Lecture23 - \documentclass[12pt,letterpaper]cfw_article

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\documentclass[12pt,letterpaper]{article} \ \usepackage{amsmath} \usepackage{amssymb} \usepackage{latexsym} \usepackage{array} \usepackage{enumerate} \usepackage{amsthm} \usepackage{amscd} \ \newtheorem{lemma}{Lemma} \newtheorem{theorem}{Theorem} \newtheorem{definition}{Definition} \ \newcommand{\bC}{{\mathbb C}} \newcommand{\bH}{{\mathbb H}} \newcommand{\bR}{{\mathbb R}} \newcommand{\ra}{\rightarrow} \newcommand{\Ra}{\Rightarrow} \newcommand{\disp}{\displaystyle} \ \date{} \ \begin{document} \ \title{\bf Lecture 23} \author{} \maketitle \ \flushleft \underline{Example}: If $A \in L(X)$ and $B \in L(X)$ commute with $[A,B]=AB-BA$ then \[e^{A}e^{B} = e^{A+B+\frac{1}{2}[A,B]}.\] Recall that $\disp f(t)=e^{tA}e^{tB}$ satisfies \[f'(t)=(S+tC)f(t)\] where $S=A+B$ and $C=[A,B]$. Now, let \[g(t)=e^{-(tS+\frac{1}{2}t^{2}C)}f(t)\] then \[g'(t)=\left[e^{-(tS+\frac{1}{2}t^{2}C)}(-S-tC)\right]f(t)+e^{-(tS+\frac{1} {2}t^{2}C)}\left[(S+tC)f(t)\right] =0.\] So, $g(t) \equiv g(0) = I$. Hence, \[e^{-(tS+\frac{1}{2}t^{2}C)}f(t) \equiv I.\] Therefore, $\disp f(t)=e^{-(tS+\frac{1}{2}t^{2}C)}$. Putting $t=1$ yields, \[e^{A}e^{B} = e^{A+B+\frac{1}{2}[A,B]}.\]\\ \begin{theorem} Let $f: \bR \ra G(X)$ be differentiable (at $0$) and satisfy
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Lecture23 - \documentclass[12pt,letterpaper]cfw_article

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