264 lines
9.9 KiB
TeX
264 lines
9.9 KiB
TeX
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\chapter{Harmonischer Oszilator}
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\section{Algebraische Lösung des Spektrums von $H$}
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\begin{align}
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H &= \frac{P^2}{2 m} + \frac{m}{2} \omega^2 X^2; \text{ mit } \hat{x} \equiv \left( \frac{m \omega}{\hbar} \right)^\frac{1}{2} X; ~ \hat{p} \equiv \left( \frac{1}{\hbar m \omega} \right)^2 P
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&= \frac{\hbar \omega}{2} \left( \hat{p}^2 + \hat{x}^2 \right)
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\end{align}
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mit
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\begin{equation}
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[\hat{x}, \hat{p}] = i
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\end{equation}
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\paragraph{Vernichtungsoperator}
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\begin{align}
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\aDs \equiv \frac{1}{\sqrt{2}} \left( \hat{x} + i \hat{p} \right)\\
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\aCr \equiv \frac{1}{\sqrt{2}} \left( \hat{x} - i \hat{p} \right)
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\end{align}
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daraus ergeben sich $\hat{x}$ und $\hat{p}$ als:
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\begin{align}
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\hat{x} &= \frac{1}{\sqrt{2}} \left( \aDs + \aCr \right)\\
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\hat{p} &= \frac{1}{\sqrt{2}} \left( \aDs - \aCr \right)
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\end{align}
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\subparagraph{Kommutator}
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\begin{align}
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[\aDs, \aCr] &= \frac{1}{2} [\hat{x} + i \hat{p}, \hat{x} - i \hat{p}]\\
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&= -i[\hat{x}, \hat{p}]\\
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&= \one = 1
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\end{align}
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eingesetzt in $H$:
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\begin{align}
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H &= \frac{\hbar \omega}{2} \left( \hat{p}^2 + \hat{x}^2 \right)\\
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&= \frac{\hbar \omega}{4} \left( -\left( \aCr\aCr - \aDs\aCr - \aDs\aCr + \aDs\aDs \right) + \left( \aDs\aDs + \aDs\aCr + \aCr\aDs + \aCr\aCr \right) \right)\\
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&= \frac{\hbar \omega}{4} \left( 2\aDs\aCr + 2\aCr\aDs \right)\\
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&= \frac{\hbar \omega}{2} \left( 2\aCr\aDs + \one \right)\\
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&= \hbar \omega \left( \aCr\aDs + \frac{\one}{2} \right)
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\end{align}
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\paragraph{Anzahloperator}
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\begin{equation}
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\nOp \equiv \aCr \aDs
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\end{equation}
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\subparagraph{Kommutatoren}
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\begin{align}
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[\nOp, \aDs] &= [\aCr\aDs, \aDs] = [\aCr, \aDs] \aDs = -\aDs\\
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[\nOp, \aCr] &= [\aCr\aDs, \aCr] = \aDs [\aDs, \aDs] = \aCr
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\end{align}
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\subparagraph*{Spektrum von $\nOp$}
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\begin{enumerate}
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\item Sei $\ket{\nu}$ Eigenvektor von $\nOp$ mit Eigenwert $\nu$:
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\begin{equation}
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\nOp \ket{\nu} = \nu \ket{\nu} \text{ mit } \braket{\nu}{\nu} > 0
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\end{equation}
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\item
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\begin{align}
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\nOp \aDs \ket{\nu} &= \aCr \aDs \aDs \ket{\nu}\\
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&= \left( \aDs \aCr - \one \right) \aDs \ket{\nu}\\
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&= \aDs \nOp \ket{\nu} - \aDs \ket{\nu}\\
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&= \aDs \cdot \nu \ket{\nu} - \aDs \ket{\nu}\\
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&= \left(\nu - 1\right) \aDs \ket{\nu}
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\end{align}
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$\rightarrow$ $\aDs\ket{\nu}$ ist Eigenvektor von $\nOp$ zum Eigenwert $\left( \nu - 1 \right)$\\
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\underline{oder}:
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\begin{equation}
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\aDs\ket{\nu} = \zero \text{ (Nullvektor)}
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\end{equation}
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\item
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\begin{equation}
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0 \leq \norm{\aDs \ket{\nu}}^2 = \braket{\nu}{\aCr \aDs \nu} = \nu \underbrace{\braket{\nu}{\nu}}_{\geq 0}
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\end{equation}
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%\begin{figure}[h]
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%\includegraphics{pdf/II/05-01-00.pdf}
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%\end{figure}
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Die obige Ungleichung wäre nach mehrfacher Anwendung von $\aDs \ket{\nu}$ verletzt wenn anfänglich $\nu$ keine ganze positive Zahl ist.
