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üb1: ergänzung

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Daniel Bahrdt 2008-07-08 12:30:05 +02:00
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ueb1.tex
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@ -87,6 +87,14 @@ Es gelte: $[A,[A,B]] = [B,[A,B]] = 0$ \\
\begin{align} \begin{align}
e^{A+B} &= e^A \cdot e^B \cdot e^{-\frac{1}{2}[A,B]} \\ e^{A+B} &= e^A \cdot e^B \cdot e^{-\frac{1}{2}[A,B]} \\
g(t) &= e^{tA} \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} \\ g(t) &= e^{tA} \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} \\
\diffTfrac{g(t)}{t} &= A \cdot e^{tA} \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} + e^{tA} \cdot (
B \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} +
e^{tB} \cdot -t \cdot e^{-\frac{t^2}{2}[A,B]} \cdot [A,B]) \\
&= A \cdot g(t) + e^{tA} \cdot B \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]}
- t \cdot [A,B] \cdot e^{tA} \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} \\
&= A \cdot g(t) - t \cdot [A,B] \cdot e^{tA} \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} +
e^{tA} \cdot B \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} \\
&= A \cdot g(t) -t \cdot g(t) \cdot [A,B] + e^{tA} \cdot B \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]}
\diffTfrac{g(t)}{t} &= A \cdot g(t) + e^{tA} \cdot B \cdot e^{-tA} \cdot g(t) - t \cdot g(t) \cdot [A,B] \\ \diffTfrac{g(t)}{t} &= A \cdot g(t) + e^{tA} \cdot B \cdot e^{-tA} \cdot g(t) - t \cdot g(t) \cdot [A,B] \\
e^{x} \cdot e^{-x} &= \one \\ e^{x} \cdot e^{-x} &= \one \\
\diffTfrac{g}{t} &= \right( A - t [A,B] + B + \frac{t}{1!} \cdot [A,B] \left) \cdot g(t) \\ \diffTfrac{g}{t} &= \right( A - t [A,B] + B + \frac{t}{1!} \cdot [A,B] \left) \cdot g(t) \\