University Quantum Mechanics II · The Fine Structure of Hydrogen and the Lamb Shift · 8.9
Midterm consolidation: six derivations from a blank page
The midterm covers eight units but tests six moves: stop a series, trace out a partner, lower a stretched state, symmetrise a pair, diagonalise inside a degenerate level, and add three α⁴ terms. This page reworks each on fresh numbers, and makes you say every time whether a result was derived here or quoted.
Build the model
Connect the measurement to the mechanism.
Units 1 to 8 look like eight topics and are one method: make an infinite problem finite, then read the physics off a small calculation. A power series that must stop gives l(l + 1) on the sphere and n = nᵣ + l + 1 in the radial equation. A two-particle space is cut down by a partial trace to what one part can show, or by exchange symmetry to what identical particles may occupy.
A product of two angular momenta splits into multiplets by lowering and orthogonality. A degenerate level becomes a g × g matrix, and an operator that commutes with H₀ and H′ makes it diagonal before it is written. Unit 8 spends all of it at once: the Bohr states come from the truncated series, the spin moment from Unit 3, the coupled basis from the Clebsch-Gordan table, and the good-state argument makes the relativistic and spin-orbit shifts diagonal, so three α⁴mc² terms sum to energies that depend on n and j alone.
The method stops where the derivations stop: non-relativistic, first order, classical fields, a point nucleus. Quoted from relativistic theory are gₑ, the spin-statistics connection, the Thomas factor, the Darwin term and the Dirac energy, and an answer that presents them as derived loses the marks. The 1057.8 MHz between 2s₁/₂ and 2p₁/₂ is a measured number no calculation here reaches: Welton's argument gives its order, α⁵mc² ln(1/α), and quantum electrodynamics its value.
- Simple definition
- Consolidating for the Quantum Mechanics II midterm means rebuilding six results of Units 1 to 8 from their starting equations, a recursion, a partial trace, a lowering operator, an exchange symmetry, a degenerate matrix and three fine-structure terms, and labelling each one derived or quoted.
- Example
- At n = 2, in units of E₂²/2mc² = 11.3 μeV, 2s₁/₂ sums to −13 + 8 = −5 and 2p₁/₂ to −7/3 − 8/3 = −5, so they coincide, while 2p₃/₂ at −7/3 + 4/3 = −1 lies 45.3 μeV (10.95 GHz) higher; 10.97 GHz is measured.
Derived. The ratio tends to 2/j, as for e(2ρ), so an unstopped series is not normalisable; stopping at j = nᵣ needs ρ₀ = 2n, so κ = 1/(na) and Eₙ = −13.6 eV/n².
ρ = κr with κ = √(−2mE)/ħ in m⁻¹; ρ₀ = 2/(κa), with a = 5.29 × 10⁻¹¹ m; c and j are pure numbers
Derived. Tr ρA² = 1 only for a product state. The singlet gives ρA = I/2, purity ½, which no measurement on A alone can tell from a 50:50 mixture.
Dimensionless; C is the dA × dB matrix of amplitudes cᵢⱼ, and Tr ρA = Σ|cᵢⱼ|² = 1
Derived by applying J₋ = L₋ + S₋ to |3/2, 3/2⟩ = |1, ↑⟩. Orthogonality then gives |½, ½⟩ = −√(1/3)|0, ↑⟩ + √(2/3)|1, ↓⟩, its sign fixed by Condon-Shortley.
Coupled |j, mⱼ⟩ on the left, uncoupled |mₗ, mₛ⟩ on the right, l = 1; ħ = 1.055 × 10⁻³⁴ J s cancels
Derived. Distinguishable particles keep only the first three terms. The fourth needs ⟨x⟩ₙₘ ≠ 0: symmetric pairs sit closer, antisymmetric ones further apart, with no force involved.
Orthonormal orbitals n ≠ m; ⟨x⟩ₙₘ = ∫ψₙ* x ψₘ dx in m; every term in m². Upper sign: symmetric spatial state
Derived. Its roots are the first-order shifts and its eigenvectors the good states. An operator commuting with H₀ and H′, with distinct eigenvalues in the level, makes W diagonal in advance.
W is the g × g matrix of H′, in J, over the g degenerate states ψᵢ⁰; δᵢⱼ is the Kronecker delta
Relativistic and spin-orbit terms derived, Thomas's ½ and the Darwin term quoted. The sum matches the quoted Dirac energy expanded to α⁴ and leaves 2s₁/₂ and 2p₁/₂ degenerate.
