Let us define our path-ordered operator $\overrightarrow{U}\left(t_1,t_2\right)$:

$$ \overrightarrow{U}\left(t_1,t_2\right)=\overrightarrow{\mathcal{P}}\exp\int_{t_1}^{t_2}dt\,\mathcal{O}\left(t\right). \tag{2.9}$$

This ordered exponential is a solution of

$$ \overrightarrow{U}\left(t_1,t_2\right)=\mathbb{1}+\int_{t_1}^{t_2}dt\,\overrightarrow{U}\left(t_1,t\right)\mathcal{O}\left(t\right). \tag{B.1}$$

The expectation value of the trace this operator is

$$ \left\langle\mathrm{tr}\overrightarrow{U}\left(t_1,t_2\right)\right\rangle~\stackrel{(B.1)}{=}~N+\int_{t_1}^{t_2}dt\,\left\langle\mathrm{tr}\left(\overrightarrow{U}\left(t_1,t\right)\mathcal{O}\left(t\right)\right)\right\rangle. \tag{*}$$

By Wick's theorem you should get

$$ \left\langle\mathrm{tr}\overrightarrow{U}\left(t_1,t_2\right)\right\rangle=N+N^{-2}\int_{t_1}^{t_2}dt\,\int_{t_1}^{t}d{t}'\left\langle\mathrm{tr}\overrightarrow{U}\left(t_1,{t}'\right)\mathrm{tr}\overrightarrow{U}\left({t}',t\right)\right\rangle\left\langle\mathrm{tr}\mathcal{O}\left(t\right)\mathcal{O}\left({t}'\right)\right\rangle. \tag{**}$$

I am having troubles to understand this apparently easy-to-get calculation. How does $\mathrm{Tr}\overrightarrow{U}$ appear again inside the correlator?


  1. D. Correa, P. Pisani, A.R. Fukelman & K. Zarembo, arXiv:1811.03552; Appendix B.
  • $\begingroup$ Possible hint : Novikov theorem $\endgroup$
    – Sunyam
    Dec 13, 2018 at 21:07

1 Answer 1


Sketched derivation of OP's eq. (**):

  1. OP's eq. (**) is basically eq. (B.4) in Ref. 1 with the opposite ordering$^1$.

  2. In OP's eq. (*) we consider the single-contractions in Wick's theorem, see eq. (B.3). The downstairs ${\cal O}(t)$ is contracted with all possible appearances of ${\cal O}(t^{\prime})$ upstairs inside the exponential $\stackrel{\rightarrow}{U}(t_1,t)$. In more detail, the downstairs ${\cal O}(t)$ is contracted with all possible $$\stackrel{\rightarrow}{U}(t^{\prime},t^{\prime}+\Delta t^{\prime})~=~ \exp\left({\cal O}(t^{\prime})\Delta t^{\prime}\right)~\approx~\mathbb{1}+{\cal O}(t^{\prime})\Delta t^{\prime} \tag{i}$$ upstairs in the exponential $$\stackrel{\rightarrow}{U}(t_1,t)~=~\stackrel{\rightarrow}{U}(t_1,t^{\prime})\stackrel{\rightarrow}{U}(t^{\prime},t^{\prime}+\Delta t^{\prime})\stackrel{\rightarrow}{U}(t^{\prime}+\Delta t^{\prime},t).\tag{ii}$$ [Here $\Delta t^{\prime}$ is assumed small.] In other words, this leads to a sum of the form $$ \sum_{t^{\prime}=t_1}^t\stackrel{\rightarrow}{U}(t_1,t^{\prime}){\cal O}(t^{\prime})\Delta t^{\prime}\stackrel{\rightarrow}{U}(t^{\prime}+\Delta t^{\prime},t)\tag{iii}$$ of possible ways to break the exponential $\stackrel{\rightarrow}{U}(t_1,t)$ into two exponentials, cf. OP's last question (v2). [It is implicitly understood that the operator ${\cal O}(t^{\prime})$ in the middle of eq. (iii) is contracted with ${\cal O}(t)$.]

  3. Next replace the sum $\sum_{t^{\prime}=t_1}^t\Delta t^{\prime}$ with an integral $\int_{t_1}^t\! dt^{\prime}$.

  4. Eq. (2.4) yields that the single-contraction/propagator is of the from $$\langle {\cal O}^i{}_j(t){\cal O}^k{}_{\ell}(t^{\prime})\rangle~=~N^{-2}\delta^i_{\ell}\delta^k_j\langle {\rm tr}{\cal O}(t){\cal O}(t^{\prime})\rangle.\tag{iv}$$ Here $i,j,k,\ell$ are color indices. The two Kronecker-deltas lead to the formation of two color traces of the two exponentials.

  5. Putting it altogether leads to the last term in OP's eq. (**). $\Box$


  1. D. Correa, P. Pisani, A.R. Fukelman & K. Zarembo, arXiv:1811.03552; eqs. (B.3)-(B.4).


$^1$ Notation: By comparing eqs. (2.9), (B.1) & (B.2) in Ref. 1, it becomes clear that $t_i\equiv t_1\leq t_2\equiv t_f$ always. Moreover, $\stackrel{\leftarrow}{U}_(t_i,t_f)$ and $\stackrel{\rightarrow}{U}(t_i,t_f)$ are (what are traditionally called) the path-ordered and anti-path-ordered exponential/Wilson-line, respectively. Confusingly Ref. 1 call them oppositely: anti-path-ordered and path-ordered exponential, respectively.


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