University Quantum Field Theory I · Vacuum Polarisation, Charge Renormalisation and the Running of alpha · 13.1

Why a cutoff fails: a photon mass from the regulator

Before computing vacuum polarisation you must decide how to cut it off, and the obvious choice fails in an instructive way. A sharp cutoff on the loop momentum hands the photon a mass that gauge invariance forbids. This topic finds that mass, traces it to the regulator, and earns the switch to dimensional regularisation.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The electron loop in the photon propagator is where QED stops letting you choose a regulator by taste. Power counting (topic 12.01) allows it a quadratic divergence; the Ward identity, through topic 12.09's Π^μν = (q²g^μν − q^μq^ν)Π(q²), forbids one, because two powers of momentum already stand outside the integral. Cut the Euclidean loop momentum off at Λ and compute anyway: beside the transverse logarithm appears −(α/2π)g^μν ∫₀¹dx Λ⁴/(Λ² + Δ), from an integrand that is positive at every Euclidean point, derived.

At q = 0 nothing else survives, and resumming it moves the photon pole to m_γ² ≈ −(α/2π)Λ², a magnitude of 34 MeV for Λ = 1 GeV against a measured bound below 10⁻¹⁸ eV. The term is not physics. Its integrand is a total derivative, so it is the flux of l^μ/(l² + Δ) through the cutoff sphere, and the same flux is what a cutoff adds when topic 12.06's proof shifts the loop momentum: q_μΠ^μν = −(α/2π)Λ²q^ν, shown.

QED conserves its current; the regulator broke the identity. A regulator that keeps shifts gives exactly zero: in dimensional regularisation the flux is a pure power Λ^{(d − 2)}, zero by continuation from d below 2, and heavy Pauli–Villars loops cancel it when Σcᵢ = 0 and ΣcᵢMᵢ² = 0. Where the cutoff stops working is the lesson: its logarithms survive a change of regulator, its powers of Λ do not.

Simple definition: A sharp momentum cutoff regulates the photon self-energy by discarding loop momenta with |l_E| above Λ; because that ball is not invariant under shifts of l, it adds a non-transverse g^μν Λ² term that acts as a photon mass.
Example: For the electron loop with Λ = 1 GeV, Π^μν(0) = −(α/2π)g^μν Λ⁴/(Λ² + m²) = −1.16 × 10⁻³ GeV² g^μν: a photon mass of magnitude 34 MeV that would spoil Coulomb's law beyond 5.8 fm, where measurement bounds it below 10⁻¹⁸ eV.

The cutoff loop: a transverse part and a leftoverΠ^μν_Λ(q) = (q²g^μν − q^μq^ν)Π_Λ(q²) − (α/2π)g^μν ∫₀¹dx Λ⁴/(Λ² + Δ)

Derived with |l_E| ≤ Λ. The first term is what topic 12.09 allows; the second has no q^μq^ν partner and survives at q = 0, so it belongs to the regulator, not to QED.

Δ = m² − x(1 − x)q²; Λ, m and √|q²| in GeV (1 GeV = 1.602 × 10⁻¹⁰ J); Π^μν in GeV²; α and Π_Λ dimensionless

The transverse logarithm with a cutoffΠ_Λ(q²) = −(2α/π) ∫₀¹dx x(1 − x)[ln(Λ²/Δ) − 1]

Derived. Its ln Λ² carries the coefficient that dimensional regularisation attaches to 2/ε in topic 13.02: a logarithm survives a change of regulator, a power does not.

Leading terms for Λ² ≫ |Δ|, dimensionless; at q = 0 it is −(α/3π)[ln(Λ²/m²) − 1], with α/3π = 7.74 × 10⁻⁴

The photon mass the leftover impliesm_γ² = −(α/2π)Λ⁴/(Λ² + m²) ≈ −(α/2π)Λ²

Derived. Λ = 1 GeV gives |m_γ| = 34 MeV against a measured bound below 10⁻¹⁸ eV, quoted. The minus sign puts the pole at spacelike q², a tachyon.

