Probablities from wave function

  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.
    * \int \sin^2(kx) dx = \frac{x}{2} - \frac{1}{4k} \sin(2 k x) +C, and

    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. Shooting Star

    Shooting Star 1,979
    Homework Helper

    sin^2(kx) = [1-cos(2kx)]/2. Now can you integrate directly?
  4. 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. Shooting Star

    Shooting Star 1,979
    Homework Helper

    (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. What is mean probability is zero?
  7. jambaugh

    jambaugh 1,798
    Science Advisor
    Gold Member

    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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