quantum mechanics - Two explanations of non-zero atomic radius

I have came across two separate explanations for why atoms have a positive atomic radius (as opposed to electrons "collapsing" into the nucleus).

1. 
   

   The first is via Heisenberg Uncertainty Principle, where decreasing the atomic radius would raise momentum and hence kinetic energy (while potential energy decreases) - the atom would "choose" a radius that minimizes energy.

   
   


2. 
   

   The second is kind of "just set-up and solve the Schrodinger equation" and obtain a certain atomic radius.

   
   

Are these two explanations just two sides of a single coin/a matter of interpretation? I have briefly heard that there is an equivalence between "Heisenberg" and "Schrodinger" formulation of QM - is this an instance of the equivalence?

Answer

The first explanation is just a quick argument to avoid doing calculations proposed in the second explanation.

As you're looking for ground state of the system, you want to minimize energy $\langle \psi | \hat{H}| \psi\rangle$. Now, from commutation relations between position and momentum operators $[\hat{q},\hat{p}] = i \hbar$ follows that $\Delta q \Delta p \geq \frac{\hbar}{2}$, no matter for which state you calculate $\Delta q$ and $\Delta p$. From this you can estimate kinetic energy to be of order $\frac{1}{(\Delta q)^2}$, while the Coulomb energy should be of order $-\frac{1}{\Delta q}$. Obviously as you shrink your candidate for ground state, sooner or later kinetic term will dominate and blow up your energy, meaning your ground state is stable against shrinking.

All these "Heisenberg" and "Schroedinger" formulations, wave-particle duality and whatever are just out-of-date jargon, which brings nothing but confusion.