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Transformation Function - Position & Momentum Operators

  1. Jun 9, 2009 #1
    1. The problem statement, all variables and given/known data
    I am currently studying for my quantum physics exam and I am trying to derive the Transformation function:
    ⟨x'│p' ⟩=Nexp{(ip' x')/ℏ}
    2. Relevant equations

    ⟨x'│p' ⟩=Nexp{(ip' x')/ℏ}

    3. The attempt at a solution

    Now I get how to get to
    p'⟨x'│p' ⟩=-iℏ d/dx' ⟨x'│p' ⟩

    but i cant work out how to get both the p' and the x' into the expontential and im not sure how to explain where the N comes from. Every example i find seems to jump from p'⟨x'│p' ⟩=-iℏ d/dx' ⟨x'│p' ⟩ straight to ⟨x'│p' ⟩=Nexp{(ip' x')/ℏ} without any explanation.
    Would anybody be able to explain how to do this?
    Thanks
     
    Last edited: Jun 9, 2009
  2. jcsd
  3. Jun 9, 2009 #2

    diazona

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    Homework Helper

    Well, you have a differential equation,
    [tex]p \langle x \vert p \rangle = -i\hbar \frac{\mathrm{d}}{\mathrm{d}x} \langle x \vert p \rangle[/tex]
    do you know how to solve it?
     
  4. Jun 9, 2009 #3
    No. Is this problem easier than im making it?
     
  5. Jun 9, 2009 #4

    Cyosis

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    Homework Helper

    Do you know how to solve [itex] a f(x)=f'(x)[/itex]? This is exactly the same.
     
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