I come aross an OPE between stress tensor components in CFT which is \begin{equation} T(z)\bar{T}(\bar{w})\sim -\frac{\pi c}{12}\partial_{z}\partial_{\bar{w}}\delta^{(2)}(z-w)+... \end{equation} I am confused about this OPE. Because in general in CFT chial part does not correlated with anti-chiral part, for example in free fermion and boson. So how to derive this OPE?

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    $\begingroup$ Possible duplicate: Weyl anomaly in 2d CFT (string theory lectures by D.Tong). The $T\bar{T}$ OPE is mentioned in a footnote. $\endgroup$
    – Qmechanic
    Dec 1, 2019 at 10:32
  • $\begingroup$ Here is the OPE of $T\bar{T}$ not OPE of $T_{z\bar{z}}T_{w\bar{w}}$ as in the linked question. So this may not be duplicate. Nevertheless, the linked question is also interest me, thanks! $\endgroup$ Dec 1, 2019 at 14:45
  • $\begingroup$ what is $\partial_{\bar w}\delta(z-w)$? shouldn't this be just $=0$? Also, the l.h.s. is holomorphic in $\bar w$, so how can the r.h.s. be holomorphic in $w$? Perhaps the r.h.s. should be $\partial_{\bar w}\delta(z-\bar w)$ $\endgroup$ Dec 1, 2019 at 16:31
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    $\begingroup$ What is the expression for the stress energy tensor which is used to compute this OPE? $\endgroup$
    – Bruce Lee
    Dec 4, 2019 at 18:50
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    $\begingroup$ where did you see such a OPE? In a lecture note? $\endgroup$
    – Nogueira
    Dec 5, 2019 at 20:01

1 Answer 1


Regarding OPEs it is assumed that $(z,\bar z)\neq (w,\bar w)$, i.e. that the points does not collide. The OPE is an expansion of $(z,\bar z)$ approaching $(w,\bar w)$, i.e. they are as close as you want, but they never collide. With this in mind is simple to see that the RHS of your equation is just zero, it vanishes. This is so because

$$ (z,\bar z)\neq (w,\bar w)\implies \delta^2(z-w;\bar z-\bar w)=0 $$

by the very definition of delta functions. This is why the holomorphic part never mixes with the anti-holomorphic part. The OPE between these two always involve some kind of anti-holomorphic derivative hitting pairs of fields that are holomorphic up to collisions. The collisions will produce these deltas. But since the points never collide, these deltas will all vanish.

It is important to keeping in mind that these operators are insertions inside some path integral. In path integral does not make sense to put $(z,\bar z)$ on top of $(w,\bar w)$. The only thing that might have sense is to do the limit where one point approach the other, and if there is divergence in this limit an appropriate renormalization is required. In your case, $T(z)$ does not have divergences with $\bar T(\bar w)$ when the points approach each other so no renormalization is required so you are free to define

$$ :T\bar T:(z,\bar z) \equiv \lim_{(w,\bar w)\rightarrow (z,\bar z)}T(w)\bar T(z) $$

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    $\begingroup$ FWIW, the $T \bar{T}$ has been studied here: hep-th/0401146 $\endgroup$
    – MannyC
    May 15, 2020 at 0:34
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    $\begingroup$ @MannyC Yes, for cases that are not CFTs this composite operator is more interesting and it mixes with the non-zero trace of the energy-momentum tensor. For CFTs however, they are not suppose to have singular OPEs with each other by the reasons I mentioned above. You can see page 7 of the paper you just mentioned, on the top. $\endgroup$
    – Nogueira
    Jun 6, 2021 at 11:20

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