LESSON NOTES · 02

Hartree–Fock from constrained variation

Prerequisites: Slater determinants and variational principle

Prerequisite lessons: Variational principle and finite bases · Separate nuclei; antisymmetrize electrons

1. A bounded single-determinant approximation

HF minimizes energy within the chosen normalized single-determinant class. Antisymmetry supplies exchange but not all correlation. Use atomic units, fixed nuclei, occupied spin orbitals \(i,j\), arbitrary \(p,q\), and spin-inclusive \(x\). Define physicists’ integral convention explicitly.

\[ \langle pq|rs\rangle=\iint\varphi_p^*(x_1)\varphi_q^*(x_2)r_{12}^{-1}\varphi_r(x_1)\varphi_s(x_2)dx_1dx_2,\quad\langle pq\Vert rs\rangle=\langle pq|rs\rangle-\langle pq|sr\rangle. \]

2. Constrain every orbital overlap

The determinant energy has one-electron terms and half the sum of antisymmetrized pair integrals. Introduce a Hermitian multiplier matrix for pairwise orthonormality. Vary each conjugate orbital: the two symmetric contributions from the interaction cancel its factor \(1/2\).

\[ \begin{aligned}E_{HF}&=\sum_i h_{ii}+\tfrac12\sum_{ij}\langle ij\Vert ij\rangle+V_{NN},\\\mathcal L&=E_{HF}-\sum_{ij}\epsilon_{ij}(\langle\varphi_i|\varphi_j\rangle-\delta_{ij}),\\\delta\mathcal L/\delta\varphi_i^*&=0\Rightarrow f\varphi_i=\sum_j\epsilon_{ij}\varphi_j,\quad f=h+\sum_j(J_j-K_j).\end{aligned} \]

3. Coulomb and exchange act differently

Coulomb multiplies an orbital by the electrostatic potential of the occupied density. Exchange mixes it nonlocally with occupied orbitals. It is not a classical force. An occupied-space unitary rotation diagonalizes the multiplier matrix, yielding canonical orbitals in a self-consistent Fock field.

\[ \begin{aligned}J_j\varphi_i(x_1)&=\left[\int\frac{|\varphi_j(x_2)|^2}{r_{12}}dx_2\right]\varphi_i(x_1),\\K_j\varphi_i(x_1)&=\left[\int\frac{\varphi_j^*(x_2)\varphi_i(x_2)}{r_{12}}dx_2\right]\varphi_j(x_1),\\f\varphi_i&=\varepsilon_i\varphi_i.\end{aligned} \]

4. Correct the eigenvalue sum

Taking occupied expectations of the Fock equation counts each pair interaction twice. Half must be subtracted to recover determinant energy. Nuclear repulsion is then added. Orbital-energy sums alone are not molecular total energies.

\[ \sum_i\varepsilon_i=\sum_i h_{ii}+\sum_{ij}\langle ij\Vert ij\rangle,\quad E_{HF}=\sum_i\varepsilon_i-\tfrac12\sum_{ij}\langle ij\Vert ij\rangle+V_{NN}. \]

Teaching schematic drawn from the equations or algorithm steps in this lesson; not measured data.

Worked self-interaction and opposite-spin checks

For one electron, \(J_1\varphi_1=K_1\varphi_1\), so self-repulsion cancels and \(E=h_{11}+V_{NN}\). For two opposite-spin electrons in one spatial \(\phi\), spin orthogonality removes their exchange: \(E=2h_{\phi\phi}+J_{\phi\phi}+V_{NN}\) while each orbital energy is \(h_{\phi\phi}+J_{\phi\phi}\). Twice the orbital energy would double count the physical repulsion.

Check yourself and limits

Is exchange present between collinear \(\alpha\) and \(\beta\) spin orbitals?

Answer and reasoning

No, spin functions are orthogonal.

Does SCF convergence establish the global HF minimum?

Answer and reasoning

No; local minima and unstable stationary solutions exist. Restricted and unrestricted references impose different constraints, and stability can matter.

References and further reading

• MSU: Hartree–Fock equations

Derivations and numerical examples here are original teaching constructions, not copied passages or reported research data.

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