2008-08-08 12:29:25 +00:00
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\chapter{Rotationsinvarianz in d=3}
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\section{Drehimpulsalgebra}
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Drehung mit dem Winkel $\phi$ um $\vec{n}$:
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\begin{equation}
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\ket{\tilde{\psi}} = D(\phi,\vec{n})\ket{\psi}
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\end{equation}
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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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2008-09-25 13:15:14 +00:00
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In (\ref{labelTransfRot}) hatten wir die Relation
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2008-08-08 12:29:25 +00:00
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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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(etc. zyclisch). Diese Vertauschungsrelation bestimmt das Spektrum der $J$-Operatoren vollständig. Wir definieren ein $J^2$ zu:
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\begin{equation}
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J^2 = \vec{J}^2 = J_x^2 + J_y^2 + J_z^2
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\end{equation}
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und daraus folgt $\forall \alpha = x,y,z$
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\begin{equation}
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\left[ J^2, J_\alpha \right] = 0
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\end{equation}
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also haben $J^2$ und $J_z$ gemeinsame Eigenvektoren.
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\begin{align}
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J^2 \ket{\alpha,\beta} &= \alpha \ket{\alpha,\beta}\\
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J_z \ket{\alpha,\beta} &= \beta \ket{\alpha,\beta}
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\end{align}
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In Anlehnung an Erzeuger und Vernichter definieren wir:
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\begin{equation}
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J_\pm \equiv J_x \pm iJ_y
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\end{equation}
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mit dem Kommutator
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\begin{align}
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[J_z, J_\pm] &= [J_z,J_x + iJ_y]\\
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&= i\hbar J_y \pm i(-i\hbar) J_x
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&= \pm \hbar J_\pm
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\end{align}
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und
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\begin{equation}
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[J^2,J_\pm] = 0
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\end{equation}
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Vergleiche Harmonischen Oszillator:
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\begin{equation}
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[\nOp,\aDs] = -\aDs; ~ [\nOp,\aCr] = \aCr
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\end{equation}
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mit $J_\pm$ ist dann
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\begin{align}
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J_z J_+ \ket{\alpha,\beta} &= (J_+ J_z + i\hbar J_+) \ket{\alpha,\beta}\\
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&= (\beta + \hbar) J_+ \ket{\alpha,\beta}\\
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J_z J_- \ket{\alpha,\beta} &= (\beta - \hbar) J_- \ket{\alpha,\beta}
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\end{align}
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und
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\begin{equation}
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J^2 J_+ \ket{\alpha,\beta} = J_+ J^2 \ket{\alpha,\beta} = \alpha J_+ \ket{\alpha,\beta}
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\end{equation}
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also
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\begin{align}
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J_+ \ket{\alpha,\beta} &= c_+(\alpha,\beta) \ket{\alpha,\beta + 1}\\
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J_- \ket{\alpha,\beta} &= c_-(\alpha,\beta) \ket{\alpha,\beta - 1}
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\end{align}
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$\beta$-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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2008-09-25 13:15:14 +00:00
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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
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$(1,-1)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $2$ \graycell \\
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\end{tabular}
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\end{center}
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\item für $J_z / \hbar$
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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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$(\frac{1}{2},\frac{1}{2})$ & $0$ & $\frac{1}{2}$ \graycell & $0$ \graycell & $0$ & $0$ & $0$ \\ \hline
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$(\frac{1}{2},-\frac{1}{2})$ & $0$ & $0$ \graycell & $-\frac{1}{2}$ \graycell & $0$ & $0$ & $0$ \\ \hline
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$(1,1)$ & $0$ & $0$ & $0$ & $1$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
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$(1,0)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
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$(1,-1)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $1$ \graycell \\
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\end{tabular}
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\end{center}
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\item für $J_x / \hbar$
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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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$(\frac{1}{2},\frac{1}{2})$ & $0$ & $0$ \graycell & $\frac{1}{2}$ \graycell & $0$ & $0$ & $0$ \\ \hline
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$(\frac{1}{2},-\frac{1}{2})$ & $0$ & $\frac{1}{2}$ \graycell & $0$ \graycell & $0$ & $0$ & $0$ \\ \hline
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$(1,1)$ & $0$ & $0$ & $0$ & $1$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
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$(1,0)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
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$(1,-1)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $1$ \graycell \\
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\end{tabular}
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\end{center}
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\item für $J_y / \hbar$
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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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$(\frac{1}{2},\frac{1}{2})$ & $0$ & $0$ \graycell & $-\frac{i}{2}$ \graycell & $0$ & $0$ & $0$ \\ \hline
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$(\frac{1}{2},-\frac{1}{2})$ & $0$ & $\frac{i}{2}$ \graycell & $0$ \graycell & $0$ & $0$ & $0$ \\ \hline
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$(1,1)$ & $0$ & $0$ & $0$ & $1$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
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$(1,0)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $0$ \graycell \\ \hline
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$(1,-1)$ & $0$ & $0$ & $0$ & $0$ \graycell & $0$ \graycell & $1$ \graycell \\
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\end{tabular}
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\end{center}
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\end{itemize}
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Erkennbar ist hier jeweils eine Blockstruktur!
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