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Expected values in infinite square well

  1. Dec 19, 2012 #1
    Ok...this must sound stupid, because i didn't found answer on the web and on my books...but i am having trouble with the infinite square well.
    I want to calculate <x>.
    V(x)=0 for 0<=x<=a
    [itex]
    <x>=\frac{2}{a}\int^{a}_{0} x \sin^2(\frac{n\pi}{a}x)dx
    [/itex]
    Doing integration by parts i got to:
    [itex]\frac{2}{a}\left[\frac{a^2}{2n}-\int^a_0\frac{a}{2n}dx\right]=0[/itex]
    What i am doing wrong?
    Thank you
     
  2. jcsd
  3. Dec 19, 2012 #2

    mfb

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    Staff: Mentor

    I would expect an error in your integration, as the first expression is not zero and the following one is.
     
    Last edited: Dec 19, 2012
  4. Dec 19, 2012 #3

    set
    [itex]
    u=\frac{n\pi}{a}x
    [/itex]

    look up
    [itex]
    \int u \sin^2(u)du
    [/itex]
    in a Table of Integrals to find
    [itex]
    \frac{u^2}{4}-\frac{u\sin(2u)}{4}-\frac{cos(2u)}{8}
    [/itex]
     
  5. Dec 19, 2012 #4
    Yeah, i got it...it is easier to use table of integrals.
    it gives the expected [itex]<x>=\frac{a}{2}[/itex] for any value of n.
    thanks
     
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