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Advanced Quantum Mechanics course sequence · Interactive lesson

The free electron's self-energy as an electromagnetic mass

Topic 13.02 left a divergence that every bound level carries in proportion to its kinetic energy. A free electron, with no spectrum to hide behind, shows what that term is: inertia lent by the electron's own field. Learn to read a p²-dependent energy as mass, and why the bare mass is never what an experiment weighs.

UAQM-13.03 · Lesson 3 of 9 in this unit. Position is not a completion score.
Starting model: The free electron's self-energy as an electromagnetic massLog-log plot of the electromagnetic mass δm/m against the photon cutoff K. Solid: the dipole model derived here, 4αK/(3πmc²), linear in K. Dashed: Weisskopf's quoted relativistic (3α/2π) ln(K/mc²), drawn only above mc². At K = 1.00 mc² the dipole model gives δm/m = 0.310 per cent and a bare mass m₀c² = 509.4 keV.K = 1.00 mc²δm/m = 1δm/m = 0.01dipole, ∝ KWeisskopf, ∝ ln K0.1110100K/mc²
Start with a picture. Change the model in the lab below.
1

Build the model

A shift that grows as p² is not an energy offset. It is inertia.

The key idea

The electromagnetic mass δm is the part of an electron's inertia supplied by its own field: coupling to transverse photons lowers the energy in proportion to p², so it changes the kinetic energy exactly as a larger mass would.

Example. With the photon cutoff at K = mc² = 511.0 keV, δm/m = 4α/(3π) = 3.10 × 10⁻³ and δm c² = 1.58 keV, so an electron with 10.0 keV of kinetic energy sits 31.0 eV lower than a bare electron of the same momentum.

The complete physical model and assumptions

Topic 13.02 found that a bound electron's second-order energy grows linearly with the photon cutoff K, level by level in proportion to ⟨p²⟩. A free electron shows what that growth is. A plane wave |p⟩ is an eigenstate of momentum, so in the dipole coupling (e/m)A⋅p the only intermediate state is the same electron plus one photon, the energy denominator is just −E, and every photon energy up to K contributes equally: δE(p) = −(p²/2m)(4αK/(3πmc²)), derived. A correction proportional to p² is not an offset of the energy zero but a change of inertia, so it is read as a mass increase δm/m = 4αK/(3πmc²), 0.31 per cent for K = mc². Planck's constant cancels from δm, which equals, as derived here, the classical field-momentum mass (4/3)UC/c² of a charge whose field stops at wavenumber K/ħc; the velocity-independent Coulomb self-energy UC sits apart, in Coulomb gauge's instantaneous term, and never reaches E(p). The model stops where its approximations do: the dipole step drops the electron's recoil, which outgrows the photon energy above 2mc², and the relativistic theory, quoted from Weisskopf (1939), replaces the linear growth by (3α/2π) ln(K/mc²). Whatever K is, the mass every experiment measures, 0.511 MeV/c², already contains δm. The bare mass m₀ is never observed, and predictions must be written in the measured one.

1One intermediate state: the free electron's shift

Units are SI throughout, with energies quoted in eV; where a four-vector appears the metric is (+, −, −, −). Topic 13.01's result is quoted, not re-derived: δEₙ = (2α/(3πm²c²)) ∫₀K dE E Σₘ |pₘₙ|²/(Eₙ − Eₘ − E), with E the photon energy. Put the electron in a periodic box with no potential. Its states are plane waves |p⟩, and the momentum operator returns p|p⟩, so pₘₙ = p when m = n and vanishes otherwise. The sum over m collapses to one term, the denominator becomes Eₙ − Eₙ − E = −E, and the E in the measure cancels it: δE(p) = −(2α/(3πm²c²)) p² ∫₀K dE = −(p²/2m)(4αK/(3πmc²)), derived. Every photon energy below K contributes the same amount, which is why the result is linear in K. Dimensions: α and K/mc² are pure numbers, so δE carries the units of p²/2m. Limits: δE vanishes as α → 0, as K → 0, and for an electron at rest.

