Merge branch 'master' of git+ssh://git@git.eluhost.de:522/qm1-script
This commit is contained in:
Oliver Groß
commit
bed5758c84
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@ -1,32 +1,90 @@
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\chapter{Notationen}
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\section{Dirac-Notation}
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\chapter{Lineare Algebra}
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\section{Allgemeines}
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\subsection*{Definitionen}
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\section{Gruppentheorie}
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\subsection{Abbildungen}
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\subsubsection*{Kommutator:}
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\begin{equation} [A,B] = AB - BA \end{equation}
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Der Kommutator von g und h ist genau dann gleich dem neutralen Element, wenn g und h kommutieren. \\
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Sei $a$, $b$ und $c$ Elemente einer assoziativen Algebra und $\lambda$ ein Skalar (Element des Grundkörpers).
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\begin{enumerate}
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\item Alternierend (antisymmetrisch):
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\begin{equation} [a,b]=-[b,a] \end{equation}
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\item Linear:
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\begin{equation} [\lambda a+b,c]=\lambda [a,c] + [b,c] \end{equation}
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\item Jacobi-Identität:
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\begin{equation} [a,[b,c]]+[b,[c,a]]+[c,[a,b]]=0 \end{equation}
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\item Leibnizregel(Produktregel):
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\begin{equation} [a,bc] = [a,b]c+b[a,c] \end{equation}
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\end{enumerate}
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Aufgrund der Eigenschaften 1, 2 und 3 wird jede assoziative Algebra $A$ mit dem Kommutator als Lie-Klammer zu einer Lie-Algebra, die teilweise mit $A^-$ bezeichnet wird. \\
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Eigenschaft 4 bedeutet, daß die Abbildung $b\mapsto [a,b]$ eine Derivation ist.
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\subsubsection*{Levi-Civita-Symbol:}
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\begin{math}
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\varepsilon_{12\dots n} = 1 \\
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\varepsilon_{ij\dots u\dots v\dots} = -\varepsilon_{ij\dots v\dots u\dots}\\
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\varepsilon_{ij\dots u\dots v\dots} = -\varepsilon_{ij\dots v\dots u\dots}\\
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\varepsilon_{ij\dots u\dots u\dots} = 0 \\
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\levicivita{i,j,k} =
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\begin{cases}
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+1, & \mbox{falls }(i,j,k,\dots) \mbox{ eine gerade Permutation von } (1,2,3,\dots) \mbox{ ist,} \\
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-1, & \mbox{falls }(i,j,k,\dots) \mbox{ eine ungerade Permutation von } (1,2,3,\dots) \mbox{ ist,} \\
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0, & \mbox{wenn mindestens zwei Indizes gleich sind.}
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\end{cases}
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\begin{cases}
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+1, & \mbox{falls }(i,j,k,\dots) \mbox{ eine gerade Permutation von } (1,2,3,\dots) \mbox{ ist,} \\
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-1, & \mbox{falls }(i,j,k,\dots) \mbox{ eine ungerade Permutation von } (1,2,3,\dots) \mbox{ ist,} \\
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0, & \mbox{wenn mindestens zwei Indizes gleich sind.}
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\end{cases} \\
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(\vec{a} \times \vec{b})_i = \sum_{j=1}^3 \sum_{k=1}^3 \levicivita{ijk} a_j b_k \\
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\vec{a} \times \vec{b} = \levicivita{ijk} a_j b_k \vec{e_i} = \levicivita{ijk} a_i b_j \vec{e_k} \\
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\det A = \levicivita{i_1 i_2 \dots i_n} A_{1i_1} A_{2i_2} \dots A_{ni_n}
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\det A = \levicivita{i_1 i_2 \dots i_n} A_{1i_1} A_{2i_2} \dots A_{ni_n}
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\end{math}
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\subsubsection*{Kronecker-Delta}
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$\krondelta{i,j}= \begin{cases} 1 & \mbox{falls } i=j \\ 0 & \mbox{falls } i \neq j \end{cases}$ \\s
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$\krondelta{i,j}= \begin{cases} 1 & \mbox{falls } i=j \\ 0 & \mbox{falls } i \neq j \end{cases}$ \\
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Die $n\times n$-Einheitsmatrix kann als $(\krondelta{ij})_{i,j\in\{1,\ldots,n\}}$ geschrieben werden.
