[vorlesung] kapitel 4 (fertig) und 5 (WIP) ohne Bilder hinzugefügt
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kapII-4.tex
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kapII-4.tex
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\chapter{Symmetrie}
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\section{Nichtentartung gebundener Zustände}
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\paragraph{Satz} Gebundene Zustände $\left( \phi(x) \xrightarrow{x \rightarrow \pm \infty} 0 \right)$ in einer Dimension sind nicht entartet
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\subparagraph{Beweis} durch Wiederspruch:
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\begin{align}
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-\frac{\hbar^2}{2m} \diffPs{x}^2 \phi_1 + V(x) \phi_1 &= E \phi_1 &\left| \phi_2 \right.\\
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-\frac{\hbar^2}{2m} \diffPs{x}^2 \phi_2 + V(x) \phi_1 &= E \phi_2 &\left| \phi_1 \right.\\[15pt]
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\rightarrow \diffPs{x}^2(\phi_1) \phi_2 + \phi_1 \diffPs{x}^2(\phi_2) &= 0\\
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\diffPs{x}\left( \diffPs{x}(\phi_1) \phi_2 - \phi_1 \diffPs{x}(\phi_2) \right)\\
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\rightarrow \diffPs{x}(\phi_1) \phi_2 - \phi_1 \diffPs{x}(\phi_2) &= \const
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&= 0 ~ \left(\text{betrachte } x = \pm \infty \right)\\
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\rightarrow \frac{\diffPs{x}(\phi_1)}{\phi_1} &= \frac{\diffPs{x}(\phi_2)}{\phi_2}\\[15pt]
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\rightarrow \phi_1(x) &= \const \cdot \phi_2(x)
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\end{align}
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\begin{flushright}
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$\square$
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\end{flushright}
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\section{Parität}
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\paragraph{Satz} Falls $V(x) = V(-x)$ können die Eigenfunktionen von $H$ als symmetrisch oder antisymmetrisch gewählt werden.
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\subparagraph{Beweis} Sei $\phi(x)$ Lösung der SG. Betrachte $\tilde{\phi}(x) \equiv \phi(-x)$:
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\begin{align}
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-\frac{\hbar^2}{2m} \diffPs{x}^2(\phi(x)) + V(x) \tilde{\phi(x)} &= -\frac{\hbar^2}{2m} \diffPs{x}^2(\tilde{\phi}(x)) + V(-x) \tilde{\phi}(x)\\[15pt]
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\rightarrow \frac{\hbar^2}{2m} \diffPs{x}^2(\phi(x)) + V(-x) \phi(-x) &= E \phi(-x)\\
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&= E \tilde{\phi}(x)
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\end{align}
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Also löst
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\begin{equation}
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\phi_{S,a}(x) \equiv \phi(x) \pm \phi(-x)
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\end{equation}
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die SG zu $E$.
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\subparagraph{Alternativer Zugang über Paritätsoperator}
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Definiere den Paritätsoperator $\Pi$ als:
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\begin{equation}
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\Pi \ket{x} \equiv \ket{-x} ~\left[~ \neq -\ket{x} ~\right]
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\end{equation}
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%\begin{figure}[h]
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%\includegraphics{pdf/II/04-02-00.pdf}
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%\caption{Beispiel für $\Pi$}
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%\end{figure}
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\begin{align}
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\Pi \ket{\psi} &= \intgr{-\infty}{+\infty}{\Pi \ket{x} \braket{x}{\psi}}{x}\\
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&= \intgr{-\infty}{+\infty}{\ket{-x} \psi(x)}{x} &(-x = y)\\
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&= \intgr{-\infty}{+\infty}{\ket{y} \psi(-y)}{(-y)}\\
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&= \intgr{-\infty}{+\infty}{\ket{y} \psi(-y)}{y} &\left| ~\bra{x} \right.\\
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\rightarrow \dirac{x}{\Pi}{\psi} &= \intgr{-\infty}{+\infty}{\braket{x}{y} \psi(-y)}{y}\\
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\braket{x}{\Pi \psi} &= \psi(-x)\\
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\left( \Pi \psi \right)(x) &= \psi(-x)
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\end{align}
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Wirkung auf Impulse:
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\begin{align}
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\dirac{x}{\Pi}{p} &= p(-x)\\
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&= \frac{1}{\sqrt{2 \pi \hbar}} e^{\frac{i p}{\hbar} (-x)}\\
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&= \frac{1}{\sqrt{2 \pi \hbar}} e^{\frac{i}{\hbar} (-p) x}\\
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&= \braket{x}{-p}\\[15pt]
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\Pi \ket{p} &= \ket{-p}
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\end{align}
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Eigenschaften von $\Pi$:
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\begin{align}