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\item
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\begin{align}
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\nOp \aCr \ket{\nu} &= \aCr \aDs \aCr \ket{\nu}\\
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&= \aCr \left( \aCr \aDs + 1 \right)\ket{\nu}\\
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&= \aCr \left( \nu + 1 \right) \ket{\nu}\\
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&= \left( \nu + 1 \right) \aCr \ket{\nu}
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\end{align}
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\item
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\begin{align}
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0 \leq \norm{\aCr \ket{\nu}}^2 &= \braket{\nu}{\aDs \aCr \nu} = \dirac{\nu}{\aCr \aDs + 1}{\nu}\\
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&= \left( \nu + 1 \right) \aCr \ket{\nu}
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\end{align}
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$\rightarrow$ kein Problem
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\end{enumerate}
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Daraus ergibt sich das Spektrum von $\nOp$:
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\begin{equation}
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\nOp \ket{n} = n \ket{n} \text{ mit } n \in \setZ^+_0
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\end{equation}
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%\begin{figure}[h]
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%\includegraphics{pdf/II/05-01-01.pdf}
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%\end{figure}
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und das Spektrum von $H$:
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\begin{equation}
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H \ket{n} = \hbar \omega \left( n + \frac{1}{2} \right) \ket{n}
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\end{equation}
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\begin{enumerate}
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\item nur diskrete Eigenwerte erlaubt: Quantisierung
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\item Grundzustandsenergie (auch Nullzustandsenergie):
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\begin{equation}
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E_0 = \frac{\hbar \omega}{2}
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\end{equation}
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\item Es gilt:
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\begin{equation}
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a \ket{0} = \ket{\zero}
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\end{equation}
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\item klassischer harmonischer Oszilator (mit $m = 1\text{kg}$; $\omega = \frac{1}{\text{sec}}$):
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\begin{align}
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\Delta E &= E_{n+1} - E_n = 10^{-34}\text{J}\\
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E_0 &= \frac{m}{2} \omega^2 x^2 = 1 \text{J}
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\end{align}
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\end{enumerate}
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\paragraph*{Matrixelemente der Erzeuger- und Vernichter-Operatoren}
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\begin{align}
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\aCr \ket{n} &= c_n \ket{n+1} ~ \left( \ket{n} \text{ seien normiert} \right)\\[15pt]
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\rightarrow \abs{c_n}^2 &= \dirac{n}{\aDs \aCr}{n}\\
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&= \dirac{n}{\aCr\aDs + 1}{n}\\
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&= (n + 1) \underbrace{\braket{n}{n}}_{1}\\[15pt]
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\rightarrow c_n &= \sqrt{n + 1} \text{ (Phase absichtlich 1 gesetzt)}
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\end{align}
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daraus ergibt sich
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\begin{equation}
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\aCr \ket{n} = \sqrt{n + 1} \ket{n + 1} \label{eqn03}
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\end{equation}
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insbesondere
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\begin{align}
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\aCr \ket{0} &= 1 \ket{1} \Rightarrow \ket{1} = \aCr \ket{0}\\
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\aCr \ket{1} &= \sqrt{2} \ket{2} \Rightarrow \ket{2} = \frac{1}{\sqrt{2}} \aCr \ket{1} = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1}} \aCr \aCr \ket{0}
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\end{align}
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und analog zu \ref{eqn03} gilt:
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\begin{equation}
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\aDs \ket{n} = \sqrt{n} \ket{n - 1}
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\end{equation}
