diff --git a/theo2.tex b/theo2.tex index 97d11e7..06a0dea 100644 --- a/theo2.tex +++ b/theo2.tex @@ -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} diff --git a/ueb1.tex b/ueb1.tex index 75fe2d1..c63f16b 100644 --- a/ueb1.tex +++ b/ueb1.tex @@ -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} diff --git a/ueb2.tex b/ueb2.tex index 1272f70..af1dc66 100644 --- a/ueb2.tex +++ b/ueb2.tex @@ -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)} diff --git a/ueb3.tex b/ueb3.tex index dd6fef9..453cbb6 100644 --- a/ueb3.tex +++ b/ueb3.tex @@ -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} diff --git a/ueb4.tex b/ueb4.tex index 8dad127..8f52ec8 100644 --- a/ueb4.tex +++ b/ueb4.tex @@ -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)} diff --git a/ueb6.tex b/ueb6.tex index 3ed3797..89e8173 100644 --- a/ueb6.tex +++ b/ueb6.tex @@ -12,11 +12,11 @@ \end{align} -$det(H-\lambda \einsmatrix) = 0 \\$ +$det(H-\lambda \einsmatrix) = 0$ \begin{align} \lambda_1 &= B\hbar^2 \\ \lambda_2 &= \frac{\hbar^2}{2} B (1 + \sqrt{1 + \frac{4 A^2}{\hbar^2 B^2}} \\ - \lambda_3 &= \frac{\hbar^2}{2} B (1-\sqrt{1 + \frac{4 A^2}{\hbar^2 B}} + \lambda_3 &= \frac{\hbar^2}{2} B (1-\sqrt{1 + \frac{4 A^2}{\hbar^2 B})} \end{align} \begin{align} @@ -40,7 +40,7 @@ $\Rightarrow$ S_x &= \hbar \Sigma_x &= \hbar \inlinematrix{0 &1 &0 \\ 1 &0 &1 \\ 0 &1 &0} \end{align} -Da $\inlinematrix{l \\ o \\ m} S_x \inlinematrix{n \\ o \\ p} = 0}$ folgt $S_x = S_y = 0$ %wo kommt das her, und wieso ist hier S_x=0, obwohl es oben != 0 ist +Da $\inlinematrix{l \\ o \\ m} S_x \inlinematrix{n \\ o \\ p} = 0$ folgt $S_x = S_y = 0$ %wo kommt das her, und wieso ist hier S_x=0, obwohl es oben != 0 ist @@ -53,7 +53,7 @@ Da $\inlinematrix{l \\ o \\ m} S_x \inlinematrix{n \\ o \\ p} = 0}$ folgt S_x = \end{align} \begin{align} - \dirac{\psi(t)}{S_z}{\psi(t)} &= \frac{1}{(a^2+b^2)^2} [(a^2-b^2)^2 + 4a^2b^2cos((E_+-E_-)t)]} + \dirac{\psi(t)}{S_z}{\psi(t)} &= \frac{1}{\sbk{a^2+b^2}^2} [(a^2-b^2)^2 + 4 a^2 b^2 \cosb{(E_+-E_-)t}] \end{align} @@ -61,15 +61,15 @@ Da\inlinematrix{l \\ o \\ m} S_x \inlinematrix{n \\ o \\ p} = 0}$folgt$S_x = \section{Aufgabe 15: Benzol-Molekül (Teil 2)} \subsection*{a)} \begin{align} - P_m H &= \norm{\braket{\phi_m}{\psi(t)}} \\ - \ket{\psi(t)} &= U(t,t_0) \ket{\psi(t)} \\ \\ - &= exp(-\frac{\i}{\hbar} t H} \\ - &= \sum{k=0}{5}{exp(-\frac{\i}{\hbar} t \lambda_k) \ket{\chi_k} \braket{\chi_k}{\phi_0}} \\ %warum 5? - &=^1 \frac{1}{6} \sum{k=0}{5}{\sum{n=0}{5}{exp(-\frac{\i}{\hbar} t (E-2Acos(\delta_k)))}} \\ %warum diese umformung? - P_m(t) &= \norm{\braket{\phi_m}{\psi(t)}} \\ - &= \frac{1}{6} \sum{k=0}{5}{\sum{n=0}{5}{exp(...)