I have read that the equivalent of the dilatation operator in 2D CFT is the sum of the Virasoro generators $L_0+\bar L_0$. The dilatation operator acting on a primary operator should give the scaling dimension of the primary, and I want to verify this in the 2-dimensional case. According to di Francesco, the OPE of the stress energy tensor with a primary operator $\phi(z, \bar z)$ is given by \begin{equation} T(z) \phi(w, \bar w) \sim \frac{h}{(z-w)^2} \phi(w, \bar w)+\frac{1}{(z-w)} \partial \phi(w, \bar w) \end{equation} where $h$ is the conformal weight (and similarly for the antichiral part).

The Virasoro generators are $L_n= \frac{1}{2\pi i}\oint dz z^{n+1} T(z) $, so I wrote \begin{align} [L_0+\bar L_0, \phi(w, \bar w)]&=\frac{1}{2 \pi i} \oint dz z^{0+1} \mathcal R T(z) \phi(w, \bar w)+ \text{anti-chiral} \\ &=\frac{1}{2 \pi i} \oint dz \bigg(\frac{hz}{(z-w)^2} \phi(w, \bar w) +\frac{z}{z-w} \partial \phi(w, \bar w) \bigg) +\text{anti-chiral} \end{align} My understanding is that the contour encircles $w$, so, using the residue theorem, I get \begin{equation} (h+ \bar h)\phi(w, \bar w) +w \partial \phi(w, \bar w)+\bar w \bar \partial \phi(w, \bar w) \end{equation} The first part $\Delta=h+\bar h$ is what I expected, however I don't understand why the other two terms are also there. I'm probably missing something super obvious, but I don't see any mistakes. Any help is appreciated. Thanks!

  • 1
    $\begingroup$ Hint: Taylor expand $z$ around $w$. $\endgroup$
    – Qmechanic
    May 8, 2021 at 19:33
  • $\begingroup$ Where though? The final expression does not contain $z$. Do you mean in order to get the last expression I've written? I think I wasn't very clear with the question, but my problem is not how to get to the last expression (I did that), but the fact that there are extra terms compared to what I expected to get. $\endgroup$
    – Xincify
    May 8, 2021 at 19:49
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    $\begingroup$ Maybe what you expected was simply not correct? $\endgroup$ May 8, 2021 at 19:53
  • $\begingroup$ @Oбжорoв I think you're right, actually. In $d>3$, the dilatation operator needs to act on a primary inserted \textbf{at the origin} to give just the scaling dimension. Acting away from the origin also gives $x^{\mu}\partial_{\mu}$, which is precisely what I have here. Setting $w=\bar w = 0$ gives the desired result. Thanks :) $\endgroup$
    – Xincify
    May 8, 2021 at 20:28
  • $\begingroup$ @Oбжорoв it is correct. See my answer for an explanation. $\endgroup$ May 9, 2021 at 14:54

1 Answer 1


Your expectation is correct, and your calculation is also correct. You are lacking an interpretation of the correct result that you obtained.

Dilatations don't simply rescale the value of the field, they also move the points of the base manifold (space-time, worldsheet, etc.) according to a very specific law. In your case, the dilatation on the worldsheet is generated by the vector field $$ v = z \partial_z + \bar{z} \partial_{\bar{z}}. $$

The infinitesimal effects of rescaling the field (parametrized by its dimension $\Delta = h + \bar{h}$) and moving the worldsheet (which induces an infinitesimal change in the field by acting on it with $v$) add together and give the formula that you obtained.


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