α = 1/137.036, dimensionless; j = l ± ½; infinite nuclear mass; complete to order α⁴mc², with 13.6 eV = α²mc²/2
A series that must stop: where l and n come from
Both of hydrogen's quantum numbers come from one move. In Unit 1, x = cos θ turns the polar equation into Legendre's, whose series obeys aₖ₊₂/aₖ = [k(k + 1) − λ]/[(k + 1)(k + 2)]; the ratio tends to 1, the series diverges logarithmically at a pole, and only λ = l(l + 1) stops it. In Unit 2, u = ρ(l+1) e(−ρ) v(ρ) leaves the recursion on the first card. Its ratio tends to 2/j, the ratio for e(2ρ), so an unstopped v makes u grow as eρ: the equation has a solution at every energy, but only a terminating one can be normalised. Stopping at j = nᵣ gives ρ₀ = 2(nᵣ + l + 1) = 2n, so κ = 1/(na) and Eₙ = −ħ²/(2ma²n²) = −13.6 eV/n², degenerate in l because only nᵣ + l enters. Checks: ħ²/(ma²) is an energy, 27.2 eV, equal to α²mc²; with the reduced mass the n = 2 → 1 line comes out at 10.199 eV, or 121.57 nm, as Lyman-α is measured.
Tracing out a partner: what a reduced density matrix keeps
Write a two-spin state as |ψ⟩ = Σ cᵢⱼ|i⟩|j⟩ and the partial trace over B is the matrix product ρA = CC†. For |ψ⟩ = cos θ|↑↓⟩ − sin θ|↓↑⟩, C is antidiagonal and ρA = diag(cos²θ, sin²θ): at θ = 30° that is diag(0.75, 0.25), with trace 1, purity Tr ρA² = 0.625 and a Bloch vector of length 0.50 along z. For a pure whole, purity below 1 is the test for entanglement. The limits bracket it: θ = 0 is a product and gives purity 1, and θ = 45° is the singlet, ρA = I/2, purity ½, the least a qubit can have. What ρA throws away is the correlation. The singlet and the MS = 0 triplet both give I/2, yet ⟨σₓ⊗σₓ⟩ is −1 in one and +1 in the other, and that discarded correlation is what breaks the CHSH inequality: the singlet reaches 2√2 = 2.83 against the local bound of 2, and the 2015 Delft loophole-free test measured 2.42 ± 0.20.
A Clebsch-Gordan table by lowering and orthogonality
Couple l = 1 to s = ½. Count first: 3 × 2 = 6 states, with mⱼ values 3/2, ½, ½, −½, −½, −3/2, so j = 3/2 takes four and j = ½ two. The top state has a single ket, |3/2, 3/2⟩ = |1, ↑⟩. Apply J₋ = L₋ + S₋, using J₋|j, m⟩ = ħ√(j(j + 1) − m(m − 1))|j, m − 1⟩: the left side gives √3ħ|3/2, ½⟩, the right √2ħ|0, ↑⟩ + ħ|1, ↓⟩, so |3/2, ½⟩ = √(2/3)|0, ↑⟩ + √(1/3)|1, ↓⟩. The j = ½ state at the same mⱼ must be orthogonal to it: |½, ½⟩ = −√(1/3)|0, ↑⟩ + √(2/3)|1, ↓⟩, with Condon-Shortley making the mₗ = l term positive. Check three ways. Each row and column of the 2 × 2 block is a unit vector. ⟨Sz⟩ in |3/2, ½⟩ is (2/3 − 1/3)ħ/2 = ħ/6, which is mⱼħ/(2l + 1). And with gₑ ≈ 2, quoted, ⟨Lz + 2Sz⟩ = ħ/3 + ħ/3 = 2ħ/3 gives g = 4/3, the partner 2/3, against 1.3342 and 0.6658 measured in sodium's 3p levels.
