142 lines
6.8 KiB
TeX
142 lines
6.8 KiB
TeX
%\includegraphics{excs/qm1_blatt04_SS08.pdf}
|
|
%\pagebreak
|
|
|
|
\chapter{Quantenmechanik I - Übungsblatt 4}
|
|
\section{Aufgabe 9: Zeitentwicklung eines allgemeinen Zweizustandssystems}
|
|
\begin{math}
|
|
H = \hbar \inlinematrix{A & B \\ B & -A} \\
|
|
A = \Omega \cosb{2 \theta} \\
|
|
B = \Omega \sinb{2 \theta} \\
|
|
E_\pm = \pm \hbar \Omega
|
|
\end{math}
|
|
|
|
\subsection*{a)}
|
|
\begin{align}
|
|
\ket{\chi_+} &= \inlinematrix{\sinb{2 \theta} \\ 1 - \cosb{2 \theta}} &= \inlinematrix{\cosb{\theta} \\ \sinb{\theta}} \\
|
|
\ket{\chi_-} &= &= \inlinematrix{\sinb{\theta} \\ -\cosb{\theta}}
|
|
\end{align}
|
|
|
|
\subsection*{b)}
|
|
\begin{align}
|
|
\i \hbar \diffPs{t} \ket{\psi} &= H \ket{\psi} \\
|
|
\ket{\dot{\psi}} &= -\i \Omega \inlinematrix{\cosb{2 \theta} & \sinb{2 \theta} \\ \sinb{2 \theta} & -\cosb{2 \theta}} \ket{\psi} \\
|
|
\ket{\psi(t)} &= c_1 \cdot e^{\i \Omega t} \ket{\chi+} + c_2 \cdot e^{-\i \Omega t} \ket{\chi-} \\
|
|
\ket{\psi(0)} &= \inlinematrix{\lambda \\ \mu} \\
|
|
\inlinematrix{c_+(t) \\ c_-(t)} &= \sbk{\lambda \cosb{\theta} + \mu \sinb{\theta}} \cdot \inlinematrix{\cosb{\theta} \\ \sinb{\theta}} \cdot e^{-\i t} +
|
|
\sbk{\lambda \sinb{\theta} - \mu \cosb{\theta}} \cdot \inlinematrix{\sinb{\theta} \\ -\cosb{\theta}} \cdot e^{\i \Omega t}
|
|
\end{align}
|
|
|
|
\subsection*{c)}
|
|
%hier stimmt evtl. was nicht
|
|
\begin{align}
|
|
\ket{\psi(0)} &= \inlinematrix{0 \\ 1} \\
|
|
\abs{\braket{+}{\psi(t)}}^2 &= \sin^2\sbk{\Omega t} \cdot \sin^2\sbk{2 \theta} \\
|
|
\abs{\braket{-}{\psi(t)}}^2 &= \cos^2\sbk{\Omega t} + \sin^2\sbk{\Omega t} \cdot \cos^2\sbk{2 \Omega}
|
|
\end{align}
|
|
|
|
|
|
\section{Aufgabe 10: Zerfall eines instabilen Zustandes}
|
|
\subsection*{a)}
|
|
\begin{align}
|
|
<H>_\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} \\
|
|
<H^2>_\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)}
|
|
\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
|
|
|
|
|