4
$\begingroup$

In http://arxiv.org/abs/hep-th/9108028 Equation (2.22), the correlation function of then energy-momentum tensor with some primary fields is

enter image description here

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 $j=1$?

|cite|improve this question
$\endgroup$
4
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Nahc Jun 13 '16 at 23:29
  • 1
    $\begingroup$ @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 $\phi_2$. $\endgroup$ – Peter Kravchuk Jun 14 '16 at 12:39
  • 1
    $\begingroup$ @Phys-Chan, and $sl_2(\mathbb C)$ descendants of $\phi_1$ are basically its derivatives. $\endgroup$ – Peter Kravchuk Jun 14 '16 at 12:41
  • 1
    $\begingroup$ @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. $\endgroup$ – Peter Kravchuk Jun 14 '16 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.