University Quantum Field Theory I · Renormalising QED: the Electron Self-Energy and the Ward-Takahashi Identity · 12.4
Mass renormalisation: δm ∝ m and only logarithmic
Topic 12.03 handed you Σ₂(p). Here you read it at the pole, find how much heavier its own field makes the electron, and then ask the better question: why so little? The answer is a symmetry that only the mass term breaks, and it is why a 0.511 MeV electron needs no fine-tuning even with a cutoff at the Planck scale.
Build the model
Connect the measurement to the mechanism.
The electron's pole sits where p̸ − m₀ − Σ(p̸) vanishes, so to order α the self-mass is δm = m − m₀ = Σ₂(p̸ = m). Read there, topic 12.03's Σ₂ gives (3α/4π) m [2/ε − γE + ln(4πμ²/m²) + 4/3] in dimensional regularisation and (3α/4π) m [ln(Λ²/m²) + ½] with a Pauli-Villars photon of mass Λ, both derived here and both finite as the photon mass goes to zero. Power counting had allowed much worse.
The self-energy has superficial degree 1, and a shift of order αΛ that ignores m is exactly what Bethe's non-relativistic electron carries (UAQM-13.03). Two facts remove that power. At m₀ = 0 the Lagrangian is invariant under ψ → e(iθγ⁵)ψ and ψ̄ψ is not, so every term of Σ that can move the pole carries a factor of m₀, at every order: a massless electron stays massless, and mass renormalisation is multiplicative.
With one power of mass spent on that factor, what is left is dimensionless and can grow only as ln Λ, Weisskopf's 1939 result, now derived; that the pair states inside SF cancel the linear growth is stated, not shown. Even a Planck-scale cutoff makes δm only 18 per cent of m. The argument has limits: it is first order in α, it cannot say why the measured m is 0.511 MeV, and above the electroweak scale the mass comes from a coupling to the Higgs field, named only, whose corrections the same chiral symmetry keeps proportional to m.
- Simple definition
- The self-mass δm = m − m₀ is how far the electron's interaction with its own photon field moves the pole of its propagator; in QED it is proportional to the mass itself and grows only as the logarithm of the cutoff.
- Example
- With a Pauli-Villars photon at Λ = 1.00 TeV, at leading logarithm, ln(Λ²/m²) = 28.97 and δm/m = 1.742 × 10⁻³ × 28.97 = 0.0505, so 25.8 keV of the electron's measured 511.0 keV is self-mass and m₀ = 485.2 keV.
Derived: on shell the x-integrals give 3, 5 and −1, the −1 being the O(ε) Dirac algebra times 2/ε; without it 4/3 would be 5/3. Finite as the photon mass μγ → 0.
m, m₀, δm and the scale μ in MeV (ħ = c = 1); α ≈ 1/137.036, ε = 4 − d and γE = 0.5772 are pure numbers
Derived from (α/2π) m ∫₀¹dx (2 − x) ln[xΛ²/((1 − x)²m²)]. The logarithm's coefficient matches the 2/ε pole's; the constants ½ and 4/3 belong to the regulators.
Λ the regulator photon's mass in MeV; Dirac algebra in four dimensions; leading logarithm (3α/2π) m ln(Λ/m)
Derived: the two phases meet across γμ and cancel, but meet directly in ψ̄ψ and add. Only the mass breaks the symmetry, so only m₀ can feed the mass shift.
θ a real angle; γ⁵ = iγ⁰γ¹γ²γ³ anticommutes with every γμ, so ψ̄γμψ and ψ̄iD̸ψ are unchanged
Derived by counting γ matrices: each chirality flip costs one m₀. So m₀ = 0 gives m = 0 at every order, mass renormalisation is multiplicative, and B grows only logarithmically, as ln Λ at one loop.
