
Advanced Quantum Mechanics course sequence · Interactive lesson
Bethe's logarithm: the mean excitation energy
Bethe's formula ends in ln(K/K₀), and K₀ is the atom's half of that ratio. Learn which weighting defines it, why that weighting puts it nearly seventy times above the binding energy, and how to compute it from hydrogen's matrix elements and know, from a sum rule, that no state is missing.
UAQM-13.06 · Lesson 6 of 9 in this unit. Position is not a completion score.Build the model
A sum over every state of the atom, folded into one logarithm.
The key idea
Bethe's mean excitation energy K₀ of a level n is the weighted geometric mean of its excitation energies |Eₘ − Eₙ|, each intermediate state weighted by |pₘₙ|²(Eₘ − Eₙ), so that the self-energy's sum of logarithms over all states becomes the single logarithm ln(K/K₀).
Example. For hydrogen's 2s, ln(K₀/Ry) = 2.812, so K₀ = 16.64 Ry = 226 eV, 67 times the 3.40 eV binding energy; with the cutoff K = mc² = 511 keV the logarithm is ln(K/K₀) = 7.72.
The complete physical model and assumptions
Once the electromagnetic mass is subtracted, a level's self-energy is a sum over every intermediate state m of wₘ ln(K/|Δₘₙ|), with Δₘₙ = Eₘ − Eₙ and weight wₘ = |pₘₙ|²Δₘₙ. The cutoff K enters every term in the same way, so the logarithms come apart: the sum is exactly (Σₘ wₘ) ln(K/K₀), where ln K₀ is the weighted mean of ln|Δₘₙ|. That defines Bethe's mean excitation energy, a property of the atom alone, since K₀/Ry contains neither K nor α, while topic 13.05's sum rule supplies Σₘ wₘ in closed form. The weighting decides everything. Since |pₘₙ|² = m²ωₘₙ²|rₘₙ|², each weight is (3/2)m fₘₙΔₘₙ², two powers of the excitation energy beyond the oscillator strength, and the cusp of an s state at the nucleus makes the continuum's weights fall only as Δ⁻³⁄². So the bound np states, which hold 65 per cent of 2s's oscillator strength, hold 2.6 per cent of its weight, and K₀ lands at 16.64 Ry = 226 eV, 67 times the binding energy: quoted from the modern evaluation, and reproduced here from hydrogen's dipole matrix elements. The construction stops where its one approximation does. Writing ln(K/Δ) for ln((K + Δ)/Δ) needs K ≫ Δ wherever the weight lies, yet 1.3 per cent of it sits above mc², in states whose electron is relativistic; and for a p state the weights add to zero, so no mean exists at all.
1From a sum of logarithms to one
Units are SI, with energies also given in eV and in rydbergs, Ry = 13.606 eV; no four-vector appears here, and where one does in this course the metric is (+, −, −, −). Topic 13.04 left the renormalised shift of level n, for K far above the atomic energies, as (2α/(3πm²c²)) Σₘ wₘ ln(K/|Δₘₙ|), with wₘ = |pₘₙ|²Δₘₙ, and topic 13.05 showed Σₘ wₘ = (ħ²/2)⟨∇²V⟩ₙ. Both are quoted. Now split each logarithm: ln(K/|Δₘₙ|) = ln(K/Ry) − ln(|Δₘₙ|/Ry). The first piece is the same for every m and multiplies Sₙ = Σₘ wₘ; the second is a weighted sum with no K in it. Calling that sum Sₙ ln(K₀/Ry) makes the result exact: δEₙ = (2α/(3πm²c²)) Sₙ ln(K/K₀). Dimensions check: Sₙ/(m²c²) is an energy and the logarithm a pure number. In the one-state limit, where a single m carries all the weight, K₀ is simply that state's excitation energy. For hydrogen's s states, |ψₙ(0)|² = 1/(πn³a₀³) turns the result into (4α⁵mc²/(3πn³)) ln(K/K₀), derived: h × 135.6 MHz × ln(K/K₀) for 2s.
