University Quantum Field Theory I · Final Examination

For a free real scalar field of mass m, in natural units ℏ = c = 1 with the metric (+, −, −, −), the Feynman propagator is D_F(x − y) = ∫ d⁴p/(2π)⁴ i e^(−ip⋅(x − y))/(p² − m² + iε), with E_p = √(|p|² + m²) for the three-momentum p and ε → 0⁺. The p⁰ integral is done first, by closing the contour in whichever half of the complex p⁰ plane makes the exponential decay. Which statement is correct?

1. AThe iε places both poles just below the real p⁰ axis, so D_F vanishes whenever x⁰ < y⁰, as the response of a causal system must.
2. BThe poles sit at p⁰ = ±(E_p − iε′); for x⁰ > y⁰ the contour closes in the lower half-plane around p⁰ = +E_p and gives ⟨0|φ(x)φ(y)|0⟩, and for x⁰ < y⁰ it closes in the upper half-plane around p⁰ = −E_p and gives ⟨0|φ(y)φ(x)|0⟩, so the placement of the iε is the time ordering.
3. CBecause it is time-ordered, D_F(x − y) vanishes when x − y is spacelike, and that is how the theory keeps measurements outside each other's light cones from interfering.
4. DFor x⁰ > y⁰ the contour closes in the upper half-plane around p⁰ = −E_p, so for that time order D_F describes an antiparticle carried from y to x.

Two theories are expanded to second order in their coupling, in natural units. In theory (i) a real scalar φ has L_int = −(g/3!)φ³. In theory (ii), the scalar Yukawa model, a complex scalar ψ and a real scalar φ have L_int = −gψ†ψφ. In each, the order-g² correction to the φ two-point function ⟨0|Tφ(x₁)φ(x₂)|0⟩ contains a one-loop diagram in which x₁ joins one vertex, x₂ joins the other, and the two vertices are linked by two internal lines; in (ii) these form a closed ψ loop. If each diagram is written with its vertex factors and propagators and an overall 1/S, what are the symmetry factors S in (i) and (ii)?

1. AS = 2 in both
2. BS = 6 in (i) and S = 1 in (ii)
3. CS = 1 in both
4. DS = 2 in (i) and S = 1 in (ii)

In φ⁴ theory, L = ½∂_μφ∂^μφ − ½m₀²φ² − (λ/4!)φ⁴ with 0 < λ ≪ 1, the single particle has physical mass m. The Källén–Lehmann representation writes the exact two-point function as ⟨Ω|Tφ(x)φ(y)|Ω⟩ = ∫₀^∞ (dM²/2π) ρ(M²) D_F(x − y; M²), with ρ(M²) = 2πZδ(M² − m²) plus a continuum. Where does the continuum begin, and what bounds Z?

1. AAt M² = 4m², the threshold for two particles, with 0 ≤ Z ≤ 1 because ρ is non-negative and the canonical commutator fixes ∫(dM²/2π)ρ(M²) = 1.
2. BAt M² = 4m², with Z ≥ 1, because interactions add multiparticle states to the one-particle state and so increase the weight of the pole.
3. CAt M² = 9m², because the symmetry φ → −φ stops φ from connecting the vacuum to states with an even number of particles; 0 ≤ Z ≤ 1, because ρ is non-negative and the canonical commutator fixes ∫(dM²/2π)ρ(M²) = 1.
4. DImmediately above M² = m², because a particle can always be accompanied by arbitrarily soft quanta, so the pole sits at the start of a branch cut.

In d spacetime dimensions a dimensionless action with a canonical kinetic term gives a scalar field the mass dimension (d − 2)/2. Which interaction has a dimensionless coupling, so that the theory is renormalisable but not super-renormalisable?

1. Agφ³ in d = 6
2. Bgφ³ in d = 4
3. Cλφ⁴ in d = 3
4. Dgφ⁶ in d = 4

A spin-zero boson φ of mass 2.40 GeV decays to a fermion–antifermion pair, the fermion being a Dirac particle of mass 0.90 GeV. Compare a scalar coupling gφψ̄ψ with a pseudoscalar coupling igφψ̄γ⁵ψ of the same real strength g. In both cases Γ = |p| Σ|M|²/(8πm_φ²), with the sum over final spins and |p| the momentum of either fermion in the rest frame of φ. What is Γ_scalar/Γ_pseudoscalar?

1. A0.66
2. B1.00
3. C2.29
4. D0.44

In a general covariant gauge the photon propagator is −i[g_μν − (1 − ξ)k_μk_ν/k²]/(k² + iε). In the tree amplitude for e⁻(p₁)e⁺(p₂) → μ⁻(p₃)μ⁺(p₄), the virtual photon carries k = p₁ + p₂ = p₃ + p₄, with k² = s. Why does the amplitude not depend on ξ?

1. ABecause the photon is massless, k² = 0, so the k_μk_ν term is multiplied by zero.
2. BBecause k_μ contracted into the electron current gives v̄(p₂)(p̸₁ + p̸₂)u(p₁) = (m_e − m_e)v̄(p₂)u(p₁) = 0 by the Dirac equation, and likewise on the muon line, so the k_μk_ν term vanishes for any ξ.
3. CBecause one may choose ξ = 1, and a choice of gauge cannot change a physical amplitude.
4. DBecause the k_μk_ν term is suppressed by m_e²/s, and vanishes once the electron mass is neglected.

At √s ≫ m_μ the reaction e⁺e⁻ → μ⁺μ⁻ by one-photon exchange proceeds through four helicity combinations, because the vector coupling joins a right-handed fermion only to a left-handed antifermion, and vice versa. The e⁻ moves along +z, and θ is the angle between the outgoing μ⁻ and the incoming e⁻. How does the amplitude for e⁻_R e⁺_L → μ⁻_L μ⁺_R depend on θ, and why?

1. AAs (1 − cos θ): the initial pair has angular momentum +1 along the beam and the final pair −1 along the μ⁻ direction, so the amplitude is proportional to the rotation matrix element d¹₁,₋₁(θ), which vanishes in the forward direction.
2. BAs (1 + cos θ): helicity is conserved at each vertex, so the μ⁻ prefers to follow the e⁻.
3. CIt vanishes at every angle: helicity is conserved along a fermion line, so a right-handed electron cannot produce a left-handed muon.
4. DAs sin θ: the virtual photon carries one unit of angular momentum, so every helicity amplitude has the angular dependence of a dipole.

A positronium level with orbital angular momentum L and total spin S has charge-conjugation parity C = (−1)^(L+S), and a state of n photons has C = (−1)^n. Landau and Yang's theorem, quoted, forbids a state of total angular momentum J = 1 from decaying into two photons. Which level can annihilate directly into two photons?

1. A1³S₁
2. B2¹P₁
3. C2³P₁
4. D2³P₂

In renormalised QED with on-shell conditions the physical charge is e = Z₂Z₃^(1/2)e₀/Z₁, where Z₁ renormalises the vertex, Z₂ the charged field and Z₃ the photon field. A proton's coupling to the photon is dressed by strong interactions that the electron's is not, yet the hydrogen atom is neutral to better than one part in 10²⁰. Why are the renormalised charges of the electron and the proton exactly opposite, given opposite bare charges?

1. AThe Ward–Takahashi identity gives Z₁ = Z₂ for every charged field, whatever its other interactions, so e = Z₃^(1/2)e₀, and Z₃ comes from the photon's own self-energy, which is common to every particle the photon meets.
2. BStrong interactions conserve charge, so they leave the proton's vertex unrenormalised.
3. CThe electron and the proton both have spin ½, so their vertex corrections are equal.
4. DThe on-shell condition is imposed separately for each particle, so each renormalised charge is set equal to the measured one by definition.

At one loop the electron loop gives the photon a self-energy iΠ^μν(q). Why is it computed in dimensional regularisation rather than with a sharp cutoff Λ on the Euclidean loop momentum?

