University Physics V · Multi-Electron Atoms · 12.3

Slater Determinants & Occupation Numbers

Antisymmetry goes in; exclusion comes out. Here the N-electron state is built as a determinant of spin-orbitals, rewritten as an occupation-number ket, and the one-electron-per-orbital ceiling is derived from the anticommutators — then we name the correlation a single determinant can never carry.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Take N electrons and N orthonormal spin-orbitals on L²(R³) ⊗ C². The symmetrisation postulate demands a state that changes sign under every transposition of coordinate labels, so apply the antisymmetriser A = (1/N!) Σ_P sgn(P) P̂ — a Hermitian projector, A² = A — to the plain product φ₁(x₁)…φ_N(x_N). What returns is a determinant, Ψ = (1/√N!) det[φᵢ(xⱼ)], orbitals down the rows and electron coordinates across the columns, and every property you need is now a fact about determinants. Swap two columns and Ψ changes sign: antisymmetry.

Repeat a row and the determinant is identically zero — not a forbidden state but no vector at all, which is why exclusion needs no separate postulate. Mix the occupied orbitals by any unitary U and Ψ picks up only the phase det U, so the individual orbitals are unobservable and only their span is physical. Second quantisation says the same thing algebraically: (aᵢ, aⱼ†) = δᵢⱼ with (aᵢ†, aⱼ†) = 0 forces (aᵢ†)² = 0, hence n̂ᵢ² = n̂ᵢ and nᵢ ∈ {0, 1}, while the bosonic [bᵢ, bⱼ†] = δᵢⱼ imposes no ceiling at all. The cost is equally sharp.

A single determinant has an idempotent one-body density matrix, so its pair density factorises up to one exchange term: a Fermi hole between parallel spins, and no hole whatever between antiparallel ones. For helium that omission is 0.042 hartree, about 1.14 eV.

Simple definition: A Slater determinant is the antisymmetrised product of N orthonormal spin-orbitals, Ψ = (1/√N!) det[φᵢ(xⱼ)] — the simplest N-fermion state, and the one whose occupation numbers are all exactly 0 or 1.
Example: For helium's ground configuration the two spin-orbitals are 1s↑ and 1s↓, so Ψ = (1/√2) 1s(r₁)1s(r₂)[α(1)β(2) − β(1)α(2)]. A third electron in 1s would repeat a row and return zero, which is why lithium starts filling 2s.

AntisymmetriserA = (1/N!) Σ_P sgn(P) P̂, A† = A, A² = A

A projector onto the totally antisymmetric sector, so applying it twice buys nothing.

P̂ permutes the N coordinate labels; sgn(P) = ±1 for even and odd permutations

Slater determinantΨ(x₁ … x_N) = (1/√N!) det[φᵢ(xⱼ)]

Rows are orbitals, columns electrons: swap two columns and Ψ flips sign; repeat a row and det = 0, so exclusion is already built in.

x = (r, σ); orthonormal spin-orbitals give ⟨Ψ|Ψ⟩ = 1, otherwise ⟨Ψ|Ψ⟩ = det S

Unitary invariance of the occupied setφ′ᵢ = Σⱼ Uⱼᵢ φⱼ ⇒ Ψ′ = det(U) Ψ

Canonical and localised orbitals are one determinant in two frames — only the occupied subspace is observable.

U unitary, so |det U| = 1 and the physical ray is untouched

Fermionic algebra(aᵢ, aⱼ†) = δᵢⱼ, (aᵢ†, aⱼ†) = 0 ⇒ (aᵢ†)² = 0

Fill an orbital twice and the algebra hands back the zero vector — exclusion with no extra postulate.

aᵢ† creates an electron in spin-orbital i; both relations are dimensionless

Occupation idempotency, and its bosonic failuren̂ᵢ = aᵢ†aᵢ, n̂ᵢ² = n̂ᵢ ⇒ nᵢ ∈ {0, 1}

The Pauli ceiling in one line, and the single sign change that removes it and allows a condensate.

bosons have [bᵢ, bⱼ†] = δᵢⱼ, so n̂² = n̂ + (b†)²b² and n = 0, 1, 2, … without limit

Pair density of one determinantP(x₁, x₂) = ρ(x₁)ρ(x₂) − |ρ₁(x₁, x₂)|²

P(x, x) = 0 exactly — a Fermi hole for parallel spins, and no correlation hole at all for antiparallel ones.

