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52
formelsammlung.tex
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@ -1,5 +1,53 @@
%fehlt:
% Hilberraum
% Erzeuger/vernichter
% Zeitentwicklungsoperator
% Schrödingergleichung
% undendlich dim. raum
% was zum henker ist ein operator
% impulsoperator
\chapter{Notationen}
\section{Dirac-Notation}
In der Bra-Ket-Notation schreibt man die Vektoren eines Vektorraums V auch außerhalb eines Skalarprodukts mit einer spitzen Klammer als Ket $\ket{v}.$
Jedem Ket $\ket{v}$ entspricht ein Bra $\bra{v}$, das dem Dualraum $\text{V}^*$ angehört, also eine lineare Abbildung von V in den zugrundeliegenden Körper K bezeichnet. Allerdings kann nicht jedes Bra aus dem algebraischen Dualraum mit einem Ket identifiziert werden. Das Ergebnis der Operation eines Bras $\bra{v}$ auf ein Ket $\ket{w}$ wird $\braket{v}{w}$ geschrieben, womit der Zusammenhang mit der konventionellen Notation des Skalarprodukts hergestellt ist.
\subsection*{Eigenschaften}
$c_1$, $c_2$, $\in \setC$; $c^*$ ist die komplex-konjugierte Zahl zu $c$, $A$, $B$ sind lineare Operatoren
\subsubsection*{Linearität}
\equationblock{\langle\phi| \; \bigg( c_1|\psi_1\rangle + c_2|\psi_2\rangle \bigg) = c_1\langle\phi|\psi_1\rangle + c_2\langle\phi|\psi_2\rangle} \\
Mit der Addition und skalaren Multiplikation von linearen Funktionalen im Dual-Raum gilt:
\equationblock{\bigg(c_1 \langle\phi_1| + c_2 \langle\phi_2|\bigg) \; |\psi\rangle = c_1 \langle\phi_1|\psi\rangle + c_2\langle\phi_2|\psi\rangle}
\subsubsection*{Assoziativität}
Given any expression involving complex numbers, bras, kets, inner products, outer products, and/or linear operators (but not addition), written in bra-ket notation, the parenthetical groupings do not matter (i.e., the [[associative property]] holds). For example:
$< \psi| (A |\phi>) = (< \psi|A)|\phi>$
$(A|\psi>)<\phi| = A(|\psi> < \phi|$
and so forth. The expressions can thus be written, unambiguously, with no parentheses whatsoever. Note that the associative property does ''not'' hold for expressions that include non-linear operators, such as the antilinear time reversal operator in physics.
\subsubsection*{Adjungierte}
\begin{itemize}
\item Die Adjungierte eines Bra ist der entsprechne Ket (und umgekehrt)
\equationblock{\bra{X^\dagger} = \ket{X}}
\item Die Adjungierte einer komplexen Zahl ist ihre komplex-konjugierte Zahl
\equationblock{c^\dagger = c^\ast}
\item Die Adjungierte einer Adjungierten von X ist X (wobei X alles mögliche sein kann)
\equationblock{\sbk{X^dagger}^\dagger = X}
\end{itemize}
\subsubsection*{Beispiele}
\begin{itemize}
\item Kets:
\equationblock{\left(c_1|\psi_1\rangle + c_2|\psi_2\rangle\right)^\dagger = c_1^* \langle\psi_1| + c_2^* \langle\psi_2|.}
\item Inner Product (übersetzen)
\equationblock{< \phi | \psi >^* = < \psi|\phi>}
\item Matrix-Elemente:
\equationblock{< \phi| A | \psi >^* = < \psi | A^\dagger |\phi >}
\equationblock{< \phi| A^\dagger B^\dagger | \psi >^* = < \psi | BA |\phi >}
\item Outer Product:
\equationblock{\left((c_1|\phi_1>< \psi_1|) + (c_2|\phi_2><\psi_2|)\right)^\dagger = (c_1^* |\psi_1>< \phi_1|) + (c_2^*|\psi_2><\phi_2|)}
\end{itemize}
\chapter{Lineare Algebra}
\section{Gruppentheorie}
@ -65,11 +113,11 @@ Die kontinuierliche Fourier-Transformation ist eine Form der Fourier-Transformat
Oft wird diese Transformation auch einfach als Fourier-Transformation bezeichnet.
\subsubsection*{Definition:}
\begin{equation}
\mathcal{F}\{f(t)\} = F(\omega)= \frac{1}{\sqrt{2 \pi}} \integrinf{f(t) e^{-\i \omega t}}{t}
\mathcal{F}\{f(t)\} = F(\omega)= \frac{1}{\sqrt{2 \pi}} \intgrinf{f(t) e^{-\i \omega t}}{t}
\end{equation}
Rücktransformation (Fouriersynthese)
\begin{equation}
\mathcal{F}^{-1}\{F(\omega)\} = f(t)= \frac{1}{\sqrt{2 \pi}} \integrinf{F(\omega) e^{\i \omega t}}{\omega}
\mathcal{F}^{-1}\{F(\omega)\} = f(t)= \frac{1}{\sqrt{2 \pi}} \intgrinf{F(\omega) e^{\i \omega t}}{\omega}
\end{equation}
Hierbei ist $F(\omega)$ das kontinuierliche Spektrum, das die Amplitude jeder Frequenz $\omega$ aus den reelle Zahlen angibt.

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2
kapI-2.tex
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@ -203,7 +203,7 @@ denn:
\item Matrixelemente $A'_{m,n} \equiv \dirac{m'}{A}{n'} = \sum_{k,l} U_{m,k}^\dagger A_{k,l} U_{l,m}$
\end{enumerate}
\section{Sektralzerlegung von hermitschen Operatoren}
\section{Spektralzerlegung von hermitschen Operatoren}
\subsection*{Satz}
\begin{align}
\text{A hermitesch} \Rightarrow &\text{(1) Eigenwerte $a_n$ sind reell}\\

27
math.tex
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@ -11,6 +11,8 @@
\newcommand{\einsmatrix}{\mathbbm{1}}
\newcommand{\bigO}{\mathbbm{O}}
\newcommand{\inlinematrix}[1]{\begin{pmatrix} #1 \end{pmatrix}}
\newcommand{\inlinearray}[2]{\left( \begin{array}{#1} #2 \end{array} \right)}
\newcommand{\inlinematrixu}[1]{\begin{matrix} #1 \end{matrix}}
\newcommand{\inlinematrixd}[2]{\left( \begin{matrix} #1\vphantom{#2} \end{matrix} ~\right| \left. \begin{matrix} \vphantom{#1}#2 \end{matrix} \right)}
\newcommand{\inlinematrixdet}[1]{\begin{vmatrix} #1 \end{vmatrix}}
@ -26,18 +28,35 @@
\newcommand{\sign}{{\text{sign}}}
\newcommand{\QED}{\begin{large}\textbf{\checkmark}\end{large}}
\newcommand{\cequiv}{\stackrel{\scriptscriptstyle\wedge}{=}}
\newcommand{\cequiv}{\stackrel{\scriptscriptstyle\wedge}{=}} %ContextEQUIvalent
\newcommand{\deq}{\stackrel{!}{=}}
\newcommand{\expval}[1]{\langle #1 \rangle}
\newcommand{\diffP}[1]{\frac{\partial}{\partial #1}}
\newcommand{\diffT}[1]{\frac{\text{d}}{\text{d} #1}}
\newcommand{\diffTfrac}[2]{\frac{\text{d} #1}{\text{d} #2}}
\newcommand{\diffTm}[3]{\diffTfrac{^{#1} #2}{#3^{#1}}}
\newcommand{\diffPfrac}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\intgr}[4]{\int_{#1}^{#2} #3 ~\text{d}#4}
\newcommand{\intgru}[2]{\int #1 ~\text{d}#2}
\newcommand{\integrinf}[2]{\int_{-\infty}^[\infty #1 \text{\,d}#2}
\newcommand{\intgr}[4]{\int_{#1}^{#2} #3 ~\text{ d}#4}
\newcommand{\intgru}[2]{\int #1 ~\text{ d}#2}
\newcommand{\intgrinf}[2]{\intgr{-\infty}{+\infty}{#1}{#2}}
\newcommand{\sbk}[1]{\left( #1 \right)}
\newcommand{\sqbk}[1]{\left[ #1 \right]}
\newcommand{\ssbk}[1]{\left< #1 \right>}
\newcommand{\detb}[1]{\det\sqbk{#1}}
\newcommand{\cosb}[1]{\cos\sbk{#1}}
\newcommand{\sinb}[1]{\sin\sbk{#1}}
\newcommand{\sinhb}[1]{\sinh\sbk{#1}}
\newcommand{\coshb}[1]{\cosh\sbk{#1}}
\newcommand{\cotb}[1]{\cot\sbk{#1}}
\newcommand{\tanb}[1]{\tan\sbk{#1}}
\newcommand{\arcsinb}[1]{\arcsin\sbk{#1}}
\newcommand{\airyb}[1]{\text{Ai}\sbk{#1}}
\newcommand{\lnb}[1]{\text{ln}\sbk{#1}}
\newcommand{\bigOb}[1]{\bigO\sbk{#1}}
\newcommand{\mtrx}[1]{\vec{#1}} %fett machen als matrix
\newcommand{\levicivita}[1]{\varepsilon_{#1}}
\newcommand{\krondelta}[1]{\delta_{#1}}

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19
theo2.tex
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@ -14,6 +14,7 @@
\include{physics}
\newcommand{\lboxed}[1]{\left\lceil #1 \right\rfloor}
\renewcommand{\i}{i}
\title{Theoretische Physik 2 Vorlesungs- und Übungsmitschrieb}
\author{Daniel Bahrdt, Oliver Groß}
@ -49,13 +50,17 @@
% \part{Übungsmitschrieb}
% \label{UE}
% \include{ueb1}
% \include{ueb2}
% \include{ueb3}
% \include{ueb4}
% \include{ueb5}
% \include{ueb6}
% \inlcude{ueb7}
\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}

52
ueb1.tex
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@ -9,49 +9,39 @@
\subsection*{b)}
\subsection*{c)}
\begin{math}
\textbf{\sigma} = (\sigma_x, \sigma_y, \sigma_z) \\
\textbf{\a},\textbf{\b} \in \setR
\vec{\sigma} = (\sigma_x, \sigma_y, \sigma_z) \\
\vec{a}, \vec{b} \in \setR
\end{math}
\begin{align}
(\textbf{a \cdot \sigma})(\textbf{b \cdot \sigma}) &= \one (\textbf{a \cdot b} + \i \textbf{\sigma} \cdot (\textbf{\a} \times \textbf{b}) \\
(\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma}) &= \one (\vec{a} \cdot \vec{b} + \i \vec{\sigma} \cdot (\vec{a} \times \vec{b}) \\
\sum a_\alpha b_\beta \sigma_\alpha \sigma_\beta &= \\
\sum a_\alpha b_\beta ( \krondelta{\alpha \beta} \one + \i \levicivita{\alpha,\beta,\gamma} \sigma_\gamma ) &= \\
\sum a_\alpha b_\beta \krondelta{\alpha \beta} \cdot \one + \i a_\alpha b_\beta \levicivita{\alpha,\beta,\gamma} \sigma_\gamma &= \\
