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Infinite Square Well

  1. Apr 28, 2009 #1
    Hi all,

    Sorry if this question is not very challenging, but Im driving myself to confusion...

    I happily derived the eigen functions for an infinite square well spanning from 0 to L and found them to be:


    ...in agreement with wikipedia.

    However my course notes derive solutions for an inf sq well between -a and +a.

    Surely L=2a, no?

    In the notes n=1,3,5... produces cosine solutions and n=2,4,6... produce sine solutions.

    Why is there now cosine solutions?

    Please help, I'm going crazy :)
  2. jcsd
  3. Apr 28, 2009 #2


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    Homework Helper
    Gold Member

    You can get the solutions on [-L/2 , L/2] from the solutions on [0, L] by replacing x by x + L/2 (i.e. moving the solutions to the left by L/2). For example, sin(πx/L) becomes sin((π/L)(x + L/2)) = sin(πx/L + π/2) = cos(πx/L).
  4. Apr 28, 2009 #3
    Thankyou dx,

    That's brilliant, it all makes sense :)
  5. Apr 28, 2009 #4


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    Staff: Mentor

    Draw a graph of the ground state (n = 1) solution for both cases. You'll see that the two graphs are identical except for the origin (x = 0) being in different places relative to the walls of the well. Likewise for n = 2, etc.
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