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Symmetric infinite well

  1. Oct 4, 2007 #1
    1. The problem statement, all variables and given/known data
    The time-independent Schrodinger equation solutions for an infinite well from 0 to a are of the form:

    [tex] \psi_n(x) = \sqrt{2/a} \sin (n \pi x/ a) [/tex]

    If you move the well over so it is now from -a/2 to a/2, then you can replace x with x-a/2 and get the new equations right?

    If I try to actually apply these new boundary conditions to the sin function, I get something different, however:


    [tex] \psi_n(x) = \sqrt{2/a} \sin (2 \pi n x/a) [/tex]

    How can these two techniques give different results?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 4, 2007 #2

    Dick

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    Probably because you did it wrong. You managed to halve the period of a wavefunction by translating it. That's not supposed to happen, is it?
     
  4. Oct 5, 2007 #3
    I just applied the boundary conditions to sin(k_n * x) by plugging in (a/2, 0) or (-a/2, 0).

    This yields k_n = 2 n pi/a?
     
  5. Oct 5, 2007 #4

    Meir Achuz

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    For the symmetric well, there are also solutions cos(n\pi x), with n odd.
     
    Last edited: Oct 5, 2007
  6. Oct 5, 2007 #5
    But I get A cos (k_n a/2) + B sin(k_n a/2) = 0 and A cos (k_n -a/2) + B sin (-k_n a/2) = 0 implies that B/A = tan (k_n a/2) = -B/A which implies that B is 0?
     
    Last edited: Oct 5, 2007
  7. Oct 5, 2007 #6
    Maybe the problem is that I am assuming that the k_n is the same for both sine and cosine?
     
  8. Oct 5, 2007 #7

    Dick

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    sin(n*pi*(x-a/2))=sin(n*pi*x-n*pi/2). Clearly this is +/- cos(n*pi*x).
     
  9. Oct 5, 2007 #8
    OK. The problem was that in post number 5 I divided by A which could have been 0.
     
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