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Probability of finding a particle in a 1D Box

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine the probability of finding a particle in a 1-D box of size L in a region of size 0.01L at the locations x = 0, 0.25L, 0.5L, 0.75L and L when it is in its ground state. As percentages

    2. Relevant equations

    [tex]\Psi[/tex](x)=[tex]\sqrt{\frac{2}{L}}[/tex]sin([tex]\frac{nxPI}{L}[/tex])
    P=[tex]\Psi[/tex](x,t)^2

    3. The attempt at a solution

    I'm not even sure if i've got the right equations there, i've tried loads of different ways of doing this and cannot get the right answers, which are 0%,1%,2%,1%,0%

    Cheers
     
  2. jcsd
  3. Feb 1, 2010 #2
    ok, finding probabilities means integrating the probability density - P=|[tex]\Psi[/tex](x,t)^2| - over parts of the box

    try integrating over the whole box first, what does this give you?...
     
  4. Feb 1, 2010 #3
    [tex]\int[/tex]|[tex]\Psi[/tex](x,t)^2| = |[tex]\Psi[/tex](x,t)^3| /3

    I think that's right.
     
    Last edited: Feb 1, 2010
  5. Feb 1, 2010 #4
    Your answer cannot be a function of x, since you are integrating over x. Try substituting your expression for [itex] \Psi(x) [/itex] from your first post before you integrate.
     
  6. Feb 1, 2010 #5
    You need to integrate |psi^2| from x=0 to x=0.01L , from x=0.245L to 0.255L, etc

    Try to plot psi, |psi^2| and the integral of |psi^2| from 0 to x (remember that the integral is the probabilty of finding the particle in the limits of integration...)
     
  7. Feb 1, 2010 #6
    Thanks guys, got a bit of extra help and the penny has dropped now. cheers
     
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