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uebungsaufgaben: kompilationsfehler korrigiert

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Daniel Bahrdt 2008-07-17 16:58:53 +02:00
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17
theo2.tex
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@ -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
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@ -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
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@ -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
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@ -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
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@ -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)}

46
ueb6.tex
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@ -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

22
ueb7.tex
View File

@ -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)}