Exchange: a correlation with an energy, not a force
Two identical particles in orbitals n ≠ m have the spatial state [ψₙ(x₁)ψₘ(x₂) ± ψₘ(x₁)ψₙ(x₂)]/√2, which differs from a distinguishable product by one cross term. In ⟨(x₁ − x₂)²⟩ that term is ∓2|⟨x⟩ₙₘ|², on the card; in a pair energy it is ±K, the exchange integral K = ∫∫ψₙ*(1)ψₘ*(2) V₁₂ ψₘ(1)ψₙ(2). Electrons need an antisymmetric total state, so a symmetric space goes with the spin singlet and an antisymmetric space with the triplet. Excited helium 1s2l is the worked case: first-order degenerate theory gives E = E₀ + J ± K, singlet above triplet by 2K. Measured, 2¹S lies 0.80 eV above 2³S and 2¹P 0.25 eV above 2³P, so the effective K is 0.40 eV and 0.13 eV; the 2p orbital reaches in to the 1s less, and its K is smaller. The Hamiltonian contains no spin at all: the splitting is Coulomb repulsion weighted by where the symmetry lets the two electrons be.
A degenerate matrix, and the operator that makes it diagonal
Perturb the two-dimensional oscillator by H′ = εmω²xy. Its first excited level holds |1,0⟩ and |0,1⟩, and with x = √(ħ/2mω)(a + a†) the only nonzero element is ⟨1,0|xy|0,1⟩ = ħ/2mω. So W = (εħω/2)[0, 1; 1, 0], the shifts are ±εħω/2, and the good states are (|1,0⟩ ± |0,1⟩)/√2. The swap x ↔ y commutes with H₀ and with H′ and takes the values ±1 on exactly those states, which is why they are good without a matrix. Check against the exact normal modes, ω√(1 ± ε): at ε = 0.10 the level moves by +0.0476ħω and −0.0526ħω against ±0.0500ħω, and the common −0.0025ħω is the second-order −ε²ħω/4. Unit 8 uses the same shortcut. Hₛₒ commutes with L², S², J² and Jz, and Hᵣₑₗ with L² and every component of L and S, so the coupled states |l, ½, j, mⱼ⟩ are good for both and no 2(2l + 1)-dimensional matrix is ever written.
Eₙⱼ from three terms, and what the Dirac energy leaves out
Work in units of Eₙ²/2mc², 2.24 μeV at n = 3. The relativistic term, derived from −p⁴/8m³c² with ⟨1/r⟩ and ⟨1/r²⟩, is −[4n/(l + ½) − 3]. Spin-orbit, derived with Thomas's ½ quoted, is 2n[j(j + 1) − l(l + 1) − ¾]/[l(l + ½)(l + 1)] for l ≥ 1. The Darwin term, quoted from the Dirac equation, is 4n for l = 0 only. At n = 3: 3s₁/₂ is −21 + 12 and 3p₁/₂ is −5 − 4, both −9; 3p₃/₂ is −5 + 2 and 3d₃/₂ is −1.8 − 1.2, both −3; 3d₅/₂ is −1.8 + 0.8 = −1. Five states make three levels, each 3 − 4n/(j + ½), which is the quoted Dirac energy expanded to α⁴mc². The Dirac theory keeps 3s₁/₂ with 3p₁/₂ at every order in α; only quantum electrodynamics separates them. Welton's estimate gives order α⁵mc² ln(1/α)/n³, for s states only, and at n = 2 the measured 1057.8 MHz, quoted, is a tenth of the 10.95 GHz fine-structure splitting.
Change one variable at a time
Make the relationship visible.
Start at d = 1, t = 0.5: 2s₁/₂ and 2p₁/₂ sit on one j = ½ line. Lower d and 2s₁/₂ falls away; raise t towards the naive 1 and 2p₁/₂ drops below it. Then set q = 1: the 1057.8 MHz Lamb shift, quoted from QED and drawn on 2s₁/₂ alone, is the one lift the three terms cannot give.
GAP 2s₁/₂ − 2p₁/₂0 MHz
GAP 2p₃/₂ − 2p₁/₂10.95 GHz
GAP 2p₃/₂ − 2s₁/₂10.949 GHz
2s₁/₂ SHIFT FROM BOHR-56.6 μeV
Live interpretationGAP 2s₁/₂ − 2p₁/₂: 0 MHz. GAP 2p₃/₂ − 2p₁/₂: 10.95 GHz. GAP 2p₃/₂ − 2s₁/₂: 10.949 GHz. 2s₁/₂ SHIFT FROM BOHR: −56.6 μeV
Catch the common trap
Explain before calculating.
A midterm question asks why hydrogen's 2s₁/₂ and 2p₁/₂ levels have the same energy after the fine-structure correction, and what finally separates them. Which answer earns full marks?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe hydrogen radial series cⱼ₊₁ = [2(j + l + 1) − ρ₀]cⱼ/[(j + 1)(j + 2l + 2)], with u(ρ) = ρ(l+1) e(−ρ) v(ρ) and v = Σ cⱼρʲ, is started at c₀ = 1 for l = 1. Find the ρ₀ that stops it after the c₁ term, write R(r), give n and the energy with the reduced mass, and test the level against the measured Balmer-α line.
