University Quantum Mechanics II · Identical Particles and Many-Electron Atoms · 7.6
Parahelium, orthohelium and the exchange integral
Helium's lines fall into two families that almost never connect, and the cause is not magnetism. Here degenerate perturbation theory, run on the 1s2s level, shows where the singlet–triplet gap comes from, why the triplet lies lower, and why one excited state survives for over two hours.
Build the model
Connect the measurement to the mechanism.
Excited helium is where identical-particle symmetry is seen to move energy levels without any new force. With the two electrons independent in the Z = 2 Coulomb field, 1s2s lies at −68.0 eV, degenerate with its own exchanged twin and with six 1s2p states. Switch on the repulsion Vₑₑ as a perturbation.
It depends only on the distance between the electrons, so it conserves total L and never mixes the L = 0 pair with the L = 1 sextet; it is symmetric in the labels, so inside the pair the exchange operator P chooses the good states, the symmetric and antisymmetric combinations, shifted by J + K and J − K. J is the classical repulsion of two charge clouds. K is the self-repulsion of the overlap charge eψ₁ₛψ₂ₛ, has no classical counterpart, and is positive.
Antisymmetry pairs symmetric space with the spin singlet, parahelium, and antisymmetric space with the triplet, orthohelium, which lies lower by 2K because its electrons avoid one another. The dipole operator ignores spin, so the two radiate as separate spectra and 2³S₁, with no triplet beneath it, lives about 8000 s. The model stops at magnitudes: bare Z = 2 orbitals put both levels 1.4 to 3.0 eV too high and the gap three times too wide, because the outer electron is screened, and ΔS = 0 fails weakly once spin–orbit coupling enters.
- Simple definition
- The exchange integral K is the matrix element of the electron–electron repulsion between ψ₁ₛ(r₁)ψ₂ₛ(r₂) and the same product with the electrons swapped; relative to E⁽⁰⁾ + J it raises the spatially symmetric 1s2s state by K and lowers the antisymmetric one by K.
- Example
- Measured, 2³S₁ lies 19.820 eV and 2¹S₀ 20.616 eV above helium's ground state, so 2K = 0.796 eV and K = 0.398 eV: a third of the 1.19 eV that bare Z = 2 orbitals give, and 2K is about 2200 times the magnetic energy scale of two spins a₀ apart.
Derived from the hydrogenic levels of Unit 2. It is the zero of first-order theory, and its eight-fold degeneracy is why Vₑₑ must be diagonalised inside a block.
R = 13.606 eV (1 eV = 1.602 × 10⁻¹⁹ J); the six 1s2p states share this energy, eight spatial states in all
The Coulomb energy of two classical clouds of charge −e|ψ₁ₛ|² and −e|ψ₂ₛ|². It lifts singlet and triplet alike, by a number times Zα²mₑc².
ψ in m(−3/2) and e²/(4πε₀r) in J, so J is an energy; 11.42 eV for bare Z = 2 orbitals, quoted, not evaluated here
Twice the self-energy of the overlap charge eψ₁ₛψ₂ₛ, whose total is zero. No classical counterpart, and positive because field energy ε₀E²/2 is.
Real orbitals; an energy in J or eV; 1.19 eV for bare Z = 2 orbitals, quoted; the measured gap implies 0.398 eV
Derived: P commutes with H₀ and Vₑₑ and has eigenvalues ±1 on the pair, so Ψ± are good states, with J on the diagonal of the 2 × 2 block and K off it.
E± belongs to Ψ± = [ψ₁ₛ(r₁)ψ₂ₛ(r₂) ± ψ₂ₛ(r₁)ψ₁ₛ(r₂)]/√2; all energies in eV
Derived from antisymmetry and the shifts; its size is measured. A positive K puts orthohelium lower, and 2K is of order α²mₑc², not α⁴mₑc².
2¹S₀ is Ψ₊ times the spin singlet, 2³S₁ is Ψ₋ times the triplet; measured gap 20.616 − 19.820 = 0.796 eV
Derived: electric-dipole lines keep ΔS = 0, so singlets and triplets form separate spectra and 2³S₁ is metastable (about 7.9 × 10³ s, measured).
