University Quantum Mechanics II · Identical Particles and Many-Electron Atoms · 7.9
Term symbols, the terms of p², and Hund's rules
The ground-state labels printed in a periodic table, ³P₀ for carbon or ⁴S₃/₂ for nitrogen, come out of this lesson. Antisymmetry cuts two p electrons down to fifteen states, a table of ML against MS sorts them into three terms, and Hund's rules, one traced to exchange, one found and one quoted, pick the level the atom sits in.
Build the model
Connect the measurement to the mechanism.
Inside one configuration the central field leaves every state degenerate; what splits them first is H₁, the repulsion the central field did not absorb. H₁ is unchanged when all positions rotate together and contains no spin, so it commutes with total L and total S: its eigenstates are terms ²ˢ⁺¹L of (2L + 1)(2S + 1) states, and spin–orbit coupling, weaker by a factor of order α² that grows with Z, then splits each term into levels J. For two electrons in one p subshell antisymmetry decides everything: it leaves 15 states, not 36, which an ML by MS table sorts into ¹D, ³P and ¹S; Unit 5's exchange sign (−1)L agrees, since L = 2 and 0 are spatially symmetric and carry the singlet while L = 1 is antisymmetric and carries the triplet.
The same antisymmetry orders the energies: the ³P angular state vanishes whenever the electrons lie on one line through the nucleus, and the Laplace expansion of 1/r₁₂ puts ³P, ¹D and ¹S at F⁰ − F²/5, F⁰ + F²/25 and F⁰ + 2F²/5. Hund's rules generalise the pattern: largest S first, traced to exchange; then largest L, stated as found; then J = |L − S| or L + S as the shell is under or over half full, quoted from the sign of spin–orbit coupling. They give carbon ³P₀, nitrogen ⁴S₃/₂ and oxygen ³P₂, and hold only for ground terms of light atoms: titanium's ¹D lies below its ³P, and in lead spin–orbit rivals the repulsion and jj coupling takes over.
- Simple definition
- A term symbol ²ˢ⁺¹LJ labels a level by its total spin S, its total orbital angular momentum L, written as a letter S, P, D, F for L = 0, 1, 2, 3, and its total angular momentum J; it describes the level only while L and S are separately conserved.
- Example
- Carbon's ground level ³P₀ has S = 1, L = 1 and J = 0, a single state. Its term also holds ³P₁ and ³P₂, measured 16.42 and 43.41 cm⁻¹ (2.04 and 5.38 meV) higher: 1 + 3 + 5 = 9 = (2L + 1)(2S + 1) states.
Derived from Unit 5's addition of L and S. It says how many levels a term can split into, and the state count catches a mislabelled level.
L, S, J dimensionless; each level holds 2J + 1 states; ³P gives J = 0, 1, 2 and 1 + 3 + 5 = 9 = 3 × 3 states
Derived: a Slater determinant is fixed by which spin-orbitals are filled, not by their order. Nonequivalent 2p3p has 6 × 6 = 36, since its electrons draw from different sets.
4l + 2 spin-orbitals labelled (mₗ, mₛ); p² gives C(6, 2) = 15, p³ gives 20, d² gives 45
Derived from Unit 5's Clebsch–Gordan exchange sign (−1)(2l − L), with 2l even. Space times spin must give −1, so even L pairs with S = 0 and odd L with S = 1.
P₁₂ exchanges the electrons; a spin state of total S has exchange sign (−1)(S+1); p² allows ¹S, ³P, ¹D
Derived from the Laplace expansion of e²/(4πε₀r₁₂). Only k = 0 and 2 survive for p², and ³P, whose electrons avoid lining up, comes out lowest.
γ: angle between r̂₁ and r̂₂; F⁰, F² greater than 0: radial Slater integrals in J (quoted in eV); P₂(x) = (3x² − 1)/2
Derived from the card above, a test with nothing adjustable; p⁴'s two holes share it, quoted. Silicon passes; carbon fails because 2p⁴ mixes into 2s²2p².
Dimensionless, free of F⁰ and F²; measured from the (2J + 1)-weighted ³P centre: Si 1.48, S 1.43, C 1.13, O 1.14
Quoted: the sign of A needs Unit 8's spin–orbit coupling. Carbon 2p² gets ³P₀ and oxygen 2p⁴ ³P₂; half-full nitrogen 2p³ has L = 0, so J = S: ⁴S₃/₂.