From −ig_μν/(q² − m_γ²) after resumming insertions; |m_γ| ≈ 0.0341Λ in GeV for Λ in GeV

The leftover integrand is a total derivative[(d − 2)l_E² + dΔ]/(l_E² + Δ)² = ∂/∂l_E^μ [l_E^μ/(l_E² + Δ)]

Derived by differentiating. Gauss's theorem turns the leftover into a flux through the cutoff sphere, π²Λ⁴/[2(Λ² + Δ)] at d = 4: a boundary term, nothing more.

Euclidean momentum l_E in GeV, d dimensions; at d = 4 the left side is 4 times the leftover integrand (l_E²/2 + Δ)/(l_E² + Δ)²

The Ward identity broken by the shiftq_μΠ^μν_Λ(q) = −(α/2π)Λ²q^ν + terms finite as Λ → ∞

Shown from the shell between the ball and its shift by q, ∮(q⋅n̂)f dS ∝ Λ²q^ν. It is q_μ times the leftover's leading term: the whole leftover is the failed shift.

Cutoff |k_E| ≤ Λ on the loop momentum k; q in GeV; the right side in GeV³, as q_μΠ^μν must be

The leftover kept in d dimensions∫₀^(Λ²) L^{(d/2 − 1)}[(1 − 2/d)L + Δ]/(L + Δ)² dL = (2/d)Λ^d/(Λ² + Δ)

Derived from the flux. A pure power Λ^{(d − 2)}: it vanishes as Λ → ∞ for d below 2, where the integral converges to zero, and continuation carries that zero to d = 4.

L = l_E² in GeV²; the angular factor Ω_d/[2(2π)^d] is set aside; d = 4 gives Λ⁴/[2(Λ² + Δ)]

Power counting allows Λ², the Ward identity forbids it

Natural units ħ = c = 1 and the metric (+, −, −, −) are used throughout. Topic 12.01 gave the photon self-energy a superficial degree D = 2, so the electron loop may diverge as Λ². Topic 12.09 said more: q_μΠ^μν = 0 forces Π^μν(q) = (q²g^μν − q^μq^ν)Π(q²), two powers of q stand outside the integral, and Π(q²) can diverge only logarithmically. Both statements concern one integral, so a regulator either respects the second or leaves a g^μν term with no −q^μq^ν/q² partner. To see which, take the loop with topic 7.04's −1 and Advanced QM's trace identities, join the propagators with one Feynman parameter and shift to l = k + xq: iΠ^μν = −4e² ∫₀¹dx ∫d⁴l/(2π)⁴ N^μν/(l² − Δ)², with Δ = m² − x(1 − x)q² and N^μν = 2l^μl^ν − g^μν l² − 2x(1 − x)q^μq^ν + g^μν[m² + x(1 − x)q²], odd terms dropped. Everything now hangs on how the l integral is cut off.

Cutting off l_E leaves a positive remainder

Cut the Euclidean momentum off at |l_E| = Λ. The region is a ball, so l^μl^ν → g^μν l²/4 and 2l^μl^ν − g^μν l² → −g^μν l²/2. Split m² + x(1 − x)q² = Δ + 2x(1 − x)q². The pieces 2x(1 − x)(q²g^μν − q^μq^ν) multiply ∫d⁴l/(2π)⁴ (l² − Δ)⁻² = (i/(16π²))[ln(Λ²/Δ) − 1], up to terms falling as Δ/Λ², and are transverse. What is left is g^μν ∫(Δ − l²/2)/(l² − Δ)², which the Wick rotation l⁰ = il_E⁰ turns into i∫d⁴l_E/(2π)⁴ (l_E²/2 + Δ)/(l_E² + Δ)². For Δ above zero, true at q = 0 and at every spacelike q, that integrand is positive everywhere, so no sharp cutoff can make it vanish. With L = l_E² and d⁴l_E = π²L dL it is exact: ∫₀^(Λ²) L(L/2 + Δ)/(L + Δ)² dL = Λ⁴/[2(Λ² + Δ)]. Collecting, Π^μν_Λ = (q²g^μν − q^μq^ν)Π_Λ(q²) − (α/2π)g^μν ∫₀¹dx Λ⁴/(Λ² + Δ), with Π_Λ = −(2α/π)∫₀¹dx x(1 − x)[ln(Λ²/Δ) − 1], derived. Dimensions check: Π^μν adds to q²g^μν in the inverse propagator, so it carries GeV², as Λ² does.

