Stress Tensor OPE in CFT

Question

In David Tong's CFT notes, he argues that the OPE for the holomorphic stress tensor with itself must take the form $$T(z)T(w)=\frac{c/2}{(z-w)^4}+2\frac{T(w)}{(z-w)^2}+\frac{\partial T(w)}{z-w}+finite$$ I'm a bit confused about this point. He argues that in a unitary CFT, no operators can have negative scaling dimensions, suppressing the higher order poles allowed by symmetry from appearing in the expansion. From this OPE, he derives the Virasoro algebra. Then later, when examining consequences of unitarity, he uses the Virasoro algebra to prove that no operators can have negative scaling dimension. Isn't this argument a little circular then? Is there a way to argue for the general form of the $TT$ OPE that doesn't invoke such circular reasoning?

Answer 1

What Tong is presumably doing is not "deriving the Virasoro algebra", but showing that the modes $T_n$ of the energy-momentum tensor as defined by $T(z) = T_n z^{-n-2}$ actually fulfill the commutation relations of the Virasoro algebra, that is, the non-trivial statement here is that given the OPE of $T(z)$ with itself, one can show its modes are the Virasoro generators (see also this question). Conversely, given $T(z) = L_n z^{-n-2}$ where the $L_n$ are known to be the Virasoro generators, one may derive the OPE of $T$ with itself.

However, the Virasoro algebra is present in a conformal field theory simply because it is a conformal field theory. You don't need to know anything about a stress-energy tensor to know that a quantum conformal field theory will carry a representation of the Virasoro algebra because the Virasoro algebra is the (centrally extended) algebra of infinitesimal conformal transformations. So the reasoning is not circular.

Answer 2

Let's start with the Ward Identities ((4.11) in Tong's lecture). In radial quantization for a 2D theory they read

$\delta_{\epsilon,\bar\epsilon}\Phi(w,\bar w)= \frac{1}{2\pi {\rm i}}\left( \oint_\gamma dz R\left[ T(z)\epsilon(z)\Phi(w,\bar w) \right] - \oint_\gamma d\bar z R\left[ \bar T(\bar z)\bar \epsilon(\bar z)\Phi(w,\bar w) \right] \right)$.

We know how a primary field transform under the a conformal transformation:

$\delta_{\epsilon}\Phi(w)= h (\partial\epsilon) \Phi(w) + \epsilon\partial\Phi(w) $.

(I have taken only the holomorphic part, a similar formula goes for the right moving sector. From now on, I omit the anti-holomorphic part.)

This, using Cauchy, we see that for a general conformal transformation, i.e. $\epsilon=z^n$, the OPE between $T$ and $\Phi$ has to be

$T(z)\Phi(w)= \frac{h}{(z-w)^2}\Phi(w) + \frac{1}{z-w}\partial \Phi(w)+\cdots$,

where it is understood that we are working in radial quantization, so we omit the $R$ in front of the OPE. Now, what happens if $\Phi$ is a quasi-primary? Quasi-primaries transform under the global part of the conformal group, $SO(3,1)$. This is the part that is preserved in $D>2$. The global part of the conformal group is given by $\epsilon=z^n$, $n=0,1,2$. In this case, what we obtain is

$T(z)\Phi(w)= \sum_{p\ge4} \frac{a_p(w)}{(z-w)^p}+\frac{h}{(z-w)^2}\Phi(w) + \frac{1}{z-w}\partial \Phi(w)+\cdots.$

What Tong is telling you now, is that $T$ is actually a quasi primary (actually, it is a descendant of the identity!) When he allows a central charge extension, he is stating that, at the quantum level, you find an anomaly determined by $c\neq0.$ classically, $T$ is a primary, but not quantically. This has nothing to do with unitarity. This is just the transformation of fields. Note that $h$ hasn't been set positive.

Now, just as you defined a symmetry transformation using Noether's currents, you can define symmetry generators for the conformal algebra (Virasoro generators) using radial quantization with $T$ being the conserved current. From there, you can compute the Virasoro algebra and you find its central-extension.

Using representation theory you can see that $h\ge0$ (or from the 2 point function and cluster decomposition) by looking at the first descendant of $\Phi$, $L_{-1}\Phi$. By looking at $L_{-2}\Phi$ you find $c\ge0$.

As you see, this is a linear argument. Not a cyclic one!