Merge branch 'master' of git+ssh://git@git.eluhost.de:522/qm1-script
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Daniel Bahrdt
commit
207cd4749b
154
kapII-0.tex
154
kapII-0.tex
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@ -41,3 +41,157 @@ Die möglichen $E$-Werte sind die Eigenwerte des $H$-Operators. Diese Form der P
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\end{equation}
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\item H-Operator ist durch das Analogon zur klassischen Hamiltonfunktion gegeben
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\end{enumerate}
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\section{Beispiel 1: $\infty$-Potentialtopf}
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%\begin{figure}[h]
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%\includegraphics{pdf/II/00-02-00.pdf}
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%\end{figure}
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\begin{equation}
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V(x) = \left\lbrace \begin{array}{ll} \infty &\text{für } \abs{x} > a\\ 0 &\text{für } \abs{x} < a \end{array} \right.
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\end{equation}
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\paragraph*{klassisch}
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$x(t_0), p(t_0) = \sqrt{2m E}$
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%\begin{figure}[h]
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%\includegraphics{pdf/II/00-02-01.pdf}
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%\end{figure}
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\paragraph*{quantal}
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\subparagraph*{Schritt 1} Stationäre Zustände
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\begin{equation}
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\left( -\frac{\hbar^2}{2m}~\partial_x^2 + V(x) \right) \phi(x) \stackrel{!}{=} E \phi(x)
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\end{equation}
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mit $V(x) = 0$ für $\abs{x} < a$.\\
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Randbedingung: $\phi(\pm a) = 0$
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\begin{equation}
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\diffPs{x}^2 \phi(x) = -\frac{2 m E}{\hbar} \phi(x)
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\end{equation}
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Lösung:
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\begin{enumerate}
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\item symmetrisch
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\begin{equation}
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\phi(x) = A \cos(kx); ~ k \equiv \sqrt{\frac{2 m E}{\hbar^2}}
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\end{equation}\\
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Rand:
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\begin{equation}
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\phi(\pm a) = A \cos k a) \stackrel{!}{=} 0
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\end{equation}\\
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daraus folgt (mit $n = 0, 2, 4, 6, ...$)
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\begin{equation}
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k_n a = \frac{\pi}{2} ( 1 + n )
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\end{equation}
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und mit $n = 0, 1, ..., \infty$ ist dann
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\begin{equation}
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E_n = \frac{\hbar^2}{2m} \frac{1}{a^2} \left(\frac{\pi}{2} \right)^2 (1 + n)^2
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\end{equation}
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\item antisymmetrisch
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\begin{equation}
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\phi(x) = A \sin(k x)
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\end{equation}
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Rand:
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\begin{equation}
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\phi(\pm a) = \pm A \sin(k a) \stackrel{!}{=} 0
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\end{equation}
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daraus folgt mit $n = 1, 3, 5, 7, 9, ...$
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\begin{equation}
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k_n a = \frac{\pi}{2} (1 + n)\\
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\end{equation}
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und mit $n = 0, 1, ..., \infty$ ist dann
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\begin{equation}
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E_n = \frac{\hbar^2}{2m} \frac{1}{a^2} \left(\frac{\pi}{2} \right)^2 (1 + n)^2
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\end{equation}
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\end{enumerate}
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\subparagraph*{Fazit}
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\begin{enumerate}
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\item Energieeigenwerte sind quantisiert.
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%\begin{figure}[h]
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%\includegraphics{pdf/II/00-02-02.pdf}
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%\end{figure}
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\item Eigenfunktionen $\phi_n(x)$ bilden ein vollständig normiertes Basissystem.
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\begin{equation}
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\phi_n = \frac{1}{\sqrt{a}} \left\lbrace \begin{array}{ll} \cos(k_n x) & n\text{ grade}\\ \sin(k_n x) & \text{sont.} \end{array} \right.
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\end{equation}
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\begin{align}
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\intgr{-\infty}{+\infty}{\phi_m(x) \phi_n(x)}{x} &= \delta_{m,n}\\
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\sum_{n=0}^{\infty} \phi_n(x) \phi_n(x') &= \delta(x - x')
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\end{align}
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d.h. jede Funktion $\psi(x)$ kann entwickelt werden in dieser Basis $\psi(x) = \sum_{n=0}^{\infty} c_n \phi_n(x)$.
