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* Vor Rechts ins System <math>{}_{S}^{R}{{H}_{I}}</math>
* Vor Rechts ins System <math>{}_{S}^{R}{{H}_{I}}</math>
* Vom System nach Links <math>{}_{L}^{S}{{H}_{I}}</math>
* Vom System nach Links <math>{}_{L}^{S}{{H}_{I}}</math>
* Vom System nach Rechts <math>{}_{R}^{S}{{H}_{I}}</math> mit <math>{}_{S}^{X}{{H}_{I}}=\sum\limits_{k,i}{{}^{X}{{V}_{k}}^{X}a_{k}^{\dagger }{{e}_{i}}}</math> und <math>{}_{X}^{S}{{H}_{I}}=\sum\limits_{k,i}{{}^{X}{{V}_{k}}^{X}{{a}_{k}}e_{i}^{\dagger }}</math>
* Vom System nach Rechts <math>{}_{R}^{S}{{H}_{I}}</math>
mit
<math>{}_{S}^{X}{{H}_{I}}=\sum\limits_{k,i}{{}^{X}{{V}_{k}}^{X}a_{k}^{\dagger }{{e}_{i}}}</math>
und
<math>{}_{X}^{S}{{H}_{I}}=\sum\limits_{k,i}{{}^{X}{{V}_{k}}^{X}{{a}_{k}}e_{i}^{\dagger }}</math>


:<math>{{e}_{i}}</math> erzeugt ein Electron im System mit Energieniveau i.
<math>{{e}_{i}}</math> erzeugt ein Electron im System mit Energieniveau i.
:<math>e_{i}^{\dagger }</math> vernichtet ...
<math>e_{i}^{\dagger }</math> vernichtet ...


==Transformation ins WW-Bild==
==Transformation ins WW-Bild==
Operator ins WWBild
Operator ins WWBild


:<math>\tilde{A}\left( t \right):=U_{0}^{\dagger }A{{U}_{0}}</math>
<math>\tilde{A}\left( t \right):=U_{0}^{\dagger }A{{U}_{0}}</math>
mit <math>{{U}_{0}}=\exp \left( -\mathfrak{i}{{H}_{0}}t \right)</math>
mit <math>{{U}_{0}}=\exp \left( -\mathfrak{i}{{H}_{0}}t \right)</math>
und <math>{{H}_{0}}={{H}_{S}}+{{H}_{B}}</math>
und <math>{{H}_{0}}={{H}_{S}}+{{H}_{B}}</math>


Starte von [[Liouville-von-Neumann-Gleichung]]
Starte von [[Liouville-von-Neumann-Gleichung]]
:<math>
<math>
\dot \rho  =  - \mathfrak{i} \left[ {H,\rho } \right]</math>
\dot \rho  =  - \mathfrak{i} \left[ {H,\rho } \right]</math>


mit der Lösung
mit der Lösung


:<math>\rho \left( t \right)={{U}^{\dagger }}{{\rho }_{0}}U</math>
<math>\rho \left( t \right)={{U}^{\dagger }}{{\rho }_{0}}U</math>


mit <math>U=\exp \left( -\mathfrak{i}Ht \right)</math>
mit <math>U=\exp \left( -\mathfrak{i}Ht \right)</math>
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Beweis
Beweis


:<math>{{\partial }_{t}}U=-\mathfrak{i}HU</math> sowie <math>{{\partial }_{t}}{{U}^{\dagger }}=\mathfrak{i}HU</math>
<math>{{\partial }_{t}}U=-\mathfrak{i}HU</math>
 
sowie
 
<math>{{\partial }_{t}}{{U}^{\dagger }}=\mathfrak{i}HU</math>


Dann ist
Dann ist
:<math>{{d}_{t}}\rho =\underbrace{-\mathfrak{i}HU{{\rho }_{0}}{{U}^{\dagger }}+U{{\rho }_{0}}\mathfrak{i}H{{U}^{\dagger }}}_{-\mathfrak{i}\left[ H,\rho  \right]}+\underbrace{U\left( {{\partial }_{t}}{{\rho }_{0}} \right){{U}^{\dagger }}}_{0}</math>
<math>{{d}_{t}}\rho =\underbrace{-\mathfrak{i}HU{{\rho }_{0}}{{U}^{\dagger }}+U{{\rho }_{0}}\mathfrak{i}H{{U}^{\dagger }}}_{-\mathfrak{i}\left[ H,\rho  \right]}+\underbrace{U\left( {{\partial }_{t}}{{\rho }_{0}} \right){{U}^{\dagger }}}_{0}</math>


beweis ende
beweis ende
Line 50: Line 58:




