OPE of stress tensor in CFT

Question

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?

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

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