91 lines
3.4 KiB
TeX
91 lines
3.4 KiB
TeX
%\includegraphics{excs/qm1_blatt07_SS08.pdf}
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%\pagebreak
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\chapter{Quantenmechanik I - Übungsblatt 7}
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\section{Aufgabe 17: Unendlich hoher Potentialtop (Ergänzungen)}
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\subsection*{a)}
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\begin{math}
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\Phi_n(p) = \intgrinf{\frac{1}{\sqrt{2 \pi \hbar}} \cdot \Phi(x) \cdot e^{-\frac{\i p x}{\hbar}}}{x}
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\end{math}
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Für n = ungerade:
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\begin{align}
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\Phi_n(p) &= \intgrinf{\frac{1}{\sqrt{2 \pi a \hbar}} \cdot \cos(\frac{(n+1) \cdot \pi x}{2 a}) \cdot e^{-\frac{\i p x}{\hbar}}}{x} \\ \\
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&= \frac{1}{\sqrt{2 \pi a \hbar}} \cdot \sbk {
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\frac{1}{\sbk{\frac{(n+1) \cdot \pi}{2 a}}-\frac{p}{\hbar}} \cdot \sin \sbk{\sbk{\frac{(n+1) \cdot \pi}{2 a}}-\frac{p}{\hbar} \cdot u} +
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\frac{1}{\sbk{\frac{(n+1) \cdot \pi}{2 a}}+\frac{p}{\hbar}} \cdot \sin \sbk{\sbk{\frac{(n+1) \cdot \pi}{2 a}}+\frac{p}{\hbar} \cdot u}}
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\Phi_n(p) &=
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\begin{cases}
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\frac{1}{\sqrt{2 \pi a \hbar}} \cdot \frac{\i^n \cdot 4 \cdot a \hbar^2 \cdot (n+1) \cdot \pi}{\hbar^2 (n+1)^2 \pi - 4 a^2 p^2} \cdot \cos\sbk{\frac{pa}{\hbar}}
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\i^{n+2} \cdot \frac{\i^n \cdot 4 \cdot a \hbar^2 \cdot (n+1) \cdot \pi}{\hbar^2 (n+1)^2 \pi - 4 a^2 p^2} \cdot \sin\sbk{\frac{pa}{\hbar}}
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\end{cases}
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\end{align}
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\subsection*{b)}
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\subsection*{c)}
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\section{Aufgabe 18: Tunneleffekt}
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\includegraphics{grafiken/U_A18_1.pdf}
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\subsection*{a)}
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I
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\begin{math}
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-\frac{\hbar^2}{2m} \diffPs{x}^2 \Phi(x) = E \Phi(x) \\
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\Phi(x) =A e^{\i k x} + B e^{-\i k x} \\
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k = \sqrt{\frac{2 E M}{\hbar^2}}
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\end{math}
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II $E < V_0$
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\begin{math}
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\Phi(x) = C e^a + D e^{-qx} \\ % ist das nen q hier?
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q = \sqrt{\frac{-2m(E-V_0)}{\hbar^2}}
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\end{math}
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III
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\begin{math}
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\Phi(x) = \tilde{E} e^{\i k x} + F e^{-\i k x}
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\end{math}
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\subsection*{b)}
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\begin{math}
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F = 0
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\Phi_I(0) = \Phi_{II}(0) \\
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A+B = C+D \\
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\diffT{x} \Phi_I(0) = \diffT{x} \Phi_{II}(0) \\ % anschlussbedingungen korrekt
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\i k A - \i k B = q C - q D \\
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\inlinematrix{1 & 1 \\\i k & -\i k} \inlinematrix{A & B} = \inlinematrix{1 & 1 \\ q & -q} \inlinematrix{C & D} \\
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\Phi_{II}(a) = \Phi_{III}(a) \\
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C e^{a \cdot a} + D e^{-a \cdot a} = \tilde{E} e^{\i k a} \\
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\diffT{x} \Phi_{II}(a) = \diffT{x} \Phi_{III}(a) \\
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q C e^{a \cdot a} - q D e^{-a \cdot a} = \i k \tilde{E} e^{\i k a} \
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\inlinematrix{e^{a \cdot a} & e^{-a \cdot a} \\ q e^{a \cdot a} & q e^{-a \cdot a}} \inlinematrix{C & D} = \inlinematrix{\tilde{E} e^{\i k a} \\ \i k \tilde{E} e^{\i k a}}
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\end{math}
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mit den \hyperlink{fs_mtrx_inv_2d}{Inversen} von: (1) und (2) % geschweifte klammern unter matrix 1 und 2 setzen
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\begin{math}
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%mit maxima berechnet
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\inlinematrix{C & D} = \inlinematrix{\frac{{e}^{{a}^{2}}}{{e}^{2 {a}^{2}}-1} & -\frac{1}{\left( {e}^{3\,{a}^{2}}-{e}^{{a}^{2}}\right) q} \\ -\frac{{e}^{{a}^{2}}}{{e}^{2{a}^{2}}-1} & \frac{{e}^{{a}^{2}}}{\left( {e}^{2 {a}^{2}}-1\right) q}} \inlinematrix{\tilde{E} e^{\i k a} \\ \i k \tilde{E} e^{\i k a}} \\
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\inlinematrix{A & B} = \inlinematrix{\frac{1}{2} & -\frac{\i}{2k}\\ \frac{1}{2} & \frac{\i}{2k}} \inlinematrix{1 & 1 \\ q & -q} \inlinematrix{C & D}
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\end{math}
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$\inlinematrix{C & D}$ eingesetzt ergibt:
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\begin{align}
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\tilde{E} &= \frac{2 k a}{e^{ika} \sbk{(q^2 - k^2) \cdot \i sinh(q \cdot a) - 2 k q cosh(qa)}} \\ % ist das richtig?
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A &= 1 \\
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T &= \sbk{\frac{E}{A}}^2 \\
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R &= 1 - T &= \sbk{\frac{B}{A}}^2 \\
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T &= \frac{1}{\sbk{\frac{a^2 - k^2}{2 q k}} \cdot 2 sinh^2(q a) + cosh^2(q a)}
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\end{align}
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\subsection*{c)}
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\section{Aufgabe 19: Doppeltes Delta-Potential}
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\subsection*{a)}
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\subsection*{b)}
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\subsection*{c)}
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\subsection*{d)} |