Merge branch 'master' of git+ssh://git@git.eluhost.de:522/qm1-script
This commit is contained in:
Oliver Groß
commit
d0db85c772
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\chapter{Lineare Algebra}
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||||
\section{Identitäten}
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||||
\begin{math}
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A^{-1} = \inlinematrix{a & b \\ c & d}^{-1} = \frac{1}{ad - bc} \inlinematrix{d & -b \\ -c & a}
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\end{math}
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math.tex
8
math.tex
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@ -7,6 +7,7 @@
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\newcommand{\setQ}{\mathbbm{Q}}
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||||
\newcommand{\setR}{\mathbbm{R}}
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||||
\newcommand{\setC}{\mathbbm{C}}
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\newcommand{\einsmatrix}{\mathbbm{1}}
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||||
\newcommand{\inlinematrix}[1]{\begin{pmatrix} #1 \end{pmatrix}}
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||||
\newcommand{\inlinematrixu}[1]{\begin{matrix} #1 \end{matrix}}
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||||
\newcommand{\inlinematrixd}[2]{\left( \begin{matrix} #1\vphantom{#2} \end{matrix} ~\right| \left. \begin{matrix} \vphantom{#1}#2 \end{matrix} \right)}
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@ -28,4 +29,9 @@
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\newcommand{\diffT}[1]{\frac{\text{d}}{\text{d} #1}}
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||||
\newcommand{\diffPfrac}[2]{\frac{\partial #1}{\partial #2}}
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||||
\newcommand{\intgr}[4]{\int_{#1}^{#2} #3 ~\text{d}#4}
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||||
\newcommand{\intgru}[2]{\int #1 ~\text{d}#2}
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||||
\newcommand{\intgru}[2]{\int #1 ~\text{d}#2}
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||||
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||||
\newcommand{\bbracket}[1]{\left #1 \right}
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||||
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||||
\newcommand{\levicivita}[1]{\varEpsilon_{#1}}
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||||
\newcommand{\krondelta}[1]{\delta_{+1}}
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62
theo2.kilepr
62
theo2.kilepr
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@ -3,7 +3,7 @@ img_extIsRegExp=false
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img_extensions=.eps .jpg .jpeg .png .pdf .ps .fig .gif
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@ -15,14 +15,23 @@ src_extensions=.tex .ltx .latex .dtx .ins
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MakeIndex=
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QuickBuild=PDFLaTeX+ViewPDF
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@ -71,11 +80,11 @@ order=-1
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[item:kapII-1.tex]
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@ -107,21 +116,21 @@ order=-1
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@ -134,10 +143,10 @@ order=-1
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@ -189,8 +198,17 @@ order=-1
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[item:ueb6.tex]
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@ -8,7 +8,7 @@
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{multirow}
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\usepackage[pdfborder={0 0 0}]{hyperref}
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\usepackage[pdfborder={0 0 0}]{hyperref} % muss immer als letztes eingebunden werden
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\include{math}
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\include{physics}
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@ -45,6 +45,11 @@
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% \include{ueb4}
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% \include{ueb5}
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% \include{ueb6}
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% \inlcude{ueb7}
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% \part{Formelsammlung}
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% \label{FS}
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% \include{formelsammlung}
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\end{document}
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110
ueb6.tex
110
ueb6.tex
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@ -4,10 +4,116 @@
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\chapter{Quantenmechanik I - Übungsblatt 6}
