210 lines
9.8 KiB
TeX
210 lines
9.8 KiB
TeX
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\chapter{Zeitabhängige Störungstheorie}
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Für $H = H_0 + \lambda_1$ mit $H_0$ exakt lösbar und $\lambda \ll 1$, lässt sich das Problem perturativ angeben.
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\section{Nichtentarteter Fall}
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gegeben:
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\begin{equation}
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\left(H_0 - E_\alpha \right) \ket{\alpha} = 0 ~ \left[ H_0 \ket{\alpha} = E_\alpha \ket{\alpha} \right]
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\end{equation}
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suche:
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\begin{equation}
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\left( H - E_a \right) \ket{a} = 0 \label{stern00}
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\end{equation}
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mit $\ket{a} \rightarrow \ket{\alpha}$ (für $x \rightarrow 0$) eindeutig, da nicht entartet.
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\paragraph{Strategie} Wir entwickeln nach $\lambda$
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\begin{equation}
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\ket{a} = c_\alpha \ket{\alpha} + \sum_{\beta \neq \alpha} d_\beta \ket{\beta} \text{ und } d_\beta = O(\lambda)
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\end{equation}
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Norm:
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\begin{align}
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\abs{c_\alpha}^2 + \sum_{\beta \neq \alpha} \abs{d_\beta}^2 &= 1\\
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\rightarrow c_\alpha &= 1 - O(\lambda^2)
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\end{align}
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einsetzen in (\ref{stern00}):
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\begin{align}
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0 &= \left( H - E_\alpha \right) \ket{\alpha}\\
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0 &= \left( H_0 - \lambda H_1 - E_\alpha \right) \left( c_\alpha \ket{\alpha} + \sum_{\beta \neq \alpha} d_\beta \ket{\beta} \right) &\left| \bra{\gamma} ~ \gamma \neq \alpha \right.\\
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&= c_\alpha \lambda \dirac{\gamma}{H_1}{\alpha} + \sum_{\beta \neq \alpha} d_\beta \left( E_\beta \krondelta{\beta,\gamma} + \lambda \dirac{\gamma}{H_1}{\beta} - E_a \krondelta{\beta,\gamma} \right)\\
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0 &= \underbrace{c_\alpha}_{1-O(\lambda^2)} \lambda \dirac{\gamma}{H_1}{\alpha} + d_\gamma \left( E_\gamma - E_a \right) + \underbrace{\lambda \sum_{\beta \neq \alpha} d_\beta \dirac{\gamma}{H_1}{\beta}}_{O(\lambda^2)}
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\end{align}
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in niedrigster Ordnung erhält man:
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\begin{align}
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\rightarrow d_\gamma &= \lambda \frac{\dirac{\gamma}{H_1}{\alpha}}{\underbrace{E_a}_{E_\alpha - O(\lambda)} - E_\gamma} + O(\lambda^2)\\
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\rightarrow d_\beta &= \lambda \frac{\dirac{\beta}{H_1}{\alpha}}{E_\alpha - E_\beta}\\
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\ket{a} = \ket{\alpha} + \lambda \sum_{\beta \neq \alpha} \frac{\dirac{\beta}{H_1}{\alpha}}{E_\alpha - E_\beta} \ket{\beta} + O(\lambda^2)
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\end{align}
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gut falls:
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\begin{equation}
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\lambda \frac{\dirac{\beta}{H_1}{\alpha}}{E_\alpha - E_\beta} \ll 1 ~ \forall \beta \neq \alpha
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\end{equation}
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Energieverschiebung:
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\begin{align}
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0 &= \left( H - E_\alpha \right) \ket{a} &\left| \ket{\alpha} \right.\\
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0 &= \bra{\alpha} \left( H_0 + \lambda_1 H_1 - E_a \right) \left( \ket{\alpha} + \lambda \sum_{\beta \neq \alpha} \frac{\dirac{\beta}{H_1}{\alpha}}{E_\alpha - E_\beta} \ket{\beta} + O(\lambda^2) \right)\\
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E_a &= E_ \alpha + \lambda \dirac{\alpha}{H_1}{\alpha} + \lambda^2 \sum_{\beta \neq \alpha} \frac{\dirac{\alpha}{H_1}{\beta} \dirac{\beta}{H_1}{\alpha}}{E_\alpha - E_\beta} + O(\lambda^3)
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\end{align}
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\paragraph{Fazit}
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\begin{enumerate}
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\item In der führenden Ordnung ist die Energieverschiebung das Matrixelement der Störung im ungestörten Zustand.
