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Integrate wave function squared - M. Chester text

  1. Jul 9, 2017 #1

    GreyNoise

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    Gold Member

    1. The problem statement, all variables and given/known data
    given: A wire loop with a circumference of L has a bead that moves freely around it. The momentum state function for the bead is ## \psi(x) = \sqrt{\frac{2}{L}} \sin \left (\frac{4\pi}{L}x \right ) ##
    find: The probability of finding the bead between ## \textstyle \frac{L}{24} ## and ## \textstyle \frac{L}{8} ##

    2. Relevant equations
    ## \int_{a}^{b}|<x| \psi >|^2 dx = \int_{a}^{b} | \psi(x) |^2 dx ##

    ## \psi(x) = \sqrt{\frac{2}{L}} \sin \left ( \frac{4\pi}{L}x \right ) \hspace{10mm}## the state function

    3. The attempt at a solution
    ## \displaystyle \int_{a}^{b}|\lt x|\psi\gt|^2 dx = \int_{a}^{b} | \psi(x) |^2 dx ##

    ## {\displaystyle \int_{\frac{L}{24}}^{\frac{L}{8}} \left [ \sqrt{\frac{2}{L}} \sin \left ( \frac{4\pi}{L}x \right ) \right ]^2 dx } \hspace{10mm} ## sub ## \textstyle \psi(x) = \sqrt{\frac{2}{L}} \sin \left ( \frac{4\pi}{L}x \right ) ## and integration limits

    ## \displaystyle \int_{\frac{L}{24}}^{\frac{L}{8}} \left [ \sqrt{\frac{2}{L}} \sin \left ( \frac{4\pi}{L}x \right ) \right ]^2 dx = \left. \frac{x}{L} - \frac{ \sin \left ( \frac{8\pi}{L}x \right )}{8\pi} \right |_{\frac{L}{24}}^{\frac{L}{8}} ##

    ## \displaystyle = \frac{1}{L}\frac{L}{8} - \frac{ \sin \left ( \frac{8\pi}{L}\frac{L}{8} \right ) }{8\pi} - \left [ \frac{1}{L}\frac{L}{24} - \frac{ \sin \left ( \frac{8\pi}{L}\frac{L}{24} \right ) }{8\pi} \right ]
    = \frac{1}{8} - \frac{1}{24} + \frac{ \sin \left ( \frac{\pi}{3} \right ) }{8\pi} ##

    ## = \displaystyle \frac{1}{12} + \frac{\sqrt{3}}{16\pi} ##

    The answer given in the text is ##\frac{1}{12} + \frac{1}{16\pi}##. I cannot shake the ## \sqrt{3} ## in the second term. I even checked my evaluation of the integral on the Wolfram site, and it returned the same integral solution as I got. The book is Primer of Quantum Mechanics by Marvin Chester (Dover Publ). It appears to be a first edition, so I guess it's plausible that the text's answer is a typo, but I felt the need to consult the community. Can anyone show me what I am doing wrong?
     
  2. jcsd
  3. Jul 9, 2017 #2

    Charles Link

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    In a quick check, I think your calculations are correct, and that the book is in error.
     
  4. Jul 10, 2017 #3

    GreyNoise

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    Gold Member

    Thnx Charles. I am confident of my own answer then.
     
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