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%\includegraphics{excs/qm1_blatt01_SS08.pdf}
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%\pagebreak
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\chapter{Quantenmechanik I - Übungsblatt 1}
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\section{Aufgabe 1: Stern-Gerlach Experimente}
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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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2008-06-30 13:18:36 +00:00
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\begin{math}
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\vec{\sigma} = (\sigma_x, \sigma_y, \sigma_z) \\
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\vec{a}, \vec{b} \in \setR
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\end{math}
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\begin{align}
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(\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma}) &= \one (\vec{a} \cdot \vec{b} + \i \vec{\sigma} \cdot (\vec{a} \times \vec{b}) \\
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\sum a_\alpha b_\beta \sigma_\alpha \sigma_\beta &= \\
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\sum a_\alpha b_\beta ( \krondelta{\alpha \beta} \one + \i \levicivita{\alpha,\beta,\gamma} \sigma_\gamma ) &= \\
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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 (\vec{a} \cdot \vec{b} + \i \sigma \cdot (a \times b)
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\end{align}
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\subsection*{d)}
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$e^{\i \frac{\i}{2} n \sigma} = \one cos(\frac{\alpha}{2}) - \i (\vec{n} \cdot \vec{\sigma}) sin(\frac{\alpha}{2})$ \\
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Mit der Reihenentwicklung von $e^x$ ergibt sich:
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Desweiteren gilt nach Aufgabe 2 c):
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\begin{align}
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(\vec{n} \cdot \vec{\sigma} &= \sigma \\
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(\vec{n} \cdot \vec{\sigma})^2 &= \one (\vec{n} \cdot \vec{n} + \underbrace{\i \vec{\sigma} \cdot (\vec{n} \times \vec{n})}_{=0} \\
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&= \one (\vec{n} \cdot \vec{n}
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\end{align}
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$\Rightarrow$
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\begin{align}
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\sum_k \sbk{- \i \frac{\alpha}{2}}^k \cdot \frac{\sbk{\vec{n} \cdot \vec{\sigma}}^k}{k!} &= \\
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\sum_k \sbk{ (-\i)^{2k} (\frac{\alpha}{2})^{2k+1} \frac{(\vec{n} \cdot \vec{\sigma})^{2k}}{(2k)!}} +
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\sum_k \sbk{ (-\i)^{2k+1} (\frac{\alpha}{2})^{2k+1} \frac{(\vec{n} \cdot \vec{\sigma})^{2k+1}}{(2k+1)!}} &= \\
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\sum_k \sbk{ (-1)^k \sbk{\frac{\alpha}{2}}^{2k+1} \frac{\sbk{\vec{n} \cdot \vec{\sigma}}^{2k}}{(2k)!}} +
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\sum_k \sbk{ \i \cdot (-1)^k \sbk{\frac{\alpha}{2}}^{2k+1} \frac{\sbk{\vec{n} \cdot \vec{\sigma}}^{2k} \cdot \sbk{\vec{n} \cdot \vec{\sigma}}}{(2k+1)!}} &= \\
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\sum_k ( (-1)^k \sbk{\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 (\vec{n} \cdot \vec{\sigma}) \cdot \frac{1}{(2k+1)!} &= \\
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\one \cos(\frac{\alpha}{2}) + \i (\vec{n} \cdot \vec{\sigma}) \sin(\frac{\alpha}{2})
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\end{align}
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2008-06-23 10:39:00 +00:00
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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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2008-06-23 10:39:00 +00:00
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\subsection*{b)}
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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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2008-07-08 10:30:05 +00:00
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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} &= \left( A - t [A,B] + B + \frac{t}{1!} \cdot [A,B] \right) \cdot g(t) \\
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\diffTfrac{g}{t} &= [A+B] \cdot g(t) \\
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\frac{\diffTfrac{g}{t}}{g} &= [A+B] \\
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g(t) &= e^{[A+B] \cdot t + c} \\
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e^A \cdot e^B &= e^B \cdot e^A \cdot e^[A,B] \\
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e^(A+B) &= e^A \cdot e^B \cdot e^{-\frac{1}{2}[A,B]} \\
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e^(B+A) &= e^B \cdot e^A \cdot e^{-\frac{1}{2}[A,B]}
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\end{align}
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