übungsblatt 4
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Daniel Bahrdt
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ueb4.tex
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ueb4.tex
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\section{Aufgabe 10: Zerfall eines instabilen Zustandes}
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\subsection*{a)}
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\begin{align}
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<H>_\psi &= \dirac{\psi(t)}{H}{\psi(t)} \\
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&= \dirac{\Phi}{e^{\frac{\i H t}{\hbar}} H e^{\frac{-\i H t}{\hbar}}}{\Phi} \\
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&mit [H,e^{H}] = 0 \\
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&= \dirac{\Phi}{H \cdot e^{\frac{\i H t}{\hbar}} \cdot e^{\frac{-\i H t}{\hbar}} \cdot H}{\Phi} %error: 2 Hs wo nur eins?
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&= \dirac{\Phi}{H}{\Phi} \\
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<H^2>_\psi &= \dirac{\Phi}{H^2}{\Phi} analog \\
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\varianz{H}{\psi}^2 &= \dirac{\Phi}{H^2}{\Phi} - \dirac{\Phi}{H}{\Phi}^2 \\
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&\Rightarrow \text{zeitunabhängig} \\
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\end{align}
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$p \deq p^2 = p \cdot p = \ket{\Phi} \underbrace{\bra{\Phi} \ket{\Phi}}_{=1} \bra{\Phi} = \ket{\Phi} \bra{\Phi} = p$
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\subsection*{b)}
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\begin{align}
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p(t) &= 1 - \frac{\varianz{H}{}^2}{\hbar^2} \cdot t^2 + \bigOb{t^3} \\
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&= \abs{\braket{\psi(t)}{\Phi}}^2 \\
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&= \abs{\braket{\Phi}{\psi(t)}}^2 - \abs{\dirac{\Phi}{e^{-\frac{\i H t}{\hbar}}}{\Phi}}^2 &\left| Taylor \right. \\
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&= \abs{\braket{\Phi}{\Phi} + \dirac{\Phi}{-\frac{\i H t}{\hbar}}{\Phi} +
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\dirac{\Phi}{\frac{1}{2} \cdot \sbk{-\frac{\i H t}{\hbar}}^2}{\Phi}}^2 + \bigOb{t^3} \\
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&= \abs{1 - \frac{\i t}{\hbar} \dirac{\Phi}{H}{\Phi} - \frac{t^2}{2 \hbar} \dirac{\Phi}{H^2}{\Phi}}^2 \\
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&= \sbk{1 - \frac{\i t}{\hbar} \ssbk{H}_\psi - \frac{t^2}{2 \hbar} \ssbk{H^2}_\psi} \cdot
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\sbk{1 + \frac{\i t}{\hbar} \ssbk{H}_\psi - \frac{t^2}{2 \hbar} \ssbk{H^2}_\psi} + \bigOb{t^3} \\
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&= 1 - \frac{H}{\hbar} t^2 + \bigOb{t^3}
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\end{align}
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\subsection*{c)}
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\begin{align}
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\diffT{t} \ssbk{0} &= \frac{\i}{\hbar} \ssbk{[H,0]} + \ssbk{\diffPfrac{0}{t}}
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\abs{\diffTfrac{p}{t}} &= \abs{\diffT{t} \ssbk{p}} \\
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&= \abs{\frac{\i}{\hbar} \ssbk{[H,p]} + \ssbk{\underbrace{\diffPfrac{p}{t}}_{=0}}} \\
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&= \frac{1}{\hbar} \abs{[H,p]} \\
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&\leq \frac{2}{\hbar} \Delta H \Delta p \\
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&= \frac{2}{\hbar} \Delta H \sqrt{\ssbk{p^2} - \ssbk{p}^2} \\
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\abs{\diffTfrac{p}{t}} &= \frac{2}{\hbar} \Delta H \sqrt{p (1 - p)}
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\end{align}
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\begin{align}
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p_0(0) &= 1 \\
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\diffTfrac{p_0(t)}{t} &\leq 0 \\
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-\diffTfrac{p_0}{t} &= \frac{2 \Delta H}{\hbar} \sqrt{p_0(t) (1 - p_0(t))} \\
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\frac{\text{d}p_0}{\sqrt{p_0 (1 - p_0)}} &= -\frac{2 \Delta H}{\hbar} \text{d}t &\left| \text{Integral drüber} \right.
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\intgr{1}{p_0}{\frac{\text{d}p'_0}{\sqrt{p'_0 (1 - p'_0)}}}{p'} &= -\frac{2 \Delta H}{\hbar} \intgr{0}{t}{}{t'} \\
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\arcsinb{2 p_0(t) -1} &= -\frac{2 \Delta H}{\hbar} t + c \\
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p_0(t) &= \frac{1}{2} \sinb{-\frac{2 \Delta H}{\hbar} t + c} + \frac{1}{2} %fehler bei 1/2?
