State with non-zero angular momentum - cannot be described by spherical harmonic? For a state with non-zero angular momentum, why is it that it cannot be described by the spherically symmetric spherical harmonic?
 A: This is because  each spherically invariant state $\psi$  must have zero angular momentum.
Indeed, by hypotheses, the state $\psi$ verifies
$$\psi(R_{\vec n}(\theta)\vec{x} ) = (e^{i \theta \vec{n}\cdot \vec{\hat{J}}} \psi)(\vec{x}) = \psi(\vec{x})\tag{1}$$
where $\vec{n}\cdot \vec{\hat{J}}$ is the self-adjoint generator of rotations $R_{\vec n}(\theta)$ around $\vec{n}$, i.e. it is the angular momentum along $\vec{n}$.
Taking the $\theta$ derivative of (1) for $\theta=0$ we have
$$\vec{n}\cdot \vec{\hat{J}} \psi =0$$
in particular, for $k=x,y,z$,
$$\hat{J}_k \psi=0\:,$$
so that $$\hat{J}^2 \psi = \hat{J}^2_x\psi + \hat{J}^2\psi + \hat{J}_z^2=0\:.$$
ADDENDUM. Actually a state is represented by a normalized vector up to a phase. A spherically symmetric state is therefore represented by a vector satisfying  version of (1) weaker than the one presented above:
$$\psi(R_{\vec n}(\theta)\vec{x} ) = (e^{i \theta \vec{n}\cdot \vec{\hat{J}}} \psi)(\vec{x}) = \chi(\theta, \vec{n})\psi(\vec{x})\tag{2}$$
where $|\chi(\theta, \vec{n})|=1$.  Taking the $\theta$ derivative for $\theta=1$ we find
$$\vec{n}\cdot \vec{\hat{J}} \psi = \alpha(\vec{n}) \psi$$
where the eigenvalue is
$$\alpha(\vec{n}) = \frac{d\chi(\theta, \vec{n})}{d\theta}|_{\theta=0}$$
which is a real  number as easily follows from $|\chi(\theta, \vec{n})|=1$.
The  common eigenvectors $\psi \neq 0$ of $\hat{J}_x,\hat{J}_y,\hat{J}_z$ have the common  eigenvalue $0$ as it can be proved by direct inspection (or by means of some straightforward theoretical argument exploiting the commutation relations $[\hat{J}_x,\hat{J}_y]= i\hat{J}_z$). We conclude that this more complete way leads to the same result as before.
