Kapitel III 3 und 4 fertig; Fixes in II 1
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Oliver Groß
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@ -146,6 +146,7 @@ $\omega_1 = \Omega$ d.h. $\omega = \omega_0$, d.h. $B_1$-Feld zirkuliert mit der
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\end{itemize}
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\section{Transformationsverhalten unter Rotationen}
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\label{labelTransfRot}
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\subsection*{klassisch}
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Vektor $\vec{a}$ wird gedreht mit Matrix $R_z(\varepsilon)$ um den Winkel $\varepsilon$ um die $z$-Achse
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\begin{align}
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10
kapII-1.tex
10
kapII-1.tex
@ -61,23 +61,23 @@ Wie verhält sich $\diffT{t} \dirac{\psi}{\hat{p}}{\psi}$ ?
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\end{align}
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\section{Ortsdatsrellung (Analogie zu Spin $\frac{1}{2}$)}
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\begin{tabular}{l|c|c}
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\begin{tabular}{l||c|c}
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& Spin & Teilchen \\ \hline\hline
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Basis & $\sigma_z \ket{z\pm} = \pm 1 \ket{z\pm}$ & $\hat{x} \ket{x} = x \ket{x}$ \\ \hline
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Orthogonalität &
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$\begin{array}[t]{rl}\braket{z+}{z+} &= 1\\ &= \braket{z-}{z-}\\ \braket{z+}{z-} &= \braket{z-}{z+}\\ &= 0\end{array}$ &
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$\begin{array}[t]{r@{\,=\,}l}\braket{z+}{z+} & 1\\ &= \braket{z-}{z-}\\ \braket{z+}{z-} & \braket{z-}{z+}\\ & 0\end{array}$ &
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$\braket{x'}{x} = \delta(x'-x)$ \\ \hline
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Zustände in Basis entwickelt & $\begin{array}[t]{rl} \ket{\psi} &= \one \ket{\psi}\\ &= \ket{z+}\braket{z+}{\psi} + \ket{z-}\braket{z-}{\psi} \end{array}$ & $\begin{array}[t]{rl} \ket{\psi} &= \one \ket{\psi}\\ &= \intgr{-\infty}{+\infty}{\ket{x}\underbrace{\braket{x}{\psi}}_{\psi(x)}}{x}\\ &= \intgr{-\infty}{+\infty}{\psi(x) \ket{x}}{x} \end{array}$
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Zustände in Basis entwickelt & $\begin{array}[t]{r@{\,=\,}l} \ket{\psi} & \one \ket{\psi}\\ & \ket{z+}\braket{z+}{\psi} + \ket{z-}\braket{z-}{\psi} \end{array}$ & $\begin{array}[t]{r@{\,=\,}l} \ket{\psi} & \one \ket{\psi}\\ & \intgr{-\infty}{+\infty}{\ket{x}\underbrace{\braket{x}{\psi}}_{\psi(x)}}{x}\\ & \intgr{-\infty}{+\infty}{\psi(x) \ket{x}}{x} \end{array}$
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\end{tabular}
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\paragraph*{Normierung}
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\begin{align}
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1 \stackrel{!}{=} \braket{\psi}{\psi} &= \intgr{-\infty}{+\infty}{\psi^* \bra{x}}{x} \intgr{-\infty}{+\infty}{\psi(x') \ket{x'}}{x'}\\
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\intgru{\intgru{\psi(x)\psi(x')\underbrace{\braket{x}{x'}}_{\delta(x-x')}}{x'}}{x}\\
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&= \intgru{\intgru{\psi(x)\psi(x')\underbrace{\braket{x}{x'}}_{\delta(x-x')}}{x'}}{x}\\
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&= \intgru{\psi^*(x)\psi(x)}{x}\\
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&= \intgru{(\psi(x))^2}{x}
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\end{align}
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$\rightarrow$ Zulässige Zustände haben eine ``Wellenfunktion'', die quadrat-integrabel ist.
