uebungsaufgaben: kompilationsfehler korrigiert

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17 theo2.tex View File

 @ -13,6 +13,7 @@ \include{physics} \newcommand{\lboxed}[1]{\left\lceil #1 \right\rfloor} \renewcommand{\i}{i} \title{Theoretische Physik 2 Vorlesungs- und Übungsmitschrieb} \author{Daniel Bahrdt, Oliver Groß} @ -44,13 +45,17 @@ % \part{Übungsmitschrieb} % \label{UE} % \include{ueb1} % \include{ueb2} % \include{ueb3} % \include{ueb4} \include{ueb1} \include{ueb2} \include{ueb3} \include{ueb4} % \include{ueb5} % \include{ueb6} % \inlcude{ueb7} \include{ueb6} \include{ueb7} % \include{ueb8} % \include{ueb9} % \include{ueb10} % \include{ueb11} \part{Formelsammlung} \label{FS}

52 ueb1.tex View File

 @ -9,49 +9,39 @@ \subsection*{b)} \subsection*{c)} \begin{math} \textbf{\sigma} = (\sigma_x, \sigma_y, \sigma_z) \\ \textbf{\a},\textbf{\b} \in \setR \vec{\sigma} = (\sigma_x, \sigma_y, \sigma_z) \\ \vec{a}, \vec{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}) \\ (\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma}) &= \one (\vec{a} \cdot \vec{b} + \i \vec{\sigma} \cdot (\vec{a} \times \vec{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) \one (\vec{a} \cdot \vec{b} + \i \sigma \cdot (a \times b) \end{align} \subsection*{d)} $e^{\i \frac{\i}{2} n \sigma} = \one cos(\frac{\alpha}{2}) - \i (\textbf{n \cdot \sigma}) sin(\frac{\alpha}{2})$ \\ $e^{\i \frac{\i}{2} n \sigma} = \one cos(\frac{\alpha}{2}) - \i (\vec{n} \cdot \vec{\sigma}) sin(\frac{\alpha}{2})$ \\ 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} (\vec{n} \cdot \vec{\sigma} &= \sigma \\ (\vec{n} \cdot \vec{\sigma})^2 &= \one (\vec{n} \cdot \vec{n} + \underbrace{\i \vec{\sigma} \cdot (\vec{n} \times \vec{n})}_{=0} \\ &= \one (\vec{n} \cdot \vec{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}) \sum_k \sbk{- \i \frac{\alpha}{2}}^k \cdot \frac{\sbk{\vec{n} \cdot \vec{\sigma}}^k}{k!} &= \\ \sum_k \sbk{ (-\i)^{2k} (\frac{\alpha}{2})^{2k+1} \frac{(\vec{n} \cdot \vec{\sigma})^{2k}}{(2k)!}} + \sum_k \sbk{ (-\i)^{2k+1} (\frac{\alpha}{2})^{2k+1} \frac{(\vec{n} \cdot \vec{\sigma})^{2k+1}}{(2k+1)!}} &= \\ \sum_k \sbk{ (-1)^k \sbk{\frac{\alpha}{2}}^{2k+1} \frac{\sbk{\vec{n} \cdot \vec{\sigma}}^{2k}}{(2k)!}} + \sum_k \sbk{ \i \cdot (-1)^k \sbk{\frac{\alpha}{2}}^{2k+1} \frac{\sbk{\vec{n} \cdot \vec{\sigma}}^{2k} \cdot \sbk{\vec{n} \cdot \vec{\sigma}}}{(2k+1)!}} &= \\ \sum_k ( (-1)^k \sbk{\frac{\alpha}{2}}^{2k+1} \one \frac{1}{(2k)!} + \i \cdot \sum_k (-1)^k (\frac{\alpha}{2})^{2k+1} \one \cdot (\vec{n} \cdot \vec{\sigma}) \cdot \frac{1}{(2k+1)!} &= \\ \one \cos(\frac{\alpha}{2}) + \i (\vec{n} \cdot \vec{\sigma}) \sin(\frac{\alpha}{2}) \end{align} @ -72,18 +62,18 @@ $f(t) = e^{tA} \cdot B e^{-tA} = B + \frac{t}{1!}[A,B]+\frac{t^2}{2!}[A,[A,B]]+\ldots$ \subsection*{b)} $\forall A,B,C : [A,B] = \i C \and [B,C] = \i A$ \\ $\forall A,B,C : [A,B] = \i C \and [B,C] = \i A$ \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 }} \\ [\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: $[A,[A,B]] = [B,[A,B]] = 0$ \\ 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]} \\ @ -97,11 +87,11 @@ Es gelte: $[A,[A,B]] = [B,[A,B]] = 0$ \\ &= 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]} \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} &= \left( A - t [A,B] + B + \frac{t}{1!} \cdot [A,B] \right) \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]} e^(B+A) &= e^B \cdot e^A \cdot e^{-\frac{1}{2}[A,B]} \end{align}

