Für
haben wir zwei Basisvektoren
und
System ist prepariert im Zustand
.
Die Wahrscheinlichkeit den Messwert
zu messen ist gegeben durch :
![{\displaystyle P_{a1}=\left(\int \Psi _{1}^{*}\Phi _{1}\ d^{3}x\right)^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e5f1efa4dd82043ae927f811dc9241778c5a367)
![{\displaystyle =\left({\frac {1}{\sqrt {5}}}\int \psi _{1}^{*}\psi _{1}+2\psi _{1}^{*}\psi _{2}\ dx\right)^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a6d121bd5152c270f983056d28248b9e38fc408)
![{\displaystyle =\left({\frac {1}{\sqrt {5}}}+{\frac {1}{\sqrt {5}}}\int 2\psi _{1}^{*}\psi _{2}\ dx\right)^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70e561c26a963a166e4e8f7bbee81baac7eb032b)
![{\displaystyle =\left({\frac {1}{\sqrt {5}}}\right)^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d45fec5eb65a2a33ba893058d93593776f60ba9e)
![{\displaystyle ={\frac {1}{5}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ee0f151036947a0035513de89c611a273d13d8b)
Die QM. Operatoren sind hermitsch
EW sind reell, hat ONB.
ist die Wahrscheinlichkeit System in
zu finden.
![{\displaystyle \int \Psi _{1}^{*}\Psi _{1}\ d^{3}x=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc268aab0a2a22ee505368fba047076378ba8c4b)
?
Durch Umformen erhalten wir :
![{\displaystyle \Psi _{1}={\frac {1}{\sqrt {5}}}\Phi _{1}+{\frac {2}{\sqrt {5}}}\Phi _{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bf62682d41a437ec71ac69878d90d66b127bb35)
![{\displaystyle \Psi _{2}={\frac {2}{\sqrt {5}}}\Phi _{1}-{\frac {1}{\sqrt {5}}}\Phi _{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b548ea64c5507e38b1c2e07c61a02ae6e747c4e)
Proba b1 après les 2 mesures
[edit]
![{\displaystyle P(b_{1})={\frac {1}{5}}*{\frac {1}{5}}+{\frac {4}{5}}*{\frac {4}{5}}={\frac {17}{25}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f74d9ca5b1bef8b40385433ed0cd4b76cf57f791)
Tipps - Aufgabe 2 c)
[edit]
Energie kleiner als
Zum Beispiel,
als der Radius eines Kreises betrachten
Ansatz :
![{\displaystyle \Psi (x,y,z)=\Psi (x)\cdot \Psi (y)\cdot \Psi (z)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cdd002da6f574e68de5ec6cf633f8aecfdbf373f)
![{\displaystyle \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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/860584f168d1a582e64b9dc100a6829adcd2d857)
Schrödinger Zeitunabhängig
[edit]
![{\displaystyle {E}\psi (x,y,z)=-{\hbar ^{2} \over 2m}\nabla ^{2}\psi (x,y,z)+V(x,y,z)\psi (x,y,z)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cff1ce744a649637f9561bdc8f1d036b14e91af)
![{\displaystyle {\hbar ^{2} \over 2m}\nabla ^{2}\psi (x,y,z)=-{E}\psi (x,y,z)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a830f4578542c29876717e0483fe299e0ccbe2f)
![{\displaystyle {\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6e55ff6b187ceecf85ebda38a448acae2a477d3)
![{\displaystyle \Psi (x)=a\cdot u_{0}(x)+b\cdot u_{1}(x)(a^{2}+b^{2}=1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1dc13981dc50bf93f2aaa25071e47be223cf9c22)
Sinnvoll zu schreiben ist
![{\displaystyle a=\cos(\varphi )\cdot e^{i\theta }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00354d9f131f7f8c7dac252cfaef3fd518b5d778)
![{\displaystyle b=\sin(\varphi )\cdot e^{i\theta }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3efc8fd6dd509873c427563b3a15be326bf55ce7)
Trouver a, b tels que
soit maximal
j'ai
![{\displaystyle \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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d449054c6007cc05106363d114c99b86d5de622)
![{\displaystyle =\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/deafe222e8d921a52f1c200eb16002be667dfb94)
Premier terme
subst. ![{\displaystyle y={\sqrt {\frac {m\omega }{\hbar }}}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17e730f203f357624b36a8ec8933b0206d86d0a1)
![{\displaystyle =\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65dc82cf654d27dcfdbf517989eeadb066224f38)
![{\displaystyle ={\frac {\cos ^{2}(\phi )}{\sqrt {\pi }}}\int _{0}^{\infty }e^{-y^{2}}dy}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78c9d87e37ae3993d1b7df348b0e3b4adbec2043)
![{\displaystyle ={\frac {\cos ^{2}(\phi )}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4800772d4c8a3d5d94745188d6e459deeff5e3c9)
Deuxième terme :