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&= \frac{1}{2} \sbk{1 + \cosb{\frac{2 \Delta H}{\hbar} t}} \\
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&= \cos^2\sbk{\frac{\Delta H}{\hbar} t} \\
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&\text{fuer} 0 \leq t \leq \frac{\pi \hbar}{2 \Delta H} \text{gilt:}
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p(t) &\geq \cos^2\sbk{\frac{\Delta H}{\hbar} t}
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\end{align}
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\subsection*{d)}
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$h \ket{n} = E_n \ket{n}$ \\
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$w(E) = \sum_n \abs{\braket{n}{\Phi}}^2 \delta\sbk{E - E_n}$
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\begin{align}
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c(t) &= \intgrinf{e^{-\frac{\i E}{\hbar}t} \cdot w(E)}{E} \\
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&= \braket{\Phi}{\psi(t)} \\
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&= \dirac{\Phi}{e^{-\frac{\i H}{\hbar}t}}{\Phi} &\left| \ket{\Phi} = \sum_n a_n \ket{n} \right. \\
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&= \dirac{\Phi}{\sum_n a_n e^{-\frac{\i E_n}{\hbar}t}}{n} \\
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&= \sum_m a_m^\ast \sum_n e^{-\frac{\i E}{\hbar} t} \braket{m}{n} a_n \\
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\sum_n \abs{a_n}^2 e^{-\frac{\i E_n t}{\hbar}}
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\end{align}
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\subsection*{e)}
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\subsection*{f)}
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\begin{align}
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w(E) &= \frac{\Gamma}{2 \pi} \cdot \frac{1}{\sbk{E - E_n}^2 + \hbar^2 \frac{\Gamma^2}{4}} \\
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c(t) &= \frac{\Gamma \hbar}{2 \pi} \intgrinf{\frac{e^{\frac{\i E}{\hbar}t}}{\sbk{E - E_n}^2 + \hbar^2 \frac{\Gamma^2}{4}}}{E} \\
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& \text{mit} Z = \frac{2 \sbk{E - E_n}}{\hbar \Gamma} \\
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&= \frac{1}{\pi} \intgrinf{\frac{e{-\frac{\i}{\hbar}t \sbk{\frac{\hbar \Gamma Z}{2} + E_0}}}{1 + Z^2}}{Z} \\
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&= e^{-\frac{\i E_0}{\hbar}t} \frac{1}{\pi} \intgrinf{\frac{e^{-\frac{\i \Gamma}{2} + Z}}{1 + Z^2}}{Z} \\
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&\text{Residuensatz} \\
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&= e^{-\frac{\i E_0}{\hbar}t} \frac{1}{\pi} \cdot - 2 \pi \i e^{-\frac{\Gamma}{2}t} \frac{1}{-2 \i} \\
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&= e^{-\frac{\i E_0}{\hbar}t} \cdot e^{-\frac{\Gamma}{2}t} \\
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p(t) &= e^{-\Gamma t}
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\end{align}
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\subsection*{f)}
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\begin{align}
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\probb{H \cequiv E_n}{\ket{\psi(t)}} &= \abs{\braket{n}{\Phi}}^2 \\
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w(E) &= \sum_n \abs{\braket{n}{\Phi}}^2 \delta\sbk{E - E_n} \\
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\end{align}
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(13) Gilt, da diese Def. vom Erwartungswert von $w(E)$ sind (Teile davon) %HÄ?
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\begin{align}
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\ssbk{H^\alpha} &= \frac{\Gamma \hbar}{2 \pi} \intgrinf{\frac{E^\alpha}{\sbk{E - E_0}^2 + \hbar^2 \frac{\Gamma^2}{4}}}{E} \\
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&= \frac{2}{\pi \hbar \Gamma} \intgrinf{\frac{E^\alpha}{1 + \sbk{\frac{2 \sbk{E - E_0}}{\hbar \Gamma}}^2}}{E} \\
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& \text{mit} Z = \frac{2 \sbk{E - E_n}}{\hbar \Gamma} \\ \\
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&= \frac{1}{\pi} \intgrinf{\frac{\sbk{\frac{\hbar \Gamma Z}{2} + E_0}^\alpha}{\sbk{1 + Z^2}}}{Z} \\
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\end{align}
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Für $\alpha = 1$
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\begin{align}
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\ssbk{H} &= \frac{1}{\pi} \intgrinf{\frac{\sbk{\frac{\hbar \Gamma Z}{2} + E_0}}{\sbk{1 + Z^2}}}{Z} \\
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&= \intgrinf{\frac{\hbar \Gamma Z}{2 \sbk{1 + Z^2}}}{Z} + \underbrace{\frac{E_0}{\hbar} \underbrace{\intgrinf{\frac{1}{1 + Z^2}}{Z}}_{=\pi}}_{E_0}
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\end{align}
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Für $\alpha = 2$
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\begin{align}
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\ssbk{H^2} &\approx \intgrinf{\frac{Z^2}{1 + Z^2}}{Z}
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\end{align}
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$\ssbk{H^2}$ ist aber nicht definiert \\
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$\Rightarrow$ Unschärfe nicht definiert \\
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$\Rightarrow$ Annahme endl. Unschärfe falsch
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