29 ueb2.tex View File

 @ -34,11 +34,11 @@ \subsection*{b)} Zu zeigen: Für $A$ hermitesch ist $U(s) = e^{-\i s A}$ unitär \\ Zu zeigen: Für $A$ hermitesch ist $U(s) = e^{-\i s A}$ unitär \begin{align} U(s) &= e^{\i s A} \\ &= \sum_{k=0}^\infty \frac{1}{k!} \cdot (-\i s A)^k \\ &= \sum_{k=0}^\infty \frac{1}{k!} \cdot (\i s A^\intercol)^k \\ &= \sum_{k=0}^\infty \frac{1}{k!} \cdot \sbk{-\i s A}^k \\ &= \sum_{k=0}^\infty \frac{1}{k!} \cdot \sbk{\i s A^{\intercal}}^k \\ &= e^{\i s A} \end{align} @ -54,7 +54,7 @@ Zu zeigen: Für $A$ hermitesch ist $U(s_1 + s_2) = U(s_1) \cdot U(s_2)$. \section{Aufgabe 5: Spur und Determinante} \subsection*{a)} Zu zeigen: $[A,BC] &= B \cdot [A,C] + [A,B] \cdot C$. Zu zeigen: $[A,BC] = B \cdot [A,C] + [A,B] \cdot C$. \begin{align} [A,BC] &= ABC -BCA \\ B \cdot [A,C] + [A,B] \cdot C &= B \cdot (AC - CA) + (AB - BA) \cdot C \\ @ -113,27 +113,25 @@ Dann ist $T^{-1}AT$ die Basistransformation von der A-Basis in die T-Basis. &= 1 + \epsilon \tr(A) + \bigO(\epsilon^2) \end{align} Zu zeigen: $\det(e^A = e^{\tr(A))$ Zu zeigen: $\detb{e^A} = e^{\tr(A)}$ \begin{align} g(t) &= \det(e^{At}) \\ &\stackrel{tailor}{=} \det(1 + At + \bigO(t^2)) \\ &= 1 + \tr(A) + \bigO(t^2) \\ &\stackrel{tailor rückwärs''}{=} e^{\tr(A)t} g(t) &= \detb{e^{At}} && \left| \text{Tailor} \right. \\ &= \detb{1 + At + \bigO(t^2)} &\\ &= 1 + \tr(A) + \bigO(t^2) && \left| \text{tailor rückwärs''} \right.\\ &= e^{\tr(A)t} & \end{align} \begin{align} g(t) &= \limits_{\epsilon \rightarrow \intfy} \frac{1}{\epsilon} \sbk{\det(e^{A(t+\epsilon)) - det(e^{At})} \\ &= g(t) \limits_{\epsilon \rightarrow \intfy} \frac{1}{\epsilon} \det(e^{A \epsilon}) \\ &= g(t) \limits_{\epsilon \rightarrow \intfy} \frac{1}{\epsilon} \sbk{\det(1 + A \epsilon + \bigO(\epsilon^2) - det(1)} \\ g(t) &= \lim_{\epsilon \rightarrow \infty} \frac{1}{\epsilon} \sbk{\detb{e^{A(t+\epsilon)}} - \detb{e^{At}}} \\ &= g(t) \lim_{\epsilon \rightarrow \infty} \frac{1}{\epsilon} \detb{e^{A \epsilon}} \\ &= g(t) \lim_{\epsilon \rightarrow \infty} \frac{1}{\epsilon} \sbk{\det(1 + A \epsilon + \bigO(\epsilon^2) - det(1)} \\ &= g(t) \tr(A) \\ &\Rightarrow g(t) = e^{\tr(A) \cdot t} \end{align} Ist A diagonalisierbar: \begin{math} \det(e^A) = \det(T^{-1} e^A T) = \det(e^\hat{A}) = \prod_i e^{\lambda_i} = e^{\sum_i \lambda_i} = e^{\tr(A)} \end{math} \section{Aufgabe 6: Hermitesche Matrizen} @ -145,7 +143,6 @@ Sei $\bra{a}$ Eigenvektor zum Eigenwert $a$ M_i^2 \ket{a} &= M_i a \cdot \bra{a} \\ \one \ket{a} &= a^2 \bra{a} \\ &\Rightarrow a = \pm 1 \end{align} \subsection*{b)}