2Three weightings of one spectrum
The same dipole matrix elements can be averaged three ways, and each average is pinned by its own sum rule. Weight by fₘₙ alone and the Thomas-Reiche-Kuhn rule makes the weights add to 1. Weight by |pₘₙ|² = (3/2)m fₘₙΔₘₙ and closure makes them add to ⟨p²⟩, the coefficient of topic 13.02's linear divergence. Weight by |pₘₙ|²Δₘₙ, Bethe's choice, and topic 13.05's rule makes them add to (ħ²/2)⟨∇²V⟩. Each extra power of Δ drags the weighted mean upward. For hydrogen's 2s, computed here, the oscillator-strength mean is 3.20 eV, close to the 3.40 eV binding energy; the |p|² mean is 7.63 eV; Bethe's mean is 226 eV. The construction with fₘₙ alone is Bethe's 1930 mean excitation energy for stopping power, 15.0 eV for ground-state hydrogen when computed the same way, against 269 eV for the 1s Lamb-shift K₀: the same name and the same sum, but different weights and an energy eighteen times smaller. Only the Δ² weighting belongs in the self-energy.
3Why the continuum carries almost everything
Take 2s → 3p, the strongest line, with fₘₙ = 0.435 at Δ = 1.89 eV. It holds 43 per cent of the oscillator strength but only 1.26 per cent of Bethe's weight, and all the bound np states together hold 64.9 per cent of fₘₙ but 2.64 per cent of the weight. The reason lies at the nucleus. An s state has a cusp there, ψ ≈ ψ(0)(1 − r/a₀), and a cusp makes the momentum-space wavefunction fall only as k⁻⁴. For Δ ≫ Ry the continuum state is nearly a plane wave of wavenumber k, and the cusp makes its overlap with ψₙ tend to 8πψ(0)/(a₀k⁴). So |pₘₙ|² ∝ k² × k⁻⁸, Δ ∝ k², and the states between k and k + dk number ∝ k² dk: the weight per unit Δ falls as Δ⁻³⁄². Integrated above E and divided by Sₙ = (ħ²e²/(2ε₀))|ψ(0)|², the factors |ψ(0)|² cancel, and the fraction above E tends to (8/π)√(Ry/E) for every s state, derived: 1.31 per cent above mc². Lower down the computed spectrum puts 25 per cent of the weight above 1 keV and 57 per cent above 100 eV. A tail like this drags the log-mean deep into the continuum.
4Computing K₀ from hydrogen's matrix elements
This is the unit's computational exercise, and it needs two kinds of state. The bound np states have closed-form radial integrals, R²(np, 2s) for 2s as quoted in topic 13.02; they are computed to n = 300 and continued as n⁻³ beyond. The continuum εp states are energy-normalised Coulomb functions, started from their power series at the origin and integrated outward by Numerov's method, then folded into the radial overlap ∫ R(εp) R(2s) r³ dr on a grid of electron wavenumbers up to k = 200/a₀, about 544 keV, beyond which the Δ⁻³⁄² tail is added analytically. Two checks decide whether the spectrum is complete before any logarithm is taken: Σₘ fₘₙ = 1, met to 4 × 10⁻⁵, and Σₘ fₘₙΔₘₙ² = (2/3)Sₙ/m, which for 2s is Eₕ²/6 with Eₕ = 27.211 eV, met to 2 × 10⁻⁶. Only then is ln(K₀/Ry) formed: 2.8118 for 2s and 2.9841 for 1s, against the quoted 2.81177 and 2.98413.
5What a factor of two in K₀ is worth
K₀ enters only through ln(K/K₀), and with K = mc² that logarithm is large: ln(511.0 keV/226.4 eV) = 7.722 for 2s. A factor of two in K₀ moves it by ln 2 = 0.693, nine per cent, or 94 MHz at 135.6 MHz per unit of logarithm. Bethe's 1947 estimate, K₀ = 17.8 Ry, was 7 per cent high, so with K = mc² his logarithm is 7.654 and the 2s shift about 1040 MHz, the value he reported, against 1047 MHz from the modern value. The cutoff is known no better, since a factor of two in K moves the result by the same 94 MHz. With the modern K₀ the 1s shift comes out at 8.19 GHz, and the two set beside the measured 1057.8 MHz 2S₁/₂–2P₁/₂ interval and 8.17 GHz 1s shift agree within one per cent, closer than the cutoff's uncertainty warrants, which topic 13.07 examines. What the logarithm cannot forgive is the wrong scale: put K₀ at the 3.40 eV binding energy and ln(K/K₀) becomes 11.92, 54 per cent too large.
