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?

  • $\begingroup$ Huh, interesting, the second one is something I haven't seen (I think). $\endgroup$
    – David Z
    Dec 16, 2012 at 7:31
  • $\begingroup$ 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. $\endgroup$ Dec 16, 2012 at 9:01
  • $\begingroup$ 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. $\endgroup$
    – Tengen
    Dec 16, 2012 at 9:22

1 Answer 1


In Quantum Field theory, for non-interacting fields, the normal ordering can be defined by requiring that the product of the two fields doesn't have the singular part. Since for non-interacting fields the singular part is nothing but the Vacuum expactation value (and is just 1 term), it is sufficient to write: $$:\phi^2:\,\,\,=\phi^2-\langle\phi\phi\rangle$$

In CFT we can't just do that. Take the energy momentum tensor. It's OPE is known to be: $$T(z)T(w)=\frac{c/2}{(z-w)^4}+\frac{2T(w)}{(z-w)^2}+\frac{\partial T(w)}{(z-w)}+regular\,terms$$ if we try to take out $\langle T(z)T(w)\rangle$, we obtain: $$T(z)T(w)-\langle T(z)T(w)\rangle=\frac{2T(w)}{(z-w)^2}+\frac{\partial T(w)}{(z-w)}+regular\,terms$$ which is still singular.

Then, instead of subtracting only the V.A.V. , every non singular term is taken out. If we have two operators with the following OPE: $$A(z)B(w)=\sum_{n-\infty}^N \frac{\{AB\}_n(w)}{(z-w)^n}$$ with $N$ positive integers (which means that the number of singular parts can be finite) and $\{AB\}_n(w)$ the resulting fields of the expansion. We then define the normal ordered product as: $$(AB)(w):=\{AB\}_0(w)$$ In fact, we can define the Contraction as: $$C(A(z)B(w)):=\sum_{n=1}^{N}\frac{\{AB\}_n(w)}{(z-w)^n}$$ And then the normal ordered product is just: $$(AB)(w)=\lim_{z\to w}[A(z)B(w)-C(A(z)B(w))]$$ since all the terms $\{AB\}_n(z-w)^n$ with $n>0$ goes to zero as $z\to w$.

In this context, we can give an integral representation of this normal ordered product as: $$(AB)(z)=\oint_z\frac{dw}{2\pi i}\frac{A(w)B(z)}{w-z}$$ where the contour integral contains the point $z$.

  • 1
    $\begingroup$ It's only for the square that free field normal ordering involves just one term. For example $:\phi^4:=\phi^4-6\langle\phi^2\rangle\phi^2+3\langle\phi^2\rangle^2$. In general the expression is a Hermite polynomial. It's true however that the OPE, even in the free case, is what governs the proper normal ordering. $\endgroup$ Oct 13, 2018 at 15:44
  • $\begingroup$ Yeah, I wrote "the product of two fields", and in the following sentence it was implicit that I was referring only to two of them. Sorry if it was unclear. $\endgroup$ Oct 13, 2018 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.