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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.

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Someone correct me, but it seems to me, that this is an assumption for the proof, rather than a general statement about arbitrary operators. – CuriousOne Aug 30 '14 at 5:08
@CuriousOne I think Ignacio may have worded this slightly ambiguously: $[A,\,[A,\,B]]=[B,\,[A,\,B]]=0$ is an assumption, and the proof Ignacio seeks is $[A,\,[A,\,B]]=[B,\,[A,\,B]]=0 \Rightarrow [A,\,F(B)]=[A,\,B]\,F(B)$ (given analyticity assumptions on $F$ and an assumption that the domains of $A,\,B,\,C$ are restricted to make the analyticity assumptions work). Is this a sound way of putting your question, Ignacio? – WetSavannaAnimal aka Rod Vance Aug 30 '14 at 7:39
@WetSavannaAnimalakaRodVance: Thanks! Now I get it. – CuriousOne Aug 30 '14 at 7:58
For the truncated BCH formula, see also and links therein. – Qmechanic Nov 20 '15 at 13:45

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 $F'(B)$.

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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 $[B,C]\neq 0$.

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It is definitely an assumption. I don't think the questioner has a problem with it being an assumption. He clearly states that he fails to see how this assumption implies $[A,F(B)]=[A,B]F'(B)$. – Martin Nov 20 '15 at 14:05

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