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Homework Help: QM: Particle in a box

  1. Dec 15, 2011 #1
    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])
  2. jcsd
  3. Dec 15, 2011 #2


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    This is right. Does the proposed wave function satisfy these conditions?

    You'll actually get both sine and cosine solutions if you solve the infinite square well problem completely, so just because this wave function has a sine in it isn't reason enough to exclude it.

  4. Dec 15, 2011 #3


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    I think your reasoning is correct. The wave function doesn't vanish at the boundaries. So it's not good. I wouldn't worry about whether the normalization is correct if the boundary conditions aren't correct.
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