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Man erhält nun aus dem Obigen die allgemeine Form für $\ket{n}$:
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\begin{equation}
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\boxed{\ket{n} = \frac{1}{\sqrt{n!}} \left( \aCr \right)^n \ket{0}}
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\end{equation}
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Die Matrixelemente von $\aCr$ sind dann:
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\begin{align}
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\dirac{n'}{\aCr}{n} &= \sqrt{n + 1} \braket{n'}{n + 1}\\
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&= \sqrt{n + 1} \krondelta{n', n + 1}
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\end{align}
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und ebenso die Matrixelemente von $a = \left( \aCr \right)^\dagger$:
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\begin{align}
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\dirac{n'}{\aDs}{n} &= \dirac{n}{\aCr}{n}\\
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&= \sqrt{n} \krondelta{n, n + 1}
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\end{align}
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als Matrix:
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\begin{align}
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\aDs &= \inlinematrix{
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0 & \sqrt{1} & 0 & 0 & 0 & \cdots \\
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0 & 0 & \sqrt{2} & 0 & 0 & \cdots \\
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0 & 0 & 0 & \sqrt{3} & 0 & \cdots \\
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0 & 0 & 0 & 0 & \ddots & \\
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\vdots & \vdots & \vdots & \vdots & &
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}\\
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\aCr &= \inlinematrix{
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0 & 0 & 0 & 0 & \cdots \\
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\sqrt{1} & 0 & 0 & 0 & \cdots \\
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0 & \sqrt{2} & 0 & 0 & \cdots \\
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0 & 0 & \sqrt{3} & 0 & \cdots \\
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0 & 0 & 0 & \ddots & \\
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\vdots & \vdots & \vdots & &
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}\\
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\aDs\aCr &= \inlinematrix{
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1 & 0 & 0 & 0 & \cdots \\
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0 & 2 & 0 & 0 & \cdots \\
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0 & 0 & 3 & 0 & \cdots \\
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0 & 0 & 0 & \ddots & \\
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\vdots & \vdots & \vdots & &
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}\\
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\aCr\aDs &= \inlinematrix{
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0 & 0 & 0 & 0 & 0 & \cdots \\
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0 & 1 & 0 & 0 & 0 & \cdots \\
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0 & 0 & 2 & 0 & 0 & \cdots \\
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0 & 0 & 0 & 3 & 0 & \cdots \\
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0 & 0 & 0 & 0 & \ddots & \\
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\vdots & \vdots & \vdots & \vdots & &
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}
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\end{align}
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\begin{equation}
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\left( \left[\aDs, \aCr \right] = \right) \aDs\aCr - \aCr\aDs = 1
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\end{equation}
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\begin{align}
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\hat{x} &= \frac{1}{\sqrt{2}} \inlinematrix{
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0 & \sqrt{1} & 0 & 0 & 0 & \cdots \\
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\sqrt{1} & 0 & \sqrt{2} & 0 & 0 & \cdots \\
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0 & \sqrt{2} & 0 & \sqrt{3} & 0 & \cdots \\
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0 & 0 & \sqrt{3} & 0 & \ddots & \\
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0 & 0 & 0 & \ddots & \ddots & \\
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\vdots & \vdots & \vdots & & &
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}\\
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\hat{p} &= \frac{i}{\sqrt{2}} \inlinematrix{
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0 & -\sqrt{1} & 0 & 0 & 0 & \cdots \\
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\sqrt{1} & 0 & -\sqrt{2} & 0 & 0 & \cdots \\
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0 & \sqrt{2} & 0 & -\sqrt{3} & 0 & \cdots \\
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0 & 0 & \sqrt{3} & 0 & \ddots & \\
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0 & 0 & 0 & \ddots & \ddots & \\
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\vdots & \vdots & \vdots & & &
