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

Derive the Fock Equations — From the energy functional to the eigenvalue equation

Exercises

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

Worked example Problem 1 of 9

From the energy functional to the eigenvalue equation

Problem

The closed-shell Hartree–Fock energy, as a functional of the occupied orbitals

{ϕ_{i}}

, is

E = 2 \sum_{i} ⟨ ϕ_{i} ∣ \hat{h} ∣ ϕ_{i} ⟩ + \sum_{ij} (2 J_{ij} - K_{ij})

. Minimize it subject to

⟨ ϕ_{i} ∣ ϕ_{j} ⟩ = δ_{ij}

, showing every term of the variation, and arrive at the canonical eigenvalue equation.

Solution

The Lagrangian. Orthonormality is a constraint, so attach a multiplier $ε_{ji}$ to each $⟨ ϕ_{i} ∣ ϕ_{j} ⟩ - δ_{ij}$ (the factor of 2 is chosen to make the final equation clean):

L = 2 i \sum ⟨ ϕ_{i} ∣ \hat{h} ∣ ϕ_{i} ⟩ + ij \sum (2 J_{ij} - K_{ij}) - 2 ij \sum ε_{ji} (⟨ ϕ_{i} ∣ ϕ_{j} ⟩ - δ_{ij})

Make $L$ stationary: vary one orbital, $ϕ_{k} \to ϕ_{k} + δ ϕ$ , keep terms first order in $δ ϕ$ , and collect the coefficient of the bra $⟨ δ ϕ ∣$ . (Treat $δ ϕ$ and $δ ϕ^{*}$ as independent.) That coefficient must vanish for every $δ ϕ$ .

One-electron term. Only the $i = k$ piece of $2 \sum_{i} ⟨ ϕ_{i} ∣ \hat{h} ∣ ϕ_{i} ⟩$ contains $ϕ_{k}$ , and its variation against the bra is one term:

δ [2 ⟨ ϕ_{k} ∣ \hat{h} ∣ ϕ_{k} ⟩] = 2 ⟨ δ ϕ ∣ \hat{h} ∣ ϕ_{k} ⟩

Coulomb term. $ϕ_{k}$ sits in $J_{ij} = \iint ∣ ϕ_{i} (1) ∣^{2} r_{12}^{- 1} ∣ ϕ_{j} (2) ∣^{2}$ in two places — as $i = k$ (electron 1) and as $j = k$ (electron 2) — and the two are equal since $J_{ij} = J_{ji}$ . Each varies $∣ ϕ_{k} ∣^{2}$ to first order:

δ [ij \sum 2 J_{ij}] = i = k 2 j \sum ⟨ δ ϕ ∣ \hat{J}_{j} ∣ ϕ_{k} ⟩ + j = k 2 i \sum ⟨ δ ϕ ∣ \hat{J}_{i} ∣ ϕ_{k} ⟩ = 4 j \sum ⟨ δ ϕ ∣ \hat{J}_{j} ∣ ϕ_{k} ⟩

with the Coulomb operator $\hat{J}_{j} (1) = \int ∣ ϕ_{j} (2) ∣^{2} r_{12}^{- 1} d 2$ — the electrostatic potential of orbital $j$ ’s charge cloud. There is the mean field, manufactured from the bare $1/ r_{12}$ by integrating the other electron out.

Exchange term. The same double appearance in $K_{ij} = \iint ϕ_{i}^{*} (1) ϕ_{j} (1) r_{12}^{- 1} ϕ_{j}^{*} (2) ϕ_{i} (2)$ gives, with the minus sign from $- K$ :

δ [- ij \sum K_{ij}] = - 2 j \sum ⟨ δ ϕ ∣ \hat{K}_{j} ∣ ϕ_{k} ⟩

where the exchange operator acts nonlocally, $\hat{K}_{j} (1) ϕ_{k} (1) = [\int ϕ_{j}^{*} (2) r_{12}^{- 1} ϕ_{k} (2) d 2] ϕ_{j} (1)$ — it swaps $ϕ_{k}$ for $ϕ_{j}$ under the integral. No classical analogue; it is the price of antisymmetry.

Constraint term. Only $i = k$ involves $ϕ_{k}$ :

δ [- 2 ij \sum ε_{ji} ⟨ ϕ_{i} ∣ ϕ_{j} ⟩] = - 2 j \sum ε_{jk} ⟨ δ ϕ ∣ ϕ_{j} ⟩

Collect. Stationarity ( $δ L = 0$ for arbitrary $⟨ δ ϕ ∣$ ) forces the bracketed coefficient to zero:

2 \hat{h} ϕ_{k} + 4 j \sum \hat{J}_{j} ϕ_{k} - 2 j \sum \hat{K}_{j} ϕ_{k} - 2 j \sum ε_{jk} ϕ_{j} = 0

Divide by 2 and group the operators into the Fock operator $\hat{F} = \hat{h} + \sum_{j} (2 \hat{J}_{j} - \hat{K}_{j})$ :

\hat{F} ϕ_{k} = j \sum ε_{jk} ϕ_{j}

Predict before reading on. This is not yet an eigenvalue equation — the right side couples all occupied orbitals through $ε_{jk}$ . What freedom remains to diagonalize the multiplier matrix?

Canonicalize. The determinant and the energy are invariant under any unitary mixing of the occupied orbitals, and $ε$ is Hermitian. Rotate to the basis that diagonalizes it, $ε_{jk} \to ε_{k} δ_{jk}$ . In those canonical orbitals:

\hat{F} ϕ_{k} = ε_{k} ϕ_{k}

The eigenvalue equation, earned. The mean field is exactly $\sum_{j} (2 \hat{J}_{j} - \hat{K}_{j})$ — produced by the one step where varying the two-electron energy integrated an electron out of $1/ r_{12}$ and left an operator acting on the other.

Result

\hat{F} ϕ_{k} = ε_{k} ϕ_{k}, \hat{F} = \hat{h} + j \sum (2 \hat{J}_{j} - \hat{K}_{j})