My notes define:
$$ L_{\pm} = L_{x} \pm i L_{y} $$
and states: $$ [L_{z},L_{\pm}] = \pm \hbar L_{\pm} $$
I'm fine with this as it's easy to show the result with some ugly algebra.
It then says:
Since each component of the angular momentum commutes with $ L^2 $ we can deduce that the action of $ L_± $ on $ |a, b> $ cannot affect the value of a relating to the magnitude of the angular momentum.
I'm happy to plug things in to prove it but I want to see how to deduce it if possible $$ $$ I understand that $ L^2$ commutes with $L_\pm$ because $L^2$ commutes with the individual $L_x$ and $L_y$ that make up $L_\pm$, but I don't see why that means it can't affect the magnitude.
Two operators which commute can be simultaneously known but that doesn't help because the ladder operator isn't Hermitian so isn't an observable.
Any help apreciated!
$$ $$ $$ $$ $$ $$
EDIT: Got it.
$$ [L^2 , L_i] = 0 $$ so $$ L^2 ( L_\pm | a, b > ) = L_\pm (L^2 |a, b>) = a L_\pm |a, b> $$
Forgot the fundamental property that commuting operators... commute.