übung 1 fast fertig

ueb1.tex

@@ -6,7 +6,8 @@
 
 \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
@@ -20,11 +21,79 @@
 	\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*{b)}
-\subsection*{c)}
-\subsection*{d)}
 
+\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)}
-\subsection*{c)}
+	$\forall A,B,C : [A,B] = \i C \and [B,C] = \i A$ \\
+	\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}