Merge branch 'master' of git+ssh://git@git.eluhost.de:522/qm1-script

formelsammlung.tex

@@ -0,0 +1,5 @@
+\chapter{Lineare Algebra}
+\section{Identitäten}
+\begin{math}
+A^{-1} = \inlinematrix{a & b \\ c & d}^{-1} = \frac{1}{ad - bc} \inlinematrix{d & -b \\ -c & a}
+\end{math}

math.tex

@@ -7,6 +7,7 @@
 \newcommand{\setQ}{\mathbbm{Q}}
 \newcommand{\setR}{\mathbbm{R}}
 \newcommand{\setC}{\mathbbm{C}}
+\newcommand{\einsmatrix}{\mathbbm{1}}
 \newcommand{\inlinematrix}[1]{\begin{pmatrix} #1 \end{pmatrix}}
 \newcommand{\inlinematrixu}[1]{\begin{matrix} #1 \end{matrix}}
 \newcommand{\inlinematrixd}[2]{\left( \begin{matrix} #1\vphantom{#2} \end{matrix} ~\right| \left. \begin{matrix} \vphantom{#1}#2 \end{matrix} \right)}
@@ -29,3 +30,8 @@
 \newcommand{\diffPfrac}[2]{\frac{\partial #1}{\partial #2}}
 \newcommand{\intgr}[4]{\int_{#1}^{#2} #3 ~\text{d}#4}
 \newcommand{\intgru}[2]{\int #1 ~\text{d}#2}
+
+\newcommand{\bbracket}[1]{\left #1 \right}
+
+\newcommand{\levicivita}[1]{\varEpsilon_{#1}}
+\newcommand{\krondelta}[1]{\delta_{+1}}

svg/U_A18_1.svg

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theo2.kilepr

@@ -3,7 +3,7 @@ img_extIsRegExp=false
 img_extensions=.eps .jpg .jpeg .png .pdf .ps .fig .gif
 kileprversion=2
 kileversion=2.0
-lastDocument=kapII-0.tex
+lastDocument=theo2.tex
 masterDocument=
 name=Theo2
 pkg_extIsRegExp=false
@@ -15,14 +15,23 @@ src_extensions=.tex .ltx .latex .dtx .ins
 MakeIndex=
 QuickBuild=PDFLaTeX+ViewPDF
 
+[item:formelsammlung.tex]
+archive=true
+column=0
+encoding=UTF-8
+highlight=LaTeX
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+open=true
+order=2
+
 [item:kapI-1.tex]
 archive=true
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+column=35
 encoding=
-highlight=
+highlight=LaTeX
 line=0
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+open=true
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+order=5
 
 [item:kapI-2.tex]
 archive=true
@@ -71,11 +80,11 @@ order=-1
 
 [item:kapII-0.tex]
 archive=true
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+column=0
 encoding=UTF-8
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+line=0
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 [item:kapII-1.tex]
@@ -107,21 +116,21 @@ order=-1
 
 [item:math.tex]
 archive=true
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+column=0
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+open=true
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 [item:physics.tex]
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-encoding=
+encoding=UTF-8
 highlight=LaTeX
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+line=10
-open=false
+open=true
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+order=4
 
 [item:theo2.kilepr]
 archive=true
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@@ -181,7 +190,7 @@ order=-1
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+order=1

theo2.tex

@@ -8,7 +8,7 @@
 \usepackage{amsfonts}
 \usepackage{amssymb}
 \usepackage{multirow}
-\usepackage[pdfborder={0 0 0}]{hyperref}
+\usepackage[pdfborder={0 0 0}]{hyperref} % muss immer als letztes eingebunden werden
 
 \include{math}
 \include{physics}
@@ -45,6 +45,11 @@
 % \include{ueb4}
 % \include{ueb5}
 % \include{ueb6}
+% \inlcude{ueb7}
+
+% \part{Formelsammlung}
+% \label{FS}
+% \include{formelsammlung}
 