1. AThe sharp cutoff leaves Π(q²) quadratically divergent, which no counterterm can absorb, whereas dimensional regularisation makes the loop finite.
2. BThe electron loop diverges in the infrared at q² = 0, which a cutoff on large momenta cannot regulate.
3. CA sharp cutoff is not invariant under shifts of the loop momentum and breaks gauge invariance: Π^μν acquires a term proportional to g^μνΛ² that is not of the transverse form (q²g^μν − q^μq^ν)Π(q²) and would give the photon a mass, whereas dimensional regularisation keeps Π^μν transverse.
4. DBoth regulators keep Π^μν transverse; dimensional regularisation is preferred only because its integrals are easier.

Work in natural units ℏ = c = 1 with the metric (+, −, −, −). In the scalar Yukawa model a complex scalar ψ of mass m_ψ and a real scalar φ of mass m_φ interact through L_int = −gψ†ψφ. Take m_φ = 0.80 GeV, m_ψ = 0.25 GeV and g = 0.40 GeV, with ℏ = 6.582 × 10⁻²⁵ GeV s and ℏc = 0.1973 GeV fm. For a particle of mass m_φ at rest decaying to two distinguishable spinless products, Γ = |p||M|²/(8πm_φ²), where M is the invariant amplitude and |p| the momentum of either product.

1. (a)

   State State the mass dimension of g, and whether the interaction is renormalisable, super-renormalisable or non-renormalisable.

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   1. 1

      In four dimensions ψ and φ each have mass dimension 1, so [g] = 4 − 3 = 1; a coupling of positive mass dimension makes the interaction super-renormalisable

2. (b)

   Show that Show that, at tree level, Γ(φ → ψψ̄) = (g²/16πm_φ)√(1 − 4m_ψ²/m_φ²).

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   1. 1

      One vertex and no propagator: contracting φ with the initial particle and ψ, ψ† with the outgoing particle and antiparticle gives iM = −ig, so |M|² = g², with no factor ½ because ψ and ψ̄ are distinguishable

   2. 2

      |p| = ½√(m_φ² − 4m_ψ²) = (m_φ/2)√(1 − 4m_ψ²/m_φ²), so Γ = |p|g²/(8πm_φ²) = (g²/16πm_φ)√(1 − 4m_ψ²/m_φ²)

3. (c)

   Calculate Calculate Γ in MeV and the mean lifetime of φ in seconds.

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   1. 1

      √(1 − 4m_ψ²/m_φ²) = √(1 − 0.25/0.64) = 0.781, so Γ = (0.16 GeV² × 0.781)/(16π × 0.80 GeV) = 3.11 × 10⁻³ GeV = 3.11 MeV

   2. 2

      τ = ℏ/Γ = (6.582 × 10⁻²⁵ GeV s)/(3.11 × 10⁻³ GeV) = 2.12 × 10⁻²² s

4. (d)

   Determine Two ψ particles scatter at low speed. From the t-channel φ-exchange diagram, determine the potential V(r) between them, allowing for the relativistic normalisation of the states, and state whether it attracts and what its range is in femtometres.

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   1. 1

      The t-channel diagram gives iM = (−ig)² i/(q² − m_φ²); for slow particles q⁰ ≈ 0 and q² ≈ −|q|², so M = g²/(|q|² + m_φ²), the u-channel diagram supplying the exchange term that identical bosons require

   2. 2

      Relativistic normalisation puts a factor 2m_ψ on each of the two particles, so the Born amplitude M/(2m_ψ)² equals −Ṽ(q), giving Ṽ(q) = −(g²/4m_ψ²)/(|q|² + m_φ²) and V(r) = −(g²/4m_ψ²)e^(−m_φr)/(4πr)

   3. 3

      V is negative, so the force attracts, with dimensionless strength g²/(16πm_ψ²) = 0.16/(16π × 0.0625) = 0.051 in the place of α, and range 1/m_φ = 0.1973 GeV fm/0.80 GeV = 0.25 fm

5. (e)

   Predict Predict what becomes of φ, and of the use of the LSZ formula with φ as an external particle, if m_φ were lowered to 0.45 GeV with m_ψ unchanged.

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      With m_φ below 2m_ψ = 0.50 GeV the decay is kinematically forbidden, so φ becomes stable and its pole is an isolated one-particle pole below the ψψ̄ continuum at 4m_ψ², as LSZ requires; at m_φ = 0.80 GeV the pole lies on the continuum and moves off the real axis by m_φΓ, so φ is not an asymptotic state

Work in natural units with the metric (+, −, −, −). A real scalar φ of mass m = 0.40 GeV has L_int = −(λ/4!)φ⁴ with λ = 0.80. Two φ particles collide head-on with centre-of-mass energy √s = 2.0 GeV. For 2 → 2 scattering of equal masses in the centre-of-mass frame, dσ/dΩ = |M|²/(64π²s). Two-body Lorentz-invariant phase space for distinguishable particles is ∫dΠ₂ = |p_cm|/(4π√s), with |p_cm| the momentum of either particle. Use (ℏc)² = 0.3894 GeV² mb. In minimal subtraction the one-loop β function of the theory is β(λ) = μ dλ/dμ = 3λ²/(16π²).

1. (a)

   Calculate Calculate the tree-level amplitude M and the total cross-section for φφ → φφ in microbarns.

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   1. 1

      The 4! ways of contracting the four external particles with the four fields at the vertex cancel the 1/4!, so iM = −iλ and M = −λ = −0.80

   2. 2

      dσ/dΩ = λ²/(64π²s) is isotropic, and the two final particles are identical, so σ = ½ × 4π × λ²/(64π²s) = λ²/(32πs)

   3. 3

      σ = 0.64/(32π × 4.0 GeV²) = 1.59 × 10⁻³ GeV⁻², and multiplying by 0.3894 GeV² mb gives 6.20 × 10⁻⁴ mb = 0.620 μb

2. (b)

   Show that At order λ² the s-channel one-loop diagram joins p₁ and p₂ at one vertex and p₃ and p₄ at the other, the two vertices being linked by two internal lines. Show by counting Wick contractions that it carries a symmetry factor ½, so that iM_s = ½(−iλ)²∫d⁴l/(2π)⁴ [i/(l² − m² + iε)][i/((k − l)² − m² + iε)], with k = p₁ + p₂.

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   1. 1

      The second-order term carries (1/2!)(1/4!)². The incoming pair can sit at either vertex (2 ways), p₁ and p₂ can take 4 × 3 = 12 choices of field at theirs, p₃ and p₄ 12 at the other, and the two remaining fields at one vertex pair with the two at the other in 2 ways: 2 × 12 × 12 × 2 = 576 contractions

   2. 2

      576 × (1/2!)(1/4!)² = 576/1152 = ½, so S = 2, the automorphism that exchanges the two internal lines, and each vertex carries −iλ

3. (c)

   Show that Show that the imaginary part of the forward amplitude at order λ², which comes from this diagram alone, is Im M = λ²β/(32π) with β = √(1 − 4m²/s), and that it satisfies the optical theorem Im M(forward) = 2E_cm|p_cm|σ_tot at √s = 2.0 GeV.

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   1. 1

      For s > 4m² the two internal lines can go on shell together, and the iε prescription gives the imaginary part by cutting the diagram, replacing each 1/(l² − m² + iε) by −2πiδ(l² − m²) with positive energy; in the forward direction the t- and u-channel loops sit at t = 0 and u = 4m² − s = −3.36 GeV², below their thresholds, and stay real

   2. 2

      The cut gives 2 Im M = ½λ²∫dΠ₂ = ½λ²|p_cm|/(4π√s), and |p_cm| = β√s/2, so Im M = λ²β/(32π); with β = √(1 − 0.64/4.0) = 0.9165, Im M = 0.64 × 0.9165/(32π) = 5.83 × 10⁻³

   3. 3

      2E_cm|p_cm|σ_tot = 2 × 2.0 GeV × 0.9165 GeV × 1.59 × 10⁻³ GeV⁻² = 5.83 × 10⁻³, equal to Im M: the loop's symmetry factor ½ does the work of the ½ for identical particles in σ_tot

4. (d)

   Explain Explain why Im M is finite although the loop integral itself diverges logarithmically.

