From 005f58500f06467a7e1863e620215bc4a7b0dde8 Mon Sep 17 00:00:00 2001
From: Daniel Bahrdt <d.bahrdt@gmx.de>
Date: Wed, 25 Jun 2008 12:22:45 +0200
Subject: [PATCH] =?UTF-8?q?=C3=BCbungsblatt=207=20aa18=20geschrieben?=
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 ueb7.tex | 65 +++++++++++++++++++++++++++++++++++++++++++++++++++-----
 1 file changed, 60 insertions(+), 5 deletions(-)

diff --git a/ueb7.tex b/ueb7.tex
index 36e1e9f..2bf8419 100644
--- a/ueb7.tex
+++ b/ueb7.tex
@@ -1,15 +1,70 @@
-%\includegraphics{excs/qm1_blatt05_SS08.pdf}
+%\includegraphics{excs/qm1_blatt07_SS08.pdf}
 %\pagebreak
 
-\chapter{Quantenmechanik I - Übungsblatt 5}
-\section{Aufgabe 11: Spin-1-Teilchen im konstanten Magnetfeld}
+\chapter{Quantenmechanik I - Übungsblatt 7}
+\section{Aufgabe 17: Unendlich hoher Potentialtop (Ergänzungen)}
 \subsection*{a)}
 \subsection*{b)}
 \subsection*{c)}
 
-\section{Aufgabe 12: Das Ethen-Molekül}
+\section{Aufgabe 18: Tunneleffekt}
+\includegraphics{grafiken/U_A18_1.pdf}
+\subsection*{a)}
 
-\section{Aufgabe 13: Das Benzol-Molekül}
+I
+\begin{math}
+	-\frac{\hbar^2}{2m} \diffPs{x}^2 \Phi(x) = E \Phi(x) \\
+	\Phi(x) =A e^{\i k x} + B e^{-\i k x} \\
+	k = \sqrt{\frac{2 E M}{\hbar^2}}
+\end{math}
+
+II $E < V_0$
+\begin{math}
+	\Phi(x) = C e^a + D e^{-qx} \\ % ist das nen q hier?
+	q = \sqrt{\frac{-2m(E-V_0)}{\hbar^2}}
+\end{math}
+
+III
+\begin{math}
+	\Phi(x) = \tilde{E} e^{\i k x} + F e^{-\i k x}
+\end{math}
+
+\subsection*{b)}
+\begin{math}
+	F = 0
+	\Phi_I(0) = \Phi_{II}(0) \\
+	A+B = C+D \\
+	\diffT{x} \Phi_I(0) = \diffT{x} \Phi_{II}(0) \\ % anschlussbedingungen korrekt
+	\i k A - \i k B = q C - q D \\
+	\inlinematrix{1 & 1 \\\i k & -\i k} \inlinematrix{A & B} = \inlinematrix{1 & 1 \\ q & -q} \inlinematrix{C & D} \\
+	\Phi_{II}(a) = \Phi_{III}(a) \\
+	C e^{a \cdot a} + D e^{-a \cdot a} = \tilde{E} e^{\i k a} \\
+	\diffT{x} \Phi_{II}(a) = \diffT{x} \Phi_{III}(a) \\
+	q C e^{a \cdot a} - q D e^{-a \cdot a} = \i k \tilde{E} e^{\i k a} \
+	\inlinematrix{e^{a \cdot a} & e^{-a \cdot a} \\ q e^{a \cdot a} & q e^{-a \cdot a}} \inlinematrix{C & D} = \inlinematrix{\tilde{E} e^{\i k a} \\ \i k \tilde{E} e^{\i k a}}
+\end{math}
+
+mit den Inversen von: (1) und (2) % geschweifte klammern unter matrix 1 und 2 setzen
+
+\begin{math}
+	%mit maxima berechnet
+	\inlinematrix{C & D} = \inlinematrix{\frac{{e}^{{a}^{2}}}{{e}^{2 {a}^{2}}-1} & -\frac{1}{\left( {e}^{3\,{a}^{2}}-{e}^{{a}^{2}}\right) q} \\ -\frac{{e}^{{a}^{2}}}{{e}^{2{a}^{2}}-1} & \frac{{e}^{{a}^{2}}}{\left( {e}^{2 {a}^{2}}-1\right) q}} \inlinematrix{\tilde{E} e^{\i k a} \\ \i k \tilde{E} e^{\i k a}} \\
+	\inlinematrix{A & B} = \inlinematrix{\frac{1}{2} & -\frac{\i}{2k}\\ \frac{1}{2} & \frac{\i}{2k}} \inlinematrix{1 & 1 \\ q & -q} \inlinematrix{C & D}
+\end{math}
+
+$\inlinematrix{C & D}$ 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)}
+\end{align}
+
+
+\subsection*{c)}
+
+\section{Aufgabe 19: Doppeltes \delta-Potential}
 \subsection*{a)}
 \subsection*{b)}
 \subsection*{c)}