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Advanced Quantum Mechanics course sequence · Interactive lesson

What only relativistic QED supplies: where QFT begins

Bethe's 1040 MHz needed a cutoff chosen by hand. This closing lesson asks what a relativistic theory of electrons and photons adds: a split that cancels, a definite constant beside the logarithm, the anomalous moment and the polarised vacuum. It ends where the systematic version of that theory begins.

UAQM-13.09 · Lesson 9 of 9 in this unit. Position is not a completion score.
Starting model: What only relativistic QED supplies: where QFT beginsWaterfall of hydrogen's one-loop 2S₁/₂ − 2P₁/₂ interval in MHz, split at Λ = 16.2 keV. Bethe's part below Λ is 578.9, the covariant part above 436.6, always 1015.6 together; g − 2 adds 67.8, vacuum polarisation takes 27.1 and 2P₁/₂'s logarithm 4.1. Total 1052.2 against 1057.8 measured.2S₁/₂ − 2P₁/₂ in MHz, Λ = 16.2 keVdashed: measured 1057.8below Λabove Λg − 2vac. pol.2P₁/₂ logtotal01000 MHz
Start with a picture. Change the model in the lab below.
1

Build the model

A cutoff that cancels, a constant beside the logarithm, and two effects Bethe left out.

The key idea

Relativistic QED is Unit 9's Dirac field coupled to Unit 11's photon field; at order α it fixes the constant that Bethe's cutoff left open and adds two effects a lone electron cannot have, the anomalous moment and vacuum polarisation.

Example. For hydrogen's 2S₁/₂ − 2P₁/₂ interval: 1015.6 MHz of cutoff-free self-energy, +67.8 MHz from g − 2, −27.1 MHz from vacuum polarisation and −4.1 MHz from 2P₁/₂'s own logarithm, 1052.2 MHz against 1057.8 MHz measured.

The complete physical model and assumptions

Bethe's calculation was a non-relativistic answer to a relativistic question, and it knew it: the coefficient of the logarithm was secure, but the cutoff, the constant beside the logarithm and anything a lone electron cannot do were left open. Relativistic QED closes them by coupling Unit 9's Dirac field to Unit 11's photon field and expanding Dyson's S-matrix in powers of e. Three things follow at order α. Pair intermediate states turn the electron's electromagnetic mass from linear in the cutoff to logarithmic, so one covariant subtraction disposes of it. Splitting the photon energies at any Λ and treating the hard photons relativistically removes the cutoff, replacing Bethe's ln(mc²/K₀) by ln(mc²/2K₀) + 11/24. And two effects with no one-electron counterpart enter: the anomalous moment, adding 3/8 to that constant on s states and lowering 2P₁/₂ by 17.0 MHz, and vacuum polarisation, subtracting 1/5. For hydrogen's n = 2 the interval comes out at 1052.2 MHz against 1057.8 MHz measured, with nothing left to adjust. The model stops at the next order. Corrections in higher powers of Zα and α, of which for heavy ions no finite number suffices, need the whole perturbation series, its diagrams and renormalisation to all orders. That is quantum field theory, and this course ends at its door.

1Where the non-relativistic answer runs out of authority

Units are SI, with xμ = (ct, x) and the metric (+, −, −, −), as throughout the course. Bethe's 2S shift, C₂ ln(Λ/K₀) with C₂/h = 135.64 MHz, rests on approximations that fail inside the range it integrates over. The dipole approximation fails at photon energies of order ħc/a₀ = αmc² = 3.73 keV, where the reduced wavelength shrinks to the Bohr radius; non-relativistic kinematics fails as the photon energy approaches mc²; and Λ = mc² is a judgement, not a result. Of the logarithm's 7.72 units, ln(1/α) = 4.92 come from photons above 3.73 keV. The coefficient C₂ is safe: the sum rule fixes it from photon energies far above the binding, where the electron is nearly free and a relativistic treatment agrees. The constant beside the logarithm is not safe: a factor of two in Λ moves 2S by C₂ ln 2 = 94.0 MHz, and nothing in a one-electron theory says which factor is right.