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      The imaginary part comes only from the region in which both internal particles are on shell with total four-momentum k, a compact two-body phase space; the ultraviolet divergence comes from large loop momenta far off shell, is a real constant independent of s, and is absorbed into δ_λ

5. (e)

   Calculate Taking λ(μ₀) = 0.80 at μ₀ = 0.40 GeV in minimal subtraction, calculate λ at μ = 100 GeV from the one-loop β function, and the scale at which the one-loop coupling would diverge.

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   1. 1

      Integrating μ dλ/dμ = 3λ²/(16π²) gives 1/λ(μ) = 1/λ(μ₀) − (3/16π²)ln(μ/μ₀) = 1.250 − 0.01900 × ln 250 = 1.250 − 0.105 = 1.145, so λ(100 GeV) = 0.873

   2. 2

      1/λ reaches zero when ln(μ/μ₀) = 16π²/(3λ(μ₀)) = 65.8, at μ = 0.40 GeV × e^(65.8) ≈ 1.5 × 10²⁸ GeV, the Landau pole: far beyond the Planck mass, and long before it the coupling is too large for one loop to be trusted

Work in natural units with the metric (+, −, −, −). In QED with an electron and a muon, e⁺e⁻ → μ⁺μ⁻ proceeds at lowest order through one-photon exchange in the s channel, with vertex −ieγ^μ and Feynman-gauge photon propagator −ig_μν/(q² + iε). Label the momenta e⁻(p₁)e⁺(p₂) → μ⁻(p₃)μ⁺(p₄), neglect the electron mass, keep the muon mass m_μ = 0.10566 GeV, and take α = e²/4π = 1/137.036 and (ℏc)² = 0.3894 GeV² mb. The tree amplitude is M = (e²/s)[v̄(p₂)γ^μu(p₁)][ū(p₃)γ_μv(p₄)], with s = (p₁ + p₂)². Use Σ u ū = p̸ + m and Σ v v̄ = p̸ − m over spins, Tr(γ^μγ^ν) = 4g^μν and Tr(γ^μγ^νγ^ργ^σ) = 4(g^μνg^ρσ − g^μρg^νσ + g^μσg^νρ); traces of an odd number of gamma matrices vanish.

1. (a)

   Show that Show that the spin-averaged squared amplitude is ¼Σ|M|² = (8e⁴/s²)[(p₁⋅p₃)(p₂⋅p₄) + (p₁⋅p₄)(p₂⋅p₃) + m_μ²(p₁⋅p₂)].

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   1. 1

      Casimir's trick turns the spin sums into traces: ¼Σ|M|² = (e⁴/4s²)Tr[p̸₂γ^μp̸₁γ^ν]Tr[(p̸₃ + m_μ)γ_μ(p̸₄ − m_μ)γ_ν]

   2. 2

      The electron trace is 4[p₂^μp₁^ν + p₂^νp₁^μ − g^μν(p₁⋅p₂)] and the muon trace 4[p₃_μp₄_ν + p₃_νp₄_μ − g_μν(p₃⋅p₄ + m_μ²)], the mass term coming from −m_μ²Tr(γ_μγ_ν) and the terms linear in m_μ vanishing as odd traces

   3. 3

      Contracting, with g^μνg_μν = 4, the (p₁⋅p₂)(p₃⋅p₄) terms cancel and the product of traces is 32[(p₁⋅p₃)(p₂⋅p₄) + (p₁⋅p₄)(p₂⋅p₃) + m_μ²(p₁⋅p₂)], so ¼Σ|M|² = (e⁴/4s²) × 32[...] = (8e⁴/s²)[...]

2. (b)

   Show that In the centre-of-mass frame, with θ the angle between p₃ and p₁ and β = √(1 − 4m_μ²/s) the muon speed, show that dσ/dΩ = (α²β/4s)(2 − β²sin²θ), and hence that σ = (4πα²/3s) × β(3 − β²)/2.

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   1. 1

      With every energy √s/2 and muon momentum β√s/2, p₁⋅p₂ = s/2, p₁⋅p₃ = p₂⋅p₄ = (s/4)(1 − β cos θ) and p₁⋅p₄ = p₂⋅p₃ = (s/4)(1 + β cos θ), so the bracket is (s²/8)(1 + β²cos²θ) + m_μ²s/2 = (s²/8)(2 − β²sin²θ), using 4m_μ²/s = 1 − β²

   2. 2

      ¼Σ|M|² = e⁴(2 − β²sin²θ) = 16π²α²(2 − β²sin²θ), and dσ/dΩ = (|p_f|/|p_i|)¼Σ|M|²/(64π²s) with |p_f|/|p_i| = β gives (α²β/4s)(2 − β²sin²θ)

   3. 3

      Integrating, ∫(2 − β² + β²cos²θ)dΩ = 2π[2(2 − β²) + (2/3)β²] = (8π/3)(3 − β²), so σ = (α²β/4s)(8π/3)(3 − β²) = (4πα²/3s)β(3 − β²)/2, which tends to 4πα²/3s as β → 1

3. (c)

   Calculate Calculate σ in nanobarns at √s = 0.300 GeV and at √s = 3.00 GeV, and, at the lower energy, the ratio of σ to the value 4πα²/3s that massless muons would give.

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   1. 1

      4πα²/3 = 2.2306 × 10⁻⁴, which with 0.3894 GeV² mb is 86.86 nb GeV²; at √s = 0.300 GeV, s = 0.0900 GeV² and β = √(1 − 0.04466/0.0900) = 0.7098, so β(3 − β²)/2 = 0.886 and σ = (86.86/0.0900) nb × 0.886 = 965 nb × 0.886 = 855 nb

   2. 2

      At √s = 3.00 GeV, β = 0.9975 and β(3 − β²)/2 = 1.0000 to four figures, so σ = 86.86/9.00 nb = 9.65 nb; the ratio at 0.300 GeV is 0.886, an 11% suppression by the muon mass that has disappeared by 3 GeV

4. (d)

   Sketch Sketch on one set of axes the angular distribution at √s = 0.300 GeV and at √s = 3.00 GeV, each normalised to its own value at cos θ = ±1, that is (dσ/dΩ)/(dσ/dΩ at cos θ = 1) against cos θ from −1 to +1, marking the value of each curve at cos θ = 0.

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   1. 1

      Both normalised curves equal 1 at cos θ = ±1, are symmetric about cos θ = 0 and have their minimum there; at 3.00 GeV the curve is (1 + cos²θ)/2 to within 0.5%, marked 0.50 at cos θ = 0

   2. 2

      At 0.300 GeV the curve is (2 − β² + β²cos²θ)/2 with β² = 0.504, flatter and lying above the 3.00 GeV curve at every angle between the ends, marked (2 − β²)/2 = 0.75 at cos θ = 0; at threshold it would become flat, the pair being produced in an s wave

5. (e)

   Determine Determine by crossing the spin-averaged |M|² for e⁻μ⁻ → e⁻μ⁻ at energies where both masses can be neglected, in the Mandelstam variables of that process, and state the sign that crossing introduces.

