$\left [(A- \langle A \rangle ),\right (B- \langle B \rangle )]=\left [A,B \right ] $

Question

Let A,B two hermitian operator I don't grasp one of the last step of the proof of Uncertainty principle. Why: $$\left [(A- \langle A \rangle ),\right (B- \langle B \rangle )]=\left [A,B \right ] $$ I'm stucked here: $$\left [(A- \langle A \rangle ),\right (B- \langle B \rangle )]=\left [AB-A\langle B \rangle - \langle A \rangle B + \langle A \rangle \langle B \rangle \right] $$

Answer 1

A commutator does not see scalar perturbations (or, more generally, perturbations by operators that commute with the operators involved). Concretely, if $J,K$ commute with $A$ and $B$ respectively, and between themselves, \begin{align} [A+K,B+J]&=(A+K)(B+J)-(B+J)(A+K)\\ \ \\ &=AB-BA+AJ+KB+KJ-BK-JA-JK\\ \ \\ &=AB-BA=[A,B]. \end{align}