From 5a540bb4bdc6824fdcf4d95b922a555a8b16a07d Mon Sep 17 00:00:00 2001 From: Daniel Bahrdt Date: Tue, 24 Jun 2008 13:49:48 +0200 Subject: [PATCH] =?UTF-8?q?=C3=BCbungen=20hinzugef=C3=BCgt,=20formelsammlu?= =?UTF-8?q?ng,=20weitere=20mathe/physik=20kommandos?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- formelsammlung.tex | 0 math.tex | 6 ++- theo2.tex | 1 + ueb6.tex | 110 ++++++++++++++++++++++++++++++++++++++++++++- ueb7.tex | 16 +++++++ 5 files changed, 130 insertions(+), 3 deletions(-) create mode 100644 formelsammlung.tex diff --git a/formelsammlung.tex b/formelsammlung.tex new file mode 100644 index 0000000..e69de29 diff --git a/math.tex b/math.tex index 5d53c6e..02b70db 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 +29,7 @@ \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{\sum}[3]{\Sigma_{#1}^{#2} #3} +\newcommand{\prod}[3]{\Pi_{#1}^{#2} #3} diff --git a/theo2.tex b/theo2.tex index 73647b4..b83b70d 100644 --- a/theo2.tex +++ b/theo2.tex @@ -45,6 +45,7 @@ % \include{ueb4} % \include{ueb5} % \include{ueb6} +% \inlcude{ueb7} \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..36e1e9f 100644 --- a/ueb7.tex +++ b/ueb7.tex @@ -0,0 +1,16 @@ +%\includegraphics{excs/qm1_blatt05_SS08.pdf} +%\pagebreak + +\chapter{Quantenmechanik I - Übungsblatt 5} +\section{Aufgabe 11: Spin-1-Teilchen im konstanten Magnetfeld} +\subsection*{a)} +\subsection*{b)} +\subsection*{c)} + +\section{Aufgabe 12: Das Ethen-Molekül} + +\section{Aufgabe 13: Das Benzol-Molekül} +\subsection*{a)} +\subsection*{b)} +\subsection*{c)} +\subsection*{d)} \ No newline at end of file