
Advanced Quantum Mechanics course sequence · Interactive lesson
Mass renormalisation: Bethe's subtraction
This is the hinge of the unit. One counterterm, fixed by the free electron before any atom is considered, turns a self-energy that grows with the cutoff into one that grows only as its logarithm, and the same algebra hands back the decay width. Learn to make the subtraction, and to say exactly what it leaves undone.
UAQM-13.04 · Lesson 4 of 9 in this unit. Position is not a completion score.Build the model
Write the Hamiltonian in the mass you measure, and the linear divergence is gone.
The key idea
Mass renormalisation rewrites the Hamiltonian in the measured electron mass, so the bare kinetic energy p²/2m₀ becomes p²/2m plus a counterterm (δm/m)p²/2m, which removes from every level the self-energy that the free electron already carries as mass.
Example. For hydrogen's 2s at K = mc², the counterterm's expectation (δm/m)⟨T⟩ = 3.097 × 10⁻³ × 3.401 eV = 10.5 meV, or +2.547 THz, cancels the −2.547 THz linear term exactly and leaves a remainder near 1 GHz that grows only as ln K.
The complete physical model and assumptions
The m in the atom's Hamiltonian is a bare mass m₀, the inertia the electron would have without its field, and topic 13.03 showed that the field adds δm/m = 4αK/(3πmc²), so every mass weighed, and hence hydrogen's measured Rydberg constant, is m = m₀ + δm. Kramers' remedy, which Bethe applied in 1947, is to write the Hamiltonian in the measured mass: p²/2m₀ becomes p²/2m plus a counterterm (δm/m)p²/2m, first order in α like the self-energy itself. Its expectation (δm/m)⟨T⟩ₙ, written by closure as a sum over the same intermediate states, puts +1 into each term's integrand, against the −1 that every term of the self-energy settles to at high photon energy. The linear divergence cancels state by state and leaves the integrand (Eₘ − Eₙ)/(E + Eₘ − Eₙ), which falls as 1/E, so what grows with the cutoff is now only a logarithm, derived. A state below the level puts a pole on the photon-energy axis: its principal value joins the logarithm, and its imaginary part is Unit 12's decay width, untouched by the real counterterm. The subtraction has no free parameter, because the free electron fixes δm once, but it does not finish the job. ln K remains, the photons near mc² it still counts lie where the dipole and non-relativistic approximations fail, and removing the cutoff takes the covariant, all-orders renormalisation of quantum electrodynamics, which this course names and does not do.
1Write the Hamiltonian in the measured mass
SI units throughout, with energies in eV and shifts quoted as frequencies ΔE/h; the electron stays non-relativistic, so the course metric (+, −, −, −) never enters. The Hamiltonian of topic 13.01 carries the bare mass m₀, and topic 13.03 derived that the field adds δm/m = 4αK/(3πmc²), quoted here. Experiments weigh m = m₀ + δm, and the Rydberg constant that spectroscopy measures is built on m. Kramers' proposal was to say so in the Hamiltonian. Expanding 1/m₀ = 1/(m − δm) gives p²/2m₀ = p²/2m + (δm/m)p²/2m + O(α²), derived. Take H₀ = p²/2m + V and treat the counterterm (δm/m)p²/2m as a perturbation beside the coupling to the field: both are of order α, the self-energy at second order in a coupling proportional to e, the counterterm at first order. Check it on a free electron. Its self-energy, −(δm/m)p²/2m from topic 13.03, and the counterterm cancel, so E(p) = p²/2m with the measured m to order α. That is the renormalisation condition, and it fixes δm once, before any atom is considered.