Rests on A in the term energy A L⋅S (A in J when L⋅S is in units of ħ²) being positive under half full and negative over
When L and S are good labels
The central field leaves every state of a configuration degenerate. What splits them first is H₁, the repulsion Σ e²/(4πε₀rᵢⱼ) minus its spherical average. H₁ depends only on distances between electrons, so it is unchanged when all positions rotate together and commutes with total L; it contains no spin operator, so it commutes with total S. By Unit 6's commuting-operator theorem the good states are then eigenstates of L² and S², and H₁ splits the configuration into terms ²ˢ⁺¹L, each (2L + 1)(2S + 1)-fold degenerate. Spin–orbit coupling, derived in Unit 8, commutes only with J = L + S, so it splits each term into levels ²ˢ⁺¹LJ of 2J + 1 states. This order, LS or Russell–Saunders coupling, holds while spin–orbit is the smaller. Term gaps are fractions of the hartree α²mₑc²; spin–orbit is of order α⁴mₑc², as in Unit 8's hydrogen, times a factor that climbs steeply with Z. Carbon's ³P spans 5.38 meV beside a ¹D − ³P gap of 1.26 eV. Lead's three lowest levels span 1.32 eV, over a third of its 3.65 eV p² spectrum, and NIST lists its ground level in jj coupling as (1/2, 1/2)₀.
Fifteen states, not thirty-six
Two electrons in different p subshells, as in carbon's excited 2p3p, have 6 × 6 = 36 product states, forming ¹S, ¹P, ¹D, ³S, ³P and ³D: 1 + 3 + 5 + 3 + 9 + 15 = 36. Put both in 2p and a Slater determinant is fixed by which two of the six spin-orbitals are filled: C(6, 2) = 15. Sort them by ML = mₗ₁ + mₗ₂ and MS = mₛ₁ + mₛ₂. With both spins up the orbitals must differ, giving ML = 1, 0, −1 once each, and likewise with both down. With opposite spins all nine pairs are allowed, giving ML = 2, 1, 0, −1, −2 with counts 1, 2, 3, 2, 1. Row by row from ML = 2 down to −2, the columns MS = +1, 0, −1 read 0 1 0, 1 2 1, 1 3 1, 1 2 1, 0 1 0. Now peel. The top entry, ML = 2 at MS = 0 only, must start L = 2, S = 0: strike one state from each row of the MS = 0 column, and ¹D is removed. The highest survivor, ML = 1 at MS = 1, starts L = 1, S = 1: strike the 3 × 3 block, ³P. One state is left, ML = MS = 0: ¹S. Check: 5 + 9 + 1 = 15.
Exchange symmetry gives the same three terms
The table is bookkeeping; symmetry is the reason. For equivalent electrons the radial part R(r₁)R(r₂) is symmetric, so the spatial state's exchange sign is its angular part's, and Unit 5's Clebsch–Gordan property gives (−1)(2l − L) = (−1)L. Check it directly. The L = 0 state, Σₘ ⟨1 m 1 −m|0 0⟩ Y₁m(r̂₁) Y₁(−m)(r̂₂), is by Unit 1's addition theorem a multiple of r̂₁ ⋅ r̂₂ = cos γ: symmetric. The three L = 1 states are the components of r̂₁ × r̂₂: antisymmetric. The five L = 2 states are the symmetric traceless products of the two unit vectors: symmetric. The singlet spin state is antisymmetric and the triplet symmetric, so the whole state is antisymmetric only for L = 0 or 2 with S = 0 and for L = 1 with S = 1: ¹S, ¹D and ³P, the rule that L + S be even. ³D fails at once, since its ML = 2, MS = 1 state puts both electrons in mₗ = 1, mₛ = +½. In 2p3p the radial part can itself be antisymmetric, and all six terms return.