2Coupling the Dirac field to the photon field

The relativistic theory is assembled rather than derived. Take Unit 9's Dirac field ψ, whose expansion carries electron annihilators b and positron creators d†, and Unit 11's transverse field A, and couple them minimally: H = ∫ψ†(cα⋅(−iħ∇) + βmc²)ψ d³x + Σ ħω a†a + HC − ∫ j⋅A d³x, with j = qcψ†αψ, q = −e, and HC the instantaneous Coulomb energy the Coulomb gauge leaves behind. The novelty sits in j⋅A. Because ψ†αψ contains b†d† and db, one photon vertex can create or destroy an electron-positron pair, which topic 13.01's A⋅p Hamiltonian cannot. In the one-electron sector, Unit 6's reduction to order (v/c)² returns the A⋅p coupling with coefficient e/m, the A² term and the spin term, so Bethe's theory is the low-energy edge of this one. Each vertex carries e, so one photon emitted and reabsorbed is order e², that is α, the order Bethe worked at.

3The S-matrix, and what pair states do to the mass

Perturbation theory in this H is organised by Dyson's S = T exp(−(i/ħ)∫HI(t) dt), taken here as a definition. At order e² it holds three processes: the electron emitting and reabsorbing a photon, the self-energy; a photon emitted before the electron scatters off the nucleus and reabsorbed after, the vertex correction; and the nucleus's Coulomb field briefly making a pair, vacuum polarisation. The self-energy is topic 13.03's object with new intermediate states containing a pair, and Weisskopf found in 1939 that these cancel the linear growth, leaving δm c² = (3α/2π)mc² ln(Λ/mc²), quoted. With Λ at the Planck energy, the non-relativistic (4α/3π)Λ is 7.4 × 10¹⁹ mc², while the relativistic δm is 0.18 m. Treated covariantly, δm is a Lorentz scalar, a shift of the rest mass alone, so subtracting it is the same in every frame, and the vacuum loop's divergence is likewise a constant times the charge. Both are absorbed into the measured m and e: renormalisation, at its first order.

4Removing the cutoff by splitting the photon energies

Choose a split Λ with K₀ ≪ Λ ≪ mc². Below it the electron is non-relativistic and Bethe's integral is right: C₂ ln(Λ/K₀), derived in topics 13.04 to 13.06. Above αmc² the dipole approximation fails, but the integrand does not, because e(ik⋅r) commutes with V and leaves the sum rule's ⟨∇²V⟩ unchanged. Above Λ the photon dwarfs the binding, so the electron is effectively free and takes a single kick from the proton's field; the needed quantity is the slope of a free electron's charge form factor, which Kroll and Lamb and, independently, French and Weisskopf computed relativistically in 1949. Their C₂[ln(mc²/2Λ) + 11/24] is quoted, but one part of it is derived here: the coefficient of ln Λ must be −C₂, because Λ marks where we changed methods and the atom cannot know it. The sum, C₂[ln(mc²/2K₀) + 11/24] = 135.644 × 7.4871 = 1015.6 MHz, contains no cutoff. Bethe's formula at Λ = mc² gives 1047.4 MHz with today's K₀ = 16.64 Ry, about 1040 MHz with his 17.8 Ry; for his formula to give 1015.6 MHz, Λ would have to be mc²e¹¹⁄²⁴/2 = 0.791 mc².

5Two effects a lone electron cannot have

The vertex correction also changes the electron's magnetic coupling, by Schwinger's a = α/2π = 0.0011614, quoted. Unit 6's Pauli term multiplies spin-orbit and Darwin by 1 + 2a, so 2S gains 2a times its Darwin energy, aα⁴mc²/8 = 50.87 MHz, and 2P₁/₂ gains 2a times its spin-orbit energy, −aα⁴mc²/24 = −16.96 MHz, computed. In the s-state bracket aα⁴mc²/n³ is (3/8)Cₙ, so self-energy and moment together carry 11/24 + 3/8 = 5/6. Vacuum polarisation is the photon's own loop. Uehling's result, quoted, is a contact potential −(4α/15)(ħ/mc)²(e²/4πε₀)δ³(r): inside ħ/mc = 386 fm the electron is within the cloud of screening pairs, meets more of the proton's charge, and an s level falls by Cₙ/5, 27.13 MHz for 2S, computed with |ψ(0)|². The contact form holds because the Bohr radius is 274 times the 193 fm range; in muonic hydrogen it is not, and polarisation becomes the largest term. Set a = 0 and drop vacuum polarisation, and 2S still lies 1011.5 MHz above 2P₁/₂: the self-energy alone breaks Dirac's degeneracy, which returns only as e → 0.

