120 lines
4.1 KiB
TeX
120 lines
4.1 KiB
TeX
%\includegraphics{excs/qm1_blatt06_SS08.pdf}
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%\pagebreak
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\chapter{Quantenmechanik I - Übungsblatt 6}
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\section{Aufgabe 14: Spin-1-Teilchen}
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\subsection*{a)}
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\begin{align}
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H &= A S_z + B S_x^2 \\
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H &= A \hbar \Sigma_z + B \hbar^2 \Sigma_x^2 \\
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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} \\
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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}}
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\end{align}
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$det(H-\lambda \einsmatrix) = 0 \\$
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\begin{align}
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\lambda_1 &= B\hbar^2 \\
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\lambda_2 &= \frac{\hbar^2}{2} B (1 + \sqrt{1 + \frac{4 A^2}{\hbar^2 B^2}} \\
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\lambda_3 &= \frac{\hbar^2}{2} B (1-\sqrt{1 + \frac{4 A^2}{\hbar^2 B}}
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\end{align}
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\begin{align}
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a &:= A + \frac{\hbar B}{2} \sqrt{1+\frac{4 A^2}{\hbar^2 B^2}} \\
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b &:= \frac{\hbar B}{2} \\
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\ket{\phi_0} &= \inlinematrix{0 \\ 1 \\ 0} &
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\ket{\phi+} &= \frac{1}{\sqrt{a^2+b^2}} \inlinematrix{a \\ 0 \\ b} &
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\ket{\phi-} &= \frac{1}{\sqrt{a^2+b^2}} \inlinematrix{-b \\ 0 \\ a} \\ \\
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\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?
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\ket{\psi(0)} &= \ket{z+} &= \inlinematrix{1 \\ 0 \\ 0} \\ \\
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\end{align}
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$\Rightarrow c_0 = 0$
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\begin{align}
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\frac{1}{\sqrt{a^2+b^2}}(c_+ a - c_- b) &\stackrel{!}{} 1 \\
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\frac{1}{\sqrt{a^2+b^2}}(c_+ b + c_- a) &\stackrel{!}{} 0
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\end{align}
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$\Rightarrow $
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\begin{align}
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c_- &= \frac{-b\sqrt{a^2+b^2}}{a^2+b^2} \\
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c_+ &= \frac{a\sqrt{a^2+b^2}}{a^2+b^2} \\
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S_x &= \hbar \Sigma_x &= \hbar \inlinematrix{0 &1 &0 \\ 1 &0 &1 \\ 0 &1 &0}
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\end{align}
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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
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\subsection*{b)}
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\begin{align}
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S_z &= \hbar \inlinematrix{1 &0 &0 \\ 0 &0 &0 \\ 0 &0 &-1}
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\end{align}
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\begin{align}
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\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)]}
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\end{align}
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\section{Aufgabe 15: Benzol-Molekül (Teil 2)}
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\subsection*{a)}
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\begin{align}
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P_m H &= \norm{\braket{\phi_m}{\psi(t)}} \\
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\ket{\psi(t)} &= U(t,t_0) \ket{\psi(t)} \\ \\
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&= exp(-\frac{\i}{\hbar} t H} \\
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&= \sum{k=0}{5}{exp(-\frac{\i}{\hbar} t \lambda_k) \ket{\chi_k} \braket{\chi_k}{\phi_0}} \\ %warum 5?
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&=^1 \frac{1}{6} \sum{k=0}{5}{\sum{n=0}{5}{exp(-\frac{\i}{\hbar} t (E-2Acos(\delta_k)))}} \\ %warum diese umformung?
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P_m(t) &= \norm{\braket{\phi_m}{\psi(t)}} \\
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&= \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?
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&=^2 \frac{1}{36} \norm{\sum{k}{biw wo?}{exp(-\frac{\i}{\hbar} t (E-2Acos(\delta_k) + \i \delta_k)}} \\ \\
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&= ?
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\end{align}
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(1)
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\begin{align}
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\braket{\chi_k}{\phi_0} &= \frac{1}{\sqrt{6}} \sum{n=0}{5}{exp(-\i n \delta_{keine ahnung}\bracket{\phi_n}{\phi_0}} \\ \\
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&= \frac{1}{\sqrt{6}} e^{-\i 0 \delta_{keine ahnung}} \\
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&= \frac{1}{\sqrt{6}}
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\end{align}
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(2)
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\begin{align}
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\braket{\phi_m}{\phi_n} = \delta_{n m}
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\end{align}
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\subsection*{b)}
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\begin{math}
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\tau = \frac{2\pi}{A}\hbar
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\end{math}
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\section{Aufgabe 16: Schrödingergleichung in Impulsraumdarstellung}
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\begin{align} \\
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%nach schrödingergleichung in ortsdarstellung gucken
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\i \hbar \partial_\tau \ket{\psi} &= H \ket{\psi} & \left| \bra{p}\right. \\
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\i \hbar \partial_\t \psi(p) &= \dirac{p}{\frac{p^2}{2m}+V(\hat{x}}{\psi}
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\end{align}
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\begin{align}}
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\int_{-\infty}^{\infty} \text{dx} \dirac{p}{V(\hat[x})}{x}\braket{x}{\psi} \\
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&= \int_{-\infty}^{\infty} \text{dx} V(x) \frac{1}{\sqrt{2\pi\hbar}}e^{-\frac{\i p}{\hbar}x}\braket{x}{\psi} \\
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&= \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} \\ \\
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&= \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') \\
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&= (2) % aus aufgabenblatt raussuchen
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\end{align}
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\begin{align}
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%wo kommt das her?
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\dirac{p}{V(\hat{x}}{\psi} &= \dirac{p}{V(\i\hbar\diffP{p}}{\psi} \\
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&= V(\i\hbar\diffP{p})\psi(p) %vertauschen von p und V: Regeln?
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\end{align}
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\i \hbar
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