Editing Beispiel des Großkanonischen Ensenbles

Jump to navigation Jump to search
Warning: You are not logged in. Your IP address will be publicly visible if you make any edits. If you log in or create an account, your edits will be attributed to your username, along with other benefits.

The edit can be undone. Please check the comparison below to verify that this is what you want to do, and then publish the changes below to finish undoing the edit.

Latest revision Your text
Line 2: Line 2:


Illustration am Anhand von
Illustration am Anhand von
:<math>\begin{align}
<math>\begin{align}
   & {{G}_{\nu }}=\left\{ H,N \right\} \\
   & {{G}_{\nu }}=\left\{ H,N \right\} \\
  & {{h}_{\alpha }}=\left\{ V \right\} \\
  & {{h}_{\alpha }}=\left\{ V \right\} \\
Line 10: Line 10:




:<math>\begin{align}
<math>\begin{align}
   & R=\frac{1}{Z}{{e}^{-\sum\limits_{\nu }{{{\lambda }_{\nu }}{{G}_{\nu }}}}} \\
   & R=\frac{1}{Z}{{e}^{-\sum\limits_{\nu }{{{\lambda }_{\nu }}{{G}_{\nu }}}}} \\
  & {{R}_{gk}}=\frac{1}{{{Z}_{gk}}}{{e}^{-{{\lambda }_{1}}H-{{\lambda }_{2}}N}}  
  & {{R}_{gk}}=\frac{1}{{{Z}_{gk}}}{{e}^{-{{\lambda }_{1}}H-{{\lambda }_{2}}N}}  
Line 18: Line 18:
oftmals <math>{{\lambda }_{1}}=\beta ,\quad {{\lambda }_{2}}=-\beta \mu </math>
oftmals <math>{{\lambda }_{1}}=\beta ,\quad {{\lambda }_{2}}=-\beta \mu </math>


:<math>\left( {{\lambda }_{1}},{{\lambda }_{2}} \right)\to \left( \beta ,\mu  \right)</math>
<math>\left( {{\lambda }_{1}},{{\lambda }_{2}} \right)\to \left( \beta ,\mu  \right)</math>


wir zeigen:
wir zeigen:
:<math>\beta =\frac{1}{kT}</math> Temperatur taucht auf muss gezeigt werden
<math>\beta =\frac{1}{kT}</math> Temperatur taucht auf muss gezeigt werden
:<math>\mu</math> = Chemisches Potential ist die Energie die man braucht um 1 Teilchen hinzu zufügen
<math>\mu</math> = Chemisches Potential ist die Energie die man braucht um 1 Teilchen hinzu zufügen




:<math>{{R}_{gk}}=\frac{1}{Z}{{e}^{-\beta \left( H-\mu N \right)}}</math>
<math>{{R}_{gk}}=\frac{1}{Z}{{e}^{-\beta \left( H-\mu N \right)}}</math>




Line 33: Line 33:
braucht man um Zustandsgleichung festzulegen
braucht man um Zustandsgleichung festzulegen


:<math>S=S\left( \left\langle {{G}_{\nu }} \right\rangle ,{{h}_{\alpha }} \right)</math>
<math>S=S\left( \left\langle {{G}_{\nu }} \right\rangle ,{{h}_{\alpha }} \right)</math>


:<math>\Rightarrow {{S}_{gk}}={{S}_{gk}}\left( \left\langle H \right\rangle ,\left\langle N \right\rangle ,V \right)</math>
<math>\Rightarrow {{S}_{gk}}={{S}_{gk}}\left( \left\langle H \right\rangle ,\left\langle N \right\rangle ,V \right)</math>


:<math>{{S}_{gk}}\left( E,\overline{N},V \right)=k\beta E-k\beta \mu \overline{N}+k\ln {{Z}_{gk}}\left( \beta \mu V \right)</math>
<math>{{S}_{gk}}\left( E,\overline{N},V \right)=k\beta E-k\beta \mu \overline{N}+k\ln {{Z}_{gk}}\left( \beta \mu V \right)</math>




Line 46: Line 46:
Beziehungen der partiellen Ableitungen aus Gibbsgleichung
Beziehungen der partiellen Ableitungen aus Gibbsgleichung