\one (\textbf{a \cdot b} + \i \sigma \cdot (a \times b)
\one (\vec{a} \cdot \vec{b} + \i \sigma \cdot (a \times b)
\end{align}
\subsection*{d)}
$e^{\i \frac{\i}{2} n \sigma} = \one cos(\frac{\alpha}{2}) - \i (\textbf{n \cdot \sigma}) sin(\frac{\alpha}{2})$ \\
$e^{\i \frac{\i}{2} n \sigma} = \one cos(\frac{\alpha}{2}) - \i (\vec{n} \cdot \vec{\sigma}) sin(\frac{\alpha}{2})$ \\
Mit der Reihenentwicklung von $e^x$ ergibt sich:
\begin{align}
e^{- \i \frac{\alpha}{2} \textbf{n \cdot \sigma} &= \sum_k \frac{(- \i \frac{\alpha}}{2} \textbf{n \cdot \sigma})^k}{k!} \\
&= \sum_k (- \i \frac{\alpha}{2})^k \cdot \frac{(\textbf{n \¢dot \sigma)^k}}{k!}
\end{align}
Desweiteren gilt nach Aufgabe 2 c):
\begin{align}
(\textbf{n \cdot \sigma} &= \sigma \\ \\
(\textbf{n \cdot \sigma}^2 &= \one (\textbf{n \cdot n} + \underbrace{\i \textbf{\sigma} \cdot (\textbf{\n} \times \textbf{n})}_{=0} \\
&= \one (\textbf{n \cdot n}
(\vec{n} \cdot \vec{\sigma} &= \sigma \\
(\vec{n} \cdot \vec{\sigma})^2 &= \one (\vec{n} \cdot \vec{n} + \underbrace{\i \vec{\sigma} \cdot (\vec{n} \times \vec{n})}_{=0} \\
&= \one (\vec{n} \cdot \vec{n}
\end{align}
$\Rightarrow$
\begin{align}
\sum_k (- \i \frac{\alpha}{2})^k \cdot \frac{(\textbf{n \cdot \sigma)^k}}{k!} &= \\
\sum_k ( (-\i)^{2k} (\frac{\alpha}{2})^{2k+1} \frac{(\textbf{n \cdot \sigma})^{2k}}{(2k)!} +
\sum_k ( (-\i)^{2k+1} (\frac{\alpha}{2})^{2k+1} \frac{(\textbf{n \cdot \sigma})^{2k+1}}{(2k+1)!} &= \\
\sum_k ( (-1)^k (\frac{\alpha}{2})^{2k+1} \frac{(\textbf{n \cdot \sigma})^{2k}}{(2k)!} +
\sum_k ( \i \cdot (-1)^k (\frac{\alpha}{2})^{2k+1} \frac{(\textbf{n \cdot \sigma})^{2k} \cdot (\textbf{n \cdot \sigma})}{(2k+1)!} &= \\
\sum_k ( (-1)^k (\frac{\alpha}{2})^{2k+1} \one \frac{1}{(2k)!} +
\i \cdot \sum_k (-1)^k (\frac{\alpha}{2})^{2k+1} \one \cdot (\textbf{n \cdot \sigma}) \cdot \frac{1}{(2k+1)!} &= \\
\one \cos(\frac{\alpha}{2}) + \i (\textbf{n \cdot \sigma}) \sin(\frac{\alpha}{2})
\sum_k \sbk{- \i \frac{\alpha}{2}}^k \cdot \frac{\sbk{\vec{n} \cdot \vec{\sigma}}^k}{k!} &= \\
\sum_k \sbk{ (-\i)^{2k} (\frac{\alpha}{2})^{2k+1} \frac{(\vec{n} \cdot \vec{\sigma})^{2k}}{(2k)!}} +
\sum_k \sbk{ (-\i)^{2k+1} (\frac{\alpha}{2})^{2k+1} \frac{(\vec{n} \cdot \vec{\sigma})^{2k+1}}{(2k+1)!}} &= \\
\sum_k \sbk{ (-1)^k \sbk{\frac{\alpha}{2}}^{2k+1} \frac{\sbk{\vec{n} \cdot \vec{\sigma}}^{2k}}{(2k)!}} +
\sum_k \sbk{ \i \cdot (-1)^k \sbk{\frac{\alpha}{2}}^{2k+1} \frac{\sbk{\vec{n} \cdot \vec{\sigma}}^{2k} \cdot \sbk{\vec{n} \cdot \vec{\sigma}}}{(2k+1)!}} &= \\
\sum_k ( (-1)^k \sbk{\frac{\alpha}{2}}^{2k+1} \one \frac{1}{(2k)!} +
\i \cdot \sum_k (-1)^k (\frac{\alpha}{2})^{2k+1} \one \cdot (\vec{n} \cdot \vec{\sigma}) \cdot \frac{1}{(2k+1)!} &= \\
\one \cos(\frac{\alpha}{2}) + \i (\vec{n} \cdot \vec{\sigma}) \sin(\frac{\alpha}{2})
\end{align}
@ -72,18 +62,18 @@
$f(t) = e^{tA} \cdot B e^{-tA} = B + \frac{t}{1!}[A,B]+\frac{t^2}{2!}[A,[A,B]]+\ldots$
\subsection*{b)}
$\forall A,B,C : [A,B] = \i C \and [B,C] = \i A$ \\
$\forall A,B,C : [A,B] = \i C \and [B,C] = \i A$
\begin{align}
e^{\i B t} \cdot B e^{-\i B t} &= A \cos(t) + C \sin(t) \\
&= A + [\i B,A] + \frac{t^2}{2}[\i B,[\i B, A]] \\
[\i B, A] &= \i [B,A] &= -\i [A,B] \\
[\i B, [\i B, A]] &= [\i B, C] &= \i \i A = -A }} \\
[\i B, [\i B, A]] &= [\i B, C] &= \i \i A = -A \\
e^{\i B t} A e^{-\i B t} &= A + C - \frac{t^2}{2} A \frac{t^3}{6} C + \frac{t^4}{24} A \\ \\
&= A \cos(t) + C \sin(t)
\end{align}
\subsection*{c)}
Es gelte: $[A,[A,B]] = [B,[A,B]] = 0$ \\
Es gelte: $[A,[A,B]] = [B,[A,B]] = 0$
\begin{align}
e^{A+B} &= e^A \cdot e^B \cdot e^{-\frac{1}{2}[A,B]} \\
g(t) &= e^{tA} \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} \\
@ -97,11 +87,11 @@ Es gelte: $[A,[A,B]] = [B,[A,B]] = 0$ \\
&= A \cdot g(t) -t \cdot g(t) \cdot [A,B] + e^{tA} \cdot B \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]}
\diffTfrac{g(t)}{t} &= A \cdot g(t) + e^{tA} \cdot B \cdot e^{-tA} \cdot g(t) - t \cdot g(t) \cdot [A,B] \\
e^{x} \cdot e^{-x} &= \one \\
\diffTfrac{g}{t} &= \right( A - t [A,B] + B + \frac{t}{1!} \cdot [A,B] \left) \cdot g(t) \\
\diffTfrac{g}{t} &= \left( A - t [A,B] + B + \frac{t}{1!} \cdot [A,B] \right) \cdot g(t) \\
\diffTfrac{g}{t} &= [A+B] \cdot g(t) \\
\frac{\diffTfrac{g}{t}}{g} &= [A+B] \\
g(t) &= e^{[A+B] \cdot t + c} \\
e^A \cdot e^B &= e^B \cdot e^A \cdot e^[A,B] \\
e^(A+B) &= e^A \cdot e^B \cdot e^{-\frac{1}{2}[A,B]} \\
e^(B+A) &= e^B \cdit e^A \cdot e^{-\frac{1}{2}[A,B]}
e^(B+A) &= e^B \cdot e^A \cdot e^{-\frac{1}{2}[A,B]}
\end{align}

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\chapter{Quantenmechanik I - Übungsblatt 10}
\section{Aufgabe 24: Sphärische symmetrische Kastenpotential}
\subsection*{a)}
$l = 0$ einsetzen und ausrechnen regibt:
\begin{math}
u_2(z) = \begin{cases}
z_{j_l}(z) = z^3 \sbk{-\frac{1}{z} \diffT{z}} \sbk{-\frac{1}{z} \diffT{z}} \frac{\sinb{z}}{z} \\
\ldots
\end{cases}
\end{math}
Dies führt dann auf:
\begin{math}
u_2(z) = \begin{cases}
-\sinb{z} - \frac{3 \cosb{z}}{z} + \frac{3 \sinb{z}}{z^2} \\
\cosb{z} - \frac{3 \sinb{z}}{z} - \frac{3 \cosb{z}}{z^2}
\end{cases}
\end{math}
Asymptotisches Verhalten:
für $z \rightarrow 0$
\begin{align}
& \lim_{z \rightarrow 0} z j_2(z)
(Taylor) & \lim_{z \rightarrow 0} 3 \frac{1}{z^2} \sbk{z - \frac{1}{3!} z^3 + \bigOb{z^5} - z + \frac{1}{2} z^3 - \bigOb{z^5}}
\gdw & 0
\end{align}
für $\lim_{z \rightarrow 0} z y_2(z)$ geht es analog.
für $z \rightarrow \infty$
\begin{align}
j_l(z) &\approx z^l \frac{1}{z^l} \sbk{- \diffT{z}}^l \sinb{z} \\
&= \begin{cases}
\frac{1}{z} \sbk{-1}^{\frac{l}{2}} \\
\frac{1}{z} \sbk{-1}^{\frac{l+1}{2}}
\end{cases}
&= \frac{\sinb{z - l \frac{\pi}{2}}}{z}
\end{align}
\subsection*{b)}
$u_l(z) = z j_l(z)$
$j_l = \frac{\sinb{z - l \frac{\pi}{2}}}{z}$
Für $z=k r = k a$
$j_l(k a) = \frac{\sinb{k a - l \frac{1}{2}}}{k a} = \deq 0$
$k a - l \frac{\pi}{2} = \lambda \pi$ mit $\lambda \in \setN$
$k a = \pi \sbk{\lambda + \frac{l}{2}}$
Energie $\sbk{l = 1}$
$E \lambda = \pi^2 \sbk{\lambda + \frac{1}{2}}^2 \frac{\hbar^2}{2 \mu a}$
\begin{align}%hier sind bestimmt fehler
E_0 &= 2,47 \\
E_1 &= 22,20 \frac{\hbar^2}{2 \mu a^2}\\
E_2 &= 61,60 \\
E_3 &= 120,90 \\
E_4 &= 199,85 \\
E_5 &= 298,55
\end{align}
\subsection*{c)}
$\frac{\hbar^2 a^2}{2 \mu} = E - V_0 > 0 z = q r$
$ \frac{\hbar^2 \kappa^2}{2 p} = -E > 0 z ? \i \kappa r$
$u_l = c \sinb{z - l \frac{\pi}{2}}$
\begin{align}
r > a u_l(\i k r) &= A \sinb{\i k r - l \frac{\pi}{2}} + B \cosb{\i k r - l \frac{\pi}{2}}
&= \frac{A}{2 \i} \sbk{e^{-k r} e^{-\i l \frac{\pi}{2}} - e^{k r} e^{\i l \frac{\pi}{2}}} +
\frac{B}{2} \sbk{e^{-k r} e^{-\i l \frac{\pi}{2}} + e^{k r} e^{\i l \frac{\pi}{2}}}
&= \frac{B}{2} \sbk{\frac{2 e^{-k r}}{e^{-\i l \frac{\pi}{2}}}}
&= B e^{-k r}
\end{align}
$l = 0$
$r \leq a u_l = l \sinb{a r}$
$r > a u_l = B e^{- k r}$
$\frac{u_l'(a-)}{u_l(a-)} = \frac{u_l'(a+)}{u_l(a+)}$
$\frac{q \cosb{q a}}{\sinb{q a}} = -k$
$\cotb{q a} = -\frac{k}{q}$
$\cotb{\sqrt{\frac{2 N}{2 \hbar} \sbk{E - V_0}} a} = -\sqrt{\frac{-E}{E - V_0}}$
%da kommt nen bild hin: doll, sagte der oliver. und dann aha. mehr aber auch nicht.
\section{Aufgabe 25: Der zweidimensionale harmonische Oszillator}
\subsection*{a)}
\begin{align}
H &= -\frac{\hbar^2}{2 m} \sbk{\diffPs{x}^2 + \diffPs{y}^2} + \frac{1}{2} m \omega^2 \sbk{x^2 + y^2} \\
&= \frac{1}{2} \underbrace{sbk{-\diffPs{x} + x^2}}_{H_1} + \frac{1}{2} \underbrace{\sbk{-\diffPs{y}^2 + y^2}}_{H_2} \\
\psi(x,y) &= \psi_x(x) \cdot \psi_y(y) \\
&= c(u_1) \cdot c(u_2) \cdot H_{n_1}(x) \cdot H_{n_2}(y) \cdot e^{-\frac{x^2 + y^3}{2}} \\
c(n) &= \sbk{\frac{1}{2^n n! \sqrt{\pi}}}^{\frac{1}{2}}
\end{align}
\subsection*{b)}
\begin{align}
\ket{2,0} &= \frac{1}{\sqrt{2}} \sbk{a_1^\dagger}^2 \ket{0,0} \\
\braket{x}{2} &= \dirac{x}{\sbk{\frac{a_1^\dagger}{\sqrt{2}}}^2}{0} \\
&= \dirac{x}{\frac{1}{\sqrt{8}} \sbk{\hat{x} - \i \hat{p}}^2}{0} \\
&= \frac{1}{\sqrt{8}} \sbk{x - \diffPs{x}}^2 \underbrace{\psi_0(x)}_{\frac{1}{\pi} \frac{1}{4} e^{-\frac{x^2}{2}}} %fehler? 4 oder n oder k oder was?
&= \frac{1}{\sqrt{2} \sqrt[4]{\pi}} \sbk{2 x^2 - 1} e^{-\frac{x^2}{2}}
\braket{x,y}{2,0} &= \frac{1}{\sqrt{2 \pi}} \sbk{2 x^2 - 1} e^{-\sbk{\frac{x^2}{2} + y^2}} \\ %fehler!