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\item Falls $\dirac{\alpha}{H_1}{\alpha}$ aus Symmetriegründen verschwindet, dann tragen alle Zustände zu nichtverschwindenden Korrekturen bei!
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\end{enumerate}
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\paragraph{Beispiel}
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\begin{align}
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H &= \frac{p^2}{2m} + \frac{m}{2} \omega^2 x^2 + \lambda \underbrace{x^4}_{H_1}\\
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H_0 \ket{n} &= \hbar \omega \left( n + \frac{1}{2} \right) \ket{n}
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\end{align}
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Grundzustandsverschiebung
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\begin{equation}
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E_0 = \hbar \omega \frac{1}{2} + \lambda \dirac{0}{x^4}{0}
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\end{equation}
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entsprechend
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\begin{equation}
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E_0 = \hbar \omega \left( n + \frac{1}{2} \right) + \lambda \dirac{n}{x^4}{n}
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\end{equation}
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%\begin{figure}[H] \centering
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%\includegraphics{pdf/III/02-01-00.pdf}
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%\end{figure}
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Konsequenz?
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\begin{itemize}
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\item für $\lambda$ negativ $\abs{\lambda} \ll 1$
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%\begin{figure}[H] \centering
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%\includegraphics{pdf/III/02-01-01.pdf}
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%\caption{gestrichelte Kurve entspricht $\frac{m}{2} \omega x^2 \abs{\lambda} x^4$}
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%\end{figure}
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\item für $\lambda$ positiv
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%\begin{figure}[H] \centering
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%\includegraphics{pdf/III/02-01-02.pdf}
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%\end{figure}
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\end{itemize}
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volle Rechnung zeigt:
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\begin{equation}
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E_n(\lambda) = \hbar \omega \left( n + \frac{1}{2} \right) + A^{(1)}_n (\lambda) + A^{(2)}_n (\lambda)
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\end{equation}
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mit
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\begin{equation}
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A^{(1)}_n (\lambda) = O(\lambda) ~ ; ~ A^{(2)}_n (\lambda) = O \left(e^{-\frac{1}{\lambda}} \right)
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\end{equation}
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\section{Entarteter Fall}
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\begin{itemize}
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\item Obige Formel wegen Energienenner nicht anwendbar bei Entartung.
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\item sehr relevant: Aufhebeung der Entartung durch Störung
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\end{itemize}
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%\begin{figure}[H] \centering
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%\includegraphics{pdf/III/02-02-00.pdf}
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%\end{figure}
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\paragraph{Ansatz}
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\begin{align}
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\ket{a} &= \sum_{\alpha \in D} c_\alpha \ket{\alpha} + \sum_{\mu \notin D} d_\mu \ket{\mu}\\[10pt]
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0 &= \left( H - E_a \right) \ket{a}\\
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\rightarrow 0 &= \left( H_0 + \lambda H_1 - E_a \right) \left( \sum_{\mu \notin D} d_\mu \ket{\mu} \right)
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\end{align}
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\paragraph*{Projektion auf $\bra{\nu} \notin D$}
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\begin{align}
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0 &= \lambda \sum_\alpha c_\alpha \dirac{\nu}{H_1}{\alpha} d_\nu \left( E_\nu - E_a \right) + \lambda \sum_{\mu \notin D} d_\mu \dirac{\nu}{H_1}{\mu}\\
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\rightarrow d_\nu &= \lambda \frac{\sum_\alpha c_a \dirac{\nu}{H_1}{\alpha}}{\underbrace{E_a}_{E_\alpha} - E_\nu} + o(\lambda^2)
