University Quantum Mechanics II · Identical Particles and Many-Electron Atoms · 7.7
The central field, penetration and quantum defects
Every alkali spectrum looks like hydrogen's with the rungs slid down, and this lesson explains the slide. Average the other electrons into one spherical potential, keep Unit 1's l and m, and let the orbitals that dive through the core record their extra binding in one number per l, the quantum defect, which sodium's levels then test.
Build the model
Connect the measurement to the mechanism.
An alkali atom is one electron outside a closed, and therefore exactly spherical, core. The central-field approximation extends that geometry to every electron: replace each electron's repulsion from the rest by its spherical average, and the Hamiltonian becomes a sum of one-electron problems in a single potential U(r), so l and m survive and only the radial equation changes. U(r) is −e²/(4πε₀r) outside the core but deepens towards −Ze²/(4πε₀r) inside it, and that ends hydrogen's l-degeneracy: an orbital with Rₙₗ ∝ rˡ near the nucleus keeps a probability of order a(2l+3) inside a radius a, so s electrons dive through the core and are bound hardest, and at fixed n the energy rises with l.
Outside the core the electron is hydrogen's again, which forces the levels into Rydberg's form with n shifted by one number per l, E = −13.606 eV/(n − δₗ)². Sodium's measured levels give δₛ = 1.37 from 3s and 1.36 from 4s, δₚ = 0.88 and δd = 0.01, and that near-constancy in n is the real test of the picture. The model stops in three places: the non-spherical remainder of the repulsion, which splits a configuration into terms; the slow drift of δ with n, from the energy dependence of the core's phase and from core polarisation; and any number not fitted to a spectrum, which needs Hartree's self-consistent field and Fock's exchange term, quoted here rather than computed.
- Simple definition
- The quantum defect δₗ is the amount by which an alkali level's effective quantum number falls short of its principal quantum number, E = −13.606 eV/(n − δₗ)², one number per l measuring how far that orbital penetrates the core.
- Example
- Potassium's 4s electron is bound by 4.341 eV, so n − δ = √(13.606/4.341) = 1.770 and δₛ = 2.23; its 4f electron, bound by 0.8533 eV, has n − δ = 3.993 and δf = 0.007.
Derived by separation exactly as in Unit 1. The angular part is untouched, so l and m stay good labels; E now depends on n and l because U is not 1/r.
ψ = (u/r)Yₗᵐ(θ, φ); U and E in J, ħ in J s, mₑ in kg, u in m(−1/2); U(r) is nucleus plus the averaged other electrons
Derived from Unit 1's addition theorem at γ = 0, where Pₗ(1) = 1. A full subshell's charge has no angular dependence, so a closed core makes an exactly central potential.
Each |Yₗᵐ|² in sr⁻¹; for l = 1 the sum is 3/(4π) = 0.239 sr⁻¹ at every angle
Derived for any spherical charge cloud. Sodium's valence electron sees Zeff = 11 at the nucleus and 1 outside the core: the two limits of U(r).
N(r): the other electrons' charge inside radius r, in units of e; the radial field there is Zeff e/(4πε₀r²) in V m⁻¹
Derived from the indicial equation, u ∝ r(l+1). Each extra unit of l costs two powers of a, so s orbitals own the core and d orbitals barely enter it.
a in m; Rₙₗ ≈ Cₙₗ rˡ near r = 0, so Cₙₗ² is in m(−2l−3); valid for a well inside the first radial node
Form derived in outline from the Coulomb exterior; δₗ as the core's phase over π is quoted from quantum-defect theory. Inverted, δₗ = n − √(R/|Eₙₗ|) reads a defect off one level.
R = α²mₑc²/2 = 13.606 eV; δₗ is dimensionless, one per l; n* = n − δₗ is the effective quantum number
Derived: ⟨nl|ΔV|nl⟩ samples |Rₙₗ|² near r = 0, which scales as n⁻³, the same power as −R/(n − δ)² + R/n² ≈ −2Rδ/n³. The n³ cancels, so δₗ carries no n.
