\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)}