Derivation of the S-matrix/Dyson's formula from David Tong's lecture notes

Question

I am studying quantum field theory from David Tong's lecture notes and I am stuck at a particular place.

In Page 52., under the heading 3.1.1 Dyson's Formula, Tong introduces an unitary operator $U(t, t_0) = T \exp(-i\int_{t_0}^{t}H_I(t') dt')$

He then introduces the usual definition of time ordered products and goes on to expand $U(t,t_0)$. I am not able to follow how he expanded the time ordered product of operators in the second-order term of the Taylor expansion of the exponential. In particular, I am unable to follow the limits being used and why why both integrals are being put in the front. Should we not get product of two integrals involving $H_I$?

The expansion of $U(t,t_0)$ is given by

$1 - i\int_{t_0}^{t}dt'H_I(t') + \frac{-i^2}{2}[\int_{t_0}^{t}dt'\int_{t'}^{t}dt''H_I(t'')H_I(t')+\int_{t_0}^{t}dt'\int_{t_0}^{t'}dt''H_I(t')H_I(t'')]+... $

Link to Course Page - David Tong: Lectures on Quantum Field Theory

Answer 1

To the second order (for example), the expansion of the time evolution operator is: $$U(t,t_0) = T\exp\left(-i\int_{t_0}^tH_I(t')dt'\right)=1-i\int_{t_0}^tH_I(t_1)dt_1+\frac{(-i)^2}{2}T(\int_{t_0}^tH_I(t')dt')^2$$ You can see that the second order term could be rewritten as: $$\int_{t_0}^tH_I(t_1)dt_1\int_{t_0}^tH_I(t_2)dt_2$$ or $$\int_{t_0}^tH_I(t_2)dt_2\int_{t_0}^tH_I(t_1)dt_1$$ depending on the value of $t_1$ and $t_2$ (Remember that the interacting Hamiltonians at different times do not commute with each other). With the aid of the time-order product, we can safely write: \begin{align} \frac{(-i)^2}{2}T(\int_{t_0}^tH_I(t')dt')^2=\frac{(-i)^2}{2}T\int_{t_0}^tH_I(t_1)dt_1\int_{t_0}^tH_I(t_2)dt_2 \\ =\frac{(-i)^2}{2}T\int_{t_0}^t\int_{t_0}^tH_I(t_1)H_I(t_2)dt_1dt_2 \end{align} The expression in the end of your post can be received from this by a change of variable, which you can see in several texts.