[vorlesung] kapitel II-5 fertig (bis auf bilder)
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Oliver Groß
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kapII-5.tex
123
kapII-5.tex
@ -260,4 +260,127 @@ $Q_n$ ist symmetrisch für $n = 2k$, antisymmetrisch für $n = 2k + 1$ und hat $
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< \hat{p} >_\ket{n} &= 0
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\end{align}
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Wegen Ehrenfest:
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\begin{align}
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\diffT{t}< \hat{x} >(t) &= < \hat{p} >(t) \frac{1}{m}\\[15pt]
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\diffT{t}< \hat{p} >(t) &= -\left< \diffTfrac{V(x)}{x} \right>\\
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&= -m \omega < \hat{x} >(t)
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\end{align}
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\paragraph{Grundzustand} Varianz:
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\begin{align}
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\varianz{x}{\ket{0}}^2 &\equiv \dirac{0}{(x - <x>)^2}{0} &\left(<x> = 0\right)\\
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&= \dirac{0}{x^2}{0}\\
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&= \dirac{0}{\frac{1}{2} \left( \aCr + \aDs \right)^2}{0}\\
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&= \frac{1}{2} \dirac{0}{\left( \aCr\ \right)^2 + \aCr\aDs + \aDs\aCr + \left( \aDs \right)^2}{0}\\
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&= \frac{1}{2} \dirac{0}{\aDs\aCr}{0}\\
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&= \frac{1}{2} \dirac{0}{\aDs}{1}\\
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&= \frac{1}{2} \braket{0}{0}\\
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&= \frac{1}{2}\\[15pt]
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\varianz{p}{\ket{0}}^2 &= \frac{1}{2} ~ \text{(genauso wie oben)}\\[15pt]
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\varianz{x}{\ket{0}}\varianz{p}{\ket{0}} &= \frac{1}{2} \geq \frac{1}{2} \abs{\dirac{0}{[x, p]}{0}} = \frac{1}{2}
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\end{align}
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\section{Darstellung durch Hermitepolynome}
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\paragraph*{Definition} Sei $H_n(x)$ ein Hermitepolynom definiert durch:
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\begin{align}
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\phi_n &\equiv \sqrt{\frac{1}{2^n n! \sqrt{n}}} e^{-\frac{x^2}{2}} H_n(x)\\[15pt]
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\rightarrow H_n(x) &= e^\frac{x^2}{2} \left( \sqrt{2} \aCr \right)^n e^{-\frac{x^2}{2}}\\
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&= e^{(x^2)} \underbrace{e^{-\frac{x^2}{2}} \left( x - \diffPs{x} \right) e^\frac{x^2}{2}}_{(-1)^n \diffPfrac{^n}{x^n}} e^{(-x^2)}\\
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&= (-1)^n e^{(x^2)} \left( \diffP{x} \right)^n e^{-\frac{x^2}{2}}
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\end{align}
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Beispiele:
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\begin{equation}
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H_0(x) = 1; ~ H_1(x) = 2x; ~ H_2(x) = 4x^2 - 2
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\end{equation}
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\subparagraph*{Eigenschaften}
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\begin{enumerate}
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\item Orthogonalität
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\begin{equation}
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\intgr{-\infty}{+\infty}{H_n(x) H_m(x) e^{(-x^2)}}{x} = \sqrt{\pi} 2^n n!
