# 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?

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maybe this answer will help: physics.stackexchange.com/questions/90173/… – Adam May 28 '14 at 14:48

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.
Well, a spherically symmetric wavefunction must be constant along all angular directions. Fixing the $z$ axis in a arbitrary direction $L_z =-i\frac{\partial}{\partial \phi}$ and thus $L_z\psi=0$ since $\psi$ is constant in $\phi$. As $z$ can be fixed in an arbitrary direction, $L_x\psi=L_y\psi = L_z\psi =0$ and thus $L^2 \psi =L_xL_x\psi + L_yL_y\psi +L_zL_z\psi=0$. – Valter Moretti May 28 '14 at 15:53
You actually said spherically symmetric spherical harmonic. However, the complete wavefunction is the product of such symmetric spherical harmonic and a function of $r$, so it is a spherically symmetric wavefunction. – Valter Moretti May 30 '14 at 6:21