The remainder is a photon mass, and a large one

At q = 0 the transverse part vanishes and only the leftover remains: Π^μν(0) = −(α/2π)g^μν Λ⁴/(Λ² + m²). Resum the insertions as in topic 12.09: a constant A g^μν in Π^μν turns −ig_μν/q² into −ig_μν/(q² − A), so the pole moves from q² = 0 to m_γ² = A ≈ −(α/2π)Λ². The positive integrand fixes the sign, which even puts the pole at spacelike q², but the size is the real failure: |m_γ| = 0.0341Λ. A cutoff of 1 GeV gives 34 MeV, and the static potential would change character beyond ħc/|m_γ| = 5.8 fm, inside a nucleus. The measured bound, quoted, is below 10⁻¹⁸ eV, a Compton length of 2 × 10¹¹ m, beyond the Earth–Sun distance. Every charged fermion loop adds a term of the same sign, so no cancellation among species helps, and renormalised QED cannot absorb it: topic 12.02's δ₃ multiplies q²g^μν − q^μq^ν and vanishes at q = 0. Limits: the term vanishes as e → 0, keeps its leading Λ² as m → 0, and grows without bound as Λ → ∞.

It is a surface term, and the shift is where it enters

In d Euclidean dimensions ∂/∂l_E^μ [l_E^μ/(l_E² + Δ)] = [(d − 2)l_E² + dΔ]/(l_E² + Δ)², which at d = 4 is 4 times the leftover integrand. By Gauss's theorem the leftover is the flux through the cutoff sphere, ¼ × 2π²Λ³ × Λ/(Λ² + Δ) = π²Λ⁴/[2(Λ² + Δ)]: the previous integral in one line. That flux is what topic 12.06's proof discards. Contract with q_μ and write q̸ = (k̸ + q̸ − m) − (k̸ − m): the loop becomes q_μ iΠ^μν = −4e² ∫d⁴k/(2π)⁴ [f^ν(k) − f^ν(k + q)] with f^ν = k^ν/(k² − m²), zero once k → k − q in the second term. With |k_E| ≤ Λ that shift slides the ball by q, and the two balls differ by a shell of thickness q⋅n̂ on which f ≈ 1/Λ, so the mismatch ∮(q⋅n̂)f dS grows as Λ³ × q × Λ⁻¹ = Λ²q. Evaluated, q_μΠ^μν_Λ = −(α/2π)Λ²q^ν, shown: q_μ times the leftover. A shift is safe only if the integrand falls faster than 1/k³, and f falls as 1/k.

Regulators that keep the shift

Dimensional regularisation keeps shifts by construction: topic 6.03 defines ∫d^dl by continuing Gaussian integrals, which are translation invariant, so a total derivative integrates to zero at every d. Explicitly, l^μl^ν → g^μν l²/d makes the leftover ∫[(2/d − 1)l² + Δ]/(l² − Δ)², and topic 6.03's master integrals give (i/(4π)^(d/2))Δ^{(d/2 − 1)}[Γ(2 − d/2) − (1 − d/2)Γ(1 − d/2)], zero since Γ(2 − d/2) = (1 − d/2)Γ(1 − d/2), derived. With the cutoff kept in d dimensions the flux is (2/d)Λ^d/(Λ² + Δ), a pure power Λ^{(d − 2)} that dies as Λ → ∞ for d below 2, where the integral converges to zero; continuation carries that zero to d = 4. So Λ² has no counterpart, while ln Λ² becomes 2/ε. Pauli–Villars, named, adds heavy fermion loops of mass Mᵢ with weights cᵢ; the leftover becomes −(α/2π)g^μν ΣcᵢΛ⁴/(Λ² + Mᵢ²) → −(α/2π)g^μν(Λ²Σcᵢ − ΣcᵢMᵢ²), so it needs Σcᵢ = 0 and ΣcᵢMᵢ² = 0 with c₀ = 1, M₀ = m: two regulators at least, one with the electron's own sign.