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}\\
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\end{align}
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\begin{align}
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\left[ \hat{x}, \hat{p} \right] &= i \one\\[15pt]
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\tr\left[ \hat{x}, \hat{p} \right] &= \tr\left( i \one \right)\\
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\tr\left( \hat{x}\hat{p} - \hat{p}\hat{x} \right) &= \tr\left( i \one \right)\\
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0 &= i \infty \text{ (falls Spur zyklisch $\leftarrow$ gilt nur für endliche Räume)}
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\end{align}
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\section{Wellenfunktion im Ortsaum}
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Gesucht:
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\begin{align}
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\phi_n(x) &= \braket{x}{n}\\
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\phi_0(x) &= \braket{x}{0}
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\end{align}
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Wir wissen:
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\begin{equation}
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\aDs \ket{0} = \zero
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\end{equation}
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daraus ergibt sich
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\begin{align}
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\frac{1}{\sqrt{2}} \left( \hat{x} + i\hat{p} \right) \ket{0} &= \zero &\left| \bra{x} \right.\\
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\dirac{x}{\hat{x} + i\hat{p}}{0} &= 0\\
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x + i(-i) \diffPs{x} \phi_0(x) &= 0 &\left(\text{denn: } \dirac{x}{\hat{p}}{\psi} = -i \hbar \diffPs{x} \psi(x) \right)\\
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\rightarrow \left(x - \diffPs{x} \right) \phi_0(x) &= 0\\
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\phi_0(x) &= c \cdot e^{-\frac{x^2}{2}}
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\end{align}
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Normierung:
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\begin{equation}
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\intgr{-\infty}{+\infty}{\phi_0(x) \phi_0^*(x)}{x} \stackrel{!}{=} 1 ~ \rightarrow ~ c = \frac{1}{\pi^\frac{1}{4}}
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\end{equation}
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%\begin{figure}[h]
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%\includegraphics{pdf/II/05-02-00.pdf}
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%\end{figure}
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\paragraph*{Angeregte Zustände}
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\begin{align}
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\ket{1} &= \aCr \ket{0} &\left| \bra{x} \right.\\[15pt]
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\phi_0(x) &= \frac{1}{\sqrt{2}} \left( x - \diffPs{x} \right) \phi_0(x)\\
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&= \frac{1}{\sqrt{2}} \left( x - \diffPs{x} \right) \frac{1}{\pi^\frac{1}{4}} e^{-\frac{x^2}{2}}\\
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&= \frac{\sqrt{2}}{\pi^\frac{1}{4}} x e^{-\frac{x^2}{2}}\\[15pt]
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\phi_2(x) &= \frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}} \left( x - \diffPs{x} \right) \left( \frac{\sqrt{2}}{\pi^\frac{1}{4}} x e^{-\frac{x^2}{2}} \right)\\
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&= \frac{1}{\pi^\frac{1}{4}} \left( 2x^2 - 1 \right) e^{-\frac{x^2}{2}}
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\end{align}
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allgemein:
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\begin{align}
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\ket{n} &= \frac{\left( \aCr \right)^n}{\sqrt{n!}} \ket{0} &\left| \bra{x} \right.\\
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\phi_n(x) &= \frac{1}{\sqrt{n!}} \frac{1}{\pi^\frac{1}{4}} \frac{1}{\sqrt{2^n}} \left( x - \diffPs{x} \right)^n e^{-\frac{x^2}{2}}
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\end{align}
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$Q_n$ ist symmetrisch für $n = 2k$, antisymmetrisch für $n = 2k + 1$ und hat $n$ Nullstellen.
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\paragraph*{Erwartungswerte}
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\begin{align}
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< \hat{x} >_\ket{n} &= \dirac{n}{\hat{x}}{n}\\
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&= \frac{1}{\sqrt{2}} \dirac{n}{\aCr + \aDs}{n}\\
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&= \frac{1}{\sqrt{2}} \bra{n} \left( \sqrt{n + 1} \ket{n + 1} + \sqrt{n} \ket{n - 1}\right)\\
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&= 0\\[15pt]
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< \hat{p} >_\ket{n} &= 0
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\end{align}
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Wegen Ehrenfest:
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