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\section{Matrix-Operationen}
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\subsection*{Inversion}
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\subsubsection*{Reihenentwicklungen}
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\begin{align}
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exp(x) = \sum_{n = 0}^{\infty} {\frac{x^n}{n!}} \\
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sin (x) = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!} \\
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cos (x) = \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}
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\end{align}
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\section{\hypertarget{trans_ft}{Fourier-Transformation}}
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\subsection*{Fourier-Reihe}
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\subsubsection*{Definitionen:}
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\subsubsection*{Eigenschaften:}
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\subsection*{Fourier-Reihe (kontinuierlich):}
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\subsubsection*{Definitionen:}
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\subsubsection*{Eigenschaften:}
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\subsection*{kontinuierliche Fourier-Transformation:}
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Die kontinuierliche Fourier-Transformation ist eine Form der Fourier-Transformation, die es erlaubt, kontinuierliche, aperiodische Vorgänge in ein kontinuierliches Spektrum zu zerlegen.
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Oft wird diese Transformation auch einfach als Fourier-Transformation bezeichnet.
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\subsubsection*{Definition:}
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\begin{equation}
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\mathcal{F}\{f(t)\} = F(\omega)= \frac{1}{\sqrt{2 \pi}} \integrinf{f(t) e^{-\i \omega t}}{t}
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\end{equation}
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Rücktransformation (Fouriersynthese)
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\begin{equation}
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\mathcal{F}^{-1}\{F(\omega)\} = f(t)= \frac{1}{\sqrt{2 \pi}} \integrinf{F(\omega) e^{\i \omega t}}{\omega}
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\end{equation}
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Hierbei ist $F(\omega)$ das kontinuierliche Spektrum, das die Amplitude jeder Frequenz $\omega$ aus den reelle Zahlen angibt.
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\subsubsection*{Eigenschaften:}
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\section{Lineare Algebra}
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\subsection{Operatoren}
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\subsubsection*{hermitesche Operatoren}
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\subsubsection*{unitäre Operatoren}
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\subsection*{Matrizen-Operationen}
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\subsubsection*{Spur}
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\subsubsection*{Determinatante}
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\subsubsection*{\hypertarget{fs_linalg_mtrx_inv}{Inversion}}
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\begin{math}
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\hypertarget{fs_mtrx_inv_2d}{A^{-1} = \inlinematrix{a & b \\ c & d}^{-1} = \frac{1}{ad - bc} \inlinematrix{d & -b \\ -c & a}}
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A^{-1} = \inlinematrix{a & b \\ c & d}^{-1} = \frac{1}{ad - bc} \inlinematrix{d & -b \\ -c & a}
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\end{math}
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8
math.tex
8
math.tex
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@ -8,6 +8,7 @@
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\newcommand{\setR}{\mathbbm{R}}
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\newcommand{\setC}{\mathbbm{C}}
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\newcommand{\einsmatrix}{\mathbbm{1}}
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\newcommand{\bigO}{\mathbbm{O}}
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\newcommand{\inlinematrix}[1]{\begin{pmatrix} #1 \end{pmatrix}}
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\newcommand{\inlinematrixu}[1]{\begin{matrix} #1 \end{matrix}}
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\newcommand{\inlinematrixd}[2]{\left( \begin{matrix} #1\vphantom{#2} \end{matrix} ~\right| \left. \begin{matrix} \vphantom{#1}#2 \end{matrix} \right)}
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@ -28,11 +29,14 @@
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\newcommand{\diffP}[1]{\frac{\partial}{\partial #1}}
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\newcommand{\diffT}[1]{\frac{\text{d}}{\text{d} #1}}
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\newcommand{\diffTfrac}[2]{\frac{\text{d} #1}{\text{d} #2}}
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\newcommand{\diffTm}[3]{\diffTfrac{^{#1} #2}{#3^{#1}}}
