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In the wiki page about Kubo formula, the expectation of some observable under weak time-dependent perturbation is derived. However, from my point of view, some crucial steps are missing. I did the derivation following the same idea, but obtained something different. Below is my derivation using interaction picture:

(Note: I will use another set of notations to make the argument easier to understand. I will use subscript $I$ to explicitly mean interaction picture quantities and quantities without sub $I$ are to be considered Schrodinger)

The system Hamiltonian is given by $H(t)=H_0+V(t)$, where $H_0$ is time independent and $V(t)$ is considered to be a small perturbation that is switched on at $t=0$. Then for some observable $A$, we have \begin{equation} <A>_t=\frac{Tr[\rho(t)A]}{Tr[\rho(t)]}=\frac{\sum_n<n(t)|e^{-\beta H(t)}A|n(t)>}{\sum_n<n(t)|e^{-\beta H(t)}|n(t)>}=\frac{\sum_ne^{-\beta E_n(t)}<n(t)|A|n(t)>}{\sum_ne^{-\beta E_n(t)}} \end{equation} Now we evaluate $<n(t)|A|n(t)>$ using interaction picture:

\begin{equation} \begin{split} <n(t)|A|n(t)>&=<n_I(t)|A_I(t)|n_I(t)>\\ &=<n_I(0)|U^{\dagger}_I(t)A_I(t)U_I(t)|n_I(0)>\\ &=<n(0)|U^{\dagger}_I(t)A_I(t)U_I(t)|n(0)> \end{split} \end{equation} where the time evolution in interaction picture has been used and also with the fact that $|n_I(t)>=e^{iH_0t}|n(t)>$, which immediately gives $|n_I(0)>=|n(0)>$.

We also know from interaction picture that $U_I(t)=e^{-i\int_0^tdt'V_I(t')}$, which upon perturbative expansion to linear order becomes $1-i\int_0^tdt'V_I(t')$. Plugging this into the above equation, we have again up to linear order \begin{equation} \begin{split} <n(t)|A|n(t)>&=<n(0)|[1+i\int_0^tdt'V_I(t')]A_I(t)[1-i\int_0^tdt'V_I(t')]|n(0)>\\ &=<n(0)|A_I(t)|n(0)>-i<n(0)|\int_0^{t}dt'[A_I(t),V_I(t')]|n(0)> \end{split} \end{equation} which after being substituted into the expression for $<A>_t$ gives us \begin{equation} \begin{split} <A>_t=&\frac{\sum_ne^{-\beta E_n(t)}<n(0)|A|n(0)>}{\sum_ne^{-\beta E_n(t)}} -i\frac{\sum_ne^{-\beta E_n(t)}<n(0)|\int_0^{t}dt'[A_I(t),V_I(t')]|n(0)>}{\sum_ne^{-\beta E_n(t)}}\\ \end{split} \end{equation} Comparing with the wiki result, it seems that the main discrepancy is the Boltzmann factor coming from the density operator. In my case, it's time-dependent while in the wiki page it's not, which is the confusing part because there seems no way to get rid of the time dependence. Any help is very much appreciated.

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  • $\begingroup$ Your first line is already wrong. $\rho=e^{-\beta H}$ only holds for systems at thermal equilibrium, which means that $H$ can not be time-dependent. $\endgroup$
    – Meng Cheng
    Jun 15, 2015 at 16:02
  • $\begingroup$ @MengCheng then how are we supposed to calculate the expectation because in density operator the Hamiltonian should be the full Hamiltonian $\endgroup$
    – M. Zeng
    Jun 16, 2015 at 2:07
  • $\begingroup$ We do not know how to calculate the exact expectation value, that is why we need to do perturbation theory. $\endgroup$
    – Meng Cheng
    Jun 16, 2015 at 2:09
  • $\begingroup$ @MengCheng so you are saying that instead of treating the full hamiltonian using an equilibrium framework, we directly approximate the full hamiltonian with the unperturbed part? I always thought perturbation has already been done through the evolution operator. $\endgroup$
    – M. Zeng
    Jun 16, 2015 at 2:18
  • $\begingroup$ @MengCheng Another thing is that if we directly replace $E_n(t)$ by $E_n(0)$, then we are basically throwing away some linear order terms in the perturbation $V$, which invalidates the perturbative expansion of $U_I$ up to linear order. $\endgroup$
    – M. Zeng
    Jun 16, 2015 at 2:36

1 Answer 1

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Here is a sketch of a derivation of the Kubo formula: Write the full Hamiltonian $H=H_0+H'e^{-\eta t}$ where $H'$ is an external (time-independent) perturbation and $\eta\rightarrow 0$. We would like to evaluate the expectation value of an operator $A(t)$: $\langle A(t)\rangle=\mathrm{Tr}[\rho(t)A(t)]$. So we need to first determine $\rho(t)$ using perturbation theory. First we write $\rho(t)=\rho_0+\rho'(t)$, and go to interaction picture (indicated by the subscript $I$ in the following), e.g. $\rho=e^{-iH_0t}\rho_I e^{iH_0t}$. Then the perturbative correction $\rho'_I(t)$ can be found, to the leading order in $H'$,

$\displaystyle \rho'_I(t)=i\int_{-\infty}^t dt'[\rho_0, H'_I(t')]$

Then we can just plug $\rho_I'(t)$ into $\langle A(t)\rangle=\mathrm{Tr}(\rho(t)A(t))$. Some straightforward algebra gives you Kubo formula.

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  • $\begingroup$ Thanks for the answer. Indeed, once we have $\rho'_I(t)$ as above, we can arrive at the desired form of Kubo formula. However, it seems to me that your $\rho'_I(t)$ is obtained by this: $i\frac{d\rho'_I(t)}{dt}=i\frac{d\rho_I(t)}{dt}=[H'_I,\rho_I(t)]\approx [H'_I,\rho_I(0)]=[H'_I,\rho_0]$, assuming the perturbation is switched on at $t=0$. Is this what you did? if so, the approximation part makes me uncomfortable. $\endgroup$
    – M. Zeng
    Jun 16, 2015 at 5:38
  • $\begingroup$ This is what I meant by "to leading order in $H'$". We are doing perturbation theory, which means we will get answer order by order in the perturbation. $[H',\rho_I]=[H',\rho_0]+[H',\rho_I']$. I should have put a small parameter $\lambda$ in $H'$, but assume that's the case, then $[H',\rho_0]$ is of the order of $\lambda$, and $[H',\rho_I']$ is of the order $\lambda^2$. So the result is correct to the order $\lambda$. $\endgroup$
    – Meng Cheng
    Jun 16, 2015 at 5:43

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