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advanced27 min studyLesson 509 in Physics

Build a ground-state density through auxiliary orbitals

Separate the exact density-functional framework from functional approximations and eigenvalue interpretation.

Before you begin
On this page

The question for today

What you’ll understand

  • Identify the terms in a Kohn–Sham energy functional.
  • Explain density self-consistency and Hartree double counting.
  • Distinguish Kohn–Sham bands from general excitation energies.

Realistic band calculations must account for electrons interacting with one another and with nuclei. Kohn–Sham density-functional theory replaces the interacting ground-state problem with auxiliary one-electron equations that reproduce a density when the exact functional is available. Practical calculations approximate part of that functional and solve a nonlinear self-consistency problem. The auxiliary eigenvalues are valuable information, but their smooth appearance and numerical convergence do not make every energy difference an exact physical excitation.

The density is the central variable

For fixed nuclei and a specified electronic Hamiltonian, ground-state density-functional theory expresses energy as a functional of electron density. In the Kohn–Sham construction, a noninteracting kinetic-energy functional is evaluated through orthonormal auxiliary orbitals. Add the external-potential energy, the classical Hartree electron interaction, and an exchange-correlation functional that contains the remaining interaction and kinetic contributions. A nuclear repulsion term is added separately when comparing complete fixed-geometry total energies.

The Hartree energy includes a factor one half because integrating density against its own Coulomb potential counts every electron-pair interaction twice. Taking its density derivative removes that half and produces the full Hartree potential in each orbital equation. Exchange-correlation is not just a correction to the external ionic potential; it compensates for the many-body physics missing from the chosen auxiliary kinetic and classical interaction terms. Its exact form is generally unknown for realistic systems.

E[n]=Ts[n]+∫vext(r)n(r) d3r+EH[n]+Exc[n],EH=12∫n(r)vH(r) d3rE[n]=T_s[n]+\int v_{\rm ext}(\mathbf r)n(\mathbf r)\,d^3r+E_H[n]+E_{\rm xc}[n],\qquad E_H=\frac12\int n(\mathbf r)v_H(\mathbf r)\,d^3r
n is electron number density in m⁻³, the v terms are one-electron potential energies in J, and every E term is an energy. Fixed nuclear repulsion is omitted here and must be restored for a full total-energy comparison.↔ Scroll wide equations horizontally.

Self-consistency closes the auxiliary problem

Vary the orbitals subject to orthonormality. The resulting equations contain the external potential, the Hartree potential of the current density, and the functional derivative of exchange-correlation energy. Fill the auxiliary states with occupations summing to the specified electron number, reconstruct density, and repeat until the input and output agree. Periodicity allows the same Bloch organization used for a one-electron Hamiltonian, but now the periodic potential depends on the density being solved for.

Density mixing can stabilize this fixed-point iteration, yet a converged residual does not by itself establish the global ground state. Competing spin, orbital, or structural configurations may yield different stationary solutions. Likewise, convergence of the nonlinear loop does not prove convergence of the basis size or reciprocal-space integration. These are separate numerical questions that should be checked while holding the physical approximation, geometry, electron number, and symmetry assumptions fixed.

[−ℏ2∇22me+vext+vH[n]+vxc[n]]ϕi=ϵiϕi,n=∑ifi∣ϕi∣2,vxc=δExcδn\left[-\frac{\hbar^2\nabla^2}{2m_e}+v_{\rm ext}+v_H[n]+v_{\rm xc}[n]\right]\phi_i=\epsilon_i\phi_i,\qquad n=\sum_i f_i|\phi_i|^2,\quad v_{\rm xc}=\frac{\delta E_{\rm xc}}{\delta n}
φi are normalized auxiliary orbitals, fi occupations whose sum is electron number, and εi auxiliary eigenvalues in matching energy units. The scalar nonrelativistic notation omits spin and nonlocal-potential extensions for clarity.↔ Scroll wide equations horizontally.

