Glauber's Formula

Question

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.

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 $F'(B)$.

Answer 2

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

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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} $$ \Rightarrow \quad Y(t) = \sum_{n=0}^{\infty} \dfrac{t^n}{n!}Y^{(n)}(t)|_{t=0}=B+t[A,B]+\dfrac{t^2}{2!}[A,[A,B]]+\dfrac{t^3}{3!}[A,[A,[A,B]]]+\cdots $$

Answer 3

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 $[B,C]\neq 0$.