University Quantum Field Theory I · Midterm Examination

With ℏ = c = 1 the action is dimensionless, so in d spacetime dimensions the Lagrangian density has mass dimension d. The kinetic terms then give a real scalar field φ the mass dimension (d − 2)/2 and a Dirac field ψ the mass dimension (d − 1)/2. For which interaction is the coupling constant dimensionless, so that the theory is renormalisable by power counting, in d = 6?

1. A(λ/4!)φ⁴
2. B(g/6!)φ⁶
3. C(g/3!)φ³
4. Dgφψ̄ψ

The Gell-Mann–Low formula expresses a correlation function in the interacting vacuum |Ω⟩ through free fields: ⟨Ω|Tφ_H(x)φ_H(y)|Ω⟩ = ⟨0|T[φ_I(x)φ_I(y) exp(−i∫H_I dt)]|0⟩ / ⟨0|T[exp(−i∫H_I dt)]|0⟩, where |0⟩ is the free vacuum, φ_I and H_I are in the interaction picture, the time integrals run from −T to T, and the limit is taken as T → ∞(1 − iε). What does taking the limit along the slightly tilted direction T(1 − iε) accomplish?

1. AIt damps every eigenstate of the full Hamiltonian above the ground state by e^(−(E_n − E₀)εT) relative to |Ω⟩, so the long evolution projects |0⟩ onto |Ω⟩, provided ⟨Ω|0⟩ ≠ 0 and there is a gap above E₀.
2. BIt switches the interaction on and off slowly, so that |0⟩ evolves into |Ω⟩ with ⟨Ω|0⟩ = 1, and the denominator is then needed only to cancel the ultraviolet divergences of the numerator.
3. CIt makes the Dyson series converge by damping the high-frequency modes of H_I, while the projection of |0⟩ onto |Ω⟩ is supplied separately by normal ordering H_I.
4. DIt makes the contractions retarded rather than Feynman propagators, so that the formula respects causality, while the time ordering itself projects |0⟩ onto |Ω⟩.

In φ⁴ theory with ℒ_int = −(λ/4!)φ⁴, two connected vacuum diagrams appear at order λ². In diagram (i) two vertices are joined by four propagators. In diagram (ii) two vertices are joined by two propagators, and each vertex also carries one propagator that begins and ends on it. Each diagram contributes its spacetime integral times (−iλ)²/S. What are the symmetry factors S of (i) and (ii)?

1. A24 and 8
2. B24 and 16
3. C48 and 8
4. D48 and 16

Work in natural units ℏ = c = 1 with the metric (+, −, −, −). A real scalar field φ, canonically normalised so that [φ(x, t), π(y, t)] = iδ³(x − y), is interacting and invariant under φ → −φ. Its lightest particle has physical mass m, and the field can create states of any odd number of these particles. Its exact two-point function has the Källén–Lehmann form ∫₀^∞ (dμ²/2π) ρ(μ²) i/(p² − μ² + iε), with ρ(μ²) = 2πZδ(μ² − m²) plus a continuum. Which statement is correct?

1. AThe continuum begins at μ² = 4m², and Z < 1 because φ also creates two-particle states.
2. BThe continuum begins at μ² = 9m², ρ(μ²) ≥ 0, and the canonical commutator fixes the total weight ∫ρ dμ²/2π = 1, so 0 ≤ Z ≤ 1.
3. CThe continuum begins at μ² = 9m², and Z may exceed 1 if the interaction is attractive, because ρ is then negative above threshold.
4. DThe continuum begins at μ² = 3m², and Z = 1 exactly, because the canonical commutator fixes the residue of the pole.

Work in natural units ℏ = c = 1 with the metric (+, −, −, −). In the scalar Yukawa model, ℒ_int = −gψ†ψφ, with ψ a complex scalar of mass m_ψ and φ a real scalar of mass m_φ < 2m_ψ. To order g² the amplitude for ψ(p₁)ψ̄(p₂) → ψ(p₃)ψ̄(p₄) is M = −g²[1/(s − m_φ²) + 1/(t − m_φ²)], with s = (p₁ + p₂)², t = (p₁ − p₃)² and u = (p₁ − p₄)², the first term from annihilation through φ and the second from φ exchange. Defining s, t and u in the same way for ψ(p₁)ψ(p₂) → ψ(p₃)ψ(p₄), what is the order-g² amplitude for that process?

1. A−g²[1/(s − m_φ²) + 1/(t − m_φ²)]
2. B−g²[1/(t − m_φ²) − 1/(u − m_φ²)]
3. C−g²[1/(s − m_φ²) + 1/(t − m_φ²) + 1/(u − m_φ²)]
4. D−g²[1/(t − m_φ²) + 1/(u − m_φ²)]

Work in natural units ℏ = c = 1 with the metric (+, −, −, −). In φ⁴ theory with ℒ_int = −(λ/4!)φ⁴ and λ = 0.60, two particles collide at a centre-of-mass energy √s = 4.0 GeV, far enough above threshold that their mass can be neglected. At tree level the amplitude for φφ → φφ is M = −λ, and for a 2 → 2 process with equal masses in and out, dσ/dΩ = |M|²/(64π²s) in the centre-of-mass frame. Take (ℏc)² = 0.3894 GeV² mb. What is the total cross-section for φφ → φφ?