2Closure turns the counterterm into +1 per term
Topic 13.02's linear term came from closure, Σₘ |pₘₙ|² = ⟨n|p²|n⟩. Run it backwards on the counterterm: ⟨n|(δm/m)p²/2m|n⟩ = (2αK/(3πm²c²))⟨n|p²|n⟩ = (2α/(3πm²c²)) Σₘ |pₘₙ|² ∫₀K dE, a sum over the same intermediate states with integrand 1. Add it term by term, using the identity E/(Eₙ − Eₘ − E) + 1 = (Eₘ − Eₙ)/(E + Eₘ − Eₙ), and ΔEₙ = (2α/(3πm²c²)) Σₘ |pₘₙ|² (Eₘ − Eₙ) ∫₀K dE/(E + Eₘ − Eₙ), derived. At K = mc² the counterterm is +10.19 THz for 1s and +2.547 THz for every n = 2 level, each cancelling that level's linear term. Dimensions: |pₘₙ|²/(m²c²) is a pure number and the rest an energy. Free limit: p is diagonal, the only term has Eₘ = Eₙ, and ΔE = 0, as the renormalisation condition demands. The cancellation needs both sums over the same states. Keep only 2s's bound states, 31.4 per cent of ⟨p²⟩ (topic 13.02), against the full counterterm, and 1.75 THz of linear divergence survives.
3What survives falls as 1/E and grows as ln K
Write Δ = Eₘ − Eₙ. The subtracted integrand Δ/(E + Δ) is 1 for photons far below Δ, where the bound term is negligible and the counterterm stands alone, and falls as Δ/E above it. For a state above the level, ∫₀K dE Δ/(E + Δ) = Δ ln(1 + K/Δ), derived: positive, so such terms raise the level, and every state with weight lies above 2s, the direction Lamb and Retherford found. Count by decades of photon energy. Before the subtraction each decade of the −1 plateau contributed ten times the decade below it; after it, each decade above Δ contributes the same Δ ln 10, which is what a logarithm is. For K far above the intervals that carry the weight, ΔEₙ → (2α/(3πm²c²)) Σₘ |pₘₙ|² (Eₘ − Eₙ) ln(K/|Eₘ − Eₙ|): its coefficient of ln K is topic 13.05's sum rule, and the weighted mean of ln|Eₘ − Eₙ| is topic 13.06's Bethe logarithm. The subtraction moved the weight down but not out of reach: for an interval of 10 eV, photons above αmc² = 3.73 keV, where e(ik⋅r) ≈ 1 fails, still supply 45 per cent of the logarithm.
4A lower state puts a pole on the photon axis
If m lies below n, Δ is negative and Δ/(E + Δ) blows up at E = |Δ| = ħωₙₘ. There |m, 1ₖₛ⟩ has exactly the energy of |n, 0⟩: the photon is real, and the integral as written has no value. Unit 12 supplied the prescription, quoted: for a state switched on adiabatically the energy denominators carry +iη with η → 0⁺, so the subtracted integrand becomes Δ/(E + Δ − iη), and 1/(E + Δ − iη) = P[1/(E + Δ)] + iπδ(E + Δ). The principal value, the integral with a vanishing symmetric gap around the pole, gives Δ ln((K − |Δ|)/|Δ|), derived: negative once K exceeds 2|Δ|, so a lower state pushes the level down, and for K ≫ |Δ| it takes the same form Δ ln(K/|Δ|) as a state above. That is why one formula in ln|Eₘ − Eₙ| covers both. For hydrogen's 2p, the 1s term at K = mc² is −172 MHz and moves by −11.0 MHz each time K doubles. The states above 2p push the other way, and topic 13.05 shows that the ln K pieces cancel for p states.