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\Pi^2 \ket{x} &= \Pi \ket{-x} = \ket{x}\\
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\rightarrow \Pi^2 &= \one\\
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\rightarrow \Pi^{-1} &= \Pi
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\rightarrow \text{Eigenwerte} &= \pm 1
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\end{align}
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Eigenfunktionen zu $+1$:
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\begin{equation}
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\Pi \ket{\psi} = +\ket{\psi}
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\end{equation}
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in Ortsdarstellung
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\begin{align}
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\braket{x \Pi}{\psi} &= + \braket{x}{\psi}
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\psi(-x) &= \psi(x)
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\end{align}
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$\Pi$ ist hermitesch und unitär.\\[15pt]
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Falls $[H, \Pi] = 0$, gibt es eine gemeinsame Eigenbasis; d.h. Eigenfunktionen von $H$ können als symmetrisch bzw. antisymmetrisch gewählt werden.\\[15pt]
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Was ist $[H, \Pi]$ ?
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\begin{enumerate}
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\item $[V(\hat{x}), \Pi]$
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\begin{align}
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\dirac{x}{V(\hat{x})\Pi - V(\hat(x)}{x'} &= (V(x) - V(x')) \underbrace{\dirac{x}{\Pi}{x}}_{\braket{x}{-x'} = \delta(-x' - x)}\\
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&= (V(x) - V(x')) \delta(x' + x)\\
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&= \left\lbrace\begin{array}{ll} 0 & \text{falls } x' \neq -x \\ \underbrace{(V(x) - V(x'))}_{= 0 \text{ falls } V(x) = V(-x)}\delta(0) & \text{falls } x' = -x \end{array}\right.
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\end{align}
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\item $[\hat{p}^2, \Pi]$
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\begin{align}
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\dirac{p}{\hat{p}^2 \Pi - \Pi \hat{p}^2}{p'} &= \left(p^2 - {p'}^2 \right) \braket{p \Pi}{p'}\\
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&= \left(p^2 - {p'}^2 \right) \braket{p}{-p'} = 0
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\end{align}
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\end{enumerate}
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\begin{flushright}
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$\square$
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\end{flushright}
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\section{Translationsoperator periodisches Potential\\und Bloch Theorem}
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\paragraph{Definition} Translationoperator
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%\begin{figure}[h]
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%\includegraphics{pdf/II/04-03-00.pdf}
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%\end{figure}
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\begin{align}
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\dirac{x}{T_a}{\psi} &\equiv \psi(x - a)\\
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&= \sum_{n=0}^{\infty} \frac{(-a)^n}{n!} \diffPfrac{^n}{x^n} \psi(x)\\
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&= e^{-a \diffP{x}} \psi(x)\\
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&= \dirac{x}{e^{-\frac{i a}{\hbar} \hat{p}}}{\psi}
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\end{align}
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\begin{align}
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\rightarrow T_a &= e^{-\frac{i a}{\hbar} \hat{p}}\\
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&\approx \one - \frac{i a}{\hbar} \hat{p}
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\end{align}
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(Vergleiche: I.5.4 $D_{x/y/z}(\varepsilon) \approx \one - \frac{i \varepsilon}{\hbar} J_{x/y/z}$)\\[15pt]
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$T_a$ unitär $\Rightarrow$ Eigenwerte sind vom Typ $\lambda_a = e^{-i \kappa a}$
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\begin{align}
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T_a \ket{\phi} &= e^{-i \kappa a} \ket{\phi} &\left| \bra{x} \right.\\
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\phi(x - a) &= e^{-i \kappa a} \phi(x)
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\end{align}
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mit $\phi(x)$, der Eigenfunktion zu
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\begin{equation}
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x_a \equiv e^{-i \kappa a}
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\end{equation}
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(mit $\kappa$ beliebig reell)
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%\begin{figure}[h]
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%\includegraphics{pdf/II/04-03-01.pdf}
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%\caption{Periodisches Potential}
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%\end{figure}
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Falls $[H, T_a] = 0$ gibt es gemeinsame Eigenfunktionen:
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\begin{enumerate}
