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A problem from Sakurai

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data
    (Sakurai 1.27)
    [...] evaluate
    [tex]\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle[/tex]
    Simplify your expression as far as you can. Note that [itex]r = \sqrt{x^2 + y^2 + z^2}[/itex], where x, y and z are operators.

    2. Relevant equations
    [tex]\langle \mathbf{x'} | \mathbf{p'} \rangle = \frac{1}{ {(2 \pi \hbar)}^{3/2} }exp(i \mathbf{p'} \cdot \mathbf{x'} / \hbar)[/tex],
    [tex]F(r) | \mathbf{x'} \rangle = F(|\mathbf{x'}|) | \mathbf{x'} \rangle[/tex]
    [tex]\langle \mathbf{x''} | \mathbf{x'} \rangle = \delta(\mathbf{x''} - \mathbf{x'})[/tex]

    3. The attempt at a solution
    Using the resolution of identity, I wrote
    [tex]\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle = \int \int \langle \mathbf{p''} | \mathbf{x''} \rangle \langle \mathbf{x''} |F(r) | \mathbf{x'} \rangle \langle \mathbf{x'} | \mathbf{p'} \rangle d \mathbf{x''} d \mathbf{x'}[/tex]
    Using the above "relevant equations", I get
    [tex]\frac{1}{(2 \pi \hbar)^3}\int exp(i (\mathbf{p'} - \mathbf{p''}) \cdot \mathbf{x'} / \hbar) F(|\mathbf{x'}|) d \mathbf{x'}[/tex]

    Is this the correct answer? (wish that answers were avaiable at the backside of the book...)
    Last edited: Oct 26, 2008
  2. jcsd
  3. Oct 29, 2008 #2
    So far this looks good: but you can simplify even further by using spherical coordinates, with the z-axis chosen in a smart way: you will end up with an integral only over r, no more angles.
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