5The imaginary part is the decay width
The δ-function adds iπΔ to each lower term, so Im ΔEₙ = −(2α/3) Σₘ |pₘₙ|² (Eₙ − Eₘ)/(m²c²) over states below n. Write the complex energy as Eₙ − iħΓₙ/2 and use pₘₙ = imωₘₙrₘₙ, quoted from Quantum Mechanics II: Γₙ = (4α/(3c²)) Σₘ ωₙₘ³ |rₘₙ|², derived, which is Unit 12's Weisskopf-Wigner rate, now the other half of one complex self-energy. Dimensions: ω³r²/c² is s⁻³ m² ÷ m² s⁻² = s⁻¹. The counterterm is real, so the subtraction never touches a width, and Γ needs no cutoff beyond K above ħωₙₘ, because a real photon has a single energy. Against measurement: for 2p, ħω = 10.20 eV and |⟨1s|r|2p⟩|² = (2¹⁵/3¹⁰)a₀² give Γ = 6.27 × 10⁸ s⁻¹, a 1.60 ns lifetime as measured, and a natural width Γ/2π = 99.8 MHz, a tenth of the Lamb shift, which is why Lamb and Retherford's resonances were so broad. 2s has no lower state with a dipole matrix element, 1s being excluded by parity, so its shift is real at this order; its 0.12 s two-photon lifetime is fourth order in e, quoted.
6One subtraction, no free parameter, and its limits
Kramers urged at the Shelter Island conference of June 1947 that the electron's electromagnetic mass be separated out and the theory written in the observed mass; Bethe worked out the non-relativistic calculation within days, and topic 13.07 recomputes his 1040 MHz. Nothing in the subtraction is adjustable. δm is fixed by the free electron, the same counterterm acts on every level, and it must use the same regulator as the bound self-energy: cut the free electron at K and the bound one at 1.01K and 2s keeps 1 per cent of its linear term, 25.5 GHz, twenty-four times the Lamb shift. What the subtraction leaves is a logarithm of K, and K stays; this course sets it at mc² by argument, where the dipole and non-relativistic treatment must fail. Removing the cutoff needs a covariant calculation, quoted in topic 13.09. Quantum electrodynamics goes further, renormalising mass and charge at every order in α, the programme of Tomonaga, Schwinger, Feynman and Dyson set out by 1949. This course names it and stops at this one subtraction.
Key equations
Read the units, assumptions and limits before substituting.
Counterterm from the bare mass
p²/2m₀ = p²/2m + (δm/m) p²/2m + O(α²), with m₀ = m − δm
Derived by expanding 1/(m − δm). The counterterm is first order in α, the order of the second-order self-energy, so its expectation value enters beside it.
m measured and m₀ bare mass in kg; p in kg m s⁻¹; δm/m = 4αK/(3πmc²) from topic 13.03, K the photon cutoff in JCounterterm as a sum over intermediate states
⟨n|(δm/m) p²/2m|n⟩ = (2α/(3πm²c²)) Σₘ |pₘₙ|² ∫₀K dE
Derived. It puts +1 in each term's integrand, against the −1 plateau of topic 13.02, so the linear divergence cancels state by state: 2.547 THz for 2s at K = mc².
closure over bound and continuum states m; pₘₙ = ⟨m|p|n⟩ in kg m s⁻¹; E the photon energy in JBethe's subtracted shift
ΔEₙ = (2α/(3πm²c²)) Σₘ |pₘₙ|² (Eₘ − Eₙ) ∫₀K dE/(E + Eₘ − Eₙ)
Derived from E/(Eₙ − Eₘ − E) + 1 = (Eₘ − Eₙ)/(E + Eₘ − Eₙ). The integrand now falls as 1/E, so only a logarithmic divergence is left.
Eₘ, Eₙ atomic energies in J; α = 1/137.036; mc² = 511.0 keV; the A² constant, equal for every level, droppedOne term, as a principal value
P∫₀K dE Δ/(E + Δ) = Δ ln|(K + Δ)/Δ|, with Δ = Eₘ − Eₙ and K above |Δ|
Derived. A state above raises the level, a state below lowers it once K exceeds 2|Δ|, and doubling K adds about Δ ln 2 per term, where the unsubtracted integral doubled.
