Confusion about Kubo Formula Derived in Interaction Picture

Question

I was reading up on "Many-body quantum theory in condensed matter physics" by Henrik Bruus and Karsten Flensberg. I'm however having some trouble understanding the meaning of the Kubo formula on pages 96-97. (I have attached relevant sections below) In the book, states in the interaction picture are denoted by $|\hat n(t) \rangle$, and Heisenberg/Schrodinger picture states are denoted by $| n \rangle$ and $| n(t) \rangle$(without a hat). Following the derivation of the book, after plugging (6.5) into (6.3a), I got to a result similar to (6.6) but with hats on the states. That is, for the first line of Eqn(6.6), I got $$\langle A(t) \rangle = 1/Z_0 \sum e^{-\beta E_n} \langle \hat n(t_0) |\hat U^+ e^{-iH_0 t} A e^{iH_0 t} \hat U | \hat n(t_0) \rangle$$ to the first order in H', $$\langle A(t) \rangle = 1/Z_0 \sum e^{-\beta E_n} \langle \hat n(t_0) |1-i\int_{t_0}^t[\hat A(t), \hat H(t')]dt' | \hat n(t_0) \rangle$$ instead of (there's no hat on $n$ in (6.6)) $$...\sum e^{-\beta E_n}\langle n(t_0) |... | n(t_0) \rangle$$

Although I think in this case they are the same, since $| \hat n(t_0) \rangle = e^{iH_0 t_0}| n(t_0) \rangle = e^{iE_n t_0}| n(t_0) \rangle $, so that

$$\langle n(t_0) |... | n(t_0) \rangle=\langle \hat n(t_0) |... | \hat n(t_0) \rangle$$

since the exponential factors cancel.

However, what does it mean to go from the first line to the second line of (6.6) My current understanding of the expression $\langle [\hat A(t), \hat H'(t')] \rangle_0$ is that it means the system's average of $[\hat A(t), \hat H'(t')]$ if all the operators (including $H'(t')) $ evolve according to $H_0$ only? Explicitly, that means $$\hat A(t) = e^{-iH_0 t} A e^{-iH_0 t}$$ $$\hat H'(t') = e^{-iH_0 t'} H'(t') e^{-iH_0 t'}$$ and the states $|n\rangle$ don't evolve and stay the eigenstates of $H_0$.

If that's true, then to me $\langle [\hat A(t), \hat H'(t')] \rangle_0$ looks like the Heisenberg picture calculation of the thermal average of $[A(t), H'(t')]$ when the system is not perturbed (doesn't have $H'(t')$ in hamiltonian). So the Kubo formula is telling us to evaluate the average of the commutator for an unperturbed system (and then to integrate it) and this gives us the response of the system to linear order.

If my understanding is correct, wouldn't it be easier for people to understand if we write the first line of (6.6) as

$$1/Z_0\sum e^{-\beta E_n}\langle \hat n(t_0) |1-i\int_{t_0}^t[\hat A(t), \hat H(to)]dt' | \hat n(t_0) \rangle$$

$$=1/Z_0\sum e^{-\beta E_n}\langle \hat n(t) |_01-i\int_{t_0}^t[\hat A(t), \hat H(t')]dt' | \hat n(t) \rangle_0$$

where $| \hat n(t) \rangle_0$ means the interaction picture state $n$ at time $t$ if there's no interaction. Then the whole expression would be in interaction picture, and we can translate it to the Heisenberg picture or the Schrodinger picture as we want. Otherwise, how could I translate the first line of (6.6) to the second line directly?

However, when I checked the wiki(Kubo Formula), the page also gives me the same expressions as (6.6). Now I'm a little bit confused- Am I understanding the Kubo formula correctly here? Or does their choice of writing the equations imply a better (or the correct) way of understanding the Kubo formula?

Below are the relevant sections from the book:

Answer 1

You are correct about $\langle\hat{n}(t_0)|...|\hat{n}(t_0)\rangle=\langle n(t_0)|...|n(t_0)\rangle$ because, as you say, the exponential factors cancel. Explicitly, $$ \langle\hat{n}(t_0)|...|\hat{n}(t_0)\rangle=\langle n|e^{-iE_nt_0}...e^{iE_n t_0}|n(t_0)=\langle n(t_0)|...|n(t_0)\rangle.$$ Now, $t_0$ is the time at which you turn on the perturbation and you can always set $t_0=0$ and should obtain the same results. This could make the algebra easier as $|\hat{n}(t_0)\rangle=|n(t_0)\rangle$.

One reason I think is Ok to use $\langle n(t_0)|...|n(t_0) \rangle$ instead of $\langle \hat{n}(t_0)|...|\hat{n}(t) \rangle$ as you propose, is that if you continue with the algebra, you will realize that the what is inside of the integral can be written as a function of $t-t'$: $$ \langle n(t_0) |[\hat{A}(t),\hat{H'}(t')]| n(t_0)\rangle =$$ $$ e^{iE_n(t-t')}\langle n(t_0) |Ae^{-iH_0(t-t')}H'| n(t_0)\rangle-e^{-iE_n(t-t')}\langle n(t_0) |H'e^{iH_0(t-t')}A| n(t_0)\rangle$$ and so here the factors $e^{iE_nt_0}$ are not really relevant. Then, you can Fourier transform and write it in frequency space (Bruus and Flensber do this in the next section).