wobei ![{\displaystyle -{\frac {k}{2}}=-{\frac {m\omega }{\hbar }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9859feb8c35498efa277e3ba1821d5873d01afd5)
![{\displaystyle =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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e23f86ccbbb4a77f520bc67e176576ecbe32d43)
![{\displaystyle =2\sin ^{2}(\phi ){\sqrt {\frac {m\omega }{\pi \hbar }}}{\frac {m\omega }{\hbar }}\cdot {\frac {1}{2k}}{\sqrt {\frac {2\pi }{k}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d5f723a893c82494c494721399869b8081ecf88)
![{\displaystyle =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 }}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/237a4cce3aa51440425f1e8fc62349e281fc1e01)
![{\displaystyle ={\frac {1}{2}}\sin ^{2}(\phi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de18aaf1aec8f0395c83bd323d5d3d178899b32a)
Troisième terme :
![{\displaystyle \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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6818f93463efe05c754731e82020a4dfaa1992f9)
![{\displaystyle =\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 }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f793672dc35ad43e016c23d4b95d17aea324e664)
![{\displaystyle ={\sqrt {2}}\cos(\phi )\sin(\phi )\left({\frac {1}{\pi }}\right)^{\frac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a30d88ea4cdc31a199ca834d9e1c560ddd6d1de5)
Après dérivation
il reste
![{\displaystyle {\sqrt {2}}(-\sin \phi )\sin \phi {\sqrt {\frac {1}{\pi }}}+{\sqrt {2}}(\cos \phi )\cos \phi {\sqrt {\frac {1}{\pi }}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fec0b7538cbc0f27b263846b5dd3946c01a9cba)
![{\displaystyle {\sqrt {\frac {2}{\pi }}}(cos^{2}\phi -\sin ^{2}\phi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15789a4fb76427f9350f36bbd4952d842e684ad8)
Reflektion :
Transmission :
Schrödingergleichungen
![{\displaystyle -{\frac {\hbar }{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}u_{I}=Eu_{I}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b253185b767e1963e942de59b59d871e39d41df)
![{\displaystyle -{\frac {\hbar }{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}u_{II}+V_{0}u_{II}=Eu_{II}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e330e3656e2f9e79a60753d15fe5329fd35153a)
Randbedingungen
Für I
![{\displaystyle -{\frac {\hbar }{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}u_{I}=Eu_{I}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b253185b767e1963e942de59b59d871e39d41df)
![{\displaystyle \Longleftrightarrow {\ddot {u}}_{I}+{\frac {2m}{\hbar ^{2}}}E\cdot u_{I}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62fcbda612ac004d737dcd06b21644caef9d99cc)
![{\displaystyle \Longrightarrow \lambda =i\pm {\sqrt {{\frac {2m}{\hbar ^{2}}}E}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d861390836adc6e01cdbb54dedd5a9b56f4c1c6)
![{\displaystyle \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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9c1b53edad21e4f78a793226beca7b0637d598c)
Für II
![{\displaystyle -{\frac {\hbar }{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}u_{I}=(V_{0}-E)u_{I}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a07c5026106f1e5637f7e431819dad53da5999d3)
...
![{\displaystyle \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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/54204ab2de2679bbe468a0591a92578d772ca2d0)
Zusätzliche Randbedingung
Also
![{\displaystyle A\ i\ k_{I}-B\ i\ k_{I}=C\ i\ k_{II}-D\ i\ k_{II}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11289b1dc2f35c570a72991e3491118e8c0e6922)
die erste Randbedingung :
![{\displaystyle A+B=C+D}](https://wikimedia.org/api/rest_v1/media/math/render/svg/283bb4c3238abfe1f9c7f451a8da47457169acce)
donc
(on a D=0)
donc
on a aussi
ce qui donne
![{\displaystyle {\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62cddd7a69797765b2ac4878b56475cd1cc82a2d)
![{\displaystyle {\frac {B}{A}}(1+{\frac {k_{I}}{k_{II}}})={\frac {k_{I}}{k_{II}}}-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ffdcee63bd5aa32cced0634b5d451a897babb50d)
donc
![{\displaystyle {\frac {B}{A}}={\frac {k_{I}-k_{II}}{k_{I}+k_{II}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7033ca93895581c44aefa7b99ad712174e31030)