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The gamma matrix trace identities kill the terms with an odd number of gamma matrices. The remaining three two-momentum terms and the term with three factors of

m

are

\frac{T _{μν}}{4 m} = m^{2} g_{μν} + l_{μ} (l - p_{2})_{ν} + (l - p_{2})_{μ} l_{ν} - g_{μν} l \cdot (l - p_{2}) + l_{μ} (l + p_{1})_{ν} + (l + p_{1})_{μ} l_{ν} - g_{μν} l \cdot (l + p_{1}) + g_{μν} (l - p_{2}) \cdot (l + p_{1}) - (l - p_{2})_{μ} (l + p_{1})_{ν} + (l + p_{1})_{μ} (l - p_{2})_{ν} .

(1)

In the

g_{μν}

coefficient the terms linear in

l \cdot p_{i}

cancel, leaving

m^{2} - l^{2} - s /2

. Among the other terms, the surviving mixed products of

l

and

p_{i}

also cancel. Consequently

T_{μν} ≃ 4 m [4 l_{μ} l_{ν} + p_{2, μ} p_{1, ν} + (m^{2} - s /2 - l^{2}) g_{μν}] .

(2)

In the second Feynman diagram, exchange the external labels and change the dummy loop momentum

l \mapsto - l

. Its projected numerator and denominator become identical to those of the first Feynman diagram. Hence

M_{μν}^{'} = 8 i N_{c} y e^{2} Q^{2} m \int \frac{d ^{4} l}{( 2 π ) ^{4}} \frac{4 l _{μ} l _{ν} + p _{2, μ} p _{1, ν} + ( m ^{2} - s /2 - l ^{2} ) g _{μν}}{D _{0} D _{1} D _{2}} .

(3)

With the common factor

8 m

extracted, the primitive integer coefficients are

A = 4, B = 1, C = 1, prefactor = 8 i N_{c} \frac{λ _{t}}{2} e^{2} e_{t}^{2} m_{t} .

(4)

Equivalently, extracting only

m

gives

(A, B, C) = (32, 8, 8)

and prefactor

i N_{c} λ_{t} e^{2} e_{t}^{2} m_{t} / 2

. A common rescaling of all three coefficients changes the prefactor inversely. The physical scattering amplitude is

ϵ_{1}^{μ} ϵ_{2}^{ν} M_{μν}^{'}

; the projected integrand is not an identity for an arbitrary uncontracted off-shell tensor.

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Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 / 2 / b / ii / Solution 2026-10-25

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Label the four-momenta as

e^{-} (p_{1}) μ^{-} (p_{2}) \to e^{-} (p_{3}) μ^{-} (p_{4})

. There is one tree-level Feynman diagram, the

t

-channel photon exchange shown in the lower-left panel. The two lepton species are distinct, so there is no exchanged-final-lepton diagram. With

s = (p_{1} + p_{2})^{2}

t = (p_{1} - p_{3})^{2}

and

u = (p_{1} - p_{4})^{2}

, the Feynman rules give

i M = \frac{i e ^{2}}{t} [\overset{u}{ˉ} (p_{3}) γ^{μ} u (p_{1})] [\overset{u}{ˉ} (p_{4}) γ_{μ} u (p_{2})] .

(1)

The massless fermion spin sum is

\sum_{r} u_{r} (p) \overset{u}{ˉ}_{r} (p) = \neq p

. Averaging the two initial spins of each lepton supplies

1/4

, so

⟨ ∣ M ∣^{2} ⟩ = \frac{e ^{4}}{4 t ^{2}} tr (\neq p_{3} γ^{μ} \neq p_{1} γ^{ν}) tr (\neq p_{4} γ_{μ} \neq p_{2} γ_{ν}) .

(2)

The gamma matrix trace identities imply, for arbitrary null external four-momenta,

tr (\neq a γ^{μ} \neq b γ^{ν}) tr (\neq c γ_{μ} \neq d γ_{ν}) = 32 [(a \cdot c) (b \cdot d) + (a \cdot d) (b \cdot c)] .