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$\rightarrow$ Zulässige Zustände haben eine ``Wellenfunktion'', die quadrat-integrabel ist.align
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\paragraph*{Erwartungswert einer Ortsmessung}
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\begin{align}
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@ -1,4 +1,5 @@
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\chapter{Rotationssymetrie im Potential in $d=2$}
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\label{labelRotSym2D}
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\section{Lösung der stationären Schrödingergleichung durch ``Separation der Variablen''}
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\label{rotSymSGL}
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Mit den Polarkoordinaten
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130
kapIII-3.tex
130
kapIII-3.tex
@ -237,5 +237,133 @@ mit
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\begin{equation}
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m(n_+,n_-) = n_+ - n_-
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\end{equation}
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%TODO...
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Für Rotationsinvarianz gilt:
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\begin{align}
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[L_z, p^2] &= 0\\
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[L_z, r^2] &= 0\\
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\rightarrow [L_z, V(r)] &= 0
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\end{align}
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(gilt analog für $L_x$ und $L_y$)
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\begin{equation}
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\rightarrow \left[ \frac{\hat{p}^2}{2m} + V(r), L^2 \right] = 0
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\end{equation}
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also vertauschen $H$, $L^2$ und $L-z$ und es existiert eine gemeinsame Eigenbasis $\set{\ket{E\,l,m}}$ mit
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\begin{align}
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H \ket{E\,l,m} &= E \ket{E\,l,m}\\
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L^2 \ket{E\,l,m} &= \hbar^2 l(l+1) \ket{E\,l,m}\\
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L_z \ket{E\,l,m} &= \hbar m \ket{E\,l,m}
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\end{align}
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Sucht man nach dem Spektrum von $H$ so muss man die möglichen $E$-Werte für festes $l$ und $m$ finden.
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\section{Radialgleichung}
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\paragraph{Ziel} $H \ket{\phi} = E \ket{\phi}$ vereinfachen mit
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\begin{equation}
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p^2 = -\frac{\hbar^2}{2\mu} \left[ \diffPs{r}^2 + \frac{2}{r} \diffPs{r} \right] + \frac{\vec{L}^2}{2\mu r^2}
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\end{equation}
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\paragraph{Beweis}
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\begin{align}
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L^2 &= \levicivita{\alpha,\beta,\gamma} x_\beta p_x \levicivita{\alpha,\mu,\nu} x_\mu p_\nu \\
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&= \left(\krondelta{\beta,\mu} \krondelta{x,\nu} - \krondelta{\beta,\nu} \krondelta{\gamma,\beta}\right) x_\beta p_\gamma x_\gamma p_\nu \\
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&= x_\beta \underbrace{p_\gamma x_\beta}_{x_\beta p_\gamma - i\hbar \krondelta{\beta,\gamma}} - x_\beta p_\gamma \underbrace{x_\gamma p_\beta}_{p_\beta x_\gamma + i\hbar \krondelta{\gamma,\beta}} \\
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&= x^2 p^2 - i\hbar x p - \left( (x \cdot p)\underbrace{p \cdot x}_{x \cdot p - 3i\hbar} + i\hbar (x \cdot p) \right) \\
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&= x^2 p^2 - (x \cdot p)^2 + i \hbar x p \label{stern00}
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\end{align}
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mit
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\begin{equation}
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\dirac{r,\theta,\phi}{x^2 p^2}{\psi} = r^2 \dirac{r,\theta,\phi}{\hat{p}^2}{\psi}
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\end{equation}
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und
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\begin{align}
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\dirac{r,\theta,\phi}{\hat{x} p}{\psi} &= \vec{x} (-i\hbar \vec{\nabla}) \psi(r,\theta,\phi) \\
