12.2: Kubo Transform Expression for the Time Correlation Function
- Page ID
- 5302
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We shall derive the following expression for the quantum time correlation function
\[ \Phi_{AB}(t) = \int_0^{\beta}d\lambda\;\langle \dot{B}(-i\hbar\lambda)A(t)\rangle _0 \nonumber \]
known as a Kubo transform relation. Since \( {\dot{B} } \) is given by the Heisenberg equation:
\[ \dot{B} = {1 \over i\hbar}[B,H_0] \nonumber \]
it follows that
\[ \dot{B}(t) = -{1 \over i\hbar}e^{iH_0t/\hbar}[H_0,B(0)]e^{-iH_0t/\hbar} \nonumber \]
Evaluating the expression at \(t=-i\hbar\lambda \) gives
\[ \dot{B}(-i\hbar\lambda) = e^{\lambda H_0}{1 \over i\hbar}[B(0),H_0]e^{-\lambda H_0} \nonumber \]
Thus,
\[ \Phi_{AB}(t) = \int_0^{\beta} d\lambda \langle e^{\lambda H_0} \left ({1 \over i\hbar}[B(0),H_0]\right)e^{-\lambda H_0}A(t)\rangle _0 \nonumber \]
By performing the trace in the basis of eigenvectors of \(H_0 \), we obtain
\[ \begin{align*} \Phi_{AB}(t) &= {1 \over Q}\int_0^{\beta}d\lambda\sum_n \langle n\vert e^{\lambda H_0} \left ({1 \over i\hbar}\right)[B(0),H_0]e^{-\lambda H_0}A(t)\vert n\rangle e^{-\beta E_n} \\[4pt] &= {1 \over Q}\int_0^{\beta}d\lambda\sum_{m,n} \langle n\vert e^{\lambda H_0}\left ( {1 \over i \hbar } \right ) \left [ B (0), H_0 \right ] e^{-\lambda H_0}\vert m\rangle \langle m\vert A(t)\vert n\rangle e^{-\beta E_n} \\[4pt] &= {1 \over Q}\int_0^{\beta}d\lambda\sum_{m,n} e^{\lambda E_n}e^{-\lambda E_m} {1 \over i \hbar } \langle n \vert [B(0),H_0]\vert m\rangle \langle m\vert A(t)\vert n\rangle e^{-\beta E_n} \\[4pt] &= {1 \over Q}\sum_{m,n} e^{-\beta E_n}{e^{\beta(E_n-E_m)}-1 \over (E_n - E_m) } {1 \over i\hbar } \langle n\vert[B(0),H_0]\vert m\rangle \langle m\vert A(t)\vert n\rangle e^{-\beta E_n} \end{align*} \]
But
\[ \langle n\vert[B(0),H_0]\vert m\rangle = \langle n\vert B(0) H_0 - H_0 B (0)\vert m\rangle = (E_m-E_n)\langle n\vert B(0)\vert m\rangle \nonumber \]
Therefore,
\[ \begin{align*} \Phi_{AB}(t) &= -{1 \over i\hbar Q}\sum_{m,n}\left(e^{-\beta E_n}-e^{-\beta E_m}\right)\langle n\vert B(0)\vert m\rangle \langle m\vert A(t)\vert n\rangle \\[4pt] &= -{1 \over i\hbar Q}\left[\sum_{m,n}e^{-\beta E_m}\langle m\vert A (t) \vert n \rangle \langle n \vert B (0) \vert m \rangle - \sum _{m,n} e^{-\beta E_n}\langle n\vert B(0)\vert m\rangle \langle m\vert A(t)\vert n\rangle \right] \\[4pt] &= { {i \over \hbar}\langle [A(t),B(0)]\rangle _0 } \end{align*} \nonumber \]
which proves the relation. The classical limit can be deduced easily from the Kubo transform relation:
\[ \Phi_{AB}(t) \longrightarrow \beta\langle \dot{B}(0)A(t)\rangle _0 \nonumber \]
Note further, by using the cylic properties of the trace, that
\[ \langle \dot{B}(-i\hbar\lambda)B(t)\rangle _0 = -{d \over dt}\langle B(-i\hbar\lambda)B(t)\rangle _0 \nonumber \]