the algebra (\[elalg\]) we observe that we need to evaluate correlators between ${\cal D}^{(q)}, {\cal D}^{(q-n)},~n=0,1,2, \dots$. We shall evaluate correlators $\langle \dots \rangle $ with respect to the free world-sheet action, since we work to leading order in the (weak) coupling $v_i$. For convenience below we shall restrict ourselves only to the time-dependent part of the operators ${\cal D}$. The incorporation of the $\partial_n X^i$ is trivial, and will be implied in what follows. With these in mind one has: $$\begin{aligned}
\label{qqn}  
\langle {\cal D}^{(q)}(z) {\cal D}^{(q-n)}(w) \rangle =
N_q N_{q-n} \int \int_{-\infty}^{+\infty}  \frac{d \omega d\omega '}{
(\omega - i\epsilon)^{q+1}(\omega' - i\epsilon)^{q-n+1}}
\langle  e^{-i\omega t(z)}~e^{-i\omega' t(w)} \rangle\end{aligned}$$ where $\epsilon \to 0^+$. As already mentioned, we work to leading order in time $t \gg \infty$, and hence we can we apply the formula (\[correl\]) for two-point correlators of the $X^\mu$ fields to write [^1] $$\begin{aligned}
&~& \langle e^{-i\omega t(z)}~e^{-i\omega' t(w)}\rangle  
=e^{-\frac{\omega^2}{2}\langle t(z) t(z)\rangle -
\frac{\omega^{'2}}{2}\langle t(w) t(w)\rangle - 
\omega \omega'\langle t(z) t(w)\rangle} = \nonumber \\
&~& = e^{-(\omega+\omega')^2{\rm ln}(L/a)^2 + 2\omega\omega'{\rm ln}(|z-w|/a)^2}~,\end{aligned}$$ where we took into account that ${\rm Lim}_{z\to w}~\langle t(z)t(w)\rangle
= -2{\rm ln}(a/L)^2$. Given that ${\rm ln}(L/a)$ is very large, one can approximate $e^{-(\omega+\omega')^2{\rm ln}(L/a)^2} \simeq 
\frac{\sqrt{\pi}}{\sqrt{{\rm ln}(L/a)^2}}\delta(\omega+\omega')$. 