University Physics IV · Atomic Physics · 11.1

Identical Particles & Exchange Symmetry

Electrons carry no serial numbers. Follow that one fact through and you get the Pauli principle for free, an apparent force that appears nowhere in the Hamiltonian, and the 0.80 eV that puts triplet helium below singlet helium.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Two electrons are not two objects wearing faint name tags. Write ψₐ(r₁)ψ_b(r₂) and you have already said more than you can measure — that the electron you called 1 is the one in orbital a. Since only |Ψ|² is observable and relabelling cannot change it, Ψ(2,1) = e^(iφ)Ψ(1,2); swapping twice must return the original state, so e^(2iφ) = 1 and in three dimensions the only options are Ψ → +Ψ and Ψ → −Ψ.

Which one a species takes is fixed by its spin — integer symmetric, half-integer antisymmetric — a result imported here from relativistic quantum field theory rather than derived. For electrons the whole state, space times spin, must change sign, and that single constraint does most of atomic physics. The antisymmetric combination vanishes identically when both orbitals are the same, which is the Pauli principle arriving as a corollary rather than a postulate.

And because a symmetric spatial factor draws the electrons together while an antisymmetric one holds them apart, the ordinary Coulomb repulsion returns two different energies for the two spin arrangements: in helium's 1s2s configuration, 0.796 eV apart with the spin triplet lower. What it costs you is correlation. A symmetrised product of orbitals is still an independent-particle guess, and the electrons' real, instant-by-instant avoidance of each other is nowhere inside it.

Simple definition: Exchange symmetry is the requirement that the state of two identical particles either keeps its sign or reverses it when the two particle labels are swapped — symmetric for bosons, antisymmetric for fermions — because the labels themselves are not observable.
Example: Helium 1s2s: the spin triplet must carry the antisymmetric spatial factor and sits 19.820 eV above the ground state, while the singlet carries the symmetric one and sits 20.616 eV up — a 0.796 eV gap produced by symmetry alone.

Exchange operator and its two eigenvaluesP₁₂Ψ(1,2) = Ψ(2,1)P₁₂² = 1 ⇒ P₁₂Ψ = ±Ψ

+1 for bosons, −1 for fermions. In three dimensions there is no third option to take.

[H, P₁₂] = 0 for identical particles, so the sign is a constant of the motion

Symmetrised two-particle spatial stateψ±(r₁, r₂) = [ψₐ(r₁)ψ_b(r₂) ± ψ_b(r₁)ψₐ(r₂)] / √2

ψ₋ vanishes identically when a = b — exclusion drops out of this line, it is not assumed.

ψₐ and ψ_b orthonormal and distinct; the 1/√2 is the wrong constant if a = b

Space and spin trade signsΨ = ψ₊⋅χ_singlet (para, S = 0) or ψ₋⋅χₜᵣᵢₚₗₑₜ (ortho, S = 1)

Only the product must be antisymmetric, so choosing S chooses the spatial symmetry.

χ_singlet = (↑↓ − ↓↑)/√2; χₜᵣᵢₚₗₑₜ = ↑↑, (↑↓ + ↓↑)/√2, ↓↓

Direct and exchange integralsJ = ⟨ab|V|ab⟩, K = ⟨ab|V|ba⟩, with V = e²/4πε₀r₁₂

J is the classical repulsion of two charge clouds; K has no classical counterpart at all.

J and K in joules or eV; r₁₂ = |r₁ − r₂| in metres; K > 0 for a Coulomb kernel

First-order energy of the two termsE± = Eₐ + E_b + J ± K, so E(singlet) − E(triplet) = 2K

Helium 1s2s measures 2K = 0.796 eV, so K = 0.398 eV and the triplet is the lower term.

+ for the symmetric space state (singlet), − for the antisymmetric one (triplet)

Mean-square separation of the pair⟨(x₁−x₂)²⟩± = ⟨x²⟩ₐ + ⟨x²⟩_b − 2⟨x⟩ₐ⟨x⟩_b ∓ 2|⟨x⟩_ab|²

Symmetric pairs sit closer, antisymmetric pairs further apart — no force, only counting.