1. A87 nb
2. B174 nb
3. C14 nb
4. D0.15 nb

Work in natural units ℏ = c = 1 with the metric (+, −, −, −). Two identical Dirac fermions of mass m scatter, ψ(p₁)ψ(p₂) → ψ(p₃)ψ(p₄), through the Yukawa interaction ℒ_int = −gφψ̄ψ, where φ is a real scalar of mass m_φ. At tree level two diagrams contribute, one with momentum transfer squared t = (p₁ − p₃)² and one with u = (p₁ − p₄)². In the non-relativistic limit ū(p₃)u(p₁) → 2m times a Kronecker delta between the spin labels. Which statement is correct?

1. AThe two diagrams add with the same sign, as they would for bosons, because the particle exchanged between the fermion lines is the boson φ.
2. BThe two diagrams have a relative minus sign, which makes scalar exchange repulsive between two fermions although it attracts a fermion and an antifermion.
3. CThe two diagrams have a relative minus sign, because their final states differ by an exchange of anticommuting operators; the t-channel term alone is the Born amplitude of an attractive Yukawa potential.
4. DOnly the t-channel diagram contributes, because the u-channel diagram counts the same final state again and is removed by the 1/2 for identical particles.

Work in natural units ℏ = c = 1 with the metric (+, −, −, −). In Feynman gauge the four photon operators of one momentum k, taken along z, obey [a^λ(k), a^λ′†(k′)] = −g^(λλ′)(2π)³2|k|δ³(k − k′) for λ, λ′ = 0, 1, 2, 3, where λ = 0 labels the timelike polarisation ε⁰ = (1, 0, 0, 0), λ = 1, 2 the transverse ones and λ = 3 the longitudinal one, ε³ = (0, 0, 0, 1). The Gupta–Bleuler condition requires the positive-frequency part of ∂_μA^μ to annihilate every physical state, which for this momentum reads (a⁰ − a³)|ψ⟩ = 0. Which statement is correct?

1. AThe condition removes timelike and longitudinal photons from every physical state, which may contain only photons created by a¹† and a²†.
2. BPhysical states may contain (a⁰† − a³†) acting on a transverse state; such a state has zero norm and adds nothing to the energy or to any gauge-invariant expectation value.
3. CA longitudinal photon a³†|0⟩ has negative norm like the timelike one, and the condition admits only states in which the two negative norms cancel.
4. DThe full condition ∂_μA^μ|ψ⟩ = 0, creation part included, would serve equally well; only its positive-frequency part is kept, to simplify the algebra.

A free real scalar field of mass m is written, in natural units ℏ = c = 1 with the metric (+, −, −, −), as φ(x) = ∫ d³p/((2π)³2E_p) [a(p)e^(−ip⋅x) + a†(p)e^(ip⋅x)], where E_p = √(|p|² + m²), p⁰ = E_p in the exponents and [a(p), a†(q)] = (2π)³2E_p δ³(p − q). Its vacuum two-point function is D(x − y) = ⟨0|φ(x)φ(y)|0⟩ = ∫ d³p/((2π)³2E_p) e^(−ip⋅(x − y)). Take ℏc = 197.327 MeV fm.

1. (a)

   Derive Starting from the definition D_F(x − y) = ⟨0|Tφ(x)φ(y)|0⟩, derive the representation D_F(x − y) = ∫ d⁴p/(2π)⁴ i e^(−ip⋅(x − y))/(p² − m² + iε), evaluating the p⁰ integral with the contour closed in whichever half-plane each time ordering requires.

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   1. 1

      p² − m² + iε = (p⁰)² − E_p² + iε vanishes at p⁰ = ±(E_p − iε′), with ε′ = ε/(2E_p): the positive-energy pole lies just below the real axis and the negative-energy pole just above it

   2. 2

      for x⁰ > y⁰ the factor e^(−ip⁰(x⁰ − y⁰)) decays in the lower half-plane, so the contour is closed below, clockwise, enclosing p⁰ = E_p: ∫dp⁰/(2π) i e^(−ip⁰(x⁰ − y⁰))/[(p⁰ − E_p)(p⁰ + E_p)] = (−2πi/2π) × i e^(−iE_p(x⁰ − y⁰))/(2E_p) = e^(−iE_p(x⁰ − y⁰))/(2E_p), and the remaining d³p integral is D(x − y)

   3. 3

      for x⁰ < y⁰ the contour is closed above, enclosing p⁰ = −E_p, which gives e^(iE_p(x⁰ − y⁰))/(2E_p); relabelling p → −p turns the d³p integral into D(y − x), so the integral equals θ(x⁰ − y⁰)D(x − y) + θ(y⁰ − x⁰)D(y − x) = ⟨0|Tφ(x)φ(y)|0⟩

2. (b)

   Sketch The retarded propagator is D_R(x − y) = θ(x⁰ − y⁰)⟨0|[φ(x), φ(y)]|0⟩, whose momentum-space form is i/((p⁰ + iε)² − |p|² − m²). Sketch the complex p⁰-plane at fixed p, marking the poles of D_F and of D_R and the contour used when x⁰ < y⁰, and state where in spacetime each propagator is non-zero.

