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In chapter 6 of Di Francesco, they introduce the normal ordering $$ (AB)(w) = \oint_w \frac{ dz }{ 2\pi i (z-w) }A(z) B(w)\ .\tag{6.130}$$

So far so good. But then starting eq (6.139) $$ \oint_w \frac{ dz }{ z-w }A(z)B(w) = \oint_{|z|>|w|} \frac{ dz }{ z-w }A(z)B(w)-\oint_{|z|<|w|} \frac{ dz }{ z-w }B(w)A(z) \tag{6.139} $$ and $$ A(z) = \sum_n (z-x)^{-n - h_A}A_n(x), \quad B(z) = \sum_m (w-x)^{-m - h_B}B_n(x) \ , \tag{6.140} $$ they try to convert the integral into sum of modes, which confuses me a lot.

The expansion coefficients/modes $A_n(x), B_n(x)$ in eq (6.140) clearly depend on the intermediate $x$. Moreover, when treating the two integrals in (6.139), two different intermediate $x$'s are needed. That makes the final conclusion eq (6.144) mysterious, $$ (AB)_m = \sum_{n \le -h_A}A_n B_{m - n }+ \sum_{n > -h_A} B_{m-n}A_n, \tag{6.144} $$ since there the dependence on $x$ is removed some how: how are the $A_n$ and $B_n$ defined concretely?

(By making the $z$ contours to be $|z| = |w| \pm \epsilon$ and take the $\epsilon \to 0$ seems to indicate that the intermediate $x \to w$: but that will render the expansion $B(w) = \sum_m (w-x)^{-m - h_B}B_m(x)$ funny.)

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OP has a point. The text in Ref. 1 above eq. (6.140) (which claims that $|x|$ should be between $|z|$ and $|w|$) is wrong. Here is a hopefully better argument: Let there be given a fixed $w\in\mathbb{C}\backslash\{0\}$. Now choose a fixed $x\in\mathbb{C}$ with $|x|<|w|$. One can for simplicity choose $x=0$, but it is enough if the following is satisfied.

  • On the outer contour $|z| > |w|$ we demand that $|z-x|>|w-x|$ for running $z$. Then we may perturbatively expand the following geometric series $$\frac{z-w}{z-x}~=~1-\frac{w-x}{z-x}\quad\Rightarrow\quad \frac{z-x}{z-w}~=~\left(1-\frac{w-x}{z-x}\right)^{-1}~=~\sum_{\ell\geq 0}\left( \frac{w-x}{z-x}\right)^{\ell}. \tag{6.141}$$

  • On the inner contour $|z| < |w|$ we demand that $|z-x|<|w-x|$ for running $z$. Then we may perturbatively expand the following geometric series $$\frac{z-w}{x-w}~=~1-\frac{z-x}{w-x}\quad\Rightarrow\quad \frac{x-w}{z-w}~=~\left(1-\frac{z-x}{w-x}\right)^{-1}~=~\sum_{\ell\geq 0}\left( \frac{z-x}{w-x}\right)^{\ell}. $$

Now the rest of the proof and the eqs. (6.142)-(6.145) in Ref. 1 are correct. [Note that eqs. (6.144)-(6.145) refer to $x=0$ but they can easily be generalized to non-zero $x$.] $\Box$

References:

  1. P. Di Francesco, P. Mathieu and D. Senechal, CFT, 1997; Section 6.5.
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  • $\begingroup$ Thanks, that solves the problem. $\endgroup$ – Lelouch Jun 24 at 11:13

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