This problem is best addressed in bra-ket notation, where you can write the angular part of the state as:

$$ \psi_{\Omega} = c_{1,1}|1,1\rangle + c_{1,-1}|1,1\rangle + c_{1,0}|1,0\rangle $$

where you have already computed the $c_{1,m}$.

It is an eigenstate $\hat L^2$ because each basis state in the expansions is an eigenstate ($l=1$) with the same eigenvalue.

It is not an eigenstate of $L_z$, though, as:

$$ L_z|1,m\rangle = m\hbar|1,m\rangle $$

so by inspection:

$$L_z|\psi_{\Omega}\rangle \propto c_{1,1}|1,1\rangle - c_{1,-1}|1,1\rangle $$

which is not proportional to $\psi_{\Omega}$.

This problem is best addressed in bra-ket notation, where you can write the angular part of the state as:

$$ \psi_{\Omega} = c_{1,1}|1,1\rangle + c_{1,-1}|1,-1\rangle + c_{1,0}|1,0\rangle $$

where you have already computed the $c_{1,m}$.

It is an eigenstate $\hat L^2$ because each basis state in the expansions is an eigenstate ($l=1$) with the same eigenvalue.

It is not an eigenstate of $L_z$, though, as:

$$ L_z|1,m\rangle = m\hbar|1,m\rangle $$

so by inspection:

$$L_z|\psi_{\Omega}\rangle \propto c_{1,1}|1,1\rangle - c_{1,-1}|1,-1\rangle $$

which is not proportional to $\psi_{\Omega}$.

This problem is best addressed in bra-ket notation, where you can write the state as:

$$ \psi = c_{1,1}|1,1\rangle + c_{1,-1}|1,1\rangle + c_{1,0}|1,0\rangle $$

where you have already computed the $c_{1,m}$.

It is an eigenstate $\hat L^2$ because each basis state in the expansions is an eigenstate ($l=1$) with the same eigenvalue.

It is not an eigenstate of $L_z$, though, as:

$$ L_z|1,m\rangle = m\hbar|1,m\rangle $$

so by inspection:

$$L_z|\psi\rangle \propto c_{1,1}|1,1\rangle - c_{1,-1}|1,1\rangle $$

which is not proportional to $\psi$.

This problem is best addressed in bra-ket notation, where you can write the angular part of the state as:

$$ \psi_{\Omega} = c_{1,1}|1,1\rangle + c_{1,-1}|1,1\rangle + c_{1,0}|1,0\rangle $$

where you have already computed the $c_{1,m}$.

It is an eigenstate $\hat L^2$ because each basis state in the expansions is an eigenstate ($l=1$) with the same eigenvalue.

It is not an eigenstate of $L_z$, though, as:

$$ L_z|1,m\rangle = m\hbar|1,m\rangle $$

so by inspection:

$$L_z|\psi_{\Omega}\rangle \propto c_{1,1}|1,1\rangle - c_{1,-1}|1,1\rangle $$

which is not proportional to $\psi_{\Omega}$.