University Quantum Mechanics II · Emission and Absorption of Radiation · 13.6

Oscillator strengths and the Thomas-Reiche-Kuhn sum rule

Line tables quote f rather than matrix elements because f is what is left of a transition once e, α and the rate's steep dependence on frequency are divided out. One double commutator then makes the strengths out of any level add to exactly 1, which audits a table, caps a polarisability and hands optics its free-electron limit.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

An oscillator strength is a dipole matrix element turned into a pure number, f_ba = (2mω_ba/3ħ)|⟨b|r|a⟩|², averaged over a's sublevels and summed over b's. It holds no e and no α, so it is of order 1 for an allowed line, and it measures the line's absorption in units of one classical electron on a spring. The Thomas-Reiche-Kuhn rule fixes the total. For H = p²/2m + V(r), [H, x] = −iħpₓ/m, so [x, [H, x]] = ħ²/m, a number.

Evaluate its expectation value in |a⟩ once directly and once by inserting Σ|b⟩⟨b|, and Σ_b f_ba = 1 follows, derived, using nothing about V but that it commutes with x. The potential decides only how the one unit is shared. Hydrogen 1s gives 0.416 to Lyman-α, derived, and 0.565 to the whole discrete series and 0.435 to photoionisation, quoted; from an excited level the downward lines count negative, so the upward ones total more than 1. Weighted by 1/ω², the same numbers build the polarisability, and far above every line they force the free-electron response, with all N electrons counted, as stated for many-electron atoms.

The rule stops where its one input fails. It needs the part of H quadratic in p to be exactly p²/2m, with nothing of higher order: spin-orbit coupling and a magnetic field's A⋅p are linear in p and leave it intact, but the relativistic −p⁴/8m³c² shifts the sum at relative order α², a non-local model potential breaks it outright, and it counts electric-dipole strength only.

Simple definition: The absorption oscillator strength f_ba of a line a → b is the pure number (2mω_ba/3ħ)|⟨b|r|a⟩|², averaged over a's sublevels and summed over b's: the line's absorption measured in units of one classical electron on a spring.
Example: Lyman-α, 1s → 2p: 3|⟨2p₀|z|1s⟩|² = 1.665a₀² and ħω = 10.20 eV = 0.375Eₕ give f = (2/3)(0.375)(1.665) = 0.416, the same for He⁺ or any hydrogen-like ion, since ω grows as Z² while |r|² shrinks as Z⁻².

Absorption oscillator strengthf_ba = (2mω_ba / 3ħ)(1/gₐ) Σ |⟨b|r|a⟩|²

A definition. It keeps one power of ω but drops e, α and the other two of the spontaneous rate's ω³, leaving a number of order 1, equal for every hydrogen-like ion. For a lower level b it is negative: gₐf_ba = −g_b f_ab.

m = 9.109 × 10⁻³¹ kg; ω_ba = (E_b − Eₐ)/ħ in rad s⁻¹; r in m; Σ over all sublevels of a and b; f has no unit

Spontaneous rate from a strengthAᵤₗ = (e²ω² / 2πε₀mc³)(gₗ/gᵤ) fₗᵤ

Derived by rewriting 13.03's A with f. The prefactor is 2α(ħω/mc²)ω, so A is of order α⁵mc²/ħ. Lyman-α's 0.416 gives 6.27 × 10⁸ s⁻¹, τ = 1.60 ns, as measured.

A in s⁻¹; u, l the upper and lower levels; ε₀ = 8.854 × 10⁻¹² F m⁻¹; c = 2.998 × 10⁸ m s⁻¹; ω in rad s⁻¹

The double commutator[x, [H, x]] = ħ²/m, for H = p²/2m + V(r)

Derived: [H, x] = −iħpₓ/m, then [x, pₓ] = iħ. The result is a number, the same in every state, and that is what makes the sum universal.

