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Associated Legendre polynomials

  1. Nov 22, 2014 #1

    m=1 and l=1

    x = cos(θ)

    What would be the solution to this?

    Last edited: Nov 22, 2014
  2. jcsd
  3. Nov 22, 2014 #2


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    Gold Member

    We won't do the calculation for you. Why can't you do it yourself? What part of it doesn't make sense to you?
  4. Nov 22, 2014 #3
    I've tried a lot of things, but I don't get it. (NOT a physics/math student)

    I get this:

    4cos(θ)^3 - 4cos(θ)

    But it should be sin(θ) since I'm applying this formula:


    and the solution for 1,1 is Sqrt(3/8π) e^((+/-)iφ) * sin(θ)
    I get how the part in italics is derived, but not how the part in bold has been derived from the associated legendre polynomial..
  5. Nov 22, 2014 #4


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    I think you've made an error in your calculation. If we plug in ##\ell = 1## and ##m=1##, we have $$P^{1}_{1}={(-1)^1 \over 2 \cdot 1!} (1-x^2)^{1/2} \cdot {d^2 \over dx^2}(x^2-1)$$
    Do you have that much? It should be easy to simplify that, then plug in ## x=\cos{\theta}##.
  6. Nov 22, 2014 #5
    Great I have it now. -(1-cos()^2)^(1/2), which is equal to sin(θ).

    Right I have it now, like you said I made a small error in not applying l+m, but instead just l and some errors on other places as well.

    Thanks for the quick reply! :)
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