2A shift proportional to p² is a change of mass

A correction that grows as p² is not an offset of the energy zero. It changes how the energy depends on momentum, and that dependence is what inertia means. Added to the bare kinetic energy, it gives p²/2m₀ − (δm/m₀)p²/2m₀ = p²/(2(m₀ + δm)) to first order in α, with δm/m = 4αK/(3πmc²), derived. Lowering E at fixed p flattens E(p), so the velocity ∂E/∂p = p/(m₀ + δm) falls: the dressed electron is heavier. With K = mc², δm/m = 4α/(3π) = 3.10 × 10⁻³ and δm c² = 1.58 keV. The A² term, first order in e², adds (e²/2m)⟨0|A²|0⟩ = αK²/(2πmc²), derived from the same mode sum, ⟨0|A²|0⟩ = Σₖₛ ħ/(2ε₀ωV) = K²/(4π²ε₀ħc³), 0.59 keV at K = mc², equally to every momentum state: it moves the zero of energy, changes no velocity, and contributes nothing to the mass. The picture is that the electron carries its transverse field along, so pushing the electron means pushing the field's momentum too.

3The classical analogue is field momentum, not Coulomb energy

ħ cancels from δm: with K = ħckₘₐₓ, δm = e²kₘₐₓ/(3π²ε₀c²), a classical quantity, and classical electrodynamics has it. A charge moving slowly with velocity v carries B = v × E/c², so its field holds momentum P = ε₀∫E × B d³r = (ε₀/c²)∫[vE² − E(E⋅v)] d³r. Averaging over directions replaces E(E⋅v) by vE²/3, giving P = (4UC/(3c²))v, derived, with UC = (ε₀/2)∫E² d³r the Coulomb self-energy. If the field has no Fourier components above kₘₐₓ, UC = e²kₘₐₓ/(4π²ε₀), and 4UC/(3c²) equals δm exactly. UC itself, 1.19 keV for kₘₐₓ = mc/ħ, does not depend on v; in Coulomb gauge it sits in the instantaneous Coulomb term as a constant, set apart from the transverse photons, and never enters E(p). The factor 4/3 between mₑₘ c² and UC is the old puzzle of Abraham's and Lorentz's charged sphere, named here: the field's energy and momentum alone do not form a four-vector, and the non-electromagnetic stresses that hold the charge together must be added.

4How big, and where the linear law fails

A cutoff K corresponds to a length ħc/K: K = mc² means 386 fm, the reduced Compton wavelength. Setting δm = m gives K = 3πmc²/(4α) = 323 mc² = 165 MeV, or ħc/K = 1.20 fm, which is (4/(3π)) times the classical electron radius rₑ = 2.82 fm. An electron whose whole mass is electromagnetic would be a femtometre or two across. The model has no right to that conclusion. The dipole step set e(ik⋅r) = 1 and so dropped the recoil: the intermediate electron really has momentum p − ħk and an energy higher by about E²/(2mc²), which equals the photon energy itself at E = 2mc². Keeping it turns the constant integrand into 1/(1 + E/(2mc²)) and the result into (8α/(3π)) ln(1 + K/(2mc²)), derived in the Hard example: linear below 2mc², logarithmic above. Measurement rules the classical picture out on its own: high-energy e⁺e⁻ scattering finds no electron structure down to below 10⁻¹⁸ m, quoted, more than a thousand times smaller than 1.20 fm.

5Weisskopf's logarithm, quoted

Near K = mc² photons carry momentum comparable to mc, and the electron's relativistic structure can no longer be left out. Weisskopf's 1939 calculation in Dirac's hole theory, quoted here and confirmed by one-loop quantum electrodynamics, gives δm/m ≈ (3α/2π) ln(K/mc²) for K ≫ mc². Two features matter. The growth is logarithmic: in Weisskopf's picture the filled negative-energy sea, through the exclusion principle, smears the electron's charge over about ħ/mc and cancels the linear term. And δm is proportional to m itself, where the dipole model's δm c² = 4αK/(3π) did not depend on m. The numbers are modest: δm/m = 0.016 at K = 100 mc², 0.024 at 1000 mc², and only 0.18 with K at the Planck energy, 1.22 × 10¹⁹ GeV. A divergence this slow keeps δm a small correction even at absurd cutoffs, so treating it at first order and absorbing it into the measured mass is self-consistent. This lesson quotes the relativistic result and derives nothing from it.

