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Probability of finding a particle?

  1. Mar 27, 2008 #1
    1. The problem statement, all variables and given/known data

    For a particle in the ground state of a rigid box, calculate the probability of finding it between x=0 and x=[tex]\frac{L}{3}[/tex]

    2. Relevant equations

    [tex]\left|\psi^{2}\right| = \frac{2}{L}Sin^{2}\left(\frac{nx\pi}{L}dx\right)[/tex]

    3. The attempt at a solution

    [tex]= \frac{2}{L}\int^{\frac{L}{3}}_{0} Sin^{2}\left(\frac{x\pi}{L}\right)[/tex]
    [tex]= \frac{2}{L}\int^{\frac{L}{3}}_{0} \frac{1-Cos \left(\frac{2x\pi}{L}\right)}{2}[/tex]

    [tex]\frac{1}{L} \int^{\frac{L}{3}}_{0} 1 - \int^{0}_{\frac{L}{3}} Cos \left(\frac{2x\pi}{L}\right)}[/tex]

    [tex]\frac{1}{L} \left[ \left|x\right|^{\frac{L}{3}}_{0} - \left|\frac{L}{2\pi}Sin\left(\frac{2x\pi}{L}\right)\right|^{\frac{L}{3}}_{0} [/tex]

    [tex]\frac{1}{L} \left[ \frac{L}{3}} - \frac{L}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right] [/tex]

    factor out the L

    [tex]\frac{1}{3}} - \frac{1}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right] [/tex]

    stuck here...when I work it out I get a negative number...am I mising something?
    Last edited: Mar 27, 2008
  2. jcsd
  3. Mar 27, 2008 #2


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    Gold Member

    [tex]\frac{1}{3}} - \frac{1}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right] [/tex]

    I get a positive number -

    sin(2pi/3) = 0.8660

    0.8660/(2*pi ) = 0.1378

    1/3 - 0.1378 = 0.1955
  4. Mar 27, 2008 #3
    oh wow...stupid math mistake....forgot the parenthesis on my calc =(

    thank you!
  5. Jan 23, 2010 #4
    what would the probability of the particle be if x=1.95 and 2.05?
  6. Jan 23, 2010 #5
    the answer is supposed to be .007, but I keep getting -.2921
  7. Apr 19, 2011 #6
    I think you should be using cos instead of sin for the ground state.
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