University Quantum Mechanics II · Identical Particles and Many-Electron Atoms · 7.5

Spin and space together: two electrons

This is where Unit 5's singlet and triplet stop being spin algebra and start setting atomic energies. The Coulomb Hamiltonian never mentions spin, yet helium's 1s2s levels sit 0.796 eV apart by total spin. We trace that to one sign rule, rewrite it as a spin coupling you can calculate with, and find where the trick stops working.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Two electrons carry two positions and two spins, and the symmetrisation postulate asks only that the whole state change sign when both are swapped together. When the Hamiltonian ignores spin, its eigenstates can be chosen as one spatial factor times one spin factor, and Unit 5 has already sorted two-spin states into a symmetric triplet and an antisymmetric singlet. The sign must come from somewhere, so a symmetric spatial state is paired with the singlet and an antisymmetric one with the triplet: choosing the spin chooses the spatial wavefunction.

That is how a Hamiltonian with no spin in it produces spin-dependent energies. The repulsion is averaged differently over the two spatial states, giving J + K and J − K, and since the triplet's spatial state vanishes where the electrons meet, the triplet lies lower. Trade the spatial parity for the spin exchange operator and the same numbers read as a coupling −(2K/ħ²)S₁⋅S₂: electrostatic in origin, of order α²mₑc², and larger than the magnetic interaction of the two spins by a factor of order 1/α².

The picture has edges. It is first order in the repulsion with fixed orbitals; the triplet is guaranteed lower only for orthogonal orbitals; spin–orbit coupling mixes spin with space at relative order α²; and for three or more electrons no spin state is antisymmetric under every swap, so the state becomes a sum of products.

Simple definition: For two electrons and a spin-independent Hamiltonian, total antisymmetry pairs a symmetric spatial state with the spin singlet and an antisymmetric one with the triplet, so energies set by spatial symmetry become energies that depend on total spin.
Example: Helium's 1s2s pair: the triplet 2³S lies 19.820 eV and the singlet 2¹S 20.616 eV above the ground state. The 0.796 eV gap is electrostatic, not magnetic: to first order in the repulsion it is 2K, because the triplet's spatial state keeps the electrons apart.

The pairing rule (derived)Ψ = ψ_S(r₁, r₂)χ₀₀ or Ψ = ψ_A(r₁, r₂)χ₁ₘ, with m = −1, 0, +1

From P₁₂ = Pᵣ P_σ and P₁₂Ψ = −Ψ. Set a = b and ψ_A vanishes: a doubly occupied orbital holds only a singlet, Pauli exclusion as a limiting case.

ψ_S, ψ_A = [φₐ(r₁)φ_b(r₂) ± φ_b(r₁)φₐ(r₂)]/√2 in m⁻³ for orthonormal orbitals a ≠ b; spin states χ are dimensionless

Singlet and triplet energies (derived, first order)E_S = Eₐ + E_b + J + K and E_T = Eₐ + E_b + J − K

With V = 0 all four states sit at Eₐ + E_b: symmetry alone costs nothing. The repulsion puts the triplet 2K lower; helium's 1s2s measures 2K ≈ 0.796 eV.

J = ⟨ab|V|ab⟩ and K = ⟨ab|V|ba⟩ ≥ 0, with |ab⟩ = φₐ(r₁)φ_b(r₂); E_S is one state, E_T three; all in J or eV

The spin exchange operator (built from Unit 5)P_σ = ½(1 + 4S₁⋅S₂/ħ²)

It swaps the spin labels only. Since Pᵣ P_σ = −1 on every allowed state, the spatial parity Pᵣ can be traded for −P_σ, a pure spin operator.

S₁⋅S₂ in J² s² and ħ = 1.055 × 10⁻³⁴ J s, so P_σ is dimensionless: +1 on the triplet, −1 on the singlet

Exchange written as a spin coupling (derived)H_eff = Eₐ + E_b + J − K/2 − (2K/ħ²)S₁⋅S₂

S₁⋅S₂ = ħ²/4 returns E_T and −3ħ²/4 returns E_S. The constant is the 3 : 1 weighted centroid, and the coupling constant 2K is electrostatic.