ħ = 1.055 × 10⁻³⁴ J s; ħ²/m = 1.221 × 10⁻³⁸ J m²; V(r) depends on position only, so it commutes with x

Thomas-Reiche-Kuhn sum ruleΣ_b f_ba = 1

Derived by evaluating ⟨a|[x, [H, x]]|a⟩ directly and through Σ|b⟩⟨b|. It fixes the total; the potential decides only how the total is shared among the lines.

b runs over every eigenstate: the bound levels below a and above it, and the continuum; each f_ba is a pure number

Integrated absorption cross-section∫ σ_ba(ω) dω = (πe² / 2ε₀mc) f_ba

Derived from 13.02's rate for incoherent light. A classical bound electron, quoted from the Lorentz model, has f = 1 whatever its frequency and damping: f counts such electrons.

σ in m², ω in rad s⁻¹; πe²/(2ε₀mc) = 2π²rₑc = 1.668 × 10⁻⁵ m² s⁻¹, with rₑ = 2.818 × 10⁻¹⁵ m

Static polarisability from strengthsαₚₒₗ = (e²/m) Σ_b f_ba / ω_ba²

Derived from Unit 9's 2e²Σ|⟨b|z|a⟩|²/(E_b − Eₐ) with |z|² = |r|²/3. The 1/ω² favours the lowest strong line, and Σf = 1 caps what the rest can add.

αₚₒₗ in C m² V⁻¹; e = 1.602 × 10⁻¹⁹ C; for an s state or a sublevel average; hydrogen 1s has 4.50(4πε₀a₀³)

A matrix element with the constants taken out

Unit 12's golden rule and 13.02's averages gave line rates proportional to e²|⟨b|r|a⟩|² times a power of ω that depends on what is measured. The oscillator strength keeps the atom's own part: f_ba = (2mω_ba/3ħ)(1/gₐ)Σ|⟨b|r|a⟩|², summed over the sublevels of both levels and averaged over the gₐ starting ones, so it does not depend on how the atom is oriented. The units cancel: kg s⁻¹ m² over J s is a pure number. In atomic units f = (2/3)(ħω/Eₕ)(Σ|r|²/a₀²), with no e and no α left, so f is of order α⁰, and it is the same for every hydrogen-like ion, since ω grows as Z² while |r|² shrinks as Z⁻². The name is literal. Dividing 13.02's absorption rate by the photon flux per unit frequency gives the line's cross-section integrated over frequency, ∫σ dω = (πe²/2ε₀mc)f = 1.668 × 10⁻⁵ m² s⁻¹ × f. A classical electron on a spring, quoted from the Lorentz model, absorbs exactly 1.668 × 10⁻⁵ m² s⁻¹ whatever its natural frequency and damping. So f says how many such electrons a line is worth: Lyman-α, at 0.416, absorbs like 0.42 of one.

One commutator, evaluated two ways

Let H = p²/2m + V(r), with V a function of position only. V commutes with x and [pₓ², x] = −2iħpₓ, so [H, x] = −iħpₓ/m. One more commutator with x, using [x, pₓ] = iħ, gives [x, [H, x]] = ħ²/m = 1.221 × 10⁻³⁸ J m²: a number, the same in every state, and dimensionally an energy times a length squared, as the left side must be. Now open the double commutator as 2xHx − x²H − Hx² and take its expectation value in the eigenstate |a⟩. Insert Σ_b|b⟩⟨b| between x and Hx in the first term and let H act on |a⟩ in the other two; the result is 2Σ_b(E_b − Eₐ)|⟨b|x|a⟩|². Equating the two evaluations gives Σ_b(2mω_ba/ħ)|⟨b|x|a⟩|² = 1. The same holds for y and z. Add the three, divide by 3, average over a's sublevels if it has several, and Σ_b f_ba = 1. The only property of V used is that it commutes with x, so the Coulomb potential, the infinite well and the oscillator all have the same total and differ only in where they put it.