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\item es gilt immer:
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\begin{equation}
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[\hat{p}^2, T_a] = 0
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\end{equation}
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\item $[v(\hat{x}), T_a]$
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\begin{align}
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\dirac{x'}{V(x) T_a - T_a V(\hat{x})}{x} &= (V(x) - V(x'))\underbrace{\dirac{x'}{T_a}{x}}_{\braket{x'}{x+a} = \delta(x' - (x - a))}\\
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&= 0 \text{ falls } V(x) = V(x + a)
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\end{align}
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\end{enumerate}
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\paragraph{Konsequenz (Bloch Theorem)} Es gibt gemeinsame Eigenfunktionen von $H$ und $T_a$:
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\begin{align}
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H \phi_\kappa(x) &= E \phi_\kappa(x)\\[15pt]
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\phi_\kappa(x) &= e^{+i \kappa a} \phi_\kappa(x - a)
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\end{align}
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d.h. SG im Intervall $[0, a]$ lösen mit Randbedingung:
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\begin{equation}
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\phi(a) = e^{-i \kappa a} \phi(0)
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\end{equation}
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\section{Bandstruktur im Beispiel ``Dirac-Kamm''}
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%\begin{figure}[h]
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%\includegraphics{pdf/II/04-04-00.pdf}
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%\end{figure}
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\begin{equation}
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V(x) = \alpha \sum_{j=-\infty}^{+\infty} \delta(x - j a)
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\end{equation}
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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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für $0 < x < a$:
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\begin{equation}
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\phi(x) = A \sin(k x) + B \cos(k x) ~, ~ k^2 = \frac{2m E}{\hbar^2}
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\end{equation}
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für $-a < x < 0$ (Bloch Theorem):
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\begin{align}
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\phi(x) &= e^{-i \kappa a} \phi(x + a)\\
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&= e^{-i \kappa A} \left[ A \sin(k (x + a)) + B \cos(k (x + a)) \right]
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\end{align}
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Anschluss bei $x = 0$:
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\begin{align}
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\phi(+\varepsilon) = \phi(-\varepsilon):~ B &= e^{-i \kappa a} \left( A \sin(k a) + B \cos(k a) \right)\\[15pt]
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\diffT{x}\phi(+\varepsilon) - \diffT{x}\phi(-\varepsilon) &= \frac{2m \alpha}{\hbar^2} \phi(0)\\
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k A - e^{-i \kappa} \left(k A \cos(k a) - k B \sin(k a)\right) &= \frac{2 m \alpha}{\hbar^2} B
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\end{align}
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Lösung falls $\det M = 0$ mit
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\begin{equation}
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M \inlinematrix{A \\ B} = 0
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\end{equation}
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\begin{equation}
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\cos(\kappa A) = \cos(k a) + \frac{m \alpha a}{\hbar^2} \frac{\sin(k a)}{k a}
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\end{equation}
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%\begin{figure}[h]
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%\includegraphics{pdf/II/04-04-01.pdf}
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%\end{figure}
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in $z$ ist erlaubt:
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\begin{equation}
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z_n(\beta) \leq z \leq n\pi
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\end{equation}
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in $E$ ist erlaubt:
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\begin{equation}
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\frac{\hbar^2}{2 m a^2} z_n(\beta) \leq E \leq \frac{\hbar^2}{2 m a^2} (\pi n)^2
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\end{equation}
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263
kapII-5.tex
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\chapter{Harmonischer Oszilator}
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\section{Algebraische Lösung des Spektrums von $H$}
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\begin{align}
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H &= \frac{P^2}{2 m} + \frac{m}{2} \omega^2 X^2; \text{ mit } \hat{x} \equiv \left( \frac{m \omega}{\hbar} \right)^\frac{1}{2} X; ~ \hat{p} \equiv \left( \frac{1}{\hbar m \omega} \right)^2 P
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&= \frac{\hbar \omega}{2} \left( \hat{p}^2 + \hat{x}^2 \right)
