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The quantum equilibrium ensembles

  • Page ID
    5255
  • [ "article:topic", "Author tag:Tuckerman", "showtoc:no" ]

    At equilibrium, the density operator does not evolve in time; thus, \({\partial \rho \over \partial t} = 0\). Thus, from the equation of motion, if this holds, then \([H,\rho]=0\), and \(\rho (t) \) is a constant of the motion. This means that it can be simultaneously diagonalized with the Hamiltonian and can be expressed as a pure function of the Hamiltonian

    \[\rho = f(H)\]

    Therefore, the eigenstates of \(\underline {\rho} \), the vectors, we called \(\vert w_k \rangle \) are the eigenvectors \(\vert E_i \rangle \) of the Hamiltonian, and we can write \(H\) and \(\underline {\rho} \) as

    \[H\] $\textstyle =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ \(\displaystyle \sum_i E_i \vert E_i\rangle \langle E_i\vert\)  
    \[\underline {\rho} \] $\textstyle =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ \(\displaystyle \sum_i f(E_i)\vert E_i\rangle \langle E_i\vert\)  


    The choice of the function \(f\) determines the ensemble.