Why ³P lies lowest: its electrons cannot line up
The angular states fix how the angle γ between r̂₁ and r̂₂ is spread. Square them, average over ML, which changes nothing since a rotational scalar has one distribution in every ML state, and normalise in cos γ: ¹S gives (3/2)cos²γ, ³P gives (3/4)sin²γ and ¹D gives 9/20 + (3/20)cos²γ, against a flat 1/2 for uncorrelated electrons. The ³P density vanishes at γ = 0, where antisymmetry forbids r̂₁ = r̂₂, and at γ = 180°, where antisymmetry combined with the even parity of p² forbids r̂₂ = −r̂₁ (parity alone does not: ¹S peaks there): the electrons are kept off any common line through the nucleus. Now expand the repulsion, e²/(4πε₀r₁₂) = (e²/4πε₀) Σₖ rₘᵢₙᵏ/rₘₐₓᵏ⁺¹ Pₖ(cos γ). For equivalent electrons the radial and angular integrals separate into EL = F⁰ + F²⟨P₂(cos γ)⟩L: odd k vanish by parity and k ≥ 4 because each density is quadratic in cos γ. With ⟨cos²γ⟩ = 1/5, 9/25 and 3/5, ⟨P₂⟩ = −1/5, +1/25 and +2/5. Every Fᵏ, an energy e²/(4πε₀) over a length, integrates positive quantities, so ³P lies below ¹D, which lies below ¹S, with gaps 6F²/25 and 9F²/25; if F² went to 0 all three would merge at F⁰. No spin force appears.
Hund's rules, and the standing of each
Hund's rules pick the ground term of one open subshell. First, the largest S lies lowest. This is the mechanism just computed, and excited helium's J − K: the largest S is the most symmetric spin state, so it goes with the most antisymmetric space, whose electrons avoid one another. For p² it is derived; in general it is traced, not proved. Second, of those terms the largest L lies lowest. It is stated as found; the picture of electrons circulating the same way and meeting less is a mnemonic, not a derivation. Third, J = |L − S| under half full and J = L + S over half full. It is quoted, because it rests on the sign of the spin–orbit constant, positive for electrons and negative for the holes of a shell more than half full. Carbon 2p²: ³P, then J = 0, with ³P₁ and ³P₂ 16.42 and 43.41 cm⁻¹ higher. Nitrogen 2p³: three parallel spins fill mₗ = 1, 0, −1, so S = 3/2 and L = 0, and J = S: ⁴S₃/₂, with ²D 2.38 eV up. Oxygen 2p⁴, two holes: ³P again, now inverted, ³P₂ lowest with ³P₁ and ³P₀ 158.27 and 226.98 cm⁻¹ higher.
Ground terms of light atoms only
The rules order the lowest term, not the rest. Titanium's 3d²4s² has two d electrons outside closed shells, and the same exchange count, L + S even, gives ¹S, ¹D, ¹G, ³P and ³F. Hund's rules correctly give ³F₂ as the ground level. But ¹D₂, measured at 0.900 eV, lies below ³P₀ at 1.046 eV, and ¹G₄ at 1.502 eV lies above both: the first rule fails among excited terms and the second among the singlets, because d² energies need F⁴ as well as F², and their ratio, not a symmetry, orders the middle of the spectrum. The model also assumes one configuration. Slater's 3/2 passes for silicon, measured 1.48, and fails for carbon, 1.13, where 2p⁴, carrying the same three terms, mixes in, a configuration mixing named here but not computed. And it assumes LS coupling. In lead the five levels and fifteen states survive, because J is exact in any scheme, but they regroup by each electron's j; the ground level still has J = 0, and calling it ³P₀ is a convention, not a description.
Change one variable at a time
Make the relationship visible.
Close the cone to 30°: 1.3% of ³P pairs lie inside, against 6.7% uncorrelated and 17.5% for ¹S, because antisymmetric space vanishes at γ = 0, Hund's first rule. Set F² to 3.18 eV, silicon: the ¹S − ¹D gap reads 1.14 eV, measured 1.13. Carbon's 5.25 eV gives 1.89 eV; it measures 1.42.
³P INSIDE THE CONE5.8 %
¹S INSIDE THE CONE32.3 %
GAP ¹D − ³P = 6F²/251.26 eV
GAP ¹S − ¹D = 9F²/251.89 eV
Live interpretation³P INSIDE THE CONE: 5.8 %. ¹S INSIDE THE CONE: 32.3 %. GAP ¹D − ³P = 6F²/25: 1.26 eV. GAP ¹S − ¹D = 9F²/25: 1.89 eV
Catch the common trap
Explain before calculating.
Sulfur's ground configuration is [Ne]3s²3p⁴. Which level do Hund's rules make its ground level?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn carbon's ³P term the angle γ between the two 2p electrons' position vectors is distributed as (3/4)sin²γ per unit cos γ, and in the ¹S term as (3/2)cos²γ. Find the probability that γ is less than 45° in each term, and compare both with uncorrelated electrons, whose density is a flat 1/2.
- The probability that γ is below 45° is the integral of the density over u = cos γ from c = cos 45° = 0.70711 to 1. Useful pieces: 1 − c = 0.29289 and 1 − c³ = 1 − 0.35355 = 0.64645.