Acts on the two spins alone, within the four states of one configuration ab, to first order in V; every term in J or eV

The magnetic spin–spin scale (derived)μ₀μ_B²/(4πa₀³) = α²Eₕ/4 = 0.362 meV

Order α⁴mₑc², against α²mₑc² for K: a magnetic coupling of the spins falls some 2200 times short of helium's 0.796 eV. The splitting is exchange.

μ_B = eħ/(2mₑ) = 9.274 × 10⁻²⁴ J T⁻¹, a₀ = 52.9 pm, Eₕ = α²mₑc² = 27.2 eV, α = 1/137.036

Antisymmetric spin states of N electrons (derived)d_A(N) = C(2, N), so d_A(2) = 1 and d_A(N) = 0 for N ≥ 3

For N = 2 that state is the singlet, so ψ_S χ₀₀ is one product. For N ≥ 3 one product survives only at maximum spin S = N/2, with totally antisymmetric space; lower S needs a sum.

d_A counts the states of N spin-halves that change sign under every swap, a pure number; the 2 is one spin's two states

Antisymmetry ties the spin label to the spatial symmetry

When H contains no spin operator, its eigenstates can be chosen as one spatial factor times one spin factor, Ψ = ψ(r₁, r₂)χ. Exchanging the electrons swaps positions and spins together, so the full exchange operator factorises, P₁₂ = Pᵣ P_σ, and the symmetrisation postulate demands P₁₂Ψ = −Ψ. Unit 5 supplies the spin half, quoted: all three triplet states χ₁ₘ are symmetric and the singlet χ₀₀ is antisymmetric. The product is therefore antisymmetric only as ψ_S χ₀₀ or ψ_A χ₁ₘ, where ψ_S and ψ_A are [φₐ(r₁)φ_b(r₂) ± φ_b(r₁)φₐ(r₂)]/√2, upper sign for ψ_S, for orthonormal orbitals a ≠ b. Count the states as a check. Four spin-orbitals a↑, a↓, b↑ and b↓ take two fermions in (4 × 3)/2 = 6 ways. The pairing gives 1 + 3 = 4 with one electron in each orbital; the other two are a² and b², where ψ_A vanishes identically, so a doubly occupied orbital exists only as a singlet. Pauli exclusion is the a = b case of the pairing rule, not an extra law, and helium obliges: its 1s² ground state is a singlet, and the lowest triplet, 1s2s ³S, lies 19.820 eV above it (measured).

One Slater determinant is usually not a spin eigenstate

Write the determinant with an up spin in a and a down spin in b: D(a↑, b↓) = [φₐ(r₁)φ_b(r₂)|↑↓⟩ − φ_b(r₁)φₐ(r₂)|↓↑⟩]/√2. It is antisymmetric, as every determinant is, but it is neither paired state. Substitute φₐ(r₁)φ_b(r₂) = (ψ_S + ψ_A)/√2 and φ_b(r₁)φₐ(r₂) = (ψ_S − ψ_A)/√2 and collect: D(a↑, b↓) = (ψ_S χ₀₀ + ψ_A χ₁₀)/√2, an equal mixture of singlet and triplet M = 0. Measure S² on it and you find 0 or 2ħ², with probability ½ each. The paired states need two determinants, ψ_S χ₀₀ = [D(a↑, b↓) − D(a↓, b↑)]/√2 and ψ_A χ₁₀ = [D(a↑, b↓) + D(a↓, b↑)]/√2, while the stretched triplet ψ_A χ₁₁ = D(a↑, b↑) needs only one. Saying which orbital holds the up spin fixes M but not S, and S is what the energy is about to follow.

A spin-free Hamiltonian gives spin-dependent energies

Take H = h(1) + h(2) + V(r₁₂), with no spin operator anywhere. Without V, ψ_S and ψ_A share the energy Eₐ + E_b: symmetry by itself costs nothing. Switch V on and average it over each paired state to first order; within one configuration they are already the good states, because V commutes with Pᵣ and with S², and the next topic runs helium through the full machinery. The result is E = Eₐ + E_b + J ± K, upper sign for the singlet. The direct integral J is the classical repulsion of the clouds |φₐ|² and |φ_b|²; the exchange integral K comes only from the cross terms of the symmetrised product. For Coulomb repulsion K is the Coulomb energy of the overlap density φₐ*φ_b interacting with its own complex conjugate, positive because the kernel's Fourier transform e²/(ε₀q²) is positive. So the triplet lies 2K lower: its spatial state vanishes at r₁ = r₂, the exchange hole of the last topic, and the pair pays less repulsion. Checks: each term is an energy; orbitals that do not overlap give K → 0 and four degenerate states; helium's triplets lie lower, by 0.796 eV in 1s2s and 0.254 eV in 1s2p, measured.