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   1. 1

      the sketch shows the real p⁰ axis with the poles of D_F at +E_p − iε′, below, and −E_p + iε′, above; the poles of D_R both below the axis, at ±E_p − iε′; and the contour for x⁰ < y⁰ closed by a large semicircle in the upper half-plane, enclosing the negative-energy pole of D_F and neither pole of D_R

   2. 2

      D_R is therefore zero for x⁰ < y⁰ and, being a commutator, also zero at spacelike separation, so it is supported on and inside the forward light cone; D_F is non-zero for both time orders and at spacelike separation, where it falls off as e^(−mr), r being the invariant spacelike distance

3. (c)

   Calculate Two static point sources, j(x) = g[δ³(x − x₁) + δ³(x − x₂)], coupled through ℒ_int = jφ and held a distance r = |x₁ − x₂| apart for a long time T, give a vacuum persistence amplitude ⟨0|0⟩ ∝ e^(−iE(r)T), with interaction energy E(r) = −(g²/(4πr))e^(−mr). For a scalar of mass m = 550 MeV, g²/(4π) = 0.80 and r = 0.70 fm, calculate E(r) in MeV and state the range of the force in femtometres.

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   1. 1

      in natural units r = 0.70 fm/(197.327 MeV fm) = 3.547 × 10⁻³ MeV⁻¹, so 1/r = 281.9 MeV, and mr = 550 × 0.70/197.327 = 1.951, with e^(−1.951) = 0.1421

   2. 2

      E = −0.80 × 281.9 MeV × 0.1421 = −32.1 MeV, negative because scalar exchange between like sources attracts; the range is 1/m = 197.327 MeV fm/550 MeV = 0.359 fm

4. (d)

   Explain D_F(x − y) is non-zero when x − y is spacelike. Explain why this does not allow a signal to travel faster than light.

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   1. 1

      whether a measurement at y can affect one at x is decided by the commutator [φ(x), φ(y)] = D(x − y) − D(y − x), not by an amplitude; for spacelike x − y a proper Lorentz transformation carries x − y into −(x − y) and leaves D invariant, so the commutator vanishes, the amplitude for propagation from y to x cancelling that from x to y, while D_F, an amplitude, need not vanish

Two real scalar fields, φ of mass m_φ and χ of mass m_χ, live in four spacetime dimensions and interact through ℒ_int = −(g/2)φχ² − (κ/2)φ∂_μχ∂^μχ, in natural units ℏ = c = 1 with the metric (+, −, −, −). Treat both couplings to the lowest order at which each process occurs. For the two-body decay of a particle of mass m_φ into particles of momentum p in its rest frame, Lorentz-invariant phase space gives ∫dΠ₂ = p/(4πm_φ) when the two final particles are distinguishable, and Γ = (1/(2m_φ))∫|M|²dΠ₂. Take ℏ = 6.582 × 10⁻²⁵ GeV s.

1. (a)

   Determine Determine the Feynman rule for the φχχ vertex, taking every momentum as incoming and p₂ and p₃ as the momenta of the two χ lines, and give the mass dimensions of g and κ.

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   1. 1

      the term −(g/2)φχ² gives i × (−g/2) × 2 = −ig, the 2 counting the ways the two χ fields meet the two χ lines; [g] = 4 − 3 = 1, a super-renormalisable coupling

   2. 2

      each ∂_μ acting on an incoming χ line of momentum p gives −ip_μ, so −(κ/2)φ∂_μχ∂^μχ gives i × (−κ/2) × 2 × (−ip₂)⋅(−ip₃) = iκ p₂⋅p₃; the full vertex is −ig + iκ p₂⋅p₃, with [κ] = 4 − 1 − 2 − 2 = −1, a non-renormalisable coupling

2. (b)

   Show that Set κ = 0. By counting the Wick contractions in the order-g² term of the Gell-Mann–Low numerator, show that the one-loop φ self-energy, a loop of two χ propagators, carries a symmetry factor of 2, and that the one-loop χ self-energy, a loop of one φ and one χ propagator, carries a symmetry factor of 1. Check each by counting the diagram's automorphisms.

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   1. 1

      the order-g² term is (1/2!)(−ig/2)²∫∫φ(z)χ(z)²φ(w)χ(w)², a prefactor (−ig)²/8; in ⟨0|Tφ(x)φ(y)…|0⟩, φ(x) joins φ(z) or φ(w) (2 ways), φ(y) the other, and the two χ(z) fields pair with the two χ(w) fields in 2 ways, so 4 contractions give (−ig)² × 4/8 = (−ig)²/2: S = 2, matching the one automorphism that swaps the two χ lines

   2. 2

      in ⟨0|Tχ(x)χ(y)…|0⟩, χ(x) joins any of the four χ fields (4 ways), χ(y) either χ field at the other vertex (2 ways), and the remaining χ and φ fields pair uniquely, so 8 contractions give (−ig)² × 8/8: S = 1, as the loop's two lines, one φ and one χ, cannot be exchanged

3. (c)

   Determine Still with κ = 0, determine the tree-level amplitude M for χ(p₁)χ(p₂) → χ(p₃)χ(p₄) in terms of the Mandelstam variables s, t and u, check its symmetry under t ↔ u and the value of s + t + u, and say why √Z = 1 may be used on each external leg at this order.

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   1. 1

      φ is exchanged in all three channels, each giving (−ig)² × i/(q² − m_φ²), so M = −g²[1/(s − m_φ²) + 1/(t − m_φ²) + 1/(u − m_φ²)], dimensionless as a 2 → 2 amplitude in four dimensions must be, since [g²] = 2

   2. 2

      M is symmetric under t ↔ u, as identical bosons in the final state require, and s + t + u = 4m_χ²; because Z = 1 + O(g²) and the mass shift is O(g²), √Z on each leg and the physical mass change M only at order g⁴

4. (d)

   Predict The masses are now chosen so that m_φ > 2m_χ. Predict what becomes of the s-channel term of the amplitude in (c) as s approaches m_φ², and of the φ particle's place in the LSZ formula, giving the reason, and say whether the t- and u-channel poles are affected.