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&= -i\hbar r \diffPs{r} \psi(r,\theta,\phi)
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\end{align}
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daraus wird (\ref{stern00})
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\begin{align}
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\dirac{r,\theta,\phi}{L^2}{\psi} &= r^2 \cdot 2 \mu \braket{r,\theta,\phi \, H_\text{kin}}{\psi} + \hbar^2 (r\diffPs{r})(r\diffPs{r})\psi(r,\theta,\phi) + \hbar^2(r\diffPs{r})\psi(r,\theta,\phi) \\[15pt]
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\rightarrow \dirac{r,\theta,\phi}{\underbrace{H_\text{kin} + V(r)}_{H}}{\psi} &\stackrel{!}{=} E \braket{r,\theta,\phi}{\psi}
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\end{align}
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für Eigenfunktion
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\begin{equation}
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\rightarrow \left( -\frac{\hbar^2}{2\mu} \left( \diffPs{r}^2 + \frac{2}{r} \diffPs{r} \right) + \frac{L^2}{2\mu r^2} + V(r) \right) \psi
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\end{equation}
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mit
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\begin{equation}
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\ket{\psi} = \ket{E\,l,m}
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\end{equation}
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ist
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\begin{align}
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\psi(r,\theta,\phi) &= R_E(r) y_{l,m}(\theta,\phi)\\
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\rightarrow \left( -\frac{\hbar^2}{2\mu} \left( \diffPs{r}^2 + \frac{2}{r} \diffPs{r} - \frac{l(l+1)}{r^2} \right) + V(r) \right) R_E(r) &= E R_E(r)
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\end{align}
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unabhängig von $m$ (nur abhängig von Gesammtdrehimpuls)!
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Als Vergleich: Rotation in 2D:
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\begin{equation}
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\left( -\frac{\hbar^2}{2\mu} \left( \diffPs{\rho}^2 + \frac{1}{\rho} \diffPs{\rho} - \frac{m^2}{\rho^2} \right) + V(\rho) \right) R_E(\rho) = E R_E(\rho)
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\end{equation}
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Nun ist
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\begin{align}
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R_E(r) &= \frac{u(r)}{r}\\
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\diffPs{r}R_E(r) &= \frac{\diffPs{r}u(r)}{r} - \frac{u(r)}{r^2}\\
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\diffPs{r}^2R_E(r) &= \frac{\diffPs{r}^2u(r)}{r} - \frac{2\diffPs{r}u(r)}{r^2} + \frac{2u}{r^3}
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\end{align}
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einsetzen:
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\begin{equation}
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\left( -\frac{\hbar^2}{2\mu} \diffPs{r}^2 + \underbrace{\frac{\hbar^2}{2\mu} \frac{l(l+1)}{r^2} + V(r)}_{V_\text{eff}(r)} \right) u(r) = E u(r) \text{ für } r \geq 0
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\end{equation}
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gleicht formal der eindimensionalen Schrödingergleichung mit effektivem Potential. Aber $r \geq 0$ beachten (mit $u(r = 0) = 0$)!
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\section{Coulomb-Problem und Wasserstoffatom}
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Potential:
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\begin{equation}
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V(r) = -\frac{Z e^2}{r}; ~~ r = a_0 y
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\end{equation}
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mit
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\begin{equation}
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a_0 = \frac{\hbar}{\mu e^2} \text{ (Bohr'scher Radius)}; ~~ E = \frac{e^2}{a_0} \varepsilon
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\end{equation}
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eingesetzt in die stationre Schrödingergleichung
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\begin{equation}
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\left( \diffPs{y}^2 - \frac{l(l+1)}{y^2} + \frac{2Z}{y} + \varepsilon \right) u(y) = 0
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\end{equation}
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\paragraph{Lösungsstrategie 1} Asymptotik bestimmen, abspalten, Potenzreihenansatz (Tailorreihe!) einsetzen, Konvergenz durch Abbruch führt auf quantisierte Energie (steht in jedem Buch).
