Is width of particles a consequence of Heisenberg's uncertainty principle? The uncertainty principle tells us that
$$\sigma_x\sigma_p \geq \frac{\hbar}{2}, $$
which means that the more precisely we measure a particle's position, the more imprecise we will know its momentum. However, the standard deviation $\sigma_x$ can't be zero, and therefore its wave function will always have some spread to it. That got me thinking, is that the reason why particles such as protons, electrons, or neutrons have size? Is their size determined by the average of the spread of their wave function's standard deviation in position space when one collapses their wave functions an infinite amount of times?
 A: In the  mainstream standard model of particle physics, all matter is made up of point particles with a fixed mass which we measure as best as we can within our experimental errors. There is no width in this masses at the table.
Protons and (and neutrons bound in a nucleus) are stable composite particles , made up by a great multitude of quarks antiquarks and gluons , plus some valence quarks, and are found as quantum mechanical solutions in a QCD lattice model. Experimentally proton decay has not been seen, so the intrinsic width of the proton mass is still a delta function , although there are models that allow baryon number non conservation. (The same is true for the free neutron, because its lifetime is such that the possible width in mass  is not measurable).
Width due to the quantum mechanical wave function is found theoretically and measured experimentally in resonances , and decaying elementary particles, as seen here. The Heisenberg uncertainty is directly connected with this width, but the width depends on the interactions allowed by the various conservation laws for the specific decay.
