How does the following commutator for measured observables and this operator relation imply the following relation? $$
\hat{\Omega}_j{(\tilde{q}_j)}=\Omega_j(\tilde{q}_j-\hat{q}_j)
$$
$$
[\hat{q}_j,\hat{q}_l]=ik_{jl}
$$
Implies
$$
[\hat{q}_j,\hat{\Omega}_l]= \frac{\partial\Omega_l(\tilde{q}_l-\hat{q}_l)}{\partial\hat{q}_l}.ik_{jl}
$$
 A: Hints to the question (v1): 


*

*Let $\hat{Z}^I$ be operators that satisfies a Heisenberg algebra 
$$\tag{1} [\hat{Z}^I,\hat{Z}^J]~=~i\hbar ~\omega^{IJ} ~{\bf 1},$$ 
where $\omega^{IJ}=-\omega^{JI}$ is an antisymmetric real matrix. The important property will be that the commutators $[\hat{Z}^I,\hat{Z}^J]$ belong to the center of the Heisenberg algebra, i.e. commute with everything.

*Let $f(\hat{Z})$ and $(\partial_{J}f)(\hat{Z})$ denote the Weyl-ordered operators corresponding to the functions/symbols $f(z)$ and 
$$\tag{2} (\partial_{J}f)(z)~:=~\frac{\partial f(z)}{\partial z^J},$$ 
respectively. 

*Then using e.g. this formula for the Weyl-ordered operators, it is possible to show$^1$ 
$$\tag{3} [\hat{Z}^I, f(\hat{Z})]~=~\sum_J [\hat{Z}^I,\hat{Z}^J]~(\partial_{J}f)(\hat{Z}).$$

*The corresponding classical formula
$$\tag{4} \{z^I, f(z)\}_{PB}~=~\sum_J \{z^I,z^J\}_{PB}~(\partial_{J}f)(z)$$
is a consequence of the Poisson property/Leibniz rule for a Poisson bracket (PB).
--
$^1$ Equation (3) actually also holds for many other operator-orderings.
