University Quantum Mechanics II · Identical Particles and Many-Electron Atoms · 7.3

Pauli exclusion and the Slater determinant

This is where the Pauli principle stops being a rule you memorise and becomes one line of algebra. You will build N-fermion states as determinants, read every sign and every zero off them, normalise bosons whose labels repeat, and see why one determinant is exact for independent electrons but only an approximation in a real atom.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Put two fermions in spin-orbitals a and b, each label fixing a spatial orbital and a spin projection together. The symmetrisation postulate, with the antisymmetric choice that the spin-statistics theorem assigns to half-integer spin (quoted from relativistic field theory, not derived here), admits only the antisymmetrised product [ψₐ(x₁)ψ_b(x₂) − ψ_b(x₁)ψₐ(x₂)]/√2. Set b = a and it vanishes identically, and the zero vector is not a state: exclusion is a theorem, not a separate law.

For N fermions the antisymmetriser builds a determinant, and the properties of determinants become physics. Swapping two particles swaps two rows and flips the sign; repeating a spin-orbital repeats a column and removes the state; mixing the occupied orbitals among themselves changes nothing but an overall factor. Bosons get the same sum with every sign positive, and a prefactor that depends on how often each label repeats.

The model stops where the Hamiltonian stops being a sum of one-particle terms. For independent particles a determinant of eigen-orbitals is an exact eigenstate. The Coulomb repulsion between electrons spoils that, so a real atom's ground state is a superposition of many determinants, and even the best single one, which the Hartree-Fock method seeks, misses a fraction of a per cent of the energy: 1.2 eV in lithium.

Simple definition: A Slater determinant is the normalised antisymmetric state of N fermions in N orthonormal spin-orbitals, (1/√(N!)) det[ψₐⱼ(xᵢ)]; it vanishes whenever two of the spin-orbitals coincide, and that vanishing is the Pauli exclusion principle.
Example: Lithium's ground configuration is the determinant of 1s↑, 1s↓ and 2s↑: six products, each of weight 1/√6 = 0.408. Put the third electron in 1s↑ instead and two columns match, so the state is zero. It must leave n = 1; in 2s it is removed by 5.39 eV, against 75.6 eV for a 1s electron of Li⁺.

Two fermions in spin-orbitals a and bΨ(x₁, x₂) = [ψₐ(x₁)ψ_b(x₂) − ψ_b(x₁)ψₐ(x₂)]/√2

Derived from the antisymmetric postulate. At b = a the bracket is zero for every x₁ and x₂, so no state puts two fermions in one spin-orbital. Exclusion needs no law of its own.

xᵢ = (rᵢ, σᵢ), position and spin projection; each ψ in m⁻³⁄² and Ψ in m⁻³; normalised when ⟨a|b⟩ = 0

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

Derived. A particle swap is a row swap, so the sign flips; a repeated spin-orbital is a repeated column, so Ψ = 0. The N! products are orthonormal, which fixes 1/√(N!).

Row i is particle i, column j is spin-orbital aⱼ; the aⱼ orthonormal; Ψ in m^(−3N/2)

Squared norm when the orbitals overlap⟨Ψ|Ψ⟩ = det[⟨aᵢ|aⱼ⟩], which for N = 2 is 1 − |⟨a|b⟩|²

Derived. It is 1 only for an orthonormal set and falls smoothly to 0 as one orbital becomes a combination of the others: exclusion is the end point of a continuous loss of norm.

Each ψ normalised but not orthogonal, the 1/√(N!) prefactor kept; the overlaps ⟨aᵢ|aⱼ⟩ are dimensionless

Mixing the occupied orbitalsψ′ⱼ = Σₖ ψₖ Uₖⱼ gives det[ψ′ⱼ(xᵢ)] = det U × det[ψⱼ(xᵢ)]

Derived from det(AB) = det A det B. A determinant records which subspace is filled, not which electron sits where, so a closed subshell is one state in any orbital basis.

