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?

  • 1
    $\begingroup$ Heisenberg and Schrodinger formulations are almost the same thing; one the evolution is in the wavefunction, and in the other in the operators. There is a third one, Dirac picture, more useful for interactions. Wiki has an explanation: en.wikipedia.org/wiki/Heisenberg_picture $\endgroup$
    – Davidmh
    May 23, 2014 at 0:57

2 Answers 2


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All these "Heisenberg" and "Schroedinger" formulations, wave-particle duality and whatever are just out-of-date jargon, which brings nothing but confusion.

  • $\begingroup$ @user47020: As a side remark, one can solve the Hydrogen atom using Heisenberg's "matrix mechanics" (i.e. working only at the operator level). Pauli did it even before Schroedinger found the same result with his equation. Dirac later showed that both approach are just two face of the same formalism. $\endgroup$
    – Adam
    May 23, 2014 at 0:56

"Are these two explanations just two sides of a single coin/a matter of interpretation?" - No, they aren't.

First of all solvin Schrödinger's equation gives correct energy for the ground state, but using Heisenberg's uncertainity principle can give only a rough estimate and it even fails if you assume that electron consists of two separate wave packets. So using Heisenberg's uncertainity principle for solvin hydrogen ground state is simply wrong.

But to be fair, it must be said that one can use Heisenberg's uncertainity principle for solvin ground state energy BUT the potential must be the right form, i.e. in the case of harmonic oscillator.

"How to find accurate hydrogen ground state energy without solvin nasty Schrödinger's equation?", you may aks. The answer is fairly simple, but one needs proper mathematical set up.

So the question is to to find the lowest expectation value for observable $ H$, that is the sum of expectation value for kinetic energy and Coulomb potential: $$\text {inf}(\psi|\hat H|\psi)=\text {inf}(\frac{{\hbar}^2}{2m}\int_{R^3} |\nabla \psi|^2dx-ke^2 \int_{R^3} \frac{1}{|x|}| \psi|^2dx)$$

"What the hell I'm gonna do next?!", you may ask again.

In the following $\sqrt{ \int_{R^3} | f|^2dx}=||f||_2$, so for wave function $||\psi||_2=1.$ Function $f$ belongs to space $L^2 (R^3)$ if $||f||_2<\infty$.

Theorem . If both $f$ and $\nabla f$ belong to space $L^2 (R^3)$ then following inequality holds:$$\int_{R^3} \frac{1}{|x|}| f|^2dx \leqq ||\nabla f||_2 || f||_2$$

You will have much fun in proving that! (It requires only calculus).

When you insert that theorem to $\text {inf}(\psi|\hat H|\psi)$ everything boils down to minimizing expression: $$\frac{{\hbar}^2}{2m}||\nabla \psi||^2_2 -ke^2||\nabla \psi||_2$$

Keeping $||\nabla \psi||_2$ as a variable you can use elementary calculus to find the fundamental:$$\text {inf}(\psi|\hat H_{\text{Hydrogen}}|\psi)=-k^2me^4/2\hbar^2.$$

EDIT: Considering your question: "explanation for non-zero atomic radius." Now you have $||\nabla \psi||_2=kme^2/\hbar^2$ and $\text {inf}(\psi|\hat H_{\text{Hydrogen}}|\psi)=-k^2me^4/2\hbar^2$, so you can put them back into expectation value equality: $$-k^2me^4/2\hbar^2=\frac{{\hbar}^2}{2m}\centerdot (kme^2/\hbar^2)^2 - ke^2 \int_{R^3} \frac{1}{|x|}| \psi|^2dx$$ You have $$<\frac{1}{ |x|}>=\int_{R^3} \frac{1}{|x|}| \psi|^2dx=kme^2/\hbar^2=1/a_0$$ That is, the Bohr's radius! $$a_0=\frac{\hbar^2}{kme^2}.$$

You can get the same result by solvin Schrödinger's equation for Coulomb potential, but it's tedious and difficult task...


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