übung 5 fertig
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Daniel Bahrdt
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ueb5.tex
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ueb5.tex
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\chapter{Quantenmechanik I - Übungsblatt 5}
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\section{Aufgabe 11: Spin-1-Teilchen im konstanten Magnetfeld}
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\subsection*{a)}
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\begin{align}
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H &= - \gamma \mtrx{B} \mtrx{\mtrx{S}} \\
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&= \gamma \hbar \mtrx{B} \mtrx{\Sigma_z} \\
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\ket{\psi(t)} &= c_+ \inlinematrix{1 \\ 0 \\ 0} \cdot e^{\i \gamma \mtrx{B} t} + c_0 \inlinematrix{0 \\ 1 \\ 0} +
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c_- \inlinematrix{0 \\ 0 \\ 1} \cdot e^{-\i \gamma \mtrx{B} t} \\
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&\deq \frac{1}{2} \inlinematrix{1 \\ \sqrt{2} \\ 1} \\
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&= \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}} \\
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\probb{\Sigma_z \cequiv 1}{\psi(t)} &= \abs{\braket{Z_1}{\psi(t)}}^2 \\
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&= \inlinematrix{1 \\ 0 \\ 0} \cdot \inlinematrix{\frac{1}{2} \cdot e^{\i \gamma \mtrx{B} t}
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\\ \frac{\sqrt{2}}{2} \\ \frac{1}{2} \cdot e^{-\i \gamma \mtrx{B} t}}
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&= \abs{\frac{1}{2} e^{\i \gamma \mtrx{B} t}}^2 \\
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&= \frac{1}{4} \\
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\probb{\Sigma_z \cequiv 0}{\psi(t)} &= \frac{1}{2} \\
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\probb{\Sigma_z \cequiv -1}{\psi(t)} &= \frac{1}{4} \\
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\probb{\Sigma_x \cequiv +1}{\psi(t)} &= \abs{\braket{x+}{\psi(t)}}^2 \\
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&= \abs{\frac{1}{2} \inlinematrix{1 \\ \sqrt{2} \\ 1} \cdot \inlinematrix{\frac{1}{2} \cdot e^{\i \gamma \mtrx{B} t} \\
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\frac{\sqrt{2}}{2} \\ \frac{1}{2} \cdot e^{-\i \gamma \mtrx{B} t}}}^2 \\
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&= \frac{1}{4} \sbk{1 + e^{\i \gamma \mtrx{B}} + e^{-\i \gamma \mtrx{B} t}}^2 \\
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&= \frac{1}{4} \sbk{1 + \cosb{\gamma \mtrx{B} t}}^2 \\
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&\Rightarrow
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\probb{\Sigma_x \cequiv +1}{\psi(t)} &= \frac{1}{2} \sin^2\sbk{\gamma \mtrx{B} t} %da stimmt wat net
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\probb{\Sigma_y \cequiv +1}{\psi(t)} &= \frac{1}{4} \sbk{1 - \cosb{\gamma \mtrx{B} t}}^2
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\end{align}
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\subsection*{b)}
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\begin{align}
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\ssbk{\Sigma_z}_\ket{\psi(t)} &= \dirac{\psi(t)}{\Sigma_z}{\psi(t)} \\
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&= 0 \\
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\ssbk{\Sigma_x}_\ket{\psi(t)} &= \cosb{\gamma \mtrx{B} t} \\
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\ssbk{\Sigma_y}_\ket{\psi(t)} &= -\sinb{\gamma \mtrx{B} t} \\
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\diffT{t} \ssbk{\mtrx{\Sigma}} &= \gamma \ssbk{\mtrx{\Sigma}} \times \mtrx{B} \\
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\diffT{t} \inlinematrix{\cosb{\gamma \mtrx{B} t} \\ -\sinb{\gamma \mtrx{B} t} \\ 0} &= \gamma \inlinematrix{\ssbk{\Sigma_y} \mtrx{B} \\ \ssbk{\Sigma_x} \mtrx{B} \\ 0} \\
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\inlinematrix{-\gamma \mtrx{B} \sinb{\gamma \mtrx{B} t} \\ -\gamma \mtrx{B} \cosb{\gamma \mtrx{B} t} \\ 0} &= \gamma \inlinematrix{\ssbk{\Sigma_y} \mtrx{B} \\ \ssbk{\Sigma_x} \mtrx{B} \\ 0}
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\end{align}
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\subsection*{c)}
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\begin{align}
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\frac{\i}{\hbar} \ssbk{\sqbk{H, \mtrx{Sigma}}} &= -\i \gamma \ssbk{\sqbk{\mtrx{B} \cdot \mtrx{\Sigma}, \mtrx{\Sigma}}} \\
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A_i &= \i \gamma \ssbk{B_j \Sigma_j \Sigma_i - \Sigma_i \Sigma_j B_j} \\
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&= -\i \gamma B_j \ssbk{\sqbk{\Sigma_j, \Sigma_j}} \\
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&= -\i \gamma B_j \ssbk{\i \levicivita{jik} \Sigma_k} \\
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&= \gamma B_j \levicivita{jik} \ssbk{\Sigma_k} \\
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&= \gamma \sbk{\ssbk{\mtrx{\Sigma}} \times \mtrx{B}} \\
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&= \diffT{t} \ssbk{\Sigma_i} \\
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&= -\gamma \levicivita{ijk} B_j \ssbk{\Sigma_k} \\
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&= \gamma \levicivita{ikj} \ssbk{\Sigma_k} B_j \\