H_2(x) &= 4 x^2 - 2 \\
\braket{x,y}{n_1,n_2} &= \braket{x}{n_1} \cdot \braket{y}{n_2} \\ %mkehr fehler
\braket{x}{n_1} &= \dirac{x}{\frac{\sbk{a^\dagger}^{n_1}}{\sqrt{n_1!}}}{0} \\
&= \frac{1}{\sqrt{2^{n_1} n_1!}} \dirac{x}{\sbk{\hat{x} - \i \hat {p}}^{n_1}}{0} \\
&= \frac{1}{\sqrt{2^{n_1} n_1!}} \sbk{x - \diffPs{x}}^{n_1} \psi_0(x) \\
&= \frac{1}{\sqrt{2^m n_1! \sqrt{n_1}}} \cdot \sbk{-1}^m e^{\frac{x^2}{2}} \sbk{\diffPs{x}}^{n_1} e^{-\frac{x^2}{2}} e^{-\frac{x^2}{2}}
\end{align}
\subsection*{c)}
\begin{align}
\braket{x,y}{2,0}_{n_+,n_-} &= \frac{1}{\sqrt{2}} \sbk{a_r^\dagger}^2 \ket{0,0} \\
&= \frac{1}{\sqrt{2 \pi}} \sbk{x^2 - y^2 + 2 \i x y} e^{- \frac{x^2 + y^2}{2}}
\end{align}
\subsection*{d)}
\begin{math}
\braket{x,y}{2,0}_{N,m} = \frac{1}{\sqrt{\pi}} \sbk{x^2 - y^2} \cdot e^{-\frac{x^2}{2} - \frac{y^2}{2}}
\end{math}

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\chapter{Quantenmechanik I - Übungsblatt 11}
\section{Aufgabe 26: Oszillator im elektrischen Feld}
\subsection*{a)}
\begin{align}
H &= \frac{p^2}{2 m} + \frac{1}{2} m \omega^2 x^2 - q E x \\
\hat{x} &= \sbk{\frac{m \omega}{\hbar}}^{\frac{1}{2}} x \\
\hat{p} &= \sbk{\frac{1}{\hbar m \omega}}^{\frac{1}{2}} p \\
E' &= q E \sbk{\frac{\hbar}{m \omega}}^{\frac{1}{2}} \frac{1}{\sqrt{2}} \\
H &= \frac{\hbar \omega}{2} \sbk{\hat{p}^2 + x^2} + \sqrt{2} E' \hat{x}
\end{align}
Mit Erzeuger- und Vernichteroperatoren:
\begin{math}
H = \underbrace{\hbar \omega \sbk{\hat{p} + \frac{1}{2}}}_{H_0} + \underbrace{E' \sbk{a + a^\dagger}}_{H_1}
\underbrace{E_a}_{\text{neu}} = \underbrace{E_\alpha}_{\text{alt}} + \lambda \dirac{\alpha}{H_1}{\alpha} + \lambda^2 \sum_{\alpha \neq \beta} \frac{\dirac{\alpha}{H_1}{\beta} \dirac{\beta}{H_1}{\alpha}}{E_\alpha - E_\beta}
\end{math}
Energie:
\begin{align}
E_n &\approx \hbar \omega \sbk{n + \frac{1}{2}} - E' \dirac{n}{a + a^\dagger}{n} + E'^2 \sum_{n \neq m} \frac{\dirac{n}{a + a^\dagger}{m} \dirac{m}{a + a^\dagger}{m}}{E_n - E_m}
&= \hbar\omega \sbk{n + \frac{1}{2}} + E'^2 \sum_{n \neq m} \frac{\sbk{\sqrt{m}\braket{n}{m-1} + \sqrt{m+1}\braket{n}{m+1}} \sbk{\sqrt{n}\braket{m}{n-1} + \sqrt{n+1}\braket{m}{n+1}}}{E_n - E_m}
&= \hbar \omega \sbk{n + \frac{1}{2}} + \sbk{\frac{n+1}{E_n - E_{n+1}} + \frac{n}{E_n - E_{n-1}}} E'^2 \\
&= \hbar \omega \sbk{n + \frac{1}{2}} - \frac{E'^2}{\hbar \omega} \\
&= \hbar \omega \sbk{n + \frac{1}{2}} - \frac{q^2 E^2}{2 m \omega^2}
\end{align}
\subsection*{b)}
\begin{align}
H &= \frac{p^2}{2 m} + \frac{1}{2} \omega^2 x^2 - q E x
\end{align}
Mit:
\begin{align}
\hat{x} &= \sbk{\frac{m \omega}{\hbar}}^{\frac{1}{2}} x
\hat{p} &= \sbk{\frac{1}{\hbar m \omega}}^{\frac{1}{2}} p
\hat{E} &= q E \sbk{\frac{1}{\hbar m \omega^3}}^{\frac{1}{2}} \cdot 2
\end{align}
\begin{align}
H &= \frac{\hbar \omega}{2} \sbk{\hat{p}^2 + \hat{x}^2 - \hat{E} x} \\
&= \frac{\hbar \omega}{2} \sbk{\hat{p}^2 + \sbk{x - \frac{1}{2} \hat{E}}^2 - \frac{1}{4} \hat{E}^2}
\end{align}
Mit $\mathfrak{H} = \hat{x} - \frac{1}{2} \hat{E}$
\begin{align}
H &= \frac{\hbar \omega}{2} \sbk{\hat{p} + \mathfrak{H}^2} - \frac{\hbar \omega}{8} \hat{E}^2 \\
E_n &= \hbar \omega \sbk{n + \frac{1}{2}} - \frac{q^2 E^2}{2 m \omega^2}
\end{align}
\subsection*{c)}
\begin{math}
H_0 = -\frac{\hbar^2}{2 m} \sbk{\diffPs{x}^2 + \diffPs{y}^2} + \frac{1}{2} m \omega^2 \sbk{x^2 + y^2}
H = H_0 + \epsilon H_1
H_1 = - 2 q x y
\end{math}
$H_0: E_\alpha:$
\begin{align}
E_0 &= \hbar \omega &\ket{00} \\
E_1 &= 2 \hbar \omega &\text{2-fach entartet} \underbrace{\ket{01}, \ket{10}}_{\ket{\alpha}} \\
\end{align}
Nach langem rechnen erhält man:
\equationblock{\sum_\alpha \dirac{\beta}{\underbrace{H_{eff} - \sbk{E_a - E_\alpha}}}{\alpha} c_\alpha = 0}
Dabei sind $c_\alpha \ket{\alpha} + \bigOb{\epsilon} = \ket{a}$ die Eigenzustände.
Matrixelemente:
$\dirac{\beta}{H_1}{\alpha}$
$x,y$ mit Erzeuger- und Vernichteroperatoren:
$H_1 = \ldots = -\frac{q \hbar}{m \omega} \sbk{a_x^\dagger a_y^\dagger + a_x^\dagger a_y + a_x a_y^\dagger + a_x a_y}$
$\ket{\alpha} = \begin{cases}
\ket{01} \\
\ket{10}
\end{cases}$
\begin{math}
\dirac{01}{H_1}{01}
\dirac{01}{H_1}{10}
\dirac{10}{H_1}{01}
\dirac{10}{H_1}{11}
\end{math}
\begin{align}
a_x^\dagger a_y^\dagger \ket{10} &= \sqrt{2} \ket{21} &\dirac{01}{a_x^\dagger a_y^\dagger}{10} &= 0 \\
a_x^\dagger a_y \ket{10} &= 0 &\dirac{01}{a_x^\dagger a_y}{10} &= 0 \\
a_x a_y^\dagger \ket{10} &= \ket{01} &\dirac{01}{a_x a_y^\dagger}{10} &= 1 \\
a_x a_y \ket{10} &= 0 &\dirac{01}{a_x a_y}{10} &= 0
\end{align}
$\Rightarrow$
\begin{align}
\dirac{01}{H_1}{10} &= -\frac{1 \hbar}{m \omega} \\
\dirac{01}{H_1}{01} &= 0 \\
\dirac{10}{H_1}{01} &= -\frac{q \hbar}{m \omega} \\
\dirac{10}{H_1}{10} &= 0
\end{align}
Matrix:
$\inlinematrix{0 & -\frac{q \hbar}{m \omega} \\ -\frac{q \hbar}{m_\omega} & 0}$
Mit $\hat{q} = -\frac{q \hbar}{m \omega}$
\begin{math}
\inlinematrix{-\Delta E & -\hat{q} \\ -\hat{q} & -\Delta E} \inlinematrix{c_{01} \\ c_{10}} = \inlinematrix{0\\0}
\Delta E = E_a - E_\alpha
\detb{\ldots} \deq 0 \Rightarrow \Delta E_+ = \pm \hat{q}
\Delta E_+ = \pm \hat{q}
\inlinematrix{c_{01} \\ c_{10}} = \frac{1}{\sqrt{2}} \inlinematrix{1 \\ \pm 1}
\end{math}
\begin{itemize}
\item $\frac{1}{\sqrt{2}} \sbk{\ket{01} \pm \ket{10}} + \bigOb{\epsilon}$
\item $E_+ = \hbar \omega \pm \hat{q} \epsilon + \bigOb{\epsilon} = \hbar \omega \pm \frac{q \hbar}{m \omega} \epsilon \bigOb{\epsilon^2}$
\end{itemize}
\subsection*{d)}
\begin{align}
V(x) &= \frac{1}{2} m \omega^2 \sbk{x^2 + y^2} - 2 q \epsilon x y &\sigma = \frac{2 q \epsilon}{m \omega^2}
&= \frac{1}{2} m \omega^2 \inlinematrix{x & y} \inlinematrix{1 & -\sigma \\ 0 & 1 + \sigma} \inlinematrix{\hat{x} \\ \hat{y}} \\
&= \frac{1}{2} m \underbrace{\omega^2 \sbk{1 + \sigma}}_{\omega_+^2} \hat{x}^2 + \frac{1}{2} m \underbrace{\omega^2 \sbk{1 - \sigma}}_{\omega_-^2} \hat{y}^2 \\
E_{n+ n-} &= \hbar \omega_+ \sbk{n_+ + \frac{1}{2}} + \hbar \omega_- \sbk{n_- + \frac{1}{2}}
\end{align}
Für $\epsilon \rightarrow 0$:
$\hbar \omega \sbk{n_+ + n_- + 1}$
\section{Aufgabe 27: Helium-Atom}
$H = \underbrace{-\frac{\hbar^2}{2 m} \sbk{\underbrace{\nabla\sigma_1^2}_{I} + \underbrace{\nabla\sigma_2^2}_{II}} - e^2 \sbk{\underbrace{\frac{2}{r_1}}_{I} + \frac{2}{r_2}}}_{H_0} \underbrace{- e^2 \sbk{\frac{1}{\abs{\vec{r_1} - \vec{r_2}}}}}_{\text{in (a) vernachlässigbar}}$
\subsection*{a)}
$E_I^n = E_{II}^n = -\frac{Z^2}{2 \omega^2} \frac{e^2}{a_0}$ (wobei $a_0$ der bohrsche' Radius ist)
$\Rightarrow Z = 2$
Grundzustandsenergie:
$E_0 = E_I^1 + E_{II}^2 = -4 \frac{e^2}{a_0}$
Grundzustandswellenfunktion:
$\psi_0\sbk{\vec{r_1}, \vec{r_2}} = \frac{Z^3}{\pi a_0^3} e^{-\frac{2 \sbk{r_1 + r_2}}{a_0}}$
\subsection*{b)}
\begin{align}
\psi_1\sbk{\vec{r_1}, \vec{r_2}} &= \frac{\hat{Z}^3}{\pi a_0^3} e^{-\frac{\hat{Z} \sbk{r_1 + r_2}}{a_0}} \\
\dirac{\psi_1}{H}{\psi_1} &= \dirac{\psi_1}{H_0}{\psi_1} + \dirac{\psi_2}{H_{int}}{\psi_1} \\
\dirac{\psi_1}{H_0}{\psi_1} &= \sbk{- 4 \hat{Z} + \hat{Z}^2} \frac{e^2}{a_0} \\
\dirac{\psi_1}{H_{int}}{\psi_1} &= \intgrinf{}{^3r_1} \intgrinf{}{^3r_2} \sbk{\frac{\hat{Z}^3}{\pi a_0^3}}^2 e^{-\frac{2 \hat{Z} \sbk{r_1 + r_2}}{a_0}}
\end{align}
Nach extrem langem rechnen erhält man:
\equationblock{\dirac{\psi_1}{H_{int}}{\psi_1} = \frac{5}{8} \frac{Z}{a_0} e^2}
Und somit:
\equationblock{\dirac{\psi_1}{H}{\psi_1} = \sbk{Z^2 - \frac{27}{8} Z} \frac{e^2}{a_0}}
Aus:
$\diffTfrac{\dirac{\psi_1}{H}{\psi_1}}{Z} = 2 Z - \frac{27}{8} \deq 0$
folgt:
$Z^\ast = \frac{27}{16}$
Und somit:
\equationblock{\dirac{\psi}{H}{\psi}_\ast = -77,5 eV}

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@ -28,17 +28,17 @@
&= a_n^\ast \braket{a_n}{a_m} a_m \\
&= e^{-\i \alpha_n} \cdot e^{\i \alpha_m} \cdot \braket{a_n}{a_m} \\
&= e^{\i ( \alpha_m - \alpha_n)} \braket{a_n}{a_m} \\
&\Rightarrow \\
\braket{a_n}{a_m} &= 0
\end{align}
Da dies für alle Eigenvektoren gelten muss, also auch für Eigenvektoren, für die $e^{\i ( \alpha_m - \alpha_n)} \neq 1$, folgt:
$\braket{a_n}{a_m} = 0$
\subsection*{b)}
Zu zeigen: Für $A$ hermitesch ist $U(s) = e^{-\i s A}$ unitär \\
Zu zeigen: Für $A$ hermitesch ist $U(s) = e^{-\i s A}$ unitär
\begin{align}
U(s) &= e^{\i s A} \\
&= \sum_{k=0}^\infty \frac{1}{k!} \cdot (-\i s A)^k \\
&= \sum_{k=0}^\infty \frac{1}{k!} \cdot (\i s A^\intercol)^k \\
&= \sum_{k=0}^\infty \frac{1}{k!} \cdot \sbk{-\i s A}^k \\
&= \sum_{k=0}^\infty \frac{1}{k!} \cdot \sbk{\i s A^{\intercal}}^k \\
&= e^{\i s A}
\end{align}
@ -54,7 +54,7 @@ Zu zeigen: Für $A$ hermitesch ist $U(s_1 + s_2) = U(s_1) \cdot U(s_2)$.
