Associated Legendre polynomials

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1. Nov 22, 2014

bvol

m=1 and l=1

x = cos(θ)

What would be the solution to this?

Thanks.

Last edited: Nov 22, 2014
2. Nov 22, 2014

ZetaOfThree

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?

3. Nov 22, 2014

bvol

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..

4. Nov 22, 2014

ZetaOfThree

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 $x=\cos{\theta}$.

5. Nov 22, 2014

bvol

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! :)