- The series stops after c₁ when c₂ = 0 with c₁ ≠ 0, so the numerator must vanish at j = 1: 2(1 + 1 + 1) − ρ₀ = 0, giving ρ₀ = 6. In general ρ₀ = 2(nᵣ + l + 1) = 2n; here nᵣ = 1 and l = 1, so n = 3 and the state is 3p.
- Then c₁ = [2(0 + 1 + 1) − 6]c₀/[(1)(0 + 4)] = −½, so v = 1 − ρ/2. With ρ₀ = 2/(κa) = 6, κ = 1/(3a) and ρ = r/(3a), so R = u/r ∝ (r/a)(1 − r/6a)e(−r/3a): one radial node, at r = 6a = 0.318 nm.
- Energy: E = −ħ²κ²/2μ = −13.5983 eV/9 = −1.5109 eV, with the reduced-mass Rydberg 13.5983 eV. Dimensions: ħ²κ²/μ is (J s)² m⁻² kg⁻¹ = J. The 3d state, l = 2 with nᵣ = 0, also needs ρ₀ = 6: the l-degeneracy comes straight out of the recursion.
- Balmer-α, n = 3 → 2: ΔE = 13.5983 eV × (1/4 − 1/9) = 1.88865 eV, so λ = hc/ΔE = 1239.842 eV nm/1.88865 eV = 656.470 nm. Measured in vacuum: 656.461 nm (656.279 nm in air). Without the reduced mass, with 13.6057 eV, the prediction is 656.11 nm, 0.35 nm off.
- Order in α: the Bohr energy is α²mc²/2n², and the 0.009 nm gap between 656.470 nm and the measured 656.461 nm, about 1.4 parts in 10⁵, is the size of α²/n² = 1.3 × 10⁻⁵ at n = 2: the relative size of the α⁴mc² fine structure, which a non-relativistic series cannot contain. Every prediction here is derived.
Answerρ₀ = 6, so n = 3 (the 3p state); v = 1 − ρ/2 and R₃₁ ∝ (r/a)(1 − r/6a)e(−r/3a), with a node at 0.318 nm; E₃ = −1.511 eV, which puts Balmer-α at 656.47 nm against 656.46 nm measured in vacuum.
MediumTwo electrons in a one-dimensional infinite well of width a = 1.00 nm occupy the orbitals n = 2 and n = 3; neglect their interaction. Find the rms separation √⟨(x₁ − x₂)²⟩ in the spin singlet and in the triplet, compare it with two distinguishable particles, and find one electron's reduced spin density matrix in the singlet and in the triplet states with MS = 1 and MS = 0.
- Single-orbital moments first: ⟨x⟩ₙ = a/2 for every n by symmetry, and ⟨x²⟩ₙ = a²[1/3 − 1/(2n²π²)], so ⟨x²⟩₂ = 0.32067a² and ⟨x²⟩₃ = 0.32770a². Distinguishable particles: ⟨(x₁ − x₂)²⟩ = 0.32067a² + 0.32770a² − 2(a/2)² = 0.14837a², an rms of 0.385a.
- The cross moment: ⟨x⟩₂₃ = (2/a)∫₀ᵃ x sin(2πx/a) sin(3πx/a) dx = −48a/(25π²) = −0.19454a, nonzero because 2 + 3 is odd; for orbitals 1 and 3 it vanishes by symmetry about a/2. The exchange term is 2|⟨x⟩₂₃|² = 0.07569a².
- The total state must be antisymmetric, so the singlet spin state goes with the symmetric spatial state and the triplet with the antisymmetric one. Singlet: 0.14837a² − 0.07569a² = 0.07268a², an rms of 0.270 nm. Triplet: 0.14837a² + 0.07569a² = 0.22406a², an rms of 0.473 nm. Distinguishable: 0.385 nm.
- Spin: the singlet (|↑↓⟩ − |↓↑⟩)/√2 has coefficient matrix C = [0, 1; −1, 0]/√2, so ρ₁ = CC† = I/2, purity ½. The MS = 0 triplet (|↑↓⟩ + |↓↑⟩)/√2 gives I/2 as well. The MS = 1 triplet |↑↑⟩ is a product state and gives |↑⟩⟨↑|, purity 1.
