OPE in the Free Boson Theory I am following the book by "Conformal Field Theory" by Di Francesco and authors. Consider a free boson theory with the action $$S = \frac{1}{2}g \int d^2x \partial_\mu \phi \partial^\mu \phi$$ Using the path integral approach, one can calculate the correlator $$ \left < \partial_z \phi(z) \partial_w \phi(w) \right > = -\frac{1}{4\pi g}\frac{1}{(z-w)^2}$$
Using the above expressions, the authors deduce the OPE as $$\partial \phi(z) \partial \phi(w) =  -\frac{1}{4\pi g}\frac{1}{(z-w)^2} + \text{regular terms}$$
My question: In the above OPE, why can't we have a term like $$\frac{\mathcal{O}(w)}{(z-w)^n}$$ where $\mathcal{O}(w)$ is some operator satisfying $\left < \mathcal{O}(w) \right >=0$, and $n$ is any positive integer? It would still satisfy that the correlation function is $\left <\partial_z \phi(z) \partial_w \phi(w) \right > = -\frac{1}{4\pi g}\frac{1}{(z-w)^2}$.
 A: If they gave you the $-\frac{1}{4\pi g} \frac{1}{(z - w)^2}$ two-point function out of the blue and didn't tell you what theory it was from then, indeed, there would be no way to determine the OPE. But for the free boson theory, there are many ways to see that this is the only singular term.
First, consider that the operator in $\frac{\mathcal{O}_n(w)}{(z - w)^n}$ needs to have dimension $2 - n$. This violates the unitarity bound for $n > 2$ so we only need to rule out a $\frac{\mathcal{O}_1(w)}{z - w}$ term. But then we can use the fact that OPEs commute for bosons. And yet, a swap of $\partial \phi(z) \leftrightarrow \partial \phi(w)$ amounts to a swap of $z \leftrightarrow w$ which makes the term in question change sign.
Alternatively, we can use the fact that we know all the operators in this CFT. So for $\frac{\mathcal{O}_1(w)}{z - w}$ to appear, there must be some dimension 1 combination of fundamental fields such that the three-point function $\left < \partial \phi(z) \partial \phi(w) \mathcal{O}_1(x) \right >$ is non-zero. But Wick contractions make it easy to prove that exponentials like $V_\alpha(x) = e^{i \alpha \phi(x)}$ do not satisfy this condition because no copies of $\partial \phi(x)$ are produced in the OPE $\partial \phi(w) V_\alpha(x)$. The only other dimension 1 operator is $\partial \phi(x)$ itself and $\left < \partial \phi(z) \partial \phi(w) \partial \phi(x) \right > = 0$ by $\mathbb{Z}_2$ symmetry of the action.
Finally, one could use the fact that the fields are derivatively coupled and therefore have a shift symmetry whose conserved current must have dimension 1. This is precisely $\partial \phi(z)$. We then get to use the OPE of affine $U(1)$ currents which is well known. This approach will probably be addressed in a later chapter.
A: Let us generalize OP's question as follows:

Why can there not be arbitrary high pole orders $\frac{1}{|z-w|^n}$ on the RHS of an OPE
$$ {\cal R}\phi_i(z)\phi_j(w)=\ldots\tag{A}$$
in a CFT?

Answer:

*

*It is enough to consider two quasi-primary operators $\phi_i(z)$ and $\phi_j(w)$, as we can get to descendant operators via differentiation.


*The dimension $\Delta_k\geq 0$ of a quasi-primary operator $\phi_k$ must be non-negative, due to unitarity bounds and cluster decomposition.


*Now can there be any descendant operator [of some quasi-primary operator $\phi_k(w)$] on the RHS of eq. (A)? Yes, but only if $\phi_k(w)$ also appears on the RHS, cf. Refs. 1 & 2. The pole order of the descendant operator is correspondingly reduced [as compared to the pole order $\Delta_i+\Delta_j-\Delta_k$ of $\phi_k(w)$].


*So the highest possible pole-order in the OPE (A) is $\Delta_i+\Delta_j$, cf. pt. 2.
For more details, see Refs. 1 & 2.
References:

*

*D. Simmons-Duffin, TASI Lectures on the Conformal Bootstrap, arXiv:1602.07982; Sections 7.3 + 8.1.


*S. Rychkov, EPFL Lectures on Conformal Field Theory in D>= 3 Dimensions, arXiv:1601.05000; Sections 3.2 + 3.3.