U is an N × N matrix acting only among the occupied spin-orbitals; |det U| = 1 when U is unitary

Boson prefactor when labels repeatC = √(nₐ! n_b! ⋯ / N!)

Derived: a sum over all N! permutations meets each distinct product nₐ! n_b! ⋯ times. Two bosons in a need C = 1; the labels a, a, b give (aab + aba + baa)/√3.

nₐ bosons in state a, with Σ nₐ = N; C multiplies each of the N!/(nₐ! n_b! ⋯) distinct products; dimensionless

Independent particles: the determinant is exactH = Σᵢ h(i) and h ψₐⱼ = εⱼ ψₐⱼ give HΨ = (ε₁ + ε₂ + ⋯ + ε_N)Ψ

Derived: a symmetric H commutes with every permutation, and each product is already an eigenstate. For lithium it gives −275.5 eV against a measured −203.5 eV.

h a one-particle Hamiltonian, energies in J or eV; fails once H holds the pair repulsion e²/(4πε₀rᵢⱼ)

Exclusion is antisymmetry meeting a repeated label

A spin-orbital fixes everything one electron carries: a spatial orbital φₙₗₘ times a spin state with mₛ = ±½, so ψₐ(x) depends on x = (r, σ). The previous topic's postulate, with the antisymmetric choice that the spin-statistics theorem assigns to spin ½ (quoted from relativistic field theory), says a two-electron state must change sign under P₁₂. The product ψₐ(x₁)ψ_b(x₂) does not; [ψₐ(x₁)ψ_b(x₂) − ψ_b(x₁)ψₐ(x₂)]/√2 does. Now set b = a. The two terms are identical and cancel for every x₁ and x₂, and the zero vector cannot be normalised, so it is no state at all. That is the exclusion principle, derived. It restricts spin-orbitals, not spatial orbitals: 1s↑ and 1s↓ are different labels, so helium's electrons may share 1s only with opposite spins. The spectrum agrees: helium has no 1s² ³S level, and its lowest triplet, 1s2s ³S₁, lies 19.82 eV above the ground state. Exclusion carries no power of α; it is exact, a statement about which vectors exist, not about any force.

A determinant antisymmetrises N particles at once

For N fermions in orthonormal spin-orbitals a₁ … a_N, write particle i into row i and spin-orbital j into column j, and take (1/√(N!)) det[ψₐⱼ(xᵢ)]. Two facts about determinants now do the physics. Exchanging particles i and j exchanges two rows, so the state changes sign for every pair. Two equal spin-orbitals make two equal columns, so the state is zero. The prefactor comes from counting: any two of the N! products differ in the orbital of at least one particle, so they are orthogonal, and each is normalised, so the unnormalised sum has ⟨Ψ|Ψ⟩ = N!. Dimensions check: each ψ carries m⁻³⁄², so Ψ carries m^(−3N/2) and |Ψ|² d³r₁ ⋯ d³r_N, summed over spins, is a pure number. Limiting cases check too: at N = 1 the determinant is ψₐ(x₁) with prefactor 1, and at N = 2 it is the pair of the previous section.

Three particles, written out and checked

Write abc for ψₐ(x₁)ψ_b(x₂)ψ_c(x₃). Expanding the 3 × 3 determinant gives Ψ = (abc − acb − bac + bca + cab − cba)/√6: the identity and the two cyclic orders carry +, the three single swaps carry −. Now test each exchange. P₁₂ swaps the first two letters of every term, pairing abc with bac, acb with cab, and bca with cba; each pair has opposite signs, so P₁₂Ψ = −Ψ. P₁₃ swaps the outer letters, pairing abc with cba, acb with bca, and bac with cab. P₂₃ pairs abc with acb, bac with bca, and cab with cba. All three give −Ψ, as the postulate requires. Now set c = a: abc and cba both become aba with opposite signs, and the other two pairs cancel the same way, so all six terms vanish. With a = 1s↑ and b = 1s↓ the 1s subshell has no third label, which is why lithium's third electron lives in n = 2.

