After reading the previous posts related to the Dyson series, I have decided to open a new thread because there is something that I am still not understanding. It concerns the expression:
$$ ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{T}[\hat{H}(t^{'})\hat{H}(t^{''})] = $$ $$ ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{H}(t^{′})\hat{H}(t^{''}) + ∫_{t_0}^{t}dt^{''}∫_{t_0}^{t^{′}}dt^{'}\hat{H}(t^{''})\hat{H}(t^{′}) $$
that is assumed in many text books. I wonder if it can be derived from the definition of the Time-ordered operator:
$$ \hat{T}[ \hat{H}(t^{′}) \hat{H}(t^{''})]=θ(t^{′}−t^{''}) \hat{H}(t') \hat{H}(t^{''}) + θ(t^{''}−t^{'}) \hat{H}(t^{''}) \hat{H}(t^{'}) $$ and its natural extension to products of integrals: $$ \hat{T}∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{H}(t^{'})\hat{H}(t^{''}) = ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{T}[\hat{H}(t^{'})\hat{H}(t^{''})] = $$ $$ = θ(t^{'}−t^{''}) ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{H}(t^{′})\hat{H}(t^{''}) + θ(t^{''}−t^{′}) ∫_{t_0}^{t}dt^{''}∫_{t_0}^{t^{′}}dt^{'}\hat{H}(t^{''})\hat{H}(t^{′}) $$
If I am right, the step-function θ$(t)$ must cancel one of the terms leading to: $$ ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{T}[\hat{H}(t^{'})\hat{H}(t^{''})] =∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{H}(t^{′})\hat{H}(t^{''}) \qquad \text{if $t'>t^{''}$} $$ or: $$ ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{T}[\hat{H}(t^{'})\hat{H}(t^{''})] =∫_{t_0}^{t}dt^{''}∫_{t_0}^{t^{′}}dt^{'}\hat{H}(t^{''})\hat{H}(t^{′}) \qquad \text{if $t'< t^{''}$} $$
but in any case it leads to the combination of both: $$ ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{H}(t^{′})\hat{H}(t^{''}) + ∫_{t_0}^{t}dt^{''}∫_{t_0}^{t^{′}}dt^{'}\hat{H}(t^{''})\hat{H}(t^{′}) $$
What I am missing?
Thanks in advance