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   1. 1

      the pole at s = m_φ² now lies inside the physical region s ≥ 4m_χ², so the tree amplitude diverges there; the φ self-energy acquires an imaginary part from the open χχ channel, and resumming it turns 1/(s − m_φ²) into 1/(s − m_φ² + im_φΓ), a Breit–Wigner peak of finite height

   2. 2

      φ is no longer stable, so it is not an asymptotic state: its pole leaves the real axis, LSZ's isolated one-particle pole is lost and φ appears only as a resonance in χχ scattering; the t- and u-channel poles at t = m_φ² and u = m_φ² stay outside the physical region, where t ≤ 0 and u ≤ 0, and still describe a force of range 1/m_φ

5. (e)

   Calculate Restore κ. With m_φ = 0.90 GeV, m_χ = 0.14 GeV, g = 0.50 GeV and κ = 0.80 GeV⁻¹, calculate the decay rate Γ(φ → χχ) in MeV and the lifetime of φ in seconds.

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   1. 1

      on shell 2p₂⋅p₃ = m_φ² − 2m_χ², so M = −g + κ(m_φ² − 2m_χ²)/2 = −0.500 GeV + 0.308 GeV = −0.192 GeV, and the χ momentum is p = √(m_φ²/4 − m_χ²) = 0.428 GeV

   2. 2

      with 1/2 for two identical χ, Γ = (1/(2m_φ)) × |M|² × ½ × p/(4πm_φ) = |M|²p/(16πm_φ²) = 3.86 × 10⁻⁴ GeV = 0.386 MeV, and τ = ℏ/Γ = 1.71 × 10⁻²¹ s

A real spin-0 boson φ of mass m_φ couples to a Dirac fermion ψ of mass m either as a scalar, ℒ_S = −gφψ̄ψ, or as a pseudoscalar, ℒ_P = −igφψ̄γ⁵ψ, in natural units ℏ = c = 1 with the metric (+, −, −, −). Use the spin sums Σu(k)ū(k) = k̸ + m and Σv(k)v̄(k) = k̸ − m; the identities tr 1 = 4 and tr(a̸b̸) = 4a⋅b; the vanishing of the trace of an odd number of γ matrices; γ⁵γ^μ = −γ^μγ⁵, (γ⁵)² = 1 and γ⁰(γ⁵)†γ⁰ = −γ⁵. For the decay φ → ψ(k₁)ψ̄(k₂) write β = √(1 − 4m²/m_φ²), the speed of either fermion in the φ rest frame; then ∫dΠ₂ = β/(8π), and Γ = (1/(2m_φ)) × (spin-summed |M|²) × ∫dΠ₂. Take ℏ = 6.582 × 10⁻²⁵ GeV s.

1. (a)

   Derive For the scalar coupling the amplitude is M_S = −g ū(k₁)v(k₂). Using Casimir's trick with the spin sums, derive the spin-summed |M_S|² in terms of g, m_φ and β.

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   1. 1

      (ū v)* = v̄ u, so Σ|M_S|² = g² Σ ū(k₁)v(k₂)v̄(k₂)u(k₁) = g² tr[(k̸₁ + m)(k̸₂ − m)] = g²(4k₁⋅k₂ − 4m²), the terms linear in m containing an odd number of γ matrices

   2. 2

      on shell 2k₁⋅k₂ = m_φ² − 2m², so Σ|M_S|² = 2g²(m_φ² − 4m²) = 2g²m_φ²β²

2. (b)

   Show that For the pseudoscalar coupling the amplitude is M_P = −g ū(k₁)iγ⁵v(k₂). Show that the spin-summed |M_P|² is 2g²m_φ², with no factor of β².

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   1. 1

      (ū iγ⁵ v)* = v̄ γ⁰(iγ⁵)†γ⁰ u = v̄(−i)(−γ⁵)u = v̄ iγ⁵ u, so Σ|M_P|² = g² tr[(k̸₁ + m)iγ⁵(k̸₂ − m)iγ⁵] = −g² tr[(k̸₁ + m)γ⁵(k̸₂ − m)γ⁵]

   2. 2

      anticommuting γ⁵ through, γ⁵(k̸₂ − m)γ⁵ = −k̸₂ − m, so Σ|M_P|² = g² tr[(k̸₁ + m)(k̸₂ + m)] = g²(4k₁⋅k₂ + 4m²) = 2g²m_φ²: the γ⁵ reverses the sign of the mass term

3. (c)

   Calculate For m_φ = 1.50 GeV, m = 0.60 GeV and g = 0.30, calculate Γ(φ → ψψ̄) and the φ lifetime in seconds for each coupling.

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   1. 1

      β = √(1 − 4 × 0.36/2.25) = 0.600, and Γ_S = (1/(2m_φ)) × 2g²m_φ²β² × β/(8π) = g²m_φβ³/(8π) = 1.16 × 10⁻³ GeV = 1.16 MeV, so τ_S = ℏ/Γ_S = 5.67 × 10⁻²² s

   2. 2

      Γ_P = g²m_φβ/(8π) = 3.22 MeV and τ_P = 2.04 × 10⁻²² s; the ratio Γ_P/Γ_S = 1/β² = 2.78

4. (d)

   Explain Explain the powers β³ and β in the two rates by the parity of φ and the orbital angular momentum of the fermion pair near threshold.