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\end{align}
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\paragraph*{Projektion auf $\bra{\beta} \in D$}
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\begin{align}
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0 &= \sum_\alpha c_\alpha \left( E_\alpha - E_a \right) \krondelta{\alpha,\beta} + \lambda \sum_\alpha c_\alpha \dirac{\beta}{H_1}{\alpha} + \lambda \sum_\mu d_\mu \dirac{\beta}{H_1}{\alpha}\\
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0 &= \sum_\alpha c_\alpha \left( \left( E_a - E_\alpha \right) \krondelta{\alpha,\beta} + \underbrace{\lambda \dirac{\beta}{H_1}{\alpha} + \lambda^2 \sum_{\mu \notin D} \frac{\dirac{\beta}{H_1}{\mu} \dirac{\mu}{H_1}{\alpha}}{E_\alpha E_\mu}}_{\equiv \dirac{\beta}{H_\text{eff}}{\alpha}} \right) + O(\lambda^3)
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\end{align}
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$\forall \beta = 1, ..., N$ mit $N$ die Dimension von $D$\\
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d.h. LGS für die $c_\alpha$ hat nichttriviale Lösung falls
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\begin{equation}
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\detb{E_\alpha \krondelta{\alpha,\beta} + \dirac{\beta}{H_\text{eff}}{\alpha} - E_\alpha \krondelta{\alpha,\beta}} = 0
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\end{equation}
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d.h. wir müssen $H_0 + H_\text{eff}$ im Unterraum $D$ diagonalisieren
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\paragraph{Fazit}
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\begin{enumerate}
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\item in Ordnung $\lambda$ reicht es $H_1$ im entarteten Unterraum zu diagonalisieren
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\item falls $H_1$ die Entartung nicht aufhebt, muss der $\lambda^2$-Term mitgenommen werden
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\end{enumerate}
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\section{Beispiel: Stark-Effekt}
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H-Atom:
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\begin{equation}
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H_0 \ket{n,l,m} = E_n \ket{n,l,m}
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\end{equation}
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mit
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\begin{equation}
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E_n = \frac{l^2}{2 a_0} \frac{1}{n^2} \text{; Entartung } g(n) = n^2
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\end{equation}
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Störung: $H_1$ sei E'feld
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\begin{equation}
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H_1 = e \abs{E} \hat{z} \text{ für } \lambda = 1
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\end{equation}
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\paragraph{Erster Grundzustand}
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\begin{equation}
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\ket{1,0,0} \rightarrow \braket{r,\theta,\varphi}{1,0,0} = \Phi_{1,0,0}(\vec{r}) = \frac{U_{1,0}}{r}Y_{0,0} = \frac{1}{\sqrt{\pi} a_0^\frac{3}{2}}e^{-\frac{r}{a_0}}
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\end{equation}
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\begin{itemize}
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\item in Ordnung $\lambda$ d.h. in $O(\abs{E})$:
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% \begin{equation}
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% \dirac{1,0,0}{H_1}{1,0,0} = \intgru{\Phi^*_{1,0,0} (\vec{r}) \cdor e \abs{E} z \Phi_{1,0,0}(\vec{r})}{r^3} = 0
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% \end{equation}
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\item in Ordnung $\abs{E}^2$:
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\begin{align}
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E_a = E_\alpha + ... &= E_1 + \sum_{\beta,\alpha} \frac{\abs{\dirac{\beta}{H_1}{\alpha}}^2}{E_\beta - E_\alpha}\\
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&= \left( E_1 \sum_{n=2}^\infty \sum_{l=0}^{n-1} \sum_{m=-l}^{+l} \frac{\abs{\dirac{n,l,m}{H_1}{1,0,0}}^2}{E_1 - E_n} + \int \text{Kontinuum} \right) + O(E^3)
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\end{align}
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mit
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\begin{align}
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\dirac{n,l,m}{\hat{z}}{1,0,0} &= \intgr{0}{-\infty}{r \intgr{-1}{+1}{\intgr{0}{2\pi}{\frac{U_{n,r}(r)}{r} Y^*_{l,m}(\theta,\varphi) r \underbrace{\cos\theta}_{Y_{1,0} \frac{1}{\sqrt{\pi} a_0^\frac{3}{2}}e^{-\frac{r}{a_0}}}}{\varphi}}{(\cos\theta)}}{r}\\
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&= \krondelta{m,0} \krondelta{l,1} \frac{1}{\sqrt{3}} \int U_{n,1}(r) r U_{1,0}(r)
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\end{align}
