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Recall the normal ordering of bosonic operators in QFT is defined by a re-arrangement of operators to put creation operators to the left of annihilation operators in the product. This is designed to avoid accidentally annihilating $|0\rangle$ when looking at an expectation value in relation to the vacuum state.

$ : \hat{b}^\dagger\hat{b} : \: =\: \hat{b}^\dagger\hat{b} \\ : \hat{b}\hat{b}^\dagger: \: = \: \hat{b}^\dagger\hat{b} $

In CFTs, I've seen defined the normal ordering of operators as the zeroth basis field of the Laurent expansion of the radial ordering product.

$\mathcal{R}(a(z)b(w)) = \sum_{n = -n_0}^\infty (z-w)^n P_n(w),$

and select

$P_0(w) = \: : a(w)b(w) : $

Is there an equivalence between these two definitions? What is the CFT analog of not annihilating the vaccuum/ how do we show this definition has that property?

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Huh, interesting, the second one is something I haven't seen (I think). – David Z Dec 16 '12 at 7:31
The radial ordering, as you defined it, is clearly a more general thing whose $P_0$ part includes the normal ordering as a tool. So the equivalence you're asking about is like the equivalence between the United States and New Jersey. The New Jersey, normal-ordered part is equivalent but the rest is not. Quite generally, radial ordering is conformally equivalent to time-ordering but time-ordering and normal ordering are pretty much different things, although they have some mathematical analogies. – LuboŇ° Motl Dec 16 '12 at 9:01
If the OPE of two fields contains only one singular term with constant coefficient such as the case of free fields, then the subtraction of the vacuum expectation value (your first definition) can serves as the right regularization procedure. If the OPE is more complicated such as the product of two energy-momentum tensors, one must subtract all singular terms as your second definition. – Tengen Dec 16 '12 at 9:22

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