\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}
