(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

V(x) = 0 if [itex]\frac{-L}{2}[/itex]<x<[itex]\frac{L}{2}[/itex] and [itex]\infty[/itex] otherwise.

Is the wave function [itex]\Psi[/itex] = (2/L)[itex]^{1/2}[/itex] (sin ([itex]\pi[/itex]x/L) an acceptable solution to this? Explain

2. Relevant equations

H[itex]\Psi[/itex]= E[itex]\Psi[/itex] , normalization: 1 = [itex]\int[/itex] wavefunction^{2}dx

3. The attempt at a solution

My logic is that I have to come up with a wave function that satisfy the boundary condition. Therefore [itex]\Psi[/itex] (x= [itex]\frac{-L}{2}[/itex]) = [itex]\Psi[/itex] (x = [itex]\frac{L}{2}[/itex]) = 0

My initial answer was that it's not because I was thinking that the wavefunction itself has to be A[/itex]cosine(bx), where b=2n[itex]\pi[/itex]/L. n = 1/4, 3/4 , 5/4 ...

I am not quite sure if that's correct. Also when I try to calculate the normalization factor (A), it turns out to be 1 = [itex]\frac{A^{2}L}{2}[/itex] + [itex]\frac{A^{2}L}{n\pi}[/itex] sin ([itex]\frac{n\pi}{2}[/itex])

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# Homework Help: QM: Particle in a box

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