User:Vort3x1337

Aufgabe 1

Für $\Phi _{1}$ haben wir zwei Basisvektoren $\Psi _{1}$ und $\Psi _{2}$

${\hat {A}}\Psi _{1}=a_{1}\Psi _{1}$

${\hat {A}}\Psi _{2}=a_{2}\Psi _{2}$

System ist prepariert im Zustand $\Phi _{1}$.

Die Wahrscheinlichkeit den Messwert $a_{1}$ zu messen ist gegeben durch :

$P_{a1}=\left(\int \Psi _{1}^{*}\Phi _{1}\ d^{3}x\right)^{2}$

$=\left({\frac {1}{\sqrt {5}}}\int \psi _{1}^{*}\psi _{1}+2\psi _{1}^{*}\psi _{2}\ dx\right)^{2}$

$=\left({\frac {1}{\sqrt {5}}}+{\frac {1}{\sqrt {5}}}\int 2\psi _{1}^{*}\psi _{2}\ dx\right)^{2}$

$=\left({\frac {1}{\sqrt {5}}}\right)^{2}$

$={\frac {1}{5}}$

Die QM. Operatoren sind hermitsch $\Longrightarrow$ EW sind reell, hat ONB.

$\Phi _{1}={\frac {1}{\sqrt {5}}}\Psi _{1}+{\frac {w}{\sqrt {5}}}\Psi _{2}$

$P_{a}^{2}$ ist die Wahrscheinlichkeit System in $\Psi _{1}$ zu finden.

$\int \Psi _{1}^{*}\Psi _{2}\ d^{3}x=0$ $\int \Psi _{1}^{*}\Psi _{1}\ d^{3}x=1$

$\Psi _{1}(\Phi _{1},\Phi _{2})$ ?

Durch Umformen erhalten wir :

$\Psi _{1}={\frac {1}{\sqrt {5}}}\Phi _{1}+{\frac {2}{\sqrt {5}}}\Phi _{2}$

$\Psi _{2}={\frac {2}{\sqrt {5}}}\Phi _{1}-{\frac {1}{\sqrt {5}}}\Phi _{2}$

Proba b1 après les 2 mesures

$P(b_{1})={\frac {1}{5}}*{\frac {1}{5}}+{\frac {4}{5}}*{\frac {4}{5}}={\frac {17}{25}}$

Tipps - Aufgabe 2 c)

Energie kleiner als $E(n=1,p=1,q=1)$

Zum Beispiel, $n^{2}+p^{2}$ als der Radius eines Kreises betrachten $\Longrightarrow E(n,p)\leq n^{2}+p^{2}$

Aufgabe 2 a)

Ansatz :

$\Psi (x,y,z)=\Psi (x)\cdot \Psi (y)\cdot \Psi (z)$

$\Longrightarrow \Psi (x,y,z)=C\left[\sin {\frac {n\pi }{a}}\cdot \sin {\frac {p\pi }{a}}\cdot \sin {\frac {q\pi }{b}}\right]$

Schrödinger Zeitunabhängig

${E}\psi (x,y,z)=-{\hbar ^{2} \over 2m}\nabla ^{2}\psi (x,y,z)+V(x,y,z)\psi (x,y,z)$

${\hbar ^{2} \over 2m}\nabla ^{2}\psi (x,y,z)=-{E}\psi (x,y,z)$

${\hbar ^{2} \over 2m}{\frac {\partial ^{2}}{\partial x^{2}}}\psi (x,y,z)+{\hbar ^{2} \over 2m}{\frac {\partial ^{2}}{\partial y^{2}}}\psi (x,y,z)+{\hbar ^{2} \over 2m}{\frac {\partial ^{2}}{\partial z^{2}}}\psi (x,y,z)=-{E}\psi (x,y,z)$

Tipps Aufgabe 3

$\Psi (x)=a\cdot u_{0}(x)+b\cdot u_{1}(x)(a^{2}+b^{2}=1)$

$\Longrightarrow$ Sinnvoll zu schreiben ist

$a=\cos(\varphi )\cdot e^{i\theta }$

$b=\sin(\varphi )\cdot e^{i\theta }$

Trouver a, b tels que

$\left(\int _{0}^{\infty }\Psi ^{*}(x)\Psi (x)dx\right)^{2}$ soit maximal

j'ai

$\int _{0}^{\infty }\Psi ^{*}(x)\Psi (x)dx=\int _{0}^{\infty }(\cos(\phi )^{2}u_{0}(x)^{2}+2\cos(\phi )\sin(\phi )u_{0}(x)u_{1}(x)+\sin(x)^{2}u_{1}(x)^{2})dx$

$=\int _{0}^{\infty }\left(\cos ^{2}(\phi )\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{2}}e^{-{\frac {m\omega }{\hbar }}x^{2}}+2\sin ^{2}(\phi )\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{2}}e^{-{\frac {m\omega }{\hbar }}x^{2}}x^{2}{\frac {m\omega }{\hbar }}+2{\sqrt {2}}\cos(\phi )\sin(\phi )\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{2}}e^{-{\frac {m\omega }{\hbar }}x^{2}}x{\sqrt {\frac {m\omega }{\hbar }}}\right)dx$

Premier terme

$\int _{0}^{\infty }\left(\cos ^{2}(\phi )\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{2}}e^{-{\frac {m\omega }{\hbar }}x^{2}}\right)dx$ subst. $y={\sqrt {\frac {m\omega }{\hbar }}}x$