6The bare mass is never observed

The Hamiltonian's m₀ is the mass the electron would have with its field switched off, and the field cannot be switched off. Every way of weighing an electron, a cyclotron frequency in a Penning trap, the bending of a beam in a magnet, the Rydberg constant of hydrogen, sees m = m₀ + δm, with mc² = 0.51099895 MeV, measured. In the dipole model m₀c² is 509.42 keV if K = mc², 507.83 keV if K = 2mc², and negative above 165 MeV: a number that moves with an arbitrary cutoff is not a property of the electron. Hydrogen shows the bookkeeping is consistent. Topic 13.02's linear term is −(δm/m)⟨n|p²/2m|n⟩, and the virial theorem gives ⟨p²/2m⟩ = Ry/n² for every l, so it shifts each level by −(δm/m)Ry/n², exactly what replacing m₀ by m₀ + δm does to Ry = α²mc²/2. At K = mc² that is 2.55 THz for every n = 2 level, 2400 times the measured 1057.8 MHz Lamb shift but identical for 2s and 2p. The measured Rydberg constant already contains it; topic 13.04 makes the subtraction.

2

Key equations

Read the units, assumptions and limits before substituting.

Free-electron shift in the dipole coupling

δE(p) = −(2α/(3πm²c²)) p² ∫₀K dE = −(p²/2m)(4αK/(3πmc²))

Derived: p is diagonal, so the only intermediate state is |p, 1ₖₛ⟩ with denominator −E. The E in the measure cancels it, and every photon energy below K adds the same amount.

p in kg m s⁻¹, m in kg, E the photon energy ħω and K its cutoff, both in J; α = e²/(4πε₀ħc) ≈ 1/137.036

Electromagnetic mass in the dipole model

δm/m = 4αK/(3πmc²)

Derived by matching p²/2m₀ + δE to p²/(2(m₀ + δm)) at first order in α. Note that δm c² = 4αK/(3π) does not depend on the electron's mass at all.

Dimensionless; 3.10 × 10⁻³ at K = mc², so δm c² = 1.58 keV and δm = 2.82 × 10⁻³³ kg

Classical field-momentum mass

mₑₘ = 4UC/(3c²)

Derived for v ≪ c from P = ε₀∫E × B d³r with B = v × E/c². With kₘₐₓ = K/(ħc) it equals δm exactly: ħ cancels, so this self-mass is classical.

UC = (ε₀/2)∫E² d³r in J, the Coulomb self-energy; e²kₘₐₓ/(4π²ε₀) if the field stops at wavenumber kₘₐₓ in m⁻¹

Bare and measured mass

m = m₀ + δm(K)

Every experiment weighs m, never m₀. m₀ moves with the unphysical K and would reach zero at K = 3πmc²/(4α) = 165 MeV, so it is not a property of the electron.

mc² = 0.51099895 MeV, measured; m₀ is the mass in the Hamiltonian, 509.42 keV/c² if K = mc² (first order in α)

Recoil restored

δm/m = (8α/(3π)) ln(1 + K/(2mc²))

Derived in the Hard example. Below 2mc² it is the linear law; above, growth is only logarithmic. At K = mc² it gives 2.51 × 10⁻³, 19 per cent below the dipole value.

Same symbols; keeps the electron's recoil energy E²/(2mc²) in the denominator, still non-relativistic

Weisskopf's relativistic self-mass

δm/m ≈ (3α/(2π)) ln(K/mc²), for K ≫ mc²

Quoted, not derived. The relativistic electron turns the linear growth into a logarithm and makes δm proportional to m itself.

Quoted: Weisskopf (1939), hole theory, confirmed by one-loop QED; 0.18 even with K at the Planck energy, 1.22 × 10¹⁹ GeV
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

0.00
−1 to 3; steps of 0.05.

