research44 min studyLesson 65 in Physics

Hartree–Fock and variational states

Optimize an antisymmetric product state, separate direct and exchange energies, and diagnose the limits of self-consistent mean field.

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What you’ll understand

  • Explain the variational upper bound and the restriction to a single Slater determinant.
  • Calculate direct and exchange contributions without double counting electron pairs.
  • Distinguish self-consistency, stationarity, stability, and physical accuracy.

Exact interacting electronic states quickly become too large to represent. Hartree–Fock restricts the trial family to normalized Slater determinants of orthonormal spin-orbitals, then minimizes the actual Hamiltonian’s expectation within that family. Each optimized orbital responds to the others’ average occupation. Understanding the optimization and the excluded correlations makes the method a useful starting point rather than a recipe for mysterious orbital energies.

The trial family makes the approximation

Expanding a normalized trial in exact energy eigenstates makes its expectation an average of eigenvalues, bounded below by the ground-state energy. A single Slater determinant automatically imposes fermionic antisymmetry but restricts the allowed many-electron states. Hartree–Fock seeks the lowest expectation among such determinants for a fixed Hamiltonian and one-particle basis. Enlarging the basis can improve that minimum; multiple determinants can capture correlations unavailable through orbital optimization alone.

The upper bound concerns the total ground-state energy of the specified Hamiltonian. It does not guarantee an upper bound for each orbital eigenvalue, an accurate excitation gap, or the correct density in every region. It also does not make a variational result for a different approximate Hamiltonian comparable without qualification. Nuclear positions are usually fixed under a Born–Oppenheimer electronic approximation, and basis truncation introduces an additional source of error. Keep those modeling choices separate from the approximation of using a single determinant.

E[Φ]=⟨Φ∣H∣Φ⟩⟨Φ∣Φ⟩≥E0,⟨ϕi∣ϕj⟩=δijE[\Phi]=\frac{\langle\Phi|H|\Phi\rangle}{\langle\Phi|\Phi\rangle}\ge E_0,\qquad \langle\phi_i|\phi_j\rangle=\delta_{ij}
The inequality applies to trial states in the domain of a Hamiltonian bounded below. Hartree–Fock restricts Φ to Slater determinants made from orthonormal spin-orbitals.↔ Scroll wide equations horizontally.

Direct and exchange terms come from the same interaction

For occupied spin-orbitals, the one-electron contribution is a sum of kinetic and external-potential expectations. Each pair also contributes a direct Coulomb integral and an exchange integral. The direct term resembles the electrostatic interaction between orbital densities. Exchange results from antisymmetrizing the state; it is not a new classical force added to Coulomb’s law. For a spin-independent interaction, spin-orthogonal orbitals have zero exchange integral because their spin overlap vanishes, although their direct repulsion remains.

The factor of one half in the double sum prevents counting every pair twice. Terms with identical orbital indices cancel between direct and exchange contributions, removing spurious self-interaction for that occupied spin-orbital in Hartree–Fock. This cancellation does not capture all correlation effects. For example, correlated electrons can avoid one another dynamically in ways unavailable to a single determinant. The exchange hole imposed by antisymmetry and the additional correlation hole are conceptually different even though both affect spatial probabilities.

EHF=∑ihii+12∑i,j occ(Jij−Kij),Jij=⟨ij∣v∣ij⟩,Kij=⟨ij∣v∣ji⟩E_{\mathrm{HF}}=\sum_i h_{ii}+\frac12\sum_{i,j\,\mathrm{occ}}(J_{ij}-K_{ij}),\qquad J_{ij}=\langle ij|v|ij\rangle,\quad K_{ij}=\langle ij|v|ji\rangle
These are spin-orbital matrix elements of a symmetric two-particle interaction. Each sum runs over occupied orbitals, and identical-index direct and exchange terms cancel.↔ Scroll wide equations horizontally.

Self-consistency is necessary but not sufficient

Varying the occupied orbitals while enforcing orthonormality produces a one-particle Fock operator that itself depends on the occupied orbitals. A self-consistent iteration builds this operator from a guessed density, solves an orbital problem, and updates the density. Convergence means that the chosen update rule has reached a fixed point within tolerance. There can be multiple stationary solutions, including saddle points and metastable states; a converged iteration need not identify the lowest determinant. Energy comparisons, orbital-gradient residuals, and stability analysis answer different questions.