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\end{align}
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mit
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\begin{equation}
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[\hat{x}, \hat{p}] = i
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\end{equation}
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\paragraph{Vernichtungsoperator}
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\begin{align}
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\aDs \equiv \frac{1}{\sqrt{2}} \left( \hat{x} + i \hat{p} \right)\\
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\aCr \equiv \frac{1}{\sqrt{2}} \left( \hat{x} - i \hat{p} \right)
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\end{align}
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daraus ergeben sich $\hat{x}$ und $\hat{p}$ als:
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\begin{align}
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\hat{x} &= \frac{1}{\sqrt{2}} \left( \aDs + \aCr \right)\\
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\hat{p} &= \frac{1}{\sqrt{2}} \left( \aDs - \aCr \right)
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\end{align}
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\subparagraph{Kommutator}
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\begin{align}
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[\aDs, \aCr] &= \frac{1}{2} [\hat{x} + i \hat{p}, \hat{x} - i \hat{p}]\\
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&= -i[\hat{x}, \hat{p}]\\
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&= \one = 1
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\end{align}
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eingesetzt in $H$:
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\begin{align}
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H &= \frac{\hbar \omega}{2} \left( \hat{p}^2 + \hat{x}^2 \right)\\
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&= \frac{\hbar \omega}{4} \left( -\left( \aCr\aCr - \aDs\aCr - \aDs\aCr + \aDs\aDs \right) + \left( \aDs\aDs + \aDs\aCr + \aCr\aDs + \aCr\aCr \right) \right)\\
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&= \frac{\hbar \omega}{4} \left( 2\aDs\aCr + 2\aCr\aDs \right)\\
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&= \frac{\hbar \omega}{2} \left( 2\aCr\aDs + \one \right)\\
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&= \hbar \omega \left( \aCr\aDs + \frac{\one}{2} \right)
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\end{align}
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\paragraph{Anzahloperator}
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\begin{equation}
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\nOp \equiv \aCr \aDs
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\end{equation}
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\subparagraph{Kommutatoren}
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\begin{align}
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[\nOp, \aDs] &= [\aCr\aDs, \aDs] = [\aCr, \aDs] \aDs = -\aDs\\
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[\nOp, \aCr] &= [\aCr\aDs, \aCr] = \aDs [\aDs, \aDs] = \aCr
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\end{align}
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\subparagraph*{Spektrum von $\nOp$}
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\begin{enumerate}
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\item Sei $\ket{\nu}$ Eigenvektor von $\nOp$ mit Eigenwert $\nu$:
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\begin{equation}
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\nOp \ket{\nu} = \nu \ket{\nu} \text{ mit } \braket{\nu}{\nu} > 0
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\end{equation}
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\item
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\begin{align}
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\nOp \aDs \ket{\nu} &= \aCr \aDs \aDs \ket{\nu}\\
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&= \left( \aDs \aCr - \one \right) \aDs \ket{\nu}\\
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&= \aDs \nOp \ket{\nu} - \aDs \ket{\nu}\\
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&= \aDs \cdot \nu \ket{\nu} - \aDs \ket{\nu}\\
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&= \left(\nu - 1\right) \aDs \ket{\nu}
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\end{align}
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$\rightarrow$ $\aDs\ket{\nu}$ ist Eigenvektor von $\nOp$ zum Eigenwert $\left( \nu - 1 \right)$\\
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\underline{oder}:
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\begin{equation}
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\aDs\ket{\nu} = \zero \text{ (Nullvektor)}
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\end{equation}
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\item
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\begin{equation}
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0 \leq \norm{\aDs \ket{\nu}}^2 = \braket{\nu}{\aCr \aDs \nu} = \nu \underbrace{\braket{\nu}{\nu}}_{\geq 0}
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\end{equation}
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%\begin{figure}[h]
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%\includegraphics{pdf/II/05-01-00.pdf}
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%\end{figure}
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Die obige Ungleichung wäre nach mehrfacher Anwendung von $\aDs \ket{\nu}$ verletzt wenn anfänglich $\nu$ keine ganze positive Zahl ist.
