University Physics IV · Atomic Physics · 11.2

Pauli Principle & Slater Determinants

Exclusion is usually handed to you as a rule to obey. Here you build the state that obeys it — one determinant whose rows are electrons and whose columns are spin-orbitals — expand it, read the Pauli principle straight off a determinant identity, and find out exactly what that single determinant still gets wrong.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Electrons are not merely alike: no measurement tells them apart, so their labels carry no physics. |Ψ|² must therefore survive the swap of any two labels, and because the swap operator squares to the identity its eigenvalues can only be +1 or −1. That sign is imported, not derived here: the spin-statistics theorem ties half-integer spin to the minus. Antisymmetrising a product of N one-electron spin-orbitals by hand then generates N! terms signed by the parity of each permutation — precisely the Leibniz expansion of a determinant.

Put electron i in row i and spin-orbital j in column j, divide by √(N!), and the bookkeeping is done. Two facts now collapse into one. Swapping two electrons swaps two rows and flips the sign, which is antisymmetry; giving two electrons the same spin-orbital repeats a column, and a determinant with two equal columns is identically zero, which is exclusion — a corollary, and an exact one, since the amplitude vanishes everywhere rather than merely shrinking.

The cost sits in that same object: one determinant is the best independent-particle state there is, so its minus sign keeps same-spin electrons apart, the Fermi hole, while leaving opposite-spin electrons statistically independent with no Coulomb hole at all. The energy lost is the correlation energy — 1.14 eV in helium, 1.4% of the atom's 79.01 eV binding, but a quarter of an H₂ bond.

Simple definition: A Slater determinant is the normalised, fully antisymmetric N-electron state built from N one-electron spin-orbitals: rows label the electrons, columns label the spin-orbitals, and the determinant supplies every sign that indistinguishability demands.
Example: For helium's 1s² ground state, Ψ = (1/√2)[1s(r₁)α(1)⋅1s(r₂)β(2) − 1s(r₁)β(1)⋅1s(r₂)α(2)] = 1s(r₁)1s(r₂)⋅(αβ − βα)/√2. Assign a third electron to 1s α and that column repeats, so the 3 × 3 determinant is zero everywhere.

Exchange operator and its eigenvaluesP₁₂ Ψ = ±Ψ, since P₁₂² = I

Only two eigenvalues exist, so a species is a boson or a fermion, with nothing in between.

P₁₂ swaps the full coordinate x = (r, s) — space and spin together. Dimensionless; electrons take the minus sign.

The Slater determinantΨ(x₁, …, x_N) = (1/√(N!)) det[χⱼ(xᵢ)]

One symbol carries every permutation with its correct sign, for any N.

Row i is electron i, column j the orthonormal spin-orbital χⱼ. The 1/√(N!) normalises Ψ; N! signed terms.

Two electrons, written outΨ = (1/√2)[χₐ(x₁)χ_b(x₂) − χ_b(x₁)χₐ(x₂)]

Exclusion appears as a determinant with two equal columns, never as an added rule.

Set χₐ = χ_b and the two terms cancel: Ψ ≡ 0 at all coordinates, not merely a small amplitude.

The Fermi holeΨ(x, x) = 0 for every antisymmetric Ψ

Same-spin electrons avoid each other with no force acting — the origin of the exchange energy.

Immediate from Ψ(x₁, x₂) = −Ψ(x₂, x₁) at x₁ = x₂. Opposite spins are different x, so they get no hole.

Energy of a single determinantE = Σᵢ hᵢᵢ + ½ Σᵢⱼ (Jᵢⱼ − Kᵢⱼ)

The i = j terms cancel exactly, Jᵢᵢ = Kᵢᵢ, so no electron repels itself.

Sums over occupied spin-orbitals; h, J, K in eV. Kᵢⱼ = 0 unless χᵢ and χⱼ share a spin projection.

Correlation energyE_corr = E_exact − E_HF

Prices what a single determinant leaves behind; only more determinants recover it.

Helium: E_exact = −79.01 eV and E_HF = −77.87 eV, so E_corr = −1.14 eV. Negative by construction.