6Where the definition stops
Three limits bound the construction. First, Sₙ ln(K/K₀) stands in for Σₘ wₘ ln((K + Δₘₙ)/Δₘₙ), which needs K ≫ Δₘₙ wherever the weight lies. With 1.31 per cent of it above mc², keeping the exact logarithm raises the 2s value from 7.722 to 7.762, half a per cent, computed; and those states have relativistic electrons that the non-relativistic spectrum misdescribes, which topic 13.09's covariant theory repairs. Second, for a p state the weights add to zero, Sₙ = 0, so no mean exists: the shift becomes −(2α/(3πm²c²)) Σₘ wₘ ln(|Δₘₙ|/Ry), with no K in it at all, and the quoted ln k₀(2p) = −0.0300, defined with the 2s value of Sₙ as its scale, makes it about +4 MHz. Third, for a hydrogen-like ion every Δ scales as Z², so the 2s K₀ = Z² × 226 eV reaches mc² near Z = 48, where the logarithm has nothing left to measure.
Key equations
Read the units, assumptions and limits before substituting.
Bethe's logarithm, defined
ln(K₀/Ry) = Σₘ wₘ ln(|Δₘₙ|/Ry) / Σₘ wₘ, with wₘ = |pₘₙ|²Δₘₙ
A definition made useful by an identity: Σₘ wₘ ln(K/|Δₘₙ|) = (Σₘ wₘ) ln(K/K₀) for every K, so the atom enters the shift through one energy. The Ry cancels from K/K₀.
Δₘₙ = Eₘ − Eₙ in J or eV; pₘₙ = ⟨m|p|n⟩ in kg m s⁻¹; m runs over bound and continuum states; Ry = 13.606 eV sets the unitEach weight in oscillator strengths
wₘ = |pₘₙ|²Δₘₙ = (3/2)m fₘₙ Δₘₙ²
Derived from pₘₙ = i m ωₘₙ rₘₙ. Two powers of Δ beyond fₘₙ: the Thomas-Reiche-Kuhn rule makes Σₘ fₘₙ = 1, but these weights lean hard towards high excitation.
fₘₙ = (2/3)(m/ħ²)Δₘₙ|rₘₙ|², dimensionless, summed over the sublevels of m; rₘₙ = ⟨m|r|n⟩ in m; m the electron mass in kgThe shift with one logarithm
δEₙ = (2α/(3πm²c²)) Sₙ ln(K/K₀), with Sₙ = Σₘ wₘ
Derived from topic 13.04's Σₘ wₘ ln(K/|Δₘₙ|), the K ≫ |Δₘₙ| form of ln((K + |Δₘₙ|)/|Δₘₙ|). For hydrogen's s states it reads (4α⁵mc²/(3πn³)) ln(K/K₀).
K the photon cutoff in J; α = 1/137.036; Sₙ = (ħ²/2)⟨∇²V⟩ₙ by topic 13.05's sum rule, quoted; Sₙ/(m²c²) is in JHydrogen's values
ln(K₀/Ry) = 2.8118 for 2s, and 2.9841 for 1s
Computed here from hydrogen's dipole matrix elements as 2.8118 and 2.9841. K₀ is 67 times the 2s binding energy and 20 times the 1s one.
Quoted (modern): K₀ = 16.64 Ry = 226.4 eV for 2s and 19.77 Ry = 269.0 eV for 1s; Bethe's 1947 estimate for 2s was 17.8 RyHow slowly the weights die away
Sₙ(Δₘₙ ≥ E)/Sₙ → (8/π)√(Ry/E), for E ≫ Ry
At E = mc² it is 4√2α/π = 1.31 per cent, the same for 1s and 2s. A tail falling only as Δ⁻³⁄² drags the log-mean far above every bound-state energy.