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\newcommand{\diffPfrac}[2]{\frac{\partial #1}{\partial #2}}
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\newcommand{\intgr}[4]{\int_{#1}^{#2} #3 ~\text{d}#4}
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\newcommand{\intgru}[2]{\int #1 ~\text{d}#2}
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\newcommand{\integrinf}[2]{\int_{-\infty}^[\infty #1 \text{\,d}#2}
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\newcommand{\sbk}[1]{\left( #1 \right)}
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\newcommand{\levicivita}[1]{\varEpsilon_{#1}}
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\newcommand{\krondelta}[1]{\delta_{+1}}
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\newcommand{\levicivita}[1]{\varepsilon_{#1}}
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\newcommand{\krondelta}[1]{\delta_{#1}}
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@ -47,9 +47,9 @@
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% \include{ueb6}
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% \inlcude{ueb7}
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% \part{Formelsammlung}
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% \label{FS}
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% \include{formelsammlung}
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\part{Formelsammlung}
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\label{FS}
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\include{formelsammlung}
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\end{document}
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|
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87
ueb1.tex
87
ueb1.tex
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@ -6,7 +6,8 @@
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\section{Aufgabe 2: Pauli Matrizen}
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\subsection*{a)}
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\subsection*{b)}
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\subsection*{c)}
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\begin{math}
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\textbf{\sigma} = (\sigma_x, \sigma_y, \sigma_z) \\
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\textbf{\a},\textbf{\b} \in \setR
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@ -20,11 +21,87 @@
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\sum a_\alpha b_\beta \krondelta{\alpha \beta} \cdot \one + \i a_\alpha b_\beta \levicivita{\alpha,\beta,\gamma} \sigma_\gamma &= \\
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\one (\textbf{a \cdot b} + \i \sigma \cdot (a \times b)
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\end{align}
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\subsection*{b)}
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\subsection*{c)}
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\subsection*{d)}
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\subsection*{d)}
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$e^{\i \frac{\i}{2} n \sigma} = \one cos(\frac{\alpha}{2}) - \i (\textbf{n \cdot \sigma}) sin(\frac{\alpha}{2})$ \\
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Mit der Reihenentwicklung von $e^x$ ergibt sich:
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\begin{align}
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e^{- \i \frac{\alpha}{2} \textbf{n \cdot \sigma} &= \sum_k \frac{(- \i \frac{\alpha}}{2} \textbf{n \cdot \sigma})^k}{k!} \\
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&= \sum_k (- \i \frac{\alpha}{2})^k \cdot \frac{(\textbf{n \¢dot \sigma)^k}}{k!}
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\end{align}
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Desweiteren gilt nach Aufgabe 2 c):
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\begin{align}
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(\textbf{n \cdot \sigma} &= \sigma \\ \\
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(\textbf{n \cdot \sigma}^2 &= \one (\textbf{n \cdot n} + \underbrace{\i \textbf{\sigma} \cdot (\textbf{\n} \times \textbf{n})}_{=0} \\
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&= \one (\textbf{n \cdot n}
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\end{align}
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$\Rightarrow$
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\begin{align}
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\sum_k (- \i \frac{\alpha}{2})^k \cdot \frac{(\textbf{n \cdot \sigma)^k}}{k!} &= \\
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\sum_k ( (-\i)^{2k} (\frac{\alpha}{2})^{2k+1} \frac{(\textbf{n \cdot \sigma})^{2k}}{(2k)!} +
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\sum_k ( (-\i)^{2k+1} (\frac{\alpha}{2})^{2k+1} \frac{(\textbf{n \cdot \sigma})^{2k+1}}{(2k+1)!} &= \\
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\sum_k ( (-1)^k (\frac{\alpha}{2})^{2k+1} \frac{(\textbf{n \cdot \sigma})^{2k}}{(2k)!} +
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\sum_k ( \i \cdot (-1)^k (\frac{\alpha}{2})^{2k+1} \frac{(\textbf{n \cdot \sigma})^{2k} \cdot (\textbf{n \cdot \sigma})}{(2k+1)!} &= \\
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\sum_k ( (-1)^k (\frac{\alpha}{2})^{2k+1} \one \frac{1}{(2k)!} +