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   1. 1

      a fermion and its antifermion have opposite intrinsic parities, so a pair with orbital angular momentum L has parity (−1)^(L+1); total J = 0 requires the pair's spin S to equal L, so S and L are both 0 or both 1

   2. 2

      the scalar coupling makes φ a 0⁺ state, which needs L odd, the ³P₀ pair with L = S = 1, and a P-wave rate carries p^(2L+1) ∝ β³; the pseudoscalar is 0⁻, which allows the ¹S₀ pair with L = 0 and a rate ∝ β, so near threshold the pseudoscalar decays faster by 1/β²

5. (e)

   Justify In the one-loop φ self-energy built from a ψ loop, Wick's theorem supplies a factor −1. Justify this factor from the anticommutation of the fields, and confirm it from the optical theorem, which requires Im M(φ → φ) = m_φΓ(φ → ψψ̄) once m_φ > 2m.

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   1. 1

      the two vertices supply ψ̄(x)ψ(x)ψ̄(y)ψ(y); writing the contractions in the order of a trace, ψ(x)ψ̄(y)ψ(y)ψ̄(x) → tr[S_F(x − y)S_F(y − x)], means moving ψ̄(x) from the front to the back past three anticommuting fields, which gives (−1)³ = −1

   2. 2

      on the cut the fermion carries k₁ and the antifermion line carries −k₂, so the numerator trace is tr[(k̸₁ + m)(−k̸₂ + m)] = −tr[(k̸₁ + m)(k̸₂ − m)] = −Σ|M_S|²/g²; only the loop's −1 makes 2 Im M(φ → φ) = ∫dΠ₂Σ|M_S|² = 2m_φΓ positive, as unitarity demands

Real scalar φ⁴ theory, ℒ = ½∂_μφ∂^μφ − ½m²φ² − (λ/4!)φ⁴, in natural units ℏ = c = 1 with the metric (+, −, −, −), is regularised dimensionally in d = 4 − 2ε dimensions, with λ replaced by λμ^(2ε) so that λ stays dimensionless, μ being an arbitrary scale. Use the Feynman-parameter identity 1/(AB) = ∫₀¹dx [xA + (1 − x)B]^(−2), the Euclidean master integral ∫d^dℓ_E/(2π)^d (ℓ_E² + Δ)^(−2) = Γ(2 − d/2)(4π)^(−d/2)Δ^(d/2 − 2), and Γ(ε) = 1/ε − γ_E + O(ε). At one loop the field-strength counterterm vanishes, so the bare coupling is λ₀ = μ^(2ε)(λ + δ_λ), where the counterterm vertex is −iδ_λ.

1. (a)

   Show that The s-channel one-loop diagram for φφ → φφ, with incoming momenta p₁ and p₂, p = p₁ + p₂ and s = p², carries a symmetry factor of 2. Show that, below the threshold s = 4m², its contribution to iM, with an overall factor μ^(2ε) removed, is (iλ²/(32π²))∫₀¹dx [1/ε − γ_E + ln 4π − ln(Δ_s/μ²)] + O(ε), where Δ_s = m² − x(1 − x)s.

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   1. 1

      two vertices, each −iλμ^(2ε), two propagators and the factor ½ give iM_s = ½(−iλμ^(2ε))²∫d^dℓ/(2π)^d [i/(ℓ² − m²)][i/((ℓ + p)² − m²)] = ½λ²μ^(4ε)∫d^dℓ/(2π)^d 1/[(ℓ² − m²)((ℓ + p)² − m²)]

   2. 2

      Feynman parameters turn the denominator into [(ℓ + xp)² − Δ_s]² with Δ_s = m² − x(1 − x)s; shifting ℓ → ℓ − xp and rotating ℓ⁰ = iℓ_E⁰, so that ℓ² = −ℓ_E² and d^dℓ = i d^dℓ_E, gives the loop integral as i∫₀¹dx Γ(ε)(4π)^(−2+ε)Δ_s^(−ε)

   3. 3

      expanding, (4π)^(−2+ε) = (1 + ε ln 4π)/(16π²) and μ^(2ε)Δ_s^(−ε) = 1 − ε ln(Δ_s/μ²), so iM_s = μ^(2ε)(iλ²/(32π²))∫₀¹dx [1/ε − γ_E + ln 4π − ln(Δ_s/μ²)] + O(ε): one μ^(2ε) belongs with the tree-level coupling and the other makes the logarithm dimensionless

2. (b)

   Show that Including the t- and u-channel diagrams and the counterterm, the renormalised coupling λ is defined by the condition M = −λ at s = 4m², t = u = 0. Show that M(s, t, u) = −λ − (λ²/(32π²))Σ∫₀¹dx ln[Δ(q²)/Δ(q₀²)], the sum running over the three channels q² = s, t, u with reference values q₀² = 4m², 0, 0 and Δ(q²) = m² − x(1 − x)q², and that neither ε nor μ survives in it.

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   1. 1

      at one loop M = −λ + (λ²/(32π²))Σ∫dx[1/ε − γ_E + ln 4π − ln(Δ(q²)/μ²)] − δ_λ, and the condition at the reference point fixes δ_λ = (λ²/(32π²))Σ∫dx[1/ε − γ_E + ln 4π − ln(Δ(q₀²)/μ²)], which carries the whole pole

   2. 2

      subtracting, 1/ε, γ_E, ln 4π and μ² cancel channel by channel, leaving M = −λ − (λ²/(32π²))Σ∫dx ln[Δ(q²)/Δ(q₀²)]: finite, independent of the regulator and of μ, and a relation between the amplitude at (s, t, u) and the coupling measured at the reference point

3. (c)

   Derive In minimal subtraction δ_λ contains only the poles of the three channels. Derive the one-loop beta function β(λ) = μ dλ/dμ from the μ-independence of the bare coupling λ₀.

