
Advanced Quantum Mechanics course sequence · Interactive lesson
The 2s shift computed: about 1040 MHz
Everything in this unit so far has been structure; here it becomes a number to hold against Lamb and Retherford's microwave line. Put the cutoff where Schrödinger's electron stops being the right particle, feed in K₀, read off 1040 MHz, and then ask how much of the agreement the method actually earned.
UAQM-13.07 · Lesson 7 of 9 in this unit. Position is not a completion score.Build the model
135.6 MHz times one logarithm, and an honest count of what the cutoff leaves open.
The key idea
Bethe's 2s shift is the rise of hydrogen's 2S₁/₂ level obtained by evaluating the mass-subtracted self-energy ΔE = (α⁵mc²/(6π)) ln(K/K₀) with the photon cutoff K set at mc² and the mean excitation energy K₀ taken from hydrogen's spectrum.
Example. With K = mc² = 511 keV and K₀ = 16.64 Ry = 226 eV, K/K₀ = 2257 and ln 2257 = 7.72, so ΔE/h = 135.6 MHz × 7.72 = 1047 MHz and ΔE = 4.33 μeV, against the measured 1057.8 MHz.
The complete physical model and assumptions
By this point the unit has turned the self-energy of an s level into ΔEₙ = (4α⁵mc²/(3πn³)) ln(K/K₀): a product of constants times the logarithm of two energies, neither of which the derivation fixes. K₀ belongs to the atom. It is the logarithmic mean of hydrogen's excitation energies weighted by the sum rule, 16.64 Ry for 2s, where Bethe estimated 17.8 Ry. K belongs to the theory. It is the photon energy past which a non-relativistic electron is the wrong description, and that happens near mc², where a photon's momentum makes the electron's recoil relativistic and pair states open. With n = 2 the constants give 135.6 MHz and Bethe's K₀ gives a logarithm of 7.654, so 2s rises by 135.6 MHz × 7.654 = 1038 MHz, his result of about 1040 MHz; the modern K₀ gives 7.72 and 1047 MHz, against 1057.8 MHz measured. The sign is right, 2S₁/₂ above 2P₁/₂ as Lamb and Retherford found, because both the logarithm and the contact term behind it are positive, and 2P₁/₂ gets no logarithm because its wavefunction vanishes at the nucleus. The method stops at its own input. The cutoff is an argument, not a result, and moving it by a factor of two moves the answer by 94 MHz, 9 per cent, so agreement at the 1 to 2 per cent level is better than the calculation can promise. The last per cent belongs to relativistic quantum electrodynamics.
1Two energies the derivation could not supply
Units are SI, with α = e²/(4πε₀ħc) = 1/137.036; everything happens in the atom's rest frame, so the course metric (+, −, −, −) never enters. The previous topics reduced the mass-subtracted self-energy of an s level to ΔEₙ = (4α⁵mc²/(3πn³)) ln(K/K₀), derived there and used here. For n = 2 the numerical factor is 4/(3π × 8) = 1/(6π). With α⁵ = 2.069 × 10⁻¹¹ and mc² = 510 999 eV, α⁵mc² = 1.057 × 10⁻⁵ eV; dividing by 6π gives 5.61 × 10⁻⁷ eV, and dividing by h = 4.136 × 10⁻¹⁵ eV s gives 135.6 MHz; from here on a shift quoted in MHz means ΔE/h, the frequency Lamb measured. The dimensions check at once: α, n and a logarithm are pure numbers, so ΔE has the dimensions of mc². What the derivation leaves open are the two energies inside the logarithm. K₀ is a property of the atom, the sum-rule-weighted logarithmic mean of hydrogen's excitation energies, computed in the previous topic. K is a property of the theory: the photon energy above which the non-relativistic calculation stops being believable. The 2s shift is 135.6 MHz times the number of factors of e between them.
