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Probablities from wave function

  1. Dec 5, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]P = \int_a^b \, \left| \psi(x) \right|^2 \, dx[/tex]

    If the particle in the box is in the second excited state (i.e., n=3), what is the probability P that it is between x=L/3 and x=L? To find this probability, you will need to evaluate the integral:

    [tex]\int_{L/3}^L \left(\sqrt{\frac{2}{L}} \sin \left( \frac{n \pi x}{L} \right) \right)^2 \, dx = \frac{2}{L} \int_{L/3}^L \sin^2 \left( \frac{n \pi x}{L} \right) \, dx .[/tex]

    2. Relevant equations
    This one I think.
    [tex]
    * \int \sin^2(kx) dx = \frac{x}{2} - \frac{1}{4k} \sin(2 k x) +C, and
    [/tex]

    3. The attempt at a solution
    Set k = n(pi)/L or np/L

    [2/L][(x/2 - 1/4k)sin(2kx)]|L and L/3

    put k back in as it is.

    [2/L]{[L/2 - L/4np]sin(2*np/L*L)]-[L/3*2 - L/4np]sin(2 * np/L * L/3)]}

    cancel out the L's and plug-in n = 3

    2{[(1/2 - 1/12p)sin(6p)] - [(1/6) - (1/12p)]sin(2p)]}

    well the answer isn't 0. but sin6p and sin2p both equal 0
    2[0 - 0] = 0

    Where did I go wrong?
     
  2. jcsd
  3. Dec 5, 2007 #2

    Shooting Star

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    sin^2(kx) = [1-cos(2kx)]/2. Now can you integrate directly?
     
  4. Dec 5, 2007 #3
    doesn't seem like it... integrating the cos(2kx) part just gives me a sin function again and again I get an integer times pi within leading to more 0. right?
     
  5. Dec 6, 2007 #4

    Shooting Star

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    (I had merely showed you how to evaluate the integral sin^2(kx), for which you have already written the formula.)

    Anyway, you have done this mistake:
    2{[(1/2 - 1/12p)sin(6p)] - [(1/6) - (1/12p)]sin(2p)]} = 2{(1/2) - (1/6)} = 2/3.
     
  6. Feb 10, 2009 #5
    What is mean probability is zero?
     
  7. Feb 10, 2009 #6

    jambaugh

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    It looks like you mistakenly are grouping the sine function of your integral outside the difference from your integral formula: (u-v)sin(w) should be (u - v sin(w) ).
     
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