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What is the momenta region of the Fourier transformed operator product expansion?

Physics Asked by poisoned tea on August 6, 2021

I guess the question is general enough but for what it’s worth I am thinking on QCD while I type this. The operator product expansion is a short distance expansion of the product of two operators

begin{equation}
langle A(x)B(y)rangle=
sum_{n}C_n^{AB}(x-y)langle mathcal{O}_n(y)rangle
qquad
{rm for}
,xsim y
,.
end{equation}

Without loss of generality we can consider the $y=0$ case.
begin{equation}
langle A(x)B(0)rangle=
sum_{n}C_n^{AB}(x)langle mathcal{O}_n(0)rangle
qquad
{rm for}
,xsim0
,.
label{OPEx0}
end{equation}

We can write a Fourier transformed version
begin{align}
int d^4x,
e^{-iqx}
langle
A(x)B(0)
rangle
=&
sum_nlanglemathcal{O}_n(0)rangle
int d^4x
,e^{-iqx}
C_{n}^{AB}(x)

equiv&
sum_n
tilde{C}_{n}^{AB}(q)
langlemathcal{O}_n(0)rangle
,.
end{align}
Nevertheless, I am confused about for which values of $q$ the equation above should be considered. If we were working in Euclidean space $xsim0$ would mean $|q|toinfty$. Nevertheless, in Minskowski space I am unsure how Fourier transforms play out. Naively speaking I would dare say $qtoinfty$, but what does this mean? the metric is not positive so that $q^2$ can be either positive or negative, so then what is exactly $q$ going to infinity?.

Summarizing then, if we have an expression that depends of the Minkowski space point $x$ going to zero, at which momenta region foes the Fourier transformed version correspond?

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