[vorlesung] WIP für Kapitel II.3
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kapII-3.tex
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kapII-3.tex
@ -25,3 +25,78 @@ der Wahscheinlichkeitsstromdichte (``Kontinuitätsgleichung''; gilt für jede Er
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\end{align}
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\end{align}
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\section{Streuung an der Potentialstufe}
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\section{Streuung an der Potentialstufe}
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%\begin{figure}[h]
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%\includegraphics{pdf/II/03-02-00.pdf}
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%\end{figure}
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\paragraph*{klassisch}
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\subparagraph*{Fall 1} $E > V_0$
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\begin{align}
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x < 0:~ & p(x < 0) = \sqrt{2m E}\\
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x > 0:~ & p(x > 0) = \sqrt{2m (E - V_0)}
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\end{align}
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Teilchen passiert die Potentialstufe, verliert Impuls
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\subparagraph*{Fall 2} $E < V_0$
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\begin{equation}
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p(x < 0) = \sqrt{2m E}
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\end{equation}
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Teilchen wird reflektiert
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\paragraph*{quantal}
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\subparagraph*{Fall 1} $E > 0$\\
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stationäre SG:
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\begin{equation}
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\left( -\frac{\hbar^2}{2m} \diffPs{x}^2 V(x) \right) \phi(x) = E \phi(x)
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\end{equation}
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links: $x < 0$
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\begin{equation}
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\diffPs{x}^2 \phi(x) = -k^2 \phi(x)
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\end{equation}
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mit
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\begin{equation}
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k = \sqrt{\frac{2m}{\hbar^2}}
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\end{equation}
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Lösung:
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\begin{equation}
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\phi(x) = A e^{i k x} + B e^{-i k x}
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\end{equation}
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rechts: $x > 0$
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\begin{equation}
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\left( -\frac{\hbar^2}{2m} \diffPs{x}^2 + V_0 \right) \phi(x) = E \phi(x)
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\end{equation}
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Lösung:
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\begin{equation}
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\phi(x) = C e^{i q x} + D e^{-i q x}
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\end{equation}
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mit
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\begin{equation}
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q = \sqrt{\frac{2m (E - V_0)}{\hbar^2}}
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\end{equation}
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Randbedinung bei $x = 0$
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\begin{align}
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\phi(-\varepsilon) &= \phi(+\varepsilon)\\
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\diffPs{x} \phi(-\varepsilon) &= \diffPs{x} \phi(+\varepsilon)\\
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\rightarrow A + B &= C + D\\
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i k (A - B) &= i q (C - D)\\[15pt]
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\inlinematrix{1 & 1 \\ i k & -i k} \inlinematrix{A \\ B} &= \inlinematrix{1 & 1 \\ i q & -i q} \inlinematrix{C \\ D}\\
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\inlinematrix{A \\ B} &= \frac{1}{2k} \inlinematrix{k+q & k-q \\ k-q & k+q} \inlinematrix{C \\ D}
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\end{align}
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$\rightarrow$ Randbedingung einer von links laufenden Welle\\
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$\Rightarrow$ keine Komponente einer von rechts einlaufenden Welle für $x > 0$ erlaubt!\\
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$\Rightarrow$ $D \equiv 0$\\
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o.B.d.A.: $A = 1$
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\begin{align}
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A &= \frac{k + q}{2k} C ~ \rightarrow C = \frac{2k}{k+q}\\
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B &= \frac{k - q}{2k} C ~ \rightarrow B = \frac{k - q}{k + q}
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\end{align}
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Strom links: $x < 0$
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\begin{align}
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j(x < 0) &= \frac{\hbar}{m} \im{\diffPs{x}\phi ~ \phi^*}\\
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&= \frac{\hbar}{m} \im{i k \left(A e^{i k x} - B e^{-i k x} \right) \left(A^* e^{- i k x} + B^* e^{i k x}\right)}\\
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&= \frac{\hbar}{m} \im{ik \left( A A^* - B B^*\right) + ik \left( A B^* e^{2 i k x} - A^* B e^{-2 i k x} \right)}\\
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&= \frac{\hbar}{m} k \left( 1 - \left( \frac{k - q}{k + q} \right)^2 \right) \equiv j_I - j_R
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\end{align}
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mit
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\begin{align}
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j_I &= \frac{\hbar k}{m} &\text{einfallend}\\
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j_R &= \frac{\hbar}{m} k \left( \frac{k - q}{k + q} \right)^2 \equiv R j_I &\text{reflectiert}
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\end{align}
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