3. Hartree–Fock from constrained variation
Position in the course: Lesson 3 of 10. Complete the preceding derivation and use the explained exercises to check understanding.
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.
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 \(1/2\).
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.
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.
Teaching schematic drawn from the equations or algorithm steps in this lesson; not measured data.
5. 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.
6. 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.
7. Reference symmetry and orbital interpretation
Restricted closed-shell HF uses the same spatial function for alpha and beta partners. Unrestricted HF allows different spatial functions, enlarging the variational class and sometimes lowering energy. The unrestricted determinant need not be an eigenfunction of total \(S^2\), however. For collinear orbitals its spin expectation is
in units of \(\hbar^2\), with \(N_\alpha\geq N_\beta\). A closed-shell matched set has \(S_z=0\) and the overlap sum equals \(N_\beta\), giving zero. Diverging orbital sets can increase the expectation above the desired spin value. Lower unrestricted energy therefore trades one constraint for another and must be interpreted alongside spin contamination and stability.
Canonical orbital eigenvalues are Lagrange multipliers of a variational problem. Koopmans' frozen-orbital estimate associates \(-\epsilon_i\) with removal of an occupied electron, neglecting orbital relaxation and correlation. It is not an exact statement that every orbital energy is a measured ionization energy. Virtual orbital energies are especially unsuitable as universal electron-affinity predictions.
Occupied unitary rotations leave the determinant density and total energy unchanged, apart from an irrelevant determinant phase. Localized occupied orbitals and canonical occupied orbitals can therefore describe the same HF state. Chemical interpretation may improve with localization, but the total state has not acquired extra correlation.
Additional exercise: In a two-electron singlet unrestricted reference with alpha/beta spatial overlap \(s\), the formula gives \(\langle S^2\rangle=1-|s|^2\). Identical orbitals give zero; nonoverlapping orbitals give one, not the pure triplet value two. The latter determinant is a mixture of spin symmetries.
8. References and further reading
Derivations and numerical examples here are original teaching constructions, not copied passages or reported research data.
9. Related theory and practice
Quantum mechanics · Density functional theory · Quantum Monte Carlo · Gaussian