PROBLEM
Suppose that $$ \sum_{m=-l}^{l} c_m Y_m^l(\theta, \phi) Y_m^l(\theta', \phi')^* $$ is rotationally invariant, then how can we show that the $c_m$'s must be all equal?
ATTEMPT AT A SOLUTION
I reformulated the rotational invariance in Dirac Notation: $$ \sum_{m=-l}^{l} c_m \hat D^{\dagger} | lm \rangle \langle lm | \hat D = \sum_{m=-l}^{l} c_m | lm \rangle \langle lm | $$ where $\hat D$ is the Unitary Representation in $L^2(S^2)$ of an arbitrary rotation in $SO(3)$.
Now, my idea was to show that we can choose $\hat D$ as a change of basis operator so that all the basis elements remain equal except two which are exchanged. If that were the case then it would be easy to show that the corresponding coefficients must be equal by the above relation.
I would like to know if my idea seems to be on the right track and how could I develop it into a complete proof.