Downward lines, the continuum, and strengths above 1

The closure sum runs over every eigenstate of H: the bound levels below a, those above it, and the continuum. For a lower level ω_ba is negative, and so is the strength; the definition gives gₐf_ba = −g_b f_ab. Hydrogen 2p has one lower level, so its downward strength is −(1/3)(0.416) = −0.139, and everything above 2p must carry 1.139. The discrete lines to ns and nd hold 0.948, quoted, with Balmer-α's 2p → 3d alone at 0.696, and the rule leaves 0.191 for photoionisation, found here by subtraction rather than by integrating a cross-section. A single strength can exceed 1: hydrogen 3d → 4f has f = 1.02. The oscillator shows the pattern at its cleanest. Its ladder operators link n only to n ± 1, which gives f = n + 1 up and −n down, summing to 1 for every n. In the ground state one line carries the whole unit, which is the classical bound electron exactly and the origin of the name.

Hydrogen's table closes the books

13.04 derived ⟨2p₀|z|1s⟩ = (128√2/243)a₀ = 0.745a₀. The three 2p sublevels together give 3 × (2¹⁵/3¹⁰)a₀² = 1.6648a₀², and with ħω = (3/8)Eₕ = 10.20 eV, f = (2/3)(3/8)(1.6648) = 2¹³/3⁹ = 0.4162, derived. Through 13.03's A, rewritten with f, that is 6.27 × 10⁸ s⁻¹ and τ = 1.60 ns, as measured. Tables print 0.4164: the proton's recoil puts the reduced mass μ in H, the double commutator becomes ħ²/μ, and strengths defined with the electron's mass sum to m/μ = 1.0005. The higher Lyman lines, quoted from Bethe and Salpeter's closed form, are 0.0791, 0.0290, 0.0139 and 0.0078 for n = 3 to 6, and the whole series sums to 0.565. The rule leaves 0.435 for photoionisation, and integrating hydrogen's photoionisation cross-section, also quoted, returns 0.435. Even the join is smooth. The products n³fₙ tend to 1.563, and the lines crowd together with spacing 2(13.6 eV)/n³, so near the series limit they hold 0.0574 eV⁻¹ of strength per unit energy, which is exactly the continuum's df/dE at threshold, quoted.

Weighted by 1/ω²: the polarisability

Unit 9 wrote the static polarisability as αₚₒₗ = 2e²Σ_b|⟨b|z|a⟩|²/(E_b − Eₐ). For an s state each level's |z|², summed over its sublevels, is a third of its Σ|r|², and E_b − Eₐ = ħω_ba, so αₚₒₗ = (e²/m)Σ_b f_ba/ω_ba²: the same strengths, now weighted by 1/ω². The weighting favours the lowest strong line. In units of 4πε₀a₀³, Lyman-α alone gives 0.4162/0.375² = 2.96 of hydrogen's exact 4.50, quoted in Unit 9; the rest of the series adds 0.70 and the continuum 0.84, both quoted. The sum rule gives a cap without a single integral: every ω_ba is at least ω₂₁, so αₚₒₗ ≤ (e²/m)/ω₂₁² = 7.11. That is looser than Unit 9's closure cap of 5.33, which also used ⟨z²⟩, but it needs only the lowest frequency, and it grows tight when one line holds nearly all the strength. In the alkali atoms one does: sodium's yellow line alone gives 98.5% of the measured polarisability.

Above every line, and where the rule bends

Driven at frequency ω, the same strengths give αₚₒₗ(ω) = (e²/m)Σ_b f_ba/(ω_ba² − ω²), stated here from first-order time-dependent theory. Far above every line this tends to −(e²/mω²)Σf, the response of free electrons. With N electrons the sum is N, stated: the double commutator of x₁ + … + x_N gives Nħ²/m. So X-ray optics counts every electron in the atom. For silicon at 8.05 keV, with 7.0 × 10²⁹ electrons per m³, the refractive index falls below 1 by δ = nₑe²/(2ε₀mω²) = 7.4 × 10⁻⁶, giving a critical angle for total external reflection of √(2δ) = 0.221°, close to the measured 0.22°; tables list 7.6 × 10⁻⁶, the inner shells not yet fully free. The rule bends where its one input fails, since [x, [x, H]] = −ħ²∂²H/∂pₓ² sees only how H depends on p. Terms linear in p, spin-orbit coupling and a magnetic field's A⋅p, drop out; the relativistic −p⁴/8m³c², quoted from the fine-structure unit, adds a term of relative order α²; and a non-local model potential, such as Hartree-Fock exchange, breaks the rule, which makes Σf a test of approximate wavefunctions.

