Glauber's Formula In the Cohen-Tannoudji Quantum Physics book, Complement BII, says:
[...] two operators $A$ and $B$ with both commute with their commutator. An argument modeled on the preceding one shows that, if we have:
\begin{align}
[A,C]=[B,C]=0
\end{align}
with $C=[A,B]$, then:
\begin{align}
[A,F(B)]=[A,B]F'(B)
\end{align}
Then this last property, is used to proof Glauber's Formula.
\begin{align*}
e^Ae^B=e^{A+B}e^{\frac{1}{2}[A,B]}
\end{align*}
I understand this proof. 

But I couldn't find a way to demonstrate,
  \begin{align}
[A,C]=[B,C]=0
\end{align}
  with $C=[A,B]$, then:
  \begin{align}
[A,F(B)]=[A,B]F'(B).
\end{align}

I would like to know to do this, so I can understand better the Glauber's Formula proof.
 A: A standard thing done in proofs of identities involving commutators is to expand things in Taylor series. Since $F(B) = \sum_{n=0}^\infty f_n B^n$, we have
$$ [A, F(B)] = \sum_{n=0}^\infty f_n (AB^n - B^nA) $$
(commutators distribute over sums, as you can check). Then take one of the parenthesized terms, say $B^nA$, and move the $A$ through to the other side of the $B$'s, one at a time. Each time picks up another term with a $C$ and $n-1$ $B$'s, which you are free to arrange however you want, because $[B,C] = 0$. You should find your $AB^n$ terms cancel, leaving
$$ [A, f(B)] = \sum_{n=0}^\infty f_n n B^{n-1} C, $$
where the $C$ can be put anywhere, including outside the sum. But the sum is just the Taylor series for $F'(B)$.
A: Here I will give some algebra method for the proof of Glauber's formula:
Assume $F(t)=e^{At} e^{Bt}$ :
$$
\dfrac{d}{d t} F(t) = A e^{A t} e^{B t} + e^{A t} B e^{B t} = (A+e^{A t}Be^{-At} ) F(t)  \tag{1}
$$
Recall that the Hadamard's lemma (Proved in appendix):
$$
\boxed{e^{A t} B e^{-A t} = B+ t[A,B]+\dfrac{t^2}{2!}[A,[A,B]]+\dfrac{t^3}{3!}[A,[A,[A,B]]]+\cdots} \tag{2}
$$
then the equation $(2)$ can be simplified as ($[A,B]=Constant$) :
$$
\dfrac{d}{d t} F(t) = (A+B+t[A,B])F(t) \tag{3}
$$
Assume $G(t) = e^{At+Bt+f(t)H(A,B)}$ :
$$
\dfrac{d}{d t}G(t) = (A+B+f'(t)H(A,B))G(t) \tag{4}
$$
Let $\dfrac{d}{d t}F(t) = \dfrac{d}{d t}G(t)$ :
\begin{align}
f'(t)  = t \Rightarrow f(t) & = \dfrac{1}{2}t^2+C \tag{5} \\ 
H(A,B) & = [A,B] \tag{6} \\
F(t) & = G(t) \tag{7}
\end{align}
$$
(5)\&(6)\&(7) \quad \Rightarrow \quad e^{A t}e^{B t} = e^{A t+B t+(\dfrac{1}{2}t^2+C)[A,B]}
$$


*

*$t=0 \Rightarrow C=0$;

*$t=1 \Rightarrow $ $$\boxed{e^A e^B=e^{A+B+\dfrac{1}{2}[A,B]}}$$



Appendix for Hadamard's lemma:
$$
\boxed{e^{A t} B e^{-A t} = B+ t[A,B]+\dfrac{t^2}{2!}[A,[A,B]]+\dfrac{t^3}{3!}[A,[A,[A,B]]]+\cdots}
$$
Assume $Y(t) =e^{A t} B e^{-At} $
\begin{align}
Y^{(1)}(t) & = e^{At}(AB-BA)e^{-At} = e^{At}[A,B]e^{-At} \\
Y^{(2)}(t) & = e^{At} A [A,B]-[A,B] A e^{-At} = e^{At}[A,[A,B]]e^{-At} \\
Y^{(3)}(t) & = e^{At} [A,[A,[A,B]]] e^{-At} \\
& \cdots\cdots \nonumber 
\end{align}
$$
\Rightarrow \quad Y(t) = \sum_{n=0}^{\infty} \dfrac{t^n}{n!}Y^{(n)}(t)|_{t=0}=B+t[A,B]+\dfrac{t^2}{2!}[A,[A,B]]+\dfrac{t^3}{3!}[A,[A,[A,B]]]+\cdots
$$
A: I think $[A,C]=[B,C]=0$ with $C=[A,B]$ is an assumption, because there exist counterexamples: for $A=\sigma_x$ are $B=\sigma_y$ Pauli matrix along $x, y$ directions respectively. then $C=2i\sigma_z$ is Pauli matrix along z direction. Obviously $[A,C]\neq 0$ and $[B,C]\neq 0$.
