Dyson series and T product (II) 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
 A: I think you problem lies in wrong integration bounds. The story goes roughly as follows. We would like to evaluate integrals such as
$$ \int_{t_0}^{t_1} dt' \int_{t_0}^{t'} dt'' H(t'') H(t').$$
The annoying feature of this integral is that we have to keep track of integration bounds that ensure $t''$ is always greater than $t'$. We can instead rewrite this as
$${1 \over 2} \left( \int_{t_0}^{t_1} dt'' \int_{t_0}^{t_1} dt' H(t'') H(t') \Theta (t'' - t') + \int_{t_0}^{t_1} dt'' \int_{t_0}^{t_1} dt' H(t') H(t'') \Theta(t' - t'') \right) = $$
$$ = {1 \over 2}  \int_{t_0}^{t_1} dt'' \int_{t_0}^{t_1} dt' T \left[ H(t'') H(t') \right]$$
because both terms are equal. This form is much friendlier since the integration bounds of both integrals are now the same.
In general one has that
$$\int_{t_0}^{t_f} dt_{n-1}' \cdots \int_{t_0}^{t_0'} dt_0' H(t_{n-1}') \cdots H(t_{0}') =
 {1 \over n!} \int_{t_0}^{t_f} dt_{n-1}' \cdots \int_{t_0}^{t_f} dt_0' T \left[ H(t_{n-1}') \cdots H(t_{0}') \right]
 $$
and finally 
$$
T \left[ \exp \left( \int_{t_0}^{t_f} dt' H(t') \right) \right] = \sum_{n=0}^{\infty} {1 \over n!} \int_{t_0}^{t_f} dt_{n-1}' \cdots \int_{t_0}^{t_f} dt_0' T \left[ H(t_{n-1}') \cdots H(t_{0}') \right]$$
which is (upto constants) the expansion of an evolution operator in quantum mechanics that can be derived from the Dyson's equation.
