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\chapter{Quantenmechanik I - Übungsblatt 1}
\section{Aufgabe 1: Stern-Gerlach Experimente}

\section{Aufgabe 2: Pauli Matrizen}
\subsection*{a)}
\subsection*{b)}
\subsection*{c)}
	\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}

\subsection*{d)}
	$e^{\i \frac{\i}{2} n \sigma} = \one cos(\frac{\alpha}{2}) - \i (\textbf{n \cdot \sigma}) sin(\frac{\alpha}{2})$ \\
	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}



	
\section{Aufgabe 3: Operator-Identitäten}
\subsection*{a)}
	$f(t) = e^{tA} \cdot B e^{-tA}$
	\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:
	$f(t) = e^{tA} \cdot B e^{-tA} = B + \frac{t}{1!}[A,B]+\frac{t^2}{2!}[A,[A,B]]+\ldots$

\subsection*{b)}
	$\forall A,B,C : [A,B] = \i C \and [B,C] = \i A$ \\
	\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]} \\
	\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}