Δ in J, of either sign; the integral is an energy and tends to Δ ln(K/|Δ|) for K ≫ |Δ|Width from the pole
Γₙ = −2 Im ΔEₙ/ħ = (4α/(3c²)) Σₘ ωₙₘ³ |rₘₙ|², over m below n
Derived, and equal to Unit 12's rate: 6.27 × 10⁸ s⁻¹ for 2p, a 1.60 ns lifetime as measured. The counterterm is real, so the subtraction never touches a width.
ωₙₘ = (Eₙ − Eₘ)/ħ in rad s⁻¹; rₘₙ = ⟨m|r|n⟩ in m; from 1/(x − iη) = P(1/x) + iπδ(x)Large-cutoff form
ΔEₙ → (2α/(3πm²c²)) Σₘ |pₘₙ|² (Eₘ − Eₙ) ln(K/|Eₘ − Eₙ|)
Derived from the principal value, which gives the same ln(K/|Δ|) for states above and below. The weighted mean of ln|Eₘ − Eₙ| becomes topic 13.06's Bethe logarithm.
K far above the intervals that carry the weight; the coefficient of ln K is evaluated in topic 13.05Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Raise K by a decade and compare readouts: the unsubtracted integral grows tenfold, the subtracted one by only about Δ ln 10. Then set Δ = −10.2 eV, 2p's 1s term: a pole appears at E = |Δ|, the principal value turns negative, and an imaginary part πΔ appears that no value of K changes.
Live results
- UNSUBTRACTED ∫₀K
- -510.889 keV
- COUNTERTERM, +K
- 510.999 keV
- SUBTRACTED, PRINCIPAL VALUE
- 110.38 eV
- IMAGINARY PART FROM THE POLE
- 0.00 eV
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. Interval Δ = Eₘ − Eₙ: 10.2 eV; Cutoff, log₁₀(K/mc²): 0.0
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting summary of renormalisation is subtracting infinity from infinity: throw away whatever diverges and keep the rest. Bethe's subtraction is not that, and the difference can be checked. A rule that dropped every divergent piece would drop the ln K term too, and with it the whole Lamb shift. What is subtracted is one operator, (δm/m)p²/2m, whose δm is fixed in advance by the free electron, the same for every level and not adjustable. Nor is it a constant offset: its expectation follows ⟨T⟩ₙ, 10.19 THz for 1s and 2.547 THz for n = 2 at K = mc², so it changes the 1S–2S interval by 7.64 THz, and it cancels the linear divergence only because closure writes it over the same intermediate states as the self-energy. What survives still depends on K, logarithmically: the subtraction makes the shift insensitive to the cutoff, not independent of it.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Hydrogen's three 2p states lie Δ = 10.204 eV above 1s and carry 23.4 per cent of ⟨p²⟩ = (αmc)² for 1s. For their term in the 1s self-energy with K = mc² = 511.0 keV, find the weight, the unsubtracted contribution, the counterterm's share, the subtracted remainder, how much of it comes from photons above αmc² = 3.73 keV, and what doubling K does.
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.
- Weight: w = (2α/(3πm²c²)) Σ|pₘₙ|² = (2α/(3π)) × 0.2341 × α² = 1.5485 × 10⁻³ × 0.2341 × 5.325 × 10⁻⁵ = 1.931 × 10⁻⁸, a pure number, as a ratio of squared momenta times α must be.
- Unsubtracted: w ∫₀K E dE/(Eₙ − Eₘ − E) = −w[K − Δ ln(1 + K/Δ)]. Here ln(1 + K/Δ) = ln(50 078) = 10.821, so Δ ln(1 + K/Δ) = 110.4 eV, and the term is −w × 510 888.5 eV = −9.863 × 10⁻³ eV, or −2.3848 THz.
- Counterterm's share, written by closure over the same states: +wK = 1.931 × 10⁻⁸ × 510 999 eV = 9.865 × 10⁻³ eV, or +2.3853 THz. The two differ by w × 110.4 eV = 2.132 × 10⁻⁶ eV, the subtracted term: +515 MHz, raising 1s.
