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 %\includegraphics{excs/qm1_blatt03_SS08.pdf}  %\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 $\ket{n_1+} \vec{n_1} = \vec{n}\sbk{\frac{\pi}{4},\frac{\pi}{4}}$ \\  $\Rightarrow$ \\  $\vec{n_1} = \inlinematrix{\cosb{\frac{\pi}{4}} \\ 1 \\ \cosb{\frac{\pi}{4}}}$  $\ket{n_1} = \inlinematrix{\cosb{\frac{\Theta}{2}} \\ e^{\i \phi} \sinb{\frac{\Theta}{2}}}$    \subsection*{a)}  $\vec{n_2} = \vec{n}\sbk{\frac{3\pi}{4},\phi}$  \begin{align}   p_+\sbk{\phi} &= \probb{\sigma_{n_2} \cequiv +1}{\ket{n_1+}} \\   &= \abs{\braket{n_2+}{n_1+}}^2 \\   &= \frac{1}{4} \sbk{1 + \cosb{\phi+\frac{\pi}{4}}} \\   p_-\sbk{\phi} &= 1 - p_+ \\   &= 1 - \frac{1}{4} \sbk{1 + \cosb{\phi+\frac{\pi}{4}}} \\   &= \frac{3}{4} - \frac{1}{4} \cosb{\phi+\frac{\pi}{4}} \\   &= \frac{1}{4} \sbk{3 - \cosb{\phi+\frac{\pi}{4}}} \\   \ket{n_1+} &= c_1 \ket{n_2+} + c_2 \ket{n_2-}  \end{align}    \subsection*{b)}   \begin{align}   \expval{sigma_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{\text{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)}  $[\Sigma_\alpha, \Sigma_\beta] = \i \Sigma_{\alpha, \beta, \gamma} \Sigma_\gamma$  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)}  \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 \vec{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 \Sigma_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}