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The Schrodinger Equation

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Die Schr"odinger Gleichung

;Herleitung ;Potentialtopf ;Photonenemmission ;Freies Teilchen ;Heisenberg ;Doppelspaltexperiment

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Die harmonische Schwingung

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Die harmonische Schwingung

$$ a = \frac{d v}{dt} = \frac{F}{m} = -\frac{Dx}{m} = \frac{d^2 x}{d^2 t}$$ ;$$ x = A\cos(\omega t) \Leftrightarrow \frac{d^2 x}{d^2 t} = -\frac{Dx}{m} = -\omega^2 * x $$ ;$$\omega = \sqrt{\frac{D}{m}}$$

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Die harmonische Schwingung

$$ x = A\cos(\sqrt{\frac{D}{m}} t) $$ ;$$ v = \frac{d x}{d t} = \omega A \sin(\sqrt{\frac{D}{m}} t)$$ ;$$ E_{kin} = \frac{1}{2} m v^2 = \frac{1}{2} D A^2 \sin^2(\omega t)$$ ;$$ E_{Feder} = \frac{1}{2} D x^2 = \frac{1}{2} D A^2 \cos^2(\omega * t)$$

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Die harmonische Schwingung

$$ E_{kin} = \frac{1}{2} m v^2 = \frac{1}{2} D A^2 \sin^2(\omega t)$$

$$ E_{Feder} = \frac{1}{2} D x^2 = \frac{1}{2} D A^2 \cos^2(\omega t)$$

;$$ E{ges} = E{kin} + E_{Feder} = \frac{1}{2} D A^2 $$ $$(\sin^2(\omega t) + \cos^2(\omega t)) = \frac{1}{2} D * A^2 $$

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Die harmonische Schwingung

$$ E_{ges} = \frac{1}{2} D A^2 $$

;$$ \Psi = \sqrt{\frac{D}{2}}\begin{pmatrix} x \ \frac{v}{\omega} \end{pmatrix} = \sqrt{\frac{D}{2}}\begin{pmatrix} A\cos(\omega t) \ \frac{\omega A \sin(\omega t)}{\omega}\end{pmatrix}$$ ;$$ \Psi = \sqrt{\frac{D}{2}}\begin{pmatrix} A \ A \end{pmatrix} \odot \circlearrowleft(\omega t) = \sqrt{\frac{D}{2}}A \circlearrowleft(\omega t) $$

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Die harmonische Schwingung

$$ \Psi = \sqrt{\frac{D}{2}}\begin{pmatrix} A \ A \end{pmatrix} \odot \circlearrowleft(\omega t) = \sqrt{\frac{D}{2}}A \circlearrowleft(\omega t) $$ $$ \textcolor{red}{\Psi=\sqrt{\frac{D}{2}} A e^{i\omegat}}$$ ;$$|\Psi|^2 = E_{ges} = \frac{DA^2}2 $$

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Die harmonische Schwingung

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Wanderwelle

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Wanderwelle

$$ \Psi = A\circlearrowleft(kx - \omega t) \textcolor{red}{= Ae^{i(kx-\omegat)}}$$

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Wanderwelle

$$ \Psi = A\circlearrowleft(kx - \omega t) \textcolor{red}{= Ae^{i(kx-\omegat)}}$$ ;$$ k = \frac{2\pi}{\lambda}, \omega = 2\pif $$ ;$$ \lambda = \frac{h}{p} \Rightarrow k = \frac{2\pip}{h} $$ ;$$ \omega = 2 \pi \frac{v}{\lambda} = \frac{2 \pi v p}{h} $$

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Wanderwelle

$$ \Psi = A\circlearrowleft(kx - \omega t) \textcolor{red}{= Ae^{i(kx-\omegat)}}$$

$$ k = \frac{2\pip}{h}, \omega = \frac{2 \pi v p}{h} $$ ;$$ \Psi = A\circlearrowleft(\frac{2 \pi p}{h}x - \frac{2 \pi v p}{h} t) $$ ;$$ \Psi = A\circlearrowleft(\frac{2\pip}{h}(x - vt)) \textcolor{red}{= Ae^{\frac{ip}{\hbar}(x-vt)}} $$

