# Spherical Harmonics and $L_+$ and $L_-$ operators

I have the spherical harmonics $Y_{m}^{l}\left(\theta,\varphi\right)$ and I want to show that the operators $L^{\pm}$ act as "creation" and "annihilation" operators such that

$$L^{\pm}Y_{m}^{l}=\sqrt{l\left(l+1\right)-m\left(m\pm1\right)}Y_{m\pm1}^{l}$$

I'm quite sure that if I substitute all the definitions and proceed analitically I will end up after 3 or 4 pages of calculations with the right solution, but I'm wondering if there's a smarter and algebric way to demonstrate this equation.

Does anybody have at least an hint to follow?

Here's the definitions I used: Spherical Harmonics $$Y_{m}^{l}\left(\theta,\varphi\right)=\left(-1\right)^{m}\sqrt{\frac{\left(2l+1\right)}{4\pi}\frac{\left(l-m\right)!}{\left(l+m\right)!}}P_{l}^{m}(\cos\theta)e^{\imath m\varphi}$$ Legendre Functions $$P_{l}^{m}(x)=\frac{1}{2^{l}l!}\left(1-x^{2}\right)^{m/2}\left(\frac{d}{dx}\right)^{l+m}\left(x^{2}-1\right)^{l},$$

Operators $$L^{+} =L_{1}+\imath L_{2}=e^{\imath\varphi}\left(-\imath\frac{\partial}{\partial\theta}+\cot\theta\frac{\partial}{\partial\varphi}\right),$$ $$L^{-} =L_{1}-\imath L_{2}=e^{-\imath\varphi}\left(\imath\frac{\partial}{\partial\theta}+\cot\theta\frac{\partial}{\partial\varphi}\right),$$ $$L_{z} =\imath L_{3}=-\imath\frac{\partial}{\partial\varphi}.$$ Commutation relations $$\left[L_{z},\,L^{+}\right]=L^{+},$$ $$\left[L_{z},\,L^{-}\right]=-L^{-},$$ $$\left[L^{+},L^{-}\right]=2L_{z}.$$ Laplacian or Casimir Operator $$L^{-}L^{+} =L_{1}^{2}+L_{2}^{2}+\imath\left[L_{1},\,L_{2}\right],$$ $$L^{+}L^{-} =L_{1}^{2}+L_{2}^{2}-\imath\left[L_{1},\,L_{2}\right],$$ $$\mathbf{L}^{2}=L_{z}^{2}+L_{z}+L^{-}L^{+}.$$