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\section{Aufgabe 14: Spin-1-Teilchen}
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\subsection*{a)}
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\begin{align}
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H &= A S_z + B S_x^2 \\
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H &= A \hbar \Sigma_z + B \hbar^2 \Sigma_x^2 \\
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H &= A \hbar \inlinematrix{1 &0 &0 \\ 0 &0 &0 \\ 0 &0 &-1} + \frac{B \hbar^2}{2} \inlinematrix{1 &0 &1 \\ 0 &2 &0 \\ 1 &0 &1} \\
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H &= \hbar \inlinematrix{A+\frac{B \hbar}{2} &0 &\frac{B \hbar}{2} \\ 0 &\frac{B \hbar}{2} &0 \\ \frac{B \hbar}{2} &0 &-A+\frac{B \hbar}{2}}
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\end{align}
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$det(H-\lambda \einsmatrix) = 0 \\$
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\begin{align}
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\lambda_1 &= B\hbar^2 \\
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\lambda_2 &= \frac{\hbar^2}{2} B (1 + \sqrt{1 + \frac{4 A^2}{\hbar^2 B^2}} \\
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\lambda_3 &= \frac{\hbar^2}{2} B (1-\sqrt{1 + \frac{4 A^2}{\hbar^2 B}}
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\end{align}
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\begin{align}
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a &:= A + \frac{\hbar B}{2} \sqrt{1+\frac{4 A^2}{\hbar^2 B^2}} \\
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b &:= \frac{\hbar B}{2} \\
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\ket{\phi_0} &= \inlinematrix{0 \\ 1 \\ 0} &
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\ket{\phi+} &= \frac{1}{\sqrt{a^2+b^2}} \inlinematrix{a \\ 0 \\ b} &
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\ket{\phi-} &= \frac{1}{\sqrt{a^2+b^2}} \inlinematrix{-b \\ 0 \\ a} \\ \\
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\ket{\psi(t)} &= c_0 e^{-\i E_0 t} \ket{\phi_0} + c_+ e^{-E_+ t} \ket{\psi_+} + c_- e^{-E_-t t} \ket{\phi-} \\ % muss da nicht e^\i?
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\ket{\psi(0)} &= \ket{z+} &= \inlinematrix{1 \\ 0 \\ 0} \\ \\
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\end{align}
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$\Rightarrow c_0 = 0$
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\begin{align}
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\frac{1}{\sqrt{a^2+b^2}}(c_+ a - c_- b) &\stackrel{!}{} 1 \\
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\frac{1}{\sqrt{a^2+b^2}}(c_+ b + c_- a) &\stackrel{!}{} 0
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\end{align}
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$\Rightarrow $
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\begin{align}
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c_- &= \frac{-b\sqrt{a^2+b^2}}{a^2+b^2} \\
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c_+ &= \frac{a\sqrt{a^2+b^2}}{a^2+b^2} \\
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S_x &= \hbar \Sigma_x &= \hbar \inlinematrix{0 &1 &0 \\ 1 &0 &1 \\ 0 &1 &0}
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\end{align}
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Da $\inlinematrix{l \\ o \\ m} S_x \inlinematrix{n \\ o \\ p} = 0}$ folgt $S_x = S_y = 0$ %wo kommt das her, und wieso ist hier S_x=0, obwohl es oben != 0 ist
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\subsection*{b)}
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\begin{align}
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S_z &= \hbar \inlinematrix{1 &0 &0 \\ 0 &0 &0 \\ 0 &0 &-1}
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\end{align}
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||||
\begin{align}
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||||
\dirac{\psi(t)}{S_z}{\psi(t)} &= \frac{1}{(a^2+b^2)^2} [(a^2-b^2)^2 + 4a^2b^2cos((E_+-E_-)t)]}
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||||
\end{align}
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||||
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||||
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||||
|
||||
\section{Aufgabe 15: Benzol-Molekül (Teil 2)}
|
||||
\subsection*{a)}
|
||||
\subsection*{b)}
|
||||
\begin{align}
|
||||
P_m H &= \norm{\braket{\phi_m}{\psi(t)}} \\
|
||||
\ket{\psi(t)} &= U(t,t_0) \ket{\psi(t)} \\ \\
|
||||
&= exp(-\frac{\i}{\hbar} t H} \\
|
||||
&= \sum{k=0}{5}{exp(-\frac{\i}{\hbar} t \lambda_k) \ket{\chi_k} \braket{\chi_k}{\phi_0}} \\ %warum 5?
|
||||
&=^1 \frac{1}{6} \sum{k=0}{5}{\sum{n=0}{5}{exp(-\frac{\i}{\hbar} t (E-2Acos(\delta_k)))}} \\ %warum diese umformung?
|
||||
P_m(t) &= \norm{\braket{\phi_m}{\psi(t)}} \\
|
||||
&= \frac{1}{6} \sum{k=0}{5}{\sum{n=0}{5}{exp(...)}\braket{\phi_m}{\phi_n}} \\ %geht die summe wirklich bis 5? und über was?
