211 lines
8.4 KiB
TeX
211 lines
8.4 KiB
TeX
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\chapter{Grundlagen}
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\section{Kommutatorrelation $[\hat{x}, \hat{p}] = i \hbar$}
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\paragraph*{klassisch} $H = \frac{p^2}{2m} + V(x)$
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\paragraph*{quantal} $\hat{x} = ?; \hat{p} = ?;$
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\begin{equation}
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H = \frac{\hat{p}^2}{2m} + V(\hat{x})
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\end{equation}
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sinnvolle Forderung (Motivation)
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\begin{equation}
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\diffT{t} \dirac{\psi}{\hat{x}}{\psi} \stackrel{!}{=} \frac{\dirac{\psi}{p}{\psi}}{m}
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\end{equation}
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(Mittelwerte für alle $\ket{\psi}$ verhalten sich klassisch: $p = m \dot{x}$)
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\begin{align}
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\diffT{t} \dirac{\psi}{\hat{x}}{\psi} &= \frac{1}{i \hbar} \dirac{\psi}{\hat{x}}{H \psi} - \dirac{\psi H}{\hat{x}}{\psi}\\
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&= -\frac{i}{\hbar} \dirac{\psi}{[\hat{x}, H]}{\psi}
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\end{align}
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\begin{align}
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[\hat{x}, H] &= \left[\hat{x}, \frac{\hat{p}^2}{2m} + V(\hat{x}) \right]\\
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&= \left[ \hat{x}, \frac{\hat{p}^2}{2m} \right]\\
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&= \frac{1}{2m} \left( \hat{p} [\hat{x}, \hat{p}] + [\hat{x}, \hat{p}]\hat{p} \right)\\
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\rightarrow [\hat{x}, \hat{p}] &= i \hbar
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\end{align}
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vergleiche klassisch $\left\lbrace q,p \right\rbrace = 1$
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\paragraph*{Lemma}
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\begin{equation}
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[f(\hat{x}), \hat{p}] = i \hbar f'(\hat{x})
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\end{equation}
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\subparagraph*{Beweis}
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für
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\begin{equation}
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f(\hat{x}) = \sum_{n=0}^{\infty}a_n \hat{x}^n
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\end{equation}
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äquivalent zu
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\begin{equation}
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[\hat{x}, \hat{p}] = n \hat{x}^{n-1} \cdot i\hbar
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\end{equation}
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vollständige Induktion:\\
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$(n := 1)$
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\begin{equation}
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[\hat{x}, \hat{p}] = i \hbar
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\end{equation}
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$(n := n+1)$
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\begin{align}
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[\hat{x}^{n+1}, \hat{p}] &= \hat{x} \underbrace{[\hat{x}^n, \hat{p}]}_{n(\hat{x}^{n+1}) i\hbar} + \underbrace{[\hat{x}, \hat{p}]}_{i\hbar} \hat{x}^n\\
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&= (n+1)\hat{x}^n i\hbar
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\end{align}
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\begin{flushright}
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$\Box$
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\end{flushright}
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Wie verhält sich $\diffT{t} \dirac{\psi}{\hat{p}}{\psi}$ ?
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\begin{align}
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\rightarrow \diffT{t} \dirac{\psi}{\hat{p}}{\psi} &= -\frac{i}{\hbar} \dirac{\psi}{[\hat{p}, H]}{\psi}\\
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&= -\frac{i}{\hbar} \dirac{\psi}{[\hat{x}, V(\hat{x})]}{\psi}\\
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&= -\dirac{\psi}{V'(\hat{x})}{\psi}\\
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&\neq -V'(<x>_\psi) \text{ falls $V(x)$ nicht quadratisch}
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\end{align}
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\section{Ortsdatsrellung (Analogie zu Spin $\frac{1}{2}$)}
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\begin{tabular}{l|c|c}
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& Spin & Teilchen \\ \hline\hline
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Basis & $\sigma_z \ket{z\pm} = \pm 1 \ket{z\pm}$ & $\hat{x} \ket{x} = x \ket{x}$ \\ \hline
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Orthogonalität &
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$\begin{array}[t]{rl}\braket{z+}{z+} &= 1\\ &= \braket{z-}{z-}\\ \braket{z+}{z-} &= \braket{z-}{z+}\\ &= 0\end{array}$ &
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$\braket{x'}{x} = \delta(x'-x)$ \\ \hline
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Zustände in Basis entwickelt & $\begin{array}[t]{rl} \ket{\psi} &= \one \ket{\psi}\\ &= \ket{z+}\braket{z+}{\psi} + \ket{z-}\braket{z-}{\psi} \end{array}$ & $\begin{array}[t]{rl} \ket{\psi} &= \one \ket{\psi}\\ &= \intgr{-\infty}{+\infty}{\ket{x}\underbrace{\braket{x}{\psi}}_{\psi(x)}}{x}\\ &= \intgr{-\infty}{+\infty}{\psi(x) \ket{x}}{x} \end{array}$
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\end{tabular}
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\paragraph*{Normierung}
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\begin{align}
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1 \stackrel{!}{=} \braket{\psi}{\psi} &= \intgr{-\infty}{+\infty}{\psi^* \bra{x}}{x} \intgr{-\infty}{+\infty}{\psi(x') \ket{x'}}{x'}\\
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\intgru{\intgru{\psi(x)\psi(x')\underbrace{\braket{x}{x'}}_{\delta(x-x')}}{x'}}{x}\\
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&= \intgru{\psi^*(x)\psi(x)}{x}\\
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&= \intgru{(\psi(x))^2}{x}
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\end{align}
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$\rightarrow$ Zulässige Zustände haben eine ``Wellenfunktion'', die quadrat-integrabel ist.
