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Quantum particle in a 1d box

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data
    A quantum particle is free to move in a infinite 1d potential well of length L excited on the second excited state n=3.What is the probability to locate the particle between 0 and L/3.


    2. Relevant equations
    ∫ψ^2=1
    ψ=sqrt(2/L)sin(n∏x/L)
    3. The attempt at a solution
    ∫ψ^2dx=∫(2/L)sin^2(3∏x/L) dx limits of integration from 0 to L/3. then I set y = 3∏x/L then dx = Ldy/3∏. So that gives (2/3∏)∫(1/2)-(1/2)cos(2y) dy then the limits change to 0 to ∏
    which gives (2/3∏)((y/2)-(1/4)sin(2y)) now use the limits 0 to pi which gives a probability of 1/3. What I don't understand is why use 0 to pi why not ∏/3 confused there . what I assume is that it has to do with the excited state the particle is in. So for instance if it was n=1 it would be 0 to ∏/3 am I correct.
     
  2. jcsd
  3. Oct 24, 2012 #2
    NVM I was dumb I figured it out it has to do with the length being defined as L it helps to read that being said I need help with a conceptual problem.
     
  4. Oct 24, 2012 #3
    Uggg ok neverming maybe my point wass valid I just did a problem that was similar but in the ground state and it used limits from 0 to npi/3 help again.
     
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