The kinematic region for the operator product expansion

Question

In Ch.18 of the textbook An Introduction to Quantum Field Theory by Peskin and Schroeder, on P.613 the operator product expansion (OPE) is introduced

$$\mathcal{O}_1(x)\mathcal{O}_2(0)\to \sum_n C_{12}^n \mathcal{O}_n(0)\ .\tag{18.64}$$

The expansion is carried out for small $x$, and therefore for large momentum transfer. Subsequently, in section 18.4, the OPE is illustrated by the $e^+e^-$ annihilation process, where the r.h.s. of the above equation corresponds the vacuum polarization:

$$i\Pi_h^{\mu\nu}(q)=-e^2\int d^4 e^{iq\cdot x}<0|T\{J^\mu(x)J^\nu(0)\}|0>\ ,\tag{18.88}$$

where $q$ is the sum of the four momenta of electron and positron, $q=k+k_+$. The OPE is specified in (18.89). Above (18.94), the text reads

Still assuming the Fourier transform of the product of currents can be computed from the OPE for the region of large time-like $q^2$, we can complete our evaluation of the cross section for $e^+e^-\to$hadrons by the vacuum expectation value ... We find

$$\sigma(e^+e^-\to\mathrm{hadrons})\\=\frac{4\pi\alpha}{s}[\mathrm{Im}\ c^1(q^2)+\mathrm{Im}\ c^{\bar q q}(q^2)<0|m\bar qq|0>+\mathrm{Im}\ c^{F^2}(q^2)<0|(F_{\alpha\beta}^2)|0>].\tag{18.94}$$

Shortly after that, it is stated that in order to calculate $\Pi_h(q^2)$ by using the leading perturbative QCD contributions shown in Fig.18.5. One should

... choose kinematic conditions such that the intermediate states that enter the computation of the product of currents are far off-shell, so that they cannot propagate far from the converging points $x$ and $0$.

I understand the following: the physical problem requires that $q^2$ is large and time-like (in the CM frame of reference and thus in any frame of reference). However, in order to evaluate the perturbative QCD contributions relatively easily, it requires that $q^2$ is large and to stay away from physical region, so that no hadron is formed, or otherwise one has to deal with soft process (to form hadron) and the problem becomes more difficult to proceed. This is the reason why the ITEP sum rule is utilized subsequently. Is it correct?

Please point me toward correct direction if I am wrong, and if my understanding is correct, are the discussions provided in the textbook a general argument? In other words, if one want to use perturbative QCD, one should always avoid the region of physical momentum to avoid significant contribution from the soft process to form intermediate hadron states. The arguments seem somewhat hand-waving, is there any alternative explanation?

Answer 1

Maybe I understand a little better now.

For a real physical process, one requires that $q^2$ is large and time-like (by evaluating $q^2$ in the CM frame of reference and thus in any frame of reference). However, in order to evaluate the perturbative QCD contributions relatively easily, it requires that $q^2$ is large. In principle, if one only considers that $x$ has to be small enough for the OPE to work reasonably well, $q$ can be time-like or space-like. But in order to stay away from singularities or cuts which might invalidate the perturbative treatment, which is also associated with the physical region, one usually goes to the space-like region of $q$ where no hadron is formed and subsequently employs perturbative QCD to evaluate things and then expands the results in terms of $1/q^2$. As one attempts to connect these two calculations, one assumes that the singularities or cuts are only related to physical states, and employ the contour integration to obtain the so-called QCD sum rule.