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Quantum Scattering

  1. May 29, 2014 #1
    I need a little more explanation about the solution discussed in the following thread.

    https://www.physicsforums.com/showthread.php?t=410830

    Do I've to lookup for orthogonal functions?

    PS. required thread is closed so i'm posting here.
     
  2. jcsd
  3. May 29, 2014 #2
    Can someone clarify how did they get equation 15.5.7?

    2gsfred.jpg
     

    Attached Files:

  4. May 29, 2014 #3
    They just compared the terms in the expansion term one by one. If the two expansions are equal to each other then each coefficient of the expansion is equal to each other. It's possible to prove that by use of the orthogonality of the functions as you guessed.
     
  5. May 31, 2014 #4
    ok after substituting values in Equation [15.5.6]

    we get

    $$\sum A_{l}\frac{e^{\iota(kr-\pi/2l+\delta_{l})}-e^{-\iota(kr-\pi/2l+\delta_{l})}}{2\iota kr}P_{l}(Cos\theta)= \sum[(2l+1)\iota^{l} \frac{e^{\iota(kr-\pi/2l+\delta_{l})}-e^{-\iota(kr-\pi/2l+\delta_{l})}}{2\iota kr}P_{l}(Cos\theta)
    + \frac{1}{r}\sum f_{l}P_{l}(Cos\theta)e^{\iota kr}]P_{l}(Cos\theta) $$

    simplifying and making [itex]P_{l}(Cos\theta) \rightarrow 1[/itex]

    we have

    [itex]A_{l}\frac{e^{\iota{kr-\pi/2l+\delta_l}}}{2\iota kr} - A_{l}\frac{e^{-\iota kr-\pi/2l+\delta_l}}{2\iota kr} = (2l+1) i^{l}\frac{e^{\iota kr-\pi/2l+\delta_l}}{2\iota kr} - (2l+1) i^{l}\frac{e^{-\iota kr-\pi/2l+\delta_l}}{2\iota kr} + f_{l} \frac{e^{\iota kr}}{r} [/itex]

    now what? please give me some hint.
     
    Last edited: May 31, 2014
  6. May 31, 2014 #5
    "Now what" what? what are you trying to calculate? Also, you can't make Pl(cosθ) equal 1. That makes no sense. It is a function of θ.
     
  7. May 31, 2014 #6
    I'm trying to follow the steps in the given solution. I don't know how to use orthogonality condition here.
     
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