I have read on Wikipedia (https://en.wikipedia.org/wiki/Compton_wavelength) that we cannot measure the position of a particle more precise than half of its Compton wavelength, since the photon we would need will be so energetic to produce electron-positron pairs.
How does the creation of electron-positron pairs lead to uncertainty? Does this this fundamentally and in principle limit our possible knowledge of a particle's position?
Consider a particle of mass m confined in a box of length L, the uncertainty in the momentum is of the order $\Delta p \geq \frac{\hbar}{L} $. Relativistically, energy and momentum are on the same footing. So, uncertainty in the energy is of the order $\Delta E \geq \frac{\hbar c}{L}$. When the uncertainty in the energy exceeds $2mc^2$ ($ L = \frac{\hbar}{mc})$ it is possible to create particle anti-particle pairs from the vacuum.
Thus, there is a high probability that we will detect particle-antiparticle pairs swarming the original particle at length scales smaller than the Compton wavelength $\frac{\hbar}{mc}$. The notion of a single, localized particle breaks down completely below the Compton wavelength.
Have a look at these lecture notes from where I have sourced this answer: http://www.damtp.cam.ac.uk/user/tong/qft/one.pdf