Indistinguishability fixes a sign, and nothing else

Two electrons are not merely alike: no measurement can distinguish them, so the numbers you write on them are bookkeeping, not physics. Define the exchange operator P₁₂ to swap the entire coordinate of two electrons — position and spin projection together, x = (r, s) — and note that for identical particles it commutes with the Hamiltonian, so exchange symmetry is conserved. Applying it twice restores the original labels, P₁₂² = I, so its eigenvalues satisfy λ² = 1 and can only be +1 or −1. That is the whole of what non-relativistic quantum mechanics can say: a state is symmetric or antisymmetric, with nothing in between and no continuous path from one to the other. Which sign a given species takes is settled elsewhere. The spin-statistics theorem, a result of relativistic quantum field theory, ties integer spin to +1 and half-integer spin to −1. Electrons are spin-½, so from here on every acceptable many-electron state must change sign under every pairwise exchange.

Antisymmetrise by hand, then recognise the determinant

Start from the naive independent-particle guess χₐ(x₁)χ_b(x₂), which claims electron 1 is in χₐ. Swapping the labels changes it, so it is illegal. Repair it by subtracting the swapped copy and renormalising: Ψ = [χₐ(x₁)χ_b(x₂) − χ_b(x₁)χₐ(x₂)]/√2, which flips sign on exchange and is normalised whenever the spin-orbitals are orthonormal. Three electrons need all 3! = 6 orderings, each carrying the parity of the permutation that produced it — plus for even, minus for odd. That signed sum over permutations is exactly the Leibniz formula for a determinant, so write it as one: build the N × N matrix whose entry in row i, column j is χⱼ(xᵢ), take the determinant, divide by √(N!). Rows are electrons, columns are spin-orbitals. The compression is free and dramatic — neon's ground state is 10! = 3 628 800 signed products, and the determinant holds them all in one symbol whose algebra then does the physics for you.

Exclusion is a property of determinants

Two determinant identities now carry the whole content of the Pauli principle. Interchange two rows and a determinant changes sign: swap two electrons and Ψ flips sign, which is antisymmetry, for free. Give a determinant two identical columns and it is zero: assign two electrons the same spin-orbital and Ψ vanishes at every set of coordinates. Not small, not suppressed — identically zero, so the occupation number of each spin-orbital is 0 or 1, and exclusion is a corollary rather than a postulate. Read the statement carefully: what may not repeat is the whole spin-orbital, space part and spin part together. The 1s orbital holds two electrons because 1s α and 1s β are different columns, and a subshell of orbital quantum number l holds 2(2l + 1) electrons, six for a p subshell. One feature is easy to miss: the determinant assigns nothing to electron 1. Mixing the occupied spin-orbitals among themselves by any unitary transformation leaves Ψ unchanged up to a phase, so which electron occupies which orbital is not a question the state answers.

The minus sign digs a Fermi hole

Set both arguments equal in Ψ(x₁, x₂) = −Ψ(x₂, x₁): you get Ψ(x, x) = −Ψ(x, x), so Ψ(x, x) = 0. Two electrons in the same spin state have zero amplitude to sit at the same point — a hole dug by a sign, with no term in the Hamiltonian doing the digging. Split the two-electron state into space and spin to see the consequence. The antisymmetric spin function is the singlet, one state with S = 0, and it must multiply a symmetric spatial function; the three symmetric spin functions form the triplet, S = 1, and they must multiply an antisymmetric spatial function that vanishes at r₁ = r₂. Parallel spins therefore keep further apart, their expectation of e²/4πε₀r₁₂ is smaller, and the triplet lies lower. Helium's 1s2s configuration shows it directly: 2³S sits 19.82 eV above the ground state while 2¹S sits 20.62 eV above it, a splitting of 0.80 eV = 2K, so the exchange integral for that pair is about 0.40 eV.