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%\begin{figure}[h]
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%\includegraphics{pdf/II/00-02-03.pdf}
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%\caption{Skizze der Eigenfunktionen}
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%\end{figure}
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\end{enumerate}
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\paragraph{Schritt 2} Dynamik\\
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Sei nun $\psi(x, t)$ beliebig gegeben durch
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\begin{equation}
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\psi(x,t) = \sum_{n=0}^{\infty} c_n(t) \phi_n(x)
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\end{equation}
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eingesetzt in die Schrödinger Gleichung
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\begin{align}
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i \hbar \sum_{n=0}^{\infty} \left( \diffPs{t} c_n(t) \right) \phi_n(x) &= \sum_{n'=0}^{\infty} \left( -\frac{\hbar^2}{2m} \diffPs{x}^2 \right) c_{n'}(t) \phi_{n'}(x)\\
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i \hbar \sum_{n=0}^{\infty} \diffPs{t} c_n(t) \phi_n(x) &= \sum_{n'=0}^{\infty} c_{n'}(t) E_{n'} \phi_{n'}(x) &\left| \intgr{-a}{+a}{\phi_m(x)}{x} \right.\\
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i\hbar \diffPs{t} c_n(t) &= E_n c_m(t)
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\end{align}
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dann ist
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\begin{equation}
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c_m(t) = e^{-\frac{i}{\hbar} E_m t}e_m(0)
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\end{equation}
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mit
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\begin{equation}
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c_m(0) = \intgr{-a}{+a}{\phi_m(x) \psi(x,t)}{x}
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\end{equation}
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und damit
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\begin{align}
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\psi(x,t) &= \intgru{\sum_n \phi_n(x) e^{-\frac{i}{\hbar} E_n (t-t_0) \phi_n(x') \psi(x,t)}}{x}\\
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&\equiv \intgru{U(x,t;x',t_0) \psi(x',t_0)}{x'}
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\end{align}
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($U(x,t;x',t_0)$ ... Zeitentwicklungsoperator in Ortsdarstellung)
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\section{Beispiel 2: $\delta$-Potentialtopf}
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%\begin{figure}[h]
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%\includegraphics{pdf/II/00-03-00.pdf}
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%\end{figure}
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Mit
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\begin{equation}
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V(x) = -\alpha \delta(x)
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\end{equation}
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ergeben sich die Stationären Zustände:
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\begin{align}
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\left[ -\frac{\hbar^2}{2m} -\alpha \delta(x) \right] \phi(x) &= E \phi(x) &\left| \intgr{-\varepsilon}{+\varepsilon}{}{x} \right.\\
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-\frac{\hbar^2}{2m} \left[ \phi'(0+\varepsilon) - \phi'(0-\varepsilon) \right] - \alpha \phi(0) &= \underbrace{2 \varepsilon E \phi(0)}_{\rightarrow 0}
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\end{align}
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$\phi'(x)$ springt bei der Null, wobei $\phi$ selbst stetig ist.
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\paragraph*{Fall 1} $E < 0$\\
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$x > 0$:
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\begin{align}
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\diffPs{x}^2 \phi(x) &= K^2 \phi(x) &K^2 \equiv \frac{\abs{E} 2m}{\hbar^2}\\[15pt]
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\phi(x) &= A e^{\pm K x}
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\end{align}
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$+K$-Lösung nicht normierbar, also:
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\begin{equation}
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\phi(x) = A_+ e^{-K x}
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\end{equation}\\[15pt]
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$x > 0$:
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\begin{equation}
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\phi(x) = A_- e^{-K \abs{x}}
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\end{equation}
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Aus der Stetigkeit von $\phi$ folgt:
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\begin{equation}
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A_+ = A_- = A
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\end{equation}
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\subparagraph*{Sprungbedingung}
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\begin{align}
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\frac{- \hbar^2}{2m} \left( \right) - \alpha A &= 0\\
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K &= \frac{m \alpha}{\hbar^2}\\
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\rightarrow E &= -\frac{\hbar^2}{2m} \left( \frac{2m}{\hbar} \right)^2 \alpha^2
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\end{align}
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$\rightarrow$ Ein gebundener Zustand.