Change one variable at a time

Make the relationship visible.

Interactive model

Starting level n2

Last line summed, n′12

Start at n = 1: one line takes 0.96 and the staircase is almost at 1 at once. Step n up to 4: downward bars open below the axis, the line to n + 1 grows past 1 to repay them, and the staircase still ends near 1. Then pull the last line summed down to n + 1 to see what the far lines hold.

UP LINE f(n → n + 1)1.868

DOWN LINE f(n → n − 1)-0.961

THE TWO NEIGHBOURS0.907

LEFT FOR ALL OTHER LINES0.093

Live interpretationUP LINE f(n → n + 1): 1.868. DOWN LINE f(n → n − 1): −0.961. THE TWO NEIGHBOURS: 0.907. LEFT FOR ALL OTHER LINES: 0.093

Catch the common trap

Explain before calculating.

Misconception check

Students read f as the probability of a transition. That picture predicts three things, and all three are false. It predicts each f lies between 0 and 1, but hydrogen 3d → 4f has f = 1.02 and the oscillator's n → n + 1 line has n + 1. It predicts every strength is positive, but a line to a lower level counts as −(g_b/gₐ) times that line's absorption strength, which is why 2p's upward strengths total 1.139. And it predicts the lines out of a level add to 1, so a student who sums hydrogen's Lyman series to 0.565 goes looking for missing lines, when the other 0.435 lies in the continuum, which the closure sum includes like any other state. f is a line's integrated cross-section divided by a classical electron's. The probabilities of where an excited atom decays are branching ratios of A coefficients, which carry ω² and the degeneracies, and it is those that sum to 1.

Hydrogen's absorption oscillator strengths from 2p to every higher bound level and into the continuum add up to 1.139. What does the Thomas-Reiche-Kuhn sum rule make of this total?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron in a one-dimensional harmonic trap with ħω = 1.50 eV sits in level n = 3. Using x = √(ħ/2mω)(a + a†) from Quantum Mechanics I, find every nonzero oscillator strength out of n = 3, check the sum rule, and give |⟨4|x|3⟩| in nm.

In one dimension only x enters, so there is no 1/3: f_ba = (2mω_ba/ħ)|⟨b|x|a⟩|². The ladder operators link 3 only to 4 and 2, with ⟨4|x|3⟩ = √(ħ/2mω)√4 and ⟨2|x|3⟩ = √(ħ/2mω)√3; every other matrix element is zero.
Up, 3 → 4: ω_ba = +ω, so f = (2mω/ħ)(ħ/2mω)(4) = 4.
Down, 3 → 2: ω_ba = −ω, so f = −(2mω/ħ)(ħ/2mω)(3) = −3.
Sum: 4 − 3 = 1, as the rule demands. The same algebra gives n + 1 and −n from any level, and at n = 0 one line carries the whole unit, exactly as a classical oscillator would.
Size: ħ/(2mω) = ħ²/(2m × ħω) = (1.0546 × 10⁻³⁴ J s)² / (2 × 9.109 × 10⁻³¹ kg × 2.403 × 10⁻¹⁹ J) = 2.540 × 10⁻²⁰ m², whose root is 0.1594 nm, so |⟨4|x|3⟩| = 2 × 0.1594 nm = 0.319 nm.

