What does $\Delta p$ in uncertainty principle mean?

Question

What does the $\Delta p$ in uncertainty principle mean? Is it standard deviation?

I have seen few videos on youtube as well as in our high school textbooks, it was proved that the electrons don't exist in nucleus using this principle. The thing which is confusing me is that they considered $\Delta p$ as the exact momentum. Some of them even tried to explain that this $\Delta p$ is simply the momentum of the electron and the inequality sign which is present in the equation is the one which is ensuring uncertainty. But the used the value of diametre of nucleus (probable diametre) as $\Delta x$. Now this thing isn't making any sense. Because when considering $\Delta p$ they considered it to be the momentum whereas the considered $\Delta x$ the probable region where they would need to be present if electrons do exist in nucleus.

Answer 1

In the uncertainty principle, $\Delta p$ is indeed the standard deviation, which is defined as $$ (\Delta p)^2 = \left<(p-⟨p⟩)^2\right> = \left<p^2\right> - \left<p\right>^2 $$ (where $⟨\cdot⟩$ represents the average value of the quantity inside the brackets).

Now, in the calculation you mentioned, we expect $⟨p⟩$ to vanish, because otherwise the electron would have a net momentum heading somewhere, and we know it is on-average stationary. Thus, the momentum uncertainty boils down to $$ (\Delta p)^2 = \left<p^2\right>, $$ i.e., the momentum uncertainty is directly related to the average kinetic energy $⟨p^2/2m⟩$. It is not the exact momentum -- it's just a direct metric of the average of the square of the momentum.