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

Derive the Fock Equations — Multipliers vs. energies

Exercises

Quantum Chemistry I › Unit 7 · The Fock problem › Derive the Fock Equations › all problems

Check Problem 9 of 9

Multipliers vs. energies

Problem

The

ε

’s entered as Lagrange multipliers, not energies. In 100–300 words: what physical meaning do they acquire (and under which choice), and why does

\sum_{k} 2 ε_{k}

nonetheless fail to be the electronic energy?

Solution

As multipliers, the $ε$ ’s are bookkeeping devices enforcing orthonormality, defined only up to the unitary freedom among occupied orbitals; nothing about them is observable. Choosing the canonical basis (diagonal $ε$ ) gives them content: by Koopmans, $- ε_{k}$ approximates the cost of removing electron $k$ with the rest frozen.

But each $ε_{k}$ already contains that electron’s full interaction with all the others, so $2 \sum_{k} ε_{k}$ counts every pair interaction twice — $E = 2 \sum_{k} ε_{k} - V_{ee}$ , the same double-counting the SCF energy formula corrects with its $\frac{1}{2}$ . Multipliers answer “what does orthonormality cost here”; only differences and removals of them answer physical questions.