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\paragraph*{Erwartungswert einer Ortsmessung}
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\begin{align}
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<\hat{x}>_\psi &= \dirac{\psi}{\hat{x}}{\psi}\\
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&= \intgru{\dirac{\psi}{\hat{x}}{x}\braket{x}{\psi}}{x}\\
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&= \intgru{\bra{\psi} \cdot x \ket{x} \braket{x}{\psi}}{x}\\
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&= \intgru{x\psi^*(x)\psi(x)}{x}\\
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&= \intgru{x\abs{\psi(x)}^2}{x}\\
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&\equiv \intgru{x \rho(x)}{x}
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\end{align}
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mit $\rho(x) \equiv \abs{\psi(x)}^2 = \psi^*(x) \psi(x)$ der Wahrscheinlichkeitsdichte das Teilchen im Zustand $\ket{\psi}$ am Ort $x$ zu messen!
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\subparagraph*{Hinweis:}
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\begin{equation}
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\intgr{-\infty}{+\infty}{\rho(x)}{x} = 1
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\end{equation}
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\paragraph*{``Matrixelemente'' des Impulsoperators in Ortsdarstellung}
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(vgl. Spins $\sigma_y = \inlinematrix{0& -i\\ i& 0}$)
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\begin{equation}
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\dirac{x'}{\hat{p}}{x} = ?
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\end{equation}
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\begin{align}
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\hat{x}\hat{p} - \hat{p}\hat{x} &= i \hbar & \left| \ket{x} \right.\\
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\hat{x}\hat{p} \ket{x}- \hat{p}\hat{x} \ket{x} &= i \hbar \ket{x} & \left| \bra{x'} \right.\\
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\dirac{x'}{\hat{x}\hat{p}}{x} - \dirac{x'}{\hat{p}\hat{x}}{x} &= i\hbar \braket{x'}{x}\\
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(x' - x) \underbrace{\dirac{x'}{p}{x}}_{\equiv i\hbar g(x' - x)} &= i\hbar \delta(x' - x)
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\end{align}
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mit $U g(u) = \delta(u)$
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%TODO: folgendes abgleichen
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% \subparagraph*{Behauptung}
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% \begin{equation}
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% g(u) = \delta'(u)
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% \end{equation}
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%
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% \subparagraph*{Beweis}
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% \begin{align}
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% -u \delta'(u) &= \delta(u)\\
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% -\intgru{u \delta'(u) f(u-u_0)}{u} &= \intgru{\delta(u) f(u-u_0)}{u}\\
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% \intgru{\delta(u)}\left(u f(u - u_0)\right)'{u} &=
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% \end{align}
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also:
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\begin{equation}
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\dirac{x'}{\hat{p}}{x} = i\hbar \delta'(x'-x)
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\end{equation}
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Wirkung $\hat{p}$ auf $\ket{\psi}$:
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\begin{align}
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\dirac{x'}{\hat{p}}{\psi} &= \intgru{\dirac{x}{\hat{p}}{x}\braket{x}{\psi}}{x}\\
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&= \intgru{(-i\hbar) \delta'(x'-x) \psi(x)}{x}\\
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&= (-i\hbar) \intgru{\delta(x'-x) \diffP{x} \psi(x)}{x}\\
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&= (-i\hbar) \psi'(x')
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\end{align}
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daher auch
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\begin{equation}
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\hat{p} \cequiv -i\hbar \diffP{x}
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\end{equation}
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\section{Schrödingergleichung in Ortsdarstellung}
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SG:
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\begin{align}
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i\hbar \partial_t \ket{\psi} &= H \ket{\psi}\\
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&= \left( \frac{\hat{p}^2}{2m} + V(\hat{x}) \right) \ket{\psi} & \left| \bra{x} \right.\\
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i\hbar \partial_t \braket{x}{\psi} &= \dirac{x}{\frac{\hat{p}^2}{2m}}{\psi}\\
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i\hbar \partial_t \psi(x,t) &= \left\lbrace \frac{1}{2m} \left( -i\hbar \diffP{x} \right)^2 + V(x) \right\rbrace \psi(x)
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\end{align}
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Falls $V(x)$ zeitabhängig ist gibt es stationäre Zustände:
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\begin{equation}
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\psi(x,t) = e^{-\frac{i}{\hbar} E_n t} \psi_n(x)
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\end{equation}
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\begin{equation}
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E_n \psi_n(x) = \left( -\frac{i\hbar}{2m} \frac{\partial^2}{\partial x^2} + V(x) \right) \psi_n(x)
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\end{equation}
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Normierbare Lösung $\intgr{-\infty}{+\infty}{\abs{\psi(x)}^2}{x} = 1$ nur für bestimmte $E_n$!
