In the process of thinking about this question, I realized that I don't understand something very fundamental about operator product expansions.

Consider a product of 3 local operators in a 2d CFT:

$$ X(x) Y(y) Z(z) = \sum_{n=-N}^{\infty} A_n(x) Z(z) (x-y)^n, $$

where we have substituted $X(x) Y(y)$ for the $XY$ OPE. This expression contains the singular terms for $x = y$.

Now because by definition of OPE $A_n(x)$ is a local operator at $x$, we can use the $A_n Z$ OPE again:

$$ X(x) Y(y) Z(z) = \sum_{n=-N}^{\infty} \sum_{m=-M}^{\infty} B_{nm} (x) (x-y)^n (x-z)^m. $$ This expression contains the singular terms for $x = y$ and $x = z$.

Question: where did the $y = z$ singular terms go?

This is likely related to the convergence of the series, but I wasn't able to formulate a convincing argument.


Let me redo the calculation, while explicitly writing OPE coefficients. Let $(A_n(z))_n$ be a basis of operators at $z$. We use the two OPEs $$ Y(y)Z(z) = \sum_n c_n(y,z) A_n(z) $$ and $$ X(x)A_n(z) = \sum_m d_{m,n}(x,z) A_m(z) $$ We end up with the result $$ X(x)Y(y)Z(z) = \sum_{m,n} c_n(y,z)d_{m,n}(x,z) A_m(z) $$ where the coefficient of $A_m(z)$ is $\sum_n c_n(y,z)d_{m,n}(x,z)$. This coefficient is an infinite sum, and it can very well be singular as $x\to y$, although this is not manifest. For instance, $$ \frac{1}{x-y} = \sum_{n=0}^\infty (y-z)^n (x-z)^{-n-1} $$ You can recover a similar result in your calculation by distinguishing more clearly the operator basis from the OPE coefficients. Your operators $B_{m,n}(x)$ should not all be linearly independent, and you should rewrite them in terms of a basis of operators.

  • $\begingroup$ That sum that you wrote down contains infinite negative powers of $x-z$. Doesn’t this contradict the axiom that OPE starts at the finite order $-N$? $\endgroup$ – Prof. Legolasov May 21 at 17:59
  • $\begingroup$ No. Each OPE $XA_n$ starts at a finite order. But we are taking a linear combination of infinitely many such OPEs. $\endgroup$ – Sylvain Ribault May 22 at 6:50

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