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   1. 1

      each channel contributes a pole λ²/(32π²ε), so δ_λ = 3λ²/(32π²ε), and μ dλ₀/dμ = 0 gives 0 = 2ε(λ + 3λ²/(32π²ε)) + β(1 + 6λ/(32π²ε))

   2. 2

      solving to order λ², β = −2ελ − 6λ²/(32π²) + 12λ²/(32π²) = −2ελ + 3λ²/(16π²), so as ε → 0, β(λ) = 3λ²/(16π²), positive: the coupling grows with the scale

4. (d)

   Calculate In a model whose coupling is λ(μ₀) = 1.5 at μ₀ = 1.0 TeV, calculate the scale at which the one-loop running coupling has doubled and the scale of its Landau pole, both in GeV.

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   1. 1

      integrating dλ/λ² = (3/(16π²))d ln μ gives 1/λ(μ) = 1/λ(μ₀) − (3/(16π²))ln(μ/μ₀); λ = 3.0 when ln(μ/μ₀) = (16π²/3)(1/1.5 − 1/3.0) = 17.55, that is at μ = 1000 GeV × e^17.55 = 4.2 × 10¹⁰ GeV

   2. 2

      1/λ vanishes when ln(μ_L/μ₀) = 16π²/(3 × 1.5) = 35.09, so the Landau pole is at μ_L = 1000 GeV × e^35.09 = 1.7 × 10¹⁸ GeV

5. (e)

   Estimate Estimate the highest scale at which this one-loop running can be trusted, stating the criterion you use, and say what the Landau pole of (d) then tells you about the theory.

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   1. 1

      a stated criterion, for example λ(μ) ≤ 4π, where the one-loop term 3λ/(16π²) has grown to about 0.24: ln(μ/μ₀) = (16π²/3)(1/1.5 − 1/(4π)) = 30.9, so μ ≈ 3 × 10¹⁶ GeV; whatever the criterion λ ≤ λ_max, the trusted range stops short of the pole by ln(μ_L/μ) = 16π²/(3λ_max), which is 4π/3 = 4.19 for λ_max = 4π, a factor e^(4.19) ≈ 66 in scale and small beside the 15 orders of magnitude between μ₀ and μ_L

   2. 2

      the pole lies beyond the scale at which the formula that predicts it holds, so one-loop running does not establish a singularity; it signals strong coupling, and with the non-perturbative evidence for triviality it means φ⁴ makes sense only as an effective theory with a cutoff below the pole

Two distinct spinless particles, a of mass 0.400 GeV and b of mass 0.600 GeV, scatter elastically, a + b → a + b, and near a centre-of-mass energy of 1.85 GeV the cross-section is dominated by an s-channel resonance X of mass m_X and total width Γ. Parts (a) to (d) treat X as a spin-0 particle coupled to ab through ℒ_int = −gXab, which gives Γ(X → ab) = g²p/(8πm_X²), where p is the momentum of a or b in the centre-of-mass frame at √s = m_X; near the peak the amplitude is M = −g²/(s − m_X² + im_XΓ). For two particles in and two out with these masses, σ = |M|²/(16πs). To the accuracy of these data the line shape is σ(√s) = σ_peak(Γ²/4)/[(√s − m_X)² + Γ²/4]. The optical theorem reads Im M(forward) = 2√s p σ_tot. Work in natural units with the metric (+, −, −, −), and take (ℏc)² = 0.3894 GeV² mb and ℏ = 6.582 × 10⁻²⁵ GeV s. The table gives the measured elastic cross-section at seven energies.

1. (a)

   Show that Show that σ_peak = (4π/p²)B², where B = Γ(X → ab)/Γ is the branching fraction to ab, so that 4π/p² is the largest cross-section an s-wave resonance can give in this channel.

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   1. 1

      at s = m_X², |M|² = g⁴/(m_X²Γ²), so σ_peak = g⁴/(16πm_X⁴Γ²)

   2. 2

      with g² = 8πm_X²Γ(X → ab)/p, g⁴ = 64π²m_X⁴Γ(X → ab)²/p², so σ_peak = (4π/p²)[Γ(X → ab)/Γ]² = (4π/p²)B², which reaches 4π/p² only for a purely elastic resonance, B = 1

2. (b)

   Determine By plotting a suitable linear function of the data, determine m_X, Γ and σ_peak.

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   1. 1

      the readings are symmetric about 1.850 GeV within their uncertainties, so m_X = 1.850 GeV, and the line shape gives 1/σ = (1/σ_peak)[1 + 4(√s − m_X)²/Γ²], linear in x = (√s − m_X)²

   2. 2

      1/σ at x = 0, 64, 256 and 576 MeV², averaging each symmetric pair, is 0.769, 1.239, 2.751 and 5.157 mb⁻¹; a straight-line fit gives an intercept of 0.768 mb⁻¹ and a slope of 7.64 × 10⁻³ mb⁻¹ MeV⁻²

   3. 3

      σ_peak = 1/(0.768 mb⁻¹) = 1.30 mb, and Γ = 2√(intercept/slope) = 2√(0.768/0.00764) MeV = 20.1 MeV, accepting 19.5 to 20.5 MeV

3. (c)

   Calculate Calculate p at √s = m_X and the s-wave unitarity limit 4π/p² in millibarns, and hence B and the lifetime of X in seconds.