:<math>k{{\lambda }_{\nu }}={{\partial }_{\left\langle {{G}_{\nu }} \right\rangle }}S;\quad k\sum\limits_{\nu }{{{\lambda }_{\nu }}{{M}_{\nu ,\alpha }}={{\partial }_{{{h}_{\alpha }}}}S}</math>
<math>k{{\lambda }_{\nu }}={{\partial }_{\left\langle {{G}_{\nu }} \right\rangle }}S;\quad k\sum\limits_{\nu }{{{\lambda }_{\nu }}{{M}_{\nu ,\alpha }}={{\partial }_{{{h}_{\alpha }}}}S}</math>
für <math>\nu=1</math>
für <math>\nu=1</math>


:<math>k{{\lambda }_{\nu }}={{\partial }_{\left\langle {{G}_{\nu }} \right\rangle }}S\Rightarrow k\beta ={{\left( \frac{\partial S}{\partial E} \right)}_{V,\bar{N}}};\quad k\sum\limits_{\nu }{{{\lambda }_{\nu }}{{M}_{\nu ,\alpha }}={{\partial }_{{{h}_{\alpha }}}}S}\Rightarrow {{\left( \frac{\partial S}{\partial N} \right)}_{E,\bar{N}}}=-k\beta \operatorname{Tr}\left( \frac{\partial H}{\partial V}R \right)</math>
<math>k{{\lambda }_{\nu }}={{\partial }_{\left\langle {{G}_{\nu }} \right\rangle }}S\Rightarrow k\beta ={{\left( \frac{\partial S}{\partial E} \right)}_{V,\bar{N}}};\quad k\sum\limits_{\nu }{{{\lambda }_{\nu }}{{M}_{\nu ,\alpha }}={{\partial }_{{{h}_{\alpha }}}}S}\Rightarrow {{\left( \frac{\partial S}{\partial N} \right)}_{E,\bar{N}}}=-k\beta \operatorname{Tr}\left( \frac{\partial H}{\partial V}R \right)</math>




:<math>\begin{align}
<math>\begin{align}
   & k{{\lambda }_{\nu }}={{\partial }_{\left\langle {{G}_{\nu }} \right\rangle }}S\Rightarrow k\beta ={{\left( \frac{\partial S}{\partial E} \right)}_{V,\bar{N}\left( \left( \text{V},\text{N sind nicht anzufassen bei der partiellen Ableitung} \right) \right)}} \\
   & k{{\lambda }_{\nu }}={{\partial }_{\left\langle {{G}_{\nu }} \right\rangle }}S\Rightarrow k\beta ={{\left( \frac{\partial S}{\partial E} \right)}_{V,\bar{N}\left( \left( \text{V},\text{N sind nicht anzufassen bei der partiellen Ableitung} \right) \right)}} \\
  & k\sum\limits_{\nu }{{{\lambda }_{\nu }}{{M}_{\nu ,\alpha }}={{\partial }_{{{h}_{\alpha }}}}S}\Rightarrow {{\left( \frac{\partial S}{\partial N} \right)}_{E,\bar{N}}}=-k\beta \operatorname{Tr}\left( \frac{\partial H}{\partial V}R \right)\quad \left( {{\partial }_{V}}N\to 0 \right) \\
  & k\sum\limits_{\nu }{{{\lambda }_{\nu }}{{M}_{\nu ,\alpha }}={{\partial }_{{{h}_{\alpha }}}}S}\Rightarrow {{\left( \frac{\partial S}{\partial N} \right)}_{E,\bar{N}}}=-k\beta \operatorname{Tr}\left( \frac{\partial H}{\partial V}R \right)\quad \left( {{\partial }_{V}}N\to 0 \right) \\
Line 59: Line 59:
für <math>\nu=2</math>
für <math>\nu=2</math>


:<math>\begin{align}
<math>\begin{align}
   & -k\beta \mu ={{\left( \frac{\partial S}{\partial E} \right)}_{V,\bar{N}}} \\
   & -k\beta \mu ={{\left( \frac{\partial S}{\partial E} \right)}_{V,\bar{N}}} \\
  & k{{\partial }_{V}}\ln {{Z}_{gk}}=k\beta p\Rightarrow p=\frac{1}{\beta }{{\partial }_{V}}\ln {{Z}_{gk}} \\
  & k{{\partial }_{V}}\ln {{Z}_{gk}}=k\beta p\Rightarrow p=\frac{1}{\beta }{{\partial }_{V}}\ln {{Z}_{gk}} \\
Line 70: Line 70:




:<math>\begin{align}
<math>\begin{align}
   & {{T}^{-1}}={{\left( \frac{\partial S}{\partial E} \right)}_{V,\bar{N}}} \\
   & {{T}^{-1}}={{\left( \frac{\partial S}{\partial E} \right)}_{V,\bar{N}}} \\
  & \mu =-T{{\left( \frac{\partial S}{\partial \bar{N}} \right)}_{V,E}} \\
  & \mu =-T{{\left( \frac{\partial S}{\partial \bar{N}} \right)}_{V,E}} \\
Line 79: Line 79:
es ist zu zeigen, dass die Temperaturdefinition sinnvoll ist
es ist zu zeigen, dass die Temperaturdefinition sinnvoll ist


:<math>{{T}^{-1}}=\left( \frac{\partial S}{\partial E} \right)</math>
<math>{{T}^{-1}}=\left( \frac{\partial S}{\partial E} \right)</math>


sonst darf man es nicht Temeratur nennen
sonst darf man es nicht Temeratur nennen


dazu zeigen:
dazu zeigen:
:<math>{{\left( \frac{\partial S}{\partial E} \right)}_{V,\bar{N}}}</math> ist als Eigenschaft bei 2 System die in Konakt über eine Grenzfläche stehen gleich
<math>{{\left( \frac{\partial S}{\partial E} \right)}_{V,\bar{N}}}</math> ist als Eigenschaft bei 2 System die in Konakt über eine Grenzfläche stehen gleich




Line 93: Line 93:
| <math>{{{\bar{N}}}_{1}},{{V}_{1}},{{E}_{1}}</math>
| <math>{{{\bar{N}}}_{1}},{{V}_{1}},{{E}_{1}}</math>
||
||
:<math>{{{\bar{N}}}_{2}},{{V}_{2}},{{E}_{2}}</math>
<math>{{{\bar{N}}}_{2}},{{V}_{2}},{{E}_{2}}</math>
|}
|}


:<math>\begin{align}
<math>\begin{align}
   & E={{E}_{1}}+{{E}_{2}} \\
   & E={{E}_{1}}+{{E}_{2}} \\
  & V={{V}_{1}}+{{V}_{2}} \\
  & V={{V}_{1}}+{{V}_{2}} \\
Line 102: Line 102:
\end{align}</math>
\end{align}</math>
Zu zeugen:
Zu zeugen:
:<math>S\overset{!}{\mathop{=}}\,{{S}_{1}}+{{S}_{2}}</math>
<math>S\overset{!}{\mathop{=}}\,{{S}_{1}}+{{S}_{2}}</math>


:<math>S\tilde{\ }\operatorname{Tr}\left( \rho \ln \rho  \right)=\operatorname{Tr}\left( {{\rho }_{1}}{{\rho }_{2}}\ln \left( {{\rho }_{1}}{{\rho }_{2}} \right) \right)</math>
<math>S\tilde{\ }\operatorname{Tr}\left( \rho \ln \rho  \right)=\operatorname{Tr}\left( {{\rho }_{1}}{{\rho }_{2}}\ln \left( {{\rho }_{1}}{{\rho }_{2}} \right) \right)</math>


statistischer Operator faktorisiert für '''kleine''' Grenzflächen
statistischer Operator faktorisiert für '''kleine''' Grenzflächen




:<math>\operatorname{Tr}\left( {{\rho }_{1}}{{\rho }_{2}}\ln \left( {{\rho }_{1}} \right) \right)+\operatorname{Tr}\left( {{\rho }_{1}}{{\rho }_{2}}\ln \left( {{\rho }_{2}} \right) \right)</math>
<math>\operatorname{Tr}\left( {{\rho }_{1}}{{\rho }_{2}}\ln \left( {{\rho }_{1}} \right) \right)+\operatorname{Tr}\left( {{\rho }_{1}}{{\rho }_{2}}\ln \left( {{\rho }_{2}} \right) \right)</math>




mit
mit
:<math>\operatorname{Tr}\overset{\wedge}{=}\left\langle  {{n}_{1}} \right|\left\langle  {{n}_{2}} \right|\ldots \left| {{n}_{1}} \right\rangle \left| {{n}_{2}} \right\rangle </math>
<math>\operatorname{Tr}\overset{\wedge}{=}\left\langle  {{n}_{1}} \right|\left\langle  {{n}_{2}} \right|\ldots \left| {{n}_{1}} \right\rangle \left| {{n}_{2}} \right\rangle </math>