Any hydrogen s state; Sₙ(Δₘₙ ≥ E) is the part of Sₙ from states excited by at least E, in eV; derived from ψₙ's cusp at r = 0Hydrogen-like ions
K₀(Z) = Z² K₀(Z = 1)
Derived by scaling r to r/Z. With K = mc², ln(K/K₀) loses 2 ln Z and reaches zero near Z = 48 for 2s: there the formula has nothing left to say.
Z the nuclear charge number; non-relativistic dipole model, in which every Δₘₙ scales as Z² and every wₘ by the same factor Z⁴Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. At j = 2, step d from 0 to 6: the solid K₀ climbs from 2.3 eV to 205 eV and is still short of the dashed 226 eV. Then set j = 0: oscillator strengths alone put the mean at 3.2 eV, beside the 3.40 eV binding energy.
Live results
- TOP OF THE SUM
- 1000000.0 eV
- K₀, ALL STATES
- 226.4 eV
- K₀ ÷ BINDING ENERGY
- 66.56
- ln(mc²/K₀), ALL STATES
- 7.72
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Weight f Δʲ, power j: 2; Continuum kept up to 10ᵈ eV, d: 6
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting picture is that K₀, an average excitation energy, must sit among the atom's own energies: near the 3.40 eV binding energy of 2s, or at the 1.89 eV of its strongest line, 2s → 3p. That would be true of an average weighted by oscillator strengths, which is 3.20 eV for 2s. Bethe's weights are |pₘₙ|²Δₘₙ = (3/2)m fₘₙΔₘₙ², and the two extra powers of Δ, together with a continuum whose weights fall only as Δ⁻³⁄², move the mean to 226 eV, 67 times the binding energy, with 97 per cent of the weight above the ionisation threshold. Inside a logarithm the error is still not small: K₀ = 3.40 eV gives ln(mc²/K₀) = 11.92 instead of 7.72, a 2s shift 54 per cent too large, about 1620 MHz instead of about 1050 MHz.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
The modern Bethe logarithms for hydrogen are ln(K₀/Ry) = 2.8118 for 2s and 2.9841 for 1s, quoted. Convert each K₀ to eV, compare it with that level's binding energy, and evaluate ln(K/K₀) with the cutoff K = mc² = 511.0 keV.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- K₀ = Ry × e(ln(K₀/Ry)). For 2s, e2.8118 = 16.64, so K₀ = 16.64 × 13.606 eV = 226.4 eV. For 1s, e2.9841 = 19.77, so K₀ = 269.0 eV.
- Binding energies are 13.606 eV/n²: 3.401 eV for 2s and 13.606 eV for 1s. So K₀ is 226.4/3.401 = 66.6 times the 2s binding energy and 269.0/13.606 = 19.8 times the 1s one.
- Since Ry = α²mc²/2, mc²/Ry = 2/α² = 3.7558 × 10⁴ and ln(mc²/Ry) = 10.5336: the argument is a ratio of two energies, as a logarithm requires.
- ln(K/K₀) = ln(mc²/Ry) − ln(K₀/Ry): 10.5336 − 2.8118 = 7.7218 for 2s, and 10.5336 − 2.9841 = 7.5495 for 1s.
- Check: 1s, bound four times more tightly, has the larger K₀ in eV, yet a smaller ratio to its own binding energy; its logarithm is smaller by 0.1723.
Answer. K₀ = 226.4 eV for 2s (66.6 times its binding energy) and 269.0 eV for 1s (19.8 times); ln(mc²/K₀) = 7.722 for 2s and 7.550 for 1s.