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Energieerhaltungssatz

$$ \Psi = A\circlearrowleft(\frac{2\pip}{h}(x - vt)) \textcolor{red}{= Ae^{\frac{ip}{\hbar}(x-vt)}} $$ ;$$ E{ges} = E{kin} + E_{pot} = \frac{p^2}{2m} + V $$ ;$$ E_{ges} \Psi = \frac{p^2}{2m} \Psi + V \Psi $$

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Energieerhaltungssatz

$$ \Psi = A\circlearrowleft(\frac{2\pip}{h}(x - vt)) \textcolor{red}{= Ae^{\frac{ip}{\hbar}(x-v*t)}} $$

$$ E_{ges} \Psi = \frac{p^2}{2m} \Psi + V \Psi $$ ;$$ \frac{\partial \Psi}{\partial x} = \circlearrowleft(90\degree) \frac{2\pip}{h} \Psi \textcolor{red}{= \frac{ip}{\hbar} \Psi} $$ ;$$ p = \circlearrowleft(-90\degree) \frac{h}{2\pi} \frac{\partial \Psi}{\partial x} \textcolor{red}{= -i \hbar * \frac{\partial \Psi}{\partial x}} $$

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Energieerhaltungssatz

$$ \Psi = A\circlearrowleft(\frac{2\pip}{h}(x - vt)) \textcolor{red}{= Ae^{\frac{ip}{\hbar}(x-v*t)}} $$

$$ E_{ges} \Psi = \circlearrowleft(180\degree) \frac{h^2}{8m\pi^2} \frac{\partial^2 \Psi}{\partial^2 x} + V \Psi $$ ;

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Energieerhaltungssatz

$$ \Psi = A\circlearrowleft(\frac{2\pip}{h}(x - vt)) \textcolor{red}{= Ae^{\frac{ip}{\hbar}(x-v*t)}} $$

;$$ E_{ph} = hf = \frac{hv}{\lambda} = \frac{hvp}{h} = vp $$ ;$$ \frac{\partial \Psi}{\partial t} = -\circlearrowleft(90\degree) \frac{2\pip v}{h} \Psi$$

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Energieerhaltungssatz

$$ \Psi = A\circlearrowleft(\frac{2\pip}{h}(x - vt)) \textcolor{red}{= Ae^{\frac{ip}{\hbar}(x-v*t)}} $$

$$ \frac{\partial \Psi}{\partial t} = -\circlearrowleft(90\degree) \frac{2\pip v}{h} * \Psi$$

;$$ pv\Psi = -\frac{\partial \Psi}{\partial t} \frac{h}{2\pi*\circlearrowleft(90\degree)}$$

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Energieerhaltungssatz

$$ \Psi = A\circlearrowleft(\frac{2\pip}{h}(x - vt)) \textcolor{red}{= Ae^{\frac{ip}{\hbar}(x-v*t)}} $$

$$ pv\Psi = -\frac{\partial \Psi}{\partial t} \frac{h}{2\pi\circlearrowleft(90\degree)}$$ ;$$ E_{ges} \Psi = \circlearrowleft(90\degree) \frac{h}{2\pi} \frac{\partial \Psi}{\partial t} \textcolor{red}{ = i\hbar*\frac{\partial \Psi}{\partial t}}$$

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Die Schr"odinger Gleichung

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Teilchen im Potentialtopf

$$ V = 0 \forall x \in [0; l], V = \infty \forall x \notin [0; l] $$

;$$ \Psi \in \mathbb{C} \forall x \in [0, l], \Psi = 0 \forall x \notin [0, l] $$ ;$$ \Psi = \Psi_x * \Psi_t $$