Try this. Start at K = mc², where δm/m is 0.310 per cent, then drag right. The solid line crosses δm/m = 1 near 323 mc², where the bare mass passes through zero, while the dashed relativistic curve has only reached 2 per cent. Below mc² the dashed marker vanishes: the logarithm is quoted only for K ≫ mc².

Live results

DIPOLE MODEL δm/m
0.310 %
δm c², DIPOLE MODEL
1.58 keV
BARE MASS m₀c²
509.4 keV
WEISSKOPF δm/m, K ≫ mc²
0.000 %

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

Interactive Advanced Quantum Mechanics modelLog-log plot of the electromagnetic mass δm/m against the photon cutoff K. Solid: the dipole model derived here, 4αK/(3πmc²), linear in K. Dashed: Weisskopf's quoted relativistic (3α/2π) ln(K/mc²), drawn only above mc². At K = 1.00 mc² the dipole model gives δm/m = 0.310 per cent and a bare mass m₀c² = 509.4 keV.K = 1.00 mc²δm/m = 1δm/m = 0.01dipole, ∝ KWeisskopf, ∝ ln K0.1110100K/mc²
Log-log plot of the electromagnetic mass δm/m against the photon cutoff K. Solid: the dipole model derived here, 4αK/(3πmc²), linear in K. Dashed: Weisskopf's quoted relativistic (3α/2π) ln(K/mc²), drawn only above mc². At K = 1.00 mc² the dipole model gives δm/m = 0.310 per cent and a bare mass m₀c² = 509.4 keV.

Predict → change one variable → check

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

Saved baseline. Cutoff, log₁₀(K/mc²): 0.00

Baseline 0.00; choose −1 to 3, in steps of 0.05.
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

The tempting readings are that a negative self-energy makes the electron lighter, by E = mc², or that it is a constant to be dropped like a zero-point energy. Both read an energy where the physics is in a slope. The transverse coupling lowers the energy by (δm/m)p²/2m, which is zero at rest and grows with p, so it changes ∂E/∂p and ∂²E/∂p²: at a given momentum the dressed electron moves more slowly, which is what a larger mass means. A non-relativistic Hamiltonian has no rest-energy term for the shift to correct. The pieces that really are constants, the A² term's αK²/(2πmc²) and the Coulomb self-energy, move every momentum state alike and leave the mass untouched. Nor can the p² term be dropped as unobservable: it is hidden only because the measured 0.511 MeV/c² already contains it, so a calculation that keeps it must start from the bare m₀.

5

Practice & worked examples

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

EasyExample 1

An electron crosses a field-free region at v = 3.00 × 10⁶ m s⁻¹. With the photon cutoff at K = mc², find δm/m, δm in kilograms, and the second-order shift of this electron's energy, then check the shift's dimensions and its limits.

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.

MediumExample 2

Model the electron classically as a thin spherical shell of charge −e and radius a = 10.0 fm. Find its Coulomb self-energy and its field-momentum mass, the photon cutoff at which the quantum dipole model gives the same δm, and the shell radius at which the whole measured mass would be electromagnetic.

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

Restore the recoil that the dipole approximation drops. For a slow electron, show that δm/m = (8α/(3π)) ln(1 + K/(2mc²)), evaluate it at K = mc² and at the cutoff where the dipole model made the whole mass electromagnetic, and compare with Weisskopf's quoted relativistic (3α/(2π)) ln(K/mc²).

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.

6

Quick check

Select an answer and explain it before checking.

Give a classical charge −e a field with no Fourier components above kₘₐₓ = mc/ħ, so that its Coulomb self-energy is UC = αmc²/π = 1.19 keV, and set it beside the quantum dipole model with K = ħckₘₐₓ = mc². Which statement about the electromagnetic mass is right?
7

Review & continue

The electromagnetic mass δm is the part of an electron's inertia supplied by its own field: coupling to transverse photons lowers the energy in proportion to p², so it changes the kinetic energy exactly as a larger mass would.

  • 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.