- Checks: every mean square is in m², and the singlet and triplet average to the distinguishable value, (0.07268 + 0.22406)/2 = 0.14837, because exchange adds and subtracts one term. No α enters: the model is non-relativistic and free of interaction. Switch on repulsion and the closer singlet pair costs more, helium's ordering, with 2¹S measured 0.80 eV above 2³S.
Answerrms separation 0.270 nm in the singlet, 0.473 nm in the triplet and 0.385 nm if distinguishable; one electron's spin is I/2 in the singlet and the MS = 0 triplet and |↑⟩⟨↑| for MS = 1, so spin alone cannot tell the first two apart, but the separation can.
HardFor the hydrogen-like ion He⁺ (Z = 2, nuclear recoil neglected), evaluate the relativistic, spin-orbit and Darwin shifts of 2s₁/₂, 2p₁/₂ and 2p₃/₂, show that the totals depend on j alone, and test the 2p₃/₂ − 2p₁/₂ splitting against the exact Dirac energy. Then scale hydrogen's measured 1057.8 MHz Lamb shift to He⁺ with Welton's Z-dependence, compare it with the measured 14.04 GHz, and label each number derived or quoted.
- Every term carries the unit Eₙ²/2mc², with Eₙ = −Z² × 13.606 eV/n². At Z = 2 and n = 2, E₂ = −13.606 eV, and the unit is (13.606 eV)²/(2 × 510 999 eV) = 181.13 μeV, or 43.80 GHz. That is (Zα)⁴mc²/128: order α⁴mc², enhanced by Z⁴ = 16 over hydrogen.
- In that unit the relativistic term −[4n/(l + ½) − 3] is −13 for 2s and −7/3 for 2p. Spin-orbit, 2n[j(j + 1) − l(l + 1) − ¾]/[l(l + ½)(l + 1)], is −8/3 for 2p₁/₂ and +4/3 for 2p₃/₂. The Darwin term is 4n = +8, for 2s only. Totals: 2s₁/₂ = −5, 2p₁/₂ = −5 and 2p₃/₂ = −1, each equal to 3 − 4n/(j + ½).
- So 2s₁/₂ and 2p₁/₂ sit together at −905.7 μeV and 2p₃/₂ at −181.1 μeV: a splitting of 4 units, 724.5 μeV or 175.19 GHz. Dimensions: Eₙ²/mc² is an energy. Limit: as Z falls every shift vanishes as Z⁴, faster than the Z² Bohr energy, so the correction is small for light ions.
- Dirac check, with the energy quoted from relativistic quantum mechanics: E = mc²[1 + (Zα/(n − j − ½ + √((j + ½)² − Z²α²)))²](−1/2). Evaluated without expanding, it gives 2p₃/₂ − 2p₁/₂ = 724.62 μeV, higher than the α⁴ value by 1.3 × 10⁻⁴, a fraction of order (Zα)² = 2.1 × 10⁻⁴: the α⁶mc² terms the expansion dropped. It keeps 2s₁/₂ and 2p₁/₂ exactly together.
- Lamb shift: Welton's estimate scales as α(Zα)⁴mc² ln(1/Zα)/n³ for s states, so from hydrogen's measured 1057.8 MHz the He⁺ value is 1057.8 MHz × 16 × ln(68.5)/ln(137.0) = 1057.8 MHz × 13.75 = 14.5 GHz. The estimate sits 3.6 per cent above the measured 14.04 GHz: the scaling holds, and the value belongs to QED. Its ratio to the fine structure falls from 0.097 in hydrogen to 0.080; the logarithm alone predicts 0.083.
- Labels: the Bohr energy and the relativistic term are derived; spin-orbit is derived apart from Thomas's ½, which is quoted; the Darwin term and the Dirac energy are quoted from relativistic quantum mechanics; the Z⁴ ln(1/Zα) scaling is Welton's estimate; both Lamb shifts are measurements, and the QED calculation that reproduces them is quoted.
AnswerIn units of 181.1 μeV, 2s₁/₂ = 2p₁/₂ = −5 and 2p₃/₂ = −1, so 2p₃/₂ lies 724.5 μeV (175.2 GHz) above the pair; the exact Dirac energy gives 724.6 μeV. Welton's scaling puts the He⁺ Lamb shift near 14.5 GHz, 3.6 per cent above the 14.04 GHz measured.