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\item
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\begin{align}
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\nOp \aCr \ket{\nu} &= \aCr \aDs \aCr \ket{\nu}\\
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&= \aCr \left( \aCr \aDs + 1 \right)\ket{\nu}\\
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&= \aCr \left( \nu + 1 \right) \ket{\nu}\\
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&= \left( \nu + 1 \right) \aCr \ket{\nu}
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\end{align}
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\item
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\begin{align}
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0 \leq \norm{\aCr \ket{\nu}}^2 &= \braket{\nu}{\aDs \aCr \nu} = \dirac{\nu}{\aCr \aDs + 1}{\nu}\\
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&= \left( \nu + 1 \right) \aCr \ket{\nu}
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\end{align}
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$\rightarrow$ kein Problem
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\end{enumerate}
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Daraus ergibt sich das Spektrum von $\nOp$:
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\begin{equation}
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\nOp \ket{n} = n \ket{n} \text{ mit } n \in \setZ^+_0
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\end{equation}
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%\begin{figure}[h]
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%\includegraphics{pdf/II/05-01-01.pdf}
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%\end{figure}
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und das Spektrum von $H$:
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\begin{equation}
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H \ket{n} = \hbar \omega \left( n + \frac{1}{2} \right) \ket{n}
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\end{equation}
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\begin{enumerate}
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\item nur diskrete Eigenwerte erlaubt: Quantisierung
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\item Grundzustandsenergie (auch Nullzustandsenergie):
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\begin{equation}
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E_0 = \frac{\hbar \omega}{2}
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\end{equation}
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\item Es gilt:
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\begin{equation}
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a \ket{0} = \ket{\zero}
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\end{equation}
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\item klassischer harmonischer Oszilator (mit $m = 1\text{kg}$; $\omega = \frac{1}{\text{sec}}$):
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\begin{align}
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\Delta E &= E_{n+1} - E_n = 10^{-34}\text{J}\\
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E_0 &= \frac{m}{2} \omega^2 x^2 = 1 \text{J}
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\end{align}
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\end{enumerate}
