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Particle in an infinite square well - interval -d/2<x<d/2

  1. Aug 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Particle is in an infinite square well of width ##L## on an interval ##-L/2<x<L/2##. The wavefunction which describes the state of this particle is of form:
    $$\psi = A_0\psi_0(x) + A_1\psi_1(x)$$
    where ##A_1=1/2## and where ##\psi_0## and ##\psi_1## are ground and first excited state. What are ##\psi_0## and ##\psi_1## like? Write them down. Calculate ##A_0##.

    2. Relevant equations
    I allways used this equation to get normalisation factors:
    $$\int\limits_{-d/2}^{d/2} \psi =1$$
    If i recall corectly the eigenfunctions are orthogonal? I remember something about this and i think there is different way too solve this problem using this orthogonality. Please someone show me, because i am interested in both ways!

    3. The attempt at a solution
    I know that normalisation factor should be ##A_0=\sqrt{3}/2## but what i got was totaly different. I think i chose ##\psi_0## and ##\psi_1## wrong. Here is how i got the wrong result:
    [Broken]

    Is it possible that I should use ##\psi_0(x) = \cos \left(\frac{(2N-1)\pi}{d}x\right)## where ##\boxed{N=1}## and ##\psi_1(x) = \sin \left(\frac{2N\pi}{d}x\right)## where ##\boxed{N=1}##?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 4, 2013 #2
    You're correct that your initial eigenfunctions are wrong, but your suggested corrections are not quite right either (but really close! :) ). [itex]N[/itex] labels the energy states, so that the ground state is [itex]N=1[/itex] and the first excited state is [itex]N=2[/itex].

    Using that in your new eigenfunctions should result in the correct answer. I didn't go in detail through your method though.

    Edit: Whoops, don't forget the [itex]\sqrt{\frac{2}{L}}[/itex] factor in those new eigenfunctions also!
     
    Last edited: Aug 4, 2013
  4. Aug 5, 2013 #3
    What i thought was: "Oh! ##A_0## and ##A_1## are already there and are playing the role of ##\sqrt{2/L}##". I guess i missread the instructions...

    I don,t understand the why N=2 though. If i plug in the N=1 in the equations i provided (i added factor of ##\sqrt{2/L}##) I get ##\psi_0(x) = \sqrt{2/L}\cos \left(\frac{1\pi}{d}x\right)## (this should be the ground state) and ##\psi_1(x) =\sqrt{2/L} \sin \left(\frac{2\pi}{d}x\right)## (this should be first excited state).
     
  5. Aug 5, 2013 #4

    vela

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    Typically, you label the states n=1, n=2, n=3, and so on. You don't use the same n for different states. The eigenfunctions would be given by
    $$\phi_n(x) = \begin{cases}
    \sqrt{\frac{2}{L}}\cos \frac{n\pi x}{L}, & n\text{ odd} \\\\
    \sqrt{\frac{2}{L}}\sin \frac{n \pi x}{L}, & n\text{ even}
    \end{cases}$$ for ##-L/2 \le x \le L/2## and 0 elsewhere. Following this convention for labeling the states, the energy of each state is given succinctly by
    $$E_n = \frac{\hbar^2 n^2 \pi^2}{2mL^2}.$$
     
  6. Aug 5, 2013 #5
    If i understand your point you wanted to say that my sugestion would have worked, but is not written in accordance to convention?
     
  7. Aug 5, 2013 #6

    vela

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    I was just clarifying what TheShrike meant. You can use whatever convention you want as long as it works for you.
     
  8. Aug 5, 2013 #7
    Yes i know that. But would my own convention work?
     
  9. Aug 5, 2013 #8
    It worked and i got the solution ##A_0 = \sqrt{3}/2##. Thank you all for your help.
     
  10. Aug 5, 2013 #9
    Oh, I see what you were doing now. A conflict of conventions, as vela stated. My mistake.

    Looks like you had the answer all along. :wink:
     
  11. Aug 5, 2013 #10
    Nope. The mistake is always a comunication :). If you were here and i could ask you in person we would solve the problem in first two sentances. But writing and reading is sometimes blah :). We should create some sort of a Skype Q&A site ;)
     
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