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   1. 1

      With the masses neglected, p₁⋅p₃ = p₂⋅p₄ = −t/2 and p₁⋅p₄ = p₂⋅p₃ = −u/2, so ¼Σ|M|² for e⁺e⁻ → μ⁺μ⁻ is 2e⁴(t² + u²)/s²

   2. 2

      Crossing the incoming e⁺ into an outgoing e⁻ and the outgoing μ⁺ into an incoming μ⁻ turns the old s, t and u into the new t, u and s; each crossed fermion multiplies Σ|M|² by −1 and two are crossed, so the sign is (−1)² = +1 and ¼Σ|M|² = 2e⁴(s² + u²)/t², positive as it must be and rising as 1/t² in the forward direction

Work in natural units with the metric (+, −, −, −). Compton scattering γ(k)e⁻(p) → γ(k′)e⁻(p′) has, at tree level, an s-channel diagram, in which the electron absorbs k before emitting k′, and a u-channel diagram, in which it emits k′ first. For an electron initially at rest and a photon of energy ω, the spin- and polarisation-summed result integrates to the Klein–Nishina total cross-section σ = 2πr_e²{[(1 + κ)/κ²][2(1 + κ)/(1 + 2κ) − ln(1 + 2κ)/κ] + ln(1 + 2κ)/(2κ) − (1 + 3κ)/(1 + 2κ)²}, with κ = ω/m and r_e = α/m. Crossing the same amplitude into e⁺e⁻ → γγ, with the factor ½ for identical photons, gives for a slow pair the spin-averaged σv_rel → πα²/m² as the relative speed v_rel → 0. Take m_e = 0.51100 MeV, m_μ = 105.658 MeV, α = 1/137.036, ℏc = 197.327 MeV fm, ℏ = 6.582 × 10⁻²² MeV s and 1 b = 100 fm².

1. (a)

   Explain Replacing the polarisation vector of the incoming photon by its momentum k, explain why neither Compton diagram alone satisfies the Ward identity k_μM^μν = 0, while their sum does.

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   1. 1

      In the s-channel diagram k̸ stands next to u(p): (p̸ + m)k̸u(p) = [2p⋅k − k̸(p̸ − m)]u(p) = 2p⋅k u(p) and k² = 0, so the numerator cancels the denominator (p + k)² − m² = 2p⋅k and leaves a term proportional to ū(p′)γ^νu(p); in the u-channel diagram k̸ stands next to ū(p′), and the same steps with (p′ − k)² − m² = −2p′⋅k leave the same term with the opposite sign

   2. 2

      So each diagram alone gives ±e²ū(p′)γ^νu(p), which does not vanish, and only their sum is zero: the Ward identity holds for the gauge-invariant set, and only for the sum can both photon polarisation sums be replaced by −g_μν

2. (b)

   Calculate Calculate the Thomson cross-section (8π/3)r_e² in barns.

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   1. 1

      r_e = αℏc/m_e = 197.327 MeV fm/(137.036 × 0.51100 MeV) = 2.818 fm, so (8π/3)r_e² = 66.52 fm² = 0.6652 b

3. (c)

   Calculate A 511 keV photon from positron annihilation enters the scintillator of a positron-emission tomograph. Calculate the Klein–Nishina cross-section per electron in barns, its ratio to the Thomson value, and the largest kinetic energy the photon can give to a free electron at rest.

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   1. 1

      κ = 0.511/0.51100 = 1.000, so the bracket is 2(4/3 − ln 3) + (ln 3)/2 − 4/9 = 0.4694 + 0.5493 − 0.4444 = 0.5743, and σ = 2π(2.818 fm)² × 0.5743 = 28.65 fm² = 0.287 b

   2. 2

      σ/σ_T = 0.287/0.665 = 0.431: at κ = 1 the cross-section has already fallen to less than half its classical value

   3. 3

      The largest transfer is in back-scattering, θ = π: ω′ = ω/(1 + 2κ) = 170.3 keV, so T_max = ω − ω′ = 2κω/(1 + 2κ) = 340.7 keV, the Compton edge of the 511 keV line

4. (d)

   Determine True muonium is the atom formed by a μ⁺ and a μ⁻. Treating it as a hydrogen-like atom bound by the Coulomb potential, determine the lifetime of its para ground state, 1¹S₀, against annihilation into two photons, which is its only significant decay.

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   1. 1

      Only the spin singlet, one of the four spin states, has C = +1 and can annihilate into two photons, so it carries the whole of the spin-averaged rate and its σv_rel is four times the average: 4πα²/m_μ²

   2. 2

      The rate is Γ = |ψ(0)|²σv_rel; with the reduced mass m_μ/2 the Bohr radius is a = 2/(m_μα), so |ψ(0)|² = 1/(πa³) = (m_μα)³/(8π) and Γ = (m_μα)³/(8π) × 4πα²/m_μ² = α⁵m_μ/2

   3. 3

      Γ = (1/137.036)⁵ × 105.658 MeV/2 = 2.069 × 10⁻¹¹ × 52.83 MeV = 1.093 × 10⁻⁹ MeV, so τ = ℏ/Γ = (6.582 × 10⁻²² MeV s)/(1.093 × 10⁻⁹ MeV) = 6.02 × 10⁻¹³ s = 0.602 ps, para-positronium's 124.5 ps scaled by m_e/m_μ

5. (e)

   Explain Explain why the two Compton diagrams enter with the same sign, whereas the two diagrams of Bhabha scattering, e⁺e⁻ → e⁺e⁻, enter with a relative minus sign.

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      The Compton diagrams differ only in which photon field is contracted with which external photon, and photon operators commute, so no sign arises; Bhabha's annihilation and exchange diagrams differ in the order in which the fermion operators are contracted with the external states, and moving one anticommuting operator past another brings a factor −1

Work in natural units ℏ = c = 1 with the metric (+, −, −, −). Charge conjugation is a unitary operator C on the Fock space of QED. On the Dirac field it acts as Cψ(x)C⁻¹ = M_Cψ̄ᵀ(x) and Cψ̄(x)C⁻¹ = −ψᵀ(x)M_C⁻¹, where ᵀ is the transpose in spinor indices and the 4 × 4 matrix M_C satisfies M_C⁻¹γ^μM_C = −(γ^μ)ᵀ; on the photon field CA_μ(x)C⁻¹ = −A_μ(x). Write (−1)^μ = +1 for μ = 0 and −1 for μ = 1, 2, 3. Under parity the Hermitian bilinear iψ̄σ^μνγ⁵ψ is multiplied by −(−1)^μ(−1)^ν and F_μν by (−1)^μ(−1)^ν; under time reversal, which is antiunitary, they are multiplied by (−1)^μ(−1)^ν and −(−1)^μ(−1)^ν respectively; under C both are multiplied by −1. Each is evaluated at the transformed point. An electric dipole moment d of the electron would add L_EDM = −(d/2)(iψ̄σ^μνγ⁵ψ)F_μν to the Lagrangian, with d real, whose non-relativistic limit is the energy −dσ⋅E. An experiment on polar molecules has found |d| < 4.1 × 10⁻³⁰ e cm. Take m_e = 0.511 MeV and ℏc = 1.973 × 10⁻¹¹ MeV cm.

1. (a)

   Show that Show that the current ψ̄γ^μψ is odd under C while ψ̄ψ is even, identifying the step at which the anticommutation of the fields supplies a sign, and say what would become of the electron's coupling to the photon if the Dirac field were quantised with commutators.

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   1. 1

      C(ψ̄γ^μψ)C⁻¹ = −ψᵀM_C⁻¹γ^μM_Cψ̄ᵀ = +ψᵀ(γ^μ)ᵀψ̄ᵀ = ψ_a(γ^μ)_ba ψ̄_b, using M_C⁻¹γ^μM_C = −(γ^μ)ᵀ

   2. 2

      Reordering ψ_aψ̄_b = −ψ̄_bψ_a, by anticommutation up to a c-number that normal ordering removes, gives −ψ̄_b(γ^μ)_baψ_a = −ψ̄γ^μψ, odd; for ψ̄ψ the same steps give −ψᵀM_C⁻¹M_Cψ̄ᵀ = −ψ_aψ̄_a = +ψ̄ψ, even

   3. 3

      With commuting fields the reordering would bring no sign, the current would be C-even, particle and antiparticle would carry the same charge, and the coupling −eψ̄γ^μψA_μ to the C-odd photon could not be C-invariant: the C-odd current, and with it the positron's opposite charge, rests on anticommutation

2. (b)

   Determine Determine whether L_EDM is even or odd under each of P, C, T and CP, and check that it is even under CPT.