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\paragraph*{Matrixelemente der Erzeuger- und Vernichter-Operatoren}
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\begin{align}
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\aCr \ket{n} &= c_n \ket{n+1} ~ \left( \ket{n} \text{ seien normiert} \right)\\[15pt]
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\rightarrow \abs{c_n}^2 &= \dirac{n}{\aDs \aCr}{n}\\
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&= \dirac{n}{\aCr\aDs + 1}{n}\\
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&= (n + 1) \underbrace{\braket{n}{n}}_{1}\\[15pt]
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\rightarrow c_n &= \sqrt{n + 1} \text{ (Phase absichtlich 1 gesetzt)}
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\end{align}
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daraus ergibt sich
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\begin{equation}
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\aCr \ket{n} = \sqrt{n + 1} \ket{n + 1} \label{eqn03}
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\end{equation}
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insbesondere
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\begin{align}
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\aCr \ket{0} &= 1 \ket{1} \Rightarrow \ket{1} = \aCr \ket{0}\\
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\aCr \ket{1} &= \sqrt{2} \ket{2} \Rightarrow \ket{2} = \frac{1}{\sqrt{2}} \aCr \ket{1} = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1}} \aCr \aCr \ket{0}
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\end{align}
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und analog zu \ref{eqn03} gilt:
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\begin{equation}
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\aDs \ket{n} = \sqrt{n} \ket{n - 1}
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\end{equation}
|
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Man erhält nun aus dem Obigen die allgemeine Form für $\ket{n}$:
|
||||
\begin{equation}
|
||||
\boxed{\ket{n} = \frac{1}{\sqrt{n!}} \left( \aCr \right)^n \ket{0}}
|
||||
\end{equation}
|
||||
Die Matrixelemente von $\aCr$ sind dann:
|
||||
\begin{align}
|
||||
\dirac{n'}{\aCr}{n} &= \sqrt{n + 1} \braket{n'}{n + 1}\\
|
||||
&= \sqrt{n + 1} \krondelta{n', n + 1}
|
||||
\end{align}
|
||||
und ebenso die Matrixelemente von $a = \left( \aCr \right)^\dagger$:
|
||||
\begin{align}
|
||||
\dirac{n'}{\aDs}{n} &= \dirac{n}{\aCr}{n}\\
|
||||
&= \sqrt{n} \krondelta{n, n + 1}
|
||||
\end{align}
|
||||
als Matrix:
|
||||
\begin{align}
|
||||
\aDs &= \inlinematrix{
|
||||
0 & \sqrt{1} & 0 & 0 & 0 & \cdots \\
|
||||
0 & 0 & \sqrt{2} & 0 & 0 & \cdots \\
|
||||
0 & 0 & 0 & \sqrt{3} & 0 & \cdots \\
|
||||
0 & 0 & 0 & 0 & \ddots & \\
|
||||
\vdots & \vdots & \vdots & \vdots & &
|
||||
}\\
|
||||
\aCr &= \inlinematrix{
|
||||
0 & 0 & 0 & 0 & \cdots \\
|
||||
\sqrt{1} & 0 & 0 & 0 & \cdots \\
|
||||
0 & \sqrt{2} & 0 & 0 & \cdots \\
|
||||
0 & 0 & \sqrt{3} & 0 & \cdots \\
|
||||
0 & 0 & 0 & \ddots & \\
|
||||
\vdots & \vdots & \vdots & &
|
||||
}\\
|
||||
\aDs\aCr &= \inlinematrix{
|
||||
1 & 0 & 0 & 0 & \cdots \\
|
||||
0 & 2 & 0 & 0 & \cdots \\
|
||||
0 & 0 & 3 & 0 & \cdots \\
|
||||
0 & 0 & 0 & \ddots & \\
|
||||
\vdots & \vdots & \vdots & &
|
||||
}\\
|
||||
\aCr\aDs &= \inlinematrix{
|
||||
0 & 0 & 0 & 0 & 0 & \cdots \\
|
||||