ρ₁(x, x′) = Σᵢ^occ φᵢ(x)φᵢ*(x′), the idempotent one-body density matrix

Antisymmetrise the product and a determinant appears

Take N orthonormal spin-orbitals φ₁ … φ_N, each a vector in L²(R³) ⊗ C², and form the Hartree product φ₁(x₁)φ₂(x₂)…φ_N(x_N) with x = (r, σ). That product is not a fermionic state: swap x₁ and x₂ and it turns into a different function rather than into minus itself. The repair is a projector. Define A = (1/N!) Σ_P sgn(P) P̂, summing over all N! permutations of the coordinate labels with sgn(P) = +1 for even and −1 for odd. A is Hermitian and idempotent, A² = A, so it projects the N-particle space onto its totally antisymmetric subspace. Apply it and restore the norm: Ψ = √(N!) A [φ₁(x₁)…φ_N(x_N)] = (1/√N!) det[φᵢ(xⱼ)]. The prefactor 1/√N! is exactly right only because the orbitals are orthonormal; if they are not, the overlaps survive and ⟨Ψ|Ψ⟩ = det S.

Rows, columns, and what is not observable

Write the determinant with orbitals down the rows and electron coordinates across the columns. Exchanging two electrons exchanges two columns, and a determinant changes sign under a column swap — antisymmetry delivered by the notation rather than imposed on it. Put the same spin-orbital in two rows and the determinant is identically zero: not a suppressed state, not a rare one, but the zero vector, which cannot be normalised and so is not a state at all. That is the whole content of the exclusion principle. One further property matters and is usually skipped. Replace the occupied orbitals by any unitary mixture φ′ᵢ = Σⱼ Uⱼᵢ φⱼ and the determinant becomes Ψ′ = det(U) Ψ, a phase. The individual orbitals are therefore not observable; the occupied subspace is, carried by the projector ρ̂₁ = Σᵢ |φᵢ⟩⟨φᵢ|. Canonical Hartree–Fock orbitals and localised bonding orbitals are the same determinant in two coordinate systems.

The same state written as an occupation-number ket

Once an ordered one-particle basis is fixed, a determinant is specified entirely by which orbitals are filled: no assignment of electrons to orbitals is left to record. So write |n₁ n₂ n₃ …⟩ with nᵢ ∈ {0, 1}, defined as (a₁†)^(n₁)(a₂†)^(n₂)… acting on the vacuum, the creation operators standing in ascending index order from the left. The ordering convention is not cosmetic. Because aᵢ†aⱼ† = −aⱼ†aᵢ†, moving a creation operator into its slot costs one sign for every occupied mode it has to cross, so aᵢ†|…⟩ = (−1)^(Σ_(j⟨i) nⱼ)|… nᵢ = 1 …⟩ — the Jordan–Wigner string. What the string counts is a parity, not a distance. With four modes, a₄† acting on |1, 1, 0, 0⟩ crosses two filled modes and returns +|1, 1, 0, 1⟩, while a₂† acting on |1, 0, 0, 0⟩ crosses one and returns −|1, 1, 0, 0⟩; the operator that travels further is the one that costs nothing. Drop the string and every matrix element you compute afterwards carries an arbitrary sign, and terms that ought to cancel do not.

Exclusion falls out of the anticommutators

The algebra is (aᵢ, aⱼ†) = δᵢⱼ, {aᵢ, aⱼ} = 0 and (aᵢ†, aⱼ†) = 0. Set i = j in the last relation: 2(aᵢ†)² = 0, so aᵢ† applied twice to anything gives the zero vector. That is exclusion in one line, with no appeal to quantum numbers. The occupation operator sharpens it. With n̂ᵢ = aᵢ†aᵢ, use aᵢ aᵢ† = 1 − aᵢ†aᵢ to get n̂ᵢ² = aᵢ†(aᵢ aᵢ†)aᵢ = aᵢ†aᵢ − (aᵢ†)²(aᵢ)² = n̂ᵢ. An operator obeying n̂² = n̂ has eigenvalues satisfying λ² = λ, so λ = 0 or 1 and nothing else. Now change one sign. Bosons obey [bᵢ, bⱼ†] = δᵢⱼ, so bᵢ bᵢ† = 1 + bᵢ†bᵢ and n̂ᵢ² = n̂ᵢ + (bᵢ†)²(bᵢ)², which is not idempotent; the eigenvalues run 0, 1, 2, … with no ceiling. The Pauli limit and Bose–Einstein condensation are one equation with a commutator in place of an anticommutator.

