Kapitel III.3 WIP
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Oliver Groß
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\chapter{Beispiel: 2-dim. harmonischer Oszillator}
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\section{Bestimmung des Spektrum: Abbildung auf 1-dim. harmonischer Oszillator}
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Teilchen mit Masse $\mu$ in $V(x,y)$:
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\begin{align}
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V(x,y) &= \frac{\mu}{2} \omega^2 (x^2 + y^2)\\
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&= \frac{\mu}{2} \omega^2 \rho^2
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\end{align}
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\paragraph{Lösung der Schrödingergleichung} Naiver Weg: Asymptotik abspalten und Potenzreihenansatz in
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\begin{equation}Darstellung
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\left( -\frac{k^2}{2\mu} \left( \diffPs{\rho}^2 + \frac{1}{\rho} \diffPs{\rho} - \frac{m^2}{\rho^2} \right) + V(\rho) \right) R_m(\rho) = E R_m(\rho)
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\end{equation}
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der abberchen muss woraus das Spektrum folgt.
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\begin{equation}
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\left( -\frac{\hbar^2}{2\mu} \left( \diffPs{x}^2 \diffPs{y}^2 \right) + \frac{\mu}{2} \omega^2 (x^2 + y^2) \right) \phi(x,y) = E \phi(x,y)
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\end{equation}
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Lösung durch Separation der Variablen: Mit
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\begin{equation}
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\phi(x,y) = \psi_1(x) \cdot \psi_2(y)
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\end{equation}
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erhält man
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\begin{equation}
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\underbrace{\frac{\left( -\frac{\hbar^2}{2\mu} \diffPs{x}^2 + \frac{\mu}{2} \omega^2 x^2 \right) \psi_1}{\psi_1}}_{\equiv E_1} +
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\underbrace{\frac{\left( -\frac{\hbar^2}{2\mu} \diffPs{y}^2 + \frac{\mu}{2} \omega^2 y^2 \right) \psi_2}{\psi_2}}_{\equiv E_2}
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\end{equation}
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daraus folgt:
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\begin{align}
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E_1 &= \hbar \omega \left( n_1 + \frac{1}{2} \right)\\
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E_2 &= \hbar \omega \left( n_2 + \frac{1}{2} \right)
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\end{align}
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Spektrum des 2-dimensionalen harmonischen Oszillators:
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\begin{align}
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E &= \hbar \omega \left( n_1 + n_2 + \frac{1}{2} + \frac{1}{2} \right) & \left( n_{1,2} = 0,1,2,... \right)\\
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\rightarrow E &= \hbar \omega (N + 1) & \left( N = 0,1,2,... \right)
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\end{align}
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Die Eigenvektoren sind dann: $\ket{n_1,n_2}$. Darüber hinaus sind alle $n_1$, $n_2$ mit $n_1 + n_2 = N$ entartet:
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%\begin{figure}[H] \centering
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%\includegraphics{pdf/III/03-01-00.pdf}
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%\end{figure}
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$\Rightarrow$ $N$ ist $(N+1)$-fach entartet!
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\paragraph{Befund} Rotationsinvarianz ist nicht sichtbar in der Darstellung$ \ket{n_1,n_2}$: Beispielsweise ist im Ortsraum
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\begin{equation}
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\braket{x,y}{0,1} = \underbrace{\braket{x}{1}}_{\phi_1(x)} \underbrace{\braket{y}{0}}_{\phi_0(y)} = \const \cdot x e^{-\frac{x^2}{2}} e^{-\frac{y^2}{2}}
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\end{equation}
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nicht rotationsinvariant.
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\section{Explizite Rotationsinvarianz}
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Wir suchen $\ket{n,m}$ mit:
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\begin{align}
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H \ket{n,m} &= E_{n,m} \ket{N,m}\\
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J_3 \ket{n,m} &= \hbar m \ket{N,m}
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\end{align}
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Wobei wir bereits $\set{\ket{n_1,n_2}}$ haben, d.h. letztlich: $\braket{n,m}{n_1,n_2}$. Die Erzeuger- und Vernichteroperatoren sind:
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\begin{itemize}
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\item Harmonischer Oszillator in $x$-Richtung:
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\begin{align}
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\aDs_1 &= \frac{1}{\sqrt{2}} (x + i p_x)\\
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\aCr_1 &= \frac{1}{\sqrt{2}} (x - i p_x)
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\end{align}
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\item Harmonischer Oszillator in $y$-Richtung:
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\begin{align}
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\aDs_2 &= \frac{1}{\sqrt{2}} (y + i p_y)\\
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\aCr_2 &= \frac{1}{\sqrt{2}} (y - i p_y)
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\end{align}
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\end{itemize}
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Daraus ergibt sich der Hamiltonoperator zu
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\begin{equation}
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H = \hbar \omega \left( \aCr_1 \aDs_1 + \aCr_2 \aDs + \frac{1}{2} + \frac{1}{2} \right)
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\end{equation}
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und mit
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\begin{equation}
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\nOp_1 = \aCr_1 \aDs_1 ; ~ \nOp_1 = \aCr_2 \aDs_2 ; ~ \ket{n_1,n_2} = \ket{n_1} \otimes \ket{n_2}
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\end{equation}
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und
