Commutation relation(s) for integrated operators

Question

Suppose I have two operators $hat{A}$ and $hat{B}$, where $hat{A}=hat{a_1}+hat{a_2}+...+hat{a_m}$, and $hat{B}=hat{b_1}+hat{b_2}+...+hat{b_n}$. Is there a necessary and sufficient condition for $hat{a_1},hat{a_2},...hat{a_m}$ and  $hat{b_1},hat{b_2},...hat{b_n}$ to make $[hat{A},hat{B}]=0$ ?
If so, how to give a formal proof? Thanks!! (I deleted my original assumption since I found a counterexample, but I'm still wondering if there's a general relation that can be proved.)

Emmy · Accepted Answer

If $A = a_1+cdots+a_m$ and $B = b_1+cdots+b_n$ we see easily that:
begin{eqnarray}
[A, B] &=& [a_1, b_1] + cdots + [a_1, b_n]
&+&
&vdots& 
&+&[a_m, b_1] + cdots + [a_m, b_n]
end{eqnarray}
so if all $a$'s commute with all $b$'s then $[A, B] = 0$ but I don't think we can say much more. In particular, I don't see why you would need the $a$'s to commute with one another and what you mean by the $b$'s having "the same commutation relation"

Commutation relation(s) for integrated operators

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