# Does the Compton wavelength put a limitation on how precise we can measure the position of a particle?

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.
• @KishoreIyer, do you mean $\Delta E \geq \frac{\hbar c}{L}$? Apr 25 '19 at 15:58