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   1. 1

      Under P the signs multiply to −(−1)^μ(−1)^ν × (−1)^μ(−1)^ν = −1, since [(−1)^μ(−1)^ν]² = 1; under T to (−1)^μ(−1)^ν × [−(−1)^μ(−1)^ν] = −1, the real coefficient −d/2 being untouched by the antiunitary T; under C to (−1)(−1) = +1: L_EDM is P-odd, T-odd and C-even

   2. 2

      Hence it is CP-odd, and under CPT the signs multiply to (−1)(+1)(−1) = +1, even, as every Hermitian, Lorentz-invariant local term must be; the non-relativistic form −dσ⋅E agrees, the spin being P-even and T-odd and E being P-odd and T-even

3. (c)

   Explain Explain, both from C invariance and from the diagrams, why the sum of the one-loop electron-loop diagrams with three external photons vanishes in QED, and what this means for the counterterms QED needs.

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   1. 1

      QED is C-invariant and A_μ is C-odd, so ⟨Ω|TA_μA_νA_ρ|Ω⟩ = (−1)³⟨Ω|TA_μA_νA_ρ|Ω⟩ = 0; diagrammatically, the two orientations of the electron loop are related by transposing the trace with M_C, which turns each γ^μ into −(γ^μ)ᵀ and reverses each propagator, so they differ by (−1)³ and cancel in pairs (Furry's theorem)

   2. 2

      The three-photon function, superficially divergent with D = 4 − B = 1, therefore needs no counterterm and no A³ term is generated; with gauge invariance this leaves the electron self-energy, the photon self-energy and the vertex as QED's divergent functions

4. (d)

   Estimate A new interaction at a mass scale Λ could generate d at one loop. Estimate the lowest Λ consistent with the experimental bound, stating your assumptions.

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   1. 1

      Assume new particles of mass Λ coupling to the electron with couplings g of order 1, a CP-violating phase of order 1, the photon attached with charge e, a loop factor g²/16π² and one power of m_e for the chirality flip that σ^μνγ⁵ requires, so that d ≈ e m_e/(16π²Λ²)

   2. 2

      In natural units d/e < 4.1 × 10⁻³⁰ cm/(1.973 × 10⁻¹¹ MeV cm) = 2.08 × 10⁻¹⁹ MeV⁻¹, so Λ² > 0.511 MeV/(16π² × 2.08 × 10⁻¹⁹ MeV⁻¹) = 1.6 × 10¹⁶ MeV² and Λ > 1.2 × 10⁸ MeV ≈ 10⁵ GeV, about 100 TeV, beyond the direct reach of any collider

5. (e)

   Predict Predict, from CPT invariance alone, the electric dipole moment of the positron if the electron's were found to be d.

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   1. 1

      CPT relates the electron to the positron with every charge and moment reversed, so the positron's dipole moment is −d, equal in size and opposite in sign, just as its magnetic moment is opposite to the electron's; a measured difference in size would violate CPT

Work in natural units with the metric (+, −, −, −). An electron of mass m bound to a nucleus of charge Ze, treated as infinitely heavy, feels at lowest order the Coulomb potential V(r) = −Zα/r, whose Fourier transform is Ṽ(q) = −4πZα/|q|². The renormalised one-loop vertex, computed with a photon mass μ_γ, has to first order in the spacelike momentum transfer q² = −|q|² the form factors F₁(q²) = 1 + (α/3π)(q²/m²)[ln(m/μ_γ) − 3/8] and F₂(0) = α/2π. For an ns state |ψ_ns(0)|² = (Zαm)³/(πn³), and states with l ≥ 1 vanish at the origin. Write U = 4α(Zα)⁴m/(3πn³). In these units, at order α(Zα)⁴: the F₂ (Pauli) term shifts ns levels by +(3/8)U and 2P_(1/2) by −(1/8)U; the Uehling (vacuum-polarisation) term shifts ns levels by −(1/5)U; Bethe's non-relativistic treatment of photons of energy below a cutoff K shifts ns levels by U ln(K/⟨E⟩_ns) and 2P_(1/2) by −U ln(⟨E⟩_2p/(Z²Ry)), where Z²Ry = (Zα)²m/2 and the Bethe logarithms are ln(⟨E⟩_2s/(Z²Ry)) = 2.8118 and ln(⟨E⟩_2p/(Z²Ry)) = −0.0300. Matching across (Zα)²m ≪ K ≪ m replaces ln(m/μ_γ) in the high-energy part by ln(m/2K) + 5/6. For hydrogen-like helium, ⁴He⁺ (Z = 2), take α = 1/137.036, m = 0.511 MeV and, for n = 2, U = 2170 MHz, reduced-mass corrections neglected. The same assembly for hydrogen gives 1051.9 MHz for the 2S_(1/2)–2P_(1/2) splitting against the measured 1057.8 MHz; for He⁺ the measured splitting is 14 041 MHz.

1. (a)

   Show that Show that the q² term of F₁ adds to the Coulomb potential a contact term δV(r) = (4Zα²/3m²)[ln(m/μ_γ) − 3/8]δ³(r), and that this raises an ns level by U[ln(m/μ_γ) − 3/8] while leaving P levels unshifted.

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   1. 1

      In a static field the vertex multiplies the Coulomb amplitude by F₁(q²), F₂ giving the separate Pauli term, so Ṽ(q) = −(4πZα/|q|²)F₁(−|q|²) = −4πZα/|q|² + (4πZα)(α/3πm²)[ln(m/μ_γ) − 3/8]

   2. 2

      The correction does not depend on q, so it transforms to a delta function, δV(r) = (4Zα²/3m²)[ln(m/μ_γ) − 3/8]δ³(r), positive, a repulsive contact term that raises the level

   3. 3

      ΔE_ns = ⟨δV⟩ = (4Zα²/3m²)[ln(m/μ_γ) − 3/8](Zαm)³/(πn³) = (4α(Zα)⁴m/3πn³)[ln(m/μ_γ) − 3/8] = U[ln(m/μ_γ) − 3/8], and zero for l ≥ 1 because those wavefunctions vanish at the origin

2. (b)

   Explain Explain why the result of (a) cannot be used on its own, and why a cutoff K with (Zα)²m ≪ K ≪ m can be found for He⁺, giving the two ends of the window in electronvolts.

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   1. 1

      The result depends on the photon mass μ_γ, an infrared regulator: the free-electron vertex treats virtual photons of every energy as though the electron were free, but photons with energies of order the binding, (Zα)²m, see a bound electron whose intermediate states are atomic levels, and there the free vertex is wrong; Bethe's dipole treatment is right there but fails for photon energies approaching m, where the electron becomes relativistic

   2. 2

      The two descriptions overlap where (Zα)²m ≪ K ≪ m: for He⁺, (Zα)²m = (2/137.036)² × 0.511 MeV = 109 eV and m = 511 000 eV, a window of more than three decades, so both hold at some K and the matching is controlled

3. (c)

   Determine Using the matching relation, determine the total order-α(Zα)⁴ shift of the He⁺ 2S_(1/2) level in units of U and in megahertz, showing that K cancels.