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\section{Impuls-Operator}
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Zu $\hat{x} \ket{x} = x\ket{x}$ bilder wir die Analogie $\hat{p}\ket{p} = p \ket{p}$.
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\paragraph*{Was ist $\braket{x}{p}$?}
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\begin{align}
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\hat{p}\ket{p} &= p \ket{p} &\left| \bra{x} \right.\\
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\dirac{x}{\hat{p}}{p} &= p \braket{x}{p}\\
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\intgr{-\infty}{+\infty}{\dirac{x}{\hat{p}}{x'}\braket{x'}{p}}{x'} &= p \braket{x}{p}\\
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\intgr{-\infty}{+\infty}{-i\hbar \delta'(x - x') \braket{x'}{p}}{x'} &= p \braket{x}{p}\\
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\rightarrow -i\hbar \diffP{x} \braket{x}{p} &= p \braket{x}{p}\\
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\braket{x}{p} &= c e^{\frac{ip}{\hbar}x}
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\end{align}
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d.h.: Die Eigenzustände des Impulsoperators in Ortsdarstellung ist eine ebene Welle.
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\paragraph*{Normierung}
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\begin{align}
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\braket{p'}{p} &\stackrel{!}{=} \delta(p'-p)\\[15pt]
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\intgr{-\infty}{+\infty}{\braket{p'}{x}\braket{x}{p}}{x} &= \intgr{-\infty}{+\infty}{c^* e^{\frac{i p' x}{\hbar}} c e^{\frac{i p x}{\hbar}}}{x}\\
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&= 2 \pi c c^* \delta\left(\frac{p - p'}{\hbar}\right)\\
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&= 2 \pi \hbar c c^* \delta(p - p')
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\end{align}
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$\rightarrow$ $c = \frac{1}{\sqrt{2 \pi \hbar}}$ ist korrekte Normierung!
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\begin{equation}
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\braket{x}{p} = \frac{1}{\sqrt{2 \pi \hbar}} e^{\frac{i p x}{\hbar}}
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\end{equation}
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\paragraph*{Erwartungswert des Impulses}
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\begin{align}
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<p>_\psi &= \dirac{\psi}{\hat{p}}{\psi}\\
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&= \intgru{\dirac{\psi}{\hat{p}}{p}\braket{p}{\psi}}{p}\\
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&= \intgru{p\abs{\braket{p}{\psi}}^2}{p}
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\end{align}
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d.h. $\rho(p) \equiv \abs{\braket{p}{\psi}}^2$ ist die Wahrscheinlichkeitsdichte im Zustand $\ket{\psi}$ den Impuls $p$ zu messen.\\[15pt]
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Für gegebenes $\braket{p}{\psi} = \psi(x)$ gilt
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\begin{align}
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\braket{p}{\psi} &= \intgr{-\infty}{+\infty}{\braket{p}{x}\braket{x}{\psi}}{x}\\
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&= \intgru{\frac{1}{\sqrt{2 \pi \hbar}}e^{\frac{-i p x}{\hbar}}\psi(x)}{x} \equiv \psi(p)
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\end{align}
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$\psi(p)$ ist (leicht anders normierte) Furiertransformierte von $\psi(x)$.
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\begin{align}
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\dirac{\psi}{p}{\psi} &= \intgru{\intgru{\braket{\psi}{x}\underbrace{\dirac{x}{\hat{p}}{x'}}_{-i \hbar \delta'(x - x')}\braket{x'}{\psi}}{x'}}{x}\\
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&= \intgru{\intgru{(-i \hbar) \delta'(x - x') \psi^*(x) \psi(x')}{x'}}{x}\\
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&= i \hbar \intgru{\intgru{\delta(x - x') {\psi^*}'(x) \psi(x')}{x'}}{x}\\
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&= i \hbar \intgru{{\psi^*}'(x') \psi(x')}{x'}\\
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<p>_\psi&= i \hbar \intgru{{\psi^*}'(x) \psi(x)}{x}
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\end{align}
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\paragraph*{Schrödingergleichung in Impulsdarstellung}
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\begin{align}
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i \hbar \partial_t\ket{\psi} &= H \ket{\psi} & \left| \bra{p} \right.\\
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i \hbar \partial_t\braket{p}{\psi} &= \dirac{p}{\frac{\hat{p}^2}{2m} + V(x)}{\psi}\\
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&= \left\lbrace \frac{p^2}{2m} \psi(p) + \dirac{p}{V(\hat{x})}{\psi} \right\rbrace
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\end{align}
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