- ³P: (3/4)∫(1 − u²) du = (3/4)[(1 − c) − (1 − c³)/3] = (3/4)(0.29289 − 0.21548) = 0.0581.
- ¹S: (3/2)∫u² du = (1/2)(1 − c³) = 0.3232. Uncorrelated: (1/2)(1 − c) = 0.1464.
- Limiting cases: at a 90° cone every term gives exactly 1/2, since each density is symmetric about γ = 90°. For a narrow cone of half-angle θ, ³P gives about 3θ⁴/16 against θ²/4 uncorrelated: the exchange node at γ = 0 costs two extra powers of the angle.
Answer5.8% for ³P and 32.3% for ¹S, against 14.6% for uncorrelated electrons: the triplet's electrons rarely share a direction, and the singlet ¹S, which favours it, lies highest.
MediumSilicon's 3p² levels, from NIST, are ³P₀ 0, ³P₁ 77.115, ³P₂ 223.157, ¹D₂ 6298.850 and ¹S₀ 15394.370 cm⁻¹. Use the ¹D − ³P gap to fix F², predict where ¹S₀ should lie, and judge the prediction.
- Remove the fine structure: the (2J + 1)-weighted ³P centre is (1 × 0 + 3 × 77.115 + 5 × 223.157)/9 = 149.68 cm⁻¹.
- E(¹D) − E(³P) = 6F²/25, so F² = (25/6)(6298.850 − 149.68) = (25/6)(6149.17) = 2.5621 × 10⁴ cm⁻¹, which is 3.18 eV at 8065.54 cm⁻¹ per eV.
- E(¹S) − E(¹D) = 9F²/25 = 9224 cm⁻¹, so ¹S₀ is predicted at 6298.85 + 9224 = 15523 cm⁻¹, or 1.925 eV.
- Measured: 15394 cm⁻¹, 1.909 eV. The predicted ¹S − ¹D gap, 9224 against 9096 cm⁻¹, is 1.4% too wide, and the spacing ratio is 9095.5/6149.2 = 1.479 against 3/2.
- Dimensions: F² is an energy and every step divides an energy by a pure number. The same test on carbon gives 1.127, a 25% miss, because 2p⁴ mixes into 2s²2p², which a single configuration cannot describe.
AnswerF² = 3.18 eV (2.562 × 10⁴ cm⁻¹); ¹S₀ predicted at 15523 cm⁻¹ (1.925 eV) against 15394 cm⁻¹ (1.909 eV) measured, the gap above ¹D 1.4% too wide.
HardTitanium's ground configuration is [Ar]3d²4s². Find the terms of two equivalent d electrons from exchange symmetry alone, check the state count, and use Hund's rules for the ground level. Then test the rules on excited terms: NIST puts ¹D₂ at 7255.355 cm⁻¹, ³P₀ at 8436.618 cm⁻¹ and ¹G₄ at 12118.393 cm⁻¹ above the ground level.
- 4s² is closed, with L = S = 0, so only 3d² matters: 4l + 2 = 10 spin-orbitals and C(10, 2) = 45 states.
- For l = 2 the spatial exchange sign is (−1)(2l − L) = (−1)L, so L + S must be even: L = 4, 2, 0 carry S = 0 and L = 3, 1 carry S = 1. The terms are ¹G, ¹D, ¹S, ³F and ³P.
- Count (2L + 1)(2S + 1): 9 + 5 + 1 + 3 × 7 + 3 × 3 = 15 + 21 + 9 = 45 ✓.
- Hund: the largest S = 1 leaves ³F and ³P; the largest L picks ³F; 2 of 10 places is under half full, so J = |L − S| = 2. Ground level ³F₂, as measured, with ³F₃ and ³F₄ 21.1 and 48.0 meV higher.
- Excited terms at 8065.54 cm⁻¹ per eV: ¹D₂ 0.900 eV, ³P₀ 1.046 eV, ¹G₄ 1.502 eV. Applied to all terms, the first rule would put ³P below every singlet and the second ¹G below ¹D; both fail. d² energies need F⁴ as well as F², and their ratio sets this order.
AnswerTerms ¹S, ¹D, ¹G, ³P, ³F (45 states); ground level ³F₂ as Hund's rules predict; but ¹D₂ lies 0.146 eV below ³P₀ and ¹G₄ lies above both, so the rules fail beyond the ground term.