Only the occupied set matters

Replace the occupied spin-orbitals by combinations of themselves, ψ′ⱼ = Σₖ ψₖUₖⱼ. The matrix of values is multiplied by U, and det(AB) = det A det B gives det[ψ′] = det U × det[ψ]: for unitary U, the same state up to a phase. A determinant records which subspace is filled, never which electron sits in which orbital, and its one-body density, n(x) = Σₐ |ψₐ(x)|², is the same in any basis of that subspace. Adding a multiple of one column to another has det U = 1 and can expose structure. For two same-spin fermions in oscillator levels 0 and 1, with ξ = x/ℓ, the determinant is π⁻¹⁄²(ξ₂ − ξ₁)e^{(−ξ₁²/2 − ξ₂²/2)} in units of ℓ⁻¹. For levels 0, 1 and 2, column operations turn each Hermite polynomial into a pure power, and the state becomes (2/√6)π⁻³⁄⁴(ξ₂ − ξ₁)(ξ₃ − ξ₁)(ξ₃ − ξ₂)e^{(−Σξ²/2)}: antisymmetry, and the zero at every coincidence, can be read off the factors.

Bosons: plus signs, and a count that depends on repeats

For identical bosons the same sum is taken with every sign +, a combination sometimes called a permanent. When all N labels differ, the N! products are orthonormal and 1/√(N!) normalises it, as for fermions. When labels repeat, the sum over permutations over-counts: with nₐ bosons in state a, each distinct product turns up nₐ! n_b! ⋯ times, so the normalised state is C = √(nₐ! n_b! ⋯ / N!) times the sum of the N!/(nₐ! n_b! ⋯) distinct products. Two bosons in one state need no symmetrising at all: the state is ψₐ(x₁)ψₐ(x₂), with C = 1, whereas the two-label formula (ab + ba)/√2 at b = a gives √2 ψₐ(x₁)ψₐ(x₂), with ⟨Ψ|Ψ⟩ = 2. Three bosons in a, a, b give (aab + aba + baa)/√3. Both limits check: every nₐ = 1 returns 1/√(N!), and all N in one state returns the bare product. Nothing caps nₐ, which is why bosons can crowd into one state and fermions cannot.

One determinant is exact only for independent particles

If H = Σᵢ h(i), every product of eigen-orbitals is an eigenstate with E = Σ εₐ, and a symmetric H commutes with every permutation, hence with the antisymmetriser. The determinant is then an exact eigenstate with the same energy: derived. Electrons are not independent. The repulsion e²/(4πε₀rᵢⱼ) acts on two coordinates at once, so acting on one determinant it produces a mixture of others, and the true ground state is a superposition of many determinants. The best single determinant, with orbitals chosen variationally so that each electron moves in the average field of the rest, is what the Hartree-Fock method seeks; the method is named here, not developed. For beryllium its energy, quoted from numerical tables, is −396.55 eV, against a measured −399.15 eV, the sum of the four ionisation energies. Almost all of the 2.6 eV gap, 0.65%, is the correlation energy. It is not relativistic: all these energies are of order α²mₑc², and corrections at α⁴mₑc² are a fraction of order (Zα)², below 10⁻³ here.

Change one variable at a time

Make the relationship visible.

Interactive model

Angle θ between orbitals a and b90 °

Particle 2 at ξ₂ = x₂/ℓ0.5

Exchange sign η (−1 same-spin fermions, +1 bosons)−1

Keep η = −1 and drag θ from 90° to 0°: the solid curve shrinks as sin θ and lies flat at θ = 0, where b = a, while the dashed product survives. That is exclusion. Then set η = +1 at θ = 0: ⟨Ψ|Ψ⟩ reads 2, so two bosons in one state need prefactor 1, not 1/√2.

