I found this statement in several paper, but I don't have a clear reasoning why. It's said that the correlation function of the energy-momentum tensor $T(z)$ like $<T(z)O_1O_2O_3>$ vanishes like $z^{-4}$ when $z$ goes to infinity. Could anyone explain why?

  • $\begingroup$ $\uparrow$ Which papers? $\endgroup$ – Qmechanic Aug 1 '16 at 14:01
  • $\begingroup$ @Qmechanic Like arxiv.org/abs/1108.4417, above equation D.18 $\endgroup$ – Nahc Aug 1 '16 at 14:03

In fact, it is true if $T$ is replaced by any other quasi-primary operator with the scaling being $|z|^{-2\Delta}$, $\Delta=h+\bar h$. Also, the same holds in higher dimensional CFT's. For $T$ in 2d you have $\Delta_T=2$. There are various ways to see this.

One way is that actually the Euclidean correlation functions can be defined on a sphere, which (with north pole removed) is conformal to the plane via the stereographic projection. This projection relates correlators on the sphere $\langle T(z)\ldots\rangle_{S^2}$ and on the plane $\langle T(z)\ldots\rangle_{R^2}$. Taking $z$ to infinity is equivalent to sending $T$ to the north pole of the sphere. On the sphere this is point is no special, and the correlator with $T$ at the north pole is regular and non-zero for a generic configuration of the remaining operators, $$ \langle T(\infty)\ldots\rangle_{S^2} \neq 0,\infty $$ When you write down the relation induced by the stereographic projection, you find something like $$ \left|\langle T(z)\ldots\rangle_{R^2}\right|\simeq |z|^{-2\Delta_T}\left|\langle T(z)\ldots\rangle_{S^2}\right|, $$ which explains the $z^{-4}$ damping. This formula follows from the standard formula for change of the correlator under a Weyl transformation (the correlators are assumed to be normalized as $\langle 1 \rangle_{S^2}=\langle 1 \rangle_{R^2}=1$, otherwize you will have to care about Weyl anomaly).

Another way is through the OPE. You move $T$ to infinity, while the remaining operators are somewhere at fixed positions. At some point you can draw a circle around all the other operators such that it will not contain $T$. It means that you can now use the OPE. In your example, you write $$ O_1(z_1,\bar z_1)O_2(z_2,\bar z_2)O_3(z_3,\bar z_3)=\sum_i (C_i(z_1,\bar z_1,z_2,\bar z_2,z_3,\bar z_3)O_i(z_1,\bar z_1)+\text{desc.}), $$ where the sum is over all quasi-primaries in the theory and "desc." denote the contribution from the $sl_2(\mathbb{C})$ descendants of $O_i$ (i.e. descendants involving $L_{-1}$ and $\bar L_{-1}$ only). Assume the basis of quasi-primaries is chosen to be diagonal, i.e. $\langle O_iO_j\rangle\propto \delta_{ij}$. Then taking the expectation value with $T$ we find $$ \langle T(z)O_1(z_1,\bar z_1)O_2(z_2,\bar z_2)O_3(z_3,\bar z_3)\rangle= \langle T(z) (C_T(z_1,\bar z_1,z_2,\bar z_2,z_3,\bar z_3)T(z_1)+\text{desc.})\rangle. $$ If the four-point function is non-zero at all, the so is $C_T$. Now, the two-point function $\left|\langle T(z) T(z_1) \rangle\right|\propto |z-z_1|^{-2\Delta_T}$. The descendant contributions fall off quicker than that since they are all proportional to the derivatives of $\langle T(z) T(z_1)\rangle$ over $z_1$.


The stress energy tensor is a quasi-primary field of dimension 2 (when the central charge vanishes). This means that in a operator product expansion $$ T(z)\phi(w) = \frac{h}{(z-w)^2}\phi(w) + \frac{1}{z-w}\partial\phi(w) + \textrm{regular terms}\ldots $$ with $\phi(w)$ being a conformal field of scaling dimension $h$. This is due to contour integration and residue theorem applied to the left hand side of the above, when calculating all possible variations (that are subject to conformal invariance).

Taking the expectation values of both sides should give back the results. For product of more than two operators just apply the rule more than once, with $\phi(w)$ being the new $T(z=z_0)\Psi(w)$. Another answer along the same lines can be found here.

  • $\begingroup$ This argument does not work since you are giving the short-distance expansion which converges only when $T$ is closer to $\phi$ than $\phi$ is to any other operator (the circle around $\phi$ and through $T$ should not contain any other ops). In particular the "regular terms" are more significant as $z\to\infty$ than the terms you have written down. Moreover, the leading term when $z\to\infty$ actually contains information about the full four-point function, and not just the three-point function as would follow from your argument. $\endgroup$ – Peter Kravchuk Aug 10 '16 at 7:21
  • $\begingroup$ And finally -- it does not expain the $z^{-4}$ behavior. $\endgroup$ – Peter Kravchuk Aug 10 '16 at 7:22
  • $\begingroup$ @PeterKravchuk "This argument does not work since you are giving the short-distance expansion which converges only when T is closer to ϕ than ϕ is to any other operator" I don't understand what that means. I'm taking the product of two operators and integrate them, and that is the result. "In particular the "regular terms" are more significant..." false, because since they are analytical the contour integral is zero due to Cauchy theorem. $\endgroup$ – gented Aug 10 '16 at 7:40
  • $\begingroup$ "not just the three-point function as would follow from your argument...." the above argument can be applied to any product of any number of operators just re-iterating the procedure: you put four, you obtain the four point function (after taking the expectation value). $\endgroup$ – gented Aug 10 '16 at 7:42
  • $\begingroup$ You are trying to use an OPE. OPE is convergent in CFT, but only if certain conditions are met. You are writing an OPE around $w$, it is convergent if and only if there are no other operators inserted closer to $w$ than $z$. If there are at least three operators in the correlator (in the OP case there are four), then this is necessarily false, since $z$ is taken to infinity. The contour integrals are used to derive the leading terms in the OPE you write, but this OPE itself is irrelevant. $\endgroup$ – Peter Kravchuk Aug 10 '16 at 7:52

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