übungsblatt 3

ueb3.tex

@@ -2,14 +2,109 @@
 %\pagebreak
 
 \chapter{Quantenmechanik I - Übungsblatt 3}
+Zugehörige Voraussetzungen:
+Operatoren
+Wahrscheinlichkeiten
+
 \section{Aufgabe 7: Eigenzustände und Erwartungswerte von Spin-1/2-Teilchen}
+$\vec{n}(\theta,\Phi) = \inlinematrix{\sin(\theta) \cos(\Phi) \\ \sin(\theta) \sin(\Phi) \\ \cos(\Phi)}$ \\
+Präparation in $\ket{n_1+} \vec{n_1} = \vec{n}(\frac{\pi}{4},\frac{\pi}{4})$ \\
+$\Rightarrow$ \\
+$\vec{n_1} = \inlinematrix{\cos(\frac{\pi}{4} \\ 1 \\ \cos(\frac{\pi}{4}}$
+$\ket{n_1} = \inlinematrix{\cos(\frac{\Thea}{2}) \\ e^{\i \phi} \sin(\frac{\Theta}{2}}$
+
 \subsection*{a)}
+$\vec{n_2} = \vec{n}(\frac{3\pi}{4},\phi)$
+\begin{align}
+	p_+(\phi)	&= \probb{\sigma_{n_2 \cequiv +1}{\ket{n_1+}}} \\
+				&= \abs{\braket{n_2+}{n_1+}}^2 \\
+				&= \frac{1}{4} \sbk{1 + \cos(\phi+\frac{\pi}{4})} \\
+	p_-(\phi)	&= 1 - p_+ \\
+				&= 1 - \frac{1}{4} \sbk{1 + \cos(\phi+\frac{\pi}{4})} \\
+				&= \frac{3}{4} - \frac{1}{4} \cos(\phi+\frac{\pi}{4}) \\
+				&= \frac{1}{4} \sbk{3 - \cos(\phi+\frac{\pi}{4})}
+	\ket{n_1+}	&= c_1 \ket{n_2+} + c_2 \ket{n_2-}
+\end{align}
+
 \subsection*{b)}
+	\begin{align}
+		\expval{şigma_z}_{\ket{n_1+}}	&= \probb{\sigma_z \cequiv +1}{\ket{n_1+}} +
+											(-1) \probb{\sigma_z \cequiv -1}{\ket{n_1+}} \\ \\
+										&= \abs{\inlinematrix{1 & 0} \inlinematrix{\cos(\frac{\pi}{8} \\ e^{\i \frac{\pi}{4}}}}^2 -
+											\abs{\inlinematrix{0 & -1} \inlinematrix{\cos(\frac{\pi}{8} \\ e^{\i \frac{\pi}{4}}}}^2 \\
+										&= \cos^2(\frac{\pi}{8}) - \sin^2(\frac{\pi}{8}) \\
+										&= \cos(\frac{\pi}{4} \\
+										&= \frac{\sqrt{2}}{2} \\
+		\sbk{\Delta \sigma_x}^2			&= \prob{\sigma_x^2} - \prob{\sigma_x}^2 \\
+		\dirac{n_1+}{\sigma_x}{n_1+}	&= \sin(\frac{\pi}{4} \cos(\frac{\pi}{4}	&= \frac{1}{2} \\
+		\dirac{n_1+}{\sigma_x^2}{n_1+}	&= \inlinematrix{\cos(\frac{\pi}{8}) & e^{-\frac{\pi}{4}} \sin(\frac{\pi}{8})} \cdot
+											\inlinematrix{1 & 0 \\ 1 & 0} \cdot
+											\inlinematrix{\cos(\frac{\pi}{8}) \\ e^{\frac{\pi}{4}} \sin(\frac{\pi}{8})} \\
+										&\Rightarrow
+		\sbk{\Delta \sigma_x}^2			&= 1 - \frac{1}{2}^2 \\
+										&= \frac{3}{4} \\
+										&\stackrel{analog}{=} \sbk{\Delta \sigma_y}^2 \\
+		\sbk{\Delta \sigma_x} \cdot \sbk{\Delta \sigma_y}	&= \frac{3}{4} \\
+		\sbk{\Delta A} \cdot \sbk{\Delta B}	&\geq \frac{1}{2} \abs{\expval{\frac{1}{\i} [A,B]}} \\
+		\frac{1}{2} \abs{\expval{\sigma_z}}	&= \frac{1}{\sqrt{2}} &\< \frac{3}{4}
+	\end{align}
+	
 \subsection*{c)}
+\begin{align}
+	\ket{\Psi}							&= N (\ket{Z+} + e^{\i \alpha} \ket{Z-}) = N \inlinematrix{1 \\ e^{\i \alpha}} \\
+	\braket{\Psi}{\Psi}					&= 1 \\
+	N^2 \sbk{1 + e^{\i \alpha - \i \alpha)}}	&\deq 1 \\
+										&\Rightarrow \\
+	N									&= \frac{1}{\sqrt{2}} \\ \\
+	P_+									&= \abs{\braket{x+}{\Psi}}^2 \\
+										&= \abs{\frac{1}{\sqrt{2}} \inlinematrix{1 \\ 1} \cdot \frac{1}{\sqrt{2}} \inlinematrix{1 \\ e^{\i \alpha}}}^2 \\
+										&= \frac{1}{4} \abs{1 + e^{\i \alpha}}^2 \\
+										&= \frac{1}{2} \sbk{1 + \cos(\alpha)}
+	
+\end{align}
+
 