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\paragraph{Lösungsstrategie 2} Abbildung auf harmonischen Oszillator in 2D:\\
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dazu:
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\begin{align}
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x^2 &= 2 \lambda y\\
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x \text{ d}x &= \lambda \text{ d}y\\
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\rightarrow \diffPfrac{}{y} = \frac{\lambda}{x} \diffPfrac{}{x}
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\end{align}
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einsetzen:
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\begin{equation}
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\left( \diffPs{x}^2 - \frac{(2l+1)^2 - \frac{1}{4}}{x^2} + \frac{4Z}{\lambda} + \frac{2\varepsilon}{\lambda^2} x2 \right) \frac{u(y)}{\sqrt{x}} = 0
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\end{equation}
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Erinnerung an 2D H.O.
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\begin{equation}
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\left( \left( \diffPs{\rho}^2 + \frac{1}{\rho} \diffPs{\rho} - \frac{m^2}{\rho^2} \right) -\frac{\mu \omega^2}{\hbar^2} \rho^2 + E \frac{2m}{\hbar^2} \right) R_{n,m}(\rho) = 0
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\end{equation}
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mit $q \equiv \sqrt{\frac{u\omega}{\hbar}}\rho$
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\begin{equation}
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\left( \diffPs{q}^2 - \frac{m^2 - \frac{1}{4}}{q^2} - q^2 + 2N + 2 \right) \sqrt{\rho} R_{n,m}(\rho) = 0
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\end{equation}
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Korrespondenz:
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\begin{center}
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\begin{tabular}{c|c}
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Coulomb & H.O. \\ \hline
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$(2l+1)^2$ & $m^2$ \\
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$\frac{4Z}{\lambda}$ & $2N+2$ \\
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$\frac{2\varepsilon}{\lambda}$ & $-1$
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\end{tabular}
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\end{center}
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Also sind:
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\begin{equation}
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\abs{m} = 2l+1 \text{ und } \lambda = \frac{2Z}{N+1} = \frac{2Z}{\abs{m} + 2n_r + 1}
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\end{equation}
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Energieeigenwerte:
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\begin{equation}
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\varepsilon = -\frac{\lambda^2}{2} = -2Z^2 \frac{1}{(\abs{m} + 2n_r + 1)^2} = -z^2 \frac{1}{(l + n_r + 1)^2}
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\end{equation}
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damit das Spektrum:
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%\begin{figure}[H] \centering
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%\includegraphics{pdf/III/03-05-00.pdf}
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%\end{figure}
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133
kapIII-4.tex
133
kapIII-4.tex
@ -8,7 +8,7 @@ mit
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\begin{equation}
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D(\phi,\vec{n}) = 1 - i\frac{\phi}{\hbar} J_{\vec{n}} + O(\phi^2)
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\end{equation}
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In () hatten wir die Relation
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In (\ref{labelTransfRot}) hatten wir die Relation
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\begin{equation}