|
||||
&=^2 \frac{1}{36} \norm{\sum{k}{biw wo?}{exp(-\frac{\i}{\hbar} t (E-2Acos(\delta_k) + \i \delta_k)}} \\ \\
|
||||
&= ?
|
||||
\end{align}
|
||||
(1)
|
||||
\begin{align}
|
||||
\braket{\chi_k}{\phi_0} &= \frac{1}{\sqrt{6}} \sum{n=0}{5}{exp(-\i n \delta_{keine ahnung}\bracket{\phi_n}{\phi_0}} \\ \\
|
||||
&= \frac{1}{\sqrt{6}} e^{-\i 0 \delta_{keine ahnung}} \\
|
||||
&= \frac{1}{\sqrt{6}}
|
||||
\end{align}
|
||||
(2)
|
||||
\begin{align}
|
||||
\braket{\phi_m}{\phi_n} = \delta_{n m}
|
||||
\end{align}
|
||||
|
||||
|
||||
|
||||
|
||||
\subsection*{b)}
|
||||
\begin{math}
|
||||
\tau = \frac{2\pi}{A}\hbar
|
||||
\end{math}
|
||||
|
||||
|
||||
\section{Aufgabe 16: Schrödingergleichung in Impulsraumdarstellung}
|
||||
|
||||
\begin{align} \\
|
||||
%nach schrödingergleichung in ortsdarstellung gucken
|
||||
\i \hbar \partial_\tau \ket{\psi} &= H \ket{\psi} & \left| \bra{p}\right. \\
|
||||
\i \hbar \partial_\t \psi(p) &= \dirac{p}{\frac{p^2}{2m}+V(\hat{x}}{\psi}
|
||||
\end{align}
|
||||
|
||||
\begin{align}}
|
||||
\int_{-\infty}^{\infty} \text{dx} \dirac{p}{V(\hat[x})}{x}\braket{x}{\psi} \\
|
||||
&= \int_{-\infty}^{\infty} \text{dx} V(x) \frac{1}{\sqrt{2\pi\hbar}}e^{-\frac{\i p}{\hbar}x}\braket{x}{\psi} \\
|
||||
&= \int_{-\infty}^{\infty} \text{dx} \int_{-\infty}^{\infty} \text{dp'} V(x)\frac{1}{\sqrt{2\pi\hbar}e^{-\frac{\i p}{\hbar}x}\braket{x}{p'}\braket{p'}{x} \\ \\
|
||||
&= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \text{dx} \text{dp'} V(x)\frac{1}{\sqrt{2\pi\hbar}e^{-\frac{i(p-p')}{\hbar}x}\psi(p') \\
|
||||
&= (2) % aus aufgabenblatt raussuchen
|
||||
\end{align}
|
||||
|
||||
\begin{align}
|
||||
%wo kommt das her?
|
||||
\dirac{p}{V(\hat{x}}{\psi} &= \dirac{p}{V(\i\hbar\diffP{p}}{\psi} \\
|
||||
&= V(\i\hbar\diffP{p})\psi(p) %vertauschen von p und V: Regeln?
|
||||
\end{align}
|
||||
|
||||
|
||||
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\i \hbar
|
||||
|
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\section{Aufgabe 16: Schrödingergleichung in Impulsraumdarstellung}
|
||||
71
ueb7.tex
71
ueb7.tex
|
|
@ -0,0 +1,71 @@
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|||
%\includegraphics{excs/qm1_blatt07_SS08.pdf}
|
||||
%\pagebreak
|
||||
|
||||
\chapter{Quantenmechanik I - Übungsblatt 7}
|
||||
\section{Aufgabe 17: Unendlich hoher Potentialtop (Ergänzungen)}
|
||||
\subsection*{a)}
|
||||
\subsection*{b)}
|
||||
\subsection*{c)}
|
||||
|
||||
\section{Aufgabe 18: Tunneleffekt}
|
||||
\includegraphics{grafiken/U_A18_1.pdf}
|
||||
\subsection*{a)}
|
||||
|
||||
I
|
||||
\begin{math}
|
||||
-\frac{\hbar^2}{2m} \diffPs{x}^2 \Phi(x) = E \Phi(x) \\
|
||||
\Phi(x) =A e^{\i k x} + B e^{-\i k x} \\
|
||||
k = \sqrt{\frac{2 E M}{\hbar^2}}
|
||||
\end{math}
|
||||
|
||||
II $E < V_0$
|
||||
\begin{math}
|
||||
\Phi(x) = C e^a + D e^{-qx} \\ % ist das nen q hier?