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   1. 1

      The high-energy part with the matching is U[ln(m/2K) + 5/6 − 3/8] and the low-energy part U ln(K/⟨E⟩_2s); their sum, U[ln(m/(2⟨E⟩_2s)) + 5/6 − 3/8], is free of K, and since 2⟨E⟩_2s = (Zα)²m e^(2.8118) the logarithm is ln(1/(Zα)²) − 2.8118 = 8.4542 − 2.8118 = 5.6424

   2. 2

      Adding the Pauli term +3/8 and the Uehling term −1/5 gives 5.6424 + 5/6 − 3/8 + 3/8 − 1/5 = 6.2757 in units of U

   3. 3

      ΔE(2S_(1/2)) = 6.2757 × 2170 MHz = 13 618 MHz, of which the Uehling term contributes −434 MHz and the Pauli term +814 MHz

4. (d)

   Calculate Calculate the shift of 2P_(1/2) and hence the 2S_(1/2)–2P_(1/2) splitting of He⁺ in megahertz.

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   1. 1

      2P_(1/2) receives no contact terms, so its shift is −U ln(⟨E⟩_2p/(Z²Ry)) − U/8 = (0.0300 − 0.1250)U = −0.0950U = −206 MHz

   2. 2

      The splitting is (6.2757 + 0.0950)U = 6.3707 × 2170 MHz = 13 824 MHz

5. (e)

   Evaluate Evaluate the agreement with the measured splittings of He⁺ and of hydrogen, and say what the comparison suggests about the largest term omitted.

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   1. 1

      For He⁺, 14 041 − 13 824 = 217 MHz, 1.5% of the measured value; for hydrogen, 1057.8 − 1051.9 = 5.9 MHz, 0.56%: in both the order-α(Zα)⁴ assembly falls short, by a fraction that grows by a factor of about 2.8 from Z = 1 to Z = 2

   2. 2

      A fractional shortfall that grows at least in proportion to Z points to an omitted term one power of Zα higher, of order α(Zα)⁵m, from a further Coulomb exchange inside the loop, about +7 MHz in hydrogen and +230 MHz in He⁺; two-loop terms are suppressed by a further α/π and do not grow with Z, and the neglected reduced-mass correction (about −1.7 MHz in hydrogen, which widens the gap) and the nuclear size (about +0.1 MHz in hydrogen) are too small to account for it

An electron–positron collider measures the total cross-section for e⁺e⁻ → μ⁺μ⁻ at six centre-of-mass energies well below the Z boson. Z exchange and initial- and final-state radiation have been corrected for, so that what remains is one-photon exchange through the vacuum-polarisation-corrected photon propagator, for which σ = 4πα(s)²/(3s), with α(s) the effective coupling at momentum transfer squared s; the muon mass is negligible at these energies. Work in natural units with α(0) = 1/137.036 and (ℏc)² = 0.3894 GeV² mb, so that 4πα(0)²/3 = 86.86 nb GeV². At one loop a fermion of charge Q_f e with N_c colours makes α run as μ dα/dμ = 2N_cQ_f²α²/(3π) while μ is well above its mass, with μ² = s. Every quoted uncertainty is 0.2% of σ, statistical and systematic combined. Lepton loops alone, run from q² = 0 with the measured lepton masses, give 1/α = 134.02 at √s = 12.0 GeV and 132.73 at √s = 91.19 GeV.

1. (a)

   Show that Show that if every fermion lighter than √s contributes at one loop, 1/α(s) is a straight line in ln s with gradient −Σ_f N_cQ_f²/(3π), and that sσ/(86.86 nb GeV²) = [α(s)/α(0)]².

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   1. 1

      d(1/α)/d ln μ = −(1/α²)μ dα/dμ = −2Σ_f N_cQ_f²/(3π), and since ln μ = ½ ln s, d(1/α)/d ln s = −Σ_f N_cQ_f²/(3π), constant while no threshold is crossed, so 1/α(s) = 1/α(s₀) − [Σ_f N_cQ_f²/(3π)]ln(s/s₀)

   2. 2

      σ = 4πα(s)²/(3s) gives sσ = [4πα(0)²/3][α(s)/α(0)]², and 4πα(0)²/3 = 2.2306 × 10⁻⁴ × 0.3894 GeV² mb = 8.686 × 10⁻⁵ GeV² mb = 86.86 nb GeV²

2. (b)

   Describe Describe how you would process the readings to test this prediction and extract the gradient, including how the uncertainty in each σ becomes an uncertainty in 1/α.

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   1. 1

      For each run form R = sσ/(86.86 nb GeV²) and 1/α(s) = 137.036/√R, then plot 1/α against ln(s/GeV²) and fit a straight line, whose gradient is −Σ_f N_cQ_f²/(3π); a straight line tests one-loop running with a fixed set of active species

   2. 2

      Since 1/α ∝ σ^(−1/2), δ(1/α) = ½(1/α)(δσ/σ) = ½ × 131 × 0.002 ≈ 0.13 at every point, equal error bars, so an unweighted fit is appropriate; the intercept at ln s = 0 has no direct meaning, since it absorbs all the running below 12 GeV

3. (c)

   Determine Determine 1/α at each energy, the gradient and intercept of the best straight line with an estimate of the uncertainty in the gradient, and hence Σ_f N_cQ_f².

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   1. 1

      sσ = 93.96, 94.54, 95.44, 96.17, 96.51 and 97.09 nb GeV², so 1/α = 131.76, 131.35, 130.73, 130.23, 130.00 and 129.62 at ln(s/GeV²) = 4.970, 5.781, 6.438, 7.111, 7.613 and 8.121

   2. 2

      Gradient = −0.695 (accept −0.66 to −0.73), with an uncertainty of about 0.05 from 0.13 error bars over the spread of ln s (accept 0.03 to 0.06), and intercept 135.25 at ln(s/GeV²) = 0 (accept 135.0 to 135.5)

   3. 3

      Σ_f N_cQ_f² = −3π × gradient = 6.6 ± 0.5 (accept 6.2 to 6.9)

4. (d)

   Evaluate Evaluate which of these the data support: the three charged leptons alone; the leptons with the five quarks u, d, s, c and b each counted once; or the leptons with those quarks in three colours.

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   1. 1

      Leptons alone give Σ = 3 and gradient −3/(3π) = −0.318; adding u and c (Q = 2/3) and d, s and b (Q = −1/3) once each gives Σ = 3 + 8/9 + 3/9 = 4.22 and gradient −0.448; with three colours Σ = 3 + 11/3 = 20/3 and gradient −0.707

   2. 2

      The measured −0.695 ± 0.05 lies about 7.5 standard deviations from the leptons-only value and 5 from uncoloured quarks, and within 0.3 standard deviations of the three-colour value, so the data require quarks in three colours

   3. 3

      The verdict survives what was neglected: the strong-interaction correction multiplies the quark term by about 1 + α_s/π ≈ 1.05, moving the prediction to about −0.73, within the error, and the top quark, at 173 GeV, is not yet active

5. (e)

   Predict Predict 1/α at √s = 91.19 GeV by extrapolating your line, and say what its difference from the lepton-only running requires of hadrons, already at 12 GeV and at the Z.

