$e^{tA}e^{tB}=e^{t(A+B)}e^{\frac{t^2}{2}([A,B])}$

Question

Prueba de ello: $$e^{tB}e^{tA}=e^{t(A+B)}e^{\frac{t^2}{2}([A,B])}$$ para todos $t\in \mathbb{R}$ si [A,[A,B]]=[B,[A,B]]=0

Pista:

1. $[A,B]=BA-AB$ y para la hipótesis $A^2B+BA^2=2ABA$ y $AB^2+B^2A=2BAB$

2. $\phi(t)=e^{-t(A+B)}e^{tB}e^{tA}$ es la solución de $X'=t[A,B]X$

Mi progreso: $\phi'(t)=-(A+B)e^{-t(A+B)}e^{tB}e^{tA}+e^{-t(A+B)}Be^{tB}e^{tA}+e^{-t(A+B)}e^{tB}Ae^{tA}$ Entonces: $$\phi'(t)=[-(A+B) + e^{-t(A+B)}Be^{t(A+B)} + e^{-t(A+B)}e^{tB}Ae^{-tB}e^{-t(A+B)}]e^{-t(A+B)}e^{tB}e^{tA}$$ $$\phi'(t)=[-(A+B) + e^{-t(A+B)}Be^{t(A+B)} + e^{-t(A+B)}e^{tB}Ae^{-tB}e^{-t(A+B)}]\phi(t)$$

Debería haber $t(BA-AB)=[-(A+B) + e^{-t(A+B)}Be^{t(A+B)} + e^{-t(A+B)}e^{tB}Ae^{-tB}e^{-t(A+B)}]$

Muchas gracias.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Consideremos $\ds{\,\mrm{U}\pars{t} \equiv \expo{-At}\expo{\pars{A + B}t}}$ .

Entonces, \begin{align} \partiald{\mrm{U}\pars{t}}{t} & = \expo{-At}\pars{-A}\expo{\pars{A + B}t} + \expo{-At}\pars{A + B}\expo{\pars{A + B}t} = \expo{-At}B\expo{\pars{A + B}t} \\[5mm] & = \pars{\expo{-At}B\expo{At}}\pars{\expo{-At}\expo{\pars{A + B}t}} \implies \bbx{\partiald{\mrm{U}\pars{t}}{t} = \,\mrm{B}\pars{t}\,\mrm{U}\pars{t}} \end{align} donde $\ds{\,\mrm{B}\pars{t} \equiv \expo{-At}B\expo{At}}$ . Además, $\ds{\,\mrm{U}\pars{t} = \mathbf{1} + \int_{0}^{t}\mrm{B}\pars{\tau}\,\mrm{U}\pars{\tau}\dd\tau}$ . $\ds{\pars{~\mathbf{1}:\ identity~}}$ . Tenga en cuenta que $$\bbx{% \partiald{\mrm{B}\pars{t}}{t} = -A\,\mrm{B}\pars{t} + \,\mrm{B}\pars{t}A = \bracks{\mrm{B}\pars{t},A}} $$

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$\Large\ds{\bracks{B,A} = 0.}$ En este caso, $\ds{\mrm{B}\pars{t} = B \implies \mrm{U}\pars{t} = \expo{Bt} \implies \expo{\pars{A + b}t} = \expo{At}\expo{Bt}}$

$\Large\ds{\bracks{B,A} = C \propto \mathbf{1} \implies \bracks{\bracks{B,A},A} = 0.}$ Tenga en cuenta que \begin{align} \partiald[2]{\mrm{B}\pars{t}}{t} & = \bracks{\partiald{\mrm{B}\pars{t}}{t},A} = \bracks{\bracks{\mrm{B}\pars{t},A},A} = 0 \\[5mm] \implies \partiald{\mrm{B}\pars{t}}{t} & = \bracks{B,A} \implies \mrm{B}\pars{t} = B + \bracks{B,A}t \\[5mm] \implies \mrm{U}\pars{t} & = \expo{Bt}\expo{\bracks{B,A}t^{2}/2} \implies \bbx{\expo{\pars{A + B}t} = \expo{At}\expo{Bt}\expo{\bracks{B,A}t^{2}/2}} \end{align}

Answer 2

En $B$ y $e^{Bt}$ conmutar, esto es $$\phi'(t)\phi(t)^{-1} =-(A+B)+e^{-t(A+B)}e^{tB}(A+B)e^{-tB}e^{t(A+B)}$$ o $$\phi'(t)\phi(t)^{-1} =-C+e^{-tC}e^{tB}Ce^{-tB}e^{tC}$$ donde $C=A+B$ . Llame a este $f(t)$ . Entonces $$f'(t)=-Cf(t)+f(t)C+e^{-tC}e^{tB}(BC-CB)e^{-tB}e^{tC}.$$ Ahora $BC-CB=BA-AB$ conmuta con $B$ y así $e^{tB}$ . Asimismo, conmuta con $e^{tC}$ . Por lo tanto $$f'(t)=-Cf(t)+f(t)C+(BA-AB).$$ Esto parece prometedor; si sólo pudiéramos probar $Cf(t)=f(t)C\ldots$ .