-or-odd-theorem", but I want to know more precise reasons!