OVERLAP ⟨a|b⟩0.00

SQUARED NORM ⟨Ψ|Ψ⟩1.000

ℓΨ AT ξ₁ = ξ₂0.000

BARE PRODUCT AT ξ₁ = ξ₂0.311

Live interpretationOVERLAP ⟨a|b⟩: 0.00. SQUARED NORM ⟨Ψ|Ψ⟩: 1.000. ℓΨ AT ξ₁ = ξ₂: 0.000. BARE PRODUCT AT ξ₁ = ξ₂: 0.311

Catch the common trap

Explain before calculating.

Misconception check

The error is to take "no two electrons share the same four quantum numbers" as the whole of the Pauli principle, so that any state avoiding a repeated label is allowed. Exclusion is only what antisymmetry says at a = b; the postulate constrains every state. The product ψ₁ₛ↑(x₁)ψ₂ₛ↑(x₂) repeats no label, yet it is illegal: it gives two same-spin electrons a nonzero probability density at one point, where the determinant vanishes. In helium the product, half symmetric and half antisymmetric in space, would put parallel spins in both 1s2s levels; the spectrum puts them only in the lower: 2³S₁, 19.82 eV above the ground state, splits three ways in a magnetic field, while 2¹S₀, 0.80 eV higher, is a singlet and does not split. The four quantum numbers are one basis: exclusion forbids a repeated spin-orbital in any orthonormal basis, because a determinant sees only the occupied subspace.

Two electrons occupy orthonormal spin-orbitals a and b in the Slater determinant Ψ. A student rebuilds the determinant from a′ = (a + b)/√2 and b′ = (a − b)/√2. How is the new state Ψ′ related to Ψ?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyBeryllium's ground state fills the spin-orbitals a = 1s↑, b = 1s↓, c = 2s↑ and d = 2s↓. How many products does its Slater determinant contain, what weight does each carry, and what is the coefficient of d(x₁)c(x₂)b(x₃)a(x₄)? What happens if a fifth electron is put in 2s↑?

A 4 × 4 determinant has 4! = 24 products, one for each way of handing the four spin-orbitals to the four electrons, and each carries weight 1/√(4!) = 1/√24 = 0.204 with a sign.
The sign is the parity of the permutation taking the order a, b, c, d to d, c, b, a. Two swaps do it, a ↔ d and then b ↔ c, so the permutation is even and the sign is +.
So the coefficient is +1/√24 = +0.204. Norm check: the 24 products are orthonormal, so the total is 24 × (0.204)² = 24 × 1/24 = 1.
A fifth electron in 2s↑ makes the 5 × 5 determinant contain column c twice, so it is identically zero: there is no such state. The fifth electron must take a spin-orbital outside the filled 1s and 2s subshells, and boron is 1s²2s²2p.
Measured check: boron's first ionisation energy, 8.30 eV, lies below beryllium's 9.32 eV, because the fifth electron sits in a new subshell. Which subshell it enters is a central-field question for a later topic; that it cannot enter 2s is exclusion.

Answer24 products of weight ±1/√24 = ±0.204; d(x₁)c(x₂)b(x₃)a(x₄) is an even permutation with coefficient +0.204. A fifth electron in 2s↑ gives a zero determinant, so it goes to a new subshell.

MediumTwo spin-up electrons are placed in hydrogen 1s orbitals centred on two protons a distance R apart: a = 1s_A↑ and b = 1s_B↑. The overlap of the two orbitals is S = e^(−R/a₀)[1 + R/a₀ + R²/(3a₀²)], quoted from the two-centre integral. At the measured H₂ bond length, 0.741 Å, find ⟨Ψ|Ψ⟩ for Ψ = [a(x₁)b(x₂) − b(x₁)a(x₂)]/√2 and the prefactor that normalises it. Repeat at R = 4.0 Å, and state the limit R → 0.

