# OPE of stress tensor in CFT

I come aross an OPE between stress tensor components in CFT which is $$$$T(z)\bar{T}(\bar{w})\sim -\frac{\pi c}{12}\partial_{z}\partial_{\bar{w}}\delta^{(2)}(z-w)+...$$$$ 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?

• Possible duplicate: Weyl anomaly in 2d CFT (string theory lectures by D.Tong). The $T\bar{T}$ OPE is mentioned in a footnote. Dec 1, 2019 at 10:32
• 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! Dec 1, 2019 at 14:45
• 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 $\partial_{\bar w}\delta(z-\bar w)$ Dec 1, 2019 at 16:31
• What is the expression for the stress energy tensor which is used to compute this OPE? Dec 4, 2019 at 18:50
• where did you see such a OPE? In a lecture note? Dec 5, 2019 at 20:01

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
$$(z,\bar z)\neq (w,\bar w)\implies \delta^2(z-w;\bar z-\bar w)=0$$
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
$$:T\bar T:(z,\bar z) \equiv \lim_{(w,\bar w)\rightarrow (z,\bar z)}T(w)\bar T(z)$$
• FWIW, the $T \bar{T}$ has been studied here: hep-th/0401146 May 15, 2020 at 0:34