\section{Aufgabe 5: Spur und Determinante}
\subsection*{a)}
Zu zeigen: $[A,BC] &= B \cdot [A,C] + [A,B] \cdot C$.
Zu zeigen: $[A,BC] = B \cdot [A,C] + [A,B] \cdot C$.
\begin{align}
[A,BC] &= ABC -BCA \\
B \cdot [A,C] + [A,B] \cdot C &= B \cdot (AC - CA) + (AB - BA) \cdot C \\
@ -113,27 +113,25 @@ Dann ist $T^{-1}AT$ die Basistransformation von der A-Basis in die T-Basis.
&= 1 + \epsilon \tr(A) + \bigO(\epsilon^2)
\end{align}
Zu zeigen: $\det(e^A = e^{\tr(A))$
Zu zeigen: $\detb{e^A} = e^{\tr(A)}$
\begin{align}
g(t) &= \det(e^{At}) \\
&\stackrel{tailor}{} \det(1 + At + \bigO(t^2)) \\
&= 1 + \tr(A) + \bigO(t^2) \\
&\stackrel{tailor ``rückwärs''}{} e^{\tr(A)t}
g(t) &= \detb{e^{At}} && \left| \text{Tailor} \right. \\
&= \detb{1 + At + \bigO(t^2)} &\\
&= 1 + \tr(A) + \bigO(t^2) && \left| \text{tailor ``rückwärs''} \right.\\
&= e^{\tr(A)t} &
\end{align}
\begin{align}
g(t) &= \limits_{\epsilon \rightarrow \intfy} \frac{1}{\epsilon} \sbk{\det(e^{A(t+\epsilon)) - det(e^{At})} \\
&= g(t) \limits_{\epsilon \rightarrow \intfy} \frac{1}{\epsilon} \det(e^{A \epsilon}) \\
&= g(t) \limits_{\epsilon \rightarrow \intfy} \frac{1}{\epsilon} \sbk{\det(1 + A \epsilon + \bigO(\epsilon^2) - det(1)} \\
g(t) &= \lim_{\epsilon \rightarrow \infty} \frac{1}{\epsilon} \sbk{\detb{e^{A(t+\epsilon)}} - \detb{e^{At}}} \\
&= g(t) \lim_{\epsilon \rightarrow \infty} \frac{1}{\epsilon} \detb{e^{A \epsilon}} \\
&= g(t) \lim_{\epsilon \rightarrow \infty} \frac{1}{\epsilon} \sbk{\det(1 + A \epsilon + \bigO(\epsilon^2) - det(1)} \\
&= g(t) \tr(A) \\
&\Rightarrow g(t) = e^{\tr(A) \cdot t}
\end{align}
Ist A diagonalisierbar:
\begin{math}
\det(e^A) = \det(T^{-1} e^A T) = \det(e^\hat{A}) = \prod_i e^{\lambda_i} = e^{\sum_i \lambda_i} = e^{\tr(A)}
\end{math}
\section{Aufgabe 6: Hermitesche Matrizen}
@ -145,7 +143,6 @@ Sei $\bra{a}$ Eigenvektor zum Eigenwert $a$
M_i^2 \ket{a} &= M_i a \cdot \bra{a} \\
\one \ket{a} &= a^2 \bra{a} \\
&\Rightarrow a = \pm 1
\end{align}
\subsection*{b)}

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@ -2,14 +2,107 @@
%\pagebreak
\chapter{Quantenmechanik I - Übungsblatt 3}
Zugehörige Voraussetzungen:
Operatoren
Wahrscheinlichkeiten
\section{Aufgabe 7: Eigenzustände und Erwartungswerte von Spin-1/2-Teilchen}
$\vec{n}(\theta,\Phi) = \inlinematrix{\sin(\theta) \cos(\Phi) \\ \sin(\theta) \sin(\Phi) \\ \cos(\Phi)}$ \\
Präparation in $\ket{n_1+} \vec{n_1} = \vec{n}\sbk{\frac{\pi}{4},\frac{\pi}{4}}$ \\
$\Rightarrow$ \\
$\vec{n_1} = \inlinematrix{\cosb{\frac{\pi}{4}} \\ 1 \\ \cosb{\frac{\pi}{4}}}$
$\ket{n_1} = \inlinematrix{\cosb{\frac{\Theta}{2}} \\ e^{\i \phi} \sinb{\frac{\Theta}{2}}}$
\subsection*{a)}
$\vec{n_2} = \vec{n}\sbk{\frac{3\pi}{4},\phi}$
\begin{align}
p_+\sbk{\phi} &= \probb{\sigma_{n_2} \cequiv +1}{\ket{n_1+}} \\
&= \abs{\braket{n_2+}{n_1+}}^2 \\
&= \frac{1}{4} \sbk{1 + \cosb{\phi+\frac{\pi}{4}}} \\
p_-\sbk{\phi} &= 1 - p_+ \\
&= 1 - \frac{1}{4} \sbk{1 + \cosb{\phi+\frac{\pi}{4}}} \\
&= \frac{3}{4} - \frac{1}{4} \cosb{\phi+\frac{\pi}{4}} \\
&= \frac{1}{4} \sbk{3 - \cosb{\phi+\frac{\pi}{4}}} \\
\ket{n_1+} &= c_1 \ket{n_2+} + c_2 \ket{n_2-}
\end{align}
\subsection*{b)}
\begin{align}
\expval{sigma_z}_{\ket{n_1+}} &= \probb{\sigma_z \cequiv +1}{\ket{n_1+}} +
(-1) \probb{\sigma_z \cequiv -1}{\ket{n_1+}} \\
&= \abs{\inlinematrix{1 & 0} \inlinematrix{\cos(\frac{\pi}{8} \\ e^{\i \frac{\pi}{4}}}}^2 -
\abs{\inlinematrix{0 & -1} \inlinematrix{\cos(\frac{\pi}{8} \\ e^{\i \frac{\pi}{4}}}}^2 \\
&= \cos^2(\frac{\pi}{8}) - \sin^2(\frac{\pi}{8}) \\
&= \cos(\frac{\pi}{4} \\
&= \frac{\sqrt{2}}{2} \\
\sbk{\Delta \sigma_x}^2 &= \prob{\sigma_x^2} - \prob{\sigma_x}^2 \\
\dirac{n_1+}{\sigma_x}{n_1+} &= \sin(\frac{\pi}{4} \cos(\frac{\pi}{4} &= \frac{1}{2} \\
\dirac{n_1+}{\sigma_x^2}{n_1+} &= \inlinematrix{\cos(\frac{\pi}{8}) & e^{-\frac{\pi}{4}} \sin(\frac{\pi}{8})} \cdot
\inlinematrix{1 & 0 \\ 1 & 0} \cdot
\inlinematrix{\cos(\frac{\pi}{8}) \\ e^{\frac{\pi}{4}} \sin(\frac{\pi}{8})} \\
&\Rightarrow
\sbk{\Delta \sigma_x}^2 &= 1 - \frac{1}{2}^2 \\
&= \frac{3}{4} \\
&\stackrel{\text{analog}}{=} \sbk{\Delta \sigma_y}^2 \\
\sbk{\Delta \sigma_x} \cdot \sbk{\Delta \sigma_y} &= \frac{3}{4} \\
\sbk{\Delta A} \cdot \sbk{\Delta B} &\geq \frac{1}{2} \abs{\expval{\frac{1}{\i} [A,B]}} \\
\frac{1}{2} \abs{\expval{\sigma_z}} &= \frac{1}{\sqrt{2}} &< \frac{3}{4}
\end{align}
\subsection*{c)}
\begin{align}
\ket{\Psi} &= N (\ket{Z+} + e^{\i \alpha} \ket{Z-}) = N \inlinematrix{1 \\ e^{\i \alpha}} \\
\braket{\Psi}{\Psi} &= 1 \\
N^2 \sbk{1 + e^{\i \alpha - \i \alpha)}} &\deq 1 \\
&\Rightarrow \\
N &= \frac{1}{\sqrt{2}} \\ \\
P_+ &= \abs{\braket{x+}{\Psi}}^2 \\
&= \abs{\frac{1}{\sqrt{2}} \inlinematrix{1 \\ 1} \cdot \frac{1}{\sqrt{2}} \inlinematrix{1 \\ e^{\i \alpha}}}^2 \\
&= \frac{1}{4} \abs{1 + e^{\i \alpha}}^2 \\
&= \frac{1}{2} \sbk{1 + \cos(\alpha)}
\end{align}
\section{Aufgabe 8: Teilchen mit Spin 1}
\subsection*{a)}
\includegraphics{grafiken/U_A6_a.pdf}
\subsection*{b)}
$[\Sigma_\alpha, \Sigma_\beta] = \i \Sigma_{\alpha, \beta, \gamma} \Sigma_\gamma$
mit allen $\Sigma_{x,y,z}$ durch testen.
\subsection*{c)}
\begin{align}
\Sigma^2 &= \Sigma_x^2 + \Sigma_y^2 + \Sigma_z^2 \\
&= 2 \one
\end{align}
\subsection*{d)}
\begin{math}
\ket{x+}, \ket{x-}, \ket{x0}, \ket{\Psi} = \ket{n_0}, n(\Theta,\Phi)
\end{math}
\begin{align}
p_0 &\Rightarrow &\ket{x0} &= \frac{1}{\sqrt{2}} \cdot \inlinematrix{1 \\ 0 \\ -1} \\
p_+ &\Rightarrow &\ket{x+} &= \frac{1}{2} \cdot \inlinematrix{1 \\ \frac{1}{\sqrt{2}} \\ 1} \\
p_- &\Rightarrow &\ket{x-} &= \frac{1}{2} \cdot \inlinematrix{-1 \\ \frac{1}{\sqrt{2}} \\ -1}
\vec{\Sigma_n} &= &\Sigma \cdot \vec{n} &= \inlinematrix{\cos(\Theta) & \frac{1}{\sqrt{2}} \sin(\Theta) e^{-\ \phi} & 0 \\
\frac{1}{\sqrt{2}} \sin(\Phi) e^{\i \phi} & 0 & \frac{1}{\sqrt{2}} \sin(\Theta) e^{-\i \phi} \\
0 & \frac{1}{\sqrt{2}} \sin(\Theta) e^{\i \phi} & \cos(\Theta)}
&\Rightarrow
\ket{n_0} & & &= \frac{1}{\sqrt{2}} \inlinematrix{-\sin(\Theta) e^{\i \Phi} \\ \sqrt{2} \cos(\Theta) \\ \sin(\Theta) e^{\i \Phi}}
\end{align}
\begin{align}
p_+ &= \abs{\braket{x_1}{n_0}}^2 &= \ldots \frac{1}{2} \cos^2(\Theta) + \frac{1}{2} \sin^2(\Theta) \sin^2(\Phi) = p_- \\
p_0 &= 1- 2 p_+ &= 1 - 2 p_-
\end{align}
\subsection*{e)}
\begin{align}
\expval{\Sigma_x}_{\ket{\Psi}} &= \dirac{n_0}{\Sigma_x}{n_0} \\
&= +1 + 1 p_+ + p_0 + (-1) p_- \\
&= 0 \\
\ket{\Psi} &= c_1 \ket{x+} + c_2 \ket{x-} + c_3 \ket{x0} \\
\sbk{\Delta \Sigma_x}^2 &= \langle \Sigma_x^2 \rangle - {\langle \Sigma_x \rangle}^2 \\
&= c_2 \inlinematrix{1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 1} %ist das richtig?