⟨x⟩_ab = ∫ψₐ*(x) x ψ_b(x) dx, in metres; the last term dies without orbital overlap

The label is yours, not nature's

A two-particle wavefunction Ψ(r₁, r₂) carries labels 1 and 2 that no apparatus can read. Define the exchange operator by P₁₂Ψ(1,2) = Ψ(2,1). For identical particles the Hamiltonian is symmetric in those labels — same mass, same charge, same interaction — so [H, P₁₂] = 0 and the exchange symmetry is a constant of the motion: decided once, kept for all time. Because swapping twice restores the original arrangement, P₁₂² = 1, so the eigenvalues are +1 and −1 and nothing else. That is why 'somewhere in between' is not on offer. A state that was neither symmetric nor antisymmetric would make |Ψ|² change when you relabelled, which is the same as saying the labels are measurable. Particles taking +1 are bosons and those taking −1 are fermions; that the choice follows the spin — integer or half-integer — is the spin-statistics theorem, proved inside relativistic quantum field theory and imported here without proof.

Two combinations, and one of them can vanish

Take two orthonormal one-particle orbitals ψₐ and ψ_b. The plain product ψₐ(r₁)ψ_b(r₂) is not an eigenstate of P₁₂, but the two combinations ψ±(r₁, r₂) = [ψₐ(r₁)ψ_b(r₂) ± ψ_b(r₁)ψₐ(r₂)]/√2 are, with eigenvalues ±1. Now set b = a. The symmetric state survives as ψₐ(r₁)ψₐ(r₂), while the antisymmetric one collapses to ψₐ(r₁)ψₐ(r₂) − ψₐ(r₁)ψₐ(r₂) = 0 at every point. There is no state at all — not a forbidden one, an absent one. That is the Pauli exclusion principle, and notice where it came from: not from a rule about how many electrons a level holds, but from antisymmetry, which came from indistinguishability. Written for N particles the same object is a Slater determinant, and a determinant with two equal rows vanishes for exactly this reason. Bosons, taking the plus sign, carry no such restriction, and pile into a single mode happily.

Space and spin trade signs

The antisymmetry demanded of electrons applies to the whole state, spin included. Two spin-half particles combine into one antisymmetric singlet, χ = (↑↓ − ↓↑)/√2 with S = 0, and three symmetric triplet states, ↑↑, (↑↓ + ↓↑)/√2 and ↓↓, with S = 1. If the state factorises as Ψ = ψ(space)⋅χ(spin), the two factors must carry opposite exchange signs, because their product has to be antisymmetric. So a spin singlet is locked to the symmetric spatial function and a spin triplet to the antisymmetric one — para and ortho helium in the older spectroscopic language, two families whose separate line series once looked like two different gases. Nothing here says that spin interacts with anything. Spin is the bookkeeping partner: fixing S fixes the spatial symmetry, and it is the spatial symmetry that the Coulomb energy actually sees.

The exchange force is a consequence of counting

Work out the mean-square separation of the two symmetrised states in one dimension and the algebra returns ⟨(x₁−x₂)²⟩± = ⟨x²⟩ₐ + ⟨x²⟩_b − 2⟨x⟩ₐ⟨x⟩_b ∓ 2|⟨x⟩_ab|², where ⟨x⟩_ab = ∫ψₐ*(x) x ψ_b(x) dx. The first three terms are what two distinguishable particles would give. The last term is new, and its sign is set by the exchange symmetry: symmetric states sit closer together, antisymmetric states further apart. Two consequences follow. First, no force has been introduced — the Hamiltonian is untouched, and what changed is which configurations the wavefunction permits. Second, the effect requires overlap, because ⟨x⟩_ab integrates over the region where both orbitals are appreciable. Two electrons in orbitals that never overlap — one here, one on the Moon — feel nothing from each other's statistics, and symmetrising them is a formality with no measurable consequence.