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[J_x,J_y] = i\hbar J_z
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\end{equation}
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@ -62,3 +62,134 @@ $\beta$-Spektrum ist eingeschränkt wegen:
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\begin{equation}
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0 \leq \dirac{\alpha,\beta}{J_x^2 + J_y^2}{\alpha,\beta} = \dirac{\alpha,\beta}{J^2-J_z^2}{\alpha,\beta} = (\alpha-\beta)^2 \underbrace{\braket{\alpha,\beta}{\alpha,\beta}}_{\geq 0}
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\end{equation}
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Also ist $\alpha \geq \beta^2$ und $\exists \beta_\text{max}$ mit
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\begin{align}
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J_+ \ket{\alpha,\beta_\text{max}} &= \ket{\zero} & \left| J_- \right.\\
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J_- J_+ \ket{\alpha,\beta_\text{max}} &= 0\\
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\left( J^2 - J_z^2 - \hbar J_z \right) \ket{\alpha,\beta_\text{max}} &= 0\\
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\alpha - \beta_\text{max}^2 \hbar^2 -\beta_\text{max} \hbar &\stackrel{!}{=} 0\\
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\rightarrow \alpha = \beta_\text{max} (\beta_\text{max} + \hbar)
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\end{align}
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entsprechend:
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\begin{align}
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J_- \ket{\alpha,\beta_\text{min}} &= \ket{\zero}\\
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... \rightarrow \alpha &= \beta_\text{min} (\beta_\text{min} - \hbar)
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\end{align}
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Daraus folgt:
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\begin{equation}
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\beta_\text{max}^2 + \beta_\text{max} \hbar = \beta_\text{min}^2 + \beta_\text{min} \hbar
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\end{equation}
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quadratische Gleichung mit den 2 Lösungen:
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\begin{align}
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\beta_\text{max} &= \beta_\text{min} - \hbar &\text{(irrelevant)}\\
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\beta_\text{max} &= -\beta_\text{min}
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\end{align}
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\begin{align}
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\left( J_+ \right)^k \ket{\alpha,\beta_\text{mix}} &= \const \ket{\alpha,\beta_\text{max}}\\
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\rightarrow \beta_\text{max} &= \beta_\text{min} + \hbar \cdot k
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\end{align}
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daraus ergibt sich mit $k = 0,1,2,...$:
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\begin{equation}
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\boxed{\beta_\text{max} = \frac{\hbar}{2} \cdot k}
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\end{equation}
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und
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\begin{equation}
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\alpha = \beta_\text{max}^2 + \beta_\text{max} \hbar = \hbar^2 \left( \frac{k}{2} + 1 \right) \frac{k}{2}
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\end{equation}
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Zusammenfassung:
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\begin{center}
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\begin{tabular}{c|c|c|c|c}
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$\frac{k}{2}$ & $\beta_\text{max}$ & $\alpha$ & $\ket{\alpha,\beta_\text{max}}$ & Anzahl $\ket{\alpha,\beta}$ \\ \hline
