“Know how to solve every problem that has been solved.” “What I cannot create, I do not understand.” — Richard Feynman

Optimizing ζ: the whole method in one variable

Learn

Lesson 2 of 24 standard ~5 min

Let's actually run the minimization for hydrogen. For the trial ψ = e^{−ζ|r|}, both expectation values have closed forms, and they fight each other: the kinetic energy rises as ζ squeezes the orbital tighter, while the nuclear attraction deepens. The energy is their tug-of-war:

Minimize like any calculus problem: dE/dζ = ζ − 1 = 0, so ζ* = 1 and E(1) = −0.5 hartree. That is the hydrogen ground state — not approximately, exactly. Too small a ζ and the orbital is too diffuse to feel the nucleus; too large and the kinetic cost wins. The balance point is the physical atom.

standardMultiple choice

For hydrogen, the trial orbital ψ(r) = e^{−ζ|r|} with ζ as a parameter makes the variational calculation a…

Why exact? Because the search family contained the answer: e^{−|r|} is the true 1s orbital. Replace the Slater form with three Gaussians (STO-3G) and the family no longer contains the truth — the same minimization now bottoms out at −0.4666 hartree, 33 mHa short. The variational principle still holds; the search space just got poorer. That gap between −0.4666 and −0.5 is pure basis-set error, and it haunts everything downstream.

standardMultiple choice

Optimizing ζ in the trial ψ = e^{−ζ|r|} for hydrogen gives E = −0.5 hartree — the exact ground-state energy, exactly. Why does the variational method land on the truth here?