(3)

To see the contraction, each trace is

4 (a^{μ} b^{ν} + a^{ν} b^{μ} - g^{μν} a \cdot b)

; the terms proportional to

(a \cdot b) (c \cdot d)

cancel in four dimensions, leaving the displayed two products. Here

p_{1} \cdot p_{2} = p_{3} \cdot p_{4} = s /2

and

p_{1} \cdot p_{4} = p_{2} \cdot p_{3} = - u /2

. Therefore

⟨ ∣ M ∣^{2} ⟩ = 2 e^{4} \frac{s ^{2} + u ^{2}}{t ^{2}} .

(4)

The requested proportionality constant is

2 e^{4} = 32 π^{2} α_{em}^{2}

, where

α_{em}

is the fine-structure constant. Both numerator terms are required by the spin sum; no final-state statistical factor enters a squared scattering amplitude for distinct leptons.

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Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 / 2 / b / i / Solution 2026-10-25

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First derive the contracted gamma matrices identities from the Clifford algebra in four dimensions. Moving the outer gamma matrix through two factors gives

γ_{ρ} γ_{α} γ_{β} γ^{ρ} = 4 g_{α β} 1 .

(1)

Moving it through three factors, and using this two-factor result, gives

γ_{ρ} γ_{α} γ_{β} γ_{γ} γ^{ρ} = 2 γ_{β} γ_{γ} γ_{α} - 4 g_{β γ} γ_{α} = - 2 γ_{γ} γ_{β} γ_{α} .

(2)

Consequently

γ^{ν} \neq b γ_{μ} \neq c γ_{ν} = - 2 \neq c γ_{μ} \neq b

and

γ_{μ} \neq b \neq d γ^{μ} = 4 (b \cdot d) 1

. Apply these in succession:

T = tr (\neq a γ^{ν} \neq b γ_{μ} \neq c γ_{ν} \neq d γ^{μ}) = - 2 tr (\neq a \neq c γ_{μ} \neq b \neq d γ^{μ}) = - 8 (b \cdot d) tr (\neq a \neq c) = - 32 (a \cdot c) (b \cdot d) .

(3)

The final step uses

tr (γ^{α} γ^{β}) = 4 g^{α β}

. This is a genuinely four-dimensional instance of the gamma matrix trace identities; using it unchanged inside a

D

-dimensional loop calculation would require separate justification.

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Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2020 / 2 / a / iii / Solution 2026-10-25

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The massive Dirac propagator is

S_{F} (x) = \int \frac{d ^{D} k}{( 2 π ) ^{D}} \frac{i ( \neq k + m ) e ^{- ik \cdot x}}{k ^{2} - m ^{2} + i 0} .

(1)

Use the gamma matrix trace identities

tr γ^{μ} = 0

and

tr (γ^{μ} γ^{ρ}) = 4 g^{μ ρ}

. They give

tr [γ^{μ} S_{F} (0)] = 4 i \int \frac{d ^{D} k}{( 2 π ) ^{D}} \frac{k ^{μ}}{k ^{2} - m ^{2} + i 0} = 0 .

(2)

The integrand is odd under

k \mapsto - k

, and dimensional regularization preserves that symmetry. The same vanishing holds for

S = i S_{F}

. One must not assert that

S (0)

itself is zero: its term proportional to

m 1

generally is nonzero and divergent, but its trace against a single gamma matrix is zero. This one-photon fermion loop is also the simplest instance of Furry's theorem.

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Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2020 / 2 / b / iii / Solution 2026-10-25

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For massless quark Compton scattering, the color average contributes one, since

(1/ N_{c}) \sum_{i, j} ∣ δ_{ji} ∣^{2} = 1

. The spin average and incoming photon polarization average supply

1/ (2 \cdot 2) = 1/4

. The Ward identities permit use of

- g_{μν}

for each photon polarization completeness relation, and the massless fermion spin sum is

\sum_{r} u_{r} (p) \overset{u}{ˉ}_{r} (p) = \neq p

Let

A_{μν} = γ_{ν} \frac{\neq q _{s}}{s} γ_{μ} + γ_{μ} \frac{\neq q _{u}}{u} γ_{ν} .

