I want to understand the operator product expansion (OPE) in the context of a relativistic non-conformally invariant theory. I wanna pose the question in a general fashion, but what I always have in the back of my mind is QCD.

First of all I wanna point out that I do am aware that in general there is no theorem guaranteeing existence or convergence or anything of the OPE. I am not looking for that. I am looking for a precise (working) definition of the OPE and specially its relation to factorization. Any reference on the topic would be greatly appreciated since I am finding that everywhere I look the topic is treated in a very lousy manner.

Ok, let's begin with the statements. To my understanding the OPE is nothing but a way to define the product of two local fields. In mathematical jargon we would say that it defines an algebra. That is, if $A(x)$ and $B(x)$ are two any local operators built out of the field degrees of freedom of your theory, and their spacetime derivatives, and with local we only mean that $A$ and $B$ only depend on a single spacetime point, the conjecture is that $$ A(x)B(y)=\sum_nC_n(x-y)P_n(y) $$ This is the way the OPE is presented in textbooks. Some caveats. It is usually stated that $x\to y$, where not too much care in specifying just what is exactly meant with $x\to y$. The way I like to state this is to assume that there is a neighborhood $U$ of $y$ such that x is included in it and $x\neq y$. The coefficients $C_n$ are believed to be distributions and the $P_n(y)$ would be some local operators that happen to be well defined. This relation is assumed to hold within brackets. So far so good. This is what the OPE is to me right now.

Now, the OPE is usually linked to factorization of scales. Being very vague we introduce a scale $\mu$ that separates or factorizes two regimes, the IR and the UV. Now it is stated usually that (in the context of QCD) the UV contribution goes into the coefficient functions but that the IR one can be absorbed in the condensates (the sandwitched $P_n(y)$). I want to clarify this. I want to understand how the picture presented in the above paragraph leads to this vaguely described factorization. I do not see the link at all, so what is it?

  • $\begingroup$ Minor comment 1: OPE coefficients are usually assumed to be analytic functions, which are much nicer than distributions. $\endgroup$
    – user1504
    Jun 22, 2017 at 18:23
  • $\begingroup$ Minor comment 2: I'm curious who's using the term 'factorization' for this separation of scales phenomenon. Some mathematicians use the term 'factorization algebra' to refer to a kind of Euclidean OPE. Seems like a hash collision. $\endgroup$
    – user1504
    Jun 22, 2017 at 18:24

1 Answer 1


I think the key point is that you want to imagine that $x$ and $y$ are very close to each other, relative to the distance scale $\hbar/\mu$ set by $\mu$.

$$|x - y| \ll \hbar/\mu$$

The OPE coefficients are meant to capture how singular the operator product becomes as $x$ approaches $y$. Since these singularities only occur when $x = y$, they need to be sensitive to UV physics.

If there's IR physics, on the other hand, it's occurring on distance scales much longer than $\hbar/\mu$. Consequently, the IR physics can't really distinguish between $x$ and $y$, so you don't change the output of your computation if you put all IR effects at $y$ instead of $x$.


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