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   1. 1

      p = √[(s − (m_a + m_b)²)(s − (m_a − m_b)²)]/(2√s) = √[(3.4225 − 1.0000)(3.4225 − 0.0400)]/3.700 GeV = 0.774 GeV, so 4π/p² = 21.0 GeV⁻² = 21.0 × 0.3894 mb = 8.18 mb

   2. 2

      B = √(σ_peak/(4π/p²)) = √(1.30/8.18) = 0.399, and τ = ℏ/Γ = 6.582 × 10⁻²⁵ GeV s/0.0201 GeV = 3.3 × 10⁻²³ s

4. (d)

   Determine Use the optical theorem to determine the total and the inelastic cross-sections at the peak, and determine Γ(X → ab) and g.

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   1. 1

      at the peak M = ig²/(m_XΓ), so Im M = g²/(m_XΓ) = 2m_Xpσ_tot gives σ_tot = g²/(2m_X²pΓ) = (4π/p²)B = 8.18 mb × 0.399 = 3.26 mb, and σ_inel = σ_tot − σ_peak = 1.96 mb

   2. 2

      Γ(X → ab) = BΓ = 0.399 × 20.1 MeV = 8.0 MeV, and g² = 8πm_X²Γ(X → ab)/p = 8π × 3.4225 GeV² × 0.0080 GeV/0.774 GeV = 0.89 GeV², so g = 0.94 GeV

5. (e)

   Evaluate A separate experiment counts the decays of X and finds B = 0.43 ± 0.05. For a resonance of spin J the peak elastic cross-section is (2J + 1)(4π/p²)B². Evaluate whether the data support J = 0 against J = 1, and say what further measurement on the same events would settle the spin directly.

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   1. 1

      the measured peak requires B = √(1.30/8.18) = 0.40 for J = 0 and B = √(1.30/(3 × 8.18)) = 0.23 for J = 1; against 0.43 ± 0.05 these lie 0.6σ and 4.0σ away, so J = 0 is supported and J = 1 excluded at about 4σ

   2. 2

      the angular distribution in the centre-of-mass frame settles it directly: a spin-0 resonance decays isotropically, so dσ/dΩ is flat at the peak, while spin 1 gives dσ/dΩ ∝ [P₁(cos θ)]² = cos²θ

The free electromagnetic field, in natural units ℏ = c = 1 and Heaviside–Lorentz units with e² = 4πα and the metric (+, −, −, −), has the Maxwell Lagrangian ℒ = −¼F_μνF^(μν), with F_μν = ∂_μA_ν − ∂_νA_μ. A complex scalar φ of mass m and charge e couples to it through ℒ_φ = (D_μφ)*(D^μφ) − m²φ*φ with D_μ = ∂_μ + ieA_μ, whose Feynman rules are a vertex −ie(p + p′)^μ, with p and p′ the incoming and outgoing scalar momenta, and a two-photon seagull vertex 2ie²g^(μν). Write Π^μ for the momentum conjugate to A_μ. Take α = 1/137.036, (ℏc)² = 0.3894 GeV² mb, and m = 139.570 MeV, the mass of the charged pion, in (f).

1. (a)

   Show that Show that the Maxwell Lagrangian gives A₀ no conjugate momentum, so that canonical quantisation of all four components fails, and that adding the gauge-fixing term −(1/(2ξ))(∂_μA^μ)² supplies one.

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   1. 1

      Π^μ = ∂ℒ/∂(∂₀A_μ) = −F^(0μ), and F^(00) = 0 by antisymmetry, so Π⁰ = 0 identically and [A₀(x), Π⁰(y)] = iδ³(x − y) cannot be imposed: A₀ enters as a Lagrange multiplier enforcing Gauss's law, not as a dynamical field

   2. 2

      with the gauge-fixing term Π^μ = −F^(0μ) − (1/ξ)g^(μ0)∂_νA^ν, so Π⁰ = −(1/ξ)∂_νA^ν, which is non-zero and makes A₀ dynamical

2. (b)

   Derive From the Euler–Lagrange equations, derive the field equation of the gauge-fixed Lagrangian and, from its quadratic action in momentum space, the photon propagator in a general covariant gauge.

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   1. 1

      ∂ℒ/∂(∂_μA_ν) = −F^(μν) − (1/ξ)g^(μν)∂⋅A, so ∂_μ[∂ℒ/∂(∂_μA_ν)] = 0 gives ∂_μF^(μν) + (1/ξ)∂^ν(∂⋅A) = 0, that is ∂²A^ν − (1 − 1/ξ)∂^ν(∂⋅A) = 0, so that at ξ = 1 each component obeys ∂²A^ν = 0

   2. 2

      integrating by parts, the action is ½∫A_μ[g^(μν)∂² − (1 − 1/ξ)∂^μ∂^ν]A_ν d⁴x, whose momentum-space kernel is −k²g^(μν) + (1 − 1/ξ)k^μk^ν, and the propagator must satisfy [−k²g^(μν) + (1 − 1/ξ)k^μk^ν]D_νρ(k) = iδ^μ_ρ

   3. 3

      the ansatz D_νρ = (−i/k²)[g_νρ + c k_νk_ρ/k²] leaves a k^μk_ρ term that cancels only if (1 − 1/ξ)(1 + c) = c, that is c = ξ − 1, so D_μν(k) = −i[g_μν − (1 − ξ)k_μk_ν/k²]/(k² + iε), reducing to −ig_μν/(k² + iε) in Feynman gauge

3. (c)

   Show that Two conserved currents, k_μj₁^μ(k) = k_μj₂^μ(k) = 0, exchange a photon with amplitude proportional to j₁^μD_μν(k)j₂^ν. Show that ξ drops out and, taking k along z with |k| = κ, that the exchange splits into an instantaneous Coulomb term i j₁⁰j₂⁰/κ² and a transverse term i j₁ᵀ⋅j₂ᵀ/(k² + iε), where jᵀ is the part of the three-vector current perpendicular to k.

