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OPEs can be used both in QFTs and CFTs, but in QFT they are only used to describe short-distance behaviour, whereas in CFT they are expected to describe the entire theory. This is supposedly due to the fact that the radius of convergence of OPEs in CFT is finite and not (generally) in non-conformal QFT.

I couldn't find any reason why the convergence between the theories should behave differently, other than some comment on bad behaviour of QFTs messing up this convergence. Can anyone give me some more details on this?

Qmechanic
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Richard122
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1 Answers1

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One argument is that $\mathcal{O}_i(x) \mathcal{O}_j(0) \left | 0 \right >$ is a state and in quantum mechanics the expansion of an arbitrary state into energy eigenstates will converge. However, it is only CFTs where all energy eigenstates can once again be written in the form $\mathcal{O}_k(0) \left | 0 \right >$. I.e. you need there to be a state-operator map.

Another argument is that a series in powers of $|x|$ will be asymptotic if you can write down small corrections that are not analytic functions around $x = 0$. The "instanton-like" terms $e^{-L/|x|}$ are perfectly sensible corrections in generic QFTs. But in CFTs there is no intrinsic length scale so there is nothing that can stand in for $L$.

Connor Behan
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