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This is about Polchinski's eq(2.3.11).

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 $z$.

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

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$ Sep 1, 2014 at 17:17
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    $\begingroup$ The equations are true when you put them inside a path integral, with arbitrary insertions (at different points). This means things like time/radial ordering are implicitly included. I suggest you carefully read through the appendix on path integrals, paying particular attention to the section on the relation to the Hilbert space formalism. $\endgroup$ Sep 1, 2014 at 17:28

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