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$0$ & $0$ & $0$ & $\ket{0,0}$ & $1$ \\ \hline
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$\frac{1}{2}$ & $\frac{1}{2}\hbar$ & $\frac{3}{4}\hbar^2$ & $\ket{\frac{3}{4},\frac{1}{2}}$ & $2$ \\ \hline
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$1$ & $\hbar$ & $2\hbar^2$ & $\ket{2,1}$ & $3$ \\ \hline
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$\frac{3}{2}$ & $\frac{3}{2}\hbar$ & $\frac{15}{4}\hbar^2$ & $\ket{\frac{15}{4},\frac{3}{2}}$ & $4$
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\end{tabular}
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\end{center}
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Wir finden halbzahlige Eigenwerte für $J_z$. Aber: in (\ref{labelRotSym2D}) hatten wir gesehen:
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\begin{equation}
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J_3 = XP_Y - YP_X
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\end{equation}
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hat Eigenwerte $m \hbar= (0,\pm 1, \pm 1, ...) \hbar$.\\
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Mit der Notation $j \equiv \frac{k}{2}$
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\begin{align}
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\alpha &= \hbar^2 j (j + 1)\\
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\beta &= \hbar m
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\end{align}
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und
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\begin{align}
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J^2 \ket{j,m} &= \hbar^2 j (j+1) \ket{j,m} & \left(j = \frac{0,1,2,3,...}{2}\right)\\
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J_z \ket{j,m} &= \hbar m \ket{j,m} & \left(m = -j, -j+1, ..., j\right)
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\end{align}
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Matrixelemente:
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\begin{align}
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J_\pm \ket{j,m} &= c_+(j,m) \ket{j,m \pm 1}\\
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\bra{j,m} J_\mp &= c_+^*(j,m) \bra{j,m \pm 1}\\[15pt]
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\dirac{j,m}{J_- J_+}{j,m} &= \abs{c_+(j,m)}^2 \underbrace{\braket{j,m+1}{j,m+1}}_1\\
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\dirac{j,m}{J^2 - J_z^2 - \hbar J_z}{j,m} &= \abs{c_+(j,m)}^2\\
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\rightarrow \hbar^2 \left( j (j+1) - m^2 - m \right) &= \abs{c_+(j,m)}^2\\
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\rightarrow c_+ &= \hbar \sqrt{j (j+1) - m^2 - m}\\
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&= \sqrt{(j + m + 1)(j - m)}
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\end{align}
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genauso
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\begin{equation}
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c_- = \hbar \sqrt{(j - m + 1)(j + m)}
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\end{equation}
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Matrixelemente von $J_x$ und $J_y$:
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\begin{align}
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\dirac{j',m'}{J_x}{j,m} &= \braket{j',m'}{\frac{J_+ + J_-}{2} - j,m}\\
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&= \krondelta{j',j} \left\lbrace \krondelta{m',m+1} c_+(j,m) + \krondelta{m'-1,m} c_-(j,m) \right\rbrace
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\end{align}
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\definecolor{lgray}{gray}{0.9}
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\newcommand{\graycell}{\cellcolor{lgray}}
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\begin{itemize}
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\item für $J^2 / \hbar^2$
