163 lines
6.2 KiB
TeX
163 lines
6.2 KiB
TeX
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\chapter{Freie quantale Teilchen}
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\section{Ebene Wellen}
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$V(x) \equiv 0$:
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\begin{equation}
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i \hbar \ket{\psi} = \frac{p^2}{2m} \ket{\psi}
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\end{equation}
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\paragraph*{stationäre Lösung}
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\begin{equation}
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E \ket{\psi} = \frac{\hat{p}^2}{2m} \ket{\psi}
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\end{equation}
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Lösungen sind Eigenzustände zu $\hat{p}$:
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$\hat{p} \ket{p} = p \ket{p}$:
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\begin{align}
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H \ket{p} &= \frac{p^2}{2m} \ket{p}\\
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&\equiv E \ket{p}
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\end{align}
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alle $E > 0$ sind möglich.\\[15pt]
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Zustände sind zweifach entartet. Zu $E > 0$ gibt es $\ket{\sqrt{2m E}}$ und $\ket{-\sqrt{2m E}}$ als mögliche Eigenzustände. (Vergleiche klassiche Mechanik: $E$ vorgegeben, dann auch $p = \pm \sqrt{2m E}$ als mögliche Bahnen)
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\paragraph*{Lösung in Ortsdarstellung}
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\begin{align}
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\braket{x}{\pm \sqrt{2m E}} &= \braket{x}{\pm p}\\
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&= \frac{1}{\sqrt{2 \pi \hbar}} e^{\pm \frac{i p x}{\hbar}}\\
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&= \frac{1}{\sqrt{2 \pi \hbar}} e^{\pm \frac{i}{\hbar}\sqrt{2m E} \cdot x}\\
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&= \psi_\pm(x)
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\end{align}
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Sei (für $p > 0$):
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\begin{equation}
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\psi(x, t = 0) = \frac{1}{\sqrt{2 \pi \hbar}} e^{\frac{i p x}{\hbar}}
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\end{equation}
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dann:
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\begin{align}
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\psi(x, t) &= e^{-\frac{i}{\hbar} E \cdot t} \psi(x, 0)\\
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&= \frac{1}{\sqrt{2 \pi \hbar}} e^{i \left( \frac{p}{\hbar} x - \frac{E}{\hbar} t \right) }\\
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&= \frac{1}{\sqrt{2 \pi \hbar}} e^{-i(k x - \omega t)}
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\end{align}
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$\rightarrow$ rechtslaufende Welle\\[15pt]
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Entsprechend für $p = -\sqrt{2m E}$ gilt:
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\begin{equation}
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\psi(x,t) = \frac{1}{\sqrt{2 \pi \hbar}} e^{i(-k x - \omega t)}
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\end{equation}
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$\rightarrow$ linkslaufende Welle
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\section{Propagator}
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$\ket{\psi(t_0)}$ gegeben als $\psi(x,t_0) = \braket{x}{\psi(t_0)}$. Gesucht: $\ket{\psi(t)}$:
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\begin{align}
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\psi(x,t) = \braket{x}{\psi(t)} &= \dirac{x}{e^{-\frac{i}{\hbar} (t - t_0) H}}{\psi(t_0)}\\
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&= \intgr{-\infty}{+\infty}{\dirac{x}{e^{-\frac{i}{\hbar} \frac{\hat{p}^2}{2m} (t - t_0)}}{p}\bra{p}\psi(t_0)}{p}\\
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&= \intgru{e^{\frac{-i p^2}{\hbar 2m} (t - t_0)} \frac{1}{\sqrt{2 \pi \hbar}} e^{\frac{i p x}{\hbar}} \psi(p, t_0)}{p}\\
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&= \intgru{\intgru{e^{-\frac{i p^2}{2m \hbar} (t - t_0)} \frac{1}{2 \pi \hbar} e^{\frac{i p}{\hbar} (x - x')} \psi(x', t_0)}{x'}}{x}\\
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&= \intgru{\sqrt{\frac{m}{2 \pi \hbar i (t - t_0))}} e^{\frac{i m (x - x')}{2 \hbar (t - t_0)}} \psi(x', 0)}{x'}\\
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&= \intgru{\dirac{x}{U(t, t_0)}{x'}\braket{x'}{\psi(t_0)}}{x'}
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\end{align}
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$\rightarrow$ Matrixelemente des Zeitentwicklungsoperators in Ortsdarstellung
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\subsection*{Analogie: Diffusion}
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Sei $c(x, t)$ die Dichte der blauen Tinte.
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%\begin{figure}[h]
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%\includegraphics{8-001.pdf}
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%\caption{$c(x,t_0)$}
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%\end{figure}
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Diffusionsgleichung:
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\begin{equation}
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\partial_t c(x, t) = D \partial_x^2 c(x, t)
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\end{equation}
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Lösung:
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\begin{equation}
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c(x, t) = \intgru{\frac{1}{\sqrt{D t 4 \pi}} c(x',t_0)}{x'}
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\end{equation}
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QM:
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\begin{align}
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i\hbar \partial_t \psi(x,t) &= -\frac{\hbar^2}{2m} \partial_x^2 \psi(x,t)\\
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\partial_t \psi(x,t) &= i \frac{\hbar}{2m} \partial_x^2 \psi(x,t)
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\end{align}
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\section{Gauss'sches Wellenpacket}
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%\begin{figure}[h]
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%\includegraphics{8-002.pdf}
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%\caption{$\abs{\psi^2(x)}$ lokalisiert um die Null mit Breite $~\Delta$}
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%\end{figure}
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\begin{equation}
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\psi(x',0) = \frac{1}{(\pi \Delta^2)^{\frac{1}{4}}}
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\end{equation}
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\begin{itemize}
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\item Die Wahrscheinlichkeit Teilchen am Ort $x'$ zu messen sei
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\begin{equation}
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\rho(x') \equiv \abs{\psi(x',0)}^2
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\end{equation}
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.
