greatitself.
Continue
advanced28 min studyLesson 457 in Physics

Hartree–Fock atomic orbitals

Build the self-consistent determinant approximation and distinguish orbital energies from total atomic energies.

Before you begin
On this page

The question for today

What you’ll understand

  • Identify the variational restriction imposed by one Slater determinant.
  • Explain direct and exchange operators and their self-interaction cancellation.
  • Compute a total energy without double-counting electron pairs.

A many-electron wavefunction depends on all electron coordinates simultaneously. Hartree–Fock makes this problem manageable by restricting it to one antisymmetric product of spin-orbitals and optimizing those orbitals. The resulting mean field depends on the very orbitals being sought, so the calculation must be self-consistent. Its success is substantial, but a converged determinant is still an approximation to an interacting atom. Neither orbital eigenvalues nor a small iterative residual prove that all physically important correlation has been captured.

Optimize an antisymmetric trial family

Begin with a fixed-nucleus nonrelativistic electronic Hamiltonian containing one-electron kinetic and nuclear-attraction terms plus pairwise Coulomb repulsion. A normalized determinant of orthonormal spin-orbitals automatically changes sign when electron labels are exchanged. Varying its energy subject to orbital orthonormality gives coupled Fock equations. The determinant restriction is the physical approximation; variation within that family is exact in principle. Basis truncation adds a separate numerical approximation that should be tested by systematic enlargement.

For the ground state, any normalized trial determinant gives an upper bound to the exact ground energy of the same Hamiltonian. A variational stationary point need not be the lowest determinant: excited occupations and unstable solutions can also converge. Symmetry restrictions further narrow the variational space. Breaking spin or spatial symmetry can lower a determinant’s energy, but its labels then require care. Comparing bounds from different nuclear models or different relativistic Hamiltonians would not be the same variational test.

EHF=∑ihii+12∑i,j(Jij−Kij)≥EexactE_{\mathrm{HF}}=\sum_i h_{ii}+\frac12\sum_{i,j}(J_{ij}-K_{ij})\geq E_{\mathrm{exact}}
h_ii, J_ij, and K_ij are energies evaluated with occupied orthonormal spin-orbitals. The inequality refers to the ground energy of the same fixed-nucleus Hamiltonian, not an arbitrary excited state.↔ Scroll wide equations horizontally.

The Fock operator depends on its own solutions

Each occupied orbital contributes a direct Coulomb operator and an exchange operator. The direct term acts through the electron density; exchange is nonlocal and depends on overlap with an occupied spin-orbital. For opposite pure spin functions, the exchange integral vanishes through spin orthogonality. Including an orbital’s own contribution gives equal direct and exchange actions on that orbital, canceling its unphysical self-interaction within Hartree–Fock. Dropping exchange would spoil both antisymmetry effects and that cancellation.

A self-consistent procedure starts with trial orbitals, builds the Fock operator, solves its eigenproblem, chooses occupations, and rebuilds until the occupied space and energy stabilize. Direct iteration can oscillate, so practical calculations use mixing or more sophisticated convergence strategies. The mathematical analogy with an iterative fixed point does not imply that a scalar gradient-descent lab implements these steps. Even after numerical convergence, electron correlation beyond the determinant and neglected relativistic terms can dominate the remaining physical error.

fχi=ϵiχi,f=h+∑j occupied(Jj−Kj)f\chi_i=\epsilon_i\chi_i,\qquad f=h+\sum_{j\ \mathrm{occupied}}(J_j-K_j)
f, h, J_j, and K_j are energy operators; epsilon_i is a canonical orbital eigenvalue. The Coulomb/exchange operators are determined by all occupied spin-orbitals, making the equations nonlinear as a system.↔ Scroll wide equations horizontally.

Keep orbital and total energies distinct

Summing occupied eigenvalues counts each electron pair twice because each orbital experiences the mean field of the other. The total-energy expression must remove one copy. For two opposite-spin electrons in the same spatial orbital, the exchange between spins vanishes, leaving a single direct pair energy. This simple case is an excellent diagnostic: adding two orbital eigenvalues without correction gives one too many copies of the repulsion and can distort an ionization-energy estimate.

