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Studying the QCD running coupling constant I ran into this figure: enter image description here

where $Q$ on the $x$ axis represents the transferred momentum. I know from a Nuclear and Subnuclear Physics course that the strong interaction coupling constant is very small at small distances, so I was wondering why "high momentum transfer" equals "small distance" in this context? I found similar questions in some physics forum and the answer was "for the Heisenberg Uncertainty Principle". I studied the Heisenberg Principle as \begin{equation} \Delta p\Delta x\ge\frac{\hbar}{2} \end{equation} so, to get a correspondence between "high momentum transfer" and "small distance" the equality should hold. If the inequality holds I could have "high momentum transfer" and "great distances" without violating the principle. Can someone explain to me why $$\Delta p\Delta x\sim\hbar/2$$ seems to hold (instead of the version with $\ge$)?

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    $\begingroup$ It's less about the HUP and more about high momenta being able to resolve small structures. Further reading on the "size" of particles: physics.stackexchange.com/q/264676/50583 $\endgroup$ – ACuriousMind Dec 7 '16 at 14:20
  • $\begingroup$ Thanks for your answer ACM. I read about experiments concerning beams of particles, foils, interference patterns and so on, and I understood the basic concepts of it. But this seems to me an entirely different thing, I can't make a connection, because it's not about resolving structures with an interference pattern, it's all about the kinematic of the particles. $\endgroup$ – Luthien Dec 12 '16 at 16:57
  • $\begingroup$ I mean, on one side I have a beam with particles with high momentum and a foil of atoms with fixed distance (and, depending on the momentum of the beam, I can resolve the distance between the atoms or the distance between the nucleons and so on...). Here I have a plasma (Quark-Gluon plasma) with gluons and quarks. So why should the momentum of the particles be connected with their distance? $\endgroup$ – Luthien Dec 12 '16 at 17:00
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It comes from the fact that momentum and positions are conjugated variables. It is best illustrated in Fourier Transform, which relates position and momentum space. $$ \psi(p) = \int dx e^{-ipx} \psi (x). $$ When $p\sim \frac{1}{x}$ the exponent is not suppressed. In the other cases, the exponent will oscillate, and the contribution to the integral will be much smaller. And this is the origin of a statement "small distance equals large momentum."

Consider an example; electron 1S wave function in hydrogen is $$ \psi(r) \sim e^{-m \alpha r}. $$ The size of hydrogen is $r\sim \frac{1}{\alpha m}$. So to probe hydrogen we should use $p\sim 1/r \sim \alpha m$. Indeed, with $$ \psi(p) \sim \frac{1}{(p^2+m^2\alpha^2)^2}. $$ probability to find momentum in region $[p,p+\Delta p]$ is proportional to $$ p^2 \psi(p)^2 \Delta p $$ which has a peak for $p\sim m\alpha$.

Note that in your question you talk about high momentum transfer, not about high momentum uncertainty. Consider an example of scattering electron with momentum $p$ on a proton. For simplicity, I will consider electron, photon, and proton to be spin 0 particles. The amplitude is $$ \mathcal{M} = (ie)^2\frac{i}{q^2}F(q^2) $$ with the momentum transfer $q=p'-p$. $F(q^2)$ is proton charge distribution function, $F(0)=1$. When you measure the differential cross-section $\frac{d\sigma}{dq^2}$ you probe the form-factor at a certain value of the momentum transfer. Note that $q$ is defined by external kinematics, $q^2=(p'-p)^2=2m_e^2-2p\cdot p'$. By Fourier transforming $F(q^2)$ you obtain charge distribution in position space - the charge density. In practice, you only know $F(q^2)$ in a certain range of $q^2$. Then, $1/q^2$ is the smallest structure you can resolve in position space. Fourier transform of $F$ is directly related to the wavefunction in position space.

On the other hand, this example is illustrative, because the off-shell propagator $\frac{i}{q^2}$ gives you an effective range of interactions. When a particle is off-shell, then it can propagate only over a short distance $\Delta x\sim \frac{1}{\Delta Q}$. This comes again from the properties of the Furier transform.

It is only partially related to the uncertainty principle. High momentum can still have large uncertainty, but you want also $$\Delta p \ll p$$ because only then the measurement is precise. In other words; $$ \Delta x \Delta p \ge \hbar/2 $$ holds always, but when $$ \Delta x \Delta p \sim \hbar/2, $$ then the measurement is most precise for fixed $p$ and $x$. So you minimize also $\frac{\Delta p}{p}$ at given $\Delta x$ and $x$.

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  • $\begingroup$ Thanks for your answer. I need some clarifications though, even if I think that this is heading to the right answer. As I'm talking about momentum TRANSFER, I think that this has something to do with the momentum of the virtual gluon (which is an off-shell particle). This has also something to do with the uncertainty principle, as virtual particles are off-shell but this lies inside the boundaries set by the uncertainty principle. What do you think? Are you thinking about a virtual particle in your dissertation or a real one? $\endgroup$ – Luthien Mar 10 '17 at 19:08
  • $\begingroup$ Strictly speaking, uncertainty principle applies only to real particles. You cannot observe virtual particle or measure its momentum. $\endgroup$ – Veritas Mar 10 '17 at 19:20
  • $\begingroup$ When you measure the structure of something e..g. of a proton, then you scatter other particles on this proton, e.g. electron. This electron will interact with proton through the exchange of a virtual photon. This photon transfers momentum to the proton. When the momentum transfer is large, you probe proton wave function at large momentum. This means that you can see small structures -- see my argument with Fourier transform. What you are probing is the charge distribution, or rather directly its Fourier transform -- the proton form-factor. $\endgroup$ – Veritas Mar 10 '17 at 19:27
  • $\begingroup$ Thank you for your efforts Veritas :) I think I get it now, plus you provided me with a lot of things worth to be studied in a deeper way. $\endgroup$ – Luthien Mar 10 '17 at 20:17
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    $\begingroup$ @Luthien You are welcome. Consider also looking into time-energy uncertainty principle and the old-fashioned perturbation theory. These concepts, often omitted in standard physics classes, may help you to better understand QFT concepts like virtual particles. $\endgroup$ – Veritas Mar 10 '17 at 20:36
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$\Delta p$ is the uncertainty in the measurement of the momentum and $\Delta x$ is the uncertainty in the position of a particle(s). The measurement resolution required to probe particle interactions at small distances requires a small particle position uncertainty. This means $\Delta x$ will necessarily be small and uncertainty in momentum, $\Delta p$ will be large. This means when you are looking at small distances, there is large uncertainty in the momentum (possible to have 'high momentum transfered' to the particles).

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