192 lines
7.0 KiB
TeX
192 lines
7.0 KiB
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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