Omitting regular terms in OPE

Question

Consider the OPE of two operators $$A(z)B(w)=\sum_{n=-\infty}^N \frac{\{AB\}_{-n}(w)}{(z-w)^n}.$$ We omit $\{AB\}_n(w)$ terms for $n>0$ since in the sense of OPE what we only care is the behavior when $z\to w$.

This however doesn't justify omitting $\{AB\}_0(w)$ (which is the definition of normal order in big yellow book), while everybody puts them in regular term without reason. This term doesn't depend on $z\to w$, and under this definition of normal order it's not obvious to me that its VEV vanishes. It's most likely that $\langle \{AB\}_0(w)\rangle=0$ but why? (For your reference the VEV or contraction of operators is defined using path integral.)

Can somebody help explain why this term is not considered in OPE? More restrictions such as A and B are primary are also welcomed.

It would be actually better if there are reasons to disregard all regular terms together. One (definitely incorrect) possibility to take away all regular terms is that VEV of operator product can be written as some contour integral around $w$, but this makes no sense for otherwise we only need the term with $n=1$.

Answer 1

"We can ignore regular terms in an OPE" is often made as a blanket statement but we need to be more careful. It's better to say that we can do this if we're trying to compute correlation functions of conserved currents (those with $h = 0$ or $\bar{h} = 0$). The most universal place to start is by looking at the stress tensor. Its OPE with a Virasoro primary is \begin{align} T(z) \phi(w, \bar{w}) = \frac{h_\phi \phi(w, \bar{w})}{(z - w)^2} + \frac{\partial \phi(w, \bar{w})}{z - w} + \dots \end{align} where we need to include regular terms. One will survive $z \to w$ and the rest will vanish. But when this is inside a correlation function, we will have \begin{align} \left < T(z) \phi_1(w_1, \bar{w}_1) \dots \phi(w_n, \bar{w}_n) \right > = \sum_i \left [ \frac{h_i}{(z - w_i)^2} + \frac{\partial_i}{z - w_i} \right ] \left < \phi_1(w_1, \bar{w}_1) \dots \phi(w_n, \bar{w}_n) \right > \end{align} and not any regular terms. This is because the variation of a Virasoro primary under a conformal transformation is given by the two expressions \begin{align} \oint_w \frac{dz}{2\pi i} \epsilon(z)T(z) \phi(w, \bar{w}), \quad h_\phi \partial \epsilon(w) \phi(w, \bar{w}) + \epsilon(w) \partial \phi(w, \bar{w}). \end{align} If you tried to add some $(z - w_i)^k \partial_i^{k + 2}$ term to the Ward identity, integrating it against a sufficiently singular $\epsilon(z)$ would produce a variation with more than the two terms above. You can then keep going to compute things like \begin{align} \left < L_{-k} \phi_1(w_1, \bar{w}_1) \dots \phi(w_n, \bar{w}_n) \right > = \oint_{w_1} \frac{dz}{2\pi i} (z - w_1)^{1 - k} \left < T(z) \phi_1(w_1, \bar{w}_1) \dots \phi(w_n, \bar{w}_n) \right >. \end{align} The best way to do this is to deform the contour so that it encircles $w_2, \dots w_n$ and use the Ward identity with only two terms near each singularity.