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\begin{align}
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\aCr_1 \ket{n_1, n_2} &= \sqrt{n_1 + 1} \ket{n_1+1,n_2} \\
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\aDs_2 \ket{n_1, n_2} &= \sqrt{n_2} \ket{n_1,n_2-1}
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\end{align}
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ist:
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\begin{equation}
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H \ket{n_1,n_2} = \hbar \omega (n_1 + n_2 + 1) \ket{n_1, n_2}
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\end{equation}
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$J_3$ in Ortsdarstellung ist
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\begin{align}
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J_3 &= \hbar \left( x \cdot p_y - y \cdot p_x \right) \\
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&= \hbar \left( \frac{1}{\sqrt{2}} \left( \aCr_1 + \aDs_1 \right) \cdot \frac{i}{\sqrt{2}} \left( \aCr_2 - \aDs_2 \right) - \frac{1}{\sqrt{2}} \left( \aCr_2 + \aDs_2 \right) \cdot \frac{i}{\sqrt{2}} \left( \aCr_1 - \aDs_1 \right) \right) \\
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&= i\hbar \left( -\aCr_1 \aDs_2 + \aCr_2 \aDs_1 \right)
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\end{align}
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und damit:
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\begin{align}
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J_3 \ket{0,0} &= i\hbar \left( -\aCr_1 \aDs_2 + \aCr_2 \aDs_1 \right) \ket{0,0} \\
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&= i\hbar \ket{\zero} \\
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&= \zero \ket{0,0}\\[15pt]
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J_3 \ket{0,1} &= i\hbar \left( -\aCr_1 \aDs_2 + \aCr_2 \aDs_1 \right) \ket{0,1} \\
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&= -i\hbar \ket{1,0} \\[15pt]
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J_3 \ket{1,0} &= i\hbar \ket{0,1}
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\end{align}
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Ansatz:
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\begin{align}
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J_3 \left( c_1 \ket{0,1} + c_2 \ket{1,0} \right) &\stackrel{!}{=} \eta \left( c_1 \ket{0,1} + c_2 \ket{1,0} \right)\\
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c_1 (-i\hbar) \ket{1,0} + c_2 (i\hbar) \ket{1,0} &\stackrel{!}{=} \eta \left( c_1 \ket{0,1} + c_2 \ket{1,0} \right)\\[15pt]
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\Rightarrow c_2 (i\hbar) &= \eta c_1\\
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c_1 (-i\hbar) &= \mu c_2\\[15pt]
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\Rightarrow \eta &= i\hbar
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\end{align}
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Eigenvektoren von $J_3$:
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\begin{itemize}
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\item zu $\eta = +i \hbar$:
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\begin{equation}
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c_1 = i c_2 \rightarrow \frac{1}{\sqrt{2}} \underbrace{\left( \ket{0,1} - i\ket{1,0} \right)}_{\ket{N=1,m=1}}
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\end{equation}
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\item zu $\eta = -i \hbar$:
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\begin{equation}
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c_2 = i c_1 \rightarrow \frac{1}{\sqrt{2}} \left( \ket{0,1} + i\ket{1,0} \right)
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\end{equation}
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\end{itemize}
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\begin{align}
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\braket{x,y}{N=1,m=1} &= \bra{x,y \frac{1}{\sqrt{2}}} \left( \ket{0,1} + i\ket{1,0} \right)\\
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&= \const (y - ix) e^{-\frac{x^2}{2}} e^{-\frac{y^2}{2}}\\
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&= \const (-i) (x + iy) e^{-\frac{x^2}{2}} e^{-\frac{y^2}{2}}\\
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&= \const (-i) \rho e^{i\phi} e^{-\frac{\rho^2}{2}} \left[ = \braket{\rho,\phi}{N=1,m=1} \right]\\
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&= \const (-i) \underbrace{e^{i\phi}}_{\chi_1(\phi)} \underbrace{\rho e^{-\frac{\rho^2}{2}}}_{R_{n=...,m=1}(\rho)}
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\end{align}
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Spektrum:\\
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Definiere neue erzeuger und Vernichter
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\begin{align}
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\aDs_\pm &\equiv \frac{1}{\sqrt{2}} \left( \aDs_1 \mp i\aDs_2 \right)\\
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\aCr_\pm &\equiv \frac{1}{\sqrt{2}} \left( \aCr_1 \pm i\aCr_2 \right)
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\end{align}
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mit Anzahloperator
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\begin{equation}
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\nOp_\pm = \aCr_\pm \aDs_\pm
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\end{equation}
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Kommutator:
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\begin{equation}
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\left[\aDs_+, \aCr_+\right] = 1 = \left[\aDs_-, \aCr_-\right] \text{ (alle anderen sind $0$)}
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\end{equation}
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Hamiltonoperator:
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\begin{equation}
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H = \hbar \omega \left( \aCr_+ \aDs_+ + \aCr_- \aDs_- + 1 \right)
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\end{equation}
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Drehoperator:
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\begin{align}
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J_3 &= \hbar \left( \aCr_+ \aDs_+ - \aCr_- \aDs_- \right)\\
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&= \hbar \left( \nOp_+ - \nOp_- \right)
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\end{align}
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und daraus ist
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\begin{equation}
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H = \hbar \omega \left( \nOp_+ + \nOp_- + 1 \right)
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\end{equation}
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d.h.