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Teilchen im Potentialtopf

$$ \Psi = \Psi_x * \Psi_t $$

;$$E\Psi = -\frac{h^2}{8m\pi^2} \frac{\partial^2 \Psi}{\partial^2 x}$$ ;$$E\Psi_t \Psi_x = -\frac{h^2}{8m\pi^2} \Psi_t \frac{\partial^2 \Psi_x}{\partial^2 x}$$ ;$$E \Psi_x = -\frac{h^2}{8m\pi^2} \frac{\partial^2 \Psi_x}{\partial^2 x}$$

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Teilchen im Potentialtopf

$$ \Psi = A\sin(kx + \alpha_0) * \Psi_t $$

$$E \Psi_x = -\frac{h^2}{8m\pi^2} \frac{\partial^2 \Psi_x}{\partial^2 x}$$ ;$$E \Psi_x = \frac{h^2}{8m\pi^2} k^2 \Psi_x$$ ;$$E = \frac{h^2k^2}{8m*\pi^2}$$

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Teilchen im Potentialtopf

$$ \Psi = A\sin(kx + \alpha_0) * \Psi_t $$

;$$ \circlearrowleft(90\degree) \frac{h}{2\pi} \frac{\partial \Psi}{\partial t} = E\Psi = \frac{h^2k^2\Psi_t}{8m\pi^2} \textcolor{red}{ = i\hbar\frac{\partial \Psi}{\partial t}} $$ ;$$ \circlearrowleft(90\degree) \frac{h}{2\pi} \frac{\partial \Psi_t}{\partial t} = E\Psi_t = \frac{h^2k^2\Psi_t}{8m\pi^2}\textcolor{red}{ = i\hbar\frac{\partial \Psi_t}{\partial t}} $$

;$$ \Psi_t = \circlearrowleft(\omega t) \textcolor{red}{=e^{i\omegat}} \Rightarrow \omega = -\frac{hk^2}{4\pim} $$

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Teilchen im Potentialtopf

$$ \Psi = A\sin(kx + \alpha_0) \circlearrowleft(-\frac{hk^2}{4\pim} t)$$ ;$$ \Psi(0) = 0 \Rightarrow \alpha_0 = 0 $$ ;$$ \Psi(l) = 0 = A\sin(kxl) \Rightarrow k = \frac{n\pi}{l} \forall n\in \mathbb{N} $$ ;$$ E_n = \frac{h^2n^2}{8ml^2}, \Psi = A\sin(\frac{n\pix}{l}) \circlearrowleft(-\frac{hk^2}{4\pim} t)$$ $$\textcolor{red}{\Psi=A\sin(\frac{n\pix}{l}) e^{-\frac{i\hbark^2t}{2m}}} $$

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"Uberlagerung von Zust"anden

$$ \Psi = \Psi_1 + \Psi_2, \rho = |\Psi|^2 = |\Psi_1 + \Psi2|^2 $$ ;$ |\Psi|^2 = (\Psi{1x}+\Psi{2x})^2 + (\Psi{1y} + \Psi_{2y})^2 $ ;$ |\Psi|^2 = A^2((\sin(k_1x)\cos(\omega_1t) + \sin(k_2x)\cos(\omega_2t))^2 + ((\sin(k_1x)\sin(\omega_1t) + \sin(k_2x)\sin(\omega_2t))^2$

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"Uberlagerung von Zust"anden

$$ |\Psi|^2 = A^2(\sin^2(k_1x) + \sin^2(k_2x) + (2\sin(k_1x) sin(k_2x) \cos((\omega_1-\omega_2)*t)$$ ;

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"Uberlagerung von Zust"anden

$$ |\Psi|^2 = A^2(\sin^2(k_1x) + \sin^2(k_2x) + (2\sin(k_1x) sin(k_2x) \cos((\omega_1-\omega_2)*t)$$

;$$\omega = \omega_1-\omega_2, f = \frac{\omega}{2\pi} = \frac{\omega_1 - \omega_2}{2\pi}$$ ;$$ f = \frac{h}{2\pi4\pim} (k_1^2 - k2^2) $$ ;$$ E{ph} = hf = \frac{h^2}{8\pi^2m} *(k_1^2 - k2^2) $$ ;$$ E{ph} = E_1 - E_2 $$