48 ueb3.tex View File

 @ -8,28 +8,28 @@ 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 $\ket{n_1+} \vec{n_1} = \vec{n}(\frac{\pi}{4},\frac{\pi}{4})$ \\ Präparation in $\ket{n_1+} \vec{n_1} = \vec{n}\sbk{\frac{\pi}{4},\frac{\pi}{4}}$ \\ $\Rightarrow$ \\ $\vec{n_1} = \inlinematrix{\cos(\frac{\pi}{4} \\ 1 \\ \cos(\frac{\pi}{4}}$ $\ket{n_1} = \inlinematrix{\cos(\frac{\Thea}{2}) \\ e^{\i \phi} \sin(\frac{\Theta}{2}}$ $\vec{n_1} = \inlinematrix{\cosb{\frac{\pi}{4}} \\ 1 \\ \cosb{\frac{\pi}{4}}}$ $\ket{n_1} = \inlinematrix{\cosb{\frac{\Theta}{2}} \\ e^{\i \phi} \sinb{\frac{\Theta}{2}}}$ \subsection*{a)} $\vec{n_2} = \vec{n}(\frac{3\pi}{4},\phi)$ $\vec{n_2} = \vec{n}\sbk{\frac{3\pi}{4},\phi}$ \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-} 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{şigma_z}_{\ket{n_1+}} &= \probb{\sigma_z \cequiv +1}{\ket{n_1+}} + (-1) \probb{\sigma_z \cequiv -1}{\ket{n_1+}} \\ \\ \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}) \\ @ -43,10 +43,10 @@ $\vec{n_2} = \vec{n}(\frac{3\pi}{4},\phi)$ &\Rightarrow \sbk{\Delta \sigma_x}^2 &= 1 - \frac{1}{2}^2 \\ &= \frac{3}{4} \\ &\stackrel{analog}{=} \sbk{\Delta \sigma_y}^2 \\ &\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} \frac{1}{2} \abs{\expval{\sigma_z}} &= \frac{1}{\sqrt{2}} &< \frac{3}{4} \end{align} \subsection*{c)} @ -60,7 +60,6 @@ $\vec{n_2} = \vec{n}(\frac{3\pi}{4},\phi)$ &= \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} @ -68,11 +67,11 @@ $\vec{n_2} = \vec{n}(\frac{3\pi}{4},\phi)$ \subsection*{a)} \includegraphics{grafiken/U_A6_a.pdf} \subsection*{b)} $[\Epsilon_\alpha, \Epsilon_\beta] = \i \Epsilon_{\alpha, \beta, \gamma} \Epsilon_\gamma$ $[\Sigma_\alpha, \Sigma_\beta] = \i \Sigma_{\alpha, \beta, \gamma} \Sigma_\gamma$ mit allen $\Sigma_{x,y,z}$ durch testen. \subsection*{c)} \begin{align} \Sigma_^2 &= \Sigma_x^2 + \Sigma_y^2 + \Sigma_z^2 \\ \Sigma^2 &= \Sigma_x^2 + \Sigma_y^2 + \Sigma_z^2 \\ &= 2 \one \end{align} @ -84,11 +83,11 @@ mit allen $\Sigma_{x,y,z}$ durch testen. 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} \\ \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}} \\ \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_- \\ @ -100,11 +99,10 @@ mit allen $\Sigma_{x,y,z}$ durch testen. \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 \\ \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}

28 ueb4.tex View File

 @ -3,9 +3,37 @@ \chapter{Quantenmechanik I - Übungsblatt 4} \section{Aufgabe 9: Zeitentwicklung eines allgemeinen Zweizustandssystems} \begin{math} H = \hbar \inlinematrix{A & B \\ B & -A} \\ A = \Omega \cosb{2 \theta} \\ B = \Omega \sinb{2 \theta} \\ E_\pm = \pm \hbar \Omega \end{math} \subsection*{a)} \begin{align} \ket{\chi_+} &= \inlinematrix{\sinb{2 \theta} \\ 1 - \cosb{2 \theta}} &= \inlinematrix{\cosb{\theta} \\ \sinb{\theta}} \\ \ket{\chi_-} &= &= \inlinematrix{\sinb{\theta} \\ -\cosb{\theta}} \end{align} \subsection*{b)} \begin{align} \i \hbar \diffPs{t} \ket{\psi} &= H \ket{\psi} \\ \ket{\dot{\psi}} &= -\i \Omega \inlinematrix{\cosb{2 \theta} & \sinb{2 \theta} \\ \sinb{2 \theta} & -\cosb{2 \theta}} \ket{\psi} \\ \ket{\psi(t)} &= c_1 \cdot e^{\i \Omega t} \ket{\chi+} + c_2 \cdot e^{-\i \Omega t} \ket{\chi-} \\ \ket{\psi(0)} &= \inlinematrix{\lambda \\ \mu} \\ \inlinematrix{c_+(t) \\ c_-(t)} &= \sbk{\lambda \cosb{\theta} + \mu \sinb{\theta}} \cdot \inlinematrix{\cosb{\theta} \\ \sinb{\theta}} \cdot e^{-\i t} + \sbk{\lambda \sinb{\theta} - \mu \cosb{\theta}} \cdot \inlinematrix{\sinb{\theta} \\ -\cosb{\theta}} \cdot e^{\i \Omega t} \end{align} \subsection*{c)} %hier stimmt evtl. was nicht \begin{align} \ket{\psi(0)} &= \inlinematrix{0 \\ 1} \\ \abs{\braket{+}{\psi(t)}}^2 &= \sin^2\sbk{\Omega t} \cdot \sin^2\sbk{2 \theta} \\ \abs{\braket{-}{\psi(t)}}^2 &= \cos^2\sbk{\Omega t} + \sin^2\sbk{\Omega t} \cdot \cos^2\sbk{2 \Omega} \end{align} \section{Aufgabe 10: Zerfall eines instabilen Zustandes} \subsection*{a)}