The following expectation of a time ordered exponential is easy to work with: $$ \left\langle n \right| T \left\lbrace e^{-i \int_{t_1}^{t_2} V_A(t’)dt’ }\right\rbrace T \left\lbrace e^{-i \int_{t_0}^{t_1} V_B(t’)dt’ }\right\rbrace \left| n \right\rangle $$ with $t_2>t_1>t_0$ and where $T$ is the time-ordering operator and $V_A, V_B$ two operators acting on the space spanned by $\left| n \right\rangle$. In this case it can easily be transformed into one time ordered exponential $T\left\lbrace e^{-i \int_{t_0}^{t_2} V(t’)dt’ }\right\rbrace $ with $V(t) = V_A(t)$ for $t_1<t<t_2$ and $V(t) = V_B(t)$ elsewhere. Then I can expand it and deal with the operators changing in function of the time variable. For example, the second order term is given by $$ \frac{(-i)^2}{2!}\int_{t_0}^{t_2}\int_{t_0}^{t_2}T\{V(t)V(t')\}dtdt',$$ which can be divided into 4 areas, then time ordered and then integrated. The problem is when the operator arise at the same time this technique cannot be used (I guess). For example, this expectation value
$$ \left\langle n \right| T \left\lbrace e^{i\int_{0}^{t} V_A(t-t’)dt’ }\right\rbrace T \left\lbrace e^{i\int_{0}^{t} V_B(t’)dt’ }\right\rbrace T\left\lbrace e^{-i\int_{0}^{t} V_A(t’-t)dt’ }\right\rbrace T \left\lbrace e^{-i\int_{0}^{t} V_A(-t’)dt’ }\right\rbrace\left| n \right\rangle $$
which arise in this paper https://arxiv.org/abs/cond-mat/0609105 eqn.6 is more complex. The two left time ordered exponential arise after the two on the right. But the two pair are operator in the same time area $[0,t]$ for the left and $[-t,0]$ for the right.
I have tried many way to regroup them to be able to expand them in a way similar to the previous expectation value but I was not able. Many questions touch the same subject especially this one Product of time ordered products (exponentials) where why my approach fail is clearly explained. But I could not find a solution.
