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


1 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.

  • $\begingroup$ Thanks for clearing this up. It seems very obvious right now. Expanding your last expression according to the definition of time ordered products gives us the desired expression. $\endgroup$
    – noir1993
    Commented Jul 10, 2017 at 18:08
  • $\begingroup$ @Lê Dũng, I still don't get how the $t'$ can be in the differential and also on the limit of the integral. Could you clarify that? $\endgroup$
    – xihiro
    Commented May 18, 2019 at 0:32
  • $\begingroup$ @xihiro For the integral limit, the appearance of T-product means there can be 2 cases $t_1>t_2$ and $t_2>t_1$ . This is why we have 2 integrals for the second-order term with 2 limits as in the question. For your first question, sorry I don't understand your question.Could you clarify it? $\endgroup$
    – Lê Dũng
    Commented Jul 9, 2019 at 14:08
  • $\begingroup$ Can you explain why we replace t with t' , t'' , t"' etc in limit of integral , can any one explain @Lê Dũng $\endgroup$
    – user212422
    Commented Jan 10, 2020 at 12:57

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