Reading the energy off a single determinant

Take the expectation of the many-electron Hamiltonian in one determinant and the Slater–Condon rules return E = Σᵢ hᵢᵢ + ½ Σᵢⱼ (Jᵢⱼ − Kᵢⱼ), the sums running over occupied spin-orbitals. The one-electron term hᵢᵢ holds kinetic energy and nuclear attraction. Jᵢⱼ is the ordinary Coulomb repulsion between two charge clouds, exactly what a classical picture would give. Kᵢⱼ has no classical analogue at all: it exists only because of the determinant's minus sign, it enters with a minus, and it vanishes unless χᵢ and χⱼ share a spin projection. Two checks follow. The i = j terms give Jᵢᵢ − Kᵢᵢ = 0 exactly, so no electron repels itself — a cancellation the older Hartree method lacks. And the counting is concrete: neon's ten electrons make 45 pairs, of which the 20 same-spin pairs carry an exchange integral and the 25 opposite-spin pairs do not. Minimising E over the orbitals themselves gives the Hartree–Fock equations.

What one determinant cannot know: correlation

A single determinant is an independent-particle state: each electron feels the average field of the rest, never their instantaneous positions. Same-spin electrons still avoid each other, because antisymmetry forces it, but opposite-spin electrons are left statistically independent, their pair density factorising with no Coulomb hole at all — in helium's ground state the two electrons are permitted to sit on top of one another. The missing energy has a name and a size. A hydrogenic trial with optimised effective charge Z′ = 27/16 gives −77.49 eV; relaxing the orbital shape while keeping one determinant reaches the Hartree–Fock limit, −77.87 eV; the exact non-relativistic value is −79.01 eV. The difference, E_corr = −1.14 eV, is only 1.4% of the total binding, yet it is a quarter of the 4.48 eV bond in H₂, so it decides chemistry. No single determinant can recover it; the cure is a sum of determinants, configuration interaction and its descendants.

Change one variable at a time

Make the relationship visible.

Interactive model

Electron 1 held at x1/L0.30

Second orbital index n2

Exchange sign s−1

Leave s at −1 and drag x1: the solid curve is pinned to exactly zero wherever electron 1 sits, for every orbital n. Now set s to 0 and the hole fills in to the faint line — that gap is the whole of what the determinant's minus sign removes.

PLAIN PRODUCT AT x2 = x10.592

THIS STATE AT x2 = x10.000

HOLE DEPTH AT x2 = x10.592

DENSITY AT MIDPOINT0.452

Live interpretationPLAIN PRODUCT AT x2 = x1: 0.592. THIS STATE AT x2 = x1: 0.000. HOLE DEPTH AT x2 = x1: 0.592. DENSITY AT MIDPOINT: 0.452

Catch the common trap

Explain before calculating.

Misconception check

Two errors travel together here. The first calls exclusion a force — a “Pauli repulsion” pushing electrons apart. Nothing has been added to the Hamiltonian; antisymmetry restricts which states exist, and the stiffness of matter is the kinetic-energy cost of promoting electrons into higher orbitals, not a new interaction. The second treats the determinant as a device you need only when two electrons might land in the same orbital, so a configuration like 1s2s looks safe without it. But the same minus sign that kills a repeated column also produces the exchange integral K, the Fermi hole, and the 0.80 eV by which helium's 2³S lies below its 2¹S — a splitting in a configuration where no two spin-orbitals are equal at all. Antisymmetry is compulsory everywhere; exclusion is only its most visible special case.

Three electrons are assigned the spin-orbitals 1s↑, 1s↓ and 1s↑, and the 3 × 3 Slater determinant is built from them. What is the resulting wave function?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyBuild the two-electron Slater determinant for helium's ground configuration from the spin-orbitals χₐ = 1s α and χ_b = 1s β. Expand it, factorise it into a space part and a spin part, and state the total spin. Then say what the determinant gives if a third electron is assigned 1s α.

Rows are electrons and columns are spin-orbitals, so Ψ = (1/√2!) × det of the 2 × 2 array with χₐ(x₁), χ_b(x₁) on the first row and χₐ(x₂), χ_b(x₂) on the second: Ψ = (1/√2)[χₐ(x₁)χ_b(x₂) − χ_b(x₁)χₐ(x₂)]. Two signed terms, because 2! = 2.
Substitute the spin-orbitals: Ψ = (1/√2)[1s(r₁)α(1)⋅1s(r₂)β(2) − 1s(r₁)β(1)⋅1s(r₂)α(2)].
The spatial factor 1s(r₁)1s(r₂) is common to both terms, so Ψ = 1s(r₁)1s(r₂) × (1/√2)[α(1)β(2) − β(1)α(2)] — a symmetric space part times an antisymmetric spin part, whose product is antisymmetric as required.
That spin function is the singlet, S = 0 with M_S = 0: one state, so the closed 1s² shell gives the single term ¹S₀.
A third electron in 1s α makes the third column a copy of the first, so the 3 × 3 determinant is identically zero and the K shell closes at two. (For scale, a six-electron determinant carries 6! = 720 signed terms.)

