Commutation Relationships and Operator Functions

[chill_factor]

There are 2 operators such that [A,B] = 0. Does [F(A),B]=0 ?

Specifically, let's say we had the Hamiltonian of a 3-D oscillator H and L^2. We know that L^2 = Lx^2+Ly^2+Lz^2, and it is known that [H,Lz] = 0. Can we say that since H and Lz commute, H and Lz^2 also commute, by symmetry H and Lx^2,Ly^2 commute also and therefore H and L^2 commute?

 

[Chopin]

If the Taylor expansion of $F(A)$ converges, then you can essentially assume that it is a polynomial in $A$, so it will commute. Your argument about $H$ and $L^2$ sounds right.