
Optional unit · institution-dependent · Interactive lesson
Hartree-Fock binding of a metal, and what correlation adds
Two terms from the last two topics and one derivative give an electron gas that holds itself together at the density of a real metal. That success is the bait. The same algebra makes cesium a magnet, and cesium is not one, so this is where a single determinant meets the correlation it leaves out.
UAQM-15.04 · Lesson 4 of 9 in this unit. Position is not a completion score.Optional unit · institution-dependent. Units 14 and 15 are the course’s optional extension into open systems and quantum gases. Unit 14 builds on radiation and decay; Unit 15 builds on second quantisation. Keep the fifteen-unit numbering and follow your institution’s required syllabus. Optional work is not a required completion gate.
Build the model
Exchange binds the electron gas; correlation decides whether it is a magnet.
The key idea
Hartree-Fock binding is the minimum of the energy per electron of jellium held in one Slater determinant, the density at which the slopes of the kinetic energy, 2.21/rₛ² Ry, and the exchange energy, −0.916/rₛ Ry, cancel.
Example. At rₛ = 4.83 the kinetic energy is 2.21/4.83² = 0.0947 Ry and the exchange −0.916/4.83 = −0.1896 Ry, so the gas is bound by 0.0949 Ry = 1.29 eV per electron, with exchange twice the kinetic energy in size.
The complete physical model and assumptions
Topics 15.02 and 15.03 gave the Fermi sea of jellium two energies per electron: a kinetic 2.21/rₛ² Ry that falls quickly as the gas expands and an exchange −0.916/rₛ Ry that falls slowly. Their sum must turn over, and a zero slope puts the minimum at rₛ = 4.83 with −0.095 Ry per electron, derived here, inside the range of measured alkali-metal densities. The same single determinant can lower its exchange further by aligning every spin, and because the kinetic price of doing so falls off faster with rₛ than the exchange reward, Hartree-Fock prefers a fully polarised gas beyond rₛ = 5.45: Bloch's ferromagnet, also derived, which cesium at rₛ = 5.62 declines to become. Both share one omission. A determinant keeps parallel spins apart through the exchange hole but lets antiparallel spins move independently, so the true ground state lies lower by the correlation energy, about −0.06 Ry at metallic densities: comparable with the binding, and nearly twice as large for the unpolarised gas as for the polarised one. Gell-Mann and Brueckner's high-density form, 0.0622 ln rₛ − 0.096 Ry, is quoted and fails before metallic densities are reached; quantum Monte Carlo energies, also quoted, move the minimum to rₛ ≈ 4.2, keep every metal's electron gas unpolarised and crystallise it only near rₛ = 100. Jellium stops short of cohesion: in a real metal the ions, not the gas, set the density.
1Where kinetic energy and exchange balance
Units are SI, with energies per electron in rydbergs, 1 Ry = ħ²/(2ma₀²) = 13.606 eV, and lengths in Bohr radii, a₀ = 52.9 pm; nothing here is relativistic, so the course's metric (+, −, −, −) never enters. From topics 15.02 and 15.03, used and not re-derived, one Slater determinant of plane waves gives E(rₛ) = 2.21/rₛ² − 0.916/rₛ Ry, the Hartree term having cancelled against the background. Set the slope to zero: −4.42/rₛ³ + 0.916/rₛ² = 0, so rₛ* = 4.42/0.916 = 4.83 and E* = −0.916²/(4 × 2.21) = −0.0949 Ry = −1.29 eV per electron. At the minimum the kinetic energy is 0.0949 Ry and the exchange −0.1898 Ry, exactly −2 times it: when one term scales as rₛ⁻² and the other as rₛ⁻¹, stationarity forces 2K + Eₓ = 0, the virial theorem of a Coulomb system. Two limits check the shape. Switch the charge off and exchange vanishes, leaving only a kinetic energy that falls monotonically as the gas expands: a free Fermi gas has no minimum and must be held in by walls. As rₛ → ∞ both terms vanish, so −0.095 Ry is counted from electrons and background spread to zero density.
