Relation of $Q^2$ with the distance of interaction In a process with only one Feynmann diagram at leading order such as $e^- \mu ^-$ scattering:

the photon propagator is
\begin{equation}
\frac{-i g_{\mu \nu}}{q^2},
\end{equation}
with $q^2 = (p_3 - p_1)^2$. Since $q^2 <0$, we sometimes define the positive value $Q^2 = -q^2$. Apparently, it seems ok to relate $Q^2$ just as a quantity related to the exchange of energy and momentum between the particles. 
But when reading about the running coupling constants in Halzen & Martin's book, they show that the running coupling constant of QCD has a small value for increasing $Q^2$, and the author relates high $Q^2$ to small distance interactions (i.e. asymptotically  freedom) and small $Q^2$ to long distance interactions.
I cannot understand how the value of the exchanged 4-momentum relates to the interaction distance. Could someone shed some light in the subject?
 A: In a sense it is based on the heisenberg uncertainty principle, HUP

In every interaction the four momentum transfer is made up of the three momentum transfer and the energy transfer. Looking at large Δp of the Q^2 transfer within the limits of the HUP the Δx gets smaller. 
A: First, note that $Q^{2}$ is always negative for non-trivial scattering. Taking into account the explicit form for $Q^{2}$, 
$$
Q^{2} = (E_{1}-E_{3})^{2}-(\mathbf p_{1}-\mathbf p_{3})^{2} < 0,
$$
from the Lorentz invariance we immediately obtain that there is possible to find the inertial frame at which $E_{1}-E_{3} = 0$. In this case,
$$
Q^{2} =-(\mathbf{p}_{1}-\mathbf p_{3})^{2}
$$
The next step is to note that the momentum $\mathbf p$ is just the fourier-conjugated coordinate for the spatial coordinate $\mathbf x$. This immediately relates the momentum $|\mathbf p|$ to the inversed coordinate $|\mathbf x|^{-1}$. Therefore, the difference $|\mathbf p_{1} -\mathbf p_{3}|$ is related to the inversed wavelength corresponding to the distance between the interacting particles.
But since this statement is true in one reference frame, then, due to the relativity principle, it is true in any another reference frame. But this is nothing but your desired result.
A: Another way of viewing it is through a Fourier transform. Let's say that you transform a picture to a frequency domain and then back to the picture. The process is invertible so you get the same picture as what you've started with. However, if you'd discard high-frequency components - the result would appear to be blurred and will lose details.
These details are small distance features, and they come from high frequency modes in the picture. Same applies to particle physics.
