From 50b57fdeb65c1883f258c2eb7a1c95134c733dd5 Mon Sep 17 00:00:00 2001 From: Daniel Bahrdt Date: Sat, 19 Jul 2008 20:58:18 +0200 Subject: [PATCH] =?UTF-8?q?=C3=BCbungsblatt=204?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ueb4.tex | 99 +++++++++++++++++++++++++++++++++++++++++++++++++++++++- 1 file changed, 98 insertions(+), 1 deletion(-) diff --git a/ueb4.tex b/ueb4.tex index 8f52ec8..f9cee9b 100644 --- a/ueb4.tex +++ b/ueb4.tex @@ -37,8 +37,105 @@ \section{Aufgabe 10: Zerfall eines instabilen Zustandes} \subsection*{a)} +\begin{align} + _\psi &= \dirac{\psi(t)}{H}{\psi(t)} \\ + &= \dirac{\Phi}{e^{\frac{\i H t}{\hbar}} H e^{\frac{-\i H t}{\hbar}}}{\Phi} \\ + &mit [H,e^{H}] = 0 \\ + &= \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? + &= \dirac{\Phi}{H}{\Phi} \\ + _\psi &= \dirac{\Phi}{H^2}{\Phi} analog \\ + \varianz{H}{\psi}^2 &= \dirac{\Phi}{H^2}{\Phi} - \dirac{\Phi}{H}{\Phi}^2 \\ + &\Rightarrow \text{zeitunabhängig} \\ +\end{align} +$p \deq p^2 = p \cdot p = \ket{\Phi} \underbrace{\bra{\Phi} \ket{\Phi}}_{=1} \bra{\Phi} = \ket{\Phi} \bra{\Phi} = p$ + \subsection*{b)} +\begin{align} + p(t) &= 1 - \frac{\varianz{H}{}^2}{\hbar^2} \cdot t^2 + \bigOb{t^3} \\ + &= \abs{\braket{\psi(t)}{\Phi}}^2 \\ + &= \abs{\braket{\Phi}{\psi(t)}}^2 - \abs{\dirac{\Phi}{e^{-\frac{\i H t}{\hbar}}}{\Phi}}^2 &\left| Taylor \right. \\ + &= \abs{\braket{\Phi}{\Phi} + \dirac{\Phi}{-\frac{\i H t}{\hbar}}{\Phi} + + \dirac{\Phi}{\frac{1}{2} \cdot \sbk{-\frac{\i H t}{\hbar}}^2}{\Phi}}^2 + \bigOb{t^3} \\ + &= \abs{1 - \frac{\i t}{\hbar} \dirac{\Phi}{H}{\Phi} - \frac{t^2}{2 \hbar} \dirac{\Phi}{H^2}{\Phi}}^2 \\ + &= \sbk{1 - \frac{\i t}{\hbar} \ssbk{H}_\psi - \frac{t^2}{2 \hbar} \ssbk{H^2}_\psi} \cdot + \sbk{1 + \frac{\i t}{\hbar} \ssbk{H}_\psi - \frac{t^2}{2 \hbar} \ssbk{H^2}_\psi} + \bigOb{t^3} \\ + &= 1 - \frac{H}{\hbar} t^2 + \bigOb{t^3} +\end{align} + \subsection*{c)} +\begin{align} + \diffT{t} \ssbk{0} &= \frac{\i}{\hbar} \ssbk{[H,0]} + \ssbk{\diffPfrac{0}{t}} + \abs{\diffTfrac{p}{t}} &= \abs{\diffT{t} \ssbk{p}} \\ + &= \abs{\frac{\i}{\hbar} \ssbk{[H,p]} + \ssbk{\underbrace{\diffPfrac{p}{t}}_{=0}}} \\ + &= \frac{1}{\hbar} \abs{[H,p]} \\ + &\leq \frac{2}{\hbar} \Delta H \Delta p \\ + &= \frac{2}{\hbar} \Delta H \sqrt{\ssbk{p^2} - \ssbk{p}^2} \\ + \abs{\diffTfrac{p}{t}} &= \frac{2}{\hbar} \Delta H \sqrt{p (1 - p)} +\end{align} + +\begin{align} + p_0(0) &= 1 \\ + \diffTfrac{p_0(t)}{t} &\leq 0 \\ + -\diffTfrac{p_0}{t} &= \frac{2 \Delta H}{\hbar} \sqrt{p_0(t) (1 - p_0(t))} \\ + \frac{\text{d}p_0}{\sqrt{p_0 (1 - p_0)}} &= -\frac{2 \Delta H}{\hbar} \text{d}t &\left| \text{Integral drüber} \right. + \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'} \\ + \arcsinb{2 p_0(t) -1} &= -\frac{2 \Delta H}{\hbar} t + c \\ + p_0(t) &= \frac{1}{2} \sinb{-\frac{2 \Delta H}{\hbar} t + c} + \frac{1}{2} %fehler bei 1/2? + &= \frac{1}{2} \sbk{1 + \cosb{\frac{2 \Delta H}{\hbar} t}} \\ + &= \cos^2\sbk{\frac{\Delta H}{\hbar} t} \\ + &\text{fuer} 0 \leq t \leq \frac{\pi \hbar}{2 \Delta H} \text{gilt:} + p(t) &\geq \cos^2\sbk{\frac{\Delta H}{\hbar} t} +\end{align} + \subsection*{d)} +$h \ket{n} = E_n \ket{n}$ \\ +$w(E) = \sum_n \abs{\braket{n}{\Phi}}^2 \delta\sbk{E - E_n}$ +\begin{align} + c(t) &= \intgrinf{e^{-\frac{\i E}{\hbar}t} \cdot w(E)}{E} \\ + &= \braket{\Phi}{\psi(t)} \\ + &= \dirac{\Phi}{e^{-\frac{\i H}{\hbar}t}}{\Phi} &\left| \ket{\Phi} = \sum_n a_n \ket{n} \right. \\ + &= \dirac{\Phi}{\sum_n a_n e^{-\frac{\i E_n}{\hbar}t}}{n} \\ + &= \sum_m a_m^\ast \sum_n e^{-\frac{\i E}{\hbar} t} \braket{m}{n} a_n \\ + \sum_n \abs{a_n}^2 e^{-\frac{\i E_n t}{\hbar}} +\end{align} + \subsection*{e)} -\subsection*{f)} \ No newline at end of file +\begin{align} + w(E) &= \frac{\Gamma}{2 \pi} \cdot \frac{1}{\sbk{E - E_n}^2 + \hbar^2 \frac{\Gamma^2}{4}} \\ + 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} \\ + & \text{mit} Z = \frac{2 \sbk{E - E_n}}{\hbar \Gamma} \\ + &= \frac{1}{\pi} \intgrinf{\frac{e{-\frac{\i}{\hbar}t \sbk{\frac{\hbar \Gamma Z}{2} + E_0}}}{1 + Z^2}}{Z} \\ + &= e^{-\frac{\i E_0}{\hbar}t} \frac{1}{\pi} \intgrinf{\frac{e^{-\frac{\i \Gamma}{2} + Z}}{1 + Z^2}}{Z} \\ + &\text{Residuensatz} \\ + &= e^{-\frac{\i E_0}{\hbar}t} \frac{1}{\pi} \cdot - 2 \pi \i e^{-\frac{\Gamma}{2}t} \frac{1}{-2 \i} \\ + &= e^{-\frac{\i E_0}{\hbar}t} \cdot e^{-\frac{\Gamma}{2}t} \\ +p(t) &= e^{-\Gamma t} +\end{align} + +\subsection*{f)} +\begin{align} + \probb{H \cequiv E_n}{\ket{\psi(t)}} &= \abs{\braket{n}{\Phi}}^2 \\ + w(E) &= \sum_n \abs{\braket{n}{\Phi}}^2 \delta\sbk{E - E_n} \\ +\end{align} +(13) Gilt, da diese Def. vom Erwartungswert von $w(E)$ sind (Teile davon) %HÄ? +\begin{align} + \ssbk{H^\alpha} &= \frac{\Gamma \hbar}{2 \pi} \intgrinf{\frac{E^\alpha}{\sbk{E - E_0}^2 + \hbar^2 \frac{\Gamma^2}{4}}}{E} \\ + &= \frac{2}{\pi \hbar \Gamma} \intgrinf{\frac{E^\alpha}{1 + \sbk{\frac{2 \sbk{E - E_0}}{\hbar \Gamma}}^2}}{E} \\ + & \text{mit} Z = \frac{2 \sbk{E - E_n}}{\hbar \Gamma} \\ \\ + &= \frac{1}{\pi} \intgrinf{\frac{\sbk{\frac{\hbar \Gamma Z}{2} + E_0}^\alpha}{\sbk{1 + Z^2}}}{Z} \\ +\end{align} +Für $\alpha = 1$ +\begin{align} + \ssbk{H} &= \frac{1}{\pi} \intgrinf{\frac{\sbk{\frac{\hbar \Gamma Z}{2} + E_0}}{\sbk{1 + Z^2}}}{Z} \\ + &= \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} +\end{align} +Für $\alpha = 2$ +\begin{align} + \ssbk{H^2} &\approx \intgrinf{\frac{Z^2}{1 + Z^2}}{Z} +\end{align} + +$\ssbk{H^2}$ ist aber nicht definiert \\ +$\Rightarrow$ Unschärfe nicht definiert \\ +$\Rightarrow$ Annahme endl. Unschärfe falsch + +