From 3d47ce3f8e404b4638135f1bf9de068504527942 Mon Sep 17 00:00:00 2001 From: Daniel Bahrdt Date: Thu, 24 Jul 2008 20:55:21 +0200 Subject: [PATCH] hbox overflow --- ueb5.tex | 15 ++++++++------- 1 file changed, 8 insertions(+), 7 deletions(-) diff --git a/ueb5.tex b/ueb5.tex index c820442..4e15195 100644 --- a/ueb5.tex +++ b/ueb5.tex @@ -1,5 +1,5 @@ -%\includegraphics{excs/qm1_blatt05_SS08.pdf} -%\pagebreak +\includegraphics{excs/qm1_blatt05_SS08.pdf} +\pagebreak \chapter{Quantenmechanik I - Übungsblatt 5} \section{Aufgabe 11: Spin-1-Teilchen im konstanten Magnetfeld} @@ -22,9 +22,11 @@ &= \abs{\frac{1}{2} \inlinematrix{1 \\ \sqrt{2} \\ 1} \cdot \inlinematrix{\frac{1}{2} \cdot e^{\i \gamma \mtrx{B} t} \\ \frac{\sqrt{2}}{2} \\ \frac{1}{2} \cdot e^{-\i \gamma \mtrx{B} t}}}^2 \\ &= \frac{1}{4} \sbk{1 + e^{\i \gamma \mtrx{B}} + e^{-\i \gamma \mtrx{B} t}}^2 \\ - &= \frac{1}{4} \sbk{1 + \cosb{\gamma \mtrx{B} t}}^2 \\ - &\Rightarrow - \probb{\Sigma_x \cequiv +1}{\psi(t)} &= \frac{1}{2} \sin^2\sbk{\gamma \mtrx{B} t} %da stimmt wat net + &= \frac{1}{4} \sbk{1 + \cosb{\gamma \mtrx{B} t}}^2 +\end{align} +$\Rightarrow$ +\begin{align} + \probb{\Sigma_x \cequiv +1}{\psi(t)} &= \frac{1}{2} \sin^2\sbk{\gamma \mtrx{B} t} \\ %da stimmt wat net \probb{\Sigma_y \cequiv +1}{\psi(t)} &= \frac{1}{4} \sbk{1 - \cosb{\gamma \mtrx{B} t}}^2 \end{align} @@ -92,7 +94,7 @@ $R \ket{\Phi} = \underbrace{e^{\i \frac{2 \pi s}{6}}}_{\text{Eigenwerte}} \ket{\ \subsection*{c)} \begin{align} - \ket{\chi_s} &= \frac{1}{\sqrt{6}} \sum_{n=0}^5 e^{\i n \delta_s} \ket{\Phi_n} + \ket{\chi_s} &= \frac{1}{\sqrt{6}} \sum_{n=0}^5 e^{\i n \delta_s} \ket{\Phi_n} \\ \detb{\mtrx{R} - \sbk{\lambda_s \one}} &\Rightarrow \\ e^{\i \delta_s} x_1 &= -x2 \\ e^{\i \delta_s} x2 &= -x3 \\ @@ -109,7 +111,6 @@ Die Eigenvektoren lauten dann: &= E_0 - 2 A \cosb{\delta_s} \\ &= E_s \end{align} - Die Gesamtenergie beträgt dann: \equationblock{E_ges = 6 E_0 - 8 A} $E_{Kekule} = 3 \sbk{E_{Ethen}} = 6 E_0 - 6 A$ (Pauli-Prinzip)