Eigenvalue sums are not total energies

Summing occupied Kohn–Sham eigenvalues includes the Hartree potential twice relative to the Hartree energy and includes the exchange-correlation potential integral rather than the exchange-correlation energy itself. Subtract and restore the appropriate terms to recover the energy functional. The same caution applies to interpreting bands: a ground-state density representation does not make all auxiliary orbital differences equal to quasiparticle addition/removal energies or optical transitions. Approximate functionals, derivative discontinuities, and electron-hole attraction can each matter to a reported gap.

The optimization lab illustrates iteration behavior only. Reset and set step size to 0.5 and initial position to 1, then Step once; predict x=0.5. Reset and set step size to 1.5 and initial position to 1, then Step once; predict x=−0.5. Both have reduced magnitude, but their signs differ. This quadratic descent is not a Kohn–Sham density map, a variational density functional, or evidence that any electronic calculation has reached its ground state.

E=∑ifiϵi−EH+Exc−∫vxcn d3rE=\sum_i f_i\epsilon_i-E_H+E_{\rm xc}-\int v_{\rm xc}n\,d^3r
This electronic-energy identity uses the same functional and orbitals as the preceding equations, with fixed nuclear repulsion still omitted. All terms have energy units; arbitrary unrelated eigenvalue sums cannot be inserted.↔ Scroll wide equations horizontally.

Work it through

Audit the double counting

For an illustrative self-consistent electronic state per cell, let Ts=3 eV, external energy −8 eV, EH=4 eV, Exc=−1 eV, and the integral of vxc n be −1.5 eV. Find the electronic energy directly and from the occupied eigenvalue sum.

  1. The density functional gives E=3−8+4−1=−2 eV per cell. This is the electronic part only; no nuclear repulsion value has been specified.

  2. The occupied eigenvalue sum is Ts plus external energy plus twice EH plus the vxc integral: 3−8+8−1.5=1.5 eV. Its positive value does not contradict the negative electronic energy because these quantities count interactions differently.

  3. Correct that sum using the energy identity: 1.5−4−1−(−1.5)=−2 eV. This reproduces the direct result and explicitly checks both the Hartree half factor and the distinct exchange-correlation energy and potential integral.

The result

The electronic energy is −2 eV per cell. The occupied eigenvalues sum to 1.5 eV and require the stated double-counting corrections.

The living notebook

Explore the model

Interactive lab

Interactive lab · optimization

Inspect a simple contraction

Reset and set η=0.5, x₀=1, then Step once; predict x₁=0.5. Reset and set η=1.5, x₀=1, then Step once; predict x₁=−0.5.

Inspect a simple contractionGradient 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 forThe instrument performs gradient descent on x²/2. It is an iteration analogy, not density mixing, an electronic energy minimizer, or a Kohn–Sham solver.

Pause & reason

Check your understanding

Why is the sum of occupied Kohn–Sham eigenvalues generally different from the electronic total energy?

Your turn

Try it yourself

A dimensionless illustrative density-map coordinate obeys xout=0.8−0.5xin. Mix with α=0.5 using xnew=(1−α)xin+αxout, starting at x=0.4. Find two iterates and the fixed point. Does convergence validate a chosen exchange-correlation functional?

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

Give me a hint

Combine the affine terms first; solve the fixed-point equation separately from the iteration.

Compare with the solution

The mixed update is xnew=0.4+0.25xin. Starting at 0.4 gives 0.5 and then 0.525. The fixed point satisfies x=0.8−0.5x, so x=8/15≈0.5333. The map is a contraction in this toy coordinate. Converging a numerical map does not establish the physical accuracy of an exchange-correlation functional or identify the global electronic minimum.

Keep the idea

Take it with you

  • Kohn–Sham orbitals are auxiliary tools for a ground-state density.
  • Hartree and exchange-correlation double counting must be corrected in eigenvalue sums.
  • Self-consistency, numerical convergence, and excitation accuracy are different questions.

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