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12.1: Perturbative solution of the Liouville equation

  • Page ID
    5300
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    As in the classical case, we assume a solution of the form

    \[ \rho(t) = \rho_0(H_0) + \Delta \rho(t) \nonumber \]

    where

    \[ [H_0,\rho_0]=0\;\;\;\;\;\Rightarrow\;\;\;\;\;{\partial \rho_0 \over \partial t}=0 \nonumber \]

    and we will assume

    \[ \rho_0(H_0) = {e^{-\beta H_0} \over Q(N,V,T)} \nonumber \]

    Substituting into the Liouville equation and working to first order in small quantities, we find

    \[ {\partial \Delta \rho \over \partial t} = {1 \over i\hbar}[H_0,\Delta \rho] -{1 \over i\hbar} [B,\rho_0]F_e(t) \nonumber \]

    which is a first order inhomogeneous equation that can be solved by using an integrating factor:

    \[ \Delta \rho(t) = -{1 \over i\hbar}\int_{-\infty}^t\;ds\;e^{-iH_0(t-s)/\hbar}[B,\rho_0]e^{iH_0(t-s)/\hbar}F_e(s) \nonumber \]

    (Note that we have chosen the origin in time to be \( {t = - \infty } \), which is an arbitrary choice.)

    For an observable \(A\), the expectation value is

    \[ \langle A(t)\rangle = {\rm Tr}(\rho A) = \langle A\rangle _0 + {\rm Tr}(\Delta \rho(t) A) \nonumber \]

    when the solution for \( \Delta \rho \) is substituted in, this becomes

    \(\langle A(t) \rangle \) \(=\)

    \(\langle A \rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\;{\rm Tr}\left[Ae^{-iH_0(t-s)/hbar}[B,\rho_0]e^{iH_0(t-s)/\hbar}\right]F_e(s)\)

     
      \(=\)

    \(\langle A \rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\;{\rm Tr}\left[e^{iH_0(t-s)/\hbar} Ae^{-iH_0(t-s)/hbar}[B,\rho_0]\right]F_e(s)\)

     
      \(=\)

    \(\langle A\rangle _0 - {1 \over i\hbar} ds\;{\rm Tr}\left[A(t-s)[B,\rho_0]\right]F_e(s) \)

     

    where the cyclic property of the trace has been used and the Heisenberg evolution for \(A\) has been substituted in. Expanding the commutator gives

    \(\langle A(t) \rangle \) \(=\)

    \(\langle A\rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\;{\rm Tr}\left[A(t-s)B\rho_0 - A(t-s)\rho_0B\right]F_e(s) \)

     
      \(=\)

    \(\langle A\rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\;{\rm Tr}\left[\rho_0\left(A(t-s)B - BA(t-s)\right)\right]F_e(s)\)

     
      \(=\)

    \(\langle A\rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\;F_e(s)\langle [A(t-s),B(0)]_0\rangle \)

     

    where the cyclic property of the trace has been used again. Define a function

    \[ \Phi_{AB}(t) = {i \over \hbar}\langle [A(t),B(0)]\rangle _0 \nonumber \]

    called the after effect function. It is essentially the antisymmetric quantum time correlation function, which involves the commutator between \(A (t) \) and \(B (0) \). Then the linear response result can be written as

    \[ \langle A(t)\rangle = \langle A \rangle _0 + \int_{-\infty}^t\;ds F_e(s)\Phi_{AB}(t-s) \nonumber \]

    which is the starting point for the theory of quantum transport coefficients. If we choose to measure the operator \(B\), then we find

    \[ \langle B(t)\rangle = \langle B\rangle _0 + \int_{-\infty}^t\;ds\;F_e(s)\Phi_{BB}(t-s) \nonumber \]


    This page titled 12.1: Perturbative solution of the Liouville equation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark Tuckerman.