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   1. 1

      the ξ-dependent term carries k_μj₁^μ and k_νj₂^ν, both zero, so j₁^μD_μνj₂^ν = −i j₁⋅j₂/(k² + iε) in every covariant gauge

   2. 2

      with k^μ = (k⁰, 0, 0, κ), conservation reads k⁰j⁰ = κj³, so j³ = (k⁰/κ)j⁰ and j₁⋅j₂ = j₁⁰j₂⁰[1 − (k⁰)²/κ²] − j₁ᵀ⋅j₂ᵀ = −(k²/κ²)j₁⁰j₂⁰ − j₁ᵀ⋅j₂ᵀ

   3. 3

      hence −i j₁⋅j₂/k² = i j₁⁰j₂⁰/κ² + i j₁ᵀ⋅j₂ᵀ/(k² + iε): the first term does not depend on k⁰, so it acts instantaneously and, as 1/κ², Fourier-transforms to the Coulomb interaction 1/(4πr); only the second has the photon pole, carried by the two transverse polarisations, which is Coulomb gauge recovered

4. (d)

   Compare A massive vector field of mass M_V, the Proca field, has the propagator −i[g_μν − k_μk_ν/M_V²]/(k² − M_V² + iε). Compare its polarisations and propagator with the photon's, and explain from the little group of a massless particle why the photon must couple to a conserved current while the Proca field need not.

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   1. 1

      a massive vector carries the three polarisations of spin 1 under its little group SO(3), with Σε_με_ν* = −g_μν + k_μk_ν/M_V², and its propagator has no gauge parameter; its k_μk_ν/M_V² term grows with energy unless it meets conserved currents, when it drops out as the photon's ξ term does

   2. 2

      a massless particle's little group is ISO(2), whose translations leave the helicity ±1 unchanged but shift ε^μ → ε^μ + ck^μ, so ε^μ is a four-vector only up to a multiple of k^μ and amplitudes are Lorentz invariant only if k_μM^μ = 0 for every external photon, which requires coupling to a conserved current; no such shift exists when M_V ≠ 0

5. (e)

   Show that For Compton scattering on the charged scalar, γ(k₁) + φ(p) → γ(k₂) + φ(p′), write iM^(μν), with μ belonging to the incoming photon and ν to the outgoing one, as the sum of the s-channel, u-channel and seagull diagrams. Show that k₁_μM^(μν) = 0 for the sum, but not for any one diagram or for the two exchange diagrams alone.

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   1. 1

      iM^(μν) = (−ie)²[(2p + k₁)^μ(2p′ + k₂)^ν i/(s − m²) + (2p′ − k₁)^μ(2p − k₂)^ν i/(u − m²)] + 2ie²g^(μν) = −ie²[(2p + k₁)^μ(2p′ + k₂)^ν/(s − m²) + (2p′ − k₁)^μ(2p − k₂)^ν/(u − m²) − 2g^(μν)], with s = (p + k₁)² and u = (p − k₂)²

   2. 2

      with k₁² = 0, k₁⋅(2p + k₁) = 2p⋅k₁ = s − m² and k₁⋅(2p′ − k₁) = 2p′⋅k₁ = m² − u, so contracting with k₁_μ cancels each propagator: the s-channel gives −ie²(2p′ + k₂)^ν, the u-channel +ie²(2p − k₂)^ν and the seagull +2ie²k₁^ν, each non-zero

   3. 3

      the two exchange diagrams together give −2ie²(p′ − p + k₂)^ν = −2ie²k₁^ν, using p + k₁ = p′ + k₂, which is not zero; the seagull's +2ie²k₁^ν cancels it exactly, so k₁_μM^(μν) = 0 holds only for the sum of all three

6. (f)

   Calculate In the rest frame of the initial scalar choose both photon polarisations with ε⁰ = 0, so that p⋅ε = k⋅ε = 0 for each photon. For photon energies ω ≪ m, where dσ/dΩ = |M|²/(64π²m²), calculate the total cross-section for a point-like scalar with the charged pion's mass, in microbarns, averaging over initial and summing over final polarisations.

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   1. 1

      with p = (m, 0) and these polarisations, (2p + k₁)⋅ε₁ = 0 and (2p − k₂)⋅ε₂* = 0, so only the seagull survives: |M|² = 4e⁴|ε₁⋅ε₂*|², dσ/dΩ = (α/m)²|ε₁⋅ε₂*|², and the polarisation average gives (α/m)²(1 + cos²θ)/2

   2. 2

      integrating over angles, σ = (8π/3)(α/m)² = (8π/3)[1/(137.036 × 0.13957 GeV)]² = 0.0229 GeV⁻², and 0.0229 × 0.3894 mb = 8.92 μb: Thomson's cross-section with the pion's mass in place of the electron's