Helium: one repulsion, two answers

Take helium's 1s2s configuration and treat the electron-electron repulsion V = e²/4πε₀r₁₂ as a first-order perturbation on ψ±. The expectation value splits into two integrals. The direct integral J = ⟨ab|V|ab⟩ is the electrostatic repulsion of the two charge clouds and is identical for both states. The exchange integral K = ⟨ab|V|ba⟩ multiplies the cross term and so enters with the exchange sign, giving E± = Eₐ + E_b + J ± K. For a positive kernel such as Coulomb repulsion K is positive, so the triplet, carrying the minus sign, lies 2K below the singlet. Experiment agrees: the 2³S level sits 19.820 eV above helium's ground state and the 2¹S sits 20.616 eV, so 2K = 0.796 eV and K = 0.398 eV. Evaluating K with unscreened hydrogenic Z = 2 orbitals gives (16/729)Z hartree = 1.19 eV — right sign, right order, three times too big, because the real 2s electron is screened toward Z_eff ≈ 1, so its orbital is about twice as wide and overlaps the 1s far less.

What the symmetrised product still leaves out

Two debts stay outstanding. The first is spin-statistics itself: nothing in non-relativistic quantum mechanics forces half-integer spin to take the minus sign. It follows from demanding a relativistic field theory with an energy bounded below whose fields commute at spacelike separation, and in two dimensions the argument fails outright, which is how anyons become possible. The second is correlation. Even after symmetrising, ψ± is assembled from single-particle orbitals, so each electron moves in an average field. Real electrons dodge one another instant by instant, digging a Coulomb hole around each that no single determinant can reproduce. The exchange hole is real but partial — it separates parallel-spin electrons and does nothing for antiparallel ones. Helium's exact non-relativistic ground state is −79.01 eV against −77.87 eV at the Hartree-Fock limit, and that 1.14 eV shortfall is precisely what the term 'correlation energy' names.

Change one variable at a time

Make the relationship visible.

Interactive model

Electron 1 position x₁/L0.30 L

Exchange mixing angle θ45 °

Set θ = 45°, the symmetric spatial factor that pairs with the spin singlet, and the solid curve doubles the dashed one exactly at x₂ = x₁. Set θ = 135°, the antisymmetric factor the triplet needs, and it drops to zero there — the exchange hole. No θ in between is an eigenstate of the swap.

PAIR DENSITY AT x₂ = x₁4.74

DIRECT PART THERE2.37

RATIO AT x₂ = x₁2.00

OVERLAP ψ₁(x₁)ψ₂(x₁)L1.54

Live interpretationPAIR DENSITY AT x₂ = x₁: 4.74. DIRECT PART THERE: 2.37. RATIO AT x₂ = x₁: 2.00. OVERLAP ψ₁(x₁)ψ₂(x₁)L: 1.54

Catch the common trap

Explain before calculating.

Misconception check

The 0.796 eV between helium's singlet and triplet is not magnetic. The temptation is to say that parallel spins are parallel magnets and magnets know about each other — but the dipole-dipole energy of two electron moments 0.1 nm apart is μ₀μ_B²/4πr³ ≈ 5 × 10⁻⁵ eV, four orders of magnitude too small, and the Hamiltonian that produced the splitting contains no spin operator at all. Spin enters only as a label: fixing S fixes the exchange symmetry of the spin factor, that forces the opposite symmetry on the spatial factor, that changes how close the two electrons sit, and that changes the ordinary electrostatic repulsion. The energy is Coulomb from start to finish. "Exchange force" names a consequence of symmetry, not a fifth interaction — and it switches off entirely once the orbitals stop overlapping, because K is an overlap integral.

In helium's 1s2s configuration the spin triplet lies 0.80 eV below the singlet. What sets that gap?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTwo electrons occupy the n = 1 and n = 2 levels of a one-dimensional infinite square well of width L = 1.0 nm, in a spin-triplet state. Write down the spatial wavefunction with the correct exchange symmetry, show what happens if both electrons are put in n = 1 instead, and give the total energy of the pair, ignoring their mutual repulsion.

A triplet spin factor is symmetric under swapping the two labels, and the total electron state must be antisymmetric, so the spatial factor is forced to carry the minus sign.
With ψₙ(x) = √(2/L) sin(nπx/L), the antisymmetric combination is ψ₋(x₁, x₂) = [ψ₁(x₁)ψ₂(x₂) − ψ₂(x₁)ψ₁(x₂)]/√2. Put x₂ = x₁ and the bracket cancels: two parallel-spin electrons are never found at the same point.
Now put both electrons in n = 1. The bracket becomes ψ₁(x₁)ψ₁(x₂) − ψ₁(x₁)ψ₁(x₂) = 0 everywhere, so there is no such state. That is exclusion, obtained rather than assumed.
Energies: Eₙ = n²h²/8mL². With L = 1.0 nm, h²/8mL² = (6.626×10⁻³⁴)² / (8 × 9.109×10⁻³¹ × 1.0×10⁻¹⁸) = 6.02×10⁻²⁰ J = 0.376 eV, so E = (1² + 2²) × 0.376 eV.