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\begin{align}
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H \ket{n_+,n_-} &= \hbar \omega (\overbrace{n_+ + n_-}^{N} + 1) \ket{n_+,n_-}\\
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J_3 \ket{n_+,n_-} &= \hbar \overbrace{(n_+ - n_-)}^{m} \ket{n_+,n_-}\\
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\end{align}
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Daraus folgt:
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\begin{enumerate}
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\item Für gegebenes $n_+ + n_- = N$ kann $m = N, N-2, N-4, ..., -N$ sein. ($(N+1)$-fache Entartung!)
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\item Für festes $m$ kann $N = \abs{m} + 2n_r$ sein.
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\end{enumerate}
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Wir hatten in (\ref{rotSymSGL}) für das Spektrum:
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%\begin{figure}[H] \centering
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%\includegraphics{pdf/III/03-02-00.pdf}
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%\end{figure}
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\paragraph{Erklärung der höheren Entartung}
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``Rotation im Iso-Spin Raum''
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\begin{align}
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H &= \hbar \omega \left( 1 + \inlinematrix{\aCr_1 & \aCr_2} \inlinematrixdet{1 & 0 \\ 0 & 1} \inlinematrixdet{\aDs_1 \\ \aDs_2} \right)\\
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&= \hbar \omega \left( 1 + \inlinematrix{\aCr_1 & \aCr_2} e^{i\phi\sigma_z} \inlinematrixdet{1 & 0 \\ 0 & 1} e^{-i\phi\sigma_z} \inlinematrixdet{\aDs_1 \\ \aDs_2} \right)\\
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&= \hbar \omega \left( 1 + \inlinematrix{h^\dagger_1 & h^\dagger_2} \inlinematrixdet{1 & 0 \\ 0 & 1} \inlinematrixdet{h_1 \\ h_2} \right)
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\end{align}
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mit
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\begin{equation}
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\inlinematrix{h_1 \\ h_2} = \left(\cos(\phi) \one - i\sin(\phi)\sigma_z\right) \inlinematrix{\aDs_1 \\ \aDs_2}
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\end{equation}
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für $\phi = \frac{\pi}{2}$ ist
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\begin{equation}
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h_{1/2} \cequiv a_\pm
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\end{equation}
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\section{Vollständiger Satz kommutierender Obselvablen (gute und schlechte Quantenzahlen)}
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\paragraph{Variante 1}
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\begin{equation}
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H = H_1 + H_2 , ~ H_1 = \frac{p_x}{2\mu} + \frac{\mu}{2} \omega^2 x^2 , ~ H_2 = \frac{p_y}{2\mu} + \frac{\mu}{2} \omega^2 y^2
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\end{equation}
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\begin{align}
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H\ket{\phi} &= E\ket{\phi}\\
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H\ket{n_1,n_2} &= E(n_1,n_2)\ket{n_1,n_2}
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\end{align}
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mit
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\begin{equation}
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E(n_1,n_2) = \hbar \omega \left( n_1 + n_2 + 1 \right)
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\end{equation}
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für $n_1, n_2 = 0, 1, 2, ...$ und den Kommutatoren
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\begin{equation}
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[H,H_2] \neq [H,H_1] \neq [H_1,H_2] = 0
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\end{equation}
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und
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\begin{equation}
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H \ket{N,n_2} = E(N,n_2)\ket{N,n_2}
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\end{equation}
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mit
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\begin{equation}
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E(N,n_2) = \hbar \omega (N+1); ~ n_2 = 0,1,2,...; ~ N = n_2,n_2+1,n_2+2,...
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\end{equation}
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VSKO (CSCO): $(H,H_1)$ , $(H,H_2)$ oder $(H_1,H_2)$\\
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Problem:
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\begin{equation}
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J_3\ket{n_1,n_2} \neq \const \ket{n_1,n_2}
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\end{equation}
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d.h.
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\begin{equation}
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[J_3,H_1] \neq 0 \neq [J_3,H_2]
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\end{equation}
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aber
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\begin{equation}
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[J_3,H] = 0
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\end{equation}
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\paragraph{Variante 2}
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\begin{align}
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H &= \hbar \omega \left( \aCr_+ \aDs_+ + \aCr_- \aDs + 1 \right)\\
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H\ket{n_+,n_-} &= E(n_+,n_-)\ket{n_+,n_-}
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\end{align}
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mit
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\begin{equation}
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E(n_+,n_-) = \hbar \omega \left( n_+ + n_- + 1 \right)
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\end{equation}
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für
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\begin{equation}
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n_+,n_- = 0,1,2,...
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\end{equation}
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und
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\begin{equation}
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J_3 \ket{n_+,n_-} = m \hbar \ket{n_+,n_-}
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\end{equation}
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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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