AnswerΨ = 1s(r₁)1s(r₂)⋅[α(1)β(2) − β(1)α(2)]/√2 — symmetric in space, singlet in spin, S = 0, term ¹S₀. A third 1s α electron gives a determinant that is zero everywhere.

MediumNeon's ground configuration 1s²2s²2p⁶ is a single Slater determinant. Find (a) how many signed terms its Leibniz expansion carries, (b) how many electron pairs appear in the two-electron part of the energy, (c) how many of those pairs contribute a non-zero exchange integral Kᵢⱼ, and (d) how many contribute a Coulomb integral only.

N = 10 electrons fill 10 spin-orbitals, so the determinant expands into N! = 10! = 3 628 800 signed products — all held in one symbol.
The two-electron part of E = Σᵢ hᵢᵢ + ½ Σᵢⱼ (Jᵢⱼ − Kᵢⱼ) runs over unordered pairs, so there are C(10, 2) = (10 × 9)/2 = 45 of them.
Neon is closed-shell: each of the five spatial orbitals 1s, 2s, 2p₋₁, 2p₀, 2p₊₁ is doubly occupied, giving 5 spin-up and 5 spin-down spin-orbitals.
Kᵢⱼ vanishes unless the two spin-orbitals share a spin projection, so the exchange-carrying pairs number C(5, 2) + C(5, 2) = 10 + 10 = 20.
The remaining 45 − 20 = 25 pairs are opposite-spin: they keep Jᵢⱼ but contribute no exchange, so nothing in this state keeps those electrons apart.

Answer(a) 10! = 3 628 800 terms; (b) 45 pairs; (c) 20 pairs carry a non-zero K; (d) 25 opposite-spin pairs carry J alone — and those 25 are exactly where the correlation energy hides.

HardA variational trial for helium puts two 1s orbitals of effective charge Z′ into one Slater determinant, giving E(Z′) = (Z′² − 3.375 Z′) hartree, with 1 hartree = 27.211 eV. (a) Minimise E and interpret Z′. (b) Convert to eV and compare with the Hartree–Fock limit, −77.87 eV. (c) Helium ionises at 24.59 eV and He⁺ at 54.42 eV: find the exact total, then the correlation energy — absolutely, as a fraction of the total, and against the 4.48 eV bond of H₂.

Minimise: dE/dZ′ = 2Z′ − 3.375 = 0, so Z′ = 1.6875 = 27/16. It sits below Z = 2 because each electron screens 0.3125 of the nuclear charge from the other.
E = (1.6875)² − 3.375 × 1.6875 = 2.8477 − 5.6953 = −2.8477 hartree = −2.8477 × 27.211 = −77.49 eV.
That is 0.38 eV above the Hartree–Fock limit of −77.87 eV, because the trial froze the orbital shape as hydrogenic; letting the shape relax, still within one determinant, recovers the rest.
Exact total: stripping both electrons costs 24.59 + 54.42 = 79.01 eV, so E_exact = −79.01 eV.
Correlation energy: E_corr = −79.01 − (−77.87) = −1.14 eV, which is 1.14/79.01 = 1.44% of the total binding.
But 1.14/4.48 = 0.25, so what one determinant misses is a quarter of an H₂ bond. A 1.4% error in the total is a 25% error in anything chemical, which is why the fix is more determinants, not a better orbital.

AnswerZ′ = 1.6875 and E = −77.49 eV; the Hartree–Fock limit is −77.87 eV and the exact value −79.01 eV, so E_corr = −1.14 eV: 1.44% of the total binding, but 25% of an H₂ bond.