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?

  • $\begingroup$ I'm not sure I understand what you mean by Tong "deriving the Virasoro algebra" from the OPE. The Virasoro algebra is just the (centrally extended) algebra of 2d conformal symmetry, it does not need to be "derived", it is by definition the symmetry algebra of a CFT. What do you mean by that? $\endgroup$ – ACuriousMind Jul 1 '17 at 23:39
  • $\begingroup$ By expanding the stress tensor in a Laurent series with the Virasoro generators as the coefficients, one may use the relationship between commutators and contour integrals of radially-ordered products to deduce the commutators of the Virasoro generators from the OPE. Are you suggesting that alternatively, one may use the Virasoro algebra to define the CFT and run it in reverse to obtain the OPE? $\endgroup$ – Spencer Tamagni Jul 1 '17 at 23:56

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.

  • 1
    $\begingroup$ You are confused: the generators of the conformal algebra are obtained using Noether theorem. This is the relation between the Virasoro generators and the stress-tensor. It is not a coincidence, and they are not set equal because they "look alike." They are set equal because they are equals. You can, if you wish, define the conformal transformations by just looking at $\delta x^\mu$, but that is not the end of the story. The relation between $L_m$ and the conserved charges related to $T$ is a deep one, not a coincidence. You are missing the role of the Ward identities in a quantum theory $\endgroup$ – CGH Jul 4 '17 at 13:31
  • $\begingroup$ @CGH No, the generators of the conformal algebra are obtained by looking at infinitesimal conformal transformations/conformal Killing vectors. Saying that a theory is a CFT means saying that algebra is a symmetry of the theory, and therefore the state of space carries a representation of it. I'm not claiming it's a "coincidence" that that rep are precisely the modes of the energy-momentum tensor, but that you can know about the algebra without the e-m tensor, making the usage of the action of that algebra to derive something about the e-m tensor non-circular, which was OP's concern. $\endgroup$ – ACuriousMind Jul 4 '17 at 13:38
  • $\begingroup$ No. You are skipping the whole construction using conserved charges. That is classical mechanics. $\endgroup$ – CGH Jul 4 '17 at 15:38

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!

  • $\begingroup$ Right, but to prove that $h$ is nonnegative, you need to use the fact that the modes of $T$ satisfy the Virasoro algebra. So what the argument looks like to me is "If the modes satisfy the Virasoro algebra, then there are no negative weight operators in a unitary theory. If there are no negative weight operators, then the $TT$ OPE takes the appropriate form. If the OPE is of this form, then the modes satisfy the Virasoro algebra". What exactly is being argued? $\endgroup$ – Spencer Tamagni Jul 4 '17 at 17:40
  • $\begingroup$ I understand that as a conformally invariant quantum theory, to be as general as possible we need to centrally extend the algebra. What I'm not getting is if what we're doing here is a proof or a definition, essentially. Are we postulating that the Noether charges of $T$ generate the symmetry, just as in ordinary quantum theory, and then showing that in order for this to be true, the $TT$ OPE must look a certain way? To postulate the form of the OPE first seems a bit out of place, as the rest of the calculation seems like a long consistency check on the Noether charges generating the symmetry. $\endgroup$ – Spencer Tamagni Jul 4 '17 at 17:46
  • $\begingroup$ 1.- There is no "if the modes satisfy the Virasoro algebra." $T$ is the Noether conserved current. The currents define generators, the $L_m$. The generators form an algebra, the Virasoro one. 2.- The Ward Identity is valid for any symmetry in a QFT. As I showed the $TT$ OPE is the most general allowed for a QFT with conformal symmetry. 3.- There is no use of unitarity here!! 4.- We do not postulate that the charges of $T$ generate the conformal symmetry. This is how you construct the algebra! $\endgroup$ – CGH Jul 4 '17 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.