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Impulsraum

$$ \Psik = A\circlearrowleft(kx) \textcolor{red}{= Ae^{i(kx)}}$$ ;$$ \Psi = \sum{i=-\infty}^{\infty} a_n \circlearrowleft(k_nx) \textcolor{red}{=\sum_{n=-\infty}^{\infty} a_n e^{ik_n*x}}$$

;$$ \Psi = \int{-\infty}^{\infty} \phi \circlearrowleft(kx)* dk \textcolor{red}{=\int{-\infty}^{\infty} \phi e^{ikx}dk}$$ ;$$ \Psi = \int{-\infty}^{\infty} \circlearrowleft(kx)dk * \int{-\infty}^{\infty} \circlearrowleft(-kx)\Psidx $$ ;$$ \phi = \int_{-\infty}^{\infty} \circlearrowleft(-kx)\Psidx \textcolor{red}{=\int_{-\infty}^{\infty} \Psi e^{-ikx}dk}$$

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Impulsraum

$$ \phi = F(\Psi) = \int{-\infty}^{\infty} \circlearrowleft(-kx)\Psi*dx \textcolor{red}{=\int{-\infty}^{\infty} \Psie^{-ikx}dk}$$

;$$\textcolor{red}{i\hbar\frac{\partial \Psi}{\partial t}} = \circlearrowleft(90\degree) \frac{h}{2\pi} \frac{\partial \Psi}{\partial t} = \frac{p^2}{2m} *\Psi $$

;$$\textcolor{red}{F(i\hbar\frac{\partial \Psi}{\partial t})} = F(\circlearrowleft(90\degree) \frac{h}{2\pi} \frac{\partial \Psi}{\partial t}) = F(\frac{p^2}{2m} *\Psi) $$

;$$\textcolor{red}{i\hbar\frac{\partial \phi}{\partial t}} = \circlearrowleft(90\degree) \frac{h}{2\pi} \frac{\partial \phi}{\partial t} = \frac{p^2}{2m} *\phi $$

;$$\textcolor{red}{i\hbar\frac{\partial \phi}{\partial t}} = \circlearrowleft(90\degree) \frac{h}{2\pi} \frac{\partial \phi}{\partial t} = \frac{(\frac{hk}{2\pi})^2}{2m} *\phi $$

;$$\textcolor{red}{i\hbar\frac{\partial \phi}{\partial t}} = \circlearrowleft(90\degree) \frac{h}{2\pi} \frac{\partial \phi}{\partial t} = \frac{h^2k^2}{8\pi^2m} *\phi $$

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Impulsraum

$$ \phi = F(\Psi) = \int{-\infty}^{\infty} \circlearrowleft(-kx)\Psi*dx \textcolor{red}{=\int{-\infty}^{\infty} \Psie^{-ikx}dk}$$

$$\textcolor{red}{i\frac{\partial \phi}{\partial t}} = \circlearrowleft(90\degree) \frac{\partial \phi}{\partial t} = \frac{hk^2}{4\pim} \phi $$

;$$\phi = \circlearrowleft(\omega t) \phi_0 = \textcolor{red}{e^{i\omegat}\phi_0}$$ ;$$\omega = -\frac{hk^2}{4\pim}$$

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Ein lokalisiertes Teilchen

$$ |\Psi|^2 = \frac{1}{\sqrt{2\pi}\sigma_x} e^{-\frac{x^2}{2\sigma_x^2}}$$

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Ein lokalisiertes Teilchen

$$ |\Psi_0|^2 = \frac{1}{\sqrt{2\pi}\sigma_x} e^{-\frac{x^2}{2\sigma_x^2}}$$ ;$$ \Psi_0 = \frac{1}{\sqrt[4]{2\pi}\sqrt{\sigma_x}} e^{-\frac{x^2}{4\sigma_x^2}}$$