- Photons above αmc² contribute Δ ln((K + Δ)/(αmc² + Δ)) = 10.204 eV × ln(511 009/3739) = 10.204 eV × 4.918 = 50.2 eV, which is 45 per cent of 110.4 eV. In the unsubtracted term the same photons supplied (K − αmc²)/K = 99.3 per cent.
- Doubling K adds Δ ln((2K + Δ)/(K + Δ)) = 10.204 eV × 0.6931 = 7.07 eV, 6.4 per cent of the subtracted term or +33.0 MHz, while the unsubtracted term and the counterterm each grow by another 2.385 THz. Limits: as K → 0 the remainder tends to wK, the counterterm alone, and every piece vanishes as α → 0.
Answer. w = 1.93 × 10⁻⁸; −2.3848 THz unsubtracted and +2.3853 THz from the counterterm leave +515 MHz, 45 per cent of it from photons above αmc², and doubling K adds only 6.4 per cent.
MediumExample 2
Hydrogen's 2p level, in one magnetic substate, has one lower state with a dipole matrix element: 1s, 10.20 eV below, with |⟨1s|r|2p⟩|² = (2¹⁵/3¹⁰)a₀² = 0.5549 a₀². With K = mc², find this term's weight, its principal-value shift in MHz, its imaginary part, and the decay rate, lifetime and natural width they imply. Compare with the measured 2p lifetime, 1.60 ns, and say what doubling K changes.
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.
- Weight: |pₘₙ|² = m²ω²|rₘₙ|², so w = (2α/(3π))(ħω/(αmc²))²|rₘₙ|²/a₀², using ħc/a₀ = αmc² = 3728.9 eV. With ħω/(αmc²) = 10.204/3728.9 = 2.737 × 10⁻³, w = 1.5485 × 10⁻³ × 7.488 × 10⁻⁶ × 0.5549 = 6.435 × 10⁻⁹.
- Here Δ = E(1s) − E(2p) = −10.204 eV, so Δ/(E + Δ) has a pole at E = 10.204 eV, where |1s, 1ₖₛ⟩ has 2p's energy and the photon is real. With E(2p) → E(2p) + iη, 1/(E + Δ − iη) = P[1/(E + Δ)] + iπδ(E + Δ).
- Real part: P∫₀K dE Δ/(E + Δ) = Δ ln((K − |Δ|)/|Δ|) = −10.204 eV × ln(50 076) = −10.204 eV × 10.821 = −110.42 eV. Times w this is −7.106 × 10⁻⁷ eV, or −171.8 MHz, pushing 2p down.
- Imaginary part: the δ-function gives w × πΔ = 6.435 × 10⁻⁹ × π × (−10.204 eV) = −2.063 × 10⁻⁷ eV. With ΔE = δE − iħΓ/2, Γ = −2 Im ΔE/ħ = 4.126 × 10⁻⁷ eV ÷ 6.582 × 10⁻¹⁶ eV s = 6.27 × 10⁸ s⁻¹, the same as (4α/(3c²))ω³|r|², Unit 12's rate.
- So τ = 1/Γ = 1.60 ns, matching the measured 1.60 ns, and Γ/2π = 99.8 MHz is the natural width of 2p, a tenth of the Lamb shift. The reduced mass would lower Γ by 0.05 per cent, below this precision.
- Doubling K adds Δ ln((2K − |Δ|)/(K − |Δ|)) = −10.204 eV × 0.6932 = −7.073 eV to the real part, or −11.0 MHz. The imaginary part is unchanged, since it needs only K above 10.2 eV, and the counterterm, being real, never enters it. Dimensions: w is a pure number, so both parts are energies.
Answer. Real part −172 MHz at K = mc², moving by −11.0 MHz per doubling of K; Im ΔE = −2.06 × 10⁻⁷ eV, so Γ = 6.27 × 10⁸ s⁻¹, τ = 1.60 ns as measured, and a natural width of 99.8 MHz.
