Is the $1s$ state of Hydrogen eigenfunction of all the three of the Hamiltonian, the orbital angular momentum $\vec L$, and $L_z$?

Question

Let $H$ be the Hamiltonian, $\vec L$ the orbital angular momentum, and $L_z$ its projection on the z-axis. Is the 1s state of Hydrogen an eigenfunction of all the three operators? I suppose it should be eigenfunction of $L_z$ and $H $ only.

Answer 1

The orbital angular momentum operator in spherical coordinates is $$\vec{L}=i\hbar\left( \frac{\hat{\theta}}{\sin\theta}\frac{\partial}{\partial\phi} -\hat{\phi}\frac{\partial}{\partial\theta} \right)$$ where $\hat{\theta}$ and $\hat{\phi}$ are the unit-vectors in $\theta$ and $\phi$ direction.

The $1s$ wavefunction of the hydrogen atom is spherically symmetric. $$\psi_{1s}(r,\theta,\phi)=C e^{-r/a_0}$$

Because this wavefunction doesn't depend on $\theta$ and $\phi$ you have $$\vec{L}\ \psi_{1s}(r,\theta,\phi)=\vec{0}$$

Hence the wavefunction is eigenfunction of $\vec{L}$ for eigenvalue $\vec{0}$.