# Expectation value of a path-ordered exponential

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?

References:

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

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. (**). $$\Box$$

References:

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

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$$^1$$ 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.