# Ward Identity in CFT

It says that $$Res_{z\rightarrow z_0}j(z)\mathcal A(z_0,\bar z_0)+\bar{Res}_{z\rightarrow z_0}\tilde j(\bar z)\mathcal A(z_0,\bar z_0)=\frac1{i\epsilon}\delta \mathcal A(z_0,\bar z_0)$$ from (2.3.10), $$\oint_{\partial R}(jdz-\tilde jd\bar z)\mathcal A(\sigma_0)=\frac{2\pi}{i\epsilon}\delta\mathcal A(\sigma_0).$$ Res'' is the coefficient of $\frac1{z-z_0}$. But how did we know that $j\sim\frac{1}{z-z_0}$, i.e., it has a pole at $z=z_0$? I know nothing about $j$ except that it is a function of $z$.

• Can you not just expand it as a Laurent series?
– Danu
Sep 1, 2014 at 15:47
• Thanks. What if it's just a regular function? Sep 1, 2014 at 15:51
• ...then you can still expand it as a Laurent series.
– Danu
Sep 1, 2014 at 16:03

Note that it's the residue of the combination $j(z)\mathcal A(z_0,\bar{z_0})$ that we need: the poles come from the OPE where the operators come together.
Now it could be that $j(z)\mathcal A(z_0,\bar{z_0})$ is more singular than $\frac{1}{z-z_0}$ at $z=z_0$. Only the simple poles contribute by the residue theorem. And if there is no simple pole (maybe $\mathcal A$ is $1$, and/or the combination is regular), then the answer will be zero.
• Thank you! I'm a little confused by why $j(z)$ is an operator. Why is it? Also, why did we insert $\mathcal A$ at the first place? Sep 1, 2014 at 17:17