2Why the cutoff goes at mc²
Lay the energies on one logarithmic ladder: the 2s binding, 3.40 eV; K₀ = 16.64 Ry = 226 eV, 67 times higher; αmc² = 3.73 keV, where a photon's wavelength equals 2πa₀; and mc² = 511 keV. A photon of energy ħω carries momentum ħω/c, and Schrödinger's electron absorbing it recoils with (ħω)²/(2mc²). That is small beside ħω only while ħω ≪ 2mc². Near mc² the recoil is relativistic, so the non-relativistic kinetic energy is wrong, and from 2mc² the intermediate states include electron-positron pairs the theory has no room for. So mc² is where the model must be cut off: this is Bethe's argument, not a derivation. Photons above αmc² have wavelengths below 2πa₀, shorter than the atom, yet they supply ln 137 = 4.92 of the 7.72, 64 per cent of the logarithm; the defence is that after the subtraction they reach the electron only through the contact term at the nucleus, a region small beside their wavelength until ħω nears mc². The covariant calculation quoted in the unit's closing topic is equivalent to his formula with K ≈ 0.79 mc² before the anomalous moment is added, inside a factor of two.
3Bethe's 1040 MHz and the modern 1047 MHz
Since Ry = α²mc²/2, the cutoff in rydbergs is mc²/Ry = 2/α² = 37 558. Bethe's own K₀ = 17.8 Ry, the estimate he made in 1947 from approximate bound and continuum sums and quoted here, gives mc²/K₀ = 2110 and ln 2110 = 7.654; then ΔE(2s) = 135.6 MHz × 7.654 = 1038 MHz, his result of about 1040 MHz. The modern K₀ = 16.64 Ry, from ln(K₀/Ry) = 2.8118 as computed in the previous topic, gives 2257, a logarithm of 7.72 and 1047 MHz. The measured 2S₁/₂–2P₁/₂ interval is 1057.8 MHz; Lamb and Retherford's first figure in 1947 was about 1000 MHz. Bethe sits 1.8 per cent low and the modern evaluation 1.0 per cent low. The logarithm is forgiving: K₀ moved by 7 per cent between the two evaluations, yet the shift moved by only 9 MHz, 0.9 per cent. In energy, 1047 MHz is 4.33 μeV, about 1.3 × 10⁻⁶ of the 2s binding energy, and its wavelength c/ν = 28.6 cm puts it in the microwave band, which is why the measurement needed microwave techniques and not an optical spectrometer.
4Sign and selection: 2s up, 2P₁/₂ left alone
Both factors of the result are positive. The logarithm is positive because K ≫ K₀, and the sum rule turned its coefficient into ⟨∇²V⟩ = (e²/ε₀)|ψ(0)|², positive for an attractive Coulomb potential, so the level rises. The picture behind the sign: the vacuum field jiggles the electron over a small region, and averaging −e²/(4πε₀r) over that region near the nucleus makes it shallower, so an electron found at the nucleus is bound less. Only s electrons are found there. A 2p wavefunction vanishes at r = 0, so its coefficient of ln K is zero and at the order of the logarithm it does not move; full QED (quoted) moves 2P₁/₂ by about −13 MHz, 1 per cent of the interval, through terms with no ln K. The Dirac degeneracy of 2S₁/₂ and 2P₁/₂, exact to all orders in Zα in Unit 7, is lifted with 2S₁/₂ on top, the order Lamb and Retherford observed. The limits are right too: the shift falls as n⁻³ with |ψ(0)|², so high Rydberg s states barely move, and it vanishes as α → 0, the free-field limit, faster than the fine structure by a factor of order α ln(1/α²).
5How much the cutoff leaves unfixed
Nothing in the argument picks mc² over mc²/2 or 2mc². Each factor of two moves the shift by 135.6 MHz × ln 2 = 94 MHz, so the honest statement is 1047 ± 94 MHz, the range 953 to 1141 MHz, a 9 per cent band. The measured 1057.8 MHz lies 10 MHz from the centre, about a ninth of the half-width. Solving for the cutoff that fits exactly gives K = 1.08 mc² with the modern K₀, but any K from about 0.55 to 2.15 mc² keeps the measurement within the band, so that fit is a choice made by the data, not a prediction. The remaining per cent is real physics this theory omits, quoted here and accounted for in the unit's closing topic: the anomalous moment raises 2s by 51 MHz, vacuum polarisation lowers it by 27 MHz, the covariant treatment of the hardest photons removes 32 MHz, and 2P₁/₂ moving down widens the interval by 13 MHz. Each is 1 to 5 per cent of the total; they net to about +5 MHz, and smaller higher-order terms supply the rest. Bethe's near-agreement is partly a cancellation among pieces he did not have.