}\braket{\phi_m}{\phi_n}} \\ %geht die summe wirklich bis 5? und über was? - &=^2 \frac{1}{36} \norm{\sum{k}{biw wo?}{exp(-\frac{\i}{\hbar} t (E-2Acos(\delta_k) + \i \delta_k)}} \\ \\ - &= ? + P_m H &= \norm{\braket{\phi_m}{\psi(t)}} \\ + \ket{\psi(t)} &= U(t,t_0) \ket{\psi(t)} \\ + &= exp{-\frac{\i}{\hbar} t H} \\ + &= \sum{k=0}{5}{exp(-\frac{\i}{\hbar} t \lambda_k) \ket{\chi_k} \braket{\chi_k}{\phi_0}} \\ %warum 5? + &=^1 \frac{1}{6} \sum{k=0}{5}{\sum{n=0}{5}{exp(-\frac{\i}{\hbar} t (E-2Acos(\delta_k)))}} \\ %warum diese umformung? + P_m(t) &= \norm{\braket{\phi_m}{\psi(t)}} \\ + &= \frac{1}{6} \sum{k=0}{5}{\sum{n=0}{5}{exp(...)}\braket{\phi_m}{\phi_n}} \\ %geht die summe wirklich bis 5? und über was? + &=^2 \frac{1}{36} \norm{\sum{k}{biw wo?}{exp(-\frac{\i}{\hbar} t (E-2Acos(\delta_k) + \i \delta_k)}} \\ \\ + &= ? \end{align} (1) \begin{align} @@ -96,24 +96,20 @@ Da $\inlinematrix{l \\ o \\ m} S_x \inlinematrix{n \\ o \\ p} = 0}$ folgt S_x = \begin{align} \\ %nach schrödingergleichung in ortsdarstellung gucken \i \hbar \partial_\tau \ket{\psi} &= H \ket{\psi} & \left| \bra{p}\right. \\ - \i \hbar \partial_\t \psi(p) &= \dirac{p}{\frac{p^2}{2m}+V(\hat{x}}{\psi} + \i \hbar \partial_t \psi(p) &= \dirac{p}{\frac{p^2}{2m}+V(\hat{x}}{\psi} \end{align} -\begin{align}} - \int_{-\infty}^{\infty} \text{dx} \dirac{p}{V(\hat[x})}{x}\braket{x}{\psi} \\ +\begin{align} + \int_{-\infty}^{\infty} \text{dx} \dirac{p}{V(\hat{x})}{x}\braket{x}{\psi} &= \int_{-\infty}^{\infty} \text{dx} V(x) \frac{1}{\sqrt{2\pi\hbar}}e^{-\frac{\i p}{\hbar}x}\braket{x}{\psi} \\ - &= \int_{-\infty}^{\infty} \text{dx} \int_{-\infty}^{\infty} \text{dp'} V(x)\frac{1}{\sqrt{2\pi\hbar}e^{-\frac{\i p}{\hbar}x}\braket{x}{p'}\braket{p'}{x} \\ \\ - &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \text{dx} \text{dp'} V(x)\frac{1}{\sqrt{2\pi\hbar}e^{-\frac{i(p-p')}{\hbar}x}\psi(p') \\ + &= \int_{-\infty}^{\infty} \text{dx} \int_{-\infty}^{\infty} \text{dp'} V(x)\frac{1}{\sqrt{2\pi\hbar}} e^{-\frac{\i p}{\hbar}x}\braket{x}{p'}\braket{p'}{x} \\ + &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \text{dx} \text{dp'} V(x)\frac{1}{\sqrt{2\pi\hbar}} e^{-\frac{i(p-p')}{\hbar}x}\psi(p') \\ &= (2) % aus aufgabenblatt raussuchen \end{align} \begin{align} - %wo kommt das her? - \dirac{p}{V(\hat{x}}{\psi} &= \dirac{p}{V(\i\hbar\diffP{p}}{\psi} \\ - &= V(\i\hbar\diffP{p})\psi(p) %vertauschen von p und V: Regeln? + %wo kommt das her + \dirac{p}{V(\hat{x})}{\psi} &= \dirac{p}{V(\i\hbar\diffP{p}}{\psi} \\ + &= V(\i \hbar \diffP{p})\psi(p) %vertauschen von p und V: Regeln? \end{align} - - -\i \hbar - diff --git a/ueb7.tex b/ueb7.tex index 7582ed5..29a1ba0 100644 --- a/ueb7.tex +++ b/ueb7.tex @@ -6,22 +6,21 @@ \subsection*{a)} \begin{math} - \Phi_n(p) = \intgrinf{\frac{1}{\sqrt{2 \pi \hbar}} \cdot \Phi(x) \cdot e^{-\frac{\i p x}{\hbar}}{x} + \Phi_n(p) = \intgrinf{\frac{1}{\sqrt{2 \pi \hbar}} \cdot \Phi(x) \cdot e^{-\frac{\i p x}{\hbar}}}{x} \end{math} Für n = ungerade: \begin{align} - \Phi_n(p) &= \integrinf{\frac{1}{\sqrt{2 \pi a \hbar}} \cdot \cos(\frac{(n+1) \cdot \pi x}{2 a}) \cdot e^{-\frac{\i p x}{\hbar}}}{x} \\ \\ + \Phi_n(p) &= \intgrinf{\frac{1}{\sqrt{2 \pi a \hbar}} \cdot \cos(\frac{(n+1) \cdot \pi x}{2 a}) \cdot e^{-\frac{\i p x}{\hbar}}}{x} \\ \\ &= \frac{1}{\sqrt{2 \pi a \hbar}} \cdot \sbk { \frac{1}{\sbk{\frac{(n+1) \cdot \pi}{2 a}}-\frac{p}{\hbar}} \cdot \sin \sbk{\sbk{\frac{(n+1) \cdot \pi}{2 a}}-\frac{p}{\hbar} \cdot u} + \frac{1}{\sbk{\frac{(n+1) \cdot \pi}{2 a}}+\frac{p}{\hbar}} \cdot \sin \sbk{\sbk{\frac{(n+1) \cdot \pi}{2 a}}+\frac{p}{\hbar} \cdot u}} \Phi_n(p) &= - \begin{case} + \begin{cases} \frac{1}{\sqrt{2 \pi a \hbar}} \cdot \frac{\i^n \cdot 4 \cdot a \hbar^2 \cdot (n+1) \cdot \pi}{\hbar^2 (n+1)^2 \pi - 4 a^2 p^2} \cdot \cos\sbk{\frac{pa}{\hbar}} \i^{n+2} \cdot \frac{\i^n \cdot 4 \cdot a \hbar^2 \cdot (n+1) \cdot \pi}{\hbar^2 (n+1)^2 \pi - 4 a^2 p^2} \cdot \sin\sbk{\frac{pa}{\hbar}} - \end{case} - + \end{cases} \end{align} @@ -76,17 +75,16 @@ mit den \hyperlink{fs_mtrx_inv_2d}{Inversen} von: (1) und (2) % geschweifte klam\inlinematrix{C & D}\$ eingesetzt ergibt: \begin{align} - \tilde{E} &= \frac{2 k a}{e^{ika} \bbracket{(q^2 - k^2) \cdot \i sinh(q \cdot a) - 2 k q cosh(qa)} \\ - A &= 1 \\ - T &= \bbracket{\frac{E}{A}}^2 \\ - R &= 1 - T &= \bbracket{\frac{B}{A}}^2 \\ - T &= \frac{1}{\bbracket{\frac{a^2 - k^2}{2 q k}} \cdot 2 sinh^2(q a) + cosh^2(q a)} + \tilde{E} &= \frac{2 k a}{e^{ika} \sbk{(q^2 - k^2) \cdot \i sinh(q \cdot a) - 2 k q cosh(qa)}} \\ % ist das richtig? + A &= 1 \\ + T &= \sbk{\frac{E}{A}}^2 \\ + R &= 1 - T &= \sbk{\frac{B}{A}}^2 \\ + T &= \frac{1}{\sbk{\frac{a^2 - k^2}{2 q k}} \cdot 2 sinh^2(q a) + cosh^2(q a)} \end{align} - \subsection*{c)} -\section{Aufgabe 19: Doppeltes \delta-Potential} +\section{Aufgabe 19: Doppeltes Delta-Potential} \subsection*{a)} \subsection*{b)} \subsection*{c)}