# Laplace equation azimuthal and non azimuthal symmetry

In general we can say that if the potential is specified in the surface of a sphere with azimuthal symmetry $(m=0)$ has a solution:

$$\Phi(r,\theta)= \sum_{l=0}^\infty \left[ A_lr^l + B_lr^{-(l+1)}\right]P_l\left(cos\theta \right)$$

and the potential for a sphere without azimuthal symmetry:

$$\Phi(r,\theta,\phi)= \sum_{l=0}^\infty \sum_{m=-l}^l \left[ A_{l,m}r^l + B_{l,m}r^{-(l+1)}\right]Y_{l,m}\left( \theta,\phi \right)$$

Lets think in the case (for azimuthal symmetry) where the sphere has potentials $\Phi=\Phi_0$ for the northern hemisphere and $\Phi=-\Phi_0$ for the southern.

That will give me certain solution.

If I rotate that sphere $\frac{\pi}{2}$ only in the $z-axis$ (left hemisphere $\Phi=-\Phi_o$ and right hemisphere $\Phi=\Phi_0$), that will give me another solution using the second ecuation.

The question is how can I get from one solution to the other using a rotation matrix?