I have one slight confusion with something that Blumenhagen & Plauschinn state in the book "Introduction to Conformal Field Theory: With Applications to String Theory". In page 24 regarding radial ordering they state:

Note that there is some ambiguity in Eq. (2.35) because we have to decide whether $w$ and $\bar w$ are inside or outside the contour ${\cal C}$. However, from quantum field theory we know that correlation functions are only defined as a time ordered product. Considering the change of coordinates (2.23), in a CFT the time ordering becomes a radial ordering and thus the product $A(z)B(w)$ does only make sense for $|z|>|w|$. To this end we define the radial ordering of two operators as $$R(A(z)B(w)):=\begin{cases}A(z)B(w),\quad \text{for $|z|>|w|$},\\ B(w)A(z),\quad\text{for $|w|>|z|$}.\end{cases}$$

I'm puzzled by their claim that "from quantum field theory we know that correlation functions are only defined as a time ordered product".

I mean, I do understand that what we really need are the time ordered correlation functions. In particular they are what naturally appear in the LSZ reduction formula which allow us to obtain the ${\cal S}$-matrix from correlation functions. Moreover, I do know that in the functional approach what we naturally get are time ordered correlation functions.

But for me, even though the non time-ordered ones are not useful and don't come out naturally in the path integral approach, that does not mean that they are ill-defined. Observe that the authors' claim is not that "we only care about time-ordered ones", he is claiming very clearly that only the time ordered ones are defined. He further confirms that when he says that, after translating to the plane, $A(z)B(w)$ only makes sense for $|z|>|w|$.

I'm clearly missing something very basic here. Why only time ordered correlation functions are defined? Why a product of two fields which is not time ordered does not make sense? In canonical quantization, for example, what would stop us from writing expressions like $\phi(x)\phi(y)$ for $x^0 < y^0$? It appears a valid product of operators in a Hilbert space to me.


1 Answer 1


I'm not sure about this (so please watch for other answers), but I suspect that the "only" wording is a mild case of careless writing. The context suggests that the authors are thinking about QFT in euclidean signature. Wick-rotating from euclidean to lorentzian signature automatically gives time-ordered correlation functions, but after we have all of the QFT's time-ordered correlation functions, we can reconstruct an operators-on-a-Hilbert-space formulation which allows non-time-ordered products.

However, the meaning of "time-ordered" depends on which coordinate we decide to use as time, such as the radial coordinate. Different choices lead to different operators-on-a-Hilbert-space formulations. So maybe the authors meant something like this: non-time-ordered correlation functions can't be defined until after we decide which coordinate we want to use as time. The euclidean CFT doesn't tell us which coordinate we should use.

  • 5
    $\begingroup$ I have to say, theoretical physics textbooks are full of little potholes like this, and I think Physics.SE's greatest contribution is supplying ways around them. If it's in the cards, you should consider writing a textbook someday and fixing this, given the clarity of your answers! $\endgroup$
    – knzhou
    Commented Jul 24, 2021 at 23:16
  • 1
    $\begingroup$ I found this comment in the big yellow book on CFT: "The time-ordering prescription may appear artificial within canonical quantization, but it is necessary to ensure convergence of the vacuum expectation values, assuming that a ground state exists with energy bounded from below." but I don't quite get this. Maybe someone can elaborate. $\endgroup$
    – M. Zeng
    Commented Oct 9, 2023 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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