diff --git a/kapI-2.tex b/kapI-2.tex index 1d7305d..daa82b7 100644 --- a/kapI-2.tex +++ b/kapI-2.tex @@ -298,10 +298,10 @@ $\rightarrow$ $B \ket{n,r}$ liegt im Untrraum zu $a_n$ \end{equation} \item[Fall (2)] $a_n$ entartet \begin{equation} - \bra{m,s} \cdot B \cdot \ket{n,r} = B_{s,r}^{(n)} + \bra{n,s} \cdot B \cdot \ket{n,r} = B_{s,r}^{(n)} \end{equation} \begin{equation} - B \cequiv \inlinematrix{\boxed{B^{(1)}} & & 0 \\ & \boxed{B^{(2)}} & & \\ & & \boxed{B} & \\ 0 & & & \boxed{B^{(4)}}} \rightarrow \text{kann diagonalisiert werden in jedem Kästchen} + B \cequiv \inlinematrix{\boxed{B^{(1)}} & & 0 \\ & \boxed{B^{(2)}} & & \\ & & \boxed{B^{(3)}} & \\ 0 & & & \boxed{B^{(4)}}} \rightarrow \text{kann diagonalisiert werden in jedem Kästchen} \end{equation} Falls $B^{(n)}$ entartet, gibt es einen dritten Opertor $C$ mit $[A,C] = [B,C] = 0$. \end{description} diff --git a/kapIV-2.tex b/kapIV-2.tex new file mode 100644 index 0000000..03d4c16 --- /dev/null +++ b/kapIV-2.tex @@ -0,0 +1,209 @@ +\chapter{Zeitabhängige Störungstheorie} +Für $H = H_0 + \lambda_1$ mit $H_0$ exakt lösbar und $\lambda \ll 1$, lässt sich das Problem perturativ angeben. +\section{Nichtentarteter Fall} +gegeben: +\begin{equation} + \left(H_0 - E_\alpha \right) \ket{\alpha} = 0 ~ \left[ H_0 \ket{\alpha} = E_\alpha \ket{\alpha} \right] +\end{equation} +suche: +\begin{equation} + \left( H - E_a \right) \ket{a} = 0 \label{stern00} +\end{equation} +mit $\ket{a} \rightarrow \ket{\alpha}$ (für $x \rightarrow 0$) eindeutig, da nicht entartet. +\paragraph{Strategie} Wir entwickeln nach $\lambda$ +\begin{equation} + \ket{a} = c_\alpha \ket{\alpha} + \sum_{\beta \neq \alpha} d_\beta \ket{\beta} \text{ und } d_\beta = O(\lambda) +\end{equation} +Norm: +\begin{align} + \abs{c_\alpha}^2 + \sum_{\beta \neq \alpha} \abs{d_\beta}^2 &= 1\\ + \rightarrow c_\alpha &= 1 - O(\lambda^2) +\end{align} +einsetzen in (\ref{stern00}): +\begin{align} + 0 &= \left( H - E_\alpha \right) \ket{\alpha}\\ + 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.\\ + &= 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)\\ + 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)} +\end{align} +in niedrigster Ordnung erhält man: +\begin{align} + \rightarrow d_\gamma &= \lambda \frac{\dirac{\gamma}{H_1}{\alpha}}{\underbrace{E_a}_{E_\alpha - O(\lambda)} - E_\gamma} + O(\lambda^2)\\ + \rightarrow d_\beta &= \lambda \frac{\dirac{\beta}{H_1}{\alpha}}{E_\alpha - E_\beta}\\ + \ket{a} = \ket{\alpha} + \lambda \sum_{\beta \neq \alpha} \frac{\dirac{\beta}{H_1}{\alpha}}{E_\alpha - E_\beta} \ket{\beta} + O(\lambda^2) +\end{align} +gut falls: +\begin{equation} + \lambda \frac{\dirac{\beta}{H_1}{\alpha}}{E_\alpha - E_\beta} \ll 1 ~ \forall \beta \neq \alpha +\end{equation} +Energieverschiebung: +\begin{align} + 0 &= \left( H - E_\alpha \right) \ket{a} &\left| \ket{\alpha} \right.