A and B dimensionless functions of p², m₀², α and the regulator, each O(α) at one loop; m and m₀ in MeV
Linear in Λ and blind to m: no ψ̄ψ to protect and no pair states. It would make the whole electron mass electromagnetic by Λ = 3πm/(4α) = 165 MeV.
Quoted from UAQM-13.03, dipole approximation; Λ the photon-energy cutoff in MeV; 1.58 keV at Λ = m
Fixed by requiring the renormalised self-energy to vanish at p̸ = m, so m stays the pole. δ₂, the slope of Σ₂ there, is topic 12.05's.
δₘ and δ₂ = Z₂ − 1 from topic 12.02's counterterm i(p̸δ₂ − δₘ); δm = Σ₂(p̸ = m) from this topic; δₘ, δm and m in MeV, δ₂ a pure number
Read Σ₂ at the pole
Natural units, ħ = c = 1, with the metric (+, −, −, −) throughout. The resummed electron propagator i/(p̸ − m₀ − Σ(p̸)) has its pole where p̸ − m₀ − Σ(p̸) = 0, so to order α the measured mass is m = m₀ + Σ₂(p̸ = m): the self-mass is the self-energy read at the pole and nowhere else. Topic 12.03 supplies, in Feynman gauge with x the Feynman parameter of the photon line, Σ₂(p) = (α/4π)∫₀¹dx [(4 − ε)m − (2 − ε)x p̸][2/ε − γE + ln(4πμ²/Δ)], with Δ = (1 − x)m² + xμγ² − x(1 − x)p². Set p̸ = m, so p² = m² and Δ = (1 − x)²m² + xμγ². Three integrals finish the job. ∫(4 − 2x)dx = 3 multiplies the pole and ln(4πμ²/m²); −∫(4 − 2x) ln(1 − x)² dx = 5 is the finite part of the logarithm; and the O(ε) pieces of the Dirac algebra, (−ε + εx)m, meet the 2/ε and leave ∫(−2 + 2x)dx = −1. Hence δm = m − m₀ = (3α/4π) m [2/ε − γE + ln(4πμ²/m²) + 4/3], derived. Four-dimensional Dirac algebra would have left 5/3: that is the scheme called dimensional reduction, a different convention rather than different physics, though the Medium example shows it moves the minimally subtracted mass by about 0.3 keV.
Infrared-safe, and the same logarithm from a cutoff
The photon mass enters only through xμγ² in Δ, and it competes with (1 − x)²m² only in a sliver 1 − x ≲ μγ/m. There ln(1 − x)² is singular but integrable, so μγ → 0 is a smooth limit: a small photon mass changes δm by −αμγ/2, derived in the Hard example, which vanishes with it. The value of Σ₂ at the pole is infrared-safe; its slope there, which fixes Z₂, is not (topic 12.05). The ultraviolet logarithm does not depend on the regulator either. Subtract instead the same graph with photon mass Λ, a Pauli-Villars regulator, in four dimensions: for Λ ≫ m, Σ₂(p̸ = m) = (α/2π) m ∫₀¹dx (2 − x) ln[xΛ²/((1 − x)²m²)]. Since ∫(2 − x)dx = 3/2 and the leftover logarithms integrate to 3/4, δm = (3α/4π) m [ln(Λ²/m²) + ½], derived: Weisskopf's (3α/2π) m ln(Λ/m) at leading logarithm. Both regulators put 3α/(4π) in front of the logarithm of a scale squared; the constants 4/3 and ½ are the regulators' own and go into m₀. Checks: δm is m times pure numbers, it vanishes as α → 0, and at Λ/m = 10 the form with ½ is already within 1.1 per cent of the exact integral.