ΔV in J: the core's extra attraction, zero beyond the core radius; valid only while δₗ ≪ 1
Average the other electrons, keep Unit 1's labels
Sodium's eleven electrons are coupled pair by pair through e²/(4πε₀rᵢⱼ), and no coordinate change separates them. The central-field approximation replaces the repulsion felt by electron i with the spherical average of the charge of all the others, giving one potential U(r) that every electron shares. Then Hc = Σᵢ [pᵢ²/2mₑ + U(rᵢ)] is a sum of one-electron operators, its eigenstates are Slater determinants of orbitals (u/r)Yₗᵐχ, and Unit 1's angular solution goes through untouched. Each (n, l) pair is a subshell of 2(2l + 1) states, and only the radial equation, with U(r) + l(l + 1)ħ²/(2mₑr²) as its effective potential, knows the other electrons exist. What is discarded, H₁ = Σ over pairs of e²/(4πε₀rᵢⱼ) − Σᵢ [U(rᵢ) + Ze²/(4πε₀rᵢ)], is the non-spherical remainder; it splits a configuration into terms and is the business of the term-symbol lesson.
Closed shells make the potential exactly central
For a filled subshell the averaging is not a guess. Unit 1's addition theorem at γ = 0 gives Σₘ |Yₗᵐ|² = (2l + 1)/(4π), so a subshell with every m occupied by both spins has density 2(2l + 1)|Rₙₗ|²/(4π), with no angle left in it. Check 2p: at θ = 0 only m = 0 contributes, (3/4π)cos²0 = 0.239 sr⁻¹; at θ = 90° only m = ±1 do, 2 × (3/8π) = 0.239 sr⁻¹. Sodium's core, 1s² 2s² 2p⁶, is therefore a spherical cloud, and Gauss's law says the field it leaves at radius r is set by the charge enclosed: Zeff(r) = Z − N(r) runs from 11 at the nucleus to 1 outside the core. For one electron beyond closed shells the direct repulsion is exactly central. Exchange with the core is not a local potential, but it is rotationally invariant, so the valence electron's l stays a good label.
Penetration lifts the l-degeneracy
In a pure 1/r potential the energy depends on n alone, an accident of that one potential. Sodium's U(r) matches −e²/(4πε₀r) outside the core and is far deeper inside, so what matters is how much time each orbital spends in there. Near the origin the centrifugal term dominates the radial equation and forces u ∝ r(l+1), so Rₙₗ ∝ rˡ and the probability inside a small radius a grows as a(2l+3). For hydrogen's n = 3 orbitals the exact fractions inside 1 a₀ are 9.9 × 10⁻³ for s, 1.3 × 10⁻³ for p and 6.5 × 10⁻⁶ for d. The s electron meets nuclear charge the p electron sees less of and the d electron almost never, so at fixed n the binding falls with l. Sodium's measured n = 3 levels, J-weighted to remove fine structure, lie at −5.139, −3.035 and −1.522 eV, against hydrogen's −1.511 eV: the d level barely notices the core.
Outside the core it is hydrogen, with n replaced by n − δ
Beyond the core the valence electron moves in exactly −e²/(4πε₀r), so there the radial equation is Unit 2's hydrogen equation. Write a bound energy as E = −R/ν². For any ν the outer equation has a solution that decays at infinity; in hydrogen only integer ν survive because that solution must also be regular at r = 0. In sodium the origin is replaced by the core, which accepts the outer solution only at certain ν. The inner wave is governed by a potential hundreds of eV deep, where a binding of a few eV is a small correction, so the phase it presents at the core boundary barely changes from level to level, and the accepted values are spaced by one: ν = n − δₗ. Quantum-defect theory, quoted here, makes δₗ exactly that phase divided by π. The data give δₛ = 1.373, 1.357, 1.353, 1.351 from 3s to 6s, δₚ = 0.883 to 0.862, δd = 0.010 to 0.013 and δf = 0.001.