|
||||
q = \sqrt{\frac{-2m(E-V_0)}{\hbar^2}}
|
||||
\end{math}
|
||||
|
||||
III
|
||||
\begin{math}
|
||||
\Phi(x) = \tilde{E} e^{\i k x} + F e^{-\i k x}
|
||||
\end{math}
|
||||
|
||||
\subsection*{b)}
|
||||
\begin{math}
|
||||
F = 0
|
||||
\Phi_I(0) = \Phi_{II}(0) \\
|
||||
A+B = C+D \\
|
||||
\diffT{x} \Phi_I(0) = \diffT{x} \Phi_{II}(0) \\ % anschlussbedingungen korrekt
|
||||
\i k A - \i k B = q C - q D \\
|
||||
\inlinematrix{1 & 1 \\\i k & -\i k} \inlinematrix{A & B} = \inlinematrix{1 & 1 \\ q & -q} \inlinematrix{C & D} \\
|
||||
\Phi_{II}(a) = \Phi_{III}(a) \\
|
||||
C e^{a \cdot a} + D e^{-a \cdot a} = \tilde{E} e^{\i k a} \\
|
||||
\diffT{x} \Phi_{II}(a) = \diffT{x} \Phi_{III}(a) \\
|
||||
q C e^{a \cdot a} - q D e^{-a \cdot a} = \i k \tilde{E} e^{\i k a} \
|
||||
\inlinematrix{e^{a \cdot a} & e^{-a \cdot a} \\ q e^{a \cdot a} & q e^{-a \cdot a}} \inlinematrix{C & D} = \inlinematrix{\tilde{E} e^{\i k a} \\ \i k \tilde{E} e^{\i k a}}
|
||||
\end{math}
|
||||
|
||||
mit den Inversen von: (1) und (2) % geschweifte klammern unter matrix 1 und 2 setzen
|
||||
|
||||
\begin{math}
|
||||
%mit maxima berechnet
|
||||
\inlinematrix{C & D} = \inlinematrix{\frac{{e}^{{a}^{2}}}{{e}^{2 {a}^{2}}-1} & -\frac{1}{\left( {e}^{3\,{a}^{2}}-{e}^{{a}^{2}}\right) q} \\ -\frac{{e}^{{a}^{2}}}{{e}^{2{a}^{2}}-1} & \frac{{e}^{{a}^{2}}}{\left( {e}^{2 {a}^{2}}-1\right) q}} \inlinematrix{\tilde{E} e^{\i k a} \\ \i k \tilde{E} e^{\i k a}} \\
|
||||
\inlinematrix{A & B} = \inlinematrix{\frac{1}{2} & -\frac{\i}{2k}\\ \frac{1}{2} & \frac{\i}{2k}} \inlinematrix{1 & 1 \\ q & -q} \inlinematrix{C & D}
|
||||
\end{math}
|
||||
|
||||
$\inlinematrix{C & D}$ eingesetzt ergibt:
|
||||
\begin{align}
|
||||
\tilde{E} &= \frac{2 k a}{e^{ika} \bbracket{(q^2 - k^2) \cdot \i sinh(q \cdot a) - 2 k q cosh(qa)} \\
|
||||
A &= 1 \\
|
||||
T &= \bbracket{\frac{E}{A}}^2 \\
|
||||
R &= 1 - T &= \bbracket{\frac{B}{A}}^2 \\
|
||||
T &= \frac{1}{\bbracket{\frac{a^2 - k^2}{2 q k}} \cdot 2 sinh^2(q a) + cosh^2(q a)}
|
||||
\end{align}
|
||||
|
||||
|
||||
\subsection*{c)}
|
||||
|
||||
\section{Aufgabe 19: Doppeltes \delta-Potential}
|
||||
\subsection*{a)}
|
||||
\subsection*{b)}
|
||||
\subsection*{c)}
|
||||
\subsection*{d)}
|
||||
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Reference in New Issue