:<math>\begin{align}
<math>\begin{align}
   & ={{\operatorname{Tr}}_{1}}\left( {{\rho }_{1}}\ln \left( {{\rho }_{1}} \right) \right)\underbrace{{{\operatorname{Tr}}_{2}}\left( \rho  \right)}_{1}+{{\operatorname{Tr}}_{2}}\left( {{\rho }_{2}}\ln \left( {{\rho }_{2}} \right) \right)\underbrace{\operatorname{Tr}\left( {{\rho }_{1}} \right)}_{1} \\
   & ={{\operatorname{Tr}}_{1}}\left( {{\rho }_{1}}\ln \left( {{\rho }_{1}} \right) \right)\underbrace{{{\operatorname{Tr}}_{2}}\left( \rho  \right)}_{1}+{{\operatorname{Tr}}_{2}}\left( {{\rho }_{2}}\ln \left( {{\rho }_{2}} \right) \right)\underbrace{\operatorname{Tr}\left( {{\rho }_{1}} \right)}_{1} \\
  & \Rightarrow S={{S}_{1}}+{{S}_{2}}  
  & \Rightarrow S={{S}_{1}}+{{S}_{2}}  
Line 124: Line 124:




:<math>\begin{align}
<math>\begin{align}
   & dE=d{{E}_{1}}+d{{E}_{2}}=0\to -d{{E}_{1}}=d{{E}_{2}} \\
   & dE=d{{E}_{1}}+d{{E}_{2}}=0\to -d{{E}_{1}}=d{{E}_{2}} \\
  & dV=d{{V}_{1}}+d{{V}_{2}}=0\to -d{{V}_{1}}=d{{V}_{2}} \\
  & dV=d{{V}_{1}}+d{{V}_{2}}=0\to -d{{V}_{1}}=d{{V}_{2}} \\
Line 138: Line 138:
nutze bei dS:
nutze bei dS:


:<math>\begin{align}
<math>\begin{align}
   & d{{S}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}=\frac{\partial {{S}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}{\partial {{V}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}d{{V}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{\partial {{S}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}{\partial {{{\bar{N}}}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}d{{{\bar{N}}}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{\partial {{S}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}{\partial {{E}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}d{{E}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}} \\
   & d{{S}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}=\frac{\partial {{S}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}{\partial {{V}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}d{{V}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{\partial {{S}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}{\partial {{{\bar{N}}}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}d{{{\bar{N}}}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{\partial {{S}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}{\partial {{E}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}d{{E}_{{}^{1}\!\!\diagup\!\!{}_{2}\;}} \\
  & d{{S}_{1}}=-d{{S}_{2}}  
  & d{{S}_{1}}=-d{{S}_{2}}  
Line 145: Line 145:




:<math>\frac{\partial {{S}_{1}}}{\partial {{V}_{1}}}d{{V}_{1}}+\frac{\partial {{S}_{1}}}{\partial {{{\bar{N}}}_{1}}}d{{{\bar{N}}}_{1}}+\frac{\partial {{S}_{1}}}{\partial {{E}_{1}}}d{{E}_{1}}=-\frac{\partial {{S}_{2}}}{\partial {{V}_{2}}}d{{V}_{2}}-\frac{\partial {{S}_{2}}}{\partial {{{\bar{N}}}_{2}}}d{{{\bar{N}}}_{2\;}}-\frac{\partial {{S}_{2}}}{\partial {{E}_{2}}}d{{E}_{2}}</math>
<math>\frac{\partial {{S}_{1}}}{\partial {{V}_{1}}}d{{V}_{1}}+\frac{\partial {{S}_{1}}}{\partial {{{\bar{N}}}_{1}}}d{{{\bar{N}}}_{1}}+\frac{\partial {{S}_{1}}}{\partial {{E}_{1}}}d{{E}_{1}}=-\frac{\partial {{S}_{2}}}{\partial {{V}_{2}}}d{{V}_{2}}-\frac{\partial {{S}_{2}}}{\partial {{{\bar{N}}}_{2}}}d{{{\bar{N}}}_{2\;}}-\frac{\partial {{S}_{2}}}{\partial {{E}_{2}}}d{{E}_{2}}</math>




Line 151: Line 151:




