I wonder why in OPE in CFT terms like

$$ \frac{:O(z) O(w):}{(z-w)^2} $$ occur, for example in the OPE of Energy-momentum tensor with itself:

$$T(z) T(w) = \frac{c/2}{(z-w)^4} + \frac{T(z)}{(z-w)^2} + \frac{\partial T}{z-w}$$

Here we have term $:\frac{T(z)}{(z-w)^2}:$, where for free scalar field $T(z) = :J(z)^2:$.

So my question is why such term $:J(z)^2:$ occurs, because all field operators are considered as correlation functions, so: $$ < T(z) T(w) > = < : J(z)^2: :J(w)^2: >$$ And with help of Wick's theorem we can decompose this correlator of 4 field operators to product of correlators like $<J(z) J(w)> <J(z) J(w)>$ and no terms with "bare" operators and one correlator ($:J(z) J(w): <J(z) J(w)>$) occurs here.

EDIT: I think that when we write OPE we don't consider field operators as operators under correlation function. And when we evaluate this correlation, only correlation functions remain and no operators. But then the question turns to be: Why do we need such OPE in terms of another operators? While evaluating correlation function of $T(z) T(w)$ all we need is the first term of OPE.

EDIT: Well, the answer is that we need OPE when we are interested in correlation functions with another operators, which are located far away from operators OPE is constructed for.

  • $\begingroup$ I don't understand your question. Can you try to explain more clearly what exactly about the $O(z)O(w)/(z-w)^2$ is worrying you? I'm not seeing a real question here currently. $\endgroup$
    – ACuriousMind
    Dec 23, 2015 at 18:04
  • $\begingroup$ Yes. Correlation function is a function, so it returns a number, not an operator such as $O(z)O(w)$ $\endgroup$
    – newt
    Dec 23, 2015 at 18:07
  • $\begingroup$ I added some edits to the question. $\endgroup$
    – newt
    Dec 23, 2015 at 18:22
  • $\begingroup$ OPEs are supposed to tell you not only the correlation functions of $O(z)O(w)$, but also $\langle O(z)O(w) A(z')B(z'')\dots\rangle$ when $z', z'', \dots$ are far away from $z$ and $w$, namely in evaluating such correlations one can replace $O(z)O(w)$ with the OPE when $z\rightarrow w$. That's why you need to have the other terms as well. $\endgroup$
    – Meng Cheng
    Dec 23, 2015 at 19:03

1 Answer 1


The question (v4) is a bit unclear, so here are some general comments, which may help:

  1. In CFT, an OPE is calculated via a Wick-like theorem changing an (implicitly written, operator-valued) radial order ${\cal R}(\hat{A}_1\ldots\hat{A}_n)$ into combinations of c-number-valued contractions and operator-valued normal order $:\ldots:$. Wick's theorem is e.g. explained in this Phys.SE post.

  2. Such calculations are organized into zero contractions, single-contractions, double-contractions, and so forth. E.g. in string theory, the central charge in the $TT$ OPE is a result of double-contractions.


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