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