\dirac{\Psi}{\Sigma_x^2}{\Psi} &= c_1^2 \cdot 1^2 + c_2^2 \cdot (-1)^2 + c_3^2 \cdot 0^2 \\
&\stackrel{c_1 = c_2}{=} 2 c_1^2 \\
&= \cos^2(\Theta) + \sin^2(\Theta) \cdot \sin^2(\phi) %Ist das richtig?
\end{align}

125
ueb4.tex
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@ -3,14 +3,139 @@
\chapter{Quantenmechanik I - Übungsblatt 4}
\section{Aufgabe 9: Zeitentwicklung eines allgemeinen Zweizustandssystems}
\begin{math}
H = \hbar \inlinematrix{A & B \\ B & -A} \\
A = \Omega \cosb{2 \theta} \\
B = \Omega \sinb{2 \theta} \\
E_\pm = \pm \hbar \Omega
\end{math}
\subsection*{a)}
\begin{align}
\ket{\chi_+} &= \inlinematrix{\sinb{2 \theta} \\ 1 - \cosb{2 \theta}} &= \inlinematrix{\cosb{\theta} \\ \sinb{\theta}} \\
\ket{\chi_-} &= &= \inlinematrix{\sinb{\theta} \\ -\cosb{\theta}}
\end{align}
\subsection*{b)}
\begin{align}
\i \hbar \diffPs{t} \ket{\psi} &= H \ket{\psi} \\
\ket{\dot{\psi}} &= -\i \Omega \inlinematrix{\cosb{2 \theta} & \sinb{2 \theta} \\ \sinb{2 \theta} & -\cosb{2 \theta}} \ket{\psi} \\
\ket{\psi(t)} &= c_1 \cdot e^{\i \Omega t} \ket{\chi+} + c_2 \cdot e^{-\i \Omega t} \ket{\chi-} \\
\ket{\psi(0)} &= \inlinematrix{\lambda \\ \mu} \\
\inlinematrix{c_+(t) \\ c_-(t)} &= \sbk{\lambda \cosb{\theta} + \mu \sinb{\theta}} \cdot \inlinematrix{\cosb{\theta} \\ \sinb{\theta}} \cdot e^{-\i t} +
\sbk{\lambda \sinb{\theta} - \mu \cosb{\theta}} \cdot \inlinematrix{\sinb{\theta} \\ -\cosb{\theta}} \cdot e^{\i \Omega t}
\end{align}
\subsection*{c)}
%hier stimmt evtl. was nicht
\begin{align}
\ket{\psi(0)} &= \inlinematrix{0 \\ 1} \\
\abs{\braket{+}{\psi(t)}}^2 &= \sin^2\sbk{\Omega t} \cdot \sin^2\sbk{2 \theta} \\
\abs{\braket{-}{\psi(t)}}^2 &= \cos^2\sbk{\Omega t} + \sin^2\sbk{\Omega t} \cdot \cos^2\sbk{2 \Omega}
\end{align}
\section{Aufgabe 10: Zerfall eines instabilen Zustandes}
\subsection*{a)}
\begin{align}
<H>_\psi &= \dirac{\psi(t)}{H}{\psi(t)} \\
&= \dirac{\Phi}{e^{\frac{\i H t}{\hbar}} H e^{\frac{-\i H t}{\hbar}}}{\Phi} \\
&mit [H,e^{H}] = 0 \\
&= \dirac{\Phi}{H \cdot e^{\frac{\i H t}{\hbar}} \cdot e^{\frac{-\i H t}{\hbar}} \cdot H}{\Phi} %error: 2 Hs wo nur eins?
&= \dirac{\Phi}{H}{\Phi} \\
<H^2>_\psi &= \dirac{\Phi}{H^2}{\Phi} analog \\
\varianz{H}{\psi}^2 &= \dirac{\Phi}{H^2}{\Phi} - \dirac{\Phi}{H}{\Phi}^2 \\
&\Rightarrow \text{zeitunabhängig} \\
\end{align}
$p \deq p^2 = p \cdot p = \ket{\Phi} \underbrace{\bra{\Phi} \ket{\Phi}}_{=1} \bra{\Phi} = \ket{\Phi} \bra{\Phi} = p$
\subsection*{b)}
\begin{align}
p(t) &= 1 - \frac{\varianz{H}{}^2}{\hbar^2} \cdot t^2 + \bigOb{t^3} \\
&= \abs{\braket{\psi(t)}{\Phi}}^2 \\
&= \abs{\braket{\Phi}{\psi(t)}}^2 - \abs{\dirac{\Phi}{e^{-\frac{\i H t}{\hbar}}}{\Phi}}^2 &\left| Taylor \right. \\
&= \abs{\braket{\Phi}{\Phi} + \dirac{\Phi}{-\frac{\i H t}{\hbar}}{\Phi} +
\dirac{\Phi}{\frac{1}{2} \cdot \sbk{-\frac{\i H t}{\hbar}}^2}{\Phi}}^2 + \bigOb{t^3} \\
&= \abs{1 - \frac{\i t}{\hbar} \dirac{\Phi}{H}{\Phi} - \frac{t^2}{2 \hbar} \dirac{\Phi}{H^2}{\Phi}}^2 \\
&= \sbk{1 - \frac{\i t}{\hbar} \ssbk{H}_\psi - \frac{t^2}{2 \hbar} \ssbk{H^2}_\psi} \cdot
\sbk{1 + \frac{\i t}{\hbar} \ssbk{H}_\psi - \frac{t^2}{2 \hbar} \ssbk{H^2}_\psi} + \bigOb{t^3} \\
&= 1 - \frac{H}{\hbar} t^2 + \bigOb{t^3}
\end{align}
\subsection*{c)}
\begin{align}
\diffT{t} \ssbk{0} &= \frac{\i}{\hbar} \ssbk{[H,0]} + \ssbk{\diffPfrac{0}{t}}
\abs{\diffTfrac{p}{t}} &= \abs{\diffT{t} \ssbk{p}} \\
&= \abs{\frac{\i}{\hbar} \ssbk{[H,p]} + \ssbk{\underbrace{\diffPfrac{p}{t}}_{=0}}} \\
&= \frac{1}{\hbar} \abs{[H,p]} \\
&\leq \frac{2}{\hbar} \Delta H \Delta p \\
&= \frac{2}{\hbar} \Delta H \sqrt{\ssbk{p^2} - \ssbk{p}^2} \\
\abs{\diffTfrac{p}{t}} &= \frac{2}{\hbar} \Delta H \sqrt{p (1 - p)}
\end{align}
\begin{align}
p_0(0) &= 1 \\
\diffTfrac{p_0(t)}{t} &\leq 0 \\
-\diffTfrac{p_0}{t} &= \frac{2 \Delta H}{\hbar} \sqrt{p_0(t) (1 - p_0(t))} \\
\frac{\text{d}p_0}{\sqrt{p_0 (1 - p_0)}} &= -\frac{2 \Delta H}{\hbar} \text{d}t &\left| \text{Integral drüber} \right. \\
\intgr{1}{p_0}{\frac{\text{d}p'_0}{\sqrt{p'_0 (1 - p'_0)}}}{p'} &= -\frac{2 \Delta H}{\hbar} \intgr{0}{t}{}{t'} \\
\arcsinb{2 p_0(t) -1} &= -\frac{2 \Delta H}{\hbar} t + c \\
p_0(t) &= \frac{1}{2} \sinb{-\frac{2 \Delta H}{\hbar} t + c} + \frac{1}{2} %fehler bei 1/2?
&= \frac{1}{2} \sbk{1 + \cosb{\frac{2 \Delta H}{\hbar} t}} \\
&= \cos^2\sbk{\frac{\Delta H}{\hbar} t} \\
&\text{fuer} 0 \leq t \leq \frac{\pi \hbar}{2 \Delta H} \text{gilt:}
p(t) &\geq \cos^2\sbk{\frac{\Delta H}{\hbar} t}
\end{align}
\subsection*{d)}
$h \ket{n} = E_n \ket{n}$ \\
$w(E) = \sum_n \abs{\braket{n}{\Phi}}^2 \delta\sbk{E - E_n}$
\begin{align}
c(t) &= \intgrinf{e^{-\frac{\i E}{\hbar}t} \cdot w(E)}{E} \\
&= \braket{\Phi}{\psi(t)} \\
&= \dirac{\Phi}{e^{-\frac{\i H}{\hbar}t}}{\Phi} &\left| \ket{\Phi} = \sum_n a_n \ket{n} \right. \\
&= \dirac{\Phi}{\sum_n a_n e^{-\frac{\i E_n}{\hbar}t}}{n} \\
&= \sum_m a_m^\ast \sum_n e^{-\frac{\i E}{\hbar} t} \braket{m}{n} a_n \\
\sum_n \abs{a_n}^2 e^{-\frac{\i E_n t}{\hbar}}
\end{align}
\subsection*{e)}
\begin{align}
w(E) &= \frac{\Gamma}{2 \pi} \cdot \frac{1}{\sbk{E - E_n}^2 + \hbar^2 \frac{\Gamma^2}{4}} \\
c(t) &= \frac{\Gamma \hbar}{2 \pi} \intgrinf{\frac{e^{\frac{\i E}{\hbar}t}}{\sbk{E - E_n}^2 + \hbar^2 \frac{\Gamma^2}{4}}}{E} \\
& \text{mit} Z = \frac{2 \sbk{E - E_n}}{\hbar \Gamma} \\
&= \frac{1}{\pi} \intgrinf{\frac{e{-\frac{\i}{\hbar}t \sbk{\frac{\hbar \Gamma Z}{2} + E_0}}}{1 + Z^2}}{Z} \\
&= e^{-\frac{\i E_0}{\hbar}t} \frac{1}{\pi} \intgrinf{\frac{e^{-\frac{\i \Gamma}{2} + Z}}{1 + Z^2}}{Z} \\
&\text{Residuensatz} \\
&= e^{-\frac{\i E_0}{\hbar}t} \frac{1}{\pi} \cdot - 2 \pi \i e^{-\frac{\Gamma}{2}t} \frac{1}{-2 \i} \\
&= e^{-\frac{\i E_0}{\hbar}t} \cdot e^{-\frac{\Gamma}{2}t} \\
p(t) &= e^{-\Gamma t}
\end{align}
\subsection*{f)}
\begin{align}
\probb{H \cequiv E_n}{\ket{\psi(t)}} &= \abs{\braket{n}{\Phi}}^2 \\
w(E) &= \sum_n \abs{\braket{n}{\Phi}}^2 \delta\sbk{E - E_n} \\
\end{align}
(13) Gilt, da diese Def. vom Erwartungswert von $w(E)$ sind (Teile davon) %HÄ?