A massless electron stays massless
Set m₀ = 0. Since γ⁵ anticommutes with every γμ, ψ → e(iθγ⁵)ψ sends ψ̄ = ψ†γ⁰ to ψ̄e(iθγ⁵), and across γμ the two phases cancel: ψ̄iD̸ψ, kinetic term and photon coupling alike, is invariant. The mass term is not, since ψ̄ψ → ψ̄e(2iθγ⁵)ψ = cos 2θ ψ̄ψ + i sin 2θ ψ̄γ⁵ψ. In chiral components ψ̄ψ = ψ̄LψR + ψ̄RψL joins left to right, which the vertex −ieγμ never does. Count γ matrices along the open electron line of any self-energy graph: n vertices and n − 1 massless propagators k̸ give 2n − 1 of them, an odd number, and with a single external momentum the only odd structure left is p̸. Each m₀ taken from a propagator removes one γ matrix. So at every order Σ(p) = p̸A(p²) + m₀B(p²), and the pole condition gives m = m₀(1 + B)/(1 − A): zero when m₀ is zero, proportional to it otherwise. At one loop, γμk̸γμ integrates to a multiple of p̸, and only m₀γμγμ = d m₀ reaches the unit matrix. The counting never uses γ⁵ itself, so dimensional regularisation keeps it, as does a Pauli-Villars photon.
One power of Λ is spent on m
Topic 12.01's power counting gives the electron self-energy D = 4 − B − (3/2)F = 1, with B = 0 external photons and F = 2 external electron lines, so a linear divergence is allowed. Σ has mass dimension 1 and p̸ already supplies one power of momentum, so A is dimensionless and at most logarithmic; only the unit-matrix part could grow as αΛ. Chiral symmetry makes that part m₀B(p²) with B(p²) dimensionless, so it too grows only logarithmically: the symmetry takes exactly one power off the superficial degree. Bethe's electron had no such protection. Its electromagnetic mass, quoted from UAQM-13.03, is δmNR = (4α/3π)Λ, linear in the cutoff and blind to m, reaching δm = m by Λ = 3πm/(4α) = 165 MeV. The relativistic electron differs twice over. Its Lagrangian has a ψ̄ψ that a symmetry forbids at m₀ = 0, and its propagator SF carries negative-energy, pair, states as well as electrons. Split by time ordering, the orderings in which a virtual pair appears cancel the linear growth of the single ordering Bethe kept, which is how Weisskopf found the logarithm in hole theory in 1939; that cancellation is stated here, not derived.
Eighteen per cent at the Planck scale
Inputs: mₑ = 0.51099895 MeV, measured, and 3α/(4π) = 1.742 × 10⁻³. Put the cutoff at the Planck energy, √(ħc⁵/G) = 1.22 × 10¹⁹ GeV, where gravity can no longer be left out. Then Λ/m = 2.39 × 10²², ln(Λ²/m²) = 103.1, and δm/m = 0.180: 91.7 keV of self-mass inside the measured 511.0 keV, and m₀ about 82 per cent of m. Bethe's linear law at the same cutoff would give δm/m = 7.4 × 10¹⁹. For the logarithm to account for the whole mass, ln(Λ²/m²) would have to reach 1/(1.742 × 10⁻³) = 574, a cutoff with Λ/m near 10¹²⁵, so the one-loop shift stays a modest correction at every cutoff anyone has proposed, and higher orders, each also proportional to m, do not change that. Set beside it one line from topic 6.05: a scalar's tadpole moves m² by λΛ²/(32π²), 4.7 × 10³⁵ GeV² for λ = 1 at the same cutoff against (125 GeV)², so that bare mass must cancel to one part in 3 × 10³¹. That is why the measured mₑ/MPl = 4.2 × 10⁻²³ is no puzzle in QED: a fermion mass that chiral symmetry protects stays small under every correction, 't Hooft's technical naturalness, quoted.
