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Integration by parts of derivative of expectation value problem

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data
    I don't know how the writer of the book took integral of the first statement and got the second statement? Can anybody clarify on this?
    Problem.png


    2. Relevant equations
    Given in the photo


    3. The attempt at a solution
    When I took the integral I just ended up with the exact same statement but without the negative sign behind the (ihbar/2m)
     
  2. jcsd
  3. Feb 20, 2013 #2

    tiny-tim

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    Welcome to PF!

    Hi Normalization! Welcome to PF! :smile:
    But that negative sign is always there in integration by parts. :confused:

    (see the last equation in your photo)
     
  4. Feb 20, 2013 #3
    Oh w8 (facepalm) they haven't actually integrated yet? Wow... I feel so stupid right now. I thought they already integrated both products inside the brackets in which case -[itex]\int[/itex][itex]\frac{\partial\Psi^*}{\partial\ x}[/itex][itex]\Psi[/itex]dx-[itex]\int[/itex][itex]\frac{\partial\Psi}{\partial\ x}[/itex]×-[itex]\Psi^*[/itex]dx Which would be the exact same thing. Alright thanks I guess...
     
  5. Feb 20, 2013 #4

    tiny-tim

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    I'm not convinced you've got this. :confused:

    They have integrated that big bracket, and so they also differentiate the x (to get 1) …

    the […] term is 0 (so they haven't written it), and that only leaves the ∫ term, which always has a minus in front of it. :smile:
     
  6. Feb 20, 2013 #5
    Yes,yes,yes I know but I thought they had already taken the integral of -int(df/dx g dx)
     
    Last edited: Feb 20, 2013
  7. Feb 20, 2013 #6
    Which by nature should be equal to the left side of the equation. So it's fine I was just being daft
     
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