Heisenberg uncertainty principle and particle physics

Question

Studying the QCD running coupling constant I ran into this figure:

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 $\ge$)?

Answer 1

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 $x$.

Answer 2

$\Delta p$ 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).