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I'm specifically working with a 2d free scalar CFT. In this case, the propagator is $$\langle X(\sigma) X(\sigma')\rangle=-\frac{\alpha'}{2}\ln(\sigma-\sigma')^2\tag{p.78}$$ while the OPE between $X(\sigma)$ and $X(\sigma')$ is $$X(\sigma)X(\sigma')=-\frac{\alpha'}{2}\ln(\sigma-\sigma')^2+\dots\tag{4.22}$$

My question is, what is the conceptual difference between these two objects? My confusion stems from the text in Tong's lecture notes around equation 4.10, the defining equation of the OPE. He writes:

We will write a lot of operator equations of the form (4.10) and it’s important to clarify exactly what they mean: they are always to be understood as statements which hold as operator insertions inside time-ordered correlation functions...Obviously it would be tedious to continually write $\langle...\rangle$. So we don’t. But it’s always implicitly there.

For reference, this is his equation (4.10): $$\mathcal{O}_i(z\bar{z})\;\mathcal{O}_j(w,\bar{w})=\sum_k C^k_{ij}(z-w, \bar{z}-\bar{w})\;\mathcal{O}_k(w,\bar{w})\tag{4.10}$$

So if I'm understanding Tong's quote correctly, then the above definition of the OPE between $\mathcal{O}_i$ and $\mathcal{O}_j$ is within an implicit $\langle...\rangle$. But if that's the case, then why does the OPE between $X(\sigma)$ and $X(\sigma')$ include the nonsingular terms that the propagator doesn't? According to Tong's passage, I would expect the two expressions to be exactly the same since when we write the OPE $X(\sigma)X(\sigma')$, we really mean $\langle X(\sigma) X(\sigma')\rangle$.

Qmechanic
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Sam
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1 Answers1

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  1. The $XX$ OPE e.g. contains a non-singular normal order term $:\!XX\!:$ that the 2-point function/propagator $\langle XX\rangle$ hasn't.

    [In contrast the normal-ordered term $:\!XX\!:$ sandwiched between the bra and ket vacuum states yields zero, cf. e.g. my Phys.SE answer here.]

  2. Generically, a 2-point function $\langle f(X,\partial X,\ldots)g(X,\partial X,\ldots)\rangle$ contains non-singular terms, but they happen to vanish for the propagator $\langle XX\rangle$.

    [The vanishing of the non-singular terms can e.g. be seen from the fact that the derivative$^1$ $\partial X$ is a $(1,0)\oplus(0,1)$ quasi-primary field, so that its 2-point function $\langle \partial X\partial X\rangle$ is fixed up to a normalization constant by conformal symmetry.]

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$^1$ In this answer the symbol $\partial$ is a shorthand for all the derivatives, e.g. both the complex derivatives $\partial$ and $\bar{\partial}$.

Qmechanic
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