;$$ \Psi0 = A{x0} e^{-b_{x0}x^2}, \frac{\partial \Psi0}{\partial x} = -2*b{x0}x\Psi_0$$ ;$$ \textcolor{red}{ikF(\Psi_0)} = \circlearrowleft(90\degree) k F(\Psi0) = -\circlearrowleft(90\degree) 2b{x0} \frac{\partial}{\partial k}F(\Psi_0) \textcolor{red}{=-2ib_{x0}*\frac{\partial}{\partial k}F(\Psi_0)}$$

;$$ k \phi_0 = -2b_{x0} \frac{\partial \phi_0}{\partial k}$$ ;$$ \frac{\partial \phi_0}{\partial k} = -\frac{k \phi0}{2*b{x0}}$$ ;$$ \phi0 = A{k} e^{-b_{k}k^2} \Rightarrow bk = \frac{1}{4*b{x0}}$$

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Ein lokalisiertes Teilchen

$$ \phi0 = A{k} e^{-\frac{k^2}{4b_{x0}}}$$ ;$$ |\phi0|^2 = A{p}^2 e^{\frac{p^2}{2\sigma_p^2}} $$ ;$$ \phi0 = A{p} e^{-\frac{p^2}{4\sigma_p^2}} $$ ;$$ \phi0 = A{k} e^{-\frac{h^2k^2}{44\pi^2\sigmap^2}} = A{k} e^{-\frac{k^2}{4b_{x0}}}$$ ;$$ \frac{h^2}{44\pi^2\sigmap^2} = \frac{1}{4*b{x0}}$$ ;$$ \frac{4\pi^2\sigmap^2}{h^2} = b{x0}$$

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Ein lokalisiertes Teilchen

$$ \frac{4\pi^2\sigmap^2}{h^2} = b{x0}$$ ;$$ \frac{4\pi^2\sigma_p^2}{h^2} = \frac{1}{4\sigma_x^2}$$ ;$$ \frac{2\pi\sigma_p}{h} = \frac{1}{2\sigma_x}$$ ;$$ \sigma_p\sigma_x = \frac{h}{4\pi}$$

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Ein lokalisiertes Teilchen

$$\phi = \circlearrowleft(\omega t) \phi_0 = \textcolor{red}{e^{i\omegat}*\phi_0}$$

;$$ \phi = A{k} e^{-\frac{k^2}{4b{x0}}} \circlearrowleft(-\frac{hk^2}{4\pim} t) = \textcolor{red}{A_{k} e^{-\frac{k^2}{4b_{x0}}}e^{-\frac{ihk^2}{4\pim} *t}}$$

;$$ \phi = A{k} e^{-\frac{k^2}{4b{x0}} + \ln(\circlearrowleft(-\frac{hk^2t}{4\pim}))}$$ ;$$ \phi = A{k} e^{-\frac{k^2}{4b{x0}} -\frac{hk^2t}{4\pim} \circlearrowleft(90\degree)}$$ ;$$ \phi = A{k} e^{-k^2(\frac{1}{4b{x0}} + \frac{ht}{4\pim} \circlearrowleft(90\degree))}\textcolor{red}{ = A_{k}e^{-k^2(\frac{1}{4b_{x0}}+\frac{i\hbart}{2*m})}}$$

;$$\phi = A_{k} e^{-ck^2}, c = \sigma_x^2+\frac{ht}{4\pim}\circlearrowleft(90\degree)\textcolor{red}{=\sigma_x^2+\frac{i\hbart}{2m}} $$

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Ein lokalisiertes Teilchen

$$\phi = A_{k} e^{-ck^2} $$ $$ \Psi = A e^{-\frac{x^2}{4c}} = Ae^{-\frac{x^2}{4\sigma_x^2+\frac{ht}{\pim}\circlearrowleft(90\degree)}}\textcolor{red}{=Ae^{-\frac{x^2}{4\sigma_x^2+\frac{2i\hbar*t}{m}}}}$$

