# Correlator of energy-momentum tensor and OPE

In http://arxiv.org/abs/hep-th/9108028 Equation (2.22), the correlation function of then energy-momentum tensor with some primary fields is We can view this as sum over the OPE of the energy-momentum tensor with each of the primary fields. I don't quite understand why we need to sum over the OPE of $T(z)$ with all the primaries. Usually when we say that we can use the OPE to reduce a n-point function to $n-1$ point functions, I think we just need to use the OPE of T(z) with $\phi_1$ in the above equations. What I am asking is why the LHS in the above equation is equal to sum of terms for $j$ from 1 to n , instead of just the term $j=1$?

Well while it has similarities with the OPE, it is more than that. In fact, it satisfies the OPE limit when $z\to w_j$ for any $j$, since the OPE you are talking about tells you only the singular terms, while there are also infinitely many non-singular terms, i.e., schematically $$T(z)\phi(w,\bar w)=\frac{h_\phi}{(z-w)^2}\phi(w,\bar w)+\frac{1}{z-w}\partial_w \phi(w,\bar w)+\sum_O(z-w)^{h_O-h_\phi-2}O(w,\bar w),$$ where $O$ runs over the Virasoro descendants of $\phi$. When you consider equation (2.22) as an OPE with $\phi_1$, what the "unwanted" $j\neq 1$ terms tell you is the summed up contribution of these non-singular terms.
• Thanks! If I consider the three point function $T\phi_1\phi_2$, where $\phi_1$ and $\phi_2$ have the same dimensions. For terms in your $O$, $\partial_3\phi_1$ and higher order derivative will not contribute, but why? I think I'm asking if I just take the OPE of $T$ and $\phi_1$, how do I know which term in $O$ will contribute, which term will not? – Nahc Jun 13 '16 at 23:29
• @Phys-Chan, an OPE always contains an infinite number of terms (in contrast to a fusion rule which can contain finite number of terms, because it only counts the primary fields). However, when you take the OPE in three-point function $\langle T\phi_1 \phi_2\rangle$, you get sum of terms of the form $\langle O \phi_2\rangle$, of which the only $O$ which can possibly contribute are the $sl_2(\mathbb C)$ descendants of $\phi_1$, since $\phi_1$ is the only quasi-primary in the OPE which has the same dimension as $\phi_2$. – Peter Kravchuk Jun 14 '16 at 12:39
• @Phys-Chan, and $sl_2(\mathbb C)$ descendants of $\phi_1$ are basically its derivatives. – Peter Kravchuk Jun 14 '16 at 12:41
• @Phys-Chan, if we diagonalize the basis of primaries so that the two points functions are $\langle \phi_i \phi_j \rangle \propto \delta_{ij}$, then comparing the known value of $\langle T\phi\phi\rangle$ three point function with the result from the OPE is actually a standard way of computing the contribution of $sl_2\mathbb C$ descendants of $\phi$ to the $T\phi$ OPE. – Peter Kravchuk Jun 14 '16 at 12:45