Dipole moment in C m; χSM is the two-electron spin state with total spin S and spin projection Mħ, the four of them orthonormal
Eight spatial states at −68.0 eV
From the hydrogen solution of Unit 2, one electron in a Z = 2 Coulomb field has Eₙ = −Z²(13.606 eV)/n², so 1s sits at −54.42 eV and every n = 2 orbital at −13.61 eV. Drop the repulsion and H₀ = h(1) + h(2) puts 1s2s at E⁽⁰⁾ = −68.03 eV. It is not alone there. Swapping the electrons gives ψ₂ₛ(r₁)ψ₁ₛ(r₂) at the same energy, the exchange degeneracy of this unit, and the Coulomb l-degeneracy adds ψ₁ₛψ₂ₚₘ and its swap for m = −1, 0, 1: eight spatial states before spin is counted. The repulsion Vₑₑ = e²/(4πε₀|r₁ − r₂|) is not small, about 10 eV against 68 eV, so first order will be rough in size; what it must get right is the structure. With eight states degenerate, non-degenerate perturbation theory is ruled out: the first job is to find the combinations in which Vₑₑ is diagonal. Spin, which H₀ + Vₑₑ never touches, is attached afterwards.
Total L keeps the 1s2p states out
Vₑₑ depends only on |r₁ − r₂|, which a common rotation of both electrons leaves unchanged, so it commutes with the generator of that rotation, L = L₁ + L₂. It does not commute with L₁ or L₂ separately, and it does not need to. Both 1s2s states have l₁ = l₂ = 0 and so L = 0; all six 1s2p states couple l = 0 to l = 1 and so have L = 1. Taking ⟨L = 1| [L², Vₑₑ] |L = 0⟩ = 0 gives (2ħ² − 0)⟨L = 1|Vₑₑ|L = 0⟩ = 0, so every matrix element between the pair and the sextet vanishes, and Lz then splits the sextet into three pairs labelled by M. Parity, even for 1s2s and odd for 1s2p, reaches the same verdict independently. The eight-by-eight problem is four two-by-two problems. This step cannot be skipped: P alone does not do it, because P = +1 is shared by four states, one from 1s2s and three from 1s2p, and a degenerate eigenvalue of P picks nothing.
Inside the pair, P picks the good states
Write |a⟩ = ψ₁ₛ(r₁)ψ₂ₛ(r₂) and |b⟩ = ψ₂ₛ(r₁)ψ₁ₛ(r₂), so P|a⟩ = |b⟩. H₀ and Vₑₑ are both symmetric in the labels, so P commutes with each, and on this pair P has two distinct eigenvalues, +1 and −1. Unit 6's commuting-operator theorem then makes the eigenstates of P, Ψ± = (|a⟩ ± |b⟩)/√2, the good states: Vₑₑ has no matrix element between them. The explicit block agrees. Relabelling the integration variables gives ⟨b|Vₑₑ|b⟩ = ⟨a|Vₑₑ|a⟩ = J, and for real orbitals ⟨a|Vₑₑ|b⟩ = ⟨b|Vₑₑ|a⟩ = K, so the block has J on the diagonal and K off it, with eigenvalue J + K for Ψ₊ and J − K for Ψ₋. Three checks: the trace, 2J, is unchanged; with no overlap K = 0 and the pair stays degenerate; and J and K are e²/(4πε₀) times an inverse length, an energy, which for hydrogenic orbitals is a pure number times Ze²/(4πε₀a₀) = Zα²mₑc², 54.4 eV for Z = 2.
J is classical, K is not
J is what a classical physicist would write: the Coulomb energy of a cloud of charge −e|ψ₁ₛ|² sitting in a second cloud −e|ψ₂ₛ|². K has the same kernel, but each coordinate appears once in ψ₁ₛ and once in ψ₂ₛ, so the charge it describes is the overlap density eψ₁ₛψ₂ₛ, which integrates to zero because the orbitals are orthogonal. Zero net charge is not zero self-energy: K is twice the electrostatic energy of that distribution, and field energy ε₀E²/2 is never negative, so K is positive. The sign is argued; the size must be computed, and for bare Z = 2 orbitals the integrals give J = 11.42 eV and K = 1.19 eV, quoted here, not evaluated. The spatial meaning of the sign is in the figure. Ψ₋ changes sign under exchange, and since both orbitals depend on radius alone it vanishes whenever |r₁| = |r₂|. The antisymmetric state empties the region where the electrons would come closest and pays less repulsion; the symmetric state piles probability there and pays more.