Unrestricted orbitals can lower a variational energy by breaking a symmetry that the exact finite-system ground state preserves. This can diagnose a tendency without proving symmetry breaking or the correct phase. Bond dissociation and near-degeneracy can expose the single-determinant restriction. The optimization lab illustrates only gradient descent on a scalar quadratic. Its stable step-size range makes a useful warning about iterative convergence, but it contains no orbital orthogonality, density matrix, exchange integrals, or self-consistent Fock operator. A lower numerical residual should never be confused with smaller model error.

Count one pair exactly once

Two occupied orthonormal spin-orbitals have h11=−5.0 eVh_{11}=-5.0\,\mathrm{eV}, h22=−3.0 eVh_{22}=-3.0\,\mathrm{eV}, direct integral J12=2.0 eVJ_{12}=2.0\,\mathrm{eV}, and exchange integral K12=0.50 eVK_{12}=0.50\,\mathrm{eV}. Find the determinant energy. Then compare with a specified alternative determinant having the same h and J but zero exchange.

  1. The one-electron sum is h11+h22=−8.0 eVh_{11}+h_{22}=-8.0\,\mathrm{eV}. Terms with equal indices cancel between J and K, so they add no self-repulsion.

  2. The two ordered cross terms are identical. Their factor of one half leaves one pair contribution: J12−K12=1.50 eVJ_{12}-K_{12}=1.50\,\mathrm{eV}. Therefore the total is −8.0+1.50=−6.50 eV-8.0+1.50=-6.50\,\mathrm{eV}.

  3. For the alternative determinant with the explicitly stipulated same one-electron and direct integrals but zero exchange, the total is −8.0+2.0=−6.0 eV-8.0+2.0=-6.0\,\mathrm{eV}. The specified exchange contribution lowers the first determinant by 0.50 eV0.50\,\mathrm{eV}.

The result

The determinant energy is −6.50 eV. This controlled comparison isolates exchange; it is not a universal rule for energy ordering after orbitals and interactions readjust.

Explore the model

Interactive lab

Interactive lab · optimization

Iteration stability before self-consistency

Inspect gradient descent on the scalar objective f(x)=x2/2f(x)=x^2/2. Compare a step size below two with one above two and observe convergence or divergence from a nonzero initial position. The lab is a numerical analogy only: it does not optimize a Slater determinant or perform Hartree–Fock self-consistency.

Iteration stability before self-consistencyGradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Current readings: Position x0: 2; Objective x²/2: 2; Iteration behavior: Stable step. All parameters are adjustable with the labeled controls below.−4−202402.174.356.528.7Parameter xf(x) = x²/2n = 0
ObjectiveDiscrete iterate
Position x0
2
Objective x²/2
2
Iteration behavior
Stable step
n = 0 · t = 0 s

Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn iteration can fail because of its update rule, and a converged update can still belong to an inaccurate trial family.

Check your understanding

What does a converged Hartree–Fock self-consistent-field calculation establish by itself?

Try it yourself

Two separate normalized determinant trials for the same Hamiltonian give total energies −10.2 eV-10.2\,\mathrm{eV} and −10.5 eV-10.5\,\mathrm{eV}. State what follows for the exact ground-state energy. Does the lower determinant prove that its orbital eigenvalues predict exact excitation energies?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Apply the variational inequality to each total expectation. Avoid converting a bound on one scalar into a statement about an entire spectrum.

Compare with the solution

The lower trial gives the stronger upper bound, E0≤−10.5 eVE_0\le-10.5\,\mathrm{eV}. Both trials could still lie substantially above the exact energy, so their difference is not an error bar on the missing correlation. The orbital eigenvalues are eigenvalues of a state-dependent effective operator rather than the full many-body Hamiltonian. They therefore do not become exact excitation energies merely because the total variational expectation is lower.

Take it with you

  • Hartree–Fock optimizes an antisymmetric single-determinant family and supplies a variational total-energy upper bound.
  • Exchange follows from fermionic antisymmetry, while the pair factor prevents double counting.
  • Self-consistency, local stability, global minimization, and physical accuracy require separate checks.

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