What the cutoff still gets right

Not everything the cutoff produced is wrong. Π_Λ(0) = −(α/3π)[ln(Λ²/m²) − 1] carries the coefficient of ln Λ² that dimensional regularisation attaches to 2/ε, because the logarithmic integral is safe to shift: its surface term falls as 1/Λ. Logarithms are what renormalisation turns into measurable running, in topics 13.05 and 13.06, so a cutoff can be trusted for them and never for a power of Λ. Compare topic 6.05's scalar, whose mass also met Λ². There no symmetry forbids a mass, and a heavy particle of mass M coupled to the scalar shifts it by order M² in every scheme, quoted; here gauge invariance forbids the photon mass in every scheme that respects it, so the Λ² has nothing physical behind it. The test is mechanical and worth making a habit: compute q_μΠ^μν with your regulator before trusting anything else it gives. If it is not zero, change the regulator, not the theory.

Change one variable at a time

Make the relationship visible.

Interactive model

Dimension d4.0

Cutoff log₁₀(Λ/mₑ)2.0

At d = 4 each doubling of Λ quadruples the leftover, so the photon mass grows with Λ. Lower d: at d = 2 the curve flattens onto 1, and below 2 it turns down and dies away, so the convergent integral there is exactly zero, the value continuation carries back to d = 4.

LEFTOVER AT Λ4999.500 mₑ^(d−2)

FACTOR IF Λ DOUBLES4.00

LARGE-Λ POWER d − 22.0

d = 4 PHOTON MASS |m_γ|1741 keV

Live interpretationLEFTOVER AT Λ: 4999.500 mₑ^(d−2). FACTOR IF Λ DOUBLES: 4.00. LARGE-Λ POWER d − 2: 2.0. d = 4 PHOTON MASS |m_γ|: 1741 keV

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that the photon self-energy really is quadratically divergent, as topic 6.05's scalar mass was, and that a photon-mass counterterm tuned to keep the pole at q² = 0 takes care of it. That is wrong twice. The term is not in QED: it is the flux through the cutoff sphere, present only because the cutoff spoils the shift the Ward identity needs, and every shift-invariant regulator returns exactly zero for it, at every order. And the cure would cost the theory: renormalised QED has no gauge-invariant counterterm that survives at q = 0, since δ₃ multiplies q²g^μν − q^μq^ν, while a mass term ½δ_γA_μA^μ breaks the gauge symmetry on which Z₁ = Z₂, charge universality and the decoupling of unphysical photons rest. A Λ² in a gauge-invariant quantity is evidence against the regulator.

A student cuts off the Euclidean loop momentum at Λ and finds that the one-loop electron loop gives Π^μν(0) = −(α/2π)Λ²g^μν at leading order. Which diagnosis is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe electron loop is regulated by a sharp cutoff Λ = 10 GeV on the Euclidean loop momentum. Find the photon mass the leftover term implies, the distance beyond which Coulomb's law would fail, and the factor by which |m_γ| exceeds the measured bound of 1 × 10⁻¹⁸ eV.

At q = 0 only the leftover survives, and resumming it gives m_γ² = −(α/2π)Λ⁴/(Λ² + m²). Here m²/Λ² = 2.6 × 10⁻⁹, so m_γ² = −(α/2π)Λ² = −1.1614 × 10⁻³ × 100 GeV² = −0.1161 GeV².
Dimensions: α is dimensionless, so m_γ² carries the GeV² of Λ², as a squared mass must. |m_γ| = √0.1161 GeV = 0.3408 GeV = 341 MeV.
The static potential changes character beyond ħc/|m_γ| = 197.327 MeV fm ÷ 340.8 MeV = 0.579 fm, smaller than a proton.
Against the bound: 3.408 × 10⁸ eV ÷ 1 × 10⁻¹⁸ eV = 3.4 × 10²⁶. The minus sign makes the mass tachyonic, but the size alone rules the regulator out.

Answer|m_γ| ≈ 341 MeV (m_γ² ≈ −0.116 GeV²), so Coulomb's law would fail beyond about 0.58 fm, and |m_γ| is 3.4 × 10²⁶ times the measured bound.

MediumKeep the cutoff Λ = 1 GeV on the electron loop and probe it at the spacelike momentum q² = −(100 MeV)². Compare the leftover coefficient (α/2π)∫₀¹dx Λ⁴/(Λ² + Δ) with the size of the transverse term, |q²Π_Λ(q²)|, using the cutoff's own Π_Λ, and say which dominates.