\begin{align}
\ssbk{H^\alpha} &= \frac{\Gamma \hbar}{2 \pi} \intgrinf{\frac{E^\alpha}{\sbk{E - E_0}^2 + \hbar^2 \frac{\Gamma^2}{4}}}{E} \\
&= \frac{2}{\pi \hbar \Gamma} \intgrinf{\frac{E^\alpha}{1 + \sbk{\frac{2 \sbk{E - E_0}}{\hbar \Gamma}}^2}}{E} \\
& \text{mit} Z = \frac{2 \sbk{E - E_n}}{\hbar \Gamma} \\ \\
&= \frac{1}{\pi} \intgrinf{\frac{\sbk{\frac{\hbar \Gamma Z}{2} + E_0}^\alpha}{\sbk{1 + Z^2}}}{Z} \\
\end{align}
Für $\alpha = 1$
\begin{align}
\ssbk{H} &= \frac{1}{\pi} \intgrinf{\frac{\sbk{\frac{\hbar \Gamma Z}{2} + E_0}}{\sbk{1 + Z^2}}}{Z} \\
&= \intgrinf{\frac{\hbar \Gamma Z}{2 \sbk{1 + Z^2}}}{Z} + \underbrace{\frac{E_0}{\hbar} \underbrace{\intgrinf{\frac{1}{1 + Z^2}}{Z}}_{=\pi}}_{E_0}
\end{align}
Für $\alpha = 2$
\begin{align}
\ssbk{H^2} &\approx \intgrinf{\frac{Z^2}{1 + Z^2}}{Z}
\end{align}
$\ssbk{H^2}$ ist aber nicht definiert \\
$\Rightarrow$ Unschärfe nicht definiert \\
$\Rightarrow$ Annahme endl. Unschärfe falsch

104
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@ -1,16 +1,116 @@
%\includegraphics{excs/qm1_blatt05_SS08.pdf}
%\pagebreak
\includegraphics{excs/qm1_blatt05_SS08.pdf}
\pagebreak
\chapter{Quantenmechanik I - Übungsblatt 5}
\section{Aufgabe 11: Spin-1-Teilchen im konstanten Magnetfeld}
\subsection*{a)}
\begin{align}
H &= - \gamma \mtrx{B} \mtrx{\mtrx{S}} \\
&= \gamma \hbar \mtrx{B} \mtrx{\Sigma_z} \\
\ket{\psi(t)} &= c_+ \inlinematrix{1 \\ 0 \\ 0} \cdot e^{\i \gamma \mtrx{B} t} + c_0 \inlinematrix{0 \\ 1 \\ 0} +
c_- \inlinematrix{0 \\ 0 \\ 1} \cdot e^{-\i \gamma \mtrx{B} t} \\
&\deq \frac{1}{2} \inlinematrix{1 \\ \sqrt{2} \\ 1} \\
&= \inlinematrix{\frac{1}{2} \cdot e^{\i \gamma \mtrx{B} t} \\ \frac{\sqrt{2}}{2} \\ \frac{1}{2} \cdot e^{-\i \gamma \mtrx{B} t}} \\
\probb{\Sigma_z \cequiv 1}{\psi(t)} &= \abs{\braket{Z_1}{\psi(t)}}^2 \\
&= \inlinematrix{1 \\ 0 \\ 0} \cdot \inlinematrix{\frac{1}{2} \cdot e^{\i \gamma \mtrx{B} t}
\\ \frac{\sqrt{2}}{2} \\ \frac{1}{2} \cdot e^{-\i \gamma \mtrx{B} t}}
&= \abs{\frac{1}{2} e^{\i \gamma \mtrx{B} t}}^2 \\
&= \frac{1}{4} \\
\probb{\Sigma_z \cequiv 0}{\psi(t)} &= \frac{1}{2} \\
\probb{\Sigma_z \cequiv -1}{\psi(t)} &= \frac{1}{4} \\
\probb{\Sigma_x \cequiv +1}{\psi(t)} &= \abs{\braket{x+}{\psi(t)}}^2 \\
&= \abs{\frac{1}{2} \inlinematrix{1 \\ \sqrt{2} \\ 1} \cdot \inlinematrix{\frac{1}{2} \cdot e^{\i \gamma \mtrx{B} t} \\
\frac{\sqrt{2}}{2} \\ \frac{1}{2} \cdot e^{-\i \gamma \mtrx{B} t}}}^2 \\
&= \frac{1}{4} \sbk{1 + e^{\i \gamma \mtrx{B}} + e^{-\i \gamma \mtrx{B} t}}^2 \\
&= \frac{1}{4} \sbk{1 + \cosb{\gamma \mtrx{B} t}}^2
\end{align}
$\Rightarrow$
\begin{align}
\probb{\Sigma_x \cequiv +1}{\psi(t)} &= \frac{1}{2} \sin^2\sbk{\gamma \mtrx{B} t} \\ %da stimmt wat net
\probb{\Sigma_y \cequiv +1}{\psi(t)} &= \frac{1}{4} \sbk{1 - \cosb{\gamma \mtrx{B} t}}^2
\end{align}
\subsection*{b)}
\begin{align}
\ssbk{\Sigma_z}_\ket{\psi(t)} &= \dirac{\psi(t)}{\Sigma_z}{\psi(t)} \\
&= 0 \\
\ssbk{\Sigma_x}_\ket{\psi(t)} &= \cosb{\gamma \mtrx{B} t} \\
\ssbk{\Sigma_y}_\ket{\psi(t)} &= -\sinb{\gamma \mtrx{B} t} \\
\diffT{t} \ssbk{\mtrx{\Sigma}} &= \gamma \ssbk{\mtrx{\Sigma}} \times \mtrx{B} \\
\diffT{t} \inlinematrix{\cosb{\gamma \mtrx{B} t} \\ -\sinb{\gamma \mtrx{B} t} \\ 0} &= \gamma \inlinematrix{\ssbk{\Sigma_y} \mtrx{B} \\ \ssbk{\Sigma_x} \mtrx{B} \\ 0} \\
\inlinematrix{-\gamma \mtrx{B} \sinb{\gamma \mtrx{B} t} \\ -\gamma \mtrx{B} \cosb{\gamma \mtrx{B} t} \\ 0} &= \gamma \inlinematrix{\ssbk{\Sigma_y} \mtrx{B} \\ \ssbk{\Sigma_x} \mtrx{B} \\ 0}
\end{align}
\subsection*{c)}
\begin{align}
\frac{\i}{\hbar} \ssbk{\sqbk{H, \mtrx{Sigma}}} &= -\i \gamma \ssbk{\sqbk{\mtrx{B} \cdot \mtrx{\Sigma}, \mtrx{\Sigma}}} \\
A_i &= \i \gamma \ssbk{B_j \Sigma_j \Sigma_i - \Sigma_i \Sigma_j B_j} \\
&= -\i \gamma B_j \ssbk{\sqbk{\Sigma_j, \Sigma_j}} \\
&= -\i \gamma B_j \ssbk{\i \levicivita{jik} \Sigma_k} \\
&= \gamma B_j \levicivita{jik} \ssbk{\Sigma_k} \\
&= \gamma \sbk{\ssbk{\mtrx{\Sigma}} \times \mtrx{B}} \\
&= \diffT{t} \ssbk{\Sigma_i} \\
&= -\gamma \levicivita{ijk} B_j \ssbk{\Sigma_k} \\
&= \gamma \levicivita{ikj} \ssbk{\Sigma_k} B_j \\
&= \gamma \levicivita{ijk} \ssbk{\Sigma_j} B_k \\
\diffT{t} l &= \left\{l,H\right\}
\end{align}
\section{Aufgabe 12: Das Ethen-Molekül}
$H = \inlinematrix{a & b \\ c & d}$
$\Rightarrow$
\begin{align}
\inlinematrix{a \\ c} &= \inlinematrix{E_0 \\ -A} \\
\inlinematrix{b \\ d} &= \inlinematrix{-A \\ E} \\
H &= \inlinematrix{E_0 & -A \\ -A & E}
\end{align}
Die Eigenwerte und Eigenvektoren von $H$ sind:
\begin{align}
\lambda_1 &= E_0 + A \\
\vec{\lambda_1} &= \frac{1}{\sqrt{2}} \inlinematrix{1 \\ -1} \\
\lambda_2 &= E_0 - A \\
\vec{\lambda_2} &= \frac{1}{\sqrt{2}} \inlinematrix{1 \\ 1}
\end{align}
Die Gesamtenergie beträgt dann:
\equationblock{E\sbk{\ket{G}} = 2 E_0 - 2 A}
Hierbei ist A die Delokalisierungsenergie (Kopplungskoeffizienten?)
\section{Aufgabe 13: Das Benzol-Molekül}
\subsection*{a)}
$H \ket{\Phi_1} = \sum \ket{\Phi_n} - A \ket{\Phi_{n-1}} - A \ket{\Phi_{n+1}}$
Kopplung besteht immer mit den benachbarten Atomen. Daher ist (1,6) bzw. (6,1) belegt (Kopplung von 6 mit 1)
\subsection*{b)}
$R = \inlinematrix{1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \ddots & 0 & 0 \\ 0 & \ddots & \ddots & 0 & 0 \\ 0 & \ddots & \ddots & 0 & 0 \\ 1 & 0 & 0 & 1 & 1}$
$\sqbk{H,R} = 0$
\begin{align}
H &= a \cdot \one + b \mtrx{R} + c \mtrx{R^\dagger} \\
H &= E_0 \one - A \mtrx{R} - A \mtrx{R^\dagger}
\end{align}
Symmetrien bekannt:
$R^6 = \one$
$\Rightarrow$
$R^6 \ket{\Phi} = 1 \ket{\Phi}$
$R \ket{\Phi} = \underbrace{e^{\i \frac{2 \pi s}{6}}}_{\text{Eigenwerte}} \ket{\Phi}$ mit $s = 0 \ldots 5$
\subsection*{c)}
\begin{align}
\ket{\chi_s} &= \frac{1}{\sqrt{6}} \sum_{n=0}^5 e^{\i n \delta_s} \ket{\Phi_n} \\
\detb{\mtrx{R} - \sbk{\lambda_s \one}} &\Rightarrow \\
e^{\i \delta_s} x_1 &= -x2 \\
e^{\i \delta_s} x2 &= -x3 \\
\vdots \\
\end{align}
Die Eigenvektoren lauten dann:
\equationblock{\frac{1}{\sqrt{6}} \inlinematrix{1 \\ -e^{\i \delta_s} \\ e^{2 \i \delta_s} \\ -e^{3 \i \delta_s} \\ \vdots }}
\subsection*{d)}
\begin{align}
H \ket{\chi_s} &= E_0 \one \ket{\chi_s} - A \mtrx{R} \ket{\chi_s} - A \mtrx{R^\dagger} \ket{\chi_s} \\
&= E_0 \ket{\chi_s} - A \lambda_s \ket{\chi_s} - A \lambda_3^\ast \ket{x_3} \\
&= E_0 - 2 A \re{\lambda_s \ket{\chi_s}} \\
&= E_0 - 2 A \cosb{\delta_s} \\
&= E_s
\end{align}
Die Gesamtenergie beträgt dann:
\equationblock{E_ges = 6 E_0 - 8 A}
$E_{Kekule} = 3 \sbk{E_{Ethen}} = 6 E_0 - 6 A$ (Pauli-Prinzip)

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@ -12,11 +12,11 @@
\end{align}
$det(H-\lambda \einsmatrix) = 0 \\$
$det(H-\lambda \einsmatrix) = 0$
\begin{align}
\lambda_1 &= B\hbar^2 \\
\lambda_2 &= \frac{\hbar^2}{2} B (1 + \sqrt{1 + \frac{4 A^2}{\hbar^2 B^2}} \\
\lambda_3 &= \frac{\hbar^2}{2} B (1-\sqrt{1 + \frac{4 A^2}{\hbar^2 B}}
\lambda_3 &= \frac{\hbar^2}{2} B (1-\sqrt{1 + \frac{4 A^2}{\hbar^2 B})}
\end{align}
\begin{align}
@ -40,7 +40,7 @@ $\Rightarrow $
S_x &= \hbar \Sigma_x &= \hbar \inlinematrix{0 &1 &0 \\ 1 &0 &1 \\ 0 &1 &0}
\end{align}
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
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
@ -53,7 +53,7 @@ Da $\inlinematrix{l \\ o \\ m} S_x \inlinematrix{n \\ o \\ p} = 0}$ folgt $S_x =
\end{align}
\begin{align}
\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)]}
\dirac{\psi(t)}{S_z}{\psi(t)} &= \frac{1}{\sbk{a^2+b^2}^2} [(a^2-b^2)^2 + 4 a^2 b^2 \cosb{(E_+-E_-)t}]
\end{align}
@ -62,8 +62,8 @@ Da $\inlinematrix{l \\ o \\ m} S_x \inlinematrix{n \\ o \\ p} = 0}$ folgt $S_x =
\subsection*{a)}
\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} \\
\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)}} \\
@ -96,24 +96,20 @@ Da $\inlinematrix{l \\ o \\ m} S_x \inlinematrix{n \\ o \\ p} = 0}$ folgt $S_x =
\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}
\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} \\
\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') \\
&= \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?
