This is a hydrogen atom — the actual cloud of where its one electron is. Drag the cloud to rotate it. The round s, the dumbbell p, the cloverleaf d: each is a real shape, and the dark gaps are nodes — surfaces where the electron is never found. The quiet rail on the left is its energy; the small inset is the radial wave whose zeros are those gaps. (ψ = R(r)·Y(θ,φ) — the angular part exact, the radial part a from-scratch eigensolve; open the badge for the proof.)
Pick 2s, read its energy; pick 2p — the same number, and the badge says coincident. The s, p, d of a shell sit at one energy (n=2 at −⅛, n=3 at −1/18), and the eigensolver was never told to — each ℓ-channel is solved separately and they land together anyway. That is the Coulomb degeneracy accident, special to the −1/r law. Now drag the screening slider: the funnel's tail gets eaten (a Yukawa −e−κr/r), and the coincident rungs fan apart — the s drops most (it dives deepest into the un-screened core), the d least. Snap ↺ and the teeth close back into one rung.
Scope & honesty. The eigensolve energies match the Rydberg ladder −1/(2n²) to a tolerance (~1e-3 relative), not the last bit — finite-box truncation + the grid's O(h²) discretization + the −1/r cusp at r→0 (worst for the 1s). The tolerance tightens as the grid refines (the self-test shows the error shrinking). The n²-fold degeneracy is two parts: the ℓ-part (s/p/d of a shell coincide) we solve here numerically; the m-part (each ℓ carries 2ℓ+1 equal-energy m-states) is fixed by spherical symmetry and is cited, not re-derived. n² = Σℓ=0n−1(2ℓ+1).
↘ next drift over · the flat-well cousins Where this came from → the Particle in a Box and the Harmonic Oscillator are the FLAT-bottomed wells. Curve the box's flat floor into −1/r and the box's single ladder splits into shells — the ℓ-degeneracy is what the curvature buys you. Same tridiagonal eigensolve, a non-constant diagonal. ↗ one wing over · The Spectroscope This ladder is where the colours come from → the gaps between these rungs are photons. Drop from n=3 to n=2 and 1/λ = RH(¼ − 1/n²) turns the very gap we solve here into Hα at 656 nm — the red line of hydrogen.