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?


1 Answer 1


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

  • $\begingroup$ Thanks. Why exactly does creation of particle-antiparticle pairs lead to uncertainty in the measurement of the position? $\endgroup$
    – Alex L
    Apr 25, 2019 at 6:28
  • $\begingroup$ The original particle loses an independent existence as is no longer localized but a swarm of particles and antiparticles. Since the particle itself does not exist (as a particle) it does not have a 'position'. $\endgroup$ Apr 25, 2019 at 7:10
  • $\begingroup$ Doesn't the measuremnt give any information about the positions of the produced pairs? $\endgroup$
    – Alex L
    Apr 25, 2019 at 15:13
  • $\begingroup$ @KishoreIyer, do you mean $\Delta E \geq \frac{\hbar c}{L}$? $\endgroup$
    – S. McGrew
    Apr 25, 2019 at 15:58
  • $\begingroup$ I'm sorry, have corrected it. Thanks for pointing it out. $\endgroup$ Apr 26, 2019 at 8:40

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