%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}
\i \hbar

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@ -6,22 +6,21 @@
\subsection*{a)}
\begin{math}
\Phi_n(p) = \intgrinf{\frac{1}{\sqrt{2 \pi \hbar}} \cdot \Phi(x) \cdot e^{-\frac{\i p x}{\hbar}}{x}
\Phi_n(p) = \intgrinf{\frac{1}{\sqrt{2 \pi \hbar}} \cdot \Phi(x) \cdot e^{-\frac{\i p x}{\hbar}}}{x}
\end{math}
Für n = ungerade:
\begin{align}
\Phi_n(p) &= \integrinf{\frac{1}{\sqrt{2 \pi a \hbar}} \cdot \cos(\frac{(n+1) \cdot \pi x}{2 a}) \cdot e^{-\frac{\i p x}{\hbar}}}{x} \\ \\
\Phi_n(p) &= \intgrinf{\frac{1}{\sqrt{2 \pi a \hbar}} \cdot \cos(\frac{(n+1) \cdot \pi x}{2 a}) \cdot e^{-\frac{\i p x}{\hbar}}}{x} \\ \\
&= \frac{1}{\sqrt{2 \pi a \hbar}} \cdot \sbk {
\frac{1}{\sbk{\frac{(n+1) \cdot \pi}{2 a}}-\frac{p}{\hbar}} \cdot \sin \sbk{\sbk{\frac{(n+1) \cdot \pi}{2 a}}-\frac{p}{\hbar} \cdot u} +
\frac{1}{\sbk{\frac{(n+1) \cdot \pi}{2 a}}+\frac{p}{\hbar}} \cdot \sin \sbk{\sbk{\frac{(n+1) \cdot \pi}{2 a}}+\frac{p}{\hbar} \cdot u}}
\Phi_n(p) &=
\begin{case}
\begin{cases}
\frac{1}{\sqrt{2 \pi a \hbar}} \cdot \frac{\i^n \cdot 4 \cdot a \hbar^2 \cdot (n+1) \cdot \pi}{\hbar^2 (n+1)^2 \pi - 4 a^2 p^2} \cdot \cos\sbk{\frac{pa}{\hbar}}
\i^{n+2} \cdot \frac{\i^n \cdot 4 \cdot a \hbar^2 \cdot (n+1) \cdot \pi}{\hbar^2 (n+1)^2 \pi - 4 a^2 p^2} \cdot \sin\sbk{\frac{pa}{\hbar}}
\end{case}
\end{cases}
\end{align}
@ -74,20 +73,59 @@ mit den \hyperlink{fs_mtrx_inv_2d}{Inversen} von: (1) und (2) % geschweifte klam
\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}
\subsection*{c)}
$\inlinematrix{C & D}$ 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)} \\
\tilde{E} &= \frac{2 k a}{e^{ika} \sbk{(q^2 - k^2) \cdot \i sinh(q \cdot a) - 2 k q cosh(qa)}} \\ % ist das richtig?
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)}
T &= \sbk{\frac{E}{A}}^2 \\
R &= 1 - T &= \sbk{\frac{B}{A}}^2 \\
T &= \frac{1}{\sbk{\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}
\section{Aufgabe 19: Doppeltes Delta-Potential}
$V(x) = - \frac{\hbar^2}{m} \kappa_0 \sbk{\delta\sbk{x -a} + \delta\sbk{x + a}}$
$E<0$
$\sbk{-\frac{\hbar^2}{2 m} \diffPs{x}^2 - \frac{\hbar^2}{m} \kappa_0 \sbk{\delta\sbk{x - a} + \delta\sbk{x + a}}} \Phi(x) = E \Phi(x)$
$\Phi'_I(-a) - \Phi'_{II}(-a) = - 2 \kappa_0 \Phi_I(-a)$
Mit $k^2 = \frac{-2 m E}{\hbar^2}$
Für I:
\begin{align}
\psi_I &= A e^{k x} \\
\psi_{II} &= B e^{-k x} + c e^{k x} \\
\psi_{III} &= D e^{-k x} \\
\abs{A} &= \abs{D} \\
A &= \pm D \\
B &= \pm C \\
0 &= B e^{2 k a} - C + A \sbk{1 - 2 \frac{k_0}{k}} \\
0 &= C e^{2 k a} - B + D \sbk{1 - 2 \frac{k_0}{k}} \\
\frac{k_0}{k} \sbk{e^{- 2 k a} \pm 1} &= \pm 1 \\
e^{-2 k a} &= \frac{k}{k_0} - 1 \\
e^{-2 k a} &= -\frac{k}{k_0} + 1 \\
\end{align}
%grafik einfügen
\begin{align}
\kappa_0 &= 1 \\
\kappa_1 &= 1,11 \\
\kappa_2 &= 0,80 \\
E &= -\frac{k^2 \hbar^2}{2 m}
\end{align}
\subsection*{a)}
\subsection*{b)}
$k_0 a \gg 1$ $\Rightarrow$ $\kappa \approx \kappa_0$
$k_\pm = k_0 \sbk{1 \pm e^{-2 \kappa_0 a}}$
\subsection*{c)}
\subsection*{d)}
$a \rightarrow 0$
\begin{align}
\kappa_+ &\approx \kappa_0 \sbk{1 + 1 - 2k + a} \\
&\approx 2 \kappa_0
\end{align}

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@ -0,0 +1,143 @@
\chapter{Quantenmechanik I - Übungsblatt 8}
\section{Aufgabe 20: Dirac-Kamm}
\subsection*{a)}
Für $0 < x < a$ und $V(x) = 0$:
$-\frac{\hbar^2}{2 m} \Phi'(x) = E \Phi(x)$
$\Rightarrow$
$\Phi(x) = A \cdot \sinb{k x} + B \cdot \cosb{k x}$ mit $k = \sqrt{\frac{2 m E}{\hbar}}$
Für $0 > x > -a$:
$\Phi(x) = e^{-\i K a} \cdot \sbk{ A \sinb{k a} + B \cosb{k a}}$
$k A - e^{-\i K a} \sbk{A \cosb{k a} + B \sinb{k a}} = -\frac{2 n \alpha B}{\hbar^2}$
$\mu \inlinematrix{A \\ B} = \inlinematrix{0 \\ 0}$
$\inlinematrix{A \\ B} = \frac{1}{\mu} \inlinematrix{0 \\ 0}$
$\Rightarrow$
$A=0$ und $B=0$
$\Rightarrow$
$\detb{\mu} = 0$ da sonst $A=B=0$.
$\cosb{K a} = \cosb{k a} - \frac{n \alpha}{\hbar k} \sinhb{a}$
$\cosb{Z} = \cosb{Z} - \beta \frac{\sinb{Z}}{Z}$
\subsection*{b)}
%grafik plotten und hier einfügen
\subsection*{c)}
\begin{align}
\cosb{K a} &= \cosb{k a} - B \frac{\sinb{k a}}{k a} &\left| k = \i k \right. \\
\cosb{K a} &= \cosb{k a} - B \frac{\sinhb{k a}}{k a}
\end{align}
\section{Aufgabe 21: Lineares Potential}
\subsection*{a)}
$V(x) = m g x$
Die Schrödingergleichung lautet:
\equationblock{\sbk{-\frac{\hbar^2}{2 m} \diffPs{x}^2 + V(x + a)} \psi(x + a) = E \psi(x + a)}
$\Rightarrow$
\equationblock{\sbk{-\frac{\hbar^2}{2 m} \diffPs{x}^2 + V(x)} \psi(x + a) = \sbk{E - V(a)} \psi(x + a)}
$E = m g a$
\subsection*{b)}
\begin{math}
\inlinematrix{kg \\ m \\ s}
m = \sqbk{kg} \inlinematrix{1 \\ 0 \\ 0}
g = \ssbk{\frac{m}{s^2}} = \inlinematrix{0 \\ 1 \\ -2}
\hbar = \ssbk{Js} = \ssbk{\frac{kg m^2}{s}} = \inlinematrix{1 \\ 2 \\ -1}
T_0 = \inlinematrix{0 \\ 0 \\ 1} \inlinematrix{1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & -2 & 1} \inlinematrix{0 \\ 0 \\ 1}
T_0 = \sqrt[3]{\frac{\hbar}{m g ^2}}
L_0 = \sqrt[3]{\frac{\hbar^2}{m^2 g}}
E_0 = \sqrt[3]{m \hbar^2 g^2}
P_0 = \sqrt[3]{m^2 \hbar g}
\end{math}
%
\begin{align}
\sbk{-\frac{\hbar^2}{2 m} J^2_{L_0y} + m g L_0 y} \frac{\psi(y)}{\sqrt{L_0}} &= \epsilon E_0 \frac{\psi(y)}{\sqrt{L_0}} \\
\sbk{-\frac{1}{2} E_0 J_y^2 + E_0 y} \psi(y) &= \epsilon E_0 \psi(y) \\
\sbk{-\frac{1}{2} J_y^2 + y} \psi(y) &= \epsilon \psi(y) \\
\psi(p) &= \frac{1}{\sqrt{p}} \Phi \\
\frac{1}{2} k^2 \Phi(k) + \i \diffPfrac{\Phi(k)}{k} &= \epsilon \Phi(k) \\
\partial y &\Rightarrow \i k \\
y &\Rightarrow \i \partial k \\
\Phi(p) &= \intgrinf{\frac{1}{2 \pi \hbar} e^{-\frac{\i}{\hbar} p x} \Phi(x)}{x} \\
\frac{\Phi(k)}{\sqrt{p_0}} &= \intgrinf{\frac{1}{L_0 2 \pi \hbar} e^{\frac{\i}{\hbar} y L_0 k p_0} \Phi(y)}{y L_0} \\
\Phi(k) &= \sqrt{\frac{p_0 L_0}{2 \pi \hbar}} \intgrinf{e^{-\frac{\i}{\hbar} L_0 I_0 k y} \Phi(y)}{y} \\
&= \frac{1}{\sqrt{2 \pi}} \intgrinf{e^{\i k y} \Phi(y)}{y}
\end{align}
%
\subsection*{c)}
$\sbk{\frac{1}{2} k^2 + \i \diffPs{k}} \Phi_0(k) = 0$
Trennung der Variable führt auf:
$\Phi_0(k) = c \cdot e^{\frac{\i}{6} k^3}$
Fourriertansformation:
\begin{align}
\psi_0(y) &= \frac{1}{\sqrt{2 \pi}} \intgrinf{e^{\i k y} \Phi(k)}{k} \\
&= \frac{c}{\sqrt{2 \pi}} \intgrinf{\sbk{\cosb{\frac{1}{6} k^3 + k y} + \i \sinb{\frac{1}{6} k^3 + k y}}}{k} \\
\end{align}
da $\intgrinf{\sinb{x}}{x} = 0$
\begin{align}
\psi_0(y) &= \frac{c}{\sqrt{2 \pi}} \intgrinf{\cosb{\frac{1}{6} k^3 + k y}}{k}
\end{align}
mit $\frac{1}{6} k^3 = \frac{1}{3} \tilde{k}^3$
\begin{align}
\psi_0(y) &= c \frac{\sqrt[6]{2^5}}{\sqrt{\pi}} \intgrinf{\cosb{\frac{1}{3} \tilde{k} + \sqrt[3]{2} \tilde{k} y}}{\tilde{k}} \\
&= 2^6 \sqrt{\pi} \airyb{\sqrt[3]{2} y} \\
&= \psi_0(y)
\end{align}
aus Teil a):
$\psi_\epsilon(x) = \psi_0(x - \frac{E}{m g})$
folgt somit:
\begin{align}
\psi_\epsilon(y) &= \sqrt{L_0} \psi_0(L_o y - \epsilon L_0) \\
&= \psi_0(y - \epsilon) \\
&= 2^\frac{5}{6} \sqrt{\pi} c \airyb{\sqrt[3]{2} \sbk{y - \epsilon}} \\
\Phi_\epsilon(t) &= e^{-\i k \epsilon} \Phi_0{k} \\
&= c \cdot e^{\i \sbk{\frac{k^3}{6} - \epsilon k}}
\end{align}
\subsection*{d)}
\begin{align}
\intgrinf{\Phi_\epsilon^\ast(k) \Phi_\epsilon(k)}{k} &= \intgrinf{\abs{A}^2 e^{\i k \sbk{\epsilon' - \epsilon}}}{k} \\
&= \intgrinf{\frac{1}{\sqrt{2 \pi}} e^{\i k \sbk{\epsilon' - \epsilon}}}{k} \\
A &= \frac{1}{\sqrt{2 \pi}} \\
\intgrinf{\Phi_\epsilon^\ast(k') \Phi_\epsilon(k)}{\epsilon} &=
\frac{1}{2 \pi} \intgrinf{e^{\i \sbk{k' - k} \epsilon} e^{-\frac{1}{6} \sbk{k'^3 - k^3}}}{\epsilon} \\