2Sodium, potassium and the limits of a coincidence
The minimum, rₛ* = 4.83, is a density of 1.43 × 10²⁸ electrons m⁻³, and the measured conduction-electron densities of the alkali metals bracket it: rₛ = 3.25 for lithium, 3.93 for sodium, 4.86 for potassium and 5.62 for cesium (quoted). In that sense exchange alone binds the gas near a real metal's density. The agreement is partly luck, and one argument shows it. Jellium has a single parameter, so it has a single equilibrium density, yet the simple metals run from aluminium at rₛ = 2.07 to cesium at 5.62: their ions, not their electron gas, choose the density. Nor is −1.29 eV a cohesive energy. Sodium's measured 1.11 eV per atom is counted from free neutral atoms, the jellium energy from electrons and background spread to zero density, and the two share no zero. Restoring the ions as point charges adds −1.8/rₛ Ry per electron, about twice the exchange term, and only the repulsion of the filled ion cores keeps the metal from shrinking to many times its measured density.
3Aligning every spin: Bloch's ferromagnet
Now put all N electrons in one spin state. Each k-state holds one electron instead of two, so the occupied sphere doubles its volume and kF grows by 2¹⁄³. Kinetic energy per electron, three fifths of the Fermi energy and so proportional to kF², rises by 2²⁄³ = 1.587. Exchange acts only between parallel spins, and the exchange energy per electron is proportional to the kF of the electron's own spin species, so it rises by 2¹⁄³ = 1.260. Hence EP = 3.508/rₛ² − 1.154/rₛ Ry, derived. Polarising costs 1.298/rₛ² of kinetic energy and earns 0.238/rₛ of exchange. The cost has over five times the coefficient but falls off faster, so the two cross at rₛ = 1.298/0.238 = 5.45, which can be written exactly as (1 + 2¹⁄³)(2.21/0.916), because (2²⁄³ − 1)/(2¹⁄³ − 1) = 1 + 2¹⁄³. Beyond it Hartree-Fock prefers every spin aligned: the ferromagnetism of the free electron gas that Bloch found in 1929. Two checks: the threshold is a ratio of coefficients, dimensionless like rₛ itself, and at rₛ* = 4.83 the polarised gas still lies 6.4 mRy higher, so the bound minimum is unpolarised.
4A first-order jump, and cesium's verdict
Partial polarisation ζ = (N↑ − N↓)/N interpolates. Each spin species fills its own Fermi sphere, so the kinetic and exchange energies acquire the factors [(1 + ζ)⁵⁄³ + (1 − ζ)⁵⁄³]/2 and [(1 + ζ)⁴⁄³ + (1 − ζ)⁴⁄³]/2, which grow as 1 + 5ζ²/9 and 1 + 2ζ²/9. Near ζ = 0 the energy is E(0) + (ζ²/9)(11.05/rₛ² − 1.832/rₛ) Ry, and the unpolarised gas stays a local minimum until rₛ = 6.03. Between 5.45 and 6.03 it is metastable while the fully polarised gas is lower, and no intermediate ζ is ever lowest, so Hartree-Fock predicts a first-order jump from ζ = 0 to ζ = 1, not a gradual onset. Cesium is the test. Its measured density gives rₛ = 5.62, past the threshold, where Hartree-Fock wants a saturated ferromagnet with μ₀M = μ₀nμB ≈ 0.11 T. Cesium is measured to be a weakly paramagnetic metal with no spontaneous moment (quoted). One determinant has answered a qualitative question wrongly.
5Correlation: what one determinant leaves out
Unit 10 defined the correlation energy, per electron here, as εc = Eexact − EHF. The plane-wave determinant is a trial state, not even the lowest single determinant (Overhauser showed in 1962 that spin-density-wave determinants lie lower, quoted), and the variational principle bounds every trial state from above, so εc ≤ 0 at every density. Its origin is the pair the determinant ignores. Parallel spins are kept apart by the exchange hole, but antiparallel spins are left uncorrelated, each blind to where the other is; the true ground state digs a Coulomb hole in the opposite-spin density too. At high density the result is Gell-Mann and Brueckner's εc = 0.0622 ln rₛ − 0.096 Ry, quoted. Plain second-order perturbation theory diverges, because the Coulomb transform e²/(ε₀q²) grows without limit at small q; only after the most divergent terms are summed, which amounts to the gas screening its own interaction, does a logarithm survive. The limit checks: as rₛ → 0 the correction grows only as ln rₛ while the kinetic energy grows as rₛ⁻², so Hartree-Fock becomes exact at high density. The formula turns positive at rₛ = e(0.096/0.0622) = 4.68, which the bound forbids.
