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!