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&= \gamma \levicivita{ijk} \ssbk{\Sigma_j} B_k \\
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\diffT{t} l &= \left\{l,H\right\}
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\end{align}
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\section{Aufgabe 12: Das Ethen-Molekül}
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$H = \inlinematrix{a & b \\ c & d}$
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$\Rightarrow$
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\begin{align}
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\inlinematrix{a \\ c} &= \inlinematrix{E_0 \\ -A} \\
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\inlinematrix{b \\ d} &= \inlinematrix{-A \\ E} \\
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H &= \inlinematrix{E_0 & -A \\ -A & E}
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\end{align}
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Die Eigenwerte und Eigenvektoren von $H$ sind:
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\begin{align}
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\lambda_1 &= E_0 + A \\
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\vec{\lambda_1} &= \frac{1}{\sqrt{2}} \inlinematrix{1 \\ -1} \\
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\lambda_2 &= E_0 - A \\
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\vec{\lambda_2} &= \frac{1}{\sqrt{2}} \inlinematrix{1 \\ 1}
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\end{align}
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Die Gesamtenergie beträgt dann:
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\equationblock{E\sbk{\ket{G}} = 2 E_0 - 2 A}
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Hierbei ist A die Delokalisierungsenergie (Kopplungskoeffizienten?)
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\section{Aufgabe 13: Das Benzol-Molekül}
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\subsection*{a)}
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$H \ket{\Phi_1} = \sum \ket{\Phi_n} - A \ket{\Phi_{n-1}} - A \ket{\Phi_{n+1}}$
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Kopplung besteht immer mit den benachbarten Atomen. Daher ist (1,6) bzw. (6,1) belegt (Kopplung von 6 mit 1)
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\subsection*{b)}
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$R = \inlinematrix{1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \ddots & 0 & 0 \\ 0 & \ddots & \ddots & 0 & 0 \\ 0 & \ddots & \ddots & 0 & 0 \\ 1 & 0 & 0 & 1 & 1}$
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$\sqbk{H,R} = 0$
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\begin{align}
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H &= a \cdot \one + b \mtrx{R} + c \mtrx{R^\dagger} \\
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H &= E_0 \one - A \mtrx{R} - A \mtrx{R^\dagger}
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\end{align}
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Symmetrien bekannt:
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$R^6 = \one$
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$\Rightarrow$
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$R^6 \ket{\Phi} = 1 \ket{\Phi}$
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$R \ket{\Phi} = \underbrace{e^{\i \frac{2 \pi s}{6}}}_{\text{Eigenwerte}} \ket{\Phi}$ mit $s = 0 \ldots 5$
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\subsection*{c)}
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\subsection*{d)}
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\begin{align}
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\ket{\chi_s} &= \frac{1}{\sqrt{6}} \sum_{n=0}^5 e^{\i n \delta_s} \ket{\Phi_n}
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\detb{\mtrx{R} - \sbk{\lambda_s \one}} &\Rightarrow \\
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e^{\i \delta_s} x_1 &= -x2 \\
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e^{\i \delta_s} x2 &= -x3 \\
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\vdots \\
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\end{align}
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Die Eigenvektoren lauten dann:
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\equationblock{\frac{1}{\sqrt{6}} \inlinematrix{1 \\ -e^{\i \delta_s} \\ e^{2 \i \delta_s} \\ -e^{3 \i \delta_s} \\ \vdots }}
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\subsection*{d)}
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\begin{align}
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H \ket{\chi_s} &= E_0 \one \ket{\chi_s} - A \mtrx{R} \ket{\chi_s} - A \mtrx{R^\dagger} \ket{\chi_s} \\
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&= E_0 \ket{\chi_s} - A \lambda_s \ket{\chi_s} - A \lambda_3^\ast \ket{x_3} \\
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&= E_0 - 2 A \re{\lambda_s \ket{\chi_s}} \\
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&= E_0 - 2 A \cosb{\delta_s} \\
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&= E_s
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\end{align}
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Die Gesamtenergie beträgt dann:
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\equationblock{E_ges = 6 E_0 - 8 A}
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$E_{Kekule} = 3 \sbk{E_{Ethen}} = 6 E_0 - 6 A$ (Pauli-Prinzip)
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