What Is the Solution to Associated Legendre Polynomials for m=1 and l=1?

[bvol]

m=1 and l=1

x = cos(θ)

What would be the solution to this?

Thanks.

 

[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?

 

[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..

 

[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 $x=\cos{\theta}$.

 

[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! :)