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It is straightforward to show that $$ \left[T_\leftarrow\exp\left(-i\int_0^tH(\tau)d\tau\right)\right]^{-1} = T_\rightarrow\exp\left(i\int_0^tH(\tau)d\tau\right), $$ where $T_\leftarrow$ and $T_\rightarrow$ impose time-ordering and anti-time-ordering, respectively.

I'm wondering what the inverse of $$ \left[T_\leftarrow\exp\left(-i\int_0^t\int_0^\tau H(\tau)H(\tau')d\tau\,d\tau'\right)\right]^{-1} = ? $$ might be. Just choosing anti-time-ordering doesn't work. Anti-time-ordering of the "first" time index ($\tau$ in this case) also fails.

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  • $\begingroup$ Is this an expression that you have encountered? Where? Which page? $\endgroup$
    – Qmechanic
    Commented Jul 30, 2020 at 11:44
  • $\begingroup$ not exactly this, but essentially equivalent expressions appear in the superoperator description of the Feynman-Vernon influence functional (see Eq (3.507) on page 192 in Breuer and Petruccione). I'm not sure the context helps much, tbh $\endgroup$
    – Daniel
    Commented Jul 30, 2020 at 11:53

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