# Application of Ward identities for OPE under scaling and rotations

I think this is a very straightforward question but I don't see it right now. In Tong's notes on String theory (http://www.damtp.cam.ac.uk/user/tong/string/four.pdf) section 4.2.3 he defines the weight of an operator under $\delta z=\epsilon z$, $\delta \bar{z}=\bar{\epsilon}\bar{z}$ in equation 4.16. Then at the end of the next page he uses the Ward identity 4.12 for that transformation. That is $$\delta O=-\mathrm{Res}[\epsilon zT(z)O(w)]=-\epsilon( hO(w)+w\partial O(w))$$ So $$\epsilon zT(z)O(w)=...+\frac{\epsilon( hO+z\partial O)}{z-w}+...$$ But the I don't see how he gets the term $(z-w)^{-2}$ which gives $$T(z)O(w)=...+\frac{ hO}{(z-w)^{2}}+\frac{\partial O}{z-w}+...$$ The naive thing is to divide by $\epsilon z$ but that doesn't give the result.

• I don't have time to look at the details of the derivation, but the 2nd equation you have written down isn't translation invariant, so they have zero chance of being correct. On the LHS of the second one, you expect $\epsilon (z-w) T(z) O(w)$, for example. If you're in doubt, you can always shift the OPE $T(z)O(w) \rightarrow T(z-w) O(0) \equiv T(z')O(0)$ by redefining coordinates. Mar 6, 2013 at 22:45

For a pole of order $n+1$, $$\mathrm{Res}[f,w]=\frac{1}{n!}\lim_{z\to w}\frac{d^n}{dz^n}(z-w)^{n+1}f(z)$$ So for a function of the type $$f(z)=\frac{g(z)}{(z-w)^{n+1}}$$ $$\mathrm{Res}[f,w]=\frac{1}{n!}g^{(n)}(w)$$ where $g$ is regular. Now the important part is to notice that $O$ depends on $w$ and it is not the result of evaluating $z$ at $w$.
Also because $\epsilon z$ is regular, we expect that $\epsilon z T(z)O(w)$ is of the form $\epsilon zh(z,w)$, where $h(z,w)$ has the singularities.
With this it becomes clear that the term $\epsilon h O(w)$ cannot arise from a simple pole as otherwise we won't get the factor $\epsilon z$. But for a pole of order 2, $$g^{(1)}(w)=\epsilon h O(w) \Rightarrow g(z)=\epsilon z h O(w)$$ On the other hand, the term $\epsilon w\partial O(w)$ can arise from a single pole: $$g(w)=\epsilon w \partial O(w)\Rightarrow g(z)=\epsilon z \partial O(w)$$ Therefore $$\epsilon z T(z)O(w)=...+\frac{\epsilon z h O(w)}{(z-w)^2}+\frac{\epsilon z \partial O(w)}{(z-w)}+...$$ Thus we get the result $$T(z)O(w)=...+\frac{ h O(w)}{(z-w)^2}+\frac{ \partial O(w)}{(z-w)}+...$$