$$\hat{E}_\text{kin}=\frac{\hat{p}^2}{2m}= \frac{1}{2m}\left( -i\hbar \frac{\partial}{\partial x}\right)\left( -i\hbar \frac{\partial}{\partial x}\right)=-\frac{\hbar^2}{2m}\frac{\partial ^2}{\partial x^2}$$

$$\hat{E}_\text{kin}=-\frac{\hbar^2}{2m}\nabla^2=-\frac{\hbar^2}{2m}\Delta$$

$$\hat{U}(x)=U(x)$$

$$\hat{E}=i\hbar \frac{\partial}{\partial t}\Leftrightarrow \frac{\partial }{\partial t} = -\frac{i}{\hbar}\hat{E} \;\text{wird noch wichtig bei nummerischen Lösung der zeitabh. SGL}$$

$$\bar{x} = \int_{-\infty} ^\infty \Psi^*(x,t)x\Psi(x,t)\text{d} x\\ \Psi_1(x,t)=\sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)e^{-\frac{i}{\hbar}Et}\\ \text{Zeitlicher Teil fällt weg: } \Psi_1^*(x,t)\Psi_1(x,t)= \psi_1(x)^*\psi_1(x)\underbrace{e^{+\frac{i}{\hbar}Et}e^{-\frac{i}{\hbar}Et}}_{=1}\\ \Rightarrow \bar{x} = \int_0^L\psi_1^*(x)x\psi_1(x)\text{d}x= \int_0^L\left( \sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)\right)x\left( \sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)\right)\text{d}x\\ =\frac{2}{L} \int_0^L x\sin^2\left(\frac{\pi x}{L}\right)\text{d}x=\frac{2}{L}\int_0^L x\frac{1-\cos\left( \frac{2\pi x}{L}\right)}{2}\text{d}x=\frac{2}{L}\frac{L^2}{4}-\frac{1}{2}\int _0^L x\cos\left(\frac{2\pi x}{L}\right) \text{d}x\\ \text{Betrachte: }\frac{1}{2}\int _0^L\underbrace{x}_u\underbrace{\cos\left(\frac{2\pi x}{L}\right) }_{v'}\text{d}x=\left.\frac{L}{2\pi}x\sin\left(\frac{2\pi x}{L}\right) \right|_0^L -\underbrace{\frac{L}{2\pi}\int _0^L \sin\left(\frac{2\pi x}{L}\right) \text{d}x}_{=0}=0\\ \Rightarrow \bar{x} = \frac{L}{2}\\$$

$$\overline{x^2}=\int_0^L\left( \sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)\right)x^2\left( \sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)\right)\text{d}x =\frac{L^2}{3}-\frac{L^2}{2\pi}= L^2\left(\frac{1}{3}- \frac{1}{2\pi}\right)\\ \Rightarrow \Delta x = \sqrt{\,\overline{x^2}-\bar{x}^2}=\sqrt{L^2\left(\frac{1}{3}- \frac{1}{2\pi}\right)-\frac{L^2}{4}}=0,181\cdot L\\ \\ \Delta p=?\\ \bar{p}= \int_0^L\psi_1^*(x)\hat{p}\psi_1(x)\text{d}x= \int_0^L\left( \sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)\right)\left(-i\hbar \frac{\text{d}}{\text{d}x} \right)\left( \sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)\right)\text{d}x\\ =c\cdot \int_0 ^L \sin\left(\frac{\pi x}{L}\right)\cos\left(\frac{\pi x}{L}\right)\text{d}x=0\\ \overline{p^2} = \int_0^L\psi_1^*(x)\hat{p}^2\psi_1(x)\text{d}x= \int_0^L\left( \sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)\right)\left(-\hbar^2 \frac{\text{d}^2}{\text{d}x^2} \right)\left( \sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)\right)\text{d}x\\ =\frac{\hbar^2\pi^2}{L^2}\int_0^L\underbrace{\left( \sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)\right)^2\text{d}x}_{=1}=\frac{\hbar^2\pi^2}{L^2}\\ \Delta p =\sqrt{\overline{p^2} - \bar{p}^2}=\frac{\hbar\pi}{L}$$

$$\Delta x\cdot \Delta p=0,181\cdot L\cdot \frac{\pi\hbar}{L}=0,568\cdot \hbar \geq \frac{\hbar}{2}\ \surd$$

$$\Psi_n(x,t)=\psi_n(x)e^{-\frac{i}{\hbar}E_nt}$$

$$\Psi(x,t)=\frac{1}{\sqrt{2}}\left(\psi_n(x)e^{-\frac{i}{\hbar}E_nt}+\psi_m(x)e^{-\frac{i}{\hbar}E_mt}\right)$$

$$P(x,t)= \Psi^*(x,t) \Psi(x,t)=\frac{1}{2}\left(\psi^*_n(x)e^{\frac{i}{\hbar}E_nt}+\psi^*_m(x)e^{\frac{i}{\hbar}E_mt}\right)\cdot\left(\psi_n(x)e^{-\frac{i}{\hbar}E_nt}+\psi_m(x)e^{-\frac{i}{\hbar}E_mt}\right)\\ =\frac{1}{2} \left[ \left|\psi_n(x)\right|+ \left|\psi_m(x)\right|+\left(\psi_n^*(x)\psi_m(x)+\psi^*_m(x)\psi_n(x)\right)\cdot \cos\left(\frac{E_n-E_m}{\hbar}\cdot t\right)\right]$$