 
 \end{document}

ueb6.tex

@@ -4,10 +4,116 @@
 \chapter{Quantenmechanik I - Übungsblatt 6}
 \section{Aufgabe 14: Spin-1-Teilchen}
 \subsection*{a)}
+\begin{align}
+	H &= A S_z + B S_x^2 \\
+	H &= A \hbar \Sigma_z + B \hbar^2 \Sigma_x^2 \\
+	H &= A \hbar \inlinematrix{1 &0 &0 \\ 0 &0 &0 \\ 0 &0 &-1} + \frac{B \hbar^2}{2} \inlinematrix{1 &0 &1 \\ 0 &2 &0 \\ 1 &0 &1} \\
+	H &= \hbar \inlinematrix{A+\frac{B \hbar}{2} &0 &\frac{B \hbar}{2} \\ 0 &\frac{B \hbar}{2} &0 \\ \frac{B \hbar}{2} &0 &-A+\frac{B \hbar}{2}}
+\end{align}
+
+
+$det(H-\lambda \einsmatrix) = 0 \\$
+\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}}
+\end{align}
+
+\begin{align}
+	a		&:= A + \frac{\hbar B}{2} \sqrt{1+\frac{4 A^2}{\hbar^2 B^2}} \\
+	b		&:= \frac{\hbar B}{2} \\
+	\ket{\phi_0}	&= \inlinematrix{0 \\ 1 \\ 0} &
+	\ket{\phi+}	&= \frac{1}{\sqrt{a^2+b^2}} \inlinematrix{a \\ 0 \\ b} &
+	\ket{\phi-}	&= \frac{1}{\sqrt{a^2+b^2}} \inlinematrix{-b \\ 0 \\ a} \\ \\
+	\ket{\psi(t)}	&= c_0 e^{-\i E_0 t} \ket{\phi_0} + c_+ e^{-E_+ t} \ket{\psi_+} + c_- e^{-E_-t t} \ket{\phi-} \\ % muss da nicht e^\i?
+	\ket{\psi(0)}	&= \ket{z+} &= \inlinematrix{1 \\ 0 \\ 0} \\ \\
+\end{align}
+$\Rightarrow c_0 = 0$
+\begin{align}
+	\frac{1}{\sqrt{a^2+b^2}}(c_+ a - c_- b) &\stackrel{!}{} 1 \\
+	\frac{1}{\sqrt{a^2+b^2}}(c_+ b + c_- a) &\stackrel{!}{} 0
+\end{align}
+$\Rightarrow $
+\begin{align}
+	c_-	&= \frac{-b\sqrt{a^2+b^2}}{a^2+b^2} \\
+	c_+	&= \frac{a\sqrt{a^2+b^2}}{a^2+b^2} \\
+	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
+
+
+
+
+
 \subsection*{b)}
 
+\begin{align}
+	S_z &= \hbar \inlinematrix{1 &0 &0 \\ 0 &0 &0 \\ 0 &0 &-1}
+\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)]}
+\end{align}
+
+
+
 \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)}} \\ \\
+			&= ?
+\end{align}
+(1)
+\begin{align}
+	\braket{\chi_k}{\phi_0} &= \frac{1}{\sqrt{6}} \sum{n=0}{5}{exp(-\i n \delta_{keine ahnung}\bracket{\phi_n}{\phi_0}} \\ \\
+				&= \frac{1}{\sqrt{6}} e^{-\i 0 \delta_{keine ahnung}} \\
+				&= \frac{1}{\sqrt{6}}
+\end{align}
+(2)
+\begin{align}
+	\braket{\phi_m}{\phi_n} = \delta_{n m}
+\end{align}
+
+
+
+
 \subsection*{b)}
+\begin{math}
+	\tau = \frac{2\pi}{A}\hbar
+\end{math}
+
 
 \section{Aufgabe 16: Schrödingergleichung in Impulsraumdarstellung}
+
+\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}
+\end{align}
+
+\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') \\
+	&= (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?
+\end{align}
+
+
+
+\i \hbar
+

ueb7.tex

@@ -0,0 +1,71 @@
+%\includegraphics{excs/qm1_blatt07_SS08.pdf}
+%\pagebreak
+
+\chapter{Quantenmechanik I - Übungsblatt 7}
+\section{Aufgabe 17: Unendlich hoher Potentialtop (Ergänzungen)}
+\subsection*{a)}
+\subsection*{b)}
+\subsection*{c)}
+
+\section{Aufgabe 18: Tunneleffekt}
+\includegraphics{grafiken/U_A18_1.pdf}
+\subsection*{a)}
+
+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)}
+\subsection*{d)}