Vorlesung_19_20171120

$\Psi (x) = \int_{-\infty} ^\infty A(k) e^{ikx}\text{d}k$

Rücktransformation: $\int_{-b}^b \Psi (x) e^{-k' x}\text{d}x=\int_{-\infty} ^\infty A(k) \int_{-b}^b e^{i(k-k')x}\text{d}x\text{d}k$

f(k)=\int_{-b}^be^{i(k-k')x}\text{d}x=\left.\frac{e^{i(k-k')x}}{i(k-k')}\right|_{-b}^b = \frac{e^{i(k-k')b}-e^{-i(k-k')b}}{i(k-k')}=2\frac{\sin((k-k')b)}{(k-k')}

$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$

\int_{-\infty}^\infty A(k) \lim_{b\rightarrow \infty}\frac{2\sin((k-k')b)}{(k-k')}\text{d}k\\ \text{l'Hospital: } \lim_{x\rightarrow x0} \frac{g(x)}{h(x)}=\left. \frac{g'(x)}{h'(x)}\right|_{x=x0}\\ \lim_{k\rightarrow k'}\frac{2\sin((k-k')b)}{(k-k')}=\frac{2b\cos(0)}{1}= 2b

Schnell oszillierende Terme kürzen sich weg.

\int_{-\infty}^\infty A(k) \lim_{b\rightarrow \infty}\frac{2\sin((k-k')b)}{(k-k')}\text{d}k=A(k')\lim_{b\rightarrow\infty}\int_{-\infty}^\infty \frac{2\sin((k-k')b)}{(k-k')}\text{d}k=A(k')2\pi\\

$\int_{-\infty}^\infty e^{i (k-k')x}dx=\underbrace{\delta(k-k')}_{\text{Distribution}} 2 \pi$

$\int_{-\infty}^\infty \delta(k)dk$ $\int_{-\infty}^\infty g(k)\delta(k-k')dk$ =g(k')

Wir haben gezeigt dass Rücktransformation sich herleitet aus:

$\Rightarrow \int_{-\infty}^\infty \Psi(x)e^{-ik'x}\text{d}x = 2\pi \int_{-\infty}^\infty A(k')\delta(k-k')dk=A(k')2\pi$

$\Rightarrow \text{Rücktransformation: } A(k)= \frac{1}{2\pi}\int_{-\infty}^\infty \Psi(x) e^{-ikx}\text{d}x$

$\frac{1}{2\pi}$ $\sqrt{\frac{1}{2\pi}}$

\left|\Psi (x) \right|^2 = \frac{1}{\sqrt{2\pi\sigma_x^2}}e^{-\frac{x^2}{2\sigma_x^2}}\\ \Psi(x)= \frac{1}{\sqrt[4]{2\pi\sigma_x^2}}e^{-\frac{x^2}{4\sigma_x^2}}\underbrace{e^{ik_0x}}_\text{für Gesamtimpuls}\\ A(k)=\int_{-\infty}^\infty \Psi(x) e^{-ikx}dx= 2^{\frac{3}{4}}\sqrt[4]{\pi}\sqrt{\sigma_x}\;e^{-\sigma_x^2(k-k_0)^2}\\ \Rightarrow \frac{1}{4\sigma_k^2}=\sigma_x^2\\ \Rightarrow \sigma _x \cdot \sigma _k = \frac{1}{2}=\Delta x\cdot \Delta k\\ \text{mit } p=\hbar k \Rightarrow \\ \boxed{\Delta x \cdot \Delta p = \frac{\hbar}{2}}

$\Rightarrow$ $\Delta x\cdot \Delta p\geq \frac{\hbar}{2}$

$\Delta \omega\cdot \Delta t \geq \frac{1}{2}$ $E=\hbar\omega$

\Rightarrow \boxed{\Delta E\cdot \Delta t \geq \frac{\hbar}{2}}

Frequenzkamm