&= e^{-\frac{1}{6} \sbk{k'^3 - k^3}} \intgrinf{\frac{e^{\i \epsilon \sbk{k' - k}}}{2 \pi}}{\epsilon} \\
&= \delta\sbk{k' - k} \\
\psi_\epsilon(x) &= \frac{\sqrt[3]{2 m \sqrt{g}}}{\hbar} \airyb{\sqrt[3]{\frac{2 g m^2}{\hbar^2}} \sbk{x - \frac{E}{m g}}}
\end{align}
\subsection*{e)}
$\sbk{-\frac{1}{2} \diffPs{y}^2 + y} \Psi(y) = 0$
\subsubsection*{i)}
$e^{f(y)}$
$f''(y) + \sbk{f'(y)}^2 = 2 y$
\subsubsection*{ii)}
$g'(y) + g^2(y) = 2 y$
$g(y) = c_1 y^{\lambda_1}$
Potenzreihenansatz:
$c_1 \lambda_1 y^{\lambda_1 -1} + c_1^2 y^{2 \lambda_1} = 2 y$
Asymptotik muss gleich sein:
\begin{align}
2 \lambda_1 &= 1 \Rightarrow \lambda_1 = \frac{1}{2}\\
c_1^2 &= 2 \Rightarrow c_1^2 = \pm \sqrt{2} \\
g_1(y) &= \pm \sqrt{2 y}
\end{align}
\subsubsection*{iii)}
$g_1(y) = g_0(y) + c_2 y^{\lambda_2} = \pm \sqrt{2 y} + c_2 y^{\lambda_2}$
$\pm \frac{1}{\sqrt{2}} y^{-\frac{1}{2}} + \lambda_2 c_2 y^{\lambda_2 -1} \pm 2 \sqrt{2} c_2 y^{\frac{1}{2} + \lambda_2} + c_2^2 y^{2 \lambda_2} = 0$
$\pm \frac{1}{\sqrt{2}} y^{-\frac{1}{2}} \sbk{1 \pm \sqrt{2} \lambda_2 c_2 y^{\lambda_2 - \frac{1}{2}} + 4 c_2 y^{\lambda_2 -1} \pm \sqrt{2 c_2^2 y^{2 \lambda + \frac{1}{2}}}} = 0$
$\lambda_2 = -1$
$c_2 = -\frac{1}{4}$
$g(y) = \pm \sqrt{2 y} - \frac{1}{4 y}$
$f(y) = - \frac{2 \sqrt{2}}{3} y^{\frac{3}{2}} - \frac{\lnb{y}}{4} + c$
$\Rightarrow$
$\psi_0(y) 2 e^{-\frac{2 \sqrt{2}}{3}} y^{\frac{3}{2}} \frac{1}{\abs{y}^4}$
\subsubsection*{iv)}
$V(x) = \begin{cases}
m g x & x > 0 \\ \infty & x \leq 0
\end{cases}$
$\psi(0) \deq 0$
$\psi_E(x) = \airyb{\frac{\sqrt[3]{2 g m^2}}{\hbar^2} \cdot \sbk{-\frac{E}{m g}}}$
$-\frac{E}{m g} \frac{\sqrt[3]{2 g m^2}}{\hbar^2} \approx -2,3$
$\Rightarrow$
$E \approx 1,41 peV$
\subsection*{f)}

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\chapter{Quantenmechanik I - Übungsblatt 9}
\section{Aufgabe 22: Elektron im ortsunabhängigen Magnetfeld:}
\subsection*{a)}
\begin{align}
H &= H_0 \otimes \one - \sum_{n=1}^2 \ket{n} \bra{n} \otimes \frac{\hbar \omega_n}{2} \sigma_z &\left| \omega_n = \gamma B_n \right. \\
&= \inlinematrix{0 & -A \one \\ -A \one & 0} - \inlinematrix{\hbar \gamma B_1 \frac{\sigma_z}{2} & 0 \\ 0 & -\hbar \gamma B_2 \frac{\sigma_2}{2}} \\
&= \inlinearray{cc|cc}{-\frac{\hbar \omega_1}{2} & -A & 0 & 0 \\ -A -\frac{\hbar \omega_2}{2} & 0 & 0 \\ \hline
0 & 0 & \frac{\hbar \omega_1}{2} & -A \\ 0 & 0 & -A & \frac{\hbar \omega_2}{2}} \\ \\
&= \inlinematrix{H\uparrow & 0 \\ 0 & H\downarrow}
\end{align}
\subsection*{b)}
\equationblock{\lambda_{1 \ldots 4} \frac{\pm \sqrt{\hbar^2 \sbk{\omega_1-\omega_2}^2 + 16 A^2} \pm \hbar \sbk{\omega_1 + \omega_2}}{4}}
\subsection*{c)}
$\pm \sqrt{x^2 + 1} \pm 2 x$
%da kommt nen bild rein
\section{Aufgabe 23: Kohärente Zustände (Glauber-Zustände):}
\subsection*{a)}
%alles richtig hier?
$a \ket{\alpha} = \alpha \ket{\alpha}$
$\hat{n} = a^\dagger a$
$\ket{n} = \frac{1}{\sqrt{n!}} \sbk{a^dagger}^n \ket{0}$
$\braket{\alpha}{\alpha} = 1$
\begin{align}
a \ket{\alpha} &= \sum_{n=0} c_n(x) a \ket{n} &\left| a \ket{n} = \sqrt{n} \ket{n-1} \right. \\
&= \sum_{n \geq 0} c_n(x) \sqrt{n} \ket{n-1} \\
&= \alpha \ket{\alpha} \\
&= \alpha \sum_{n \geq 0} c_n(\alpha) \ket{n} \\
&= \sum_{n \geq 0} \alpha \braket{n(\alpha)}{n} \\
&= \sum_{n \geq 0} \alpha \braket{n-\i \alpha}{n - 1} \\ %HÄ?
\end{align}
$\Rightarrow$
\begin{align}
c_n(\alpha) \sqrt{n} &= \alpha c_{n-1}(\alpha) \\
c_n(\alpha) &= \frac{\alpha}{\sqrt{n}} c_{n-1}(\alpha) \\
&= \frac{\alpha^2}{\sqrt{n!}} c_0(alpha) \\
\ket{\alpha} &= \sbk{\sum_{n \geq 0} \frac{\alpha^n}{\sqrt{n!}} \ket{n}} \cdot c_0(\alpha) &\left| \bra{\alpha} \right. \\
c_0(\alpha) &= e^{-\abs{\alpha}^2 \frac{1}{2}} \\
\sum_{n \geq 0} \frac{\alpha^n}{\sqrt{n!}} \ket{n} &= \sum_{n \geq 0} \frac{\alpha^n}{\sqrt{n!}} \frac{1}{\sqrt{n!}} \sbk{a^\dagger}^n \ket{0} \\
&= e^{\alpha a^\dagger} \ket{0} \\
\ket{\alpha} &= \underbrace{e^{-\frac{1}{2} \abs{\alpha}^2} e^{\alpha a^\dagger}}_{c(\alpha)} \ket{0}
\end{align}
\subsection*{b)}
\begin{align}
\dirac{\alpha}{\hat{n}}{\alpha} &= \dirac{\alpha}{a^\dagger a}{\alpha} \\
&= \braket{a \alpha}{a x} \\
&= \braket{\alpha}{\alpha} \alpha^\ast \alpha \\
&= \abs{\alpha}^2 \\
\sbk{\Delta \hat{n}}^2 &= \dirac{\alpha}{\hat{n}^2}{\alpha} - \abs{\alpha}^4 \\
&= \dirac{\alpha}{a^\dagger a a^\dagger a}{x} - \abs{\alpha}^4 \\
&= \abs{\alpha}^2 \dirac{\alpha}{a a^\dagger}{\alpha} - \abs{\alpha}^4 \\
&= \abs{\alpha}^2 \sbk{\abs{\alpha}^2 + 1} - \abs{\alpha}^4 \\
&= \abs{\alpha}^2 \\
&= \ssbk{\hat{n}}
\end{align}
%fehlt da was?
\subsection*{c)}
\begin{align} %komisch hier. sicher viel falsch
\ket{\psi(0)} &= \sum_{n=0}^{\infty} c_r \ket{n}
\ket{\psi(t)} &= e^{-\frac{\i}{\hbar} H t} \\
&= \sum_{n=0}^{\infty} e^{-\i \sbk{n + \frac{1}{2}} \omega t} \\
&= e^{-\frac{\abs{\alpha}^2}{\hbar}} e^{-\frac{\i}{2} \omega t} \sum_{n=0}^{\infty} \frac{\alpha^n e^{-\i \omega t n}}{\sqrt{n!}} \ket{n} \\
&= e^{\alpha \i \frac{1}{2} e^{-\i \omega t} I^2} \sum_{n=0}^\infty \frac{\sbk{\alpha e^{-\i \omega t}}^2}{\sqrt{n!}} \ket{n}
\ket{\alpha} &= e^{-\frac{\abs{\alpha}^2}{2}} \sum_n \frac{\alpha^n}{\sqrt{n!}} \ket{n} \\
\ket{\alpha(t)} &= \sum_n e^{-\i \sbk{n + \frac{1}{2}} \omega t} \frac{\alpha^n}{\sqrt{n!}} \ket{n} c_0
\end{align}
mit $\tilde{\alpha} = \alpha e^{-\i \omega t}$
%anderes blatt
\begin{align}
\left<x\right> &= \dirac{\tilde{\alpha}(t)}{\frac{1}{\sqrt{2}} \sbk{a + a^\dagger}}{\tilde{\alpha}(t)}
\ket{\psi(t)} &= e^{-\frac{\i H t}{\hbar}} \ket{\alpha_0} \\
&= \sum_n c_0 \frac{\alpha_0^2}{\sqrt{n!}} e^{-\frac{\i \sbk{n +\frac{1}{2}} \omega t}{\hbar}} \ket{n} \\
&= e^{-\i \frac{1}{2} \omega t} \ket{\alpha(t)} \\
\left<x\right> &= \dirac{\alpha(t)}{\frac{1}{\sqrt{2}} \sbk{a + a^\dagger}}{\alpha(t)} \\
&= \frac{1}{\sqrt{2}} \sbk{\alpha^\ast(t) + \alpha(t)}
\end{align}
mit $\alpha(t) = \alpha e^{-\i \omega t}$
\begin{align}
\ssbk{x} &= \frac{\alpha_0}{\sqrt{2}} \sbk{e^{\i \omega t} + e^{-\i \omega t}} \\
&= \frac{a_0}{\sqrt{2}} 2 \cosb{\omega t} \\
\ssbk{p} &= \dirac{\alpha(t)}{\frac{\i}{\sqrt{2}} \sbk{a - a^\dagger}}{\alpha(t)} \\
&= - \sqrt{2} a_0 \sinb{\omega t}
\ssbk{x^2} &= \dirac{\alpha(t)}{\frac{1}{2} \sbk{a a + a^\dagger a^\dagger - a a^\dagger + a^\dagger a}}{\alpha(t)} \\
&= \frac{1}{2} \sbk{\sbk{\alpha^\ast + \alpha}^2 + 1} \\
&= \sbk{1 + \ssbk{\hat{x}}^2} \frac{1}{2}
\ssbk{p^2} &= \sbk{\ssbk{p}^2 + 1} \frac{1}{2}
\sbk{\Delta x}^2 &= \left<x^2\right> - \left<x\right>^2 &= \frac{1}{2} \\
\sbk{\Delta p}^2 &= \frac{1}{2} \\
\Delta p \Delta x &= \frac{1}{2}
\end{align}
\subsection*{d)}
\begin{align}
\braket{\alpha'}{\alpha} &= \sum_m \sum_n c_m^\ast(\alpha) c_n(\alpha) \braket{m}{n} \\
&= c_0^\ast(\alpha) c_0^\ast(\alpha!) \sum_n \frac{\sbk{\alpha^\ast \alpha}^n}{n!} \\
&= e^{-\frac{1}{2} \abs{\alpha}^2} - \frac{1}{2} \abs{\alpha}^2 + \alpha^\ast \alpha \\
\intgrinf{\ket{\alpha} \braket{\alpha}}{^2\alpha} &= \sum_m \sum_n \frac{1}{\sqrt{m!} \sqrt{n!}} \ket{n} \bra{m} \intgrinf{\sbk{\alpha^ast}^m \sbk{\alpha^\ast}^n e^{- \abs{\alpha}^2}}{^2\alpha} \\
&= \sum_m \sum_n \frac{1}{\sqrt{m!} \sqrt{n!}} \ket{n} \bra{m} \intgrinf{\abs{\alpha}^{m+n} e^{\i \sbk{n - m} \phi} e^{-\abs{\alpha}^2}}{\alpha} \\
&= \sum_m \sum_n \frac{1}{\sqrt{m!} \sqrt{n!}} \ket{n} \bra{m}
\int_\infty^\infty ~\text{ d}\alpha \int_\infty^\infty ~\text{ d}\phi e^{\i \sbk{n -m} \phi} \abs{\alpha}^{m+n} e^{-\abs{\alpha}^2}
&= \Gamma \frac{m x_1 + 1}{4}
\end{align}