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\begin{center}
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\begin{tabular}{c||c|c|c|c|c|c}
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$j',m' \backslash j,m$ & $(0,0)$ & $(\frac{1}{2},\frac{1}{2})$ & $(\frac{1}{2},-\frac{1}{2})$ & $(1,1)$ & $(1,0)$ & $(1,-1)$ \\ \hline\hline
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$(0,0)$ & $0$ \graycell & $0$ & $0$ & $0$ & $0$ & $0$ \\ \hline
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$(\frac{1}{2},\frac{1}{2})$ & $0$ & $\frac{3}{4}$ \graycell & $0$ \graycell & $0$ & $0$ & $0$ \\ \hline
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$(\frac{1}{2},-\frac{1}{2})$ & $0$ & $0$ \graycell & $\frac{3}{4}$ \graycell & $0$ & $0$ & $0$ \\ \hline
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$(1,1)$ & $0$ & $0$ & $0$ & $2$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
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$(1,0)$ & $0$ & $0$ & $0$ & $0$ \graycell & $2$ \graycell & $0$ \graycell \\ \hline
|
||||
$(1,-1)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $2$ \graycell \\
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\item für $J_z / \hbar$
|
||||
\begin{center}
|
||||
\begin{tabular}{c||c|c|c|c|c|c}
|
||||
$j',m' \backslash j,m$ & $(0,0)$ & $(\frac{1}{2},\frac{1}{2})$ & $(\frac{1}{2},-\frac{1}{2})$ & $(1,1)$ & $(1,0)$ & $(1,-1)$ \\ \hline\hline
|
||||
$(0,0)$ & $0$ \graycell & $0$ & $0$ & $0$ & $0$ & $0$ \\ \hline
|
||||
$(\frac{1}{2},\frac{1}{2})$ & $0$ & $\frac{1}{2}$ \graycell & $0$ \graycell & $0$ & $0$ & $0$ \\ \hline
|
||||
$(\frac{1}{2},-\frac{1}{2})$ & $0$ & $0$ \graycell & $-\frac{1}{2}$ \graycell & $0$ & $0$ & $0$ \\ \hline
|
||||
$(1,1)$ & $0$ & $0$ & $0$ & $1$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
|
||||
$(1,0)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
|
||||
$(1,-1)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $1$ \graycell \\
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\item für $J_x / \hbar$
|
||||
\begin{center}
|
||||
\begin{tabular}{c||c|c|c|c|c|c}
|
||||
$j',m' \backslash j,m$ & $(0,0)$ & $(\frac{1}{2},\frac{1}{2})$ & $(\frac{1}{2},-\frac{1}{2})$ & $(1,1)$ & $(1,0)$ & $(1,-1)$ \\ \hline\hline
|
||||
$(0,0)$ & $0$ \graycell & $0$ & $0$ & $0$ & $0$ & $0$ \\ \hline
|
||||
$(\frac{1}{2},\frac{1}{2})$ & $0$ & $0$ \graycell & $\frac{1}{2}$ \graycell & $0$ & $0$ & $0$ \\ \hline
|
||||
$(\frac{1}{2},-\frac{1}{2})$ & $0$ & $\frac{1}{2}$ \graycell & $0$ \graycell & $0$ & $0$ & $0$ \\ \hline
|
||||
$(1,1)$ & $0$ & $0$ & $0$ & $1$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
|
||||
$(1,0)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
|
||||
$(1,-1)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $1$ \graycell \\
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\item für $J_y / \hbar$
|
||||
\begin{center}
|
||||
\begin{tabular}{c||c|c|c|c|c|c}
|
||||
$j',m' \backslash j,m$ & $(0,0)$ & $(\frac{1}{2},\frac{1}{2})$ & $(\frac{1}{2},-\frac{1}{2})$ & $(1,1)$ & $(1,0)$ & $(1,-1)$ \\ \hline\hline
|
||||
$(0,0)$ & $0$ \graycell & $0$ & $0$ & $0$ & $0$ & $0$ \\ \hline
|
||||
$(\frac{1}{2},\frac{1}{2})$ & $0$ & $0$ \graycell & $-\frac{i}{2}$ \graycell & $0$ & $0$ & $0$ \\ \hline
|
||||
$(\frac{1}{2},-\frac{1}{2})$ & $0$ & $\frac{i}{2}$ \graycell & $0$ \graycell & $0$ & $0$ & $0$ \\ \hline
|
||||
$(1,1)$ & $0$ & $0$ & $0$ & $1$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
|
||||
$(1,0)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
|
||||
$(1,-1)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $1$ \graycell \\
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\end{itemize}
|
||||
Erkennbar ist hier jeweils eine Blockstruktur!
|
||||
|
||||
4
math.tex
4
math.tex
@ -37,8 +37,8 @@
|
||||
\newcommand{\diffTfrac}[2]{\frac{\text{d} #1}{\text{d} #2}}
|
||||
\newcommand{\diffTm}[3]{\diffTfrac{^{#1} #2}{#3^{#1}}}
|
||||
\newcommand{\diffPfrac}[2]{\frac{\partial #1}{\partial #2}}
|
||||
\newcommand{\intgr}[4]{\int_{#1}^{#2} #3 ~\text{ d}#4}
|
||||
\newcommand{\intgru}[2]{\int #1 ~\text{ d}#2}
|
||||
\newcommand{\intgr}[4]{\int_{#1}^{#2} #3 \,\text{d}#4}
|
||||