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\i \cdot \sum_k (-1)^k (\frac{\alpha}{2})^{2k+1} \one \cdot (\textbf{n \cdot \sigma}) \cdot \frac{1}{(2k+1)!} &= \\
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\one \cos(\frac{\alpha}{2}) + \i (\textbf{n \cdot \sigma}) \sin(\frac{\alpha}{2})
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\end{align}
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\section{Aufgabe 3: Operator-Identitäten}
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\subsection*{a)}
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$f(t) = e^{tA} \cdot B e^{-tA}$
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\begin{align}
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\diffTm{}{f(t)}{t} &= [A,f(t)] \\
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\diffTm{2}{f(t)}{t} &= [A,[A,f(t)]] \\ \\
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\diffTm{}{f(t)}{t} &= A e^{tA} B e^{-tA} t - e^{tA} B e^{-tA} A &= [A,f(t)] = A \cdot f(t) - f(t) \cdot A \\
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\diffT{t} [A,f(t)] &= \diffT{t} A \cdot f(t) - \diffT{t} f(t) \cdot A \\
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&= A \cdot [A,f(t)] - [A,f(t)] \cdot A \\
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&= [A,[A,f(t)]]
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\end{align}
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Die Taylorreihenentwicklung lautet dann:
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$f(t) = e^{tA} \cdot B e^{-tA} = B + \frac{t}{1!}[A,B]+\frac{t^2}{2!}[A,[A,B]]+\ldots$
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\subsection*{b)}
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\subsection*{c)}
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$\forall A,B,C : [A,B] = \i C \and [B,C] = \i A$ \\
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\begin{align}
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e^{\i B t} \cdot B e^{-\i B t} &= A \cos(t) + C \sin(t) \\
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&= A + [\i B,A] + \frac{t^2}{2}[\i B,[\i B, A]] \\
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[\i B, A] &= \i [B,A] &= -\i [A,B] \\
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[\i B, [\i B, A]] &= [\i B, C] &= \i \i A = -A }} \\
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e^{\i B t} A e^{-\i B t} &= A + C - \frac{t^2}{2} A \frac{t^3}{6} C + \frac{t^4}{24} A \\ \\
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&= A \cos(t) + C \sin(t)
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\end{align}
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\subsection*{c)}
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Es gelte: $[A,[A,B]] = [B,[A,B]] = 0$ \\
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\begin{align}
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e^{A+B} &= e^A \cdot e^B \cdot e^{-\frac{1}{2}[A,B]} \\
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g(t) &= e^{tA} \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} \\
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\diffTfrac{g(t)}{t} &= A \cdot e^{tA} \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} + e^{tA} \cdot (
|
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B \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} +
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e^{tB} \cdot -t \cdot e^{-\frac{t^2}{2}[A,B]} \cdot [A,B]) \\
|
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&= A \cdot g(t) + e^{tA} \cdot B \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]}
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- t \cdot [A,B] \cdot e^{tA} \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} \\
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&= A \cdot g(t) - t \cdot [A,B] \cdot e^{tA} \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} +
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e^{tA} \cdot B \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]} \\
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&= A \cdot g(t) -t \cdot g(t) \cdot [A,B] + e^{tA} \cdot B \cdot e^{tB} \cdot e^{-\frac{t^2}{2}[A,B]}
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\diffTfrac{g(t)}{t} &= A \cdot g(t) + e^{tA} \cdot B \cdot e^{-tA} \cdot g(t) - t \cdot g(t) \cdot [A,B] \\
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e^{x} \cdot e^{-x} &= \one \\
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\diffTfrac{g}{t} &= \right( A - t [A,B] + B + \frac{t}{1!} \cdot [A,B] \left) \cdot g(t) \\
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||||
\diffTfrac{g}{t} &= [A+B] \cdot g(t) \\
|
||||
\frac{\diffTfrac{g}{t}}{g} &= [A+B] \\
|
||||
g(t) &= e^{[A+B] \cdot t + c} \\
|
||||
e^A \cdot e^B &= e^B \cdot e^A \cdot e^[A,B] \\
|
||||
e^(A+B) &= e^A \cdot e^B \cdot e^{-\frac{1}{2}[A,B]} \\
|
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e^(B+A) &= e^B \cdit e^A \cdot e^{-\frac{1}{2}[A,B]}
|
||||
\end{align}
|
||||
|
|
|
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143
ueb2.tex
143
ueb2.tex
|
|
@ -3,15 +3,158 @@
|
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|
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\chapter{Quantenmechanik I - Übungsblatt 2}
|
||||
\section{Aufgabe 4: Unitäre Operatoren}
|
||||
Es gilt: $U^{-1} = U^\intercal$.
