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Daniel Bahrdt 2008-07-24 20:55:21 +02:00
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ueb5.tex
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@ -1,5 +1,5 @@
%\includegraphics{excs/qm1_blatt05_SS08.pdf} \includegraphics{excs/qm1_blatt05_SS08.pdf}
%\pagebreak \pagebreak
\chapter{Quantenmechanik I - Übungsblatt 5} \chapter{Quantenmechanik I - Übungsblatt 5}
\section{Aufgabe 11: Spin-1-Teilchen im konstanten Magnetfeld} \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} \\ &= \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{\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 + e^{\i \gamma \mtrx{B}} + e^{-\i \gamma \mtrx{B} t}}^2 \\
&= \frac{1}{4} \sbk{1 + \cosb{\gamma \mtrx{B} t}}^2 \\ &= \frac{1}{4} \sbk{1 + \cosb{\gamma \mtrx{B} t}}^2
&\Rightarrow \end{align}
\probb{\Sigma_x \cequiv +1}{\psi(t)} &= \frac{1}{2} \sin^2\sbk{\gamma \mtrx{B} t} %da stimmt wat net $\Rightarrow$
\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 \probb{\Sigma_y \cequiv +1}{\psi(t)} &= \frac{1}{4} \sbk{1 - \cosb{\gamma \mtrx{B} t}}^2
\end{align} \end{align}
@ -92,7 +94,7 @@ $R \ket{\Phi} = \underbrace{e^{\i \frac{2 \pi s}{6}}}_{\text{Eigenwerte}} \ket{\
\subsection*{c)} \subsection*{c)}
\begin{align} \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 \\ \detb{\mtrx{R} - \sbk{\lambda_s \one}} &\Rightarrow \\
e^{\i \delta_s} x_1 &= -x2 \\ e^{\i \delta_s} x_1 &= -x2 \\
e^{\i \delta_s} x2 &= -x3 \\ e^{\i \delta_s} x2 &= -x3 \\
@ -109,7 +111,6 @@ Die Eigenvektoren lauten dann:
&= E_0 - 2 A \cosb{\delta_s} \\ &= E_0 - 2 A \cosb{\delta_s} \\
&= E_s &= E_s
\end{align} \end{align}
Die Gesamtenergie beträgt dann: Die Gesamtenergie beträgt dann:
\equationblock{E_ges = 6 E_0 - 8 A} \equationblock{E_ges = 6 E_0 - 8 A}
$E_{Kekule} = 3 \sbk{E_{Ethen}} = 6 E_0 - 6 A$ (Pauli-Prinzip) $E_{Kekule} = 3 \sbk{E_{Ethen}} = 6 E_0 - 6 A$ (Pauli-Prinzip)