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Schrodinger equation help

  1. Jan 8, 2013 #1
    1. The problem statement, all variables and given/known data
    What is the probability that a particle in the ground state will be found between L/2 and 2L/3?
    im new guys so go easy :)

    2. Relevant equations
    ∫ψ(x)^2 dx = ∫2/L (sin(πx/L))^2 dx
    in attachment

    3. The attempt at a solution
    The answer should be 30.44%
    i got 66.66% and sometimes a negative number
    please show me the steps too :/
     

    Attached Files:

  2. jcsd
  3. Jan 8, 2013 #2
    So you have the integral:[tex]\int{\psi \psi^{*} dx} = \int^{\frac{L}{2}}_{\frac{2L}{3}}{\sqrt{\frac{2}{L}} \sin{\frac{n \pi x}{L}} \sqrt{\frac{2}{L}} \sin{\frac{n \pi x}{L}} dx} = \int^{\frac{L}{2}}_{\frac{2L}{3}}{\frac{2}{L} \sin^{2}{\frac{n \pi x}{L}} dx}[/tex] You can get the integral of sine squared from an integral table:[tex]\int{\sin^{2}{ax} dx} = \frac{x}{2} - \frac{\sin{2ax}}{4a}[/tex] Keep in mind that both [itex]\frac{x}{2}[/itex] and [itex]\frac{\sin{2ax}}{4a}[/itex] are evaluated at the limits of integration.

    Doing all of this I obtained 30% for the answer when I plugged in n = 1 (for the ground state).

    You're probably just messing up the minus sign on one of the 4 terms that come about when you evaluate the [itex]\frac{x}{2} - \frac{\sin{2ax}}{4a}[/itex] term at the limits of integration. You 4 terms should be [tex]\int{\psi \psi^{*} dx} = \frac{2}{L} (\frac{L}{3} - \frac{L}{4} - \frac{\sin{(2\frac{n \pi}{L}\frac{2L}{3}})}{4 \frac{n \pi}{L}} + \frac{\sin{(2 \frac{n \pi}{L} \frac{L}{2}})}{4 \frac{n \pi}{L}})[/tex]

    Edit*** After a 3rd check, when n = 1 the answer is indeed 30%. Looks like we are both susceptible to math errors on this one >_< (I had edited my post thinking the answer was 60% when I double checked my original answer...)
     
    Last edited: Jan 8, 2013
  4. Jan 8, 2013 #3
    I love you :')
     
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