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\end{itemize}
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am meisten trägt $n=2$ bei
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\begin{align}
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\dirac{2,1,0}{\hat{z}}{1,0,0} &= \frac{a_0}{\sqrt{3}} \intgr{0}{\infty}{\frac{r^2 e^{-\frac{r}{2}}}{2 \sqrt{6}} r \left( 2r e^{-r} \right)}{r}\\
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&= \frac{2^7 \sqrt{2}}{3^5} a_0
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\end{align}
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und damit ist
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\begin{equation}
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E_{1,0,0} = E_1 - 1,48 a_0^3 \abs{E}^2
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\end{equation}
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und für alle weiteren $n$
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\begin{equation}
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E_{1,0,0} \rightarrow E_1 - \frac{9}{4} a_0^3 \abs{E}^2
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\end{equation}
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\paragraph{linerarer Stark-Effekt für n=2}
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\begin{align}
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\braket{r,\theta,\varphi}{2,0,0} &= \Phi_{2,0,0}(\vec{r}) = \frac{1}{\sqrt{2 a_0^3}} \left(1 - \frac{r}{2a_0} \right) e^{-\frac{r}{2a_0}} Y_{0,0}\\
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\braket{r,\theta,\varphi}{2,1,m} &= \Phi_{2,1,m} (\vec{r}) = \frac{1}{24 a_0^3} \frac{r}{a_0} e^{-\frac{r}{2a_0}} Y_{l,m}
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\end{align}
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im entartetetn Unterraum $\setCond{\bra{\alpha}}{\alpha = 1,2,3,4}$ mit
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\begin{equation}
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\bra{1} = \bra{1,0,0} ;~ \bra{2} = \bra{2,1,0} ;~ \bra{3} = \bra{2,1,1} ;~ \bra{4} = \bra{2,1,-1}
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\end{equation}
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diagonalisieren: $\dirac{\beta}{H_1}{\alpha}$
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\begin{enumerate}
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\item Diagonalelemente
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\begin{equation}
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\dirac{\alpha}{H_1}{\alpha} \tilde \intgru{\cos\theta\abs{\Phi_{n,l,m}}^2}{(\cos\theta)} \equiv 0
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\end{equation}
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\item Nichtdiagonalelemente
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\begin{equation}
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\dirac{n',l',m'}{\hat{z}}{n,l,m} \tilde \intgru{e^{i(m-m')\phi}}{\phi} \tilde \krondelta{m,m'}
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\end{equation}
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mit
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\begin{equation}
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\dirac{1}{H_1}{2} = \dirac{2}{H_1}{1} = \dirac{2,0,0}{e \abs{E} z}{2,1,0} \equiv V_0
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\end{equation}
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\end{enumerate}
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d.h. das Eigenwertproblem lautet
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\begin{equation}
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\inlinematrix{0& V_0& 0& 0\\ V_0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0} \inlinematrix{c_1\\ c_2\\ c_3\\ c_4} = \left( E_a - E_\alpha \right) \inlinematrix{c_1\\ c_2\\ c_3\\ c_4}
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\end{equation}
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triviale Eigenvektoren $\inlinematrix{0\\ 0\\ 1\\ 0}$ und $\inlinematrix{0\\ 0\\ 0\\ 1}$ zum Eigenwert $E_a = E_\alpha$ reduziert auf
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\begin{equation}
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\inlinematrix{0& V_0\\ V_0& 0} \inlinematrix{c_1\\ c_2} = \left( E_a - E_\alpha \right) \inlinematrix{c_1\\ c_2}
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\end{equation}
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also
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\begin{align}
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\frac{1}{\sqrt{2}} \inlinematrix{1\\ 1\\ 0\\ 0} &\text{ zu } E_a - E_\alpha = +V_0\\
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\frac{1}{\sqrt{2}} \inlinematrix{1\\ -1\\ 0\\ 0} &\text{ zu } E_a - E_\alpha = -V_0\\
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\end{align}
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mit
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\begin{equation}
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V_0 = -3 e \abs{E} a_0
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\end{equation}
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