 \section{Aufgabe 8: Teilchen mit Spin 1}
 \subsection*{a)}
+\includegraphics{grafiken/U_A6_a.pdf}
 \subsection*{b)}
+$[\Epsilon_\alpha, \Epsilon_\beta] = \i \Epsilon_{\alpha, \beta, \gamma} \Epsilon_\gamma$
+mit allen $\Sigma_{x,y,z}$ durch testen.
 \subsection*{c)}
+\begin{align}
+	\Sigma_^2	&= \Sigma_x^2 + \Sigma_y^2 + \Sigma_z^2 \\
+				&= 2 \one
+\end{align}
+
 \subsection*{d)}
-\subsection*{e)}
+\begin{math}
+	\ket{x+}, \ket{x-}, \ket{x0}, \ket{\Psi} = \ket{n_0}, n(\Theta,\Phi)
+\end{math}
+\begin{align}
+	p_0					&\Rightarrow	&\ket{x0}				&= \frac{1}{\sqrt{2}} \cdot \inlinematrix{1 \\ 0 \\ -1} \\
+	p_+					&\Rightarrow	&\ket{x+}				&= \frac{1}{2} \cdot \inlinematrix{1 \\ \frac{1}{\sqrt{2}} \\ 1} \\
+	p_-					&\Rightarrow	&\ket{x-}				&= \frac{1}{2} \cdot \inlinematrix{-1 \\ \frac{1}{\sqrt{2}} \\ -1}
+	\vec{\Sigma_n}		&= 				&\Sigma \cdot \vev{n}	&= \inlinematrix{\cos(\Theta) & \frac{1}{\sqrt{2}} \sin(\Theta) e^{-\ \phi} & 0 \\
+																					\frac{1}{\sqrt{2}} \sin(\Phi) e^{\i \phi} & 0 & \frac{1}{\sqrt{2}} \sin(\Theta) e^{-\i \phi} \\
+																					0	& \frac{1}{\sqrt{2}} \sin(\Theta) e^{\i \phi} & \cos(\Theta)}
+						&\Rightarrow
+	\ket{n_0}			&				&						&= \frac{1}\sqrt{2}} \inlinematrix{-\sin(\Theta) e^{\i \Phi} \\ \sqrt{2} \cos(\Theta) \\ \sin(\Theta) e^{\i \Phi}} \\
+\end{align}
+\begin{align}
+	p_+		&= \abs{\braket{x_1}{n_0}}^2		&= \ldots \frac{1}{2} \cos^2(\Theta) + \frac{1}{2} \sin^2(\Theta) \sin^2(\Phi) = p_- \\
+	p_0		&= 1- 2 p_+							&= 1 - 2 p_-
+\end{align}
+
+\subsection*{e)}
+\begin{align}
+	\expval{\Sigma_x}_{\ket{\Psi}}	&= \dirac{n_0}{\Sigma_x}{n_0} \\
+									&= +1 + 1 p_+ + p_0 + (-1) p_- \\
+									&= 0 \\
+	\ket{\Psi}						&= c_1 \ket{\x+} + c_2 \ket{x-} + c_3 \ket{x0} \\
+	\sbk{\Delta \Epsilon_x}^2		&= \langle \Sigma_x^2 \rangle - {\langle \Sigma_x \rangle}^2 \\
+									&= c_2 \inlinematrix{1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 1} %ist das richtig?
+	\dirac{\Psi}{\Sigma_x^2}{\Psi}	&= c_1^2 \cdot 1^2 + c_2^2 \cdot (-1)^2 + c_3^2 \cdot 0^2 \\
+									&\stackrel{c_1 = c_2}{=} 2 c_1^2 \\
+									&= \cos^2(\Theta) + \sin^2(\Theta) \cdot \sin^2(\phi) %Ist das richtig?
+									
+\end{align}