übungsblatt 11
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Daniel Bahrdt
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ueb11.tex
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ueb11.tex
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\chapter{Quantenmechanik I - Übungsblatt 11}
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\section{Aufgabe 26: Oszillator im elektrischen Feld}
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\subsection*{a)}
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\begin{align}
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H &= \frac{p^2}{2 m} + \frac{1}{2} m \omega^2 x^2 - q E x \\
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\hat{x} &= \sbk{\frac{m \omega}{\hbar}}^{\frac{1}{2}} x \\
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\hat{p} &= \sbk{\frac{1}{\hbar m \omega}}^{\frac{1}{2}} p \\
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E' &= q E \sbk{\frac{\hbar}{m \omega}}^{\frac{1}{2}} \frac{1}{\sqrt{2}} \\
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H &= \frac{\hbar \omega}{2} \sbk{\hat{p}^2 + x^2} + \sqrt{2} E' \hat{x}
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\end{align}
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Mit Erzeuger- und Vernichteroperatoren:
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\begin{math}
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H = \underbrace{\hbar \omega \sbk{\hat{p} + \frac{1}{2}}}_{H_0} + \underbrace{E' \sbk{a + a^\dagger}}_{H_1}
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\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}
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\end{math}
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Energie:
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\begin{align}
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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}
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&= \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}
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&= \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 \\
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&= \hbar \omega \sbk{n + \frac{1}{2}} - \frac{E'^2}{\hbar \omega} \\
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&= \hbar \omega \sbk{n + \frac{1}{2}} - \frac{q^2 E^2}{2 m \omega^2}
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\end{align}
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\subsection*{b)}
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\begin{align}
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H &= \frac{p^2}{2 m} + \frac{1}{2} \omega^2 x^2 - q E x
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\end{align}
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Mit:
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\begin{align}
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\hat{x} &= \sbk{\frac{m \omega}{\hbar}}^{\frac{1}{2}} x
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\hat{p} &= \sbk{\frac{1}{\hbar m \omega}}^{\frac{1}{2}} p
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\hat{E} &= q E \sbk{\frac{1}{\hbar m \omega^3}}^{\frac{1}{2}} \cdot 2
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\end{align}
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\begin{align}
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H &= \frac{\hbar \omega}{2} \sbk{\hat{p}^2 + \hat{x}^2 - \hat{E} x} \\
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&= \frac{\hbar \omega}{2} \sbk{\hat{p}^2 + \sbk{x - \frac{1}{2} \hat{E}}^2 - \frac{1}{4} \hat{E}^2}
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\end{align}
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Mit $\mathfrak{H} = \hat{x} - \frac{1}{2} \hat{E}$
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\begin{align}
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H &= \frac{\hbar \omega}{2} \sbk{\hat{p} + \mathfrak{H}^2} - \frac{\hbar \omega}{8} \hat{E}^2 \\
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E_n &= \hbar \omega \sbk{n + \frac{1}{2}} - \frac{q^2 E^2}{2 m \omega^2}
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\end{align}
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\subsection*{c)}
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\begin{math}
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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}
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H = H_0 + \epsilon H_1
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H_1 = - 2 q x y
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\end{math}
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$H_0: E_\alpha:$
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\begin{align}
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E_0 &= \hbar \omega &\ket{00} \\
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E_1 &= 2 \hbar \omega &\text{2-fach entartet} \underbrace{\ket{01}, \ket{10}}_{\ket{\alpha}} \\
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\end{align}
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Nach langem rechnen erhält man:
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\equationblock{\sum_\alpha \dirac{\beta}{\underbrace{H_{eff} - \sbk{E_a - E_\alpha}}}{\alpha} c_\alpha = 0}
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Dabei sind $c_\alpha \ket{\alpha} + \bigOb{\epsilon} = \ket{a}$ die Eigenzustände.