The same energies, written as a spin coupling

Unit 5's values of S₁⋅S₂, ħ²/4 on the triplet and −3ħ²/4 on the singlet, build the spin exchange operator P_σ = ½(1 + 4S₁⋅S₂/ħ²), which is +1 on the triplet and −1 on the singlet. On every allowed state Pᵣ P_σ = −1, so Pᵣ = −P_σ there, and the energy Eₐ + E_b + J + K Pᵣ becomes H_eff = Eₐ + E_b + J − K/2 − (2K/ħ²)S₁⋅S₂. Check it: S₁⋅S₂ = ħ²/4 on the triplet returns Eₐ + E_b + J − K, and −3ħ²/4 on the singlet returns Eₐ + E_b + J + K. The constant is not the midpoint of the two levels but their 3 : 1 weighted mean, because S₁⋅S₂ is traceless. H_eff acts on spins alone, yet every number in it is electrostatic, and it moves spins as well as sorting them: its flip-flop part turns D(a↑, b↓) into D(a↓, b↑) with probability sin²(Kt/ħ), a full swap after πħ/(2K), or 2.60 fs for helium's measured 2K = 0.796 eV. This Heisenberg–Dirac coupling is the starting point of the Heisenberg model of magnetism; spread over an open shell, the same preference is Hund's first rule, which puts carbon's 2p² ³P 1.26 eV below its ¹D (measured). Both are named here, not developed.

Not magnetism: a check by orders of α

Could the 0.796 eV be the magnetic interaction of the two spin moments instead? Estimate it. Two moments of one Bohr magneton, μ_B = eħ/(2mₑ), a Bohr radius apart interact with an energy of order μ₀μ_B²/(4πa₀³). Put μ₀ = 1/(ε₀c²) and a₀ = ħ/(mₑcα) and it becomes α²Eₕ/4 = 0.362 meV, with Eₕ = α²mₑc² = 27.2 eV. The magnetic term is of order α⁴mₑc², while J and K, Coulomb integrals over atomic orbitals, are of order α²mₑc²: helium's gap is some 2200 times too large to be magnetic. The spectrum confirms the ranking. The 2¹P − 2³P exchange gap is 0.254 eV, while the whole fine structure of 2³P, spin–orbit and spin–spin together, spans 31.9 GHz, or 1.32 × 10⁻⁴ eV, both measured and quoted from tables. Solids follow the same arithmetic: the dipole energy between neighbouring iron moments of 2.2μ_B, 0.248 nm apart, is only of order k_B × 0.20 K, yet iron stays ferromagnetic up to 1043 K.

Three electrons break the factorisation

The pairing leaned on a fact about two spins: the swap sorts their states into just two symmetry types, and these coincide with the two multiplets, triplet symmetric and singlet antisymmetric. Three spin-halves span 2³ = 8 states: a symmetric quartet with S = 3/2 and two doublets with S = ½, whose states can be symmetric or antisymmetric under one swap but never under all three. No spin state is antisymmetric under all three swaps, because antisymmetrising three factors drawn from a two-state space gives zero. So only the quartet can multiply a single antisymmetric spatial function, and every doublet state is a sum of products. Expand lithium's ground state D(a↑, a↓, b↑), with a = 1s and b = 2s: its six terms group, by which electron is down, into [φ_b(r₁)φₐ(r₂) − φₐ(r₁)φ_b(r₂)]φₐ(r₃)|↑↑↓⟩ and its two cyclic relabellings, each weighted 1/√6 and each with a different spatial function. Spin still labels the spatial symmetry, but through the permutation group, which this course does not develop.

Change one variable at a time

Make the relationship visible.

