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

The Variational Principle by Hand — Reading a physical constant out of the optimum

Exercises

Quantum Chemistry I › Unit 2 · The variational game › The Variational Principle by Hand › all problems

Practice Problem 5 of 8

Reading a physical constant out of the optimum

Problem

From the helium result

ζ^{⋆} = 27/16

, the effective nuclear charge each electron feels is

Z_{eff} = ζ^{⋆}

. What ground-state energy does a single electron in a

Z_{eff}

-charged hydrogenic 1s orbital have, and why is twice that not the helium energy?

Solution

E_{1 elec} = - \frac{1}{2} Z_{eff}^{2} = - \frac{1}{2} (1.6875)^{2} = - 1.424 hartree

Twice is $- 2.848$ , which matches $E (ζ^{⋆})$ — but only because the screened one-electron picture has already folded the repulsion into $Z_{eff}$ . Adding the bare $⟨ V_{ee} ⟩$ on top would double-count it. $Z_{eff}$ is a bookkeeping device, not an independent extra energy.