Is the time-ordered exponential

$$ \mathcal{T}\exp\left\{-i\int_{t_0}^tdt'V(t')\right\}\tag{1} $$

just a mnemonic device for the series

$$ \begin{aligned} 1 + (-i)\int_{t_0}^tdt_1 \, V(t_1) +{} & (-i)^2\int_{t_0}^t dt_1 \int_{t_0}^{t_1}dt_2 \, V(t_1)V(t_2) \\&+(-i)^3 \int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \int_{t_0}^{t_2} dt_3 \, V(t_1)V(t_2)V(t_3)+ \cdots \end{aligned}\tag{2} $$

or is there more to it?

The series one gets when expanding the time-ordered exponential is full of time-ordered products and not useful as such. Are there other advantages to the first expression except for compactness?

  • 2
    $\begingroup$ Another representation that I find useful is that, if you call $h_{t_2\leftarrow t_1}$ the evolution from time $t_1$ to $t_1$, then the time ordered exponential is $\lim_{\epsilon\to0}\prod_{k=1}^{1/\epsilon}h_{t -(k-1)\epsilon \delta\leftarrow t -k\epsilon \delta}$, having called $\delta = t-t_0$. I particularly like this representation for Wilson lines because it makes it easier to prove the transformation properties. Might this be useful for you? $\endgroup$
    – MannyC
    Jan 5, 2019 at 16:17

1 Answer 1


The time-ordered exponential (1) is the unique solution $U(t,t_0)$ to the first-order initial value problem: $$ i\frac{d}{dt} U(t,t_0)~=~V(t)U(t,t_0), \qquad t~\geq ~t_0, \qquad U(t_0,t_0)~=~\mathbb{1}. \tag{A} $$ The ODE (A) is sometimes useful. See also this related Phys.SE post.

  • 1
    $\begingroup$ Well, that is exactly why I’m asking. The way to prove that (1) is a solution to (A) is actually to prove that (1) = (2) and then showing by explicit differentiation that (2) is a solution to (A). Seems kinda cumbersome to remember (1) and how to get to (2), I’d rather just remember (2). $\endgroup$
    – Andrea
    Jan 5, 2019 at 14:44

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