6Monte Carlo numbers: the minimum, the magnet and the crystal
At metallic densities correlation has to be computed, and the standard numbers are Ceperley and Alder's quantum Monte Carlo energies of 1980, quoted through Perdew and Zunger's fit: εc = −0.2846/(1 + 1.0529√rₛ + 0.3334rₛ) Ry for the unpolarised gas. At rₛ = 4 that is −0.064 Ry, 70 per cent of the Hartree-Fock energy there and six and a half times the Gell-Mann–Brueckner value of −0.0098 Ry. Because it weakens as rₛ grows, it tilts the curve and pulls the minimum in from 4.83 to rₛ = 4.18, at −0.155 Ry. It also decides the magnetism. For the polarised gas the fit is −0.1686/(1 + 1.3981√rₛ + 0.2611rₛ) Ry, far smaller, because the exchange hole had already kept parallel spins apart: at rₛ = 5.45 the two are −0.054 and −0.030 Ry, and a 0.024 Ry preference for the unpolarised gas swamps Hartree-Fock's millirydberg preference for alignment. The gas stays unpolarised across the metallic range and far beyond it. Only near rₛ = 100 does the potential energy win outright and freeze the electrons into Wigner's crystal (quoted).
Key equations
Read the units, assumptions and limits before substituting.
Equilibrium density in Hartree-Fock
rₛ* = 2(2.21)/0.916 = 4.83, from dE/drₛ = −4.42/rₛ³ + 0.916/rₛ² = 0
Derived. Exchange falls as rₛ⁻¹, more slowly than the kinetic rₛ⁻², so it must win at low density and a minimum must exist: n = 1.43 × 10²⁸ electrons m⁻³.
rₛ = r₀/a₀ is dimensionless, with (4π/3)r₀³ = 1/n and a₀ = 52.9 pm; E is in Ry per electronBinding at the minimum
E* = −0.916²/(4 × 2.21) Ry = −0.0949 Ry = −1.29 eV
Derived. At rₛ* the exchange is exactly −2 times the kinetic energy, the virial theorem for terms scaling as rₛ⁻¹ and rₛ⁻², so E* = −K* = Eₓ*/2.
per electron, measured from the gas spread to infinite rₛ; 1 Ry = 13.606 eV = 2.180 × 10⁻¹⁸ JFully polarised gas, ζ = 1
EP = 2²⁄³(2.21)/rₛ² − 2¹⁄³(0.916)/rₛ = 3.508/rₛ² − 1.154/rₛ Ry
Derived. Kinetic energy per electron goes as kF², exchange per electron as the kF of the electron's own spin species, so aligning the spins raises them by different factors.
ζ = (N↑ − N↓)/N; one spin species holding all N electrons has a Fermi wave number kF larger by 2¹⁄³Bloch's Hartree-Fock ferromagnetism
EP − EU = 1.298/rₛ² − 0.238/rₛ Ry, negative beyond rₛ = (1 + 2¹⁄³)(2.21/0.916) = 5.45
Derived. The kinetic cost has the larger coefficient but falls faster. Cesium, at rₛ = 5.62, is past the threshold, yet it is measured to be a paramagnet.
EU and EP are the energies per electron of the unpolarised and fully polarised gas at the same density, in RyStiffness against partial polarisation
E(ζ) − E(0) ≈ (ζ²/9)(11.05/rₛ² − 1.832/rₛ) Ry, negative beyond rₛ = 6.03
Derived. Between rₛ = 5.45 and 6.03 the unpolarised gas is locally stable but not lowest, so the Hartree-Fock transition is a first-order jump from ζ = 0 to ζ = 1.
small ζ; expands the spin-resolved factors (1 ± ζ)⁵⁄³ and (1 ± ζ)⁴⁄³ to second orderGell-Mann–Brueckner correlation energy
εc = 0.0622 ln rₛ − 0.096 Ry per electron, for rₛ ≪ 1
Quoted (1957). The variational principle makes εc ≤ 0, so its turning positive at rₛ = 4.68 shows the expansion has already failed; metals need Monte Carlo values.
εc = Eexact − EHF per electron; 0.0622 Ry = 2(1 − ln 2)/π² Ry, halved to 0.0311 ln rₛ when fully polarisedInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. At λ = 0, slide rₛ through 4.83, where the pressure changes sign, and past 5.45, where the dashed curve dips below the solid: Hartree-Fock makes cesium a magnet. Then raise λ to 1: the minimum moves in to about 4.2 and the dashed curve stays above the solid everywhere.
Live results
- E, UNPOLARISED
- -0.0900 Ry
- E, FULLY POLARISED
- -0.0665 Ry
- POLARISED MINUS UNPOLARISED
- 23.5 mRy
- PRESSURE, UNPOLARISED GAS
- 1.02 GPa
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Density parameter rₛ: 3.93; Correlation energy added, fraction λ: 0.0
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Plugging rₛ = 4 into Gell-Mann and Brueckner's formula gives εc = 0.0622 ln 4 − 0.096 = −0.0098 Ry, about a tenth of the Hartree-Fock energy, and invites the conclusion that correlation is a small correction and Hartree-Fock's predictions stand. Both halves are wrong. The formula is the leading term of an expansion valid for rₛ ≪ 1; at rₛ = 4 it is outside its range, and beyond rₛ = 4.68 it even turns positive, which the variational bound εc ≤ 0 forbids. The quoted Monte Carlo value at rₛ = 4 is −0.064 Ry, 70 per cent of the Hartree-Fock energy. And a correction can be small against the total yet large against the difference that settles a question: near cesium's density Hartree-Fock separates the polarised and unpolarised gas by about a millirydberg, while correlation separates them by 0.024 Ry, in the opposite direction.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Potassium's measured conduction-electron density is 1.40 × 10²⁸ m⁻³. Find rₛ, the Hartree-Fock kinetic, exchange and total energies per electron of jellium at that density, and decide whether the Hartree-Fock gas would be in equilibrium there.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- r₀ = (3/(4πn))¹⁄³ = (3/(4π × 1.40 × 10²⁸ m⁻³))¹⁄³ = 2.574 × 10⁻¹⁰ m, so rₛ = r₀/a₀ = 2.574 × 10⁻¹⁰ m / 5.292 × 10⁻¹¹ m = 4.86, a pure number as it must be.
- Kinetic: 2.21/4.86² = 0.0936 Ry. Exchange: −0.916/4.86 = −0.1885 Ry. Total: −0.0949 Ry, or −0.0949 × 13.606 eV = −1.29 eV per electron.
- Equilibrium test: a stationary point needs 2K + Eₓ = 0, that is K/|Eₓ| = 1/2. Here K/|Eₓ| = 0.09357/0.18848 = 0.496, and 2K + Eₓ = −0.0013 Ry, about 1% of either term.
- Equivalently, rₛ = 4.86 lies less than 1% beyond rₛ* = 4.83, and the energy matches E* = −0.0949 Ry to four decimal places, as it must near a minimum, where E changes only to second order in the displacement.
Answer. rₛ = 4.86 and E = −0.0949 Ry = −1.29 eV per electron (kinetic 0.0936 Ry, exchange −0.1885 Ry): potassium's density is the Hartree-Fock jellium minimum to within 1%.
MediumExample 2
Cesium's measured conduction-electron density is 0.91 × 10²⁸ m⁻³. (a) Which spin state does Hartree-Fock jellium prefer there, by how much per electron, and what magnetisation would follow? (b) Add the quoted Monte Carlo correlation energies at that density, −0.0530 Ry unpolarised and −0.0292 Ry fully polarised, and decide again.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- rₛ = (3/(4π × 0.91 × 10²⁸ m⁻³))¹⁄³/a₀ = 2.971 × 10⁻¹⁰ m / 5.292 × 10⁻¹¹ m = 5.62, past Bloch's threshold of 5.45.
- Unpolarised: 2.21/5.62² − 0.916/5.62 = 0.06997 − 0.16299 = −0.09302 Ry. Polarised: 3.508/5.62² − 1.154/5.62 = 0.11107 − 0.20534 = −0.09427 Ry.
- Hartree-Fock prefers ζ = 1, by 1.25 mRy = 17 meV per electron. Every conduction electron then carries μB the same way: M = nμB = 0.91 × 10²⁸ m⁻³ × 9.27 × 10⁻²⁴ J T⁻¹ = 8.4 × 10⁴ A m⁻¹, so μ₀M = 0.11 T.
- With correlation: unpolarised −0.09302 − 0.0530 = −0.1460 Ry; polarised −0.09427 − 0.0292 = −0.1235 Ry. The unpolarised gas is now lower, by 0.023 Ry = 0.31 eV per electron.
- The reversal comes from the 0.0238 Ry by which correlation favours the unpolarised gas, 19 times Hartree-Fock's 1.25 mRy preference for alignment. Measured cesium has no spontaneous magnetisation: correlation, not exchange alone, gets the sign right.
Answer. rₛ = 5.62. Hartree-Fock: polarised lower by 1.25 mRy (17 meV) per electron, a ferromagnet with μ₀M ≈ 0.11 T. With correlation: unpolarised lower by 0.023 Ry (0.31 eV), as measured.
HardExample 3
Restore sodium's ions, each Na⁺ a point charge centred in a Wigner-Seitz sphere of radius rₛa₀ holding one evenly spread electron. (a) Find the cell's electrostatic energy and the new Hartree-Fock minimum. (b) Model the 2p core by switching the ion's attraction off inside radius rc, find the rc putting the minimum at the measured rₛ = 3.93, and compare the energy with the measured 6.25 eV per atom needed to split the metal into Na⁺ ions and electrons.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- A uniform sphere of charge −e has potential −3e/(8πε₀R) at its centre, so the ion-electron energy is −(3/2)e²/(4πε₀R); the cloud's self-energy is +(3/5)e²/(4πε₀R). Neutral spherical cells do not interact, so each cell holds −(9/10)e²/(4πε₀R) = −1.8/rₛ Ry, using e²/(4πε₀a₀) = 2 Ry. The exact bcc lattice sum, −1.792/rₛ Ry (quoted), agrees to 0.5%.
- Total: E = 2.21/rₛ² − (0.916 + 1.8)/rₛ = 2.21/rₛ² − 2.716/rₛ Ry. Its minimum is at rₛ = 4.42/2.716 = 1.63, with E = −2.716²/(4 × 2.21) = −0.834 Ry = −11.4 eV: a density 14 times sodium's, bound nearly twice as strongly as measured. Point ions overbind.
- Empty core: setting the attraction −e²/(4πε₀r) to zero inside rc mimics the core's Pauli repulsion. For the uniform density n = 3/(4πR³) it removes the attraction n∫e²/(4πε₀r) 4πr² dr taken from r = 0 to rc, which is (3/2)(rc²/R³)e²/(4πε₀): the energy rises by 3rc²/rₛ³ Ry, with rc in units of a₀.
- Now dE/drₛ = −4.42/rₛ³ + 2.716/rₛ² − 9rc²/rₛ⁴ = 0 at rₛ = 3.93 gives rc² = (2.716 × 3.93² − 4.42 × 3.93)/9 = 2.73, so rc = 1.65 a₀ = 0.087 nm, comparable with the Na⁺ ionic radius of about 0.10 nm (quoted).
- Energy there: 0.1431 − 0.2331 − 0.4580 + 0.1350 = −0.4130 Ry = −5.62 eV per atom, against the measured −(1.11 + 5.14) eV = −6.25 eV, the cohesive plus first ionisation energy. The 0.63 eV shortfall is of the size of the correlation energy at this density, −0.065 Ry = −0.88 eV (quoted), which one determinant omits.
Answer. Point ions give −1.8/rₛ Ry per cell and a minimum at rₛ = 1.63 with −0.834 Ry, far too dense. An empty core of rc = 1.65 a₀ = 0.087 nm restores rₛ = 3.93, at −0.413 Ry = −5.6 eV per atom against the measured −6.25 eV.
Quick check
Select an answer and explain it before checking.
Review & continue
Hartree-Fock binding is the minimum of the energy per electron of jellium held in one Slater determinant, the density at which the slopes of the kinetic energy, 2.21/rₛ² Ry, and the exchange energy, −0.916/rₛ Ry, cancel.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.