:<math>d{{E}_{1}}=-d{{E}_{2}},-d{{{\bar{N}}}_{1}}=d{{{\bar{N}}}_{2}},-d{{V}_{1}}=d{{V}_{2}}</math>
<math>d{{E}_{1}}=-d{{E}_{2}},-d{{{\bar{N}}}_{1}}=d{{{\bar{N}}}_{2}},-d{{V}_{1}}=d{{V}_{2}}</math>






:<math>\begin{align}
<math>\begin{align}
   & \left( \frac{\partial {{S}_{1}}}{\partial {{E}_{1}}}-\frac{\partial {{S}_{2}}}{\partial {{E}_{2}}} \right)d{{E}_{2}}=0 \\
   & \left( \frac{\partial {{S}_{1}}}{\partial {{E}_{1}}}-\frac{\partial {{S}_{2}}}{\partial {{E}_{2}}} \right)d{{E}_{2}}=0 \\
  & \left( \frac{\partial {{S}_{1}}}{\partial {{{\bar{N}}}_{1}}}-\frac{\partial {{S}_{2}}}{\partial {{{\bar{N}}}_{2}}} \right)d{{{\bar{N}}}_{2}}=0 \\
  & \left( \frac{\partial {{S}_{1}}}{\partial {{{\bar{N}}}_{1}}}-\frac{\partial {{S}_{2}}}{\partial {{{\bar{N}}}_{2}}} \right)d{{{\bar{N}}}_{2}}=0 \\
Line 162: Line 162:


weil N,V,E unabhängig variiert werden können gilt für alle
weil N,V,E unabhängig variiert werden können gilt für alle
:<math>d{{E}_{2}}</math>,<math>d{{{\bar{N}}}_{2}}</math>,
<math>d{{E}_{2}}</math>,<math>d{{{\bar{N}}}_{2}}</math>,
:<math>d{{V}_{2}}</math>
<math>d{{V}_{2}}</math>


→ folgende Eigenschaften des Systems im Kontakt sind gleich:
→ folgende Eigenschaften des Systems im Kontakt sind gleich:


:<math>\begin{align}
<math>\begin{align}
   & {{\left( \frac{\partial {{S}_{1}}}{\partial {{E}_{1}}} \right)}_{{{V}_{1}},{{{\bar{N}}}_{1}}}}={{\left( \frac{\partial {{S}_{2}}}{\partial {{E}_{2}}} \right)}_{{{V}_{2}},{{{\bar{N}}}_{2}}}} \\
   & {{\left( \frac{\partial {{S}_{1}}}{\partial {{E}_{1}}} \right)}_{{{V}_{1}},{{{\bar{N}}}_{1}}}}={{\left( \frac{\partial {{S}_{2}}}{\partial {{E}_{2}}} \right)}_{{{V}_{2}},{{{\bar{N}}}_{2}}}} \\
  & {{\left( \frac{\partial {{S}_{1}}}{\partial {{{\bar{N}}}_{1}}} \right)}_{{{V}_{1}},{{E}_{1}}}}={{\left( \frac{\partial {{S}_{2}}}{\partial {{{\bar{N}}}_{2}}} \right)}_{{{V}_{2}},{{E}_{2}}}} \\
  & {{\left( \frac{\partial {{S}_{1}}}{\partial {{{\bar{N}}}_{1}}} \right)}_{{{V}_{1}},{{E}_{1}}}}={{\left( \frac{\partial {{S}_{2}}}{\partial {{{\bar{N}}}_{2}}} \right)}_{{{V}_{2}},{{E}_{2}}}} \\
Line 180: Line 180:




:<math>\beta =\frac{1}{kT}</math>
<math>\beta =\frac{1}{kT}</math>
beides muss am Experiment verifiziert werden
beides muss am Experiment verifiziert werden


Line 194: Line 194:
===Dichtematrixdynamik und Zustandsgleichung===
===Dichtematrixdynamik und Zustandsgleichung===
Dichtematrixdynamik für 2Niveausystem: 1 Teilchen =
Dichtematrixdynamik für 2Niveausystem: 1 Teilchen =
:<math>{\bar{N}}</math>
<math>{\bar{N}}</math>
Besetzungszahldarstellung
Besetzungszahldarstellung


Please note that all contributions to testwiki are considered to be released under the Creative Commons Attribution (see Testwiki:Copyrights for details). If you do not want your writing to be edited mercilessly and redistributed at will, then do not submit it here.
You are also promising us that you wrote this yourself, or copied it from a public domain or similar free resource. Do not submit copyrighted work without permission!
Cancel Editing help (opens in new window)