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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 at 5:08
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@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 at 7:39
    
@WetSavannaAnimalakaRodVance: Thanks! Now I get it. –  CuriousOne Aug 30 at 7:58

1 Answer 1

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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