Why one δ serves every n, and where it stops
Treat the core's extra attraction ΔV, zero beyond a radius rc, as a perturbation on a hydrogenic level. The first-order shift ⟨nl|ΔV|nl⟩ = ∫ ΔV |Rₙₗ|² r² dr collects only r below rc, where Rₙₗ ≈ Cₙₗ rˡ, and every Cₙₗ² scales as n⁻³ once n ≫ l; for s states Cₙ₀² = 4/(n³a₀³). Expand the defect formula: −R/(n − δ)² = −R/n² − 2Rδ/n³ + O(δ²). Shift and formula share n⁻³, so δₗ = −n³⟨nl|ΔV|nl⟩/(2R) has no n in it. The argument needs δ ≪ 1, which sodium's s and p defects violate, so the phase statement carries the rest. Exact constancy fails at low n: dE/dδ = 2R/n*³ is 6.3 eV per unit δ at 3s, so a drift of 0.016 moves 3s by 0.10 eV, and defects fitted to 4s and 4p put the D line at 605 nm instead of 589 nm. δ is α⁰, pure electrostatics, while R = α²mₑc²/2. Fine structure, removed here by J-weighting, enters at relative order α², enhanced inside the core: sodium's 3p splitting, 2.13 meV, is 7 × 10⁻⁴ of its binding, 13 times α².
Hartree and Fock: computing U(r), and what is quoted
So far U(r) has been read off spectra. Hartree's self-consistent field computes it: guess the orbitals, build the averaged charge of all the other electrons, solve each radial equation in the resulting U(r), rebuild the charge, and iterate until the potential reproduces itself. Fock added what the symmetrisation postulate demands. Varying a Slater determinant rather than a simple product adds an exchange term, a non-local operator acting only between electrons of the same spin: the exchange correlation of the earlier lessons, now inside the atom. Neither calculation is done here and their results are quoted. Hartree–Fock tables put sodium's 3s orbital energy near −4.95 eV, whose magnitude by Koopmans' theorem estimates the ionisation energy, against the measured 5.139 eV: about 4% short; the rest is correlation, chiefly the valence electron polarising the core, which also supplies most of the small d defect that penetration alone cannot.
Change one variable at a time
Make the relationship visible.
At l = 0, slide δ until the rings close on the dots: one value near 1.37 catches 3s, 4s and 5s together, the n-independence under test. Then set l to 1 and to 2: the p series needs about 0.87 and the d series almost none, because the centrifugal barrier keeps d out of the core.
δ FROM n = 3 DATA1.373
δ FROM n = 5 DATA1.353
TRIAL E AT n = 3-2.177 eV
MEASURED E AT n = 3-5.139 eV
Live interpretationδ FROM n = 3 DATA: 1.373. δ FROM n = 5 DATA: 1.353. TRIAL E AT n = 3: −2.177 eV. MEASURED E AT n = 3: −5.139 eV
Catch the common trap
Explain before calculating.
Sodium's 3s electron is bound by 5.139 eV, which gives δₛ = 1.37 in E = −13.606 eV/(n − δₛ)². Using the central-field picture, predict the binding energy of the 6s electron.
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyLithium's 2s electron is bound by 5.392 eV and its 3s electron by 2.019 eV. Find the quantum defect from each level, then use the 2s value to predict the 4s binding energy, whose measured value is 1.051 eV.
- Invert E = −R/(n − δ)² with R = 13.606 eV: n* = √(R/Eb). For 2s, n* = √(13.606/5.392) = √2.5234 = 1.5885, so δₛ = 2 − 1.5885 = 0.411.
- For 3s, n* = √(13.606/2.019) = √6.7390 = 2.5960, so δₛ = 3 − 2.5960 = 0.404. The two differ by 0.007, under 2%: the defect is close to independent of n, as the model requires.
- Predict 4s from the 2s defect: Eb = 13.606/(4 − 0.411)² = 13.606/12.881 = 1.056 eV.
- Compare with the measured 1.051 eV: the prediction is 0.5% high, the size expected from δ drifting towards 0.40. Limiting case: with δ = 0 the formula returns hydrogen's 0.850 eV, so even at n = 4 the 1s² core adds 0.2 eV of binding.
Answerδₛ = 0.411 from 2s and 0.404 from 3s; the predicted 4s binding is 1.056 eV against a measured 1.051 eV.
MediumNear the nucleus the hydrogenic n = 3 radial functions for Z = 1 start as R₃₀ → [2/(3√3)] a₀(−3/2), R₃₁ → [8/(27√6)](r/a₀) a₀(−3/2) and R₃₂ → [4/(81√30)](r/a₀)² a₀(−3/2). Estimate the probability that each is found inside r = 0.5 a₀, about the size of a light atom's core, and the factor by which each falls if that radius is halved.