$=\int _{0}^{\infty }\left(\cos ^{2}(\phi )\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{2}}{\sqrt {\frac {\hbar }{m\omega }}}e^{-y^{2}}\right)dy$

$={\frac {\cos ^{2}(\phi )}{\sqrt {\pi }}}\int _{0}^{\infty }e^{-y^{2}}dy$

$={\frac {\cos ^{2}(\phi )}{2}}$

Deuxième terme :

$2\sin ^{2}(\phi ){\sqrt {\frac {m\omega }{\pi \hbar }}}{\frac {m\omega }{\hbar }}\int _{0}^{\infty }e^{-{\frac {k}{2}}x^{2}}x\cdot x$ wobei $-{\frac {k}{2}}=-{\frac {m\omega }{\hbar }}$

$=2\sin ^{2}(\phi ){\sqrt {\frac {m\omega }{\pi \hbar }}}{\frac {m\omega }{\hbar }}\cdot {\frac {1}{k}}\int _{0}^{\infty }e^{-{\frac {k}{2}}x^{2}}$

$=2\sin ^{2}(\phi ){\sqrt {\frac {m\omega }{\pi \hbar }}}{\frac {m\omega }{\hbar }}\cdot {\frac {1}{2k}}{\sqrt {\frac {2\pi }{k}}}$

$=2\sin ^{2}(\phi ){\sqrt {\frac {m\omega }{\pi \hbar }}}{\frac {m\omega }{\hbar }}\cdot {\frac {\hbar }{4m\omega }}{\sqrt {\frac {\pi \hbar }{m\omega }}}$

$={\frac {1}{2}}\sin ^{2}(\phi )$

Troisième terme :

$\int _{0}^{\infty }\left(2{\sqrt {2}}\cos(\phi )\sin(\phi )\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{2}}e^{-{\frac {m\omega }{\hbar }}x^{2}}x{\sqrt {\frac {m\omega }{\hbar }}}\right)dx$

$=\left[-{\sqrt {2}}\cos(\phi )\sin(\phi )\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{2}}e^{-{\frac {m\omega }{\hbar }}x^{2}}{\sqrt {\frac {\hbar }{m\omega }}}\right]_{0}^{\infty }$

$={\sqrt {2}}\cos(\phi )\sin(\phi )\left({\frac {1}{\pi }}\right)^{\frac {1}{2}}$

Après dérivation ${\frac {d}{d\phi }}$ il reste

${\sqrt {2}}(-\sin \phi )\sin \phi {\sqrt {\frac {1}{\pi }}}+{\sqrt {2}}(\cos \phi )\cos \phi {\sqrt {\frac {1}{\pi }}}$

${\sqrt {\frac {2}{\pi }}}(cos^{2}\phi -\sin ^{2}\phi )$

Tipp Aufgabe 4

Reflektion : $e^{ik_{I}x}+R\cdot e^{-ik_{I}x}$

Transmission : $T\cdot e^{ik_{II}x}$

Aufgabe 4

Schrödingergleichungen

$-{\frac {\hbar }{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}u_{I}=Eu_{I}$

$-{\frac {\hbar }{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}u_{II}+V_{0}u_{II}=Eu_{II}$

Randbedingungen $u_{I}(0)=u_{II}(0)$

Für I

$-{\frac {\hbar }{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}u_{I}=Eu_{I}$

$\Longleftrightarrow {\ddot {u}}_{I}+{\frac {2m}{\hbar ^{2}}}E\cdot u_{I}=0$

$\Longrightarrow \lambda =i\pm {\sqrt {{\frac {2m}{\hbar ^{2}}}E}}$

$\Longrightarrow u_{I}=A\cdot e^{i{\sqrt {{\frac {2m}{\hbar ^{2}}}E}}x}+B\cdot e^{-i{\sqrt {{\frac {2m}{\hbar ^{2}}}E}}x}$

Für II

$-{\frac {\hbar }{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}u_{I}=(V_{0}-E)u_{I}$

...

$\Longrightarrow u_{I}=C\cdot e^{i{\sqrt {{\frac {2m}{\hbar ^{2}}}(V_{0}-E)}}x}+D\cdot e^{-i{\sqrt {{\frac {2m}{\hbar ^{2}}}(V_{0}-E)}}x}$

Zusätzliche Randbedingung ${\frac {\partial }{\partial x}}u_{I}(0)={\frac {\partial }{\partial x}}u_{II}(0)$

Also

$A\ i\ k_{I}-B\ i\ k_{I}=C\ i\ k_{II}-D\ i\ k_{II}$

die erste Randbedingung :

$A+B=C+D$

donc $B=C-A$ (on a D=0)

donc ${\frac {B}{A}}={\frac {C}{A}}-1$

on a aussi $C={\frac {A\ k_{I}-B\ k_{I}}{k_{II}}}$

ce qui donne

${\frac {B}{A}}={\frac {C}{A}}-1={\frac {A\ k_{I}-B\ k_{I}}{A\ k_{II}}}-1={\frac {k_{I}}{k_{II}}}-{\frac {B}{A}}{\frac {k_{I}}{k_{II}}}-1$

${\frac {B}{A}}(1+{\frac {k_{I}}{k_{II}}})={\frac {k_{I}}{k_{II}}}-1$

donc

${\frac {B}{A}}={\frac {k_{I}-k_{II}}{k_{I}+k_{II}}}$