Spin labels the levels; electrostatics splits them
The full state must be antisymmetric, so Ψ₊ pairs with the spin singlet and Ψ₋ with the triplet: 2¹S₀, parahelium, at E⁽⁰⁾ + J + K, and 2³S₁, orthohelium, at E⁽⁰⁾ + J − K, lower by 2K. Measurement agrees on the order: 2³S₁ lies 19.820 eV and 2¹S₀ 20.616 eV above the ground state, a gap of 0.796 eV. Taken at face value that gives effective values J = 9.24 eV and K = 0.398 eV, a third of the bare-orbital K, mainly because the 2s electron is screened by the 1s and overlaps it less. With the last lesson's spin-exchange operator the result reads E = E⁽⁰⁾ + J − (K/2)(1 + 4S₁⋅S₂/ħ²), a coupling of the spins that is Coulomb energy in disguise. Its size shows where it comes from: two electron spins a Bohr radius apart have a magnetic energy of order μ₀μB²/(4πa₀³) = α²Eₕ/4 = α⁴mₑc²/4 = 0.36 meV, about 2200 times smaller than the measured gap, which is of order α²mₑc².
Two spectra, and a level that waits two hours
An electric-dipole line needs ⟨f| −e(r₁ + r₂) |i⟩ ≠ 0, and the dipole operator acts on positions only. The spin factor comes out as ⟨χS′M′|χSM⟩ = δS′S δM′M, so lines join singlets to singlets and triplets to triplets: the two systems of lines once taken for two different gases. The ground configuration 1s² puts both electrons in one orbital, so its spatial state is symmetric and it exists only as a singlet; there is no 1³S. That leaves 2³S₁ as the lowest triplet, with nothing below it to reach by an allowed line, and its decay to 1¹S₀ is doubly forbidden, by ΔS = 0 and by parity. A weak relativistic magnetic-dipole process empties it with a lifetime of about 7.9 × 10³ s, quoted from measurement; 2¹S₀ escapes by two-photon emission in about 20 ms. ΔS = 0 is exact only while H ignores spin: spin–orbit coupling, Unit 8's, mixes a trace of singlet into 2³P₁, and its 59.1 nm line to 1¹S₀ runs about 10⁷ times slower than the 58.4 nm singlet line.
Change one variable at a time
Make the relationship visible.
At r₁ = 0.4 a₀ the singlet piles electron 2 onto electron 1's radius while the triplet empties it: that hole is why the triplet repels less. At r₁ = 1.0 a₀, the 2s node, the curves merge. Then take J to 9.24 eV and K to 0.40 eV to land on the measured lines.
2¹S₀ PARA, ABOVE GROUND23.59 eV
2³S₁ ORTHO, ABOVE GROUND21.21 eV
GAP 2K2.38 eV
ELECTRON 2 IN 1s, GIVEN r₁9 %
Live interpretation2¹S₀ PARA, ABOVE GROUND: 23.59 eV. 2³S₁ ORTHO, ABOVE GROUND: 21.21 eV. GAP 2K: 2.38 eV. ELECTRON 2 IN 1s, GIVEN r₁: 9 %
Catch the common trap
Explain before calculating.
At zeroth order helium's 1s2s and 1s2p spatial states share one energy, −68.0 eV, eight states in all. Why can the 1s2s pair be treated on its own, with [ψ₁ₛ(r₁)ψ₂ₛ(r₂) ± ψ₂ₛ(r₁)ψ₁ₛ(r₂)]/√2 as the good states?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTreat helium's two electrons as independent in the Z = 2 Coulomb field. Find the energy of 1s2s, list the spatial states that share it and the blocks Vₑₑ splits them into, and count the physical states once spin is added and the full state is made antisymmetric.
- Hydrogenic levels Eₙ = −Z²(13.606 eV)/n²: 1s at −54.42 eV and n = 2 at −13.61 eV for 2s and 2p alike, so E⁽⁰⁾ = −68.03 eV for 1s2s and for 1s2p. The Coulomb l-degeneracy survives into the two-electron problem.
- Spatial products at that energy: ψ₁ₛ(r₁)ψ₂ₛ(r₂) and its swap, 2 states; ψ₁ₛ(r₁)ψ₂ₚₘ(r₂) and its swap for m = −1, 0, 1, 6 states. Eight in all.
- Vₑₑ conserves total L and Lz: the L = 0 pair cannot mix with the L = 1 sextet, and the sextet falls into three pairs of fixed M. The 8 × 8 matrix becomes four 2 × 2 blocks, each diagonalised by P.
- Spin: 8 × 4 = 32 products, of which the antisymmetric ones are symmetric space times singlet plus antisymmetric space times triplet. For 1s2s that is 1 + 3 = 4 states; for 1s2p, 3 × 1 + 3 × 3 = 12.
- Check by spin-orbitals: one electron in 1s (2 choices) and one at n = 2 (8 choices), always different orbitals, so 2 × 8 = 16 antisymmetric states, exactly half of 32.
