# $i\epsilon$ in CFT correlation functions

M. Luescher in his talk on p.6 writes that the 2-point correlation function of a Hermitian local field $O_k$ of scaling dimension $d=3-k$ looks like $$\langle 0| O_k(x) O_k(y) |0\rangle = A_k (x-y-i \epsilon)^{2k-6},\quad A_k\in\mathbb{C}.$$ I am not really sure why is it not just $$A_k(x-y)^{2k-6}\quad?$$ Is $-i\epsilon$ simply added to show that one needs to integrate over the upper half-plane?

• Yes. See this other SE question for details. – DanielSank Feb 25 '16 at 18:23
• @DanielSank thank you, but I still don't get it. The talk I am citing is on axiomatic QFT, so there should be no room for arbitrary choices. Since we have $\frac{1}{x\pm i\epsilon} = \mathbb{p.v.}\left(\frac{1}{x}\right) \mp i\pi\delta (x)$ and similarly for other powers as distributions, the two choices $\frac{1}{x\pm i\epsilon}$ are not equivalent. OK, so maybe one could disregard $\frac{1}{x+ i\epsilon}$ because the fields can be extended to the upper-half plane, but then why choose $\frac{1}{x-i\epsilon}$ over $\mathbb{p.v.}\left(\frac{1}{x}\right)$? – Gytis Feb 26 '16 at 9:31
• I don't know anything about QFT but I guarantee you the $i\epsilon$ is implicitly a choice of boundary condition for some Green function (you call it a 2-point function but it's the same thing). As explained in the linked post, the different choices correspond to different behaviors of the fields at infinity (i.e. damping means fields infinitely far from sources are zero). Note that either choice of Green function works just fine; both can be used to express the solutions to any problem, but one may be more convenient (i.e. physical) than the other. – DanielSank Feb 26 '16 at 9:41

First it means that it is a distribution, which you have to smear with test functions and then take the limit $\varepsilon \to 0$. Without the $\varepsilon$ description it is not well defined.
Now the precise $i\varepsilon$ comes from the positivity of energy, i.e. the spectrum of the generator of translations $P$ is positive.
Roughly speaking, in your case if you calculate the Fourier transformation $\hat W(p)$ of $W(x-y)=\langle \Omega,\phi(x)\phi(y)\Omega\rangle$ then the $i\varepsilon$ gives you that this is a function with support on the positive real line. The support of $\hat W(p)$ is the spectrum of $P$. Namely, $\hat W(p) \sim \Theta(p) p^{2d-1}$ where $d$ is the scaling dimension.
The general Fourier transformation is formula (2.13) in https://arxiv.org/abs/hep-th/0009004 Note that for higher dimensions it must be $i\varepsilon e_0$ where $e_0$ is a timelike vector.