# The minimum uncertainty of a particle's position

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It is stated that the minimum uncertainty $\Delta x$ of the position of some particle with the mass $m$ at rest is
$$\Delta x \sim \frac{\hbar}{mc}$$
For the particle with energy $E$ corresponding statement is
$$\Delta x \sim \frac{\hbar c}{E}$$
But I don't understand how to obtain these results (except, of course, thinking in the spirit of dimensional analysis).

Could you please explain how to obtain them quantitatively, from the ground of relativistic physics?

edited Aug 31, 2017

The particle position uncertainty depends on the state in which you prepare the particle. A plane wave with energy $E$ may have a huge wave-packet size. What you wrote looks like the De Broglie wave-length estimation rather than the position uncertainty. The latter is determined with the wave-packet size.  Note that $mc$ and $E$ have different dimensions.

@VladimirKalitvianski : in the source (Landau, Vol. 4, paragraph 1) where I've seen these results there is a comparison with DeBroglie wavelength only after obtaining the results. I'll also correct the misprint.

Relativistic quantum theory/experiment demonstrates a multi-particle states after collisions. So if you measure the particle position with help of collisions, then you can encounter a situation with many particles and anti-particles in the final state, so you may not ascribe any of them to your initial particle. The threshold of the transferred energy for pair creation is $2mc^2$. If the transferred energy is too huge, you can obtain a "shower" of particles in the final state, so your second formula is not about the particle position uncertainty.

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