6The budget, and where quantum field theory begins

For 2S: 1015.57 + 50.87 − 27.13 = 1039.31 MHz. For 2P₁/₂: −16.96 MHz from the moment and +4.07 MHz from its own Bethe logarithm, ln(K₀/Ry) = −0.0300, quoted, so −12.89 MHz. The interval is 1052.2 MHz, the 1949 result evaluated with today's constants, against 1057.8 MHz measured: 0.53% short with every piece fixed. Dimensions check, since α is pure and each term is a multiple of α⁵mc²; the free-field limit e → 0 removes all of it. The remaining 5.6 MHz is higher order, quoted: the next power of Zα adds 7.1 MHz, the reduced mass removes 3 parts in 1837 of the total, 1.7 MHz, and recoil, two-loop and proton-size terms close the rest. Those need the S-matrix beyond order e², or the bound electron's propagator to all orders in Zα, and at uranium's Zα = 0.67 nothing less works. Doing that systematically, drawing every term as a Feynman diagram and proving that every divergence is absorbed into m and e at all orders, is quantum field theory, and this course stops there.

2

Key equations

Read the units, assumptions and limits before substituting.

The QED Hamiltonian in Coulomb gauge

H = ∫ψ†(cα⋅(−iħ∇) + βmc²)ψ d³x + Σ ħω a†a + HC − ∫ j⋅A d³x

Assembled, not derived: two quantised fields coupled minimally. ψ also holds positron creators, so a single j⋅A vertex can make a pair, which no one-electron theory allows.

ψ is Unit 9's field, in m(−3/2); A is Unit 11's, in V s m⁻¹; j = qcψ†αψ in A m⁻², q = −e; HC is the Coulomb energy, in J.

Dyson's S-matrix

S = T exp(−(i/ħ) ∫ HI(t) dt)

A definition, quoted. Its order-e² terms hold the self-energy, the vertex correction and vacuum polarisation; organising every order is the work of Feynman diagrams.

HI(t) = e(iH₀t/ħ) Hᵢₙₜ e(−iH₀t/ħ), in J; T puts later times to the left; S is dimensionless and unitary.

Electromagnetic mass with pair states (quoted)

δm c² = (3α/2π) mc² ln(Λ/mc²) + finite terms

Topic 13.03's non-relativistic δm c² = (4α/3π)Λ grows linearly; pair states cancel that growth, and the δm left is a Lorentz scalar, subtracted once for every frame.

Λ is the photon cutoff, in J; Weisskopf, 1939. With Λ at the Planck energy, 1.22 × 10²⁸ eV, δm/m = 0.18.

The s-state self-energy with the cutoff removed

ΔESE(nS) = Cₙ ln(Λ/K₀) + Cₙ[ln(mc²/2Λ) + 11/24] = Cₙ[ln(mc²/2K₀) + 11/24]

First term Bethe's, derived; the second's −ln Λ is forced, its 11/24 quoted. Λ cancels: 1015.6 MHz for 2S, as if Bethe had cut off at 0.791 mc².

Cₙ = 4α⁵mc²/(3πn³), C₂/h = 135.64 MHz; K₀ = 16.64 Ry = 226.4 eV for 2S; any split with K₀ ≪ Λ ≪ mc².

The anomalous moment's share of an s state

ΔEₐ(nS) = aα⁴mc²/n³ = (3/8)Cₙ, with a = α/2π

This is the 3/8 in the QED constant, so the self-energy and moment together carry 11/24 + 3/8 = 5/6. At a = 0 Dirac's g = 2 returns and the share vanishes.

a is dimensionless, Schwinger's 0.0011614, quoted; it is Unit 6's Darwin energy times 2a. 2S rises 50.87 MHz, computed.

Vacuum polarisation as a contact term

ΔEVP(nS) = −(4α/15)(ħ/mc)²(e²/4πε₀)|ψₙ₀₀(0)|² = −Cₙ/5

Pairs screen the proton beyond ħ/mc = 386 fm, so an s electron inside meets more charge and binds harder. Valid while a₀ is far above the 193 fm range: here 274 times.

Uehling's coefficient quoted; |ψₙ₀₀(0)|² = 1/(πn³a₀³) in m⁻³, a₀ = 52 918 fm. 2S falls 27.13 MHz, computed; 2P is untouched.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

−1.5
−2 to 0; steps of 0.1.
1.00
0 to 2; steps of 0.25.
1
0 to 1; steps of 1.

Try this. Drag the split Λ: the two self-energy bars trade megahertz and their sum never moves, which is what removing the cutoff means. At Λ = mc², beyond the split's range, the lower part is Bethe's 1047.4 MHz. At a = 0 the total falls 73.5 MHz short; without vacuum polarisation it overshoots by 21.5.