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   1. 1

      ln(s/GeV²) = 9.026 at 91.19 GeV, so 1/α ≈ 135.25 − 0.695 × 9.026 = 128.98, about 129.0 (accept 128.8 to 129.2), against 132.73 from leptons alone

   2. 2

      Hadrons must supply Δ(1/α) ≈ 3.7 at the Z, of which 2.3 (134.02 − 131.76) is already present at 12 GeV and comes from hadronic states below 12 GeV, resonances included, where quark loops with ill-defined masses cannot be trusted; it is obtained instead from a dispersion relation over the measured e⁺e⁻ → hadrons cross-section, which the optical theorem ties to Im Π

Work in natural units with the metric (+, −, −, −), in QED with an electron of physical mass m and bare mass m₀, in Feynman gauge with a small photon mass μ_γ. At one loop the electron self-energy is −iΣ₂(p) = (−ie)²∫d^dk/(2π)^d γ^μ[i(k̸ + m)/(k² − m² + iε)]γ_μ[−i/((p − k)² − μ_γ² + iε)], continued to d = 4 − ε dimensions with e² → e²μ^ε to keep e dimensionless. You may use: in d dimensions γ^μγ_μ = d and γ^μk̸γ_μ = −(d − 2)k̸; Feynman's parametrisation 1/(AB) = ∫₀¹dx 1/[xA + (1 − x)B]²; the Minkowski master integral ∫d^dl/(2π)^d 1/(l² − Δ + iε)² = iΓ(2 − d/2)Δ^(d/2 − 2)/(4π)^(d/2), integrals odd in l vanishing; and, as ε → 0, Γ(ε/2)(4πμ²/Δ)^(ε/2) = 2/ε − γ_E + ln(4πμ²/Δ) + O(ε). The physical mass is the pole of the full propagator, m = m₀ + Σ₂(p̸ = m) at this order. With the full vertex Γ^μ (its factor −ie removed) and the full inverse propagator S̃⁻¹(p) = p̸ − m₀ − Σ(p), the Ward–Takahashi identity reads q_μΓ^μ(p + q, p) = S̃⁻¹(p + q) − S̃⁻¹(p). Take α = 1/137.036, m = 0.511 MeV and the Planck mass 1.22 × 10¹⁹ GeV.

1. (a)

   Derive Derive, by counting powers of loop momentum, the superficial degree of divergence D of a QED diagram in terms of the number B of its external photons and the number F of its external electron lines, and hence the value of D for the electron self-energy.

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   1. 1

      With L loops, P_e internal electron and P_γ internal photon lines and V vertices, D = 4L − P_e − 2P_γ and L = P_e + P_γ − V + 1, and each vertex ends two electron lines and one photon line: 2V = 2P_e + F and V = 2P_γ + B

   2. 2

      Substituting, D = 3P_e + 2P_γ − 4V + 4 = 3(V − F/2) + (V − B) − 4V + 4 = 4 − B − (3/2)F, independent of V, so finitely many functions diverge; the self-energy, with B = 0 and F = 2, has D = 1

2. (b)

   Explain Explain why, although D = 1, the mass shift can diverge only logarithmically and must be proportional to m, and contrast this with what happens to a scalar's mass.

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   1. 1

      At m₀ = 0 the QED Lagrangian is invariant under the chiral rotation ψ → e^(iθγ⁵)ψ, which forbids a ψ̄ψ term, so Σ at p̸ = m must vanish as m → 0: δm is proportional to m, which takes one power of the cutoff out of D = 1 and leaves a logarithm

   2. 2

      No symmetry protects a scalar's mass, so its shift is of order (coupling/16π²)Λ², quadratically sensitive to high scales; the electron mass is natural because a small m is protected by the symmetry that appears when it vanishes

3. (c)

   Show that Show that Σ₂(p) = [e²μ^ε/(4π)^(d/2)]∫₀¹dx[(4 − ε)m − (2 − ε)xp̸]Γ(2 − d/2)Δ^(d/2 − 2), with Δ = (1 − x)m² + xμ_γ² − x(1 − x)p².

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   1. 1

      In d dimensions the numerator is γ^μ(k̸ + m)γ_μ = −(d − 2)k̸ + dm = −(2 − ε)k̸ + (4 − ε)m

   2. 2

      With x multiplying the photon denominator, x[(p − k)² − μ_γ²] + (1 − x)(k² − m²) = (k − xp)² − Δ, with Δ = (1 − x)m² + xμ_γ² − x(1 − x)p²

   3. 3

      Shifting to l = k − xp, the term linear in l vanishes and the numerator becomes (4 − ε)m − (2 − ε)xp̸

   4. 4

      The master integral gives iΓ(2 − d/2)Δ^(d/2 − 2)/(4π)^(d/2), and the vertices and propagator numerators give (−ie)² × i × (−i) = −e², with e² → e²μ^ε, so −iΣ₂ = −e²μ^ε × i[...]/(4π)^(d/2) and Σ₂ = [e²μ^ε/(4π)^(d/2)]∫₀¹dx[(4 − ε)m − (2 − ε)xp̸]Γ(2 − d/2)Δ^(d/2 − 2)

4. (d)

   Derive Derive the self-mass δm = m − m₀ at one loop, showing that it is proportional to m, logarithmic and finite as μ_γ → 0, and find its finite part.

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   1. 1

      On shell p̸ = m and p² = m², and as μ_γ → 0, Δ = (1 − x)²m², whose logarithm is integrable at x = 1, so δm = Σ₂(p̸ = m) needs no photon mass; with e²μ^ε(4π)^(−d/2)Γ(ε/2)Δ^(−ε/2) = (α/4π)Γ(ε/2)(4πμ²/Δ)^(ε/2), δm = (α/4π)m∫₀¹dx[(4 − ε) − (2 − ε)x][2/ε − γ_E + ln(4πμ²/((1 − x)²m²))]

   2. 2

      The pole term gives (2/ε)∫(4 − 2x)dx = 6/ε; the O(ε) part of the numerator, −ε(1 − x), times 2/ε gives −2∫(1 − x)dx = −1; and −∫(4 − 2x)ln[(1 − x)²]dx = −2∫₀¹(2 + 2y)ln y dy = +5

   3. 3

      So δm = (3α/4π)m[2/ε − γ_E + ln(4πμ²/m²) + 4/3]: proportional to m and only logarithmically dependent on the regulator, the 1/ε pole standing for ln Λ²

5. (e)

   Compare Interpreting the pole as ln(Λ²/m²) for a cutoff Λ, so that δm ≈ (3α/4π)m ln(Λ²/m²), compare δm/m for Λ equal to the Planck mass with the shift of order αΛ that a linearly divergent electromagnetic mass would give.

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      ln(Λ²/m²) = 2 ln(1.22 × 10¹⁹ GeV/5.11 × 10⁻⁴ GeV) = 103.1, so δm/m = [3/(4π × 137.036)] × 103.1 = 0.0017421 × 103.1 = 0.18: an 18% shift even with the Planck mass as cutoff

   2. 2

      A linear shift αΛ = 1.22 × 10¹⁹ GeV/137.036 = 8.9 × 10¹⁶ GeV would be 1.7 × 10²⁰ times m, so the observed mass would need a cancellation to twenty decimal places; the relativistic theory, with its antiparticles, turns the linear divergence into a logarithm and makes the electron's small mass natural

6. (f)

   Show that From the Ward–Takahashi identity show that Z₁ = Z₂, where near the mass shell S̃⁻¹(p) → Z₂⁻¹(p̸ − m) and, between on-shell spinors as q → 0, Γ^μ(p, p) → Z₁⁻¹γ^μ.

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      Expanding the identity to first order in q, q_μΓ^μ(p + q, p) = q_μ∂S̃⁻¹(p)/∂p_μ + O(q²), so Γ^μ(p, p) = ∂S̃⁻¹(p)/∂p_μ, Ward's original identity

   2. 2

      Near the mass shell ∂S̃⁻¹/∂p_μ → Z₂⁻¹γ^μ, so Z₁⁻¹γ^μ = Z₂⁻¹γ^μ and Z₁ = Z₂: the vertex and field renormalisations cancel in e = Z₂Z₃^(1/2)e₀/Z₁, and the charge is renormalised by Z₃ alone