HardExample 3
Model a valence electron as an isotropic harmonic oscillator with ħω₀ = 2.10 eV, the energy of sodium's yellow lines. Make Bethe's subtraction with K = mc² for the ground level and for the excited state |1, 0, 0⟩: show that the linear terms cancel, find the ground-level shift, the shift of the transition energy with and without the subtraction, and the excited state's width, and compare its lifetime with sodium's measured 3p lifetime, 16.2 ns.
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.
- p links level N only to N ± 1, with |⟨N + 1|pₓ|N⟩|² = (N + 1)mħω₀/2 along each axis. The ground level reaches three states, all at Δ = +ħω₀, and Σ|pₘ₀|² = (3/2)mħω₀ = ⟨p²⟩, so closure is exact with one intermediate level. The linear term −(2αK/(3πm²c²)) × (3/2)mħω₀ and the counterterm (4αK/(3πmc²))(3ħω₀/4) are equal and opposite, 4.878 meV each.
- Ground level: ΔE₀ = (2α/(3πm²c²)) × (3/2)mħω₀ × ħω₀ ln(1 + K/ħω₀) = (α/(πmc²))(ħω₀)² ln(1 + K/ħω₀) = 2.3228 × 10⁻³ × 8.630 × 10⁻⁶ eV × 12.402 = 2.486 × 10⁻⁷ eV, or +60.1 MHz.
- |1, 0, 0⟩ reaches |0, 0, 0⟩ below, weight mħω₀/2 at Δ = −ħω₀, and three states above with weights mħω₀, mħω₀/2 and mħω₀/2. The coefficient of ln K, Σ|pₘₙ|²Δ, is (3/2)m(ħω₀)² for both levels, as topic 13.05's sum rule requires for a constant ∇²V, so ΔE₁ − ΔE₀ = (α/(3πmc²))(ħω₀)² ln((K + ħω₀)/(K − ħω₀)) = 5.5 × 10⁻¹⁴ eV, 13 Hz, vanishing as K → ∞.
- Without the subtraction each level keeps −(δm/m)⟨T⟩, with ⟨T⟩ = (N + 3/2)ħω₀/2 on level N, so the transition would move by −(δm/m)ħω₀/2, which at K = mc² is −(2α/(3π))ħω₀ = −3.252 meV, or −786 GHz: 0.155 per cent, linear in K, exactly the shift of √(k/m) if the mass were m + δm. The counterterm removes it.
- Width: the lower term's pole gives Im ΔE₁ = (2α/(3πm²c²))(mħω₀/2)π(−ħω₀) = −(α/(3mc²))(ħω₀)² = −2.099 × 10⁻⁸ eV, so Γ = −2 Im ΔE₁/ħ = (2α/3)ħω₀²/(mc²) = 6.38 × 10⁷ s⁻¹ and τ = 15.7 ns. That is exactly the classical damping rate e²ω₀²/(6πε₀mc³) of a charged oscillator, the correspondence limit.
- Against measurement: sodium's 3s to 3p lines carry 0.96 of the one-electron oscillator strength that the model's single transition carries in full, so the model predicts 15.7 ns ÷ 0.96 = 16.3 ns against the measured 16.2 ns. Limits: as ω₀ → 0 every shift and the width vanish, the free electron again.
Answer. The linear terms cancel exactly, 4.878 meV each; the ground level rises 60.1 MHz; the transition moves 13 Hz after the subtraction against −786 GHz without it; Γ = 6.38 × 10⁷ s⁻¹, τ = 15.7 ns, or 16.3 ns for sodium's lines against 16.2 ns measured.
Quick check
Select an answer and explain it before checking.
Review & continue
Mass renormalisation rewrites the Hamiltonian in the measured electron mass, so the bare kinetic energy p²/2m₀ becomes p²/2m plus a counterterm (δm/m)p²/2m, which removes from every level the self-energy that the free electron already carries as mass.
- 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.