Interactive model

Exchange integral K0.40 eV

Time since preparing D(a↑, b↓)2.6 fs

At K = 0.40 eV, near helium's 1s2s, the spins swap fully at about 2.6 fs. Raise K to 0.60 eV: the levels spread further and the swap comes at 1.72 fs. Drag K to 0.02 eV and the four levels merge while the swap takes 52 fs: orbitals that barely overlap barely exchange.

GAP E_S − E_T = 2K0.80 eV

S₁⋅S₂/ħ² COEFFICIENT-0.80 eV

FIRST FULL SWAP πħ/(2K)2.58 fs

P(SWAPPED) AT t1.000

Live interpretationGAP E_S − E_T = 2K: 0.80 eV. S₁⋅S₂/ħ² COEFFICIENT: −0.80 eV. FIRST FULL SWAP πħ/(2K): 2.58 fs. P(SWAPPED) AT t: 1.000

Catch the common trap

Explain before calculating.

Misconception check

The tempting shorthand is that the singlet means opposite spins and the triplet parallel ones, so an up spin in orbital a with a down spin in b gets called a singlet. It is not one. The determinant D(a↑, b↓) fixes M = 0 but not S: it equals (ψ_S χ₀₀ + ψ_A χ₁₀)/√2, half singlet and half triplet, because the M = 0 triplet is antiparallel along z as well. The difference lies along the other two axes: in χ₁₀ the spins are correlated along x and y, so ⟨S₁⋅S₂⟩ = ħ²/4, while the singlet is anticorrelated along every axis, giving −3ħ²/4. In the z basis the two differ only in the relative sign of |↑↓⟩ and |↓↑⟩, and antisymmetry hands that sign to the spatial state, which is what the repulsion sees. So D(a↑, b↓) is not stationary: with helium's 2K = 0.796 eV it turns wholly into D(a↓, b↑) after 2.60 fs, which no energy eigenstate could do.

In helium the 1s2s triplet 2³S lies 19.820 eV and the singlet 2¹S 20.616 eV above the ground state. Which operator, acting on the two electron spins alone, reproduces both levels?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTwo electrons occupy orthonormal orbitals a and b of a model atom, with Eₐ + E_b = −9.60 eV. Their repulsion gives the direct integral J = 5.20 eV and the exchange integral K = 0.35 eV. Pair each spatial state with its spin state, find each level's energy and degeneracy to first order, and write the spin operator that reproduces them.

Antisymmetry pairs the symmetric spatial state ψ_S with the singlet χ₀₀ and the antisymmetric ψ_A with the three triplet states χ₁ₘ: 1 + 3 = 4 states, the same as the 2 × 2 ways of giving each orbital a spin.
Singlet: E_S = Eₐ + E_b + J + K = −9.60 + 5.20 + 0.35 = −4.05 eV, one state. Triplet: E_T = −9.60 + 5.20 − 0.35 = −4.75 eV, three states. The triplet lies lower by 2K = 0.70 eV.
Spin operator: H_eff = Eₐ + E_b + J − K/2 − (2K/ħ²)S₁⋅S₂ = −4.575 eV − (0.70 eV)S₁⋅S₂/ħ². Triplet: −4.575 − 0.70 × ¼ = −4.75 eV. Singlet: −4.575 + 0.70 × ¾ = −4.05 eV.
Checks: the constant equals the weighted mean [3 × (−4.75) + (−4.05)]/4 = −4.575 eV, as a traceless S₁⋅S₂ requires; every term is in eV; and as K → 0 all four states merge at Eₐ + E_b + J = −4.40 eV.

AnswerSinglet ψ_S χ₀₀ at −4.05 eV, one state; triplet ψ_A χ₁ₘ at −4.75 eV, three states; H_eff = −4.575 eV − (0.70 eV)S₁⋅S₂/ħ².

MediumTwo electrons, one in each of orthonormal orbitals a and b, start at t = 0 in the determinant D(a↑, b↓). The Hamiltonian is spin-independent and puts the singlet ΔE = 0.796 eV above the triplet, helium's measured 1s2s gap. Find the probability of D(a↓, b↑) at time t, its value at 1.00 fs, and the first full-swap time. How long would a swap take if the splitting were a magnetic 0.362 meV?