|
||||
\subsection*{a)}
|
||||
Zu zeigen: Der Eigenwert ist von der Form $a_n = e^{\i \alpha_n}$:
|
||||
Sei $U$ ein unitärer Operator und $\ket{a}$ Eigenvektor zum Eigenwert $a$.
|
||||
\begin{align}
|
||||
U \ket{a} &= a \cdot \ket{a} \\
|
||||
\bra{a} U^\intercal &= a^\ast \cdot \bra{a} \\
|
||||
\braket{a}{a} &= \dirac{a}{1}{a} \\
|
||||
&= \dirac{a}{U^\intercal U}{a}
|
||||
&= a^\ast \braket{a}{a} a \\
|
||||
&= \abs{a}^2 \cdot \braket{a}{a} \\
|
||||
&\Rightarrow \\
|
||||
\abs{a} &= 1 \\
|
||||
&\Rightarrow \\
|
||||
\abs{e^{\i \alpha_n}} &= 1
|
||||
\end{align}
|
||||
|
||||
Zu zeigen: Die Eigenvektoren sind orthogonal:
|
||||
Sei $U$ ein unitärer Operator und $\ket{a_n}$ bzw. $\ket{a_m}$ Eigenvektor zum Eigenwert $a_n$ bzw. $a_m$.
|
||||
\begin{align}
|
||||
\braket{a_n}{a_m} &= \dirac{a_n}{1}{a_m} \\
|
||||
&= \dirac{a_n}{U^\intercal U}{a_m} \\
|
||||
&= a_n^\ast \braket{a_n}{a_m} a_m \\
|
||||
&= e^{-\i \alpha_n} \cdot e^{\i \alpha_m} \cdot \braket{a_n}{a_m} \\
|
||||
&= e^{\i ( \alpha_m - \alpha_n)} \braket{a_n}{a_m} \\
|
||||
&\Rightarrow \\
|
||||
\braket{a_n}{a_m} &= 0
|
||||
\end{align}
|
||||
|
||||
|
||||
\subsection*{b)}
|
||||
Zu zeigen: Für $A$ hermitesch ist $U(s) = e^{-\i s A}$ unitär \\
|
||||
\begin{align}
|
||||
U(s) &= e^{\i s A} \\
|
||||
&= \sum_{k=0}^\infty \frac{1}{k!} \cdot (-\i s A)^k \\
|
||||
&= \sum_{k=0}^\infty \frac{1}{k!} \cdot (\i s A^\intercol)^k \\
|
||||
&= e^{\i s A}
|
||||
\end{align}
|
||||
|
||||
Zu zeigen: Für $A$ hermitesch ist $U(s_1 + s_2) = U(s_1) \cdot U(s_2)$.
|
||||
\begin{align}
|
||||
U(s_1 + s_2) &= e^{-\i (s_1 + s_2) A}\\
|
||||
&= e^{-\i s_1 A - \i s_2 A} \\
|
||||
&= e^{-\i s_1 A} \cdot e^{-\i s_2 A} \\
|
||||
&= U(s_1) \cdot U(s_2)
|
||||
\end{align}
|
||||
|
||||
|
||||
|
||||
\section{Aufgabe 5: Spur und Determinante}
|
||||
\subsection*{a)}
|
||||
Zu zeigen: $[A,BC] &= B \cdot [A,C] + [A,B] \cdot C$.