Squared norm: ⟨Ψ|Ψ⟩ = ½[⟨a|a⟩⟨b|b⟩ − ⟨a|b⟩⟨b|a⟩ − ⟨b|a⟩⟨a|b⟩ + ⟨b|b⟩⟨a|a⟩] = 1 − S², with S real here and the spin factor ⟨↑|↑⟩ = 1 in every term. S is dimensionless, as a ratio of lengths inside it must make it.
At R = 0.741 Å, R/a₀ = 0.741/0.5292 = 1.400, so S = e^(−1.400) × (1 + 1.400 + 0.653) = 0.2466 × 3.053 = 0.753.
Then ⟨Ψ|Ψ⟩ = 1 − 0.753² = 1 − 0.567 = 0.433. The 1/√2 prefactor leaves the state 57% short, and the correct prefactor is 1/√(2 × 0.433) = 1.07 in place of 0.707.
At R = 4.0 Å, R/a₀ = 7.56 and S = e^(−7.56) × (1 + 7.56 + 19.05) = 5.21 × 10⁻⁴ × 27.6 = 0.0144, so ⟨Ψ|Ψ⟩ = 1 − 2.1 × 10⁻⁴ = 0.9998. Antisymmetrising two electrons whose orbitals barely overlap changes almost nothing, and the effect falls off roughly as e^(−2R/a₀).
As R → 0 the bracket tends to 1, so S → 1 and ⟨Ψ|Ψ⟩ → 0: both electrons would share one 1s↑ spin-orbital, and the state disappears continuously. Exclusion is the end point of a smooth loss of norm.

AnswerAt 0.741 Å: S = 0.753, ⟨Ψ|Ψ⟩ = 0.433 and the normalising prefactor is 1.07, not 0.707. At 4.0 Å: S = 0.0144 and ⟨Ψ|Ψ⟩ = 0.9998. As R → 0 the norm goes to zero.

HardModel lithium (Z = 3) first without electron repulsion, H₀ = Σᵢ h(i). Show that the determinant of 1s↑, 1s↓ and 2s↑ hydrogenic orbitals is an exact eigenstate of H₀ and find its energy; find what exclusion costs; then compare with the measured ionisation energies 5.392, 75.640 and 122.454 eV and with the Hartree-Fock energy, −7.43273 hartree, quoted from numerical tables.

H₀ is unchanged by relabelling electrons, so it commutes with every permutation and with the antisymmetriser. Each of the six products is an eigenstate of H₀ with E = ε₁ₛ + ε₁ₛ + ε₂ₛ, so their signed sum is too: the determinant is exact for H₀. Derived.
Hydrogenic levels: εₙ = −Z² Ry/n², with Ry = α²mₑc²/2 = 13.606 eV, so ε₁ₛ = −122.45 eV and ε₂ₛ = −30.61 eV. Then E = 2(−122.45) − 30.61 = −275.5 eV. It is an energy, and of order α²mₑc², as a non-relativistic Coulomb energy must be.
Without exclusion, distinguishable particles or bosons could all take 1s: E = 3(−122.45) = −367.4 eV. Exclusion alone, with no force between the electrons, costs 367.4 − 275.5 = 91.8 eV.
Measured: removing all three electrons costs 5.392 + 75.640 + 122.454 = 203.486 eV: E = −203.49 eV. The model misses by 72.0 eV: e²/(4πε₀rᵢⱼ) is not a sum of one-body terms. Limit: Li²⁺ has one electron and no repulsion; hydrogenic 122.451 eV lies 0.003 eV, or 2.5 × 10⁻⁵, below the measured 122.454 eV, the net of quoted fractional shifts: reduced mass −8 × 10⁻⁵, relativistic (Zα)²/4 = +1.2 × 10⁻⁴, Lamb −1.6 × 10⁻⁵.
Hartree-Fock, the best single determinant: −7.43273 × 27.2114 eV = −202.25 eV, above the measured −203.49 eV as a variational energy must be. The remaining 1.23 eV, 0.61% of the total, is correlation, reachable only by superposing determinants.

AnswerExact eigenstate of H₀ with E = −275.5 eV (−367.4 eV if exclusion were lifted, a cost of 91.8 eV). Measured E = −203.49 eV; the best single determinant gives −202.25 eV, 1.23 eV (0.61%) too high.