AnswerE⁽⁰⁾ = −68.0 eV; 8 spatial states in four 2 × 2 blocks; 16 physical states: 2¹S₀ (1), 2³S₁ (3), 2¹P₁ (3) and 2³P (9).
MediumHelium's 2³S₁ and 2¹S₀ levels are measured at 19.820 eV and 20.616 eV above the ground state, which lies 79.005 eV below He²⁺ plus two free electrons. Treat E⁽⁰⁾ + J ± K, with E⁽⁰⁾ = −68.028 eV, as exact, extract effective J and K, compare them with the bare Z = 2 values J = 11.42 eV and K = 1.19 eV, and compare the gap with the magnetic spin–spin energy μ₀μB²/(4πa₀³).
- Absolute energies: −79.005 + 19.820 = −59.185 eV for 2³S₁ and −79.005 + 20.616 = −58.389 eV for 2¹S₀. The triplet is lower, as a positive K requires, so it is E⁽⁰⁾ + J − K.
- Mean: (−59.185 − 58.389)/2 = −58.787 eV, so J = −58.787 + 68.028 = 9.241 eV. Half the gap: K = (−58.389 + 59.185)/2 = 0.398 eV.
- Against the bare values: J is 9.24/11.42 = 0.81 of its bare value and K only 0.398/1.19 = 0.33. The bare first-order levels, −55.42 and −57.80 eV, sit 23.59 and 21.21 eV above the ground state, 2.97 and 1.39 eV too high. The screened 2s electron spreads out and overlaps the 1s cloud less, and K, built from the overlap product itself, suffers most. The ordering survives.
- Magnetic scale: μ₀μB²/(4πa₀³) = 10⁻⁷ × (9.274 × 10⁻²⁴)² / (5.292 × 10⁻¹¹)³ J = 5.80 × 10⁻²³ J = 3.62 × 10⁻⁴ eV, which is α²Eₕ/4 as it must be. Units: T m A⁻¹ × J² T⁻² ÷ m³ = J, since T A m² = J.
- Ratio: 0.796 eV ÷ 3.62 × 10⁻⁴ eV = 2.2 × 10³. The gap is of order α²mₑc² and the spin–spin energy of order α⁴mₑc².
AnswerJ ≈ 9.24 eV and K ≈ 0.398 eV, against 11.42 and 1.19 eV from bare orbitals; the 0.796 eV gap is about 2.2 × 10³ times the magnetic spin–spin energy, so it is electrostatic.
HardThe measured singlet–triplet gaps of helium's 1sns levels are 6421.5, 1628.0, 642.1 and 316.5 cm⁻¹ for n = 2, 3, 4 and 5. Argue from the form of K how the gap should scale with n at large n, test the law on these data, and use it to predict the n = 5 gap from the n = 4 one.
- Kₙ = ∫∫ ψ₁ₛ(r₁)ψₙₛ(r₁) e²/(4πε₀|r₁ − r₂|) ψ₁ₛ(r₂)ψₙₛ(r₂) d³r₁ d³r₂. Every ψₙₛ is multiplied by ψ₁ₛ ∝ e(−2r/a₀), so only the part of the ns orbital within about a₀ of the nucleus contributes.
- Near the nucleus the potential dwarfs the small binding energy, so the ns orbital has an n-independent shape there, ψₙₛ(r) ≈ ψₙₛ(0)f(r). Then Kₙ ∝ |ψₙₛ(0)|², and from the hydrogen solution |ψₙₛ(0)|² ∝ n⁻³. So the gap 2Kₙ should fall as n⁻³ at large n.
- Convert with 1 cm⁻¹ = 1.23984 × 10⁻⁴ eV: gaps of 0.7962, 0.2018, 0.07961 and 0.03924 eV. Then n³ × gap = 6.37, 5.45, 5.10 and 4.91 eV: not constant, but the fall shrinks from 14% to 6.5% to 3.7% per step, as a large-n law should.
- Prediction: 642.1 cm⁻¹ × (4/5)³ = 328.8 cm⁻¹, against 316.5 cm⁻¹ measured, 3.9% high. The same step from n = 2 to 3 overshoots by 17%, because the 2s orbital is compact enough that its binding energy still shapes it near the core.
- Limit: as n → ∞, K → 0 and the ¹S and ³S series converge on the same limit, He⁺(1s) plus a free electron, 24.587 eV above the ground state. Exchange needs overlap, and a distant electron has almost none.
AnswerThe gap should fall as n⁻³; n³ × gap drifts from 6.37 to 4.91 eV and levels off. Predicted n = 5 gap 329 cm⁻¹ against 316.5 cm⁻¹ measured, 4% high.