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删除989字节 、 2021年11月26日 (五) 14:46
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where x, p is a normalised Gaussian wavepacket centered at
 
where x, p is a normalised Gaussian wavepacket centered at
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其中 x,p 是一个被位置 x 和动量 p包围的正态高斯波包被
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其中 x,p 是一个被位置 x 和动量 p包围的正态高斯波包
          
<math>
 
<math>
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《数学》
      
  Z = \int \operatorname{tr} \left( \mathrm{e}^{-\beta\hat{H}} |x, p\rangle \langle x, p| \right) \frac{dx \,dp}{h}
 
  Z = \int \operatorname{tr} \left( \mathrm{e}^{-\beta\hat{H}} |x, p\rangle \langle x, p| \right) \frac{dx \,dp}{h}
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</math>
 
</math>
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  = \int \langle x,p| \mathrm{e}^{-\beta\hat{H}} |x, p\rangle \frac{dx \,dp}{h}.
      
A coherent state is an approximate eigenstate of both operators <math> \hat{x} </math> and <math> \hat{p} </math>, hence also of the Hamiltonian Ĥ, with errors of the size of the uncertainties. If Δx and Δp can be regarded as zero, the action of Ĥ reduces to multiplication by the classical Hamiltonian, and Z reduces to the classical configuration integral.
 
A coherent state is an approximate eigenstate of both operators <math> \hat{x} </math> and <math> \hat{p} </math>, hence also of the Hamiltonian Ĥ, with errors of the size of the uncertainties. If Δx and Δp can be regarded as zero, the action of Ĥ reduces to multiplication by the classical Hamiltonian, and Z reduces to the classical configuration integral.
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<math>
 
<math>
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《数学》
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  p_i = \frac{\Omega_B(E - E_i)}{\Omega_{(S,B)}(E)}.
 
  p_i = \frac{\Omega_B(E - E_i)}{\Omega_{(S,B)}(E)}.
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P _ i = frac { Omega _ b (e-e _ i)}{ Omega _ {(s,b)}(e)}.
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:<math>
      
</math>
 
</math>
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p_i = \frac{\Omega_B(E - E_i)}{\Omega_{(S,B)}(E)}.
      
Assuming that the heat bath's internal energy is much larger than the energy of S (E ≫ E<sub>i</sub>), we can Taylor-expand <math>\Omega_B</math> to first order in E<sub>i</sub> and use the thermodynamic relation <math>\partial S_B/\partial E = 1/T</math>, where here <math>S_B</math>, <math>T</math> are the entropy and temperature of the bath respectively:
 
Assuming that the heat bath's internal energy is much larger than the energy of S (E ≫ E<sub>i</sub>), we can Taylor-expand <math>\Omega_B</math> to first order in E<sub>i</sub> and use the thermodynamic relation <math>\partial S_B/\partial E = 1/T</math>, where here <math>S_B</math>, <math>T</math> are the entropy and temperature of the bath respectively:
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<math>
 
<math>
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《数学》
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\begin{align}
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开始{ align }
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:<math>
      
  k \ln p_i &= k \ln \Omega_B(E - E_i) - k \ln \Omega_{(S,B)}(E) \\[5pt]
 
  k \ln p_i &= k \ln \Omega_B(E - E_i) - k \ln \Omega_{(S,B)}(E) \\[5pt]
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K ln p _ i & = k ln Omega _ b (e-e _ i)-k ln Omega _ (s,b)}(e)[5 pt ]
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\begin{align}
      
   &\approx -\frac{\partial\big(k \ln \Omega_B(E)\big)}{\partial E} E_i + k \ln\Omega_B(E) - k \ln \Omega_{(S,B)}(E)  
 
   &\approx -\frac{\partial\big(k \ln \Omega_B(E)\big)}{\partial E} E_i + k \ln\Omega_B(E) - k \ln \Omega_{(S,B)}(E)  
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大约-frac { partial big (k ln Omega _ b (e) big)}{ partial e } e _ i + k ln Omega _ b (e)-k ln Omega _ {(s,b)}(e))
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k \ln p_i &= k \ln \Omega_B(E - E_i) - k \ln \Omega_{(S,B)}(E) \\[5pt]
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\\[5pt]
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[5 pt ]
      
   &\approx -\frac{\partial\big(k \ln \Omega_B(E)\big)}{\partial E} E_i + k \ln\Omega_B(E) - k \ln \Omega_{(S,B)}(E)  
 
   &\approx -\frac{\partial\big(k \ln \Omega_B(E)\big)}{\partial E} E_i + k \ln\Omega_B(E) - k \ln \Omega_{(S,B)}(E)  
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  &\approx -\frac{\partial S_B}{\partial E} E_i + k \ln \frac{\Omega_B(E)}{\Omega_{(S,B)}(E)} \\[5pt]
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大约-frac { partial s _ b }{ partial e } e _ i + k ln frac { Omega _ b (e)}{ Omega _ {(s,b)}(e)}[5 pt ]
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\\[5pt]
      
   &\approx -\frac{E_i}{T} + k \ln \frac{\Omega_B(E)}{\Omega_{(S,B)}(E)}  
 
   &\approx -\frac{E_i}{T} + k \ln \frac{\Omega_B(E)}{\Omega_{(S,B)}(E)}  
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约-frac { e _ i }{ t } + k ln frac { Omega _ b (e)}{ Omega _ {(s,b)}(e)}
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  &\approx -\frac{\partial S_B}{\partial E} E_i + k \ln \frac{\Omega_B(E)}{\Omega_{(S,B)}(E)} \\[5pt]
      
\end{align}
 
\end{align}
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