qm1-script/ueb1.tex

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%\includegraphics{excs/qm1_blatt01_SS08.pdf}
%\pagebreak
\chapter{Quantenmechanik I - Übungsblatt 1}
\section{Aufgabe 1: Stern-Gerlach Experimente}
\section{Aufgabe 2: Pauli Matrizen}
\subsection*{a)}
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\subsection*{b)}
\subsection*{c)}
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\begin{math}
\textbf{\sigma} = (\sigma_x, \sigma_y, \sigma_z) \\
\textbf{\a},\textbf{\b} \in \setR
\end{math}
\begin{align}
(\textbf{a \cdot \sigma})(\textbf{b \cdot \sigma}) &= \one (\textbf{a \cdot b} + \i \textbf{\sigma} \cdot (\textbf{\a} \times \textbf{b}) \\
\sum a_\alpha b_\beta \sigma_\alpha \sigma_\beta &= \\
\sum a_\alpha b_\beta ( \krondelta{\alpha \beta} \one + \i \levicivita{\alpha,\beta,\gamma} \sigma_\gamma ) &= \\
\sum a_\alpha b_\beta \krondelta{\alpha \beta} \cdot \one + \i a_\alpha b_\beta \levicivita{\alpha,\beta,\gamma} \sigma_\gamma &= \\
\one (\textbf{a \cdot b} + \i \sigma \cdot (a \times b)
\end{align}
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\subsection*{d)}
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$e^{\i \frac{\i}{2} n \sigma} = \one cos(\frac{\alpha}{2}) - \i (\textbf{n \cdot \sigma}) sin(\frac{\alpha}{2})$ \\
Mit der Reihenentwicklung von $e^x$ ergibt sich:
\begin{align}
e^{- \i \frac{\alpha}{2} \textbf{n \cdot \sigma} &= \sum_k \frac{(- \i \frac{\alpha}}{2} \textbf{n \cdot \sigma})^k}{k!} \\
&= \sum_k (- \i \frac{\alpha}{2})^k \cdot \frac{(\textbf{n dot \sigma)^k}}{k!}
\end{align}
Desweiteren gilt nach Aufgabe 2 c):
\begin{align}
(\textbf{n \cdot \sigma} &= \sigma \\ \\
(\textbf{n \cdot \sigma}^2 &= \one (\textbf{n \cdot n} + \underbrace{\i \textbf{\sigma} \cdot (\textbf{\n} \times \textbf{n})}_{=0} \\
&= \one (\textbf{n \cdot n}
\end{align}
$\Rightarrow$
\begin{align}
\sum_k (- \i \frac{\alpha}{2})^k \cdot \frac{(\textbf{n \cdot \sigma)^k}}{k!} &= \\
\sum_k ( (-\i)^{2k} (\frac{\alpha}{2})^{2k+1} \frac{(\textbf{n \cdot \sigma})^{2k}}{(2k)!} +
\sum_k ( (-\i)^{2k+1} (\frac{\alpha}{2})^{2k+1} \frac{(\textbf{n \cdot \sigma})^{2k+1}}{(2k+1)!} &= \\
\sum_k ( (-1)^k (\frac{\alpha}{2})^{2k+1} \frac{(\textbf{n \cdot \sigma})^{2k}}{(2k)!} +
\sum_k ( \i \cdot (-1)^k (\frac{\alpha}{2})^{2k+1} \frac{(\textbf{n \cdot \sigma})^{2k} \cdot (\textbf{n \cdot \sigma})}{(2k+1)!} &= \\
\sum_k ( (-1)^k (\frac{\alpha}{2})^{2k+1} \one \frac{1}{(2k)!} +
\i \cdot \sum_k (-1)^k (\frac{\alpha}{2})^{2k+1} \one \cdot (\textbf{n \cdot \sigma}) \cdot \frac{1}{(2k+1)!} &= \\
\one \cos(\frac{\alpha}{2}) + \i (\textbf{n \cdot \sigma}) \sin(\frac{\alpha}{2})
\end{align}
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\section{Aufgabe 3: Operator-Identitäten}
\subsection*{a)}
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$f(t) = e^{tA} \cdot B e^{-tA}$
\begin{align}
\diffTm{}{f(t)}{t} &= [A,f(t)] \\
\diffTm{2}{f(t)}{t} &= [A,[A,f(t)]] \\ \\
\diffTm{}{f(t)}{t} &= A e^{tA} B e^{-tA} t - e^{tA} B e^{-tA} A &= [A,f(t)] = A \cdot f(t) - f(t) \cdot A \\
\diffT{t} [A,f(t)] &= \diffT{t} A \cdot f(t) - \diffT{t} f(t) \cdot A \\
&= A \cdot [A,f(t)] - [A,f(t)] \cdot A \\
&= [A,[A,f(t)]]
\end{align}
Die Taylorreihenentwicklung lautet dann:
$f(t) = e^{tA} \cdot B e^{-tA} = B + \frac{t}{1!}[A,B]+\frac{t^2}{2!}[A,[A,B]]+\ldots$
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\subsection*{b)}
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$\forall A,B,C : [A,B] = \i C \and [B,C] = \i A$ \\
\begin{align}
e^{\i B t} \cdot B e^{-\i B t} &= A \cos(t) + C \sin(t) \\
&= A + [\i B,A] + \frac{t^2}{2}[\i B,[\i B, A]] \\
[\i B, A] &= \i [B,A] &= -\i [A,B] \\
[\i B, [\i B, A]] &= [\i B, C] &= \i \i A = -A }} \\
e^{\i B t} A e^{-\i B t} &= A + C - \frac{t^2}{2} A \frac{t^3}{6} C + \frac{t^4}{24} A \\ \\
&= A \cos(t) + C \sin(t)
\end{align}
\subsection*{c)}
Es gelte: $[A,[A,B]] = [B,[A,B]] = 0$ \\
\begin{align}
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]} \\
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\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]}
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\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 \\
\diffTfrac{g}{t} &= \right( A - t [A,B] + B + \frac{t}{1!} \cdot [A,B] \left) \cdot g(t) \\
\diffTfrac{g}{t} &= [A+B] \cdot g(t) \\
\frac{\diffTfrac{g}{t}}{g} &= [A+B] \\
g(t) &= e^{[A+B] \cdot t + c} \\
e^A \cdot e^B &= e^B \cdot e^A \cdot e^[A,B] \\
e^(A+B) &= e^A \cdot e^B \cdot e^{-\frac{1}{2}[A,B]} \\
e^(B+A) &= e^B \cdit e^A \cdot e^{-\frac{1}{2}[A,B]}
\end{align}