Key equations
Read the units, assumptions and limits before substituting.
Bethe's shift of an s level
ΔEₙ = (4α⁵mc²/(3πn³)) ln(K/K₀), for l = 0
Derived earlier in this unit by the mass subtraction and the commutator sum rule, and only evaluated here. For l ≥ 1 the coefficient of the logarithm is zero.
α = 1/137.036, mc² = 8.187 × 10⁻¹⁴ J (511 keV), n the principal quantum number; cutoff K and mean excitation K₀ both in JThe n = 2 prefactor
α⁵mc²/(6πh) = 135.6 MHz
The whole 2s shift is this number times one logarithm: every factor of e between K₀ and the cutoff adds another 135.6 MHz.
h = 6.626 × 10⁻³⁴ J s turns the energy into the frequency Lamb measured; 4/(3π × 2³) = 1/(6π). Electron mass, not reducedThe logarithm with the cutoff at mc²
ln(mc²/K₀) = ln(2/α²) − ln(K₀/Ry)
The atom enters only through K₀, and only logarithmically: Bethe's K₀, 7 per cent too high, costs just 0.9 per cent in the shift.
Ry = α²mc²/2 = 13.606 eV, so 2/α² = 37 558. K₀ = 16.64 Ry gives 7.72, Bethe's 17.8 Ry gives 7.65What a factor of two in the cutoff moves
δ(ΔE)/h = ±(α⁵mc²/(6πh)) ln 2 = ±94 MHz
This is the error bar a cutoff calculation carries. Agreement with experiment closer than ±94 MHz is not something the method itself can claim.
K doubled or halved; ln 2/ln(mc²/K₀) = 0.693/7.72 makes it ±9.0 per cent of the 2s shiftSize against the n = 2 fine structure
ΔE(2s)/ΔEfs = (16α/(3π)) ln(mc²/K₀) ≈ 0.096
One extra factor α/π, enlarged by a logarithm of 7.7 and by 16/3, is why Lamb's line sits a decade below the fine structure.
ΔEfs = α⁴mc²/32 = h × 10.95 GHz separates 2P₃/₂ from 2P₁/₂. Measured: 1057.8 MHz/10 969 MHz = 0.096Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set K₀ to Bethe's 17.8 Ry and the dot drops to 1038 MHz. Then, with K₀ back at 16.64 Ry, slide K: the measured line stays inside the factor-of-two box for every cutoff from 0.55 to 2.15 mc², so the data cannot pin K down.
Live results
- 2s SHIFT ΔE
- 1047 MHz
- LOGARITHM ln(K/K₀)
- 7.722
- OFF MEASURED 1057.8 MHz
- -1.0 %
- SPREAD FROM K ×2 OR ÷2
- 9.0 %
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. Photon cutoff K: 1.00 mc²; Mean excitation energy K₀: 16.64 Ry
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
It is tempting to read Bethe's 1.8 per cent, or the modern 1.0 per cent, as the accuracy of the non-relativistic theory. It is not. The calculation leaves the cutoff free to within a factor of about two, and through the logarithm that freedom is worth ±94 MHz, ±9 per cent. Any K from mc²/2 to 2mc² is as defensible as mc², and the one that matches the data exactly, 1.08 mc², is chosen by the answer rather than by the theory. The 1 per cent is a cancellation among omitted pieces each of 1 to 5 per cent. The opposite overreaction is also wrong: an arbitrary cutoff does not make the number arbitrary. A factor of two moves a logarithm of 7.7 by 0.69, where the unsubtracted linear divergence would have doubled the answer.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Hydrogen's 3s level has a Bethe mean excitation energy K₀(3s) = 15.92 Ry, quoted from the same kind of spectral sum as for 2s. With the cutoff at mc², compute Bethe's 3s shift in MHz and compare it with the measured 3S₁/₂–3P₁/₂ interval of about 315 MHz.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- The prefactor 4α⁵mc²/(3πn³h) at n = 3 is the n = 2 value scaled by (2/3)³ = 8/27: 135.64 MHz × 8/27 = 40.19 MHz.