\newcommand{\intgru}[2]{\int #1 \,\text{d}#2}
|
||||
\newcommand{\intgrinf}[2]{\intgr{-\infty}{+\infty}{#1}{#2}}
|
||||
|
||||
\newcommand{\sbk}[1]{\left( #1 \right)}
|
||||
|
||||
55
theo2.kilepr
55
theo2.kilepr
@ -3,7 +3,7 @@ img_extIsRegExp=false
|
||||
img_extensions=.eps .jpg .jpeg .png .pdf .ps .fig .gif
|
||||
kileprversion=2
|
||||
kileversion=2.0
|
||||
lastDocument=kapIII-4.tex
|
||||
lastDocument=kapIV-1.tex
|
||||
masterDocument=
|
||||
name=Theo2
|
||||
pkg_extIsRegExp=false
|
||||
@ -65,7 +65,7 @@ archive=true
|
||||
column=7
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=82
|
||||
line=148
|
||||
open=false
|
||||
order=6
|
||||
|
||||
@ -84,15 +84,15 @@ column=12
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=122
|
||||
open=true
|
||||
open=false
|
||||
order=1
|
||||
|
||||
[item:kapII-1.tex]
|
||||
archive=true
|
||||
column=6
|
||||
encoding=
|
||||
column=113
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=110
|
||||
line=67
|
||||
open=false
|
||||
order=-1
|
||||
|
||||
@ -120,7 +120,7 @@ column=13
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=39
|
||||
open=true
|
||||
open=false
|
||||
order=2
|
||||
|
||||
[item:kapII-5.tex]
|
||||
@ -129,7 +129,7 @@ column=29
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=92
|
||||
open=true
|
||||
open=false
|
||||
order=3
|
||||
|
||||
[item:kapIII-0.tex]
|
||||
@ -138,7 +138,7 @@ column=8
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=0
|
||||
open=true
|
||||
open=false
|
||||
order=4
|
||||
|
||||
[item:kapIII-1.tex]
|
||||
@ -147,42 +147,51 @@ column=15
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=39
|
||||
open=true
|
||||
open=false
|
||||
order=5
|
||||
|
||||
[item:kapIII-2.tex]
|
||||
archive=true
|
||||
column=49
|
||||
column=20
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=0
|
||||
open=true
|
||||
line=1
|
||||
open=false
|
||||
order=6
|
||||
|
||||
[item:kapIII-3.tex]
|
||||
archive=true
|
||||
column=50
|
||||
column=108
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=0
|
||||
open=true
|
||||
line=338
|
||||
open=false
|
||||
order=7
|
||||
|
||||
[item:kapIII-4.tex]
|
||||
archive=true
|
||||
column=69
|
||||
column=0
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=8
|
||||
open=true
|
||||
line=195
|
||||
open=false
|
||||
order=8
|
||||
|
||||
[item:kapIV-1.tex]
|
||||
archive=true
|
||||
column=104
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=53
|
||||
open=true
|
||||
order=1
|
||||
|
||||
[item:math.tex]
|
||||
archive=true
|
||||
column=43
|
||||
column=34
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=37
|
||||
line=40
|
||||
open=false
|
||||
order=2
|
||||
|
||||
@ -206,10 +215,10 @@ order=-1
|
||||
|
||||
[item:theo2.tex]
|
||||
archive=true
|
||||
column=17
|
||||
column=24
|
||||
encoding=UTF-8
|
||||
highlight=LaTeX
|
||||
line=47
|
||||
line=70
|
||||
open=true
|
||||
order=0
|
||||
|
||||
|
||||
30
theo2.tex
30
theo2.tex
@ -9,7 +9,7 @@
|
||||
\usepackage{float}
|
||||
\usepackage{pdflscape}
|
||||
\usepackage[pdfborder={0 0 0}]{hyperref} % muss immer als letztes eingebunden werden
|
||||
|
||||
\usepackage{colortbl}
|
||||
\include{math}
|
||||
\include{physics}
|
||||
|
||||
@ -48,23 +48,27 @@
|
||||
\include{kapIII-3}
|
||||
\include{kapIII-4}
|
||||
|
||||
\part{Näherungsmethoden}
|
||||
\label{IV}
|
||||
\include{kapIV-1}
|
||||
\include{kapIV-2}
|
||||
|
||||
% \part{Übungsmitschrieb}
|
||||
% \label{UE}
|
||||
\include{ueb1}
|
||||
\include{ueb2}
|
||||
\include{ueb3}
|
||||
\include{ueb4}
|
||||
\include{ueb5}
|
||||
\include{ueb6}
|
||||
\include{ueb7}
|
||||
\include{ueb8}
|
||||
\include{ueb9}
|
||||
\include{ueb10}
|
||||
\include{ueb11}
|
||||
% \include{ueb1}
|
||||
% \include{ueb2}
|
||||
% \include{ueb3}
|
||||
% \include{ueb4}
|
||||
% \include{ueb5}
|
||||
% \include{ueb6}
|
||||
% \include{ueb7}
|
||||
% \include{ueb8}
|
||||
% \include{ueb9}
|
||||
% \include{ueb10}
|
||||
% \include{ueb11}
|
||||
|
||||
\part{Formelsammlung}
|
||||
\label{FS}
|
||||
\include{formelsammlung}
|
||||
|
||||
|
||||
\end{document}
|
||||
|
||||
Loading…
Reference in New Issue
Block a user