0 & 1 & 0 & 0 & 0 & \cdots \\
|
||||
0 & 0 & 2 & 0 & 0 & \cdots \\
|
||||
0 & 0 & 0 & 3 & 0 & \cdots \\
|
||||
0 & 0 & 0 & 0 & \ddots & \\
|
||||
\vdots & \vdots & \vdots & \vdots & &
|
||||
}
|
||||
\end{align}
|
||||
\begin{equation}
|
||||
\left( \left[\aDs, \aCr \right] = \right) \aDs\aCr - \aCr\aDs = 1
|
||||
\end{equation}
|
||||
\begin{align}
|
||||
\hat{x} &= \frac{1}{\sqrt{2}} \inlinematrix{
|
||||
0 & \sqrt{1} & 0 & 0 & 0 & \cdots \\
|
||||
\sqrt{1} & 0 & \sqrt{2} & 0 & 0 & \cdots \\
|
||||
0 & \sqrt{2} & 0 & \sqrt{3} & 0 & \cdots \\
|
||||
0 & 0 & \sqrt{3} & 0 & \ddots & \\
|
||||
0 & 0 & 0 & \ddots & \ddots & \\
|
||||
\vdots & \vdots & \vdots & & &
|
||||
}\\
|
||||
\hat{p} &= \frac{i}{\sqrt{2}} \inlinematrix{
|
||||
0 & -\sqrt{1} & 0 & 0 & 0 & \cdots \\
|
||||
\sqrt{1} & 0 & -\sqrt{2} & 0 & 0 & \cdots \\
|
||||
0 & \sqrt{2} & 0 & -\sqrt{3} & 0 & \cdots \\
|
||||
0 & 0 & \sqrt{3} & 0 & \ddots & \\
|
||||
0 & 0 & 0 & \ddots & \ddots & \\
|
||||
\vdots & \vdots & \vdots & & &
|
||||
}\\
|
||||
\end{align}
|
||||
\begin{align}
|
||||
\left[ \hat{x}, \hat{p} \right] &= i \one\\[15pt]
|
||||
\tr\left[ \hat{x}, \hat{p} \right] &= \tr\left( i \one \right)\\
|
||||
\tr\left( \hat{x}\hat{p} - \hat{p}\hat{x} \right) &= \tr\left( i \one \right)\\
|
||||
0 &= i \infty \text{ (falls Spur zyklisch $\leftarrow$ gilt nur für endliche Räume)}
|
||||
\end{align}
|
||||
|
||||
\section{Wellenfunktion im Ortsaum}
|
||||
Gesucht:
|
||||
\begin{align}
|
||||
\phi_n(x) &= \braket{x}{n}\\
|
||||
\phi_0(x) &= \braket{x}{0}
|
||||
\end{align}
|
||||
Wir wissen:
|
||||
\begin{equation}
|
||||
\aDs \ket{0} = \zero
|
||||
\end{equation}
|
||||
daraus ergibt sich
|
||||
\begin{align}
|
||||
\frac{1}{\sqrt{2}} \left( \hat{x} + i\hat{p} \right) \ket{0} &= \zero &\left| \bra{x} \right.\\
|
||||
\dirac{x}{\hat{x} + i\hat{p}}{0} &= 0\\
|
||||
x + i(-i) \diffPs{x} \phi_0(x) &= 0 &\left(\text{denn: } \dirac{x}{\hat{p}}{\psi} = -i \hbar \diffPs{x} \psi(x) \right)\\
|
||||
\rightarrow \left(x - \diffPs{x} \right) \phi_0(x) &= 0\\
|
||||
\phi_0(x) &= c \cdot e^{-\frac{x^2}{2}}
|
||||
\end{align}
|
||||
Normierung:
|
||||
\begin{equation}
|
||||
\intgr{-\infty}{+\infty}{\phi_0(x) \phi_0^*(x)}{x} \stackrel{!}{=} 1 ~ \rightarrow ~ c = \frac{1}{\pi^\frac{1}{4}}
|
||||
\end{equation}
|
||||
%\begin{figure}[h]
|
||||
%\includegraphics{pdf/II/05-02-00.pdf}
|
||||
%\end{figure}
|
||||
\paragraph*{Angeregte Zustände}
|
||||
\begin{align}
|
||||
\ket{1} &= \aCr \ket{0} &\left| \bra{x} \right.\\[15pt]
|
||||
\phi_0(x) &= \frac{1}{\sqrt{2}} \left( x - \diffPs{x} \right) \phi_0(x)\\
|
||||
&= \frac{1}{\sqrt{2}} \left( x - \diffPs{x} \right) \frac{1}{\pi^\frac{1}{4}} e^{-\frac{x^2}{2}}\\
|
||||
&= \frac{\sqrt{2}}{\pi^\frac{1}{4}} x e^{-\frac{x^2}{2}}\\[15pt]
|
||||
\phi_2(x) &= \frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}} \left( x - \diffPs{x} \right) \left( \frac{\sqrt{2}}{\pi^\frac{1}{4}} x e^{-\frac{x^2}{2}} \right)\\
|
||||
&= \frac{1}{\pi^\frac{1}{4}} \left( 2x^2 - 1 \right) e^{-\frac{x^2}{2}}
|
||||
\end{align}
|
||||
allgemein:
|
||||
\begin{align}
|
||||
\ket{n} &= \frac{\left( \aCr \right)^n}{\sqrt{n!}} \ket{0} &\left| \bra{x} \right.\\
|
||||
\phi_n(x) &= \frac{1}{\sqrt{n!}} \frac{1}{\pi^\frac{1}{4}} \frac{1}{\sqrt{2^n}} \left( x - \diffPs{x} \right)^n e^{-\frac{x^2}{2}}
|
||||
\end{align}
|
||||
$Q_n$ ist symmetrisch für $n = 2k$, antisymmetrisch für $n = 2k + 1$ und hat $n$ Nullstellen.
