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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?

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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.

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  • $\begingroup$ You are using the general form of 3-point functions in CFTs. That is a particular case in which $T \bar{T}$ appears within a correlation function with only one more field. You could use instead, with the same lack of generality, the general form of 2-point functions in CFTs to compute $\langle T \bar{T} \rangle$. I feel there must be a general proof which doesn't make use of the CFT form of $n$-point functions. $\endgroup$ – MBolin Apr 30 at 22:05
  • $\begingroup$ @MBolin, I am sorry, but your comment makes no sense to me. In CFT studying the OPE $\mathcal{O}_1\times \mathcal{O}_2$ is equivalent to studying $\langle\mathcal{O}_1\mathcal{O}_2\mathcal{O}\rangle$ for all possible $\mathcal{O}$, which is what I have done. There is no lack of generality. $\endgroup$ – Peter Kravchuk May 1 at 17:06
  • $\begingroup$ As far as I know, an OPE $ \mathcal{O}_1 \mathcal{O}_2$ is equivalent to $ \langle \mathcal{O}_1 \mathcal{O}_2 \mathcal{O} \rangle $ where $\mathcal{O}$ is any set of operators, not just one operator. $\endgroup$ – MBolin May 2 at 8:22
  • $\begingroup$ @MBolin, if a primary $\mathcal{O}$ appears in $\mathcal{O}_1\times \mathcal{O}_2$ OPE, then the coefficient with which it appears is given by $\langle\mathcal{O}_1\mathcal{O}_2\mathcal{O}\rangle$. $\endgroup$ – Peter Kravchuk May 2 at 18:31
  • $\begingroup$ Are you saying that $\mathcal{O}_1 \mathcal{O}_2 \sim \quad ... + \langle \mathcal{O}_1 \mathcal{O}_2 \mathcal{O} \rangle \mathcal{O} + ... \quad $ ? If you substitute this in the 3-point function you find $\langle \mathcal{O}_1 \mathcal{O}_2 \mathcal{O} \rangle = ... + \langle \mathcal{O}_1 \mathcal{O}_2 \mathcal{O} \rangle \langle \mathcal{O} \mathcal{O} \rangle+ ... \quad $. I'm not sure, but that doesn't seem correct to me. Is it correct? $\endgroup$ – MBolin May 5 at 17:20
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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*}

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  • $\begingroup$ Your argument that $C^k$ is a constant is wrong. The formula in my answer gives a counter-example. If you differentiate it with respect to $w$ or $\bar z$ you will find $0$ both on the left and on the right, but $C^k$ are not constants. Also, a constant is holomorphic and anti-holomorphic. And the sum that you write does not vanish, contrary to what you say. $\endgroup$ – Peter Kravchuk May 2 at 18:35
  • $\begingroup$ Also, $\partial_{\bar z} T(z,\bar z)=0$ as an operator equation. This means that it is zero in Wightman functions or in time-ordered correlators at separeted points. If you include coincident points, then it is equal to contact terms in time-ordered correlators. It vanishes in Wightman functions even at coincident points. $\endgroup$ – Peter Kravchuk May 2 at 18:37
  • $\begingroup$ Actually I don't understand what you do in your answer when you define $(T\bar{T} ) (z, \bar{z} ) = T(z) \bar{T} (\bar{z})$. As far as I understand, it is not that easy to define the composite operator. See for instance this paper by Zamolodchikov: arxiv.org/abs/hep-th/0401146 . $\endgroup$ – MBolin May 5 at 17:29
  • $\begingroup$ The paper you link to defines $T\bar T$ in a general 2d QFT, not in CFT. In this more general case the construction is indeed more subtle, as I mention in the end of my answer. $\endgroup$ – Peter Kravchuk May 5 at 17:41
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    $\begingroup$ You say that since there is no dependence on $w$ or $\bar z$ in the left hand side, then the coefficient functions $C$ do not depend them either. This conclusion is wrong, it would only be true if you knew that $\mathcal{O}_k(w,\bar w)$ did not depend on $w$. This is not so. In fact, my answer contains the correct OPE and even if you do not believe this, it still provides a counter example to your conclusion, because you can check that derivative of the rhs with respect to $w$ vanishes. $\endgroup$ – Peter Kravchuk May 5 at 20:48

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