 
<math>\begin{align}
:<math>\begin{align}
   & {{d}_{t}}\tilde{\rho }={{d}_{t}}\left( U_{0}^{\dagger }\rho {{U}_{0}} \right) \\  
   & {{d}_{t}}\tilde{\rho }={{d}_{t}}\left( U_{0}^{\dagger }\rho {{U}_{0}} \right) \\
  & =\mathfrak{i}{{H}_{0}}U_{0}^{\dagger }\rho {{U}_{0}}-iU_{0}^{\dagger }\rho {{H}_{0}}{{U}_{0}}+U_{0}^{\dagger }{{d}_{t}}\left( \rho  \right){{U}_{0}} \\  
  & =\mathfrak{i}{{H}_{0}}U_{0}^{\dagger }\rho {{U}_{0}}-iU_{0}^{\dagger }\rho {{H}_{0}}{{U}_{0}}+U_{0}^{\dagger }{{d}_{t}}\left( \rho  \right){{U}_{0}} \\
  & =\mathfrak{i}\left[ {{H}_{0}},\tilde{\rho } \right]-\mathfrak{i}U_{0}^{\dagger }\left[ H,\rho  \right]{{U}_{0}} \\  
  & =\mathfrak{i}\left[ {{H}_{0}},\tilde{\rho } \right]-\mathfrak{i}U_{0}^{\dagger }\left[ H,\rho  \right]{{U}_{0}} \\
  & =\mathfrak{i}\left[ {{H}_{0}},\tilde{\rho } \right]-\mathfrak{i}U_{0}^{\dagger }\left[ {{H}_{0}}+{{H}_{SB}},\rho  \right]{{U}_{0}} \\  
  & =\mathfrak{i}\left[ {{H}_{0}},\tilde{\rho } \right]-\mathfrak{i}U_{0}^{\dagger }\left[ {{H}_{0}}+{{H}_{I}},\rho  \right]{{U}_{0}} \\
  & =\mathfrak{i}\left[ {{H}_{0}},\tilde{\rho } \right]-\mathfrak{i}\left[ {{H}_{0}},\tilde{\rho } \right]-\mathfrak{i}U_{0}^{\dagger }\left[ {{H}_{SB}},\rho  \right]{{U}_{0}} \\  
  & =\mathfrak{i}\left[ {{H}_{0}},\tilde{\rho } \right]-\mathfrak{i}\left[ {{H}_{0}},\tilde{\rho } \right]-\mathfrak{i}U_{0}^{\dagger }\left[ {{H}_{I}},\rho  \right]{{U}_{0}} \\
  & =-\mathfrak{i}\left[ {{{\tilde{H}}}_{SB}},\tilde{\rho } \right] \\  
  & =-\mathfrak{i}\left[ {{{\tilde{H}}}_{I}},\tilde{\rho } \right] \\
\end{align}</math>
===Lösung===
Integrieren
:<math>\tilde{\rho}=\rho_0 - \mathfrak{i} \int_0^t [\tilde{H_I},\tilde{\rho}]\,dt'</math>
auf rechter Seite einsetzen
 
:<math>\begin{align}
  & {{d}_{t}}\tilde{\rho }=-\mathfrak{i}\left[ {{{\tilde{H}}}_{I}},{{\rho }_{0}}-\mathfrak{i}\int_{0}^{t}{[{{{\tilde{H}}}_{I}},\tilde{\rho }]}\,d{t}' \right] \\
& =-\mathfrak{i}\left[ {{{\tilde{H}}}_{I}},{{\rho }_{0}} \right]-\left[ {{{\tilde{H}}}_{I}},\int_{0}^{t}{[{{{\tilde{H}}}_{I}},\tilde{\rho }]}\,d{t}' \right] \\
& =-\mathfrak{i}\left[ {{{\tilde{H}}}_{I}},{{\rho }_{0}} \right]-\int_{0}^{t}{\left[ {{{\tilde{H}}}_{I}},\left[ {{{\tilde{H}}}_{I}},\,\tilde{\rho } \right] \right]}d{t}'
\end{align}</math>
==System Dichteoperator==
Der Dichteoperator des Systems ist die Spur über das Bad
 
:<math>{{\rho }_{S}}={{\operatorname{Tr}}_{B}}\left[ \rho  \right]</math>
 
 
:<math>{{{\tilde{\rho }}}_{S}}=U_{S}^{\dagger }{{\rho }_{S}}{{U}_{S}}</math>
 
 
:<math>{{U}_{S}}=\exp \left( -\mathsf{\mathfrak{i}}{{H}_{S}}t \right)</math>
 
damit folgt für
:<math>\begin{align}
  & {{d}_{t}}\tilde{\rho_S }=-\mathfrak{i} \operatorname{Tr}_B \left[ {{{\tilde{H}}}_{I}},{{\rho }_{0}} \right]-\int_{0}^{t}{\operatorname{Tr}_B  \left[ {{{\tilde{H}}}_{I}},\left[ {{{\tilde{H}}}_{I}},\,\tilde{\rho } \right] \right]}d{t}'
\end{align}</math>
\end{align}</math>
==Annahmen==
* WW zur Zeit t=0 eingeschaltet
* no korrelation beteween System and Bath at t=0
-->
:<math>{{\tilde{\rho }}_{0}}={{\rho }_{0}}={{\rho }_{S,0}}{{R}_{B,0}}</math>
* Kopplung Reservoiroperatoren ans System in Zustand R_0 liefern keinen Beitrag.
-->
:<math>{{\operatorname{Tr}}_{S}}\left[ {{{\tilde{H}}}_{I}}{{R}_{B,0}} \right]=0\Rightarrow \left[ {{{\tilde{H}}}_{I}},{{\rho }_{0}} \right]=0</math>
*Dichtematrix zu t=0 Sperabel
*Schwache Kopplung zwischen System und Bad H_I
*Systemgröße von B größer als S daher B nicht beeinflusst
:<math>\tilde{\rho }={{{\tilde{\rho }}}_{S,0}}{{R}_{B,0}}+O\left( {{H}_{I}} \right)</math>
===Bornsche Näherung===
* Jetzt vernachlässigen von Termen mit Ordnung von H_I>2
:<math>{{d}_{t}}{{{\tilde{\rho }}}_{S}}=-\int_{0}^{t}{{{\operatorname{Tr}}_{B}}\left[ {{{\tilde{H}}}_{I}},\left[ \tilde{H}{{'}_{I}},\,\tilde{\rho }{{'}_{S}}{{R}_{B,0}} \right] \right]}d{t}'</math>
===Markov Näherung===
* Zukunft hängt nur von aktuellem Zustand ab
:<math>{{\rho }_{S}}=\rho {{'}_{S}}</math>
[[Kategorie:Thermodynamik]]
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