|
||||
\begin{align}
|
||||
[A,BC] &= ABC -BCA \\
|
||||
B \cdot [A,C] + [A,B] \cdot C &= B \cdot (AC - CA) + (AB - BA) \cdot C \\
|
||||
&= BAC - BCA + ABC - BAC \\
|
||||
&= ABC - BCA \\
|
||||
&= [A,BC]
|
||||
\end{align}
|
||||
\subsection*{b)}
|
||||
Zu zeigen: Für endliche Operatoren gilt:
|
||||
$\tr(AB) = \tr(BA)$
|
||||
|
||||
\begin{align}
|
||||
\tr(AB) &= \sum_{i=1}^N \sum_{k=1}^N a_{ik} b_{ki} \\
|
||||
&= \sum_{i=1}^N \sum_{k=1}^N b_{ik} a_{ki} \\
|
||||
&= \tr(BA)
|
||||
\end{align}
|
||||
|
||||
Zu zeigen: Die Spur ist invariant unter zyklischen Vertauschungen:
|
||||
\begin{align}
|
||||
\tr(ABC) &= \tr(A \cdot (BC)) \\
|
||||
&= \tr((BC) \cdot A) \\
|
||||
&= \tr(BCA) \\
|
||||
&= \tr(B \cdot (CA)) \\
|
||||
&= \tr((CA) \cdot B) \\
|
||||
&= \tr(CAB)
|
||||
\end{align}
|
||||
|
||||
Zu zeigen: Die Spur ist unabhänging von der Basis:
|
||||
Sei hierzu $T^{-1}$ die Matrix der neuen Basisverktoren dargestellt in der alten Basis.
|
||||
Dann ist $T^{-1}AT$ die Basistransformation von der A-Basis in die T-Basis.
|
||||
\begin{equation}
|
||||
\tr(T^{-1}AT) = \tr(T^{-1}TA) = \tr(A)
|
||||
\end{equation}
|
||||
|
||||
\subsection*{c)}
|
||||
\begin{align}
|
||||
\diffT{t}A(t) &= A(t) \cdot B \\
|
||||
\frac{\diffT{t}A(t)}{A(t)} &= B \\
|
||||
\ln(A(t)) &= B \cdot t + c \\
|
||||
A(t) &= e^{Bt+c} \\
|
||||
&= A(0) \cdot e^{Bt} \\
|
||||
&= A_0 \cdot B \cdot e^{Bt} \\
|
||||
&= A_0 \cdot e^{Bt} \cdot B \\
|
||||
\diffT{t}A_2(t) &= B \cdot A_2(t)\\
|
||||
A_2(t) &= e^{Bt} \cdot A_0
|
||||
\end{align}
|
||||
|
||||
\subsection*{d)}
|
||||
\begin{align}1 + \epsilon \tr(A) + O(\epsilon^2)
|
||||
det(1 + \epsilon A) &= \inlinematrixdet{1+\epsilon A_{11} & \ldots & 0+\epsilon A_{1n} \\
|
||||
\vdots & 1+\epsilon A_{ii} & \vdots \\
|
||||
0+\epsilon A_{n1} & \ldots & 1+\epsilon A_{nn}
|
||||
} \\
|
||||
&= \prod_i (1 + \epsilon A_{ii}) + \bigO(\epsilon^2) \\
|
||||
&= 1 + \epsilon \sum A_{ii} + \bigO(\epsilon^2)\\
|
||||
&= 1 + \epsilon \tr(A) + \bigO(\epsilon^2)
|
||||
\end{align}
|
||||
|
||||
Zu zeigen: $\det(e^A = e^{\tr(A))$
|
||||
\begin{align}
|
||||
g(t) &= \det(e^{At}) \\
|
||||
&\stackrel{tailor}{} \det(1 + At + \bigO(t^2)) \\
|
||||
&= 1 + \tr(A) + \bigO(t^2) \\
|
||||
&\stackrel{tailor ``rückwärs''}{} e^{\tr(A)t}
|
||||
\end{align}
|
||||
|
||||
\begin{align}
|
||||
g(t) &= \limits_{\epsilon \rightarrow \intfy} \frac{1}{\epsilon} \sbk{\det(e^{A(t+\epsilon)) - det(e^{At})} \\