- With Rₙₗ ≈ Cₙₗ rˡ, P(r ≤ a) = ∫ Cₙₗ² r(2l+2) dr from 0 to a = Cₙₗ² a(2l+3)/(2l + 3). In units of a₀(−2l−3): C₃₀² = 4/27 = 0.1481, C₃₁² = 64/4374 = 0.01463, C₃₂² = 16/196 830 = 8.13 × 10⁻⁵.
- At a = 0.5 a₀: Pₛ ≈ 0.1481 × 0.125/3 = 6.2 × 10⁻³; Pₚ ≈ 0.01463 × 0.03125/5 = 9.1 × 10⁻⁵; Pd ≈ 8.13 × 10⁻⁵ × 0.00781/7 = 9.1 × 10⁻⁸. Each is dimensionless, as a probability must be.
- Ratios s : p : d ≈ 1 : 1.5 × 10⁻² : 1.5 × 10⁻⁵. At this radius the step from s to p costs about two orders of magnitude and the step from p to d three.
- Halving a divides P by 2(2l+3): by 8 for s, 32 for p and 128 for d. The smaller the core, the more completely it belongs to the s electrons.
- Check against the exact integrals, done numerically: 2.8 × 10⁻³, 6.0 × 10⁻⁵ and 6.8 × 10⁻⁸. The leading term overestimates by factors of 2.2, 1.5 and 1.3 because it drops the falling polynomial and exponential, but the ranking and the steepness survive.
AnswerP ≈ 6.2 × 10⁻³ (s), 9.1 × 10⁻⁵ (p) and 9.1 × 10⁻⁸ (d), exact 2.8 × 10⁻³, 6.0 × 10⁻⁵ and 6.8 × 10⁻⁸; halving the radius divides them by 8, 32 and 128.
HardSodium's 6s and 7s levels lie 36 372.618 and 38 012.042 cm⁻¹ above the 3s ground state. Assuming one quantum defect serves both, with R = 109 734.7 cm⁻¹ for sodium, find δₛ and the ionisation energy, compare with the tabulated 41 449.451 cm⁻¹, and use the tabulated limit to test whether δₛ is the same at 3s.
- Each level sits at Tₙ = I − R/(n − δ)² above the ground state, so subtracting two eliminates the unknown limit I: R[1/(6 − δ)² − 1/(7 − δ)²] = 38 012.042 − 36 372.618 = 1639.424 cm⁻¹. One equation, one unknown.
- Newton's method from δ = 1.35: the left side is 5075.02 − 3437.53 = 1637.49 cm⁻¹, short by 1.93 cm⁻¹. Its slope, 2R[1/(6 − δ)³ − 1/(7 − δ)³], is 966.0 cm⁻¹ per unit δ, so δ = 1.35 + 1.93/966.0 = 1.3520; a second step changes nothing in the fourth decimal.
- Then I = T₇ + R/(7 − δ)² = 38 012.042 + 109 734.7/(5.6480)² = 38 012.04 + 3439.97 = 41 452.01 cm⁻¹. At 1.23984 × 10⁻⁴ eV per cm⁻¹ that is 5.1394 eV.
- The tabulated limit is 41 449.451 cm⁻¹, so the fit is 2.6 cm⁻¹ (0.32 meV) high, 6 parts in 10⁵. The same fit to 4s and 5s gives δ = 1.3597 and a limit 32 cm⁻¹ high: the higher pair wins because δ has almost stopped drifting there.
- With the tabulated limit, 3s has n* = √(109 734.7/41 449.451) = 1.6271 and δₛ = 1.3729. The defect grows by 0.021 from the 6s–7s fit down to 3s, the energy dependence of the core's phase, which the Ritz expansion, quoted rather than derived, absorbs.
Answerδₛ = 1.352 and I = 41 452.0 cm⁻¹ = 5.139 eV, 2.6 cm⁻¹ above the tabulated limit; at 3s the defect is 1.373, so δₛ grows by 0.02 from the 6s–7s pair down to 3s.