advanced28 min studyLesson 456 in Physics

Many-electron atoms and Hund’s rules

Use antisymmetry, exchange, and state counting to interpret the empirical ordering of atomic LS terms.

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What you’ll understand

  • Count the microstates of an open equivalent-electron subshell.
  • Explain exchange without introducing an additional classical force.
  • Apply Hund’s ordering rules only after identifying Pauli-allowed terms.

Hydrogenic orbitals provide a basis for many-electron atoms, but electron repulsion changes both radial structure and the ordering of angular-momentum terms. Exclusion restricts which states exist; it does not by itself rank all their energies. Hund’s rules summarize a frequent ordering pattern for a fixed configuration in approximately LS-coupled atoms. They become useful predictions only after the allowed terms have been constructed, and they do not replace an energy calculation when configurations or coupling schemes compete.

Count states before assigning energies

A subshell of orbital quantum number l has 2(2l+1)2(2l+1) spin-orbitals. Choosing N occupied spin-orbitals gives a binomial number of Slater determinants, with each complete spin-orbital occupied at most once. A closed subshell has total L and S zero, so open subshells carry the angular-momentum structure. Configuration parity is (−1)∑ili(-1)^{\sum_i l_i}, which is unchanged by coupling the same occupied orbitals into different terms. Parity is separate from spin multiplicity.

For two equivalent p electrons there are six available spin-orbitals and fifteen determinants. Their allowed terms are 1S^1S, 1D^1D, and 3P^3P, with dimensions one, five, and nine. The sum is fifteen. A tempting 3D^3D term from naive addition is forbidden by electron antisymmetry, rather than merely high in energy. This check is stronger than drawing two arrows in boxes: it verifies the entire angular-momentum state count and exposes terms that cannot occur.

Nmicro=(2(2l+1)N),p2:1+5+9=15N_{\mathrm{micro}}=\binom{2(2l+1)}N,\qquad p^2:\quad 1+5+9=15
The binomial count concerns one equivalent-electron subshell. Each allowed LS term contributes (2L+1)(2S+1) magnetic microstates before spin-orbit splitting.↔ Scroll wide equations horizontally.

Exchange ties spin symmetry to spatial structure

For two electrons in distinct orthonormal spatial orbitals, symmetric and antisymmetric spatial combinations have different Coulomb expectation values. To make the complete fermionic wavefunction antisymmetric, the spin singlet uses the symmetric spatial combination and the spin triplet uses the antisymmetric one. The direct integral J is shared, while the exchange integral K enters with opposite signs. For a repulsive Coulomb interaction in this simple fixed-orbital comparison, positive K lowers the triplet relative to the singlet.

This energy difference does not arise from a new force that recognizes parallel arrows. The same Coulomb interaction acts in both states; antisymmetry changes their spatial pair probability. Also, a singlet is not synonymous with two opposite definite spins, because its spin wavefunction is a coherent antisymmetric superposition. Orbital relaxation and correlation change quantitative energies. An exchange explanation is a useful mechanism, but claiming that one fixed-orbital argument proves every Hund ordering would extend it beyond its assumptions.

Esinglet=E0+J+K,Etriplet=E0+J−KE_{\mathrm{singlet}}=E_0+J+K,\qquad E_{\mathrm{triplet}}=E_0+J-K
E_0 is the sum of one-electron energies and J, K are Coulomb and exchange energies for two distinct fixed orthonormal orbitals. This is not the complete many-configuration atomic spectrum.↔ Scroll wide equations horizontally.

Apply the rules within their domain

For a given configuration, the first Hund rule usually favors the largest allowed total spin. Among terms with that spin, the second favors the largest allowed L. For the resulting ground term, the third favors the smallest J for a subshell less than half full and the largest J for one more than half full. At half filling the maximum-spin term often has L zero. These rules concern energy ordering, whereas triangle rules and Pauli restrictions concern existence.

The p-squared example therefore favors 3P^3P and, below half filling, its J equal to zero level. The p-to-the-fourth configuration has two holes and the same term types, but the ground 3P^3P level has J equal to two under the third rule. Strong spin-orbit coupling, configuration mixing, or accidental near degeneracy can weaken this picture. A simple two-state phase instrument can distinguish singlet-like and triplet-like spin combinations in a restricted subspace, but it cannot enforce orbital antisymmetry or calculate exchange energies.

p2: 3P0,p4: 3P2(Hund ground-level predictions)p^2:\ {}^3P_0,\qquad p^4:\ {}^3P_2\quad\text{(Hund ground-level predictions)}
These assignments assume the indicated configuration is the ground configuration and LS coupling gives a useful description. They do not determine a numerical excitation energy.↔ Scroll wide equations horizontally.

Check the p-squared ground level

For an isolated atom whose open configuration is p2p^2, verify the microstate count and predict its ground term and ground J using Hund’s rules.

  1. The six p spin-orbitals admit (62)=15\binom62=15 two-electron determinants. The Pauli-allowed terms contribute 11 from 1S^1S, 55 from 1D^1D, and 99 from 3P^3P.

  2. The largest allowed spin is one, so the triplet P term is favored over the two singlets by the first rule. Its J values are zero, one, and two, whose degeneracies total nine.

  3. Two electrons occupy less than half of a six-state subshell. The third rule therefore favors its smallest J, giving 3P0^3P_0. This prediction presumes that the chosen configuration and LS approximation are appropriate.

The result

The fifteen states are exhausted by the three allowed terms, and Hund’s rules predict the ground level ³P₀.

Explore the model

Interactive lab

Interactive lab · quantum

Two-spin symmetry in a restricted basis

Interpret basis zero and one as the two spin products up-down and down-up. Reset and set θ=90∘\theta=90^\circ, ϕ=0∘\phi=0^\circ for their symmetric superposition; repeat with both controls 90∘90^\circ, 180∘180^\circ for the antisymmetric superposition. Predict equal probabilities but opposite Bloch x. The lab omits spatial orbitals, the other triplet substates, and exchange energies.

Two-spin symmetry in a restricted basisA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forRelative phase distinguishes symmetric and antisymmetric combinations even when product-basis populations agree.

Check your understanding

Why is a ³D term absent from the p² equivalent-electron configuration?

Try it yourself

For two fixed distinct orbitals let E0=−20.0 eVE_0=-20.0\,\mathrm{eV}, J=4.0 eVJ=4.0\,\mathrm{eV}, and K=0.30 eVK=0.30\,\mathrm{eV}. Find the singlet and triplet energies and their separation. Explain whether opposite-spin product populations alone specify either eigenstate.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Retain the exchange sign in each spatial-symmetry sector; consider the phase between the two opposite-spin product states.

Compare with the solution

The singlet energy is −20+4+0.30=−15.70 eV-20+4+0.30=-15.70\,\mathrm{eV}; the triplet is −20+4−0.30=−16.30 eV-20+4-0.30=-16.30\,\mathrm{eV}. The triplet is lower by 2K=0.60 eV2K=0.60\,\mathrm{eV}. Equal probabilities of up-down and down-up do not select a singlet or triplet: their relative phase distinguishes the antisymmetric and symmetric spin combinations. The corresponding spatial symmetry must also be included.

Take it with you

  • Pauli restrictions determine allowed terms before empirical ordering rules are applied.
  • Exchange is a consequence of antisymmetry acting through the Coulomb interaction.
  • Hund’s rules assume a fixed configuration and a useful LS-coupling description.

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