Koopmans’ approximation identifies the negative occupied canonical orbital energy with an ionization energy when the remaining orbitals are frozen. Actual ionization allows relaxation and changes correlation, so measured removal energies need not equal those eigenvalues. Virtual orbital eigenvalues have still weaker claims to measured electron affinities. A spectral transition requires a difference between appropriate many-electron states, potentially with different orbital optimization. Hartree–Fock supplies a disciplined starting point, not a declaration that electrons move independently in the exact atom.

EHF=∑iϵi−12∑i,j(Jij−Kij)E_{\mathrm{HF}}=\sum_i\epsilon_i-\frac12\sum_{i,j}(J_{ij}-K_{ij})
This identity uses canonical occupied Hartree–Fock spin-orbitals and the same integrals as the first equation. The one-half removes double counting; no nuclear–nuclear term occurs for one fixed atomic nucleus.↔ Scroll wide equations horizontally.

Work it through

Remove pair double counting

Two opposite-spin electrons occupy one spatial orbital with haa=−2.0 Ehh_{aa}=-2.0\,E_h and spatial Coulomb integral Jaa=1.2 EhJ_{aa}=1.2\,E_h. Find each occupied orbital energy and the determinant’s total energy; EhE_h denotes one hartree.

  1. Each spin-orbital’s self-interaction cancels between direct and exchange terms. Its other occupied partner has opposite spin, so their mutual exchange term is zero while the direct repulsion remains.

  2. Each occupied eigenvalue is therefore ϵ=−2.0+1.2=−0.8 Eh\epsilon=-2.0+1.2=-0.8\,E_h. Their sum, −1.6 Eh-1.6\,E_h, includes the same physical pair repulsion twice.

  3. The total energy contains two one-electron contributions and one pair: E=2(−2.0)+1.2=−2.8 EhE=2(-2.0)+1.2=-2.8\,E_h. Equivalently subtract one 1.2 Eh1.2\,E_h pair term from the eigenvalue sum.

The result

Each occupied eigenvalue is −0.8 hartree, while the total determinant energy is −2.8 hartree.

The living notebook

Explore the model

Interactive lab

Interactive lab · optimization

Convergence is a separate question

Reset and set step size η=0.5\eta=0.5, initial position x0=2x_0=2; after one Step predict x1=1x_1=1. Repeat from Reset with both controls η=2.5\eta=2.5, x0=2x_0=2; predict x1=−3x_1=-3. The lab iterates xn=(1−η)nx0x_n=(1-\eta)^nx_0 on a scalar quadratic, not a Fock operator. It illustrates why an iteration rule needs its own convergence analysis.

Convergence is a separate questionGradient 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 forA numerical iteration can converge or diverge independently of the physical adequacy of its chosen approximation.

Pause & reason

Check your understanding

Why is the Hartree–Fock total energy generally different from the sum of occupied orbital eigenvalues?

Your turn

Try it yourself

A converged two-electron calculation gives equal occupied eigenvalues −0.90 Eh-0.90\,E_h and one opposite-spin spatial Coulomb pair integral 1.10 Eh1.10\,E_h. Reconstruct its total energy and frozen-orbital ionization energy.

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

Give me a hint

Subtract one pair integral from the eigenvalue sum; Koopmans’ estimate uses the negative occupied eigenvalue.

Compare with the solution

The total energy is 2(−0.90)−1.10=−2.90 Eh2(-0.90)-1.10=-2.90\,E_h. The frozen-orbital ionization estimate is 0.90 Eh0.90\,E_h, not half the magnitude of the total energy. Removing an electron also permits the remaining orbital to relax, so an independently optimized ion energy can give a different removal energy. Correlation beyond Hartree–Fock supplies another physical correction even if the SCF iteration is fully converged.

Keep the idea

Take it with you

  • A single determinant enforces antisymmetry but omits correlation beyond that trial family.
  • Self-consistency solves coupled orbital equations and does not establish exact physical accuracy.
  • Total energies require a pair-counting correction to the occupied eigenvalue sum.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.