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QM: Infinite Square Well -a/2 to a/2

  1. Oct 25, 2011 #1
    I have read similar threads about this problem but I wasn't able to make progress using them.

    1. The problem statement, all variables and given/known data

    Consider an infinite square-well potential of width a, but with the coordinate system shifted so that the infinite potential barriers lie at x=[itex]\frac{-a}{2}[/itex] and x=[itex]\frac{a}{2}[/itex].

    Solve the Schrodinger equation for this case to calculate the normalized wave function [itex]\psi[/itex]n(x) and the corresponding energies En

    2. Relevant equations

    Time Independent Schrodinger Equation (V(x)=0 within the well)
    -\hbar2[itex]\frac{1}{2m}[/itex] ([itex]\frac{∂}{∂x}[/itex])2[itex]\psi[/itex]=E[itex]\psi[/itex]

    3. The attempt at a solution

    Skipping ahead to the general solution for [itex]\psi[/itex](x) , I get:

    [itex]\psi[/itex](x) = Aeikx + Be-ikx, k = [itex]\frac{\sqrt{2mE}}{\hbar}[/itex]

    Using the boundary conditions,

    [itex]\psi[/itex]([itex]\frac{a}{2}[/itex])=Aeika/2 + Be-ika/2=0
    [itex]\psi[/itex]([itex]\frac{-a}{2}[/itex])=Ae-ika/2 + Beika/2=0,

    together with some substitution I am able to prove that,

    k =[itex]\frac{\pi}{a}[/itex]n

    En =([itex]\frac{n\pi\hbar}{a}[/itex])2[itex]\frac{1}{2m}[/itex]

    which I believe to be correct for infinite square well.

    I run into trouble trying to solve for the constants A and B. My approach was to try and normalize the wavefunction and then use substitution to solve for A or B, but I can't seem to get anywhere after a certain point.
    I would expect to get cosines for n=1,3,5... and sines for n=2,4,6... based on what I know about the shape of the wavefunction, but I am unsure how to prove this mathemetically.
    Any tips would be appreciated.
     
    Last edited: Oct 25, 2011
  2. jcsd
  3. Oct 26, 2011 #2

    vela

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    Try plugging in your allowed values of k into the boundary condition equations. You'll get a different relationship between A and B depending on whether n is odd or even.
     
  4. Oct 26, 2011 #3
    Ah, that's the ticket. It's probably easier without using exponential form, but I was able to prove that A=B for odd n (giving me a cosine) and A=-B for even n (giving me a sine). Thanks!
     
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