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\end{equation}
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denn:
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\begin{equation}
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\intgr{-\infty}{+\infty}{\phi_n^*(x)\phi_m(x)}{x} = \krondelta{n,m}
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\end{equation}
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\item Vollständigkeit
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\begin{equation}
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\sum_{n = 0}^{\infty} \phi_n(x)\phi_n(x') = \delta(x - x')
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\end{equation}
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\item DGL:
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\begin{equation}
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\left( \diffPs{x}^2 - 2x \diffPs{x} + 2n \right) H_n(x) = 0
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\end{equation}
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\item Erzeugende Funktion
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\begin{equation}
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\sum_{n = 0}^{\infty} \frac{t^n}{n!} H_n(x) = e^{-t^2 + 2 t x}
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\end{equation}
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\end{enumerate}
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\section{Spektrum von $H$ aus der DGL}
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Die stationäre Schrödingergleichung ist wiefolgt:
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\begin{equation}
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\left( \frac{-\hbar^2}{2m} \diffPs{X}^2 + \frac{m}{2} X^2 \right) \Phi(X) = E \Phi(X)
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\end{equation}
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mit
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\begin{equation}
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x = \frac{X}{X_0}; ~ p = P \cdot X_0; ~ X_0 = \sqrt{\frac{\hbar}{m \omega_0}}
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\end{equation}
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Also ergibt sich:
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\begin{equation}
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\rightarrow \left( \right) \phi(x) = \varepsilon \phi(x)
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\end{equation}
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Wir suchen normierbare Lösungen $\left(\intgr{-\infty}{+\infty}{\phi^2(x)}{x} < \infty\right)$ für $x \rightarrow \pm \infty$. Wir verwenden den Ansatz
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\begin{equation}
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\phi(x) \tilde e^{-\alpha x^m}
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\end{equation}
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und erhalten
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\begin{equation}
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-\frac{1}{2} \left( (-\alpha m) (-\alpha (m - 1)) \alpha x^{m - 2} + \alpha^2 m^2 x^{2(m - 1)}\right) + \frac{1}{2}x^2 = 0
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\end{equation}
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für $x \rightarrow \infty$:
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\begin{equation}
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\phi(x) \rightarrow e^{-\frac{x^2}{2}}
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\end{equation}
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neuer Ansatz:
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\begin{equation}
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\phi(x) = e^{-\frac{x^2}{2}} \cdot u(x)
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\end{equation}
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eingesetzt in die statische Schrödingergleichung:
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\begin{align}
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-\frac{1}{2} \diffPs{x}^2(u) + \diffPs{x}(u) \cdot x + \frac{1}{2} u x^2 &= \varepsilon u ~ \text{(exakt)}\\
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\diffPs{x}^2(u) - 2x \diffPs{x}(u) + (2\varepsilon - 1) u &= 0
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\end{align}
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mit dem Ansatz
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\begin{equation}
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u(x) = \sum_{n = 0}^\infty b_n x^n
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\end{equation}
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ergibt sich
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\begin{align}
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\sum_{n=2}^\infty b_n n(n-1) x^{n-2} - 2x\sum_{n=1}^\infty b_n n x^{n-1} + (2 \varepsilon - 1) \sum_{n = 0}^\infty b^n x^n &= 0\\
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\sum_{n=0}^\infty b_{n+2} (n+2)(n+1) x^n + \sum_{n=0}^\infty b_n \left[ (2\varepsilon - 1) - 2n \right] x^n &= 0
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\end{align}
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und damit
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\begin{equation}
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b_{n+2} = \frac{2n - (2\varepsilon - 1)}{(n+2)(n+1)} b_n
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\end{equation}
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Scheinbar lässt sich für alle $\varepsilon$ eine Lösung für gegebene $b_0, b_1$ finden.\\
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\underline{Aber:} Die Lösung muss normierbar sein.
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\begin{equation}
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\frac{b_{n+2}}{b_n} \rightarrow \frac{2}{n} ~ \text{für} ~ n \rightarrow \infty
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\end{equation}
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Mit
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\begin{equation}
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e^{\left( x^2 \right)} = \sum_k \frac{1}{k!} x^{2k} = \sum_n \frac{1}{\left(\frac{n}{2}\right)!} x^n
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\end{equation}
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für $b_n$ erhält man
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\begin{equation}
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\frac{b_{n+2}}{b_n} = \frac{\left(\frac{n}{2}\right)!}{\left(\frac{n+2}{2}\right)!} = \frac{2}{n+2} \rightarrow \frac{2}{n}
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\end{equation}
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\underline{Also:} Rekursion muss abbrechen, d.h. $b_{\tilde{n}} = 0$ für irgendein $\tilde{n}$.
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\begin{align}
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2n - (2\varepsilon - 1) &= 0\\
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\rightarrow \varepsilon &= n + \frac{1}{2}
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\end{align}
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(Quantisierung und Eigenfunktionen wir vorhin!)
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