# conformal symmetry vs time ordering

In Quantum field theories we generally calculate time ordered correlation functions. But it seems that in a conformal field theory I can use conformal symmetry to destroy time ordering.

Let me look at the 3 point function in d+1 dimensions : $$\langle \phi(t_1,x_1)\phi(t_2,x_2)\phi(t_3,x_3)\rangle$$

We assume this is time ordered $t_1 \geq t_2 \geq t_3$. We also assume that all the points are time-like separated.

But we know that using conformal symmetry we can fix any 3 points. So we can for instance map $t_1 \to 0, t_2 \to \infty, t_3 \to 1$. This destroys the time ordering of the correlation function.

'Edit: Even more simply, consider a radially ordered n point function in a 2d CFT. An inversion will reverse the radial ordering.'

So it seems to follow that in CFTs time (/radial) ordering has no co-ordinate invariant meaning (as in it is not conformally invariant). (Note that in usual Poincare invariant field theory I can certainly change the time ordering of spacelike separated operators, but this does not lead to any contradiction because such operators commute).

So how does the demand of conformal covariance of time ordered n point functions gel with the fact that time ordering itself can be screwed up by conformal transformations?

• @bianchira The point is I can take a time ordered correlation function and make a conformal transformation that does not preserve the time ordering. Then is there a meaningful, coordinate independent way of talking about time ordered correlation functions? – Nirmalya Kajuri Jun 4 '16 at 23:23

There is a certain difference between the Euclidean and Lorenzian cases.

In Euclidean QFT the correlation functions, the Schwinger functions, are all in a sense time-ordered (see Osterwalder–Schrader axioms). Intuitively one can say that this is because all separations are space-like. Say, if you consider the correlation function of one scalar, they are permutation symmetric. Usually you cannot even define out-of-time ordered functions, since these will involve evolution like $e^{+Ht}$, which is an unbounded operator and will cause these correlation functions to be infinite. (To be more precise, Euclidean QFT's can have different quantizations, i.e. radial space slices around different centers or even flat time slices. To compute a Schwinger function, you should take the ordered expectation value appropriate for your quantization.)

In this case, the conformal symmetry relates the time-ordered correlation functions, i.e. a transformation which changes the radial order of operators still relates the two (now differently) radial-ordered correlation functions.

In Lorentzian QFT there are many differently-ordered correlation functions (Wightman functions), not just the time-ordered. It may seem that this contains more information than Euclidean correlators, but in fact these all are different analytic continuations of the Schwinger function of Euclidean QFT. So in principle the idea that conformal transformations mess with the ordering is not so bad on itself. But I think that they actually don't.

I am not an expert on the structure of Lorentzian conformal transformations, but consider the following argument. Suppose we have a sequence of operators that has to be in a specific order in the time-ordered correlator, i.e. we have the fields $\phi(x_1),\ldots,\phi(x_n)$, $x_1<\ldots<x_n$, where the relation is that $x_1$ is in the absolute past of $x_2$, and so on. Then we can draw a smooth time-like curve which connects all these points. Conformal transformations in the identity component of conformal group change the metric by a positive factor, so they leave this curve time-like. They also cannot change the direction of the curve (i.e. it will again go from past to future), so the ordering of $x_i$ is preserved. Now, what happens when a transformation is not connected to the identity? A familiar case is the time reflection symmetry, which explicitly changes the ordering of coordinates. At the same time, it is represented by an anti-unitary operator and it also changes the ordering of operators, so that in the end the correlator remains time-ordered. I believe that in the same way conformal transformations actually do not spoil the ordering. The same mechanism works with your example of inversion, since inversion is precisely the time reflection in radial-ordered quantization.

• Consider two time like separated points (t1,x1) and (t2,x2) with t1<t2. An inversion will certainly change the time ordering. The fact that conformal transformations allow us to map any three points to any other three points already tells you that we can mess with the ordering. However I think the answer to my question is that time ordering simply means that one reshuffles the order after the conformal transformation has been performed. – Nirmalya Kajuri Jun 7 '16 at 11:11
• @NirmalyaKajuri As I said, the conformal transformations disconnected from the identity may change the order of operators. Inversion is essentially Hermitian conjugation in radial quantization, and it changes the order of operators in the inner product, so that they end up being in still radially ordered. Please read the last paragraph on my answer carefully. – Peter Kravchuk Jun 7 '16 at 12:36
• If you look at page 25 of these notes on CFT arxiv.org/abs/1602.07982 there is an example of how the choice of origin in radial quantization can reverse radial ordering. In the notes it is said that this is completely equivalent to changing frames in Lorentz Invariant theories. But I don't see how, because in those theories those operators for whom time ordering can be messed up are spacelike separated and always commute. This need not be the case here. – Nirmalya Kajuri Jun 18 '16 at 21:52