|
||||
|
||||
\paragraph*{Erwartungswerte}
|
||||
\begin{align}
|
||||
< \hat{x} >_\ket{n} &= \dirac{n}{\hat{x}}{n}\\
|
||||
&= \frac{1}{\sqrt{2}} \dirac{n}{\aCr + \aDs}{n}\\
|
||||
&= \frac{1}{\sqrt{2}} \bra{n} \left( \sqrt{n + 1} \ket{n + 1} + \sqrt{n} \ket{n - 1}\right)\\
|
||||
&= 0\\[15pt]
|
||||
< \hat{p} >_\ket{n} &= 0
|
||||
\end{align}
|
||||
Wegen Ehrenfest:
|
||||
|
||||
64
theo2.kilepr
64
theo2.kilepr
@ -3,7 +3,7 @@ img_extIsRegExp=false
|
||||
img_extensions=.eps .jpg .jpeg .png .pdf .ps .fig .gif
|
||||
kileprversion=2
|
||||
kileversion=2.0
|
||||
lastDocument=theo2.tex
|
||||
lastDocument=kapII-5.tex
|
||||
masterDocument=
|
||||
name=Theo2
|
||||
pkg_extIsRegExp=false
|
||||
@ -20,17 +20,17 @@ archive=true
|
||||
column=0
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=6
|
||||
open=true
|
||||
line=0
|
||||
open=false
|
||||
order=2
|
||||
|
||||
[item:kapI-1.tex]
|
||||
archive=true
|
||||
column=35
|
||||
column=0
|
||||
encoding=
|
||||
highlight=LaTeX
|
||||
line=0
|
||||
open=true
|
||||
open=false
|
||||
order=5
|
||||
|
||||
[item:kapI-2.tex]
|
||||
@ -98,39 +98,57 @@ order=-1
|
||||
|
||||
[item:kapII-2.tex]
|
||||
archive=true
|
||||
column=10936
|
||||
encoding=
|
||||
highlight=
|
||||
column=33
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=0
|
||||
open=false
|
||||
order=-1
|
||||
order=2
|
||||
|
||||
[item:kapII-3.tex]
|
||||
archive=true
|
||||
column=0
|
||||
encoding=
|
||||
highlight=
|
||||
line=0
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=30
|
||||
open=false
|
||||
order=-1
|
||||
order=3
|
||||
|
||||
[item:math.tex]
|
||||
[item:kapII-4.tex]
|
||||
archive=true
|
||||
column=0
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=34
|
||||
line=183
|
||||
open=true
|
||||
order=4
|
||||
|
||||
[item:kapII-5.tex]
|
||||
archive=true
|
||||
column=11
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=132
|
||||
open=true
|
||||
order=1
|
||||
|
||||
[item:math.tex]
|
||||
archive=true
|
||||
column=1
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=0
|
||||
open=true
|
||||
order=3
|
||||
|
||||
[item:physics.tex]
|
||||
archive=true
|
||||
column=37
|
||||
column=23
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=10
|
||||
line=7
|
||||
open=true
|
||||
order=4
|
||||
order=2
|
||||
|
||||
[item:theo2.kilepr]
|
||||
archive=true
|
||||
@ -143,10 +161,10 @@ order=-1
|
||||
|
||||
[item:theo2.tex]
|
||||
archive=true
|
||||
column=39
|
||||
column=0
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=10
|
||||
line=48
|
||||
open=true
|
||||
order=0
|
||||
|
||||
@ -206,9 +224,9 @@ order=-1
|
||||
|
||||
[item:ueb7.tex]
|
||||
archive=true
|
||||
column=44
|
||||
column=0
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=46
|
||||
open=true
|
||||
line=0
|
||||
open=false
|
||||
order=1
|
||||
|
||||
@ -2,9 +2,7 @@
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage{ngerman}
|
||||
\usepackage{graphics}
|
||||
%\usepackage{pstricks}
|
||||
\usepackage[a4paper,left=2.5cm,right=2.5cm,top=2cm,bottom=5cm]{geometry}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{multirow}
|
||||
@ -36,6 +34,8 @@
|
||||
\include{kapII-1}
|
||||
\include{kapII-2}
|
||||
\include{kapII-3}
|
||||
\include{kapII-4}
|
||||
\include{kapII-5}
|
||||
|
||||
% \part{Übungsmitschrieb}
|
||||
% \label{UE}
|
||||
|
||||
Loading…
Reference in New Issue
Block a user