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Perturbation theory infinite well

  1. Oct 6, 2012 #1
    in the infinite well with small potential shown in the attachment.
    I calculated the total energy by using the time independent Schrodinger equation and adding the correction energy to the equation of the slope k=(Vo/L)x.

    E=h^2/8mL^2 +∫ ψkψ dx

    ψ=√(2/L) sin⁡(∏/L x)

    when integrating ∫ ψkψ dx between 0 and L
    I got Zero, ∫ ψkψ dx=0
    ∴ total energy=h^2/8mL^2 +0

    so what i dont understand is when adding a small potential it doesnt affect the total energy of the system? that is what it shows when i integrated it.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 6, 2012 #2

    TSny

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    I don't believe that integral is zero. Check your work to see if you might have made a mistake.
     
  4. Oct 7, 2012 #3
    i have looked over and did it again, i still get zero.
    could you please try out the integral
    ∫ ψkψ dx

    where ψ=√(2/L) sin⁡((∏/L) x)
    and
    k=(Vo/L)x
     
  5. Oct 7, 2012 #4

    TSny

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    Maybe you can show some of the steps of how you integrated [itex]\int_0^L sin^2(\pi x/L)\; x \;dx[/itex]. Note that the integrand is positive throughout the range of integration. So, it's got to yield a positive result.
     
  6. Oct 7, 2012 #5
    the pictures in the attachment shows some of my working
     

    Attached Files:

  7. Oct 7, 2012 #6
    looking over it again i see where i went wrong, i used integration in the product rule. sorry for wasting your time and thank you very much!
     
  8. Oct 7, 2012 #7
    sorry, please don't read my previous comment. i see where i acutely went wrong, i have to use integration by parts. i have a problem of mixing up integration and differentiation.
    hope im correct this time
    sorry and thank you
     
  9. Oct 7, 2012 #8

    TSny

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    No need to be sorry. Glad to be of help.
     
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