Answerψ₋(x₁, x₂) = [ψ₁(x₁)ψ₂(x₂) − ψ₂(x₁)ψ₁(x₂)]/√2, which vanishes at x₁ = x₂ and vanishes identically if both electrons are given n = 1. E = 5h²/8mL² = 1.88 eV.

MediumHelium's 2³S₁ level lies 19.820 eV above the ground state and its 2¹S₀ level lies 20.616 eV above it; both belong to the 1s2s configuration. Find the exchange integral K for this configuration, then compare it with the unscreened hydrogenic prediction K(1s,2s) = (16/729)Z hartree, taking 1 hartree = 27.2114 eV.

Both terms come from the same configuration, so they share the same unperturbed energies and the same direct integral J. First-order perturbation theory gives E± = E₁s + E₂s + J ± K.
Subtracting kills everything except the exchange term: E(singlet) − E(triplet) = (J + K) − (J − K) = 2K. The measured difference is 20.616 − 19.820 = 0.796 eV.
So K = 0.796/2 = 0.398 eV, and it comes out positive — which is why the triplet, taking the minus sign, is the lower of the two terms.
Hydrogenic prediction with the bare nuclear charge Z = 2: K = (16/729)(2) = 32/729 = 0.043896 hartree = 0.043896 × 27.2114 = 1.194 eV.
That is 1.194/0.398 = 3.0 times the measured value. The 1s electron screens the nucleus, so the 2s electron sees an effective charge nearer 1 than 2; its orbital is about twice as wide, it overlaps the 1s far less, and K — an overlap integral — shrinks with it.

AnswerK = 0.398 eV, so the exchange splitting is 2K = 0.796 eV with the triplet lower. The unscreened hydrogenic estimate of 1.194 eV overshoots by a factor of three because it ignores screening of the 2s orbital.

HardTwo electrons sit in the n = 1 and n = 2 levels of an infinite well of width L = 1.0 nm. Using ⟨x⟩ₙ = L/2, ⟨x²⟩ₙ = L²[1/3 − 1/(2n²π²)] and ⟨x⟩₁₂ = −16L/9π², find the root-mean-square separation for the unsymmetrised product, for the symmetric spatial state and for the antisymmetric one, and say which spin state each belongs to.

Use ⟨(x₁−x₂)²⟩± = ⟨x²⟩₁ + ⟨x²⟩₂ − 2⟨x⟩₁⟨x⟩₂ ∓ 2|⟨x⟩₁₂|², upper sign for the symmetric state. Dropping the final term leaves the unsymmetrised product.
⟨x²⟩₁ = L²(1/3 − 1/2π²) = 0.28267L² and ⟨x²⟩₂ = L²(1/3 − 1/8π²) = 0.32067L²; also 2⟨x⟩₁⟨x⟩₂ = 2(L/2)(L/2) = 0.5L².
Unsymmetrised: 0.28267 + 0.32067 − 0.5 = 0.10334L², so the rms separation is √0.10334 L = 0.3215L = 0.321 nm.
Exchange term: ⟨x⟩₁₂ = −16L/9π² = −0.18013L, so 2|⟨x⟩₁₂|² = 2 × 0.032445 L² = 0.06489L².
Symmetric: 0.10334 − 0.06489 = 0.03845L², giving 0.1961L = 0.196 nm. Antisymmetric: 0.10334 + 0.06489 = 0.16823L², giving 0.4102L = 0.410 nm.
The symmetric spatial state pairs with the spin singlet and the antisymmetric one with the triplet. The triplet holds the electrons 2.1 times further apart, so it pays less Coulomb repulsion — helium's ordering, though a one-dimensional box overstates the size of the effect.

AnswerUnsymmetrised 0.321 nm; symmetric (singlet) 0.196 nm; antisymmetric (triplet) 0.410 nm. The triplet's electrons sit 2.1 times further apart, which is why the exchange integral pushes that term below the singlet.