$T \bar{T}$ OPE

Question

In page 157 of Di Francesco (Conformal Field Theory) it is said that the holomorphic and antiholomorphic components of the energy-momentum tensor have the trivial OPE

$T(z) \bar{T}(\bar{w}) \sim 0$.

I don't know if this was explained earlier in the book and I missed it or it is trivial and I don't see it. Do you have a proof?

Answer 1

This OPE is not trivial, it just doesn't have singular terms.

Suppose there is a quasi-primary operator $\mathcal{O}$ with weight $(h,\bar h)$ appearing in the right-hand side. We can compute the coefficient with which it appears by looking at three point function $$ \langle T(z)\bar T(\bar w)\mathcal{O}(x,\bar x)\rangle=\frac{f_{T\bar T\mathcal{O}}}{(z-w)^{2-h}(\bar z-\bar w)^{2-\bar h}(x-w)^{h-2}(\bar x-\bar w)^{\bar h+2}(z-x)^{2+h}(\bar z-\bar x)^{\bar h-2}}. $$ The right-hand side is fixed by global conformal invariance up to the coefficient $f_{T\bar T\mathcal{O}}$. However, the left-hand side only depends on $z$ and not $\bar z$, so we must conclude $\bar h=2$. Similarly because it only depends on $\bar w$ and not $w$ we must conclude $h=2$. This means that no singular terms can appear in the OPE because these must necessarily have $h+\bar h<2$. But then we can define the operator $$ (T\bar T)(z,\bar z) \equiv T(z)\bar T(\bar z). $$ It is a quasi-primary and has dimensions $(h,\bar h)=(2,2)$. It is in fact the only quasi-primary that appears in the OPE. The OPE takes the form simply $$ T(z)\bar T(\bar w) = (T\bar T)(z,\bar w)=\sum_{n=0}^\infty \frac{1}{n!}(z-w)^n\partial^n_w(T\bar T)(w,\bar w). $$

The operator $T\bar T$ can in fact be defined in any 2d QFT, not necessarily conformal, but the argument is more subtle. Presently there is a great deal of research into the theories one gets by adding $T\bar T$ to the Lagrangian. (Note that this is an irrelevant deformation.) Search for "$T\bar T$-deformation".

Added: Per request of the OP, here is a crash course in OPEs in conformal field theories. Since this answer only requires global conformal invariance, I will not discuss implications of Virasoro symmetry. Because of this, the below applies (with small modifications to accommodate general spin) in CFTs in $d\geq 2$. Virasoro symmetry also leads to straightforward modifications.

Below $x_i$ denote space-time points.

Any CFT possesses operator product expansion that is convergent in vacuum state. That is, $$ \mathcal{O}_1(x_1)\mathcal{O}_1(x_2)|0\rangle=\sum_i f_{\mathcal{O}_1\mathcal{O}_2\mathcal{O}_i} C_{12i}(x_1,x_2,x_3,\partial_{x_3})\mathcal{O}(x_3)|0\rangle. $$ Point $x_3$ is in principle arbitrary and often taken to be $x_3=x_2$. Here the differential operator $C_{12i}(x_1,x_2,x_3,\partial_{x_3})$ is completely fixed by conformal symmetry. It depends only on the quantum numbers of opeartors $\mathcal{O}_1,\mathcal{O}_2,\mathcal{O}_i$. The coefficient $f_{\mathcal{O}_1\mathcal{O}_2\mathcal{O}_i}$ is not fixed by conformal symmetry and represents the dynamical information about the theory.

This expansion is exact and covergent. It is often written by omitting the vacuum state $|0\rangle$. This is because it is often used inside Euclidean correlation functions, where one doesn't necessarily have to talk about a particular quantization. In Euclidean correlation functions one interprets vacuum state in radial quantization around point $x_3$. The OPE is applicable in a Euclidean $n$-point correlation function if there exists a sphere around $x_3$ which only contains the operators $\mathcal{O}_1,\mathcal{O}_2$ at $x_1$ and $x_2$ and no other operators.