Here Δ = m² + x(1 − x)Q² with Q² = −q² = 0.01 GeV², and Δ ≪ Λ² everywhere, so Λ⁴/(Λ² + Δ) = Λ² − Δ to good accuracy. With ∫₀¹Δ dx = m² + Q²/6 = 1.667 × 10⁻³ GeV², the leftover is (α/2π) × 0.99833 GeV² = 1.1595 × 10⁻³ GeV².
Since Q ≫ m, set Δ ≈ x(1 − x)Q² inside the logarithm: ∫₀¹x(1 − x)ln(Λ²/Δ)dx = (1/6)ln(Λ²/Q²) − ∫₀¹x(1 − x)ln[x(1 − x)]dx = 0.7675 + 0.2778 = 1.0453.
Subtract ∫₀¹x(1 − x)dx = 1/6 for the −1, giving 0.8786. So Π_Λ(q²) = −(2α/π) × 0.8786 = −4.646 × 10⁻³ × 0.8786 = −4.08 × 10⁻³, dimensionless, and |q²Π_Λ| = 0.01 GeV² × 4.08 × 10⁻³ = 4.08 × 10⁻⁵ GeV².
Ratio: 1.1595 × 10⁻³ ÷ 4.08 × 10⁻⁵ = 28.4. In general it is (3/2)Λ²/(Q²[ln(Λ²/Q²) + 2/3]), which exceeds 1 for every Q between m and Λ: the artefact outweighs the whole transverse loop wherever the cutoff theory claims to apply.

AnswerLeftover 1.16 × 10⁻³ GeV² against |q²Π_Λ| = 4.1 × 10⁻⁵ GeV²: the regulator's term is about 28 times the entire transverse loop.

HardAdd two Pauli–Villars fermions, M₁ = 50 GeV and M₂ = 100 GeV with weights c₁ and c₂, to the electron loop (c₀ = 1, M₀ = m), and keep a cutoff Λ = 10 TeV. Fix c₁ and c₂, evaluate the leftover (α/2π)ΣcᵢΛ⁴/(Λ² + Mᵢ²) at q = 0 and the photon mass it leaves, compare with the electron loop alone and with M₁ alone, and find the effective cutoff in Π(0).

Killing Λ² needs Σcᵢ = 0 and killing the finite remainder needs ΣcᵢMᵢ² = 0: c₁ + c₂ = −1 and 2500c₁ + 10⁴c₂ = −m² ≈ 0 in GeV², so c₂ = 1/3 and c₁ = −4/3. The heavier regulator enters with the electron's own sign.
Since Λ⁴/(Λ² + M²) = Λ² − M² + M⁴/(Λ² + M²), both conditions leave ΣcᵢMᵢ⁴/(Λ² + Mᵢ²) = −0.08333 + 0.33330 = 0.24997 GeV², the electron's m⁴ term being negligible. Times α/2π this is 2.903 × 10⁻⁴ GeV², so |m_γ| = 17.0 MeV.
The electron loop alone gives (α/2π)Λ² = 1.161 × 10⁵ GeV², |m_γ| = 341 GeV, so the regulators cut m_γ² by 4.0 × 10⁸. The remainder is (α/2π)ΣcᵢMᵢ⁴/Λ² with ΣcᵢMᵢ⁴ = 2.5 × 10⁷ GeV⁴: it falls as 1/Λ² and vanishes in the limit.
With M₁ alone (c₁ = −1) only Σcᵢ = 0 holds, and the leftover tends to (α/2π)(M₁² − m²) = 2.90 GeV² for any Λ: |m_γ| = 1.70 GeV, growing as M₁ is sent to infinity. One regulator is not enough.
The transverse part, Π(0) = −(α/3π)Σcᵢ[ln(Λ²/Mᵢ²) − 1], loses Λ and the −1 because Σcᵢ = 0, leaving −(α/3π)ln(Λ_eff²/m²) with Λ_eff = ∛(M₁⁴/M₂) = 39.7 GeV. Then ln(Λ_eff²/m²) = 22.52 and Π(0) = −7.743 × 10⁻⁴ × 22.52 = −0.0174.

Answerc₁ = −4/3, c₂ = 1/3; leftover 2.9 × 10⁻⁴ GeV², |m_γ| ≈ 17 MeV and falling as 1/Λ (341 GeV unregulated, 1.70 GeV with M₁ alone); Λ_eff = 39.7 GeV and Π(0) ≈ −0.0174.