To derive Dyson's Series, we use the fact that for every operator $V$, we have

$$\int_0^tdt_1\int_0^{t_1}dt_2\dots\int_0^{t_n}dt_nV(t_1)V(t_2)\dots V(t_n)= \frac{1}{n!}T\left(\left({\int_0^{t}dt'V(t')}\right)^n\right)$$

where $T$ is the time-ordering operator. This can be found e.g. here on page 29

But how do we interpret the RHS of the equation $$T\left(\left({\int_0^{t}dt'V(t')}\right)^n\right)$$? If we evaluate $$\left({\int_0^{t}dt'V(t')}\right)^n$$ it is only dependent on one time variable $t$. What role would the operator $T$ then play?


1 Answer 1


You act with the time-ordering operator before performing any integrals. For instance \begin{align} T \left( \int_0^t dt' V(t') \right)^2 &= \int_0^t dt' \int_0^t dt'' T ( V(t') V(t'')) \\ &= \int_0^t dt' \int_0^t dt'' \left[ V(t') V(t'') \theta(t'-t'') + V(t'') V(t') \theta(t''-t) \right] \end{align} In the second term, we can interchange $t' \leftrightarrow t''$ and it then equals the first term. It follows that \begin{align} T \left( \int_0^t dt' V(t') \right)^2 &= 2 \int_0^t dt' \int_0^{t'} dt'' V(t') V(t'') . \end{align} I hope this makes it clear.

  • $\begingroup$ I see your point, thanks! But then how is the compact form of the interaction picture with dependent Hamiltonian helpful at all? $\endgroup$
    – J Chen
    Jun 20, 2021 at 19:00
  • $\begingroup$ If you keep on reading, you will see how time-ordering of local operators can be related to normal ordering via Wick contraction. Further, Wick contractions simplify to the Feynman propagator so this gives a simple way of doing a perturbative expansion in QFT. Also, inside time-ordering, all operators commute so that's a bonus. Finally, the path integral in QFT directly gives us a time-ordered product as well so in fact, the time-ordered product is the EASIEST thing to calculate in a quantum theory. $\endgroup$
    – Prahar
    Jun 20, 2021 at 19:03

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