\\ + 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)\\ + 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) +\end{align} +\paragraph{Fazit} +\begin{enumerate} + \item In der führenden Ordnung ist die Energieverschiebung das Matrixelement der Störung im ungestörten Zustand. + \item Falls $\dirac{\alpha}{H_1}{\alpha}$ aus Symmetriegründen verschwindet, dann tragen alle Zustände zu nichtverschwindenden Korrekturen bei! +\end{enumerate} +\paragraph{Beispiel} +\begin{align} + H &= \frac{p^2}{2m} + \frac{m}{2} \omega^2 x^2 + \lambda \underbrace{x^4}_{H_1}\\ + H_0 \ket{n} &= \hbar \omega \left( n + \frac{1}{2} \right) \ket{n} +\end{align} +Grundzustandsverschiebung +\begin{equation} + E_0 = \hbar \omega \frac{1}{2} + \lambda \dirac{0}{x^4}{0} +\end{equation} +entsprechend +\begin{equation} + E_0 = \hbar \omega \left( n + \frac{1}{2} \right) + \lambda \dirac{n}{x^4}{n} +\end{equation} +%\begin{figure}[H] \centering +%\includegraphics{pdf/III/02-01-00.pdf} +%\end{figure} +Konsequenz? +\begin{itemize} + \item für $\lambda$ negativ $\abs{\lambda} \ll 1$ + %\begin{figure}[H] \centering + %\includegraphics{pdf/III/02-01-01.pdf} + %\caption{gestrichelte Kurve entspricht $\frac{m}{2} \omega x^2 \abs{\lambda} x^4$} + %\end{figure} + \item für $\lambda$ positiv + %\begin{figure}[H] \centering + %\includegraphics{pdf/III/02-01-02.pdf} + %\end{figure} +\end{itemize} +volle Rechnung zeigt: +\begin{equation} + E_n(\lambda) = \hbar \omega \left( n + \frac{1}{2} \right) + A^{(1)}_n (\lambda) + A^{(2)}_n (\lambda) +\end{equation} +mit +\begin{equation} + A^{(1)}_n (\lambda) = O(\lambda) ~ ; ~ A^{(2)}_n (\lambda) = O \left(e^{-\frac{1}{\lambda}} \right) +\end{equation} + +\section{Entarteter Fall} +\begin{itemize} + \item Obige Formel wegen Energienenner nicht anwendbar bei Entartung. + \item sehr relevant: Aufhebeung der Entartung durch Störung +\end{itemize} +%\begin{figure}[H] \centering +%\includegraphics{pdf/III/02-02-00.pdf} +%\end{figure} +\paragraph{Ansatz} +\begin{align} + \ket{a} &= \sum_{\alpha \in D} c_\alpha \ket{\alpha} + \sum_{\mu \notin D} d_\mu \ket{\mu}\\[10pt] + 0 &= \left( H - E_a \right) \ket{a}\\ + \rightarrow 0 &= \left( H_0 + \lambda H_1 - E_a \right) \left( \sum_{\mu \notin D} d_\mu \ket{\mu} \right) +\end{align} +\paragraph*{Projektion auf $\bra{\nu} \notin D$} +\begin{align} + 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}\\ + \rightarrow d_\nu &= \lambda \frac{\sum_\alpha c_a \dirac{\nu}{H_1}{\alpha}}{\underbrace{E_a}_{E_\alpha} - E_\nu} + o(\lambda^2) +\end{align} +\paragraph*{Projektion auf $\bra{\beta} \in D$} +\begin{align} + 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}\\ + 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) +\end{align} +$\forall \beta = 1, ..., N$ mit $N$ die Dimension von $D$\\ +d.h. LGS für die $c_\alpha$ hat nichttriviale Lösung falls +\begin{equation} + \detb{E_\alpha \krondelta{\alpha,\beta} + \dirac{\beta}{H_\text{eff}}{\alpha} - E_\alpha \krondelta{\alpha,\beta}} = 0 +\end{equation} +d.h. wir müssen $H_0 + H_\text{eff}$ im Unterraum $D$ diagonalisieren +\paragraph{Fazit} +\begin{enumerate} + \item in Ordnung $\lambda$ reicht es $H_1$ im entarteten Unterraum zu diagonalisieren + \item falls $H_1$ die Entartung nicht aufhebt, muss der $\lambda^2$-Term mitgenommen werden +\end{enumerate} + +\section{Beispiel: Stark-Effekt} +H-Atom: +\begin{equation} + H_0 \ket{n,l,m} = E_n \ket{n,l,m} +\end{equation} +mit +\begin{equation} + E_n = \frac{l^2}{2 a_0} \frac{1}{n^2} \text{; Entartung } g(n) = n^2 +\end{equation} +Störung: $H_1$ sei E'feld +\begin{equation} + H_1 = e \abs{E} \hat{z} \text{ für } \lambda = 1 +\end{equation} +\paragraph{Erster Grundzustand} +\begin{equation} + \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}} +\end{equation} +\begin{itemize} + \item in Ordnung $\lambda$ d.h. in $O(\abs{E})$: +% \begin{equation} +% \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 +% \end{equation} + \item in Ordnung $\abs{E}^2$: + \begin{align} + E_a = E_\alpha + ... &= E_1 + \sum_{\beta,\alpha} \frac{\abs{\dirac{\beta}{H_1}{\alpha}}^2}{E_\beta - E_\alpha}\\ + &= \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) + \end{align} + mit + \begin{align} + \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}\\ + &= \krondelta{m,0} \krondelta{l,1} \frac{1}{\sqrt{3}} \int U_{n,1}(r) r U_{1,0}(r) + \end{align} +\end{itemize} +am meisten trägt $n=2$ bei +\begin{align} + \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}\\ + &= \frac{2^7 \sqrt{2}}{3^5} a_0 +\end{align} +und damit ist +\begin{equation} + E_{1,0,0} = E_1 - 1,48 a_0^3 \abs{E}^2 +\end{equation} +und für alle weiteren $n$ +\begin{equation} + E_{1,0,0} \rightarrow E_1 - \frac{9}{4} a_0^3 \abs{E}^2 +\end{equation} +\paragraph{linerarer Stark-Effekt für n=2} +\begin{align} + \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}\\ + \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} +\end{align} +im entartetetn Unterraum $\setCond{\bra{\alpha}}{\alpha = 1,2,3,4}$ mit +\begin{equation} + \bra{1} = \bra{1,0,0} ;~ \bra{2} = \bra{2,1,0} ;~ \bra{3} = \bra{2,1,1} ;~ \bra{4} = \bra{2,1,-1} +\end{equation} +diagonalisieren: $\dirac{\beta}{H_1}{\alpha}$ +\begin{enumerate} + \item Diagonalelemente + \begin{equation} + \dirac{\alpha}{H_1}{\alpha} \tilde \intgru{\cos\theta\abs{\Phi_{n,l,m}}^2}{(\cos\theta)} \equiv 0 + \end{equation} + \item Nichtdiagonalelemente + \begin{equation} + \dirac{n',l',m'}{\hat{z}}{n,l,m} \tilde \intgru{e^{i(m-m')\phi}}{\phi} \tilde \krondelta{m,m'} + \end{equation} + mit + \begin{equation} + \dirac{1}{H_1}{2} = \dirac{2}{H_1}{1} = \dirac{2,0,0}{e \abs{E} z}{2,1,0} \equiv V_0 + \end{equation} +\end{enumerate} +d.h. das Eigenwertproblem lautet +\begin{equation} + \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} +\end{equation} +triviale Eigenvektoren $\inlinematrix{0\\ 0\\ 1\\ 0}$ und $\inlinematrix{0\\ 0\\ 0\\ 1}$ zum Eigenwert $E_a = E_\alpha$ reduziert auf +\begin{equation} + \inlinematrix{0& V_0\\ V_0& 0} \inlinematrix{c_1\\ c_2} = \left( E_a - E_\alpha \right) \inlinematrix{c_1\\ c_2} +\end{equation} +also +\begin{align} + \frac{1}{\sqrt{2}} \inlinematrix{1\\ 1\\ 0\\ 0} &\text{ zu } E_a - E_\alpha = +V_0\\ + \frac{1}{\sqrt{2}} \inlinematrix{1\\ -1\\ 0\\ 0} &\text{ zu } E_a - E_\alpha = -V_0\\ +\end{align} +mit +\begin{equation} + V_0 = -3 e \abs{E} a_0 +\end{equation} diff --git a/theo2.kilepr b/theo2.kilepr index 7e2b8f3..6ae912c 100644 --- a/theo2.kilepr +++ b/theo2.kilepr @@ -1,13 +1,14 @@ [General] +def_graphic_ext= img_extIsRegExp=false img_extensions=.eps .jpg .jpeg .png .pdf .ps .fig .gif kileprversion=2 -kileversion=2.0 -lastDocument=kapIV-1.tex +kileversion=2.0.85 +lastDocument=theo2.tex masterDocument= name=Theo2 pkg_extIsRegExp=false -pkg_extensions=.cls .sty +pkg_extensions=.cls .sty .bbx .cbx .lbx src_extIsRegExp=false src_extensions=.tex .ltx .latex .dtx .ins @@ -15,12 +16,37 @@ src_extensions=.tex .ltx .latex .dtx .ins MakeIndex= QuickBuild=PDFLaTeX+ViewPDF +[document-settings,item:kapIV-1.tex] +Bookmarks= +Encoding=UTF-8 +Highlighting=LaTeX +Indentation Mode=normal +Mode=LaTeX +ReadWrite=true + +[document-settings,item:kapIV-2.tex] +Bookmarks= +Encoding=UTF-8 +Highlighting=LaTeX +Indentation Mode=normal +Mode=LaTeX +ReadWrite=true + +[document-settings,item:theo2.tex] +Bookmarks= +Encoding=UTF-8 +Highlighting=LaTeX +Indentation Mode=normal +Mode=LaTeX +ReadWrite=true + [item:formelsammlung.tex] archive=true column=115 encoding=UTF-8 highlight=LaTeX line=42 +mode= open=false order=6 @@ -30,6 +56,7 @@ column=17 encoding=UTF-8 highlight=LaTeX line=103 +mode= open=false order=3 @@ -39,6 +66,7 @@ column=13 encoding=UTF-8 highlight=LaTeX line=163 +mode= open=false order=4 @@ -48,6 +76,7 @@ column=0 encoding= highlight=LaTeX line=0 +mode= open=false order=-1 @@ -57,6 +86,7 @@ column=6 encoding=UTF-8 highlight=LaTeX line=192 +mode= open=false order=5 @@ -66,6 +96,7 @@ column=7 encoding=UTF-8 highlight=LaTeX line=148 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\part{Übungsmitschrieb} -% \label{UE} -% \include{ueb1} -% \include{ueb2} -% \include{ueb3} -% \include{ueb4} -% \include{ueb5} -% \include{ueb6} -% \include{ueb7} -% \include{ueb8} -% \include{ueb9} -% \include{ueb10} -% \include{ueb11} +\part{Übungsmitschrieb} +\label{UE} +\include{ueb1} +\include{ueb2} +\include{ueb3} +\include{ueb4} +\include{ueb5} +\include{ueb6} +\include{ueb7} +\include{ueb8} +\include{ueb9} +\include{ueb10} +\include{ueb11} \part{Formelsammlung} \label{FS}