- #1
GreyNoise
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Homework Statement
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} ##
Homework 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
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?