What the counterterm absorbs, and what is measured
In topic 12.02's renormalised Lagrangian the electron's counterterm vertex is i(p̸δ₂ − δₘ). Requiring the renormalised self-energy to vanish at p̸ = m fixes δₘ = mδ₂ − δm, with δm from this topic and δ₂, the slope of Σ₂ at the pole, from topic 12.05. The pole in ε, the regulator's constant and ln μ all go into it, and none of them survives in an amplitude written in terms of m. Off shell Σ₂ depends on the gauge parameter ξ, but its value at the pole does not, stated: the pole mass is gauge-independent, which is why it is the mass to renormalise to. Neither δm nor m₀ is measured. A Penning trap weighs m = 0.51099895 MeV, while m₀ changes with the regulator, a pole in one scheme and a logarithm of Λ in another. What experiment tests is the subtraction itself: written in the measured m, every amplitude is finite and predictive, and Unit 15 assembles the Lamb shift at about 1052 MHz against the measured 1057.8 MHz, quoted. Bethe cut off his photon integral near mc² in 1947, expecting relativity to tame what lies above; for the free electron's mass, this lesson shows how.
Change one variable at a time
Make the relationship visible.
Drop the mass to 0.01 mₑ: the solid curve and the dashed δm = m line sink about two decades together while Bethe's line stays put, which is chiral symmetry at work. Then, back at m = mₑ, sweep Λ from 1 GeV to the Planck line: δm/m only climbs from 0.027 to 0.18.
δm/m, COVARIANT0.180
SELF-MASS δm91.83 keV
BETHE log₁₀(δm/m)19.8
BARE m₀/m0.820
Live interpretationδm/m, COVARIANT: 0.180. SELF-MASS δm: 91.83 keV. BETHE log₁₀(δm/m): 19.8. BARE m₀/m: 0.820
Catch the common trap
Explain before calculating.
A student power-counts the electron self-energy, finds D = 1, and concludes that with a Pauli-Villars photon of mass Λ = 1.22 × 10¹⁹ GeV the bare mass m₀ must cancel a shift of order αΛ. What does the one-loop calculation actually give, and why?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyWith a Pauli-Villars photon of mass Λ = 100 GeV, find at leading logarithm the self-mass of the electron (0.5110 MeV) and of the muon (105.66 MeV). Check the dimensions and the m → 0 limit, and test whether δm is proportional to m.
- Natural units: δm = (3α/4π) m ln(Λ²/m²) with 3α/(4π) = 3/(4π × 137.036) = 1.742 × 10⁻³. α and the logarithm are pure numbers, so δm is a mass, as it must be.
- Electron: Λ/m = 1.000 × 10⁵ MeV ÷ 0.5110 MeV = 1.957 × 10⁵, so ln(Λ²/m²) = 2 × 12.18 = 24.37 and δm/m = 1.742 × 10⁻³ × 24.37 = 0.0425, giving δm = 0.0425 × 511.0 keV = 21.7 keV.
- Muon: Λ/m = 1.000 × 10⁵ MeV ÷ 105.66 MeV = 946.4, so ln(Λ²/m²) = 13.71 and δm/m = 0.02388, giving δm = 0.02388 × 105.66 MeV = 2.52 MeV.
- Proportionality: the muon's δm over the electron's is 2.52 MeV ÷ 21.7 keV = 116, the mass ratio 206.8 times the logarithm ratio 13.705/24.369 = 0.562. Proportional to m up to a slowly varying logarithm, which is what chiral symmetry allows.
- Limit: at fixed Λ, δm = m ln(Λ²/m²) × 1.742 × 10⁻³ goes to zero as m → 0, because m ln m → 0. A massless fermion acquires no mass.
AnswerElectron: δm/m = 4.25 × 10⁻², δm = 21.7 keV. Muon: δm/m = 2.39 × 10⁻², δm = 2.52 MeV. Both vanish as m → 0 at fixed Λ.
MediumStarting from topic 12.03's Σ₂(p), carry out the x-integrals at p̸ = m to obtain the constant 4/3, say what dropping the O(ε) Dirac terms would give, and find how far the MS-bar mass m̄(μ) at μ = m lies below the electron's pole mass.