MediumExample 2
The 2s → 3p line of hydrogen has oscillator strength f = 0.4349 at Δ = 1.889 eV. Topic 13.05's sum rule gives Σₘ fₘₙΔₘₙ² = Eₕ²/6 for 2s, with Eₕ = 27.211 eV. Find the share of Bethe's weight that 3p carries, then the K₀ and ln(mc²/K₀) of a model in which 3p is the only state, and the factor by which that model misstates the 2s shift.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- The weights are proportional to fₘₙΔₘₙ², so shares can be taken in those units. Δ = 1.889 eV/27.211 eV = 0.06942 Eₕ, so Δ² = 4.819 × 10⁻³ Eₕ² and fΔ² = 0.4349 × 4.819 × 10⁻³ = 2.096 × 10⁻³ Eₕ².
- Share: 2.096 × 10⁻³ ÷ (1/6) = 0.01258, or 1.26 per cent of the weight, against 43.5 per cent of Σf = 1. The factor Δ² penalises a line at 1.9 eV against states at tens and hundreds of eV.
- One-state limit: a weighted mean over a single state is that state's energy, so K₀ = 1.889 eV, 120 times smaller than the true 226.4 eV.
- ln(mc²/K₀) = ln(5.110 × 10⁵ eV/1.889 eV) = ln(2.705 × 10⁵) = 12.508, against 7.722 for the full spectrum.
- The shift is proportional to ln(K/K₀), so the model overstates it by 12.508/7.722 = 1.620, putting 2s near 1700 MHz instead of about 1050 MHz.
Answer. 3p carries 1.26 per cent of Bethe's weight but 43.5 per cent of the oscillator strength. A 3p-only model gives K₀ = 1.889 eV and ln(mc²/K₀) = 12.51, overstating the 2s shift by a factor of 1.62.
HardExample 3
A computation of hydrogen's 2s spectrum groups Bethe's weights by excitation energy, giving each group's share of Sₙ and its log-mean energy Ē, the exponential of its weighted mean of ln Δ: bound np, 2.64 per cent at 2.346 eV; continuum to 100 eV, 40.06 per cent at 25.54 eV; 100 eV to 10 keV, 48.43 per cent at 605.2 eV; above 10 keV, 8.87 per cent at 78.28 keV. Check completeness, find K₀, and find what stopping the sum at 10 keV would do.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Completeness: 2.64 + 40.06 + 48.43 + 8.87 = 100.00 per cent of Sₙ, the total fixed by topic 13.05's sum rule, so no state is missing.
- ln(K₀/eV) = Σ share × ln(Ē/eV) = 0.0264 × 0.8527 + 0.4006 × 3.2402 + 0.4843 × 6.4056 + 0.0887 × 11.2680 = 0.0225 + 1.2980 + 3.1022 + 0.9995 = 5.4222.
- K₀ = e5.4222 eV = 226.4 eV, matching the quoted ln(K₀/Ry) = 2.8118; with K = mc² = 511.0 keV, ln(K/K₀) = 7.722.
- Stop at 10 keV: the first three groups hold 0.9113 of the weight, so ln(K₀/eV) = (0.0225 + 1.2980 + 3.1022)/0.9113 = 4.8532, K₀ = 128.1 eV, and ln(K/K₀) = 8.291.
- The truncated logarithm is 0.569 too large, 7.4 per cent, or 77 MHz at 135.6 MHz per unit of logarithm: the 8.87 per cent above 10 keV moves the result because each of its states has a long lever arm in ln Δ.
- Tail check: (8/π)√(Ry/E) at E = 10 keV gives 9.4 per cent above, and a Δ⁻³⁄² tail has log-mean e² × 10 keV = 73.9 keV; the computed 8.87 per cent and 78.28 keV differ by the Coulomb corrections still present at 10 keV.
Answer. The shares close the sum rule at 100.00 per cent and give K₀ = 226 eV, ln(mc²/K₀) = 7.72. Stopping at 10 keV gives K₀ = 128 eV and a logarithm 0.57 (7.4 per cent, about 77 MHz) too large.
Quick check
Select an answer and explain it before checking.
Review & continue
Bethe's mean excitation energy K₀ of a level n is the weighted geometric mean of its excitation energies |Eₘ − Eₙ|, each intermediate state weighted by |pₘₙ|²(Eₘ − Eₙ), so that the self-energy's sum of logarithms over all states becomes the single logarithm ln(K/K₀).
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.