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\item Die Wahrscheinlichkeit Teilchen mit Impuls $p$ zu messen sei
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\begin{equation}
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\rho(p) \equiv \abs{\psi(p,0)}^2
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\end{equation}
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.
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\end{itemize}
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\begin{align}
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\psi(p,0) \equiv \braket{p}{\psi(0)} &= \intgr{-\infty}{+\infty}{\braket{p}{x'} \braket{x'}{\psi(0)}}{x'}\\
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&= \intgr{-\infty}{+\infty}{\frac{e^\frac{-i p x'}{\hbar}}{\sqrt{2 \pi \hbar}} \frac{1}{\left(\pi \Delta^2\right)^\frac{1}{4}} e^\frac{i p_0 x'}{\hbar} e^\frac{-{x'}^2}{2 \Delta^2}}{x'}\\
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&= \frac{\Delta^\frac{1}{2}}{\pi^\frac{1}{4} \hbar^\frac{1}{2}} e^\frac{-(p - p_0)^2 \Delta^2}{2 \hbar}
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\end{align}
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Also ergibt sich für $\rho(p)$
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\begin{equation}
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\rho(p) = \frac{\Delta}{\pi^\frac{1}{2} \hbar} e^\frac{-(p - p_0)^2}{\hbar^2 \Delta^{-2}}
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\end{equation}
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%\begin{figure}[h]
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%\includegraphics{8-002.pdf}
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%\caption{$\abs{\psi^2(x)}$ lokalisiert um die Null mit Breite $~\Delta$}
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%\end{figure}
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\paragraph*{Dispersion, Unschärfe}
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\begin{align}
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(\Delta x)_\ket{\psi(0)} &= \sqrt{<x^2> - <x>^2} = \sqrt{\dirac{\psi(0)}{\hat{x}^2}{\psi(0)}}\\
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&= \frac{\Delta}{\sqrt{2}}\\[15pt]
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(\Delta p)_\ket{\psi(0)} &= \sqrt{<p^2> - <p>^2} = ...\\
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&= \frac{\hbar}{\sqrt{2} \Delta}
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\end{align}
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Heisenberg:
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\begin{equation}
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(\Delta x)_\ket{\psi(0)} (\Delta p)_\ket{\psi(0)} \geq \frac{1}{2}\dirac{\psi_0}{[x,p]}{\psi_0} = \frac{\hbar}{2}
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\end{equation}
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für Gauss'sches Wellenpacket ist Gleichheit erreicht.
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\paragraph*{Dynamik}
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\begin{align}
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\psi(x,t) &= \intgr{-\infty}{+infty}{U(x,t; x',t_0)}{x'} &\left| t_0 = 0; ~ U(x,t; x',t_0) = \dirac{x}{U(t,t_0}{x'} \right.\\
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&= \left( \sqrt{\pi} \left( \Delta + \frac{i \hbar t}{m \Delta} \right) \right)^{-\frac{1}{2}} e^\frac{-\left(x - \frac{p_0 t}{m} \right)^2}{2 \Delta^2 \left( 1 + i \hbar \frac{t}{m \Delta^2} \right)} e^{\frac{i p_0}{\hbar} \left( x - \frac{p_0 t}{m} \right)}
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\end{align}
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\begin{equation}
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\rho(x,t) = \abs{\psi(x,t)}^2
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\end{equation}
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%\begin{figure}[h]
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%\includegraphics{8-002.pdf}
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%\caption{$\abs{\psi^2(x)}$ lokalisiert um die Null mit Breite $~\Delta$}
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%\end{figure}
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\begin{equation}
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\Delta(t) = \Delta^{(0)} \sqrt{1 + \frac{\hbar^2 t^2}{m^2 \Delta^4}}
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\end{equation}
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\paragraph{Fazit}
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\begin{enumerate}
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\item \begin{equation}
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<x>(t) = p_0 \frac{t}{m} = <p>(t) \frac{t}{m} \text{ und } <p>(t) = p_0
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\end{equation}
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Mittelwerte verhalten sich klassisch (Ehrenfest!).
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\item \begin{equation}
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(\Delta x)(t) = \frac{\Delta}{\sqrt{2}} \left( 1 + \frac{\hbar^2 t^2}{m^2 \Delta^4} \right)^\frac{1}{2}
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\end{equation}
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Breite im Ort läuft auseinander. Sie ändert sich sinifikant auf der Zeitskala:
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\begin{equation}
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t^* = \frac{m}{\hbar} \Delta
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\end{equation}
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Beispiel makroskopisch:
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\begin{align}
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m &=0,1kg; ~ \Delta = 10^{-3}m; ~ \hbar = 10^{-23}Js\\[15pt]
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t^* &= \frac{10^{-1} 10^{-6}}{10^{-34}}s = 10^{27}s
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\end{align}
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Elektron:
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\begin{align}
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m_e &= 10^{-30}kg; ~ \Delta = 10^{-10}m; ~ \hbar = 10^{-23}Js\\[15pt]
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t^* &= \frac{10^{-30} 10^{-20}}{10^{-34}}s = 10^{-16}s
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\end{align}
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\item Impulse verbreitern sich analog.
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\end{enumerate}
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