Answerf(3 → 4) = 4, f(3 → 2) = −3, all others zero, total 1; |⟨4|x|3⟩| = 0.319 nm. The strengths do not depend on ħω; the matrix element does.

MediumA table lists hydrogen's absorption strengths from 1s to np as 0.4162, 0.0791, 0.0290, 0.0139 and 0.0078 for n = 2 to 6, and notes that n³fₙ falls steadily towards 1.563 as n grows. With Σ n⁻³ over n ≥ 7 equal to 0.01177, bracket the strength of all the lines above n = 6, then use the sum rule to find the share of photoionisation.

Running sum of the tabulated lines: 0.4162, 0.4953, 0.5243, 0.5382, 0.5460. The first line alone holds 42% of the ground state's strength.
Every later line has fₙ = (n³fₙ)/n³, with n³fₙ below its value at n = 6, 6³ × 0.0078 = 1.685, and above its limit 1.563. So the lines with n ≥ 7 hold between 1.563 × 0.01177 = 0.0184 and 1.685 × 0.01177 = 0.0198.
The whole discrete series therefore holds between 0.5460 + 0.0184 = 0.5644 and 0.5460 + 0.0198 = 0.5658.
The sum rule includes the continuum, so photoionisation carries 1 minus that: between 0.434 and 0.436.
Check: integrating hydrogen's photoionisation cross-section, a quoted result, gives 0.435, inside the bracket. The table and the rule agree to better than one part in 10³.

AnswerLines with n ≥ 7 hold 0.018 to 0.020, the discrete series 0.564 to 0.566, and photoionisation 0.434 to 0.436, matching the quoted 0.435.

HardSodium's 3p level decays only to 3s, at 589.3 nm, with a measured lifetime of 16.3 ns. Treat sodium as one valence electron outside a closed core. Find f(3s → 3p) from Aᵤₗ = (e²ω²/2πε₀mc³)(gₗ/gᵤ)fₗᵤ, estimate αₚₒₗ/(4πε₀) from this line alone, cap what the other valence lines can add given that the next one, 3s → 4p, lies at 330.3 nm, and compare with the measured 24.11 Å³.

ω = 2πc/λ = 3.196 × 10¹⁵ rad s⁻¹ and A = 1/τ = 6.135 × 10⁷ s⁻¹. The prefactor is e²ω²/(2πε₀mc³) = 1.921 × 10⁸ s⁻¹, and gᵤ/gₗ = 3 for p over s, the spin factors cancelling.
f = (gᵤ/gₗ)A/(1.921 × 10⁸ s⁻¹) = 3 × 6.135 × 10⁷ / (1.921 × 10⁸) = 0.958. One line holds 96% of the valence electron's unit.
With e²/(4πε₀m) = 253.3 m³ s⁻² and ω² = 1.0217 × 10³¹ s⁻², αₚₒₗ/(4πε₀) = 253.3 × 0.958 / (1.0217 × 10³¹) = 2.375 × 10⁻²⁹ m³ = 23.75 Å³ from the D line alone.
For one electron in a local potential the valence strengths total 1, stated here as the model, so the other lines share at most 1 − 0.958 = 0.042, all at ω ≥ 2πc/(330.3 nm) = 5.703 × 10¹⁵ rad s⁻¹. They add at most 253.3 × 0.042 / (5.703 × 10¹⁵)² = 3.3 × 10⁻³¹ m³, or 0.33 Å³.
The valence electron thus gives 23.75 to 24.08 Å³. The closed Na⁺ core adds about 0.15 Å³, quoted, for a prediction of 23.9 to 24.2 Å³.
The measured 24.11 ± 0.08 Å³ sits inside. The D line alone accounts for 98.5% of it, because its 1/ω² dwarfs every other term's.

Answerf(3s → 3p) = 0.958; the D line alone gives αₚₒₗ/(4πε₀) = 23.8 Å³, the other valence lines at most 0.33 Å³ more, so with the core the prediction is 23.9 to 24.2 Å³, bracketing the measured 24.11 Å³.