;$$ \Psi = Ae^{-\frac{x^2(4\sigma_x^2-\frac{ht}{\pim}\circlearrowleft(90\degree))}{(4\sigma_x^2+\frac{ht}{\pim}\circlearrowleft(90\degree))(4\sigma_x^2-\frac{ht}{\pim}*\circlearrowleft(90\degree))}} $$

;$$ \Psi = Ae^{-\frac{x^2(4\sigma_x^2-\frac{ht}{\pim}\circlearrowleft(90\degree))}{16\sigma_x^4+\frac{h^2t^2}{\pi^2*m^2}}} $$

;$$ \Psi = Ae^{-\frac{4x^2\sigma_x^2}{16\sigma_x^4+\frac{h^2t^2}{\pi^2m^2}}+\frac{\frac{x^2ht}{\pim}\circlearrowleft(90\degree))}{16\sigma_x^4+\frac{h^2t^2}{\pi^2*m^2}}} $$

;$$ \Psi = Ae^{-\frac{x^2}{4\sigma_x^2+\frac{h^2t^2}{16\pi^2m^2\sigma_x^2}}}\circlearrowleft(\frac{\frac{x^2ht}{\pim}}{16\sigma_x^4+\frac{h^2t^2}{\pi^2*m^2}}) $$

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Ein lokalisiertes Teilchen

$$ \Psi = Ae^{-\frac{x^2}{4\sigma_x^2+\frac{h^2t^2}{16\pi^2m^2\sigma_x^2}}}\circlearrowleft(\frac{x^2}{\frac{16\sigma_x^4\pim}{ht}+\frac{ht}{\pi*m}}) $$

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Das Doppelspaltexperiment

$$ \Psi = \Psi{lok}(x-\frac{d}{2}) + \Psi{lok}(x+\frac{d}{2})$$

;$$ \Delta\alpha = 2\pin \forall n \in \mathbb{Z} $$

;$$ \Psi_{lok} = Ae^{-\frac{x^2}{4\sigma_x^2+\frac{h^2t^2}{16\pi^2m^2\sigma_x^2}}}\circlearrowleft(\frac{x^2}{\frac{16\sigma_x^4\pim}{ht}+\frac{ht}{\pi*m}}) $$

;$$ \frac{(x+\frac{d}{2})^2}{\frac{16\sigma_x^4\pim}{ht}+\frac{ht}{\pim}} = \frac{(x-\frac{d}{2})^2}{\frac{16\sigma_x^4\pim}{ht}+\frac{ht}{\pim}} + 2\pin$$

;$$ \frac{xd}{\frac{16\sigma_x^4\pim}{ht}+\frac{ht}{\pim}} = \frac{-xd}{\frac{16\sigma_x^4\pim}{ht}+\frac{ht}{\pim}} + 2\pin$$

;$$ \frac{2xd}{\frac{16\sigma_x^4\pim}{ht}+\frac{ht}{\pim}} = 2\pin$$

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Das Doppelspaltexperiment

$$ \frac{2xd}{\frac{16\sigma_x^4\pim}{ht}+\frac{ht}{\pim}} = 2\pin$$ ;$$ 2xd = 2\pin (\frac{16\sigma_x^4\pim}{ht}+\frac{ht}{\pim})$$ ;$$ x = \frac{\pin}{d} (\frac{16\sigma_x^4\pim}{ht}+\frac{ht}{\pi*m})$$

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Das Doppelspaltexperiment

$$ x = \frac{\pin}{d} (\frac{16\sigma_x^4\pim}{ht}+\frac{ht}{\pim})$$ ;$$ \lim_{t\rightarrow \infty},v = \frac{s}{t} \Rightarrow t = \frac{s}{v} $$ ;$$ x = \frac{\pin}{d} \frac{ht}{\pim} = \frac{nht}{dm} = \frac{nhs}{dmv}$$ ;$$ \lambda = \frac{h}{p} = \frac{h}{mv} $$ ;$$ x = \frac{ns\lambda}{d} $$

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Das Doppelspaltexperiment

$$ x = \frac{ns\lambda}{d} $$

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Vielen Dank f"ur eure Aufmerksamkeit