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%\begin{figure}[h]
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%\includegraphics{pdf/II/00-03-01.pdf}
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%\end{figure}
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\subparagraph*{Normierung}
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\begin{equation}
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\phi_0(x) = \frac{1}{\sqrt{K}} e^{-K \abs{x}}
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\end{equation}
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\paragraph*{Fall 2} $E > 0$: Streuzustände (nicht normierbar)
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@ -121,7 +121,7 @@ Heisenberg:
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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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\psi(x,t) &= \intgr{-\infty}{+infty}{U(x,t; x',t_0)}{x'} &\left| \begin{array}{l} t_0 = 0;\\ U(x,t; x',t_0) = \dirac{x}{U(t,t_0}{x'} \end{array} \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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77
kapII-3.tex
77
kapII-3.tex
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@ -24,4 +24,79 @@ der Wahscheinlichkeitsstromdichte (``Kontinuitätsgleichung''; gilt für jede Er
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% &= \frac{1}{\sqrt{2 \pi \hbar}} \frac{p}{}
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\end{align}
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\section{Streuung an der Potentialstufe}
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\section{Streuung an der Potentialstufe}
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%\begin{figure}[h]
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%\includegraphics{pdf/II/03-02-00.pdf}
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%\end{figure}
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\paragraph*{klassisch}
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\subparagraph*{Fall 1} $E > V_0$
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\begin{align}
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x < 0:~ & p(x < 0) = \sqrt{2m E}\\
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x > 0:~ & p(x > 0) = \sqrt{2m (E - V_0)}
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\end{align}
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Teilchen passiert die Potentialstufe, verliert Impuls
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\subparagraph*{Fall 2} $E < V_0$
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\begin{equation}
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p(x < 0) = \sqrt{2m E}
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\end{equation}
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Teilchen wird reflektiert
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\paragraph*{quantal}
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\subparagraph*{Fall 1} $E > 0$\\
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stationäre SG:
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\begin{equation}
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\left( -\frac{\hbar^2}{2m} \diffPs{x}^2 V(x) \right) \phi(x) = E \phi(x)
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\end{equation}
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links: $x < 0$
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\begin{equation}
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\diffPs{x}^2 \phi(x) = -k^2 \phi(x)
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\end{equation}
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mit
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\begin{equation}
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k = \sqrt{\frac{2m}{\hbar^2}}
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\end{equation}
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Lösung:
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\begin{equation}
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\phi(x) = A e^{i k x} + B e^{-i k x}
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\end{equation}
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rechts: $x > 0$
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\begin{equation}
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\left( -\frac{\hbar^2}{2m} \diffPs{x}^2 + V_0 \right) \phi(x) = E \phi(x)
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\end{equation}
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Lösung:
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\begin{equation}
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\phi(x) = C e^{i q x} + D e^{-i q x}
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\end{equation}
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mit
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\begin{equation}
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q = \sqrt{\frac{2m (E - V_0)}{\hbar^2}}
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\end{equation}
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Randbedinung bei $x = 0$
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\begin{align}
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\phi(-\varepsilon) &= \phi(+\varepsilon)\\
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\diffPs{x} \phi(-\varepsilon) &= \diffPs{x} \phi(+\varepsilon)\\
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\rightarrow A + B &= C + D\\
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i k (A - B) &= i q (C - D)\\[15pt]
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\inlinematrix{1 & 1 \\ i k & -i k} \inlinematrix{A \\ B} &= \inlinematrix{1 & 1 \\ i q & -i q} \inlinematrix{C \\ D}\\
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\inlinematrix{A \\ B} &= \frac{1}{2k} \inlinematrix{k+q & k-q \\ k-q & k+q} \inlinematrix{C \\ D}
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\end{align}
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$\rightarrow$ Randbedingung einer von links laufenden Welle\\
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$\Rightarrow$ keine Komponente einer von rechts einlaufenden Welle für $x > 0$ erlaubt!\\
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$\Rightarrow$ $D \equiv 0$\\
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o.B.d.A.: $A = 1$
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\begin{align}
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A &= \frac{k + q}{2k} C ~ \rightarrow C = \frac{2k}{k+q}\\
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B &= \frac{k - q}{2k} C ~ \rightarrow B = \frac{k - q}{k + q}
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\end{align}
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Strom links: $x < 0$
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\begin{align}
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j(x < 0) &= \frac{\hbar}{m} \im{\diffPs{x}\phi ~ \phi^*}\\
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&= \frac{\hbar}{m} \im{i k \left(A e^{i k x} - B e^{-i k x} \right) \left(A^* e^{- i k x} + B^* e^{i k x}\right)}\\
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&= \frac{\hbar}{m} \im{ik \left( A A^* - B B^*\right) + ik \left( A B^* e^{2 i k x} - A^* B e^{-2 i k x} \right)}\\
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&= \frac{\hbar}{m} k \left( 1 - \left( \frac{k - q}{k + q} \right)^2 \right) \equiv j_I - j_R
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\end{align}
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mit
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\begin{align}
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j_I &= \frac{\hbar k}{m} &\text{einfallend}\\
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j_R &= \frac{\hbar}{m} k \left( \frac{k - q}{k + q} \right)^2 \equiv R j_I &\text{reflectiert}
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\end{align}
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2
math.tex
2
math.tex
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@ -33,7 +33,7 @@
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\newcommand{\intgr}[4]{\int_{#1}^{#2} #3 ~\text{d}#4}
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\newcommand{\intgru}[2]{\int #1 ~\text{d}#2}
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\newcommand{\bbracket}[1]{\left #1 \right}
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\newcommand{\sbk}[1]{\left( #1 \right)}
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\newcommand{\levicivita}[1]{\varepsilon_{#1}}
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\newcommand{\krondelta}[1]{\delta_{#1}}
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2
ueb6.tex
2
ueb6.tex
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@ -73,7 +73,7 @@ Da $\inlinematrix{l \\ o \\ m} S_x \inlinematrix{n \\ o \\ p} = 0}$ folgt $S_x =
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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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\braket{\chi_k}{\phi_0} &= \frac{1}{\sqrt{6}} \sum{n=0}{5}{exp(-\i n \delta_{keine ahnung}\braket{\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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