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Particle in a box question

  1. Nov 22, 2007 #1
    1. The problem statement, all variables and given/known data

    Hey all!
    A particle in a box of side L has wavefunction [tex]\psi[/tex](x) = Asin([Pi*x/L) for the ground state.
    a) Determine the value of A for a properly normalised wavefunction
    b) Determine the probability that x>0.75L

    2. Relevant equations

    -

    3. The attempt at a solution

    Ok so I started off by normalising the wavefunction

    [tex]\int[/tex][tex]\psi[/tex]*[tex]\psi[/tex]dx with the limits of L and 0

    = A[tex]^{2}[/tex][tex]\int[/tex]sin[tex]^{2}[/tex](Pi*x/L)dx which would equal 1.

    Integrating...

    ∫ sin^{2} pi x/L) dx

    ∫[1 - cos(2pi x/L)]/2 dx

    ∫ dx - 1/2(L/2pi) ∫ cos(2pi x/L) (2pi/L)dx

    x/2 - (L/4 pi) sin(2pi x/L)

    applying limits 0 and L

    [L/2 - 0] - (L/4pi)[sin(2pi L/L) - 0

    L/2 - L/4 pi (0)

    L/2.

    Ok so I need a value though. So because it is in the ground state, the value of L would be [tex]\pi[/tex] I think, or is that incorrect?

    So then I had:

    A[tex]^{2}[/tex]*Pi/2 = 1
    A = 1/(\sqrt{Pi/2})
    A = 0.798

    Is this correct?

    For the second part, b)
    I'm totally confused, should I do the same as I did in part a), and integrate, but put in the limits of L and 0.75L?
     
  2. jcsd
  3. Nov 22, 2007 #2

    nrqed

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    Homework Helper
    Gold Member

    Do not use any value for L. Just leave it as L. It's ok to have your wavefunction containing L. In the second part, yes, integrate from 0.75 L to L. You will find out that all L dependence will cancel out in your result, leaving you with an actual number.
     
  4. Nov 22, 2007 #3
    I got:

    Let y = pix/L. So the limit becomes pi and 3pi/4
    ∫sin^2(y) dy
    = 0.5∫(1-cos2y) dy
    = 0.5(y - 0.5sin2y), y from 3pi/4 to pi
    =0.5(pi/4-0.5)

    1 = A^2*0.5(pi/4-0.5)

    I'm not sure if I did that right though...

    And how do I go from this to finding the probability that x>0.75L I'm a bit confused...
     
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