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Matrixelemente:
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$\dirac{\beta}{H_1}{\alpha}$
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$x,y$ mit Erzeuger- und Vernichteroperatoren:
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$H_1 = \ldots = -\frac{q \hbar}{m \omega} \sbk{a_x^\dagger a_y^\dagger + a_x^\dagger a_y + a_x a_y^\dagger + a_x a_y}$
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$\ket{\alpha} = \begin{cases}
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\ket{01} \\
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\ket{10}
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\end{cases}$
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\begin{math}
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\dirac{01}{H_1}{01}
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\dirac{01}{H_1}{10}
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\dirac{10}{H_1}{01}
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\dirac{10}{H_1}{11}
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\end{math}
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\begin{align}
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a_x^\dagger a_y^\dagger \ket{10} &= \sqrt{2} \ket{21} &\dirac{01}{a_x^\dagger a_y^\dagger}{10} &= 0 \\
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a_x^\dagger a_y \ket{10} &= 0 &\dirac{01}{a_x^\dagger a_y}{10} &= 0 \\
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a_x a_y^\dagger \ket{10} &= \ket{01} &\dirac{01}{a_x a_y^\dagger}{10} &= 1 \\
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a_x a_y \ket{10} &= 0 &\dirac{01}{a_x a_y}{10} &= 0
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\end{align}
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$\Rightarrow$
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\begin{align}
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\dirac{01}{H_1}{10} &= -\frac{1 \hbar}{m \omega} \\
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\dirac{01}{H_1}{01} &= 0 \\
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\dirac{10}{H_1}{01} &= -\frac{q \hbar}{m \omega} \\
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\dirac{10}{H_1}{10} &= 0
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\end{align}
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Matrix:
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$\inlinematrix{0 & -\frac{q \hbar}{m \omega} \\ -\frac{q \hbar}{m_\omega} & 0}$
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Mit $\hat{q} = -\frac{q \hbar}{m \omega}$
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\begin{math}
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\inlinematrix{-\Delta E & -\hat{q} \\ -\hat{q} & -\Delta E} \inlinematrix{c_{01} \\ c_{10}} = \inlinematrix{0\\0}
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\Delta E = E_a - E_\alpha
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\detb{\ldots} \deq 0 \Rightarrow \Delta E_+ = \pm \hat{q}
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\Delta E_+ = \pm \hat{q}
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\inlinematrix{c_{01} \\ c_{10}} = \frac{1}{\sqrt{2}} \inlinematrix{1 \\ \pm 1}
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\end{math}
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\begin{itemize}
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\item $\frac{1}{\sqrt{2}} \sbk{\ket{01} \pm \ket{10}} + \bigOb{\epsilon}$
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\item $E_+ = \hbar \omega \pm \hat{q} \epsilon + \bigOb{\epsilon} = \hbar \omega \pm \frac{q \hbar}{m \omega} \epsilon \bigOb{\epsilon^2}$
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\end{itemize}
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\subsection*{d)}
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\begin{align}
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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}
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&= \frac{1}{2} m \omega^2 \inlinematrix{x & y} \inlinematrix{1 & -\sigma \\ 0 & 1 + \sigma} \inlinematrix{\hat{x} \\ \hat{y}} \\
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&= \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 \\
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E_{n+ n-} &= \hbar \omega_+ \sbk{n_+ + \frac{1}{2}} + \hbar \omega_- \sbk{n_- + \frac{1}{2}}
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\end{align}
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Für $\epsilon \rightarrow 0$:
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$\hbar \omega \sbk{n_+ + n_- + 1}$
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\section{Aufgabe 27: Helium-Atom}
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$H = \underbrace{-\frac{\hbar^2}{2 m} \sbk{\underbrace{\nabla\sigma_1^2}_{I} + \underbrace{\nabla\sigma_2^2}_{II}} - e^2 \sbk{\underbrace{\frac{2}{r_1}}_{I} + \frac{2}{r_2}}}_{H_0} \underbrace{- e^2 \sbk{\frac{1}{\abs{\vec{r_1} - \vec{r_2}}}}}_{\text{in (a) vernachlässigbar}}$
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\subsection*{a)}
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$E_I^n = E_{II}^n = -\frac{Z^2}{2 \omega^2} \frac{e^2}{a_0}$ (wobei $a_0$ der bohrsche' Radius ist)
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$\Rightarrow Z = 2$
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Grundzustandsenergie:
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$E_0 = E_I^1 + E_{II}^2 = -4 \frac{e^2}{a_0}$
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Grundzustandswellenfunktion:
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$\psi_0\sbk{\vec{r_1}, \vec{r_2}} = \frac{Z^3}{\pi a_0^3} e^{-\frac{2 \sbk{r_1 + r_2}}{a_0}}$
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\subsection*{b)}
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\begin{align}
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\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}} \\
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\dirac{\psi_1}{H}{\psi_1} &= \dirac{\psi_1}{H_0}{\psi_1} + \dirac{\psi_2}{H_{int}}{\psi_1} \\
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\dirac{\psi_1}{H_0}{\psi_1} &= \sbk{- 4 \hat{Z} + \hat{Z}^2} \frac{e^2}{a_0} \\
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\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}}
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\end{align}
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Nach extrem langem rechnen erhält man:
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\equationblock{\dirac{\psi_1}{H_{int}}{\psi_1} = \frac{5}{8} \frac{Z}{a_0} e^2}
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Und somit:
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\equationblock{\dirac{\psi_1}{H}{\psi_1} = \sbk{Z^2 - \frac{27}{8} Z} \frac{e^2}{a_0}}
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Aus:
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$\diffTfrac{\dirac{\psi_1}{H}{\psi_1}}{Z} = 2 Z - \frac{27}{8} \deq 0$
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folgt:
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$Z^\ast = \frac{27}{16}$
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Und somit:
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\equationblock{\dirac{\psi}{H}{\psi}_\ast = -77,5 eV}
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