What one determinant buys: Slater–Condon and the Fermi hole

Because the state is a single determinant, every expectation value collapses to sums over occupied orbitals. For Ĥ = Σᵢ ĥ(i) + ½ Σ_(i≠j) 1/rᵢⱼ the diagonal rule reads ⟨Ψ|Ĥ|Ψ⟩ = Σᵢ hᵢᵢ + ½ Σ_(i, j) (Jᵢⱼ − Kᵢⱼ), with direct integral Jᵢⱼ = ⟨ij|ij⟩ and exchange integral Kᵢⱼ = ⟨ij|ji⟩. Kᵢⱼ carries a spin overlap ⟨σᵢ|σⱼ⟩², so it vanishes unless the two spin-orbitals share a spin. Geometrically this is the Fermi hole: the pair density of a determinant is P(x₁, x₂) = ρ(x₁)ρ(x₂) − |ρ₁(x₁, x₂)|², and at coincident space-and-spin points ρ₁(x, x) = ρ(x), so P(x, x) = 0 exactly. Parallel-spin electrons never meet. Nothing here referred to a magnetic force between spins — the hole is a consequence of the determinant, and it is why the triplet of a configuration lies below the singlet.

What it costs: a determinant is uncorrelated by construction

The algebra that bounds n̂ᵢ also exposes the limitation. For a single determinant the one-body density matrix is idempotent, ρ̂₁² = ρ̂₁, so its eigenvalues — the natural occupation numbers — are exactly 0 or 1. No exact interacting ground state passes that test: helium's leading natural orbital holds about 1.98 electrons of the 2, not 2.00. Equivalently P(x₁, x₂) factorises into ρρ up to one exchange term, so antiparallel electrons in the same determinant stay statistically independent exactly where 1/r₁₂ diverges — a Fermi hole but no Coulomb hole. The deficit is defined as E_corr = E_exact − E_HF. For helium E_HF = −2.8617 Eₕ against the exact non-relativistic −2.9037 Eₕ, giving E_corr = −0.0420 Eₕ = −1.14 eV: only 1.4% of the total energy, but a quarter of a typical covalent bond. Recovering it means adding determinants to the wavefunction, and there is no shortage of them to add: 10 electrons in 20 spin-orbitals already furnish C(20,10) = 184 756.

Change one variable at a time

Make the relationship visible.

Interactive model

Reference electron x₁ / L0.26 L

Exchange weight s (−1 = fermion)−1.00

Start at s = −1 (parallel spins) and drag x₁: the solid curve touches zero exactly at the marker, and the hole travels with it. Slide to s = 0, where the spin integral kills exchange, and the hole fills in. The area under the curve never changes — exchange only moves probability.

DIRECT TERM at x₂ = x₁0.693

EXCHANGE TERM at x₂ = x₁-0.693

TOTAL P₂(x₁ | x₁)0.000

P₂ AT x₂ = 0.75 L1.348

Live interpretationDIRECT TERM at x₂ = x₁: 0.693. EXCHANGE TERM at x₂ = x₁: −0.693. TOTAL P₂(x₁ | x₁): 0.000. P₂ AT x₂ = 0.75 L: 1.348

Catch the common trap

Explain before calculating.

Misconception check

The usual sentence — no two electrons may share all four quantum numbers — gets treated as an extra law bolted onto the Schrödinger equation, with the determinant as bookkeeping that enforces it. Both halves are wrong. Antisymmetry is the input, fixed by the symmetrisation postulate; exclusion is the output, because a determinant with a repeated row is the zero vector, and a state you cannot write down is not a state you are forbidden to write down. The second error is reading the leading diagonal as an assignment — electron 1 in φ₁, electron 2 in φ₂. No such assignment survives antisymmetrisation: every electron appears in every occupied orbital with equal weight, and mixing those orbitals by any unitary U leaves the physical state untouched, Ψ → det(U) Ψ. What the state fixes is the occupied subspace, not a labelling of electrons.

In the occupation-number representation the electron operators obey (aᵢ, aⱼ†) = δᵢⱼ and {aᵢ, aⱼ} = 0. Which line actually forces the eigenvalues of n̂ᵢ = aᵢ†aᵢ to be 0 or 1, rather than merely permitting it?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA minimal basis for beryllium keeps five spatial orbitals (1s, 2s, 2pₓ, 2p_y, 2p_z), hence ten spin-orbitals, and Be has four electrons. (a) How many distinct Slater determinants can be built from them? (b) How many of those place two electrons in one spin-orbital? (c) How many states would four identical spinless bosons have in the same ten modes?

A determinant is fixed by which spin-orbitals are occupied — no assignment of electrons to orbitals is recorded — so the count is a choice of 4 from 10 with no ordering: C(10,4).
C(10,4) = (10 × 9 × 8 × 7)/(4 × 3 × 2 × 1) = 5040/24 = 210.
(b) None. A repeated spin-orbital puts two identical rows in the determinant, which is then the zero vector — so C(10,4) never counted them, because there was nothing there to exclude.
(c) Bosons may repeat a mode, so the count is of multisets: C(N + M − 1, N) = C(13,4) = (13 × 12 × 11 × 10)/24 = 17160/24 = 715, which is 3.4 times as many.

