I think you are looking for the Wigner-Eckart theorem. The operators $x,y,z$ themselves form a representation of the rotation group with $l=1$. That is if we take the operator $T^0$ to be $z$ and $T^\pm$ to be $\mp (x\pm i y)$, and have $R$ be a member of the rotation group, and $U(R)$ the unitary rotation operator

$$U(R) T^i U(R)^{-1} = {R^{(1)}}^{i}_{\,\,\,j} T^j$$

where $R(1)$ is the irreducible representation with $l=1$.

The Wigner-Eckart theorem pretty much says you can add the angular momentum quantum numbers of the operator and the state in the usual way, so an operator $x,y,z$ acting on a $l=0$ state has $l=1$.

It's not so difficult to see why:

$$U(R) (T^i |l,m\rangle) =U(R) T^i U(R)^{-1} (U(R)|l,m\rangle) = {R^{(1)}}^{i}_{\,\,\,j}{R^{(l)}}^{m^\prime}_{\,\,\,m} T^j |l,m^\prime\rangle$$

This looks just like the tensor product of two representations, which we know leads to the addition of angular momenta.

I think you are looking for the Wigner-Eckart theorem. The operators $x,y,z$ themselves form a representation of the rotation group with $l=1$. That is if we take the operator $T^0$ to be $z$ and $T^{\pm 1}$ to be $\mp (x\pm i y)$, and have $R$ be a member of the rotation group, and $U(R)$ the unitary rotation operator

$$U(R) T^i U(R)^{-1} = {R^{(1)}}^{i}_{\,\,\,j} T^j$$

where $R^{(1)}$ is the irreducible representation with $l=1$.

The Wigner-Eckart theorem pretty much says you can add the angular momentum quantum numbers of the operator and the state in the usual way, so an operator $x,y,z$ acting on a $l=0$ state has $l=1$.

It's not so difficult to see why:

$$U(R) (T^i |l,m\rangle) =U(R) T^i U(R)^{-1} (U(R)|l,m\rangle) = {R^{(1)}}^{i}_{\,\,\,j}{R^{(l)}}^{m^\prime}_{\,\,\,m} T^j |l,m^\prime\rangle$$

This looks just like the tensor product of two representations, which we know leads to the addition of angular momenta.