= Solution
The <orbital angular momentum commutation relations> give
$$
[L_z,L_x]=i\hbar L_y,
\qquad
[L_z,L_y]=-i\hbar L_x.
$$
Consequently
$$
\begin{aligned}
[L_z,L_\pm]
&=[L_z,L_x]\pm i[L_z,L_y]\\
&=i\hbar L_y\pm\hbar L_x
=\boxed{\pm\hbar L_\pm}.
\end{aligned}
$$
The same commutation relations imply $[L^2,L_x]=[L^2,L_y]=0$, and hence
$$
\boxed{[L^2,L_\pm]=0}.
$$
These are the defining identities for the <angular momentum ladder operator>[angular momentum ladder operators].
If
$$
L_z\phi=m\hbar\phi,
\qquad
L^2\phi=\ell(\ell+1)\hbar^2\phi,
$$
then, whenever $L_\pm\phi\ne0$,
$$
\begin{aligned}
L_z(L_\pm\phi)
&=([L_z,L_\pm]+L_\pm L_z)\phi
=(m\pm1)\hbar L_\pm\phi,\\
L^2(L_\pm\phi)
&=L_\pm L^2\phi
=\ell(\ell+1)\hbar^2L_\pm\phi.
\end{aligned}
$$
\b[Thus $L_\pm\phi$ has quantum numbers $(\ell,m\pm1)$.]
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