Decompose, as derived above: D(a↑, b↓) = (ψ_S χ₀₀ + ψ_A χ₁₀)/√2 and D(a↓, b↑) = (−ψ_S χ₀₀ + ψ_A χ₁₀)/√2.
Each paired state is stationary, so Ψ(t) = [e^{(−iE_S t/ħ)}ψ_S χ₀₀ + e^{(−iE_T t/ħ)}ψ_A χ₁₀]/√2.
Project: ⟨D(a↓, b↑)|Ψ(t)⟩ = ½[e^{(−iE_T t/ħ)} − e^{(−iE_S t/ħ)}], whose squared modulus is ½[1 − cos(ΔE t/ħ)] = sin²(ΔE t/(2ħ)). A Hamiltonian with no spin in it moves the up spin from a to b.
At t = 1.00 fs: ΔE t/(2ħ) = (0.796 eV × 1.00 × 10⁻¹⁵ s)/(2 × 6.582 × 10⁻¹⁶ eV s) = 0.605 rad, so P = sin²(0.605) = 0.323. A full swap needs ΔE t/(2ħ) = π/2, so t = h/(2ΔE) = 4.136 × 10⁻¹⁵ eV s/(1.592 eV) = 2.60 × 10⁻¹⁵ s.
Magnetic only: t = 4.136 × 10⁻¹⁵ eV s/(2 × 3.62 × 10⁻⁴ eV) = 5.71 × 10⁻¹² s, about 2200 times slower. Checks: eV s over eV is s, and ΔE → 0, as for electrons whose orbitals never overlap, means no swap at all.

AnswerP(t) = sin²(ΔE t/(2ħ)), which is 0.323 at 1.00 fs; the spins first swap fully at 2.60 fs. A 0.362 meV magnetic splitting would need 5.71 ps.

HardLithium's ground configuration is 1s²2s. Write its M_S = +½ state as the determinant D(a↑, a↓, b↑), with a = 1s and b = 2s, and expand it. Show that it cannot be one spatial function times one spin function, find the probability that electron 1 is spin-down and ⟨S₁z⟩, and explain why this configuration has no S = 3/2 state.

The six terms, each weighted 1/√6, group by which electron is down: Ψ = [f₃|↑↑↓⟩ + f₂|↑↓↑⟩ + f₁|↓↑↑⟩]/√6, with f₃ = [φ_b(r₁)φₐ(r₂) − φₐ(r₁)φ_b(r₂)]φₐ(r₃), f₂ = φₐ(r₂)[φₐ(r₁)φ_b(r₃) − φ_b(r₁)φₐ(r₃)] and f₁ = φₐ(r₁)[φ_b(r₂)φₐ(r₃) − φₐ(r₂)φ_b(r₃)].
One product ψχ would make f₁, f₂ and f₃ multiples of a single function. They are not: f₃ vanishes whenever r₁ = r₂, while f₂(r, r, r₃) = φₐ(r)[φₐ(r)φ_b(r₃) − φ_b(r)φₐ(r₃)] does not. Nor are they independent, since f₁ + f₂ + f₃ = 0.
Each fᵢ is a difference of two orthonormal products, so ⟨fᵢ|fᵢ⟩ = 2 and each spin pattern carries probability 2/6 = 1/3; the three sum to 1.
Electron 1 is down only in |↓↑↑⟩, so P = 1/3, and ⟨S₁z⟩ = (2/3)(ħ/2) + (1/3)(−ħ/2) = ħ/6 = 1.76 × 10⁻³⁵ J s. Check: three electrons give 3 × ħ/6 = ħ/2, which is M_S ħ.
S = 3/2 contains |↑↑↑⟩, which is symmetric, so its spatial part would have to be antisymmetric in all three electrons. With two of them in orbital a, that antisymmetrised product has two equal columns and vanishes, so the configuration is a doublet only, as lithium's measured ²S ground term confirms.

AnswerΨ is a sum of three different space-times-spin products, so no single product exists; P(electron 1 down) = 1/3 and ⟨S₁z⟩ = ħ/6 = 1.76 × 10⁻³⁵ J s; a quartet needs a totally antisymmetric spatial state, which vanishes with two 1s electrons.