Daniel Bahrdt cfe4db4633
2008-07-24 20:55:46 +02:00

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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 &= \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 &= \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 - \left^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}