Natural exponential function with noncommutative exponents.

Question

I've been studying Quantum Mechanics, and I'm stuck in this problem:

Show that if $[A, B] = k$ then $e^{A} e^{B} = e^{A + B +\frac{1}{2} [A, B]}$, where k is a constant and A and B are operators.

There's a tip to write $e^{\lambda A} e^{\lambda B}$, differentiate, then integrate and set $\lambda = 1$, but I can't get to the result.

Thanks.

Answer 1

Let $C(\lambda) = e^{\lambda A}e^{\lambda B}$. Then \begin{align*} \frac{d}{d\lambda} C &= A C + C B \\ &= A C + B C + [C,B]. \end{align*} But, since $[XY,Z]=X[Y,Z]+[X,Z]Y$, \begin{align*} [C,B] &= [e^{\lambda A}e^{\lambda B},B] \\ &= e^{\lambda A}[e^{\lambda B},B] + [e^{\lambda A},B]e^{\lambda B} \\ &= [e^{\lambda A},B]e^{\lambda B}. \end{align*} Prove by induction that $[A^n,B]=n A^{n-1}[A,B]$. (This is a standard calculation.) Then \begin{align*} [e^{\lambda A},B] &= \sum_{n=0}^\infty \frac{\lambda^n}{n!}[A^n,B] \\ &= \sum_{n=0}^\infty \frac{\lambda^n}{n!}n A^{n-1}[A,B] \\ &= \lambda \sum_{m=0}^\infty \frac{\lambda^m}{m!}A^m [A,B] \\ &= \lambda e^{\lambda A}[A,B]. \end{align*} Thus, \begin{align*} \frac{d}{d\lambda} C &= (A+B)C + \lambda e^{\lambda A}[A,B]e^{\lambda B} \\ &= (A+B+\lambda [A,B])C, \end{align*} if $[A,B]=k$. Therefore, up to a multiplicative constant that may be determined to be the identity, $$C(\lambda)=e^{\lambda(A+B)+\frac{\lambda^2}{2}[A,B]}.$$ Since $C(1)=e^A e^B$, $$e^A e^B = e^{A+B+\frac{1}{2}[A,B]},$$ as claimed.

Answer 2

You may e.g. show first that $A^n B = B A^n + k n A^{n-1}$ so that $e^{\lambda A} B = B e^{\lambda A} + k \lambda e^{\lambda A}$ Then $f(\lambda)=e^{\lambda A} e^{\lambda B}$ has as derivative $$ f'(\lambda) = A f(\lambda) + f(\lambda) B = (A + B + k \lambda ) f(\lambda)$$ from which $$f(\lambda)= \exp (\lambda A+\lambda B+k \lambda^2/2 )$$ and you are done.

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{f}\pars{\lambda} \equiv \expo{-\lambda A}\expo{\lambda\pars{A + B}}\,,\qquad \bracks{A,B} = k = \mbox{constant}}$.

\begin{align} \totald{\mrm{f}\pars{\lambda}}{\lambda} & = -\expo{-\lambda A}A\expo{\lambda\pars{A + B}} + \expo{-\lambda A}\pars{A + B}\expo{\lambda\pars{A + B}} = \expo{-\lambda A}B\expo{\lambda A}\expo{-\lambda A}\expo{\lambda\pars{A + B}} \\[5mm] & = B\pars{\lambda}\mrm{f}\pars{\lambda}\label{1}\tag{1} \\& \end{align} where $\ds{B\pars{\lambda} \equiv \expo{-\lambda A}B\expo{\lambda A}}$.

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Moreover, \begin{align} \totald{B\pars{\lambda}}{\lambda} & = -\expo{-\lambda A}AB\expo{\lambda A} +\expo{-\lambda A}BA\expo{\lambda A} = -\expo{-\lambda A}\bracks{A,B}\expo{\lambda A} = -\expo{-\lambda A}k\expo{\lambda A} = -k \\[5mm] \implies & \quad B\pars{\lambda} = B - k\lambda \end{align}

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Eqtn. \eqref{1} becomes: $$ \totald{\mrm{f}\pars{\lambda}}{\lambda} - \pars{B - \lambda k}\mrm{f}\pars{\lambda} = 0\quad\implies\quad \totald{}{\lambda}\bracks{\exp\pars{-B\lambda + \half\,k\lambda^{2}}\mrm{f}\pars{\lambda}} = 0 $$ Then, \begin{align} \exp\pars{-B\lambda + \half\,k\lambda^{2}}\mrm{f}\pars{\lambda} & = \mrm{f}\pars{0} = \mathbf{1}\,,\qquad\forall\ \lambda \\[5mm] \implies\quad\mathbf{1} = \expo{-B + k/2}\,\,\, \overbrace{\expo{-A}\expo{A + B}}^{\ds{=\ \mrm{f}\pars{1}}}\ &\ =\ \expo{-B}\expo{-A}\,\,\expo{A + B + k/2} \end{align}

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$$ \color{#f00}{\expo{A}\expo{B}} = \expo{A + B + k/2} = \color{#f00}{\expo{A + B + \bracks{A,B}/2}}\,\,\,\,,\qquad \bracks{A,B} = \mbox{constant} $$