Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
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

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 $F'(B)$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.