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?

• Eeeeek.... "width" is a technical resonance term and not "size", which you are apparently asking about... Can you supplant the first with the second? And what do you do with pointlike particles like the μ? Aug 7, 2019 at 22:34
• I'm aware of the term 'pointlike' in physics, but that is exactly why I'm asking, as if my hypothesis behind my question were true, µ would have a size, eventhough it's negligible.
– Cazo
Aug 7, 2019 at 22:48
• Related: physics.stackexchange.com/questions/495576/… . 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 and infinite amount of times? No, wavefunction collapse isn't even a part of standard quantum mechanics. It only exists in the Copenhagen interpretation.
– user4552
Aug 7, 2019 at 23:07
• What I mean is that, when a wave function is influenced by another wave function, say light hitting an electron, the wave function localizes. Is the width of that location (bell curve), related to the size of the electron?
– Cazo
Aug 7, 2019 at 23:46