Live results

SELF-ENERGY BELOW Λ
578.9 MHz
SELF-ENERGY ABOVE Λ
436.6 MHz
ONE-LOOP INTERVAL
1052.2 MHz
MEASURED − ONE LOOP
5.7 MHz

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Advanced Quantum Mechanics modelWaterfall of hydrogen's one-loop 2S₁/₂ − 2P₁/₂ interval in MHz, split at Λ = 16.2 keV. Bethe's part below Λ is 578.9, the covariant part above 436.6, always 1015.6 together; g − 2 adds 67.8, vacuum polarisation takes 27.1 and 2P₁/₂'s logarithm 4.1. Total 1052.2 against 1057.8 measured.2S₁/₂ − 2P₁/₂ in MHz, Λ = 16.2 keVdashed: measured 1057.8below Λabove Λg − 2vac. pol.2P₁/₂ logtotal01000 MHz
Waterfall of hydrogen's one-loop 2S₁/₂ − 2P₁/₂ interval in MHz, split at Λ = 16.2 keV. Bethe's part below Λ is 578.9, the covariant part above 436.6, always 1015.6 together; g − 2 adds 67.8, vacuum polarisation takes 27.1 and 2P₁/₂'s logarithm 4.1. Total 1052.2 against 1057.8 measured.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Split Λ, as log₁₀(Λ/mc²): −1.5; Anomalous moment a, in units of α/2π: 1.00; Vacuum polarisation: off 0, on 1: 1

Baseline −1.5; choose −2 to 0, in steps of 0.1.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

That QED's agreement is bought by choosing the cutoff: Bethe picked mc², a relativistic theory merely blesses the choice, and 1057.8 MHz is fitted. It is the other way round. In the covariant calculation no photon integral is cut off in the answer. Λ survives only inside δm and the vacuum's charge correction, and both are fixed by measuring the electron's mass and charge, after which the 2S shift has no free parameter. The split Λ used to organise the calculation is bookkeeping, and its value cancels. The equivalent Bethe cutoff, 0.791 mc² for the self-energy or 0.942 mc² with the moment and vacuum polarisation included, is an output. Bethe's mc² happened to land 8.1 MHz from it; a factor of two either way would have cost 94 MHz, while the QED value cannot move at all.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

Uehling's result, quoted, is that the polarised vacuum adds a contact potential V(r) = −(4α/15)(ħ/mc)²(e²/4πε₀)δ³(r) to hydrogen's Coulomb attraction. Find its first-order shift of the 1S, 2S and 2P₁/₂ levels in MHz, and check that treating it as a contact term is justified.

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.

MediumExample 2

Split the 2S photon energies at Λ. Below, Bethe's non-relativistic integral gives C₂ ln(Λ/K₀); above, the relativistic free-electron part, quoted, gives C₂[ln(mc²/2Λ) + 11/24], with C₂/h = 135.644 MHz and K₀ = 226.4 eV. Evaluate both at Λ = 5 keV and 50 keV, add the moment's (3/8)C₂ and vacuum polarisation's −C₂/5, and find the one cutoff with which Bethe's formula alone would give the same 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.

HardExample 3

For a hydrogen-like ion every leading term scales as Z⁴, the logarithm becomes ln(1/(Zα)²) − ln k₀, with ln k₀ = ln(K₀/Z²Ry) = 2.81177 for 2S at every Z, and the constant stays 19/30. With C₂/h = 135.644 MHz and hydrogen's 2P₁/₂ pieces (+4.072 MHz from its logarithm, −16.955 MHz from the moment), predict He⁺'s 2S₁/₂ − 2P₁/₂ interval, neglecting nuclear motion, and compare with the measured 14 041 MHz and with hydrogen's 1052.2 against 1057.8 MHz.

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.

6

Quick check

Select an answer and explain it before checking.

Bethe's cutoff-dependent logarithm acts only on s states, so it is tempting to say that 2P₁/₂ stays where Dirac put it. At order α(Zα)⁴mc², what does relativistic QED do to hydrogen's 2P₁/₂ level?
7

Review & continue

Relativistic QED is Unit 9's Dirac field coupled to Unit 11's photon field; at order α it fixes the constant that Bethe's cutoff left open and adds two effects a lone electron cannot have, the anomalous moment and vacuum polarisation.

  • 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.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.