(1)

Reversing the gamma matrices order gives its Dirac-conjugate matrix

\overset{ˉ}{A}^{μν}

. The squared scattering amplitude is

⟨ ∣ M ∣^{2} ⟩ = \frac{e ^{4} Q _{u}^{4}}{4} tr (\neq p_{3} A_{μν} \neq p_{1} \overset{ˉ}{A}^{μν}) .

(2)

For clarity evaluate the four traces before dividing by their denominators. The Mandelstam variables obey

s + t + u = 0

, with

t = (p_{1} - p_{3})^{2}

p_{1} \cdot p_{2} = s /2

p_{1} \cdot p_{4} = u (- 1/2)

and

p_{1} \cdot p_{3} = - t /2

. For the squared

s

-channel contribution, the gamma matrix trace identities imply

T_{ss} = tr (\neq p_{3} γ_{ν} \neq q_{s} γ_{μ} \neq p_{1} γ^{μ} \neq q_{s} γ^{ν}) = - 2 tr (\neq p_{3} γ_{ν} \neq q_{s} \neq p_{1} \neq q_{s} γ^{ν}) = - 2 s tr (\neq p_{3} γ_{ν} \neq p_{2} γ^{ν}) = 4 s tr (\neq p_{3} \neq p_{2}) = - 8 s u .

(3)

Here

\neq q_{s} \neq p_{1} \neq q_{s} = 2 (q_{s} \cdot p_{1}) \neq q_{s} - q_{s}^{2} \neq p_{1} = s \neq p_{2}

and

p_{3} \cdot p_{2} = - u /2

. For the squared

u

-channel contribution,

\neq q_{u} \neq p_{1} \neq q_{u} = - u \neq p_{4}

gives

T_{uu} = tr (\neq p_{3} γ_{μ} \neq q_{u} γ_{ν} \neq p_{1} γ^{ν} \neq q_{u} γ^{μ}) = - 16 u (p_{3} \cdot p_{4}) = - 8 s u .

(4)

One interference trace reduces to

T_{s u} = tr (\neq p_{3} γ_{ν} \neq q_{s} γ_{μ} \neq p_{1} γ^{ν} \neq q_{u} γ^{μ}) = - 2 tr (\neq p_{3} γ_{ν} \neq q_{s} \neq q_{u} γ^{ν} \neq p_{1}) = - 32 (q_{s} \cdot q_{u}) (p_{3} \cdot p_{1}) = 0,

(5)

since

q_{s} \cdot q_{u} = (s + t + u) /2 = 0

. The other interference trace is its complex conjugate and also vanishes. Therefore

⟨ ∣ M ∣^{2} ⟩ = - 2 e^{4} Q_{u}^{4} (\frac{u}{s} + \frac{s}{u}) = - \frac{32 e ^{4}}{81} \frac{s ^{2} + u ^{2}}{s u}, C = \frac{32 e ^{4}}{81} .

(6)

Equivalently

C = 512 π^{2} α_{em}^{2} /81

in terms of the fine-structure constant. In the physical noncollinear region

s > 0

u < 0

, so the result is positive. The singular massless endpoint

u = 0

reflects a collinear divergence; it is not a negative squared scattering amplitude.

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Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 62 / 3 / Solution Created 2026-10-03 Updated 2026-10-14

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Use the Minkowski metric

g^{μν} = diag (1, - 1, - 1, - 1)

and Feynman slash notation

\neq p = γ^{μ} p_{μ}

. The Pauli matrices satisfy

σ_{j} σ_{k} + σ_{k} σ_{j} = 2 δ_{jk} I_{