- The logarithm: mc²/K₀ = 37 558/15.92 = 2359, and ln 2359 = 7.766.
- The shift: 40.19 MHz × 7.766 = 312.1 MHz, positive, so 3s is raised above 3P₁/₂.
- Checks: the result is a frequency, like the prefactor; it is 0.9 per cent below 315 MHz and well inside the cutoff band ±40.19 MHz × ln 2 = ±27.9 MHz; and the ratio 1047.4/312.1 = 3.36 is close to the pure n⁻³ value 3.375, because K₀ barely moves with n.
Answer. ΔE(3s) ≈ 312 MHz, 0.9 per cent below the measured 315 MHz and inside its own ±28 MHz cutoff band.
MediumExample 2
Treat the cutoff as the unknown. What K makes Bethe's formula reproduce the measured 1057.8 MHz, first with the modern K₀ = 16.64 Ry and then with Bethe's 17.8 Ry? Give each in keV and in units of mc² = 511.0 keV, and say whether either counts as a prediction.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Invert ΔE = 135.64 MHz × ln(K/K₀): ln(K/K₀) = 1057.8/135.64 = 7.7986, so K/K₀ = e(7.7986) = 2437.
- Modern K₀: K = 2437 × 16.64 Ry = 4.055 × 10⁴ Ry, and with 1 Ry = 13.606 eV that is 551.7 keV, or 1.080 mc².
- Bethe's K₀: K = 2437 × 17.8 Ry = 4.338 × 10⁴ Ry = 590.2 keV, or 1.155 mc².
- Both lie well inside the range mc²/2 to 2mc² that the argument for the cutoff allows, so the measurement picks a value the theory had left open. That is a one-parameter fit, not a test.
Answer. K = 551.7 keV (1.080 mc²) with the modern K₀ and 590.2 keV (1.155 mc²) with Bethe's; both are fits inside the factor-of-two band, not predictions.
HardExample 3
Apply Bethe's formula to the 2s level of He⁺ (Z = 2). Every atomic energy, K₀ included, scales as Z², and the prefactor as Z⁴; keep the cutoff at mc². Compare with the measured He⁺ 2S₁/₂–2P₁/₂ interval of about 14.04 GHz, find the cutoff band, and find the Z at which the formula's logarithm would vanish.
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.
- Prefactor: 4α(Zα)⁴mc²/(3πn³h) = Z⁴ × 135.64 MHz = 16 × 135.64 MHz = 2170 MHz. The Z⁴ is Z³ from |ψ(0)|² and Z from ∇²V.
- Logarithm: K₀(He⁺) = 4 × 16.64 Ry = 66.56 Ry, so mc²/K₀ = 37 558/66.56 = 564.3 and ln 564.3 = 6.336, shorter than hydrogen's 7.722 by ln 4 = 1.386.
- Shift: 2170 MHz × 6.336 = 1.375 × 10⁴ MHz = 13.75 GHz, 2.1 per cent below the measured 14.04 GHz. The prefactor grew 16 times but the shift only 13.1 times, because the logarithm shrank.
- Cutoff band: ±2170 MHz × ln 2 = ±1.50 GHz, ±10.9 per cent, wider than hydrogen's ±9.0 per cent because the logarithm is shorter.
- The logarithm vanishes when Z²K₀ = mc², at Z = √2257 = 47.5, where Zα = 0.35. Well before that the premise K₀ ≪ K fails and the bound electron is itself relativistic, so the formula is a low-Z result.
Answer. ΔE(He⁺ 2s) ≈ 13.75 GHz with a cutoff band of ±1.50 GHz, 2.1 per cent below the measured 14.04 GHz; the logarithm would reach zero near Z = 48.
Quick check
Select an answer and explain it before checking.
Review & continue
Bethe's 2s shift is the rise of hydrogen's 2S₁/₂ level obtained by evaluating the mass-subtracted self-energy ΔE = (α⁵mc²/(6π)) ln(K/K₀) with the photon cutoff K set at mc² and the mean excitation energy K₀ taken from hydrogen's spectrum.
- 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.