Work in natural units with the metric (+, −, −, −). Between on-shell spinors the full electron–photon vertex is ū(p′)Γ^μu(p), with q = p′ − p and σ^μν = (i/2)[γ^μ, γ^ν]; the Gordon identity reads ū(p′)γ^μu(p) = ū(p′)[(p′ + p)^μ/2m + iσ^μνq_ν/2m]u(p). At one loop in Feynman gauge, after Feynman parameters, the shift ℓ = k + yq − zp and use of the Dirac equation, the correction is ū(p′)δΓ^μu(p) = 2ie²∫d⁴ℓ/(2π)⁴∫₀¹dx dy dz δ(x + y + z − 1)(2/D³)ū(p′)N^μu(p), with D = ℓ² − Δ + iε, Δ = −xyq² + (1 − z)²m² and N^μ = γ^μ[−ℓ²/2 + (1 − x)(1 − y)q² + (1 − 4z + z²)m²] + (iσ^μνq_ν/2m)[2m²z(1 − z)]. Use ∫d⁴ℓ/(2π)⁴ 1/(ℓ² − Δ)³ = −i/(2(4π)²Δ). The measured electron anomaly is a_e = 1.159 652 180 59(13) × 10⁻³, and the QED series is a_e = α/2π + C₂(α/π)² + C₃(α/π)³ + C₄(α/π)⁴ + ..., with C₂ = −0.328 478 965, C₃ = 1.181 241 456 and C₄ = −1.912 245 764; in this question take α⁻¹ = 137.035 999, from atom-recoil measurements. For elastic scattering of an electron of energy E through angle θ by a recoil-free nucleus, −q² = 4E²sin²(θ/2); the order-α virtual correction multiplies the Born cross-section by 1 − (α/π)f_IR(q²)ln(−q²/μ_γ²), μ_γ being a photon mass, and the probability of emitting one soft photon with energy between μ_γ and E_l is (α/π)f_IR(q²)ln(E_l²/μ_γ²), where for −q² ≫ m², f_IR(q²) = ln(−q²/m²) − 1. Take m = 0.511 MeV.

1. (a)

   Justify Justify writing the on-shell vertex as Γ^μ = γ^μF₁(q²) + (iσ^μνq_ν/2m)F₂(q²).

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   1. 1

      Between on-shell spinors the vertex can be built only from γ^μ, p^μ and p′^μ, that is γ^μ, (p′ + p)^μ and q^μ, multiplied by functions of the one invariant q², since p² = p′² = m²; structures with γ⁵ are excluded because QED conserves parity, and the Gordon identity trades (p′ + p)^μ for γ^μ and iσ^μνq_ν

   2. 2

      A q^μ term is excluded by the Ward identity q_μΓ^μ = 0: q_μγ^μ vanishes between on-shell spinors by the Dirac equation and q_μσ^μνq_ν = 0 by antisymmetry, but q_μq^μ = q² does not, so its coefficient must vanish and two form factors remain

2. (b)

   Show that Show that in a slowly varying magnetic field the vertex gives the electron a magnetic moment with g = 2[F₁(0) + F₂(0)].

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   1. 1

      In a static vector potential with Fourier transform A^i(q) the amplitude contains ū(p′)Γ^iu(p)A^i(q); in the non-relativistic expansion of the spinors the spin-dependent part of ū(p′)γ^iu(p) is 2m ξ′†(−iε^ijkq^jσ^k/2m)ξ, through the lower components, and ū(p′)(iσ^iνq_ν/2m)u(p) gives the same structure, so the spin term is weighted by F₁(0) + F₂(0)

   2. 2

      Since ε^ijkq^jA^k(q) is, up to a factor ±i set by the Fourier convention, the transform of the magnetic field B = ∇ × A, the spin term is the Born amplitude of the energy −μ⋅B with μ = (e/m)[F₁(0) + F₂(0)]S, that is g = 2[F₁(0) + F₂(0)]; F₁(0) = 1 defines the charge, so g − 2 = 2F₂(0) and a_e = F₂(0)

3. (c)

   Derive Derive F₂(q²) as an integral over Feynman parameters, evaluate F₂(0), and explain why F₂ needs neither an ultraviolet nor an infrared regulator.

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   1. 1

      Only the σ^μν term of N^μ contributes to F₂, giving 2ie² × 2 × 2m²z(1 − z) × [−i/(2(4π)²Δ)] under the parameter integrals, that is F₂(q²) = (α/2π)∫dx dy dz δ(x + y + z − 1) 2m²z(1 − z)/[(1 − z)²m² − xyq²]

   2. 2

      At q² = 0 the integrand depends only on z, and ∫dx dy δ(x + y + z − 1) = 1 − z, so F₂(0) = (α/2π)∫₀¹(1 − z) × 2z(1 − z)/(1 − z)² dz = (α/2π)∫₀¹2z dz = α/2π

   3. 3

      The σ^μν term carries no ℓ² in its numerator, so ∫d⁴ℓ/D³ converges in the ultraviolet and F₂ has no counterterm; at z → 1 the factor z(1 − z) and the measure's 1 − z cancel the (1 − z)² of Δ, so the integral is finite without a photon mass: F₂(0) is a prediction

4. (d)

   Calculate Calculate the residual, the measured a_e minus the QED series truncated after one, two and three terms, and compare each residual with the size of the next term.

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   1. 1

      α/2π = 1.161 409 73 × 10⁻³, and a_e − α/2π = −1.7576 × 10⁻⁶; divided by (α/π)² = 5.3955 × 10⁻⁶ this is −0.326, close to C₂

   2. 2

      C₂(α/π)² = −1.7723 × 10⁻⁶, leaving +1.475 × 10⁻⁸; divided by (α/π)³ = 1.2533 × 10⁻⁸ this is 1.18, close to C₃

   3. 3

      C₃(α/π)³ = 1.4804 × 10⁻⁸, leaving −5.2 × 10⁻¹¹, the size of the four-loop term C₄(α/π)⁴ = −5.57 × 10⁻¹¹: each order shrinks the residual by a factor of order α/π and confirms the next coefficient

5. (e)

   Estimate Estimate the fractional precision with which this measurement of a_e would determine α if the QED series were known exactly, stating your assumption.

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   1. 1

      Assuming the theoretical series, hadronic and weak terms included, is exact, a_e ≈ α/2π to leading order, so δα/α ≈ δa_e/(α/2π) = 1.3 × 10⁻¹³/1.1614 × 10⁻³ = 1.1 × 10⁻¹⁰, about 0.1 parts per billion

6. (f)

   Show that Show that in the cross-section measured by a detector that cannot register photons below E_l the photon mass cancels at order α, leaving the factor 1 − (α/π)f_IR(q²)ln(−q²/E_l²).

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   1. 1

      The detector counts elastic events and events with one undetected photon below E_l together, so at order α the measured cross-section is the Born value times [1 − (α/π)f_IR ln(−q²/μ_γ²)] + (α/π)f_IR ln(E_l²/μ_γ²)

   2. 2

      Since ln(−q²/μ_γ²) − ln(E_l²/μ_γ²) = ln(−q²/E_l²), μ_γ cancels and the factor is 1 − (α/π)f_IR(q²)ln(−q²/E_l²): the infrared divergences of the virtual and real corrections cancel in the measured rate, as Bloch and Nordsieck showed

7. (g)

   Calculate Calculate this correction factor for 4.0 GeV electrons scattered through 20° with E_l = 20 MeV, and compare it with its exponentiated form exp[−(α/π)f_IR ln(−q²/E_l²)].

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   1. 1

      −q² = 4 × (4.0 GeV)² × sin²10° = 1.930 GeV², so ln(−q²/m²) = ln[1.930/(5.11 × 10⁻⁴)²] = 15.82 and f_IR = 14.82, and ln(−q²/E_l²) = ln(1.930/0.0004) = 8.48

   2. 2

      (α/π)f_IR ln(−q²/E_l²) = 0.0023228 × 14.82 × 8.48 = 0.292, so the order-α factor is 0.708 while the exponentiated form gives e^(−0.292) = 0.747: the correction is so large that the difference of about 0.04 between them, the α² terms, matters at this resolution

8. (h)

   Predict Predict what happens, as E_l → 0, to the rate of scattering with no photon above E_l and to the inclusive rate.

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   1. 1

      The exponentiated factor (−q²/E_l²)^(−(α/π)f_IR) tends to zero as E_l → 0, so scattering with no photon at all is impossible, every scattered electron radiating a cloud of soft photons; the inclusive rate, summed over all emissions, stays finite and close to the Born value