- On shell Δ = (1 − x)²m² + xμγ², which tends to (1 − x)²m² as μγ → 0 with an integrable logarithm at x = 1, and the numerator is m[(4 − ε) − (2 − ε)x] = m[(4 − 2x) − ε(1 − x)].
- Multiply out: (α/4π) m ∫₀¹dx ((4 − 2x)[2/ε − γE + ln(4πμ²/m²) − ln(1 − x)²] − 2(1 − x)), the last term being −ε(1 − x) times 2/ε.
- Integrals: ∫(4 − 2x)dx = 3. With y = 1 − x, ∫(4 − 2x) ln(1 − x)² dx = 2∫(2 + 2y) ln y dy = 2(−2 − ½) = −5, so the logarithm contributes +5. And −2∫(1 − x)dx = −1.
- So δm = (α/4π) m [3(2/ε − γE + ln(4πμ²/m²)) + 5 − 1] = (3α/4π) m [2/ε − γE + ln(4πμ²/m²) + 4/3]. Without the O(ε) terms the −1 is lost and 4/3 becomes 5/3.
- MS-bar puts 2/ε − γE + ln 4π into the counterterm, so m − m̄(μ) = (3α/4π) m [ln(μ²/m²) + 4/3]. At μ = m that is αm/π = 2.323 × 10⁻³ × 511.0 keV = 1.19 keV; the 5/3 would have given 1.48 keV.
- Check: QCD's quoted relation, pole mass = m̄(m̄)[1 + 4αₛ/(3π)], is this result with α multiplied by the colour factor 4/3. Dimensions: a mass times pure numbers.
Answerδm = (3α/4π) m [2/ε − γE + ln(4πμ²/m²) + 4/3]; m̄(m) lies αm/π = 1.19 keV below the 511.0 keV pole mass, and 1.48 keV if the O(ε) terms were dropped.
HardKeep a photon mass μγ ≪ m in topic 12.03's Σ₂ at p̸ = m. Show that it changes δm by −αμγ/2, recognise that number as a classical self-energy, and evaluate it for μγ = 1.00 keV.
- Only −ln Δ feels μγ, so δm(μγ) − δm(0) = −(α/4π) m ∫₀¹(4 − 2x) ln[1 + xμγ²/((1 − x)²m²)] dx.
- The bracket differs from 1 only where 1 − x ≲ μγ/m. There set x = 1 in the smooth factors, so 4 − 2x → 2 and xμγ² → μγ², and with y = 1 − x and a = μγ/m extend the integral to infinity: 2∫₀^∞ ln(1 + a²/y²) dy.
- By parts, ∫₀^∞ ln(1 + a²/y²) dy = [y ln(1 + a²/y²)]₀^∞ + ∫₀^∞ 2a² dy/(y² + a²) = 0 + πa. So the change is −(α/4π) m × 2πμγ/m = −αμγ/2. Direct quadrature of the full integral gives 0.999 of this at μγ/m = 2 × 10⁻³.
- Classical reading: at rest the electron couples only to A⁰, and a photon mass turns its Coulomb potential e/(4πr) into the Yukawa potential e e(−μγ r)/(4πr). The self-energy ½eφ(0) changes by (e²/8π) lim (e(−μγ r) − 1)/r = −e²μγ/(8π) = −αμγ/2 as r → 0: soft photons see a static point charge.
- Numbers: μγ = 1.00 keV gives −(7.297 × 10⁻³/2) × 1.00 keV = −3.65 eV, a fraction 7.1 × 10⁻⁶ of m, linear in μγ and gone as μγ → 0. δ₂ instead carries ln μγ² (topic 12.05) and has no such limit.
Answerδm(μγ) = δm(0) − αμγ/2 for μγ ≪ m: −3.65 eV for μγ = 1.00 keV, 7.1 × 10⁻⁶ of the electron mass, so δm is finite as μγ → 0.