One can compute the coefficient $f_{\mathcal{O}_1\mathcal{O}_2\mathcal{O}_i}$ by looking at three-point function $\langle\mathcal{O}_1\mathcal{O}_2\mathcal{O}_i\rangle$ and using the OPE inside the three-point function. Since the two-point functions are canonically chosen to be diagonal $\langle\mathcal{O}_i\mathcal{O}_j\rangle\propto \delta_{i,j}$, we have $$ \langle\mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\mathcal{O}_i(x_3)\rangle=f_{\mathcal{O}_1\mathcal{O}_2\mathcal{O}_i} C_{12i}(x_1,x_2,x'_3,\partial_{x'_3})\langle\mathcal{O}_i(x'_3)\mathcal{O}_i(x_3)\rangle. $$ Again, often one uses $x'_3=x_2$. Since $C_{12i}(x_1,x_2,x'_3,\partial_{x'_3})\langle\mathcal{O}(x'_3)\mathcal{O}(x_3)\rangle$ is fixed by conformal symmetry and canonical normalization of two-pt functions, the coefficient $f_{\mathcal{O}_1\mathcal{O}_2\mathcal{O}_i}$ is computed by three-point functions. However, this coefficient appears in the OPE, and the OPE is applicable in all $n$-point correlation functions, so there is no lack of generality the OP seems to be worried about.

Answer 2

I think I found what the answer might be. Let us start with Eq.(5.41) from Di Francesco: \begin{align*} \langle T(z, \bar{z}) X \rangle = \sum_{i=1}^n \left( \frac{1}{z-w_i} \partial_{w_i} \langle X \rangle + \frac{h_i}{(z-w_i)^2} \langle X \rangle \right) + \text{reg.} \, , \end{align*} where $X$ is any set of quasi-primary fields and "reg." is something holomorphic and regular (it would not be written in the OPE). From this equation we can see that the dependence of $T(z, \bar{z})$ in $\bar{z}$ is killed when computing expectation values with other quasi-primary fields, and therefore in these cases we can write \begin{align*} T(z, \bar{z}) = T(z), \end{align*} just like in the classical theory. As far as I understand, while in the classical theory we have $\bar{\partial} T = 0$, so this component of the energy momentum tensor is identically holomorphic, in the quantum theory (since the fields are not on shell and the e.o.m.s are not satisfied) this only holds under the conditions explained above.

A similar argument works for $\bar{T} (\bar{z})$. Now having this we can easily compute the OPE we wanted. Following the general formula for an OPE \begin{align*} T(z, \bar{z}) \bar{T}(w, \bar{w}) \sim \sum_k C^k (z-w, \bar{z}- \bar{w}) \mathcal{O}_k(w, \bar{w}) , \end{align*} we can see that since $T$ and $\bar{T}$ are quasi-primaries, they both behave as holomorphic and antiholomorphic, respectively. So the LHS does not depend on $\bar{z}$ nor $w$, which in turn implies that \begin{align*} C^k (z-w, \bar{z}- \bar{w}) = C^k \end{align*} is just a constant. Since a constant has neither singular nor antiholomorphic terms, the sum $\sum_k C^k (z-w, \bar{z}- \bar{w}) \mathcal{O}_k(w, \bar{w})$ vanishes in OPEs and we have \begin{align*} T(z) \bar{T}(\bar{w}) \sim 0. \end{align*}

Answer 3

In a general 2D CFT, the author defines the OPE of two operators $A$ and $B$ as

$$ A(z,\bar{z})B(w,\bar{w}) = \sum_n C_n(z-w,\bar{z}-\bar{w}) O_n (w,\bar{w}). $$

where $ O_n $ are regular operators. In particular, if $A(z,\bar{z})=T(z)$ and $B(w,\bar{w}) = \bar{T} (\bar{w})$,

$$ T(z) \bar{T} (\bar{w}) = \sum_n C_n(z-w,\bar{z}-\bar{w}) O_n (w,\bar{w}). $$

Given that the l.h.s. does not depend on $\bar{z}$ and $w$, the following limit is well defined

$$ T(z) \bar{T} (\bar{w}) = \lim_{\bar{z} \to \bar{w} \\ w \to z} T(z) \bar{T} (\bar{w}). $$

Therefore, using the OPE definition,

$$ T(z) \bar{T} (\bar{w}) = \sum_n C_n(0,0) O_n (z,\bar{w}), $$

which is regular. In the Di Francesco's notation, you can write

$$ T(z) \bar{T} (\bar{w}) \sim 0. $$