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Probability of a particle in a box in the first excited state.

  1. Apr 5, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle is in the first excited state of a box of length L. Find the probability of finding the particle in the interval ∆x = 0.007L at x = 0.55L.

    2. Relevant equations
    P = ∫ ψ*ψdx from .543L to .557L


    3. The attempt at a solution
    Normalizing ψ gives ψ=√(2/L)sin(nπx/L)
    P = ∫ ψ*ψdx = ∫(2/L)sin^2(nπx/L)dx from .543L to .557L
    The integration simplifies to
    P = x/L - sin(4πx/L)/4
    so P = [.557L/L - sin(4π*.557L/L)/4] - [.543L/L - sin(4π*.543L/L)/4]
    P = 0.0132 or 1.32%

    This is wrong though and the hint given afterwords was that because the Δx is so small, there is no need for integration. This just confuses me because abs(ψ)^2 will have a 1/L factor in it. Any help will be useful. Thanks!
     
  2. jcsd
  3. Apr 5, 2013 #2

    BruceW

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    Ah, ok, think of how you would approximate the area under a small section of a graph. (What is the simplest shape you can use?) And your 'graph' is abs(ψ)^2 against x, with Δx= 0.007L
     
  4. Apr 5, 2013 #3
    Thanks

    Thanks for reminding me that dx can be approximated as Δx. For some reason I didn't make that jump
     
  5. Apr 6, 2013 #4

    BruceW

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    hehe yeah, that's alright. It is quite unusual to do an 'approximate integral' in this way. You could maybe even calculate the fractional error (to first order), by calculating the derivative of abs(ψ)^2, and finding the difference that this makes to the approximation.

    (but that's not part of the question, so whatever).
     
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