|
||||
&= g(t) \limits_{\epsilon \rightarrow \intfy} \frac{1}{\epsilon} \det(e^{A \epsilon}) \\
|
||||
&= g(t) \limits_{\epsilon \rightarrow \intfy} \frac{1}{\epsilon} \sbk{\det(1 + A \epsilon + \bigO(\epsilon^2) - det(1)} \\
|
||||
&= g(t) \tr(A) \\
|
||||
&\Rightarrow g(t) = e^{\tr(A) \cdot t}
|
||||
\end{align}
|
||||
|
||||
|
||||
Ist A diagonalisierbar:
|
||||
\begin{math}
|
||||
\det(e^A) = \det(T^{-1} e^A T) = \det(e^\hat{A}) = \prod_i e^{\lambda_i} = e^{\sum_i \lambda_i} = e^{\tr(A)}
|
||||
\end{math}
|
||||
|
||||
|
||||
\section{Aufgabe 6: Hermitesche Matrizen}
|
||||
\subsection*{a)}
|
||||
Es gilt: $M_i^2 = \one$
|
||||
Sei $\bra{a}$ Eigenvektor zum Eigenwert $a$
|
||||
\begin{align}
|
||||
M_i \ket{a} &= a \cdot \bra{a} &| M_i von links
|
||||
M_i^2 \ket{a} &= M_i a \cdot \bra{a} \\
|
||||
\one \ket{a} &= a^2 \bra{a} \\
|
||||
&\Rightarrow a = \pm 1
|
||||
|
||||
\end{align}
|
||||
|
||||
\subsection*{b)}
|
||||
Für $i \neq j$ folgt mit $M_i M_j + M_j M_i = 2 \krondelta{ij} \one$ : $M_i M_j = - M_j M_i$
|
||||
Dann gilt:
|
||||
\begin{align}
|
||||
\tr(M_i M_j + M_j M_i) &= tr(2 \krondelta{ij} \one) \\
|
||||
\tr(M_i M_j) + \tr(M_j M_i) &= tr(\mathbb{0}) \\
|
||||
\tr(M_i M_j) &= -tr(M_j M_i) \\
|
||||
\tr(M_i M_j)
|
||||
\end{align}
|
||||
|
||||
|
|
|
|||
22
ueb7.tex
22
ueb7.tex
|
|
@ -4,9 +4,31 @@
|
|||
\chapter{Quantenmechanik I - Übungsblatt 7}
|
||||
\section{Aufgabe 17: Unendlich hoher Potentialtop (Ergänzungen)}
|
||||
\subsection*{a)}
|
||||
|
||||
\begin{math}
|
||||
\Phi_n(p) = \intgrinf{\frac{1}{\sqrt{2 \pi \hbar}} \cdot \Phi(x) \cdot e^{-\frac{\i p x}{\hbar}}{x}
|
||||
\end{math}
|
||||
|
||||
Für n = ungerade:
|
||||
|
||||
\begin{align}
|
||||
\Phi_n(p) &= \integrinf{\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} \\ \\
|
||||
&= \frac{1}{\sqrt{2 \pi a \hbar}} \cdot \sbk {
|
||||
\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} +
|
||||
\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}}
|
||||
\Phi_n(p) &=
|
||||
\begin{case}
|
||||
\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}}
|
||||
\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}}
|
||||
\end{case}
|
||||
|
||||
\end{align}
|
||||
|
||||
|
||||
\subsection*{b)}
|
||||
\subsection*{c)}
|
||||
|
||||
|
||||
\section{Aufgabe 18: Tunneleffekt}
|
||||
\includegraphics{grafiken/U_A18_1.pdf}
|
||||
\subsection*{a)}
|
||||
|
|
|
|||
Loading…
Reference in New Issue