diff --git a/excs/qm1_blatt07_SS08.pdf b/excs/qm1_blatt07_SS08.pdf new file mode 100644 index 0000000..e627803 Binary files /dev/null and b/excs/qm1_blatt07_SS08.pdf differ diff --git a/excs/qm1_blatt08_SS08.pdf b/excs/qm1_blatt08_SS08.pdf new file mode 100644 index 0000000..12c68f2 Binary files /dev/null and b/excs/qm1_blatt08_SS08.pdf differ diff --git a/formelsammlung.tex b/formelsammlung.tex new file mode 100644 index 0000000..03b751a --- /dev/null +++ b/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} diff --git a/grafiken/U_A18_1.pdf b/grafiken/U_A18_1.pdf new file mode 100644 index 0000000..1edac69 Binary files /dev/null and b/grafiken/U_A18_1.pdf differ diff --git a/math.tex b/math.tex index 5d53c6e..bc28c08 100644 --- a/math.tex +++ b/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)} @@ -28,4 +29,9 @@ \newcommand{\diffT}[1]{\frac{\text{d}}{\text{d} #1}} \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} \ No newline at end of file +\newcommand{\intgru}[2]{\int #1 ~\text{d}#2} + +\newcommand{\bbracket}[1]{\left #1 \right} + +\newcommand{\levicivita}[1]{\varEpsilon_{#1}} +\newcommand{\krondelta}[1]{\delta_{+1}} \ No newline at end of file diff --git a/svg/U_A18_1.svg b/svg/U_A18_1.svg new file mode 100644 index 0000000..5817219 --- /dev/null +++ b/svg/U_A18_1.svg @@ -0,0 +1,117 @@ + + + + + + + + + + + + + image/svg+xml + + + + + + + + + I + II + III + + + diff --git a/theo2.kilepr b/theo2.kilepr index ddadb96..4ce0695 100644 --- a/theo2.kilepr +++ b/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 +line=6 +open=true +order=2 + [item:kapI-1.tex] archive=true -column=10936 +column=35 encoding= -highlight= +highlight=LaTeX line=0 -open=false -order=-1 +open=true +order=5 [item:kapI-2.tex] archive=true @@ -71,11 +80,11 @@ order=-1 [item:kapII-0.tex] archive=true -column=98 +column=0 encoding=UTF-8 highlight=LaTeX -line=46 -open=true +line=0 +open=false order=1 [item:kapII-1.tex] @@ -107,21 +116,21 @@ order=-1 [item:math.tex] archive=true -column=42 +column=0 encoding=UTF-8 highlight=LaTeX -line=29 -open=false +line=34 +open=true order=3 [item:physics.tex] archive=true -column=22 -encoding= +column=37 +encoding=UTF-8 highlight=LaTeX -line=1 -open=false -order=-1 +line=10 +open=true +order=4 [item:theo2.kilepr] archive=true @@ -134,10 +143,10 @@ order=-1 [item:theo2.tex] archive=true -column=0 +column=39 encoding=UTF-8 highlight=LaTeX -line=46 +line=10 open=true order=0 @@ -181,7 +190,7 @@ order=-1 archive=true column=0 encoding= -highlight= +highlight=LaTeX line=0 open=false order=-1 @@ -189,8 +198,17 @@ order=-1 [item:ueb6.tex] archive=true column=0 -encoding= -highlight= -line=0 +encoding=UTF-8 +highlight=LaTeX +line=119 open=false order=-1 + +[item:ueb7.tex] +archive=true +column=44 +encoding=UTF-8 +highlight=LaTeX +line=46 +open=true +order=1 diff --git a/theo2.tex b/theo2.tex index 73647b4..befdb0d 100644 --- a/theo2.tex +++ b/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} diff --git a/ueb6.tex b/ueb6.tex index a281369..6f015ea 100644 --- a/ueb6.tex +++ b/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)} -\subsection*{b)} +\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 -\section{Aufgabe 16: Schrödingergleichung in Impulsraumdarstellung} \ No newline at end of file diff --git a/ueb7.tex b/ueb7.tex index e69de29..2bf8419 100644 --- a/ueb7.tex +++ b/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)} \ No newline at end of file