Answer(a) 210 determinants. (b) None — a repeated spin-orbital gives the zero vector, not a suppressed state. (c) 715 bosonic states, 3.4 times more.

(a) Modes 1 and 3 each carry one electron and the operators already stand in ascending order, so |Φ⟩ = |1, 0, 1, 0⟩.
(b) a₂†|Φ⟩ = a₂†a₁†a₃†|vac⟩. One anticommutation moves a₂† into place: a₂†a₁† = −a₁†a₂†, so a₂†|Φ⟩ = −|1, 1, 1, 0⟩. The sign is (−1)^(n₁), the Jordan–Wigner string counting filled modes below mode 2.
(c) a₃†|Φ⟩ = a₃†a₁†a₃†|vac⟩ = −a₁†(a₃†)²|vac⟩ = 0, because (a₃†)² = 0. Exclusion, with no separate rule invoked.
(d) Act first with a₃: a₃a₁†a₃†|vac⟩ = −a₁†a₃a₃†|vac⟩ = −a₁†(1 − a₃†a₃)|vac⟩ = −a₁†|vac⟩, using a₃|vac⟩ = 0.
Then n̂₃|Φ⟩ = a₃†(−a₁†|vac⟩) = −a₃†a₁†|vac⟩ = +a₁†a₃†|vac⟩ = |Φ⟩, so the eigenvalue is 1, exactly as the ket |1, 0, 1, 0⟩ advertised.

Answer|Φ⟩ = |1, 0, 1, 0⟩; a₂†|Φ⟩ = −|1, 1, 1, 0⟩; a₃†|Φ⟩ = 0; n̂₃|Φ⟩ = |Φ⟩ with eigenvalue 1.

HardFor a two-electron Slater determinant built from orthonormal spin-orbitals φₐ and φ_b, the pair density is P(x₁, x₂) = 2|Ψ(x₁, x₂)|². (a) Show that P = ρ(x₁)ρ(x₂) − |ρ₁(x₁, x₂)|², with ρ₁(x, x′) = Σᵢ φᵢ(x)φᵢ*(x′). (b) Evaluate P at x₁ = x₂. (c) Take φₐ = ψ₁α and φ_b = ψ₂β, and evaluate P at one position but opposite spins. (d) Name the energy this construction misses in helium.

Ψ = (1/√2)[φₐ(1)φ_b(2) − φ_b(1)φₐ(2)], so P = 2|Ψ|² = |φₐ(1)|²|φ_b(2)|² + |φ_b(1)|²|φₐ(2)|² − 2Re[φₐ*(1)φ_b(1)φ_b*(2)φₐ(2)].
Expand the claim: ρ(x) = |φₐ(x)|² + |φ_b(x)|², and |ρ₁(1,2)|² = |φₐ(1)|²|φₐ(2)|² + |φ_b(1)|²|φ_b(2)|² + 2Re[φₐ(1)φₐ*(2)φ_b*(1)φ_b(2)]. The like-orbital squares cancel between ρρ and |ρ₁|², and the surviving cross term is the complex conjugate of step 1's, which Re cannot tell apart.
(b) Put x₁ = x₂ = x. Then ρ₁(x, x) = ρ(x), so P(x, x) = ρ(x)² − ρ(x)² = 0, exactly and for every determinant. Two electrons never share a point of space and spin — that is the Fermi hole.
(c) With φₐ = ψ₁α and φ_b = ψ₂β, ρ₁((r,↑),(r,↓)) = |ψ₁(r)|²α(↑)α*(↓) + |ψ₂(r)|²β(↑)β*(↓) = 0, since α*(↓) = β(↑) = 0. So P = ρ(r,↑)ρ(r,↓), a plain product: no hole at all between antiparallel electrons.
(d) Helium's ground state is 1s², an antiparallel pair, so the determinant leaves the two independent exactly where 1/r₁₂ diverges. E_HF = −2.8617 Eₕ against the exact non-relativistic −2.9037 Eₕ, so E_corr = −0.0420 Eₕ = −0.0420 × 27.211 eV = −1.14 eV.

AnswerP = ρρ − |ρ₁|²; P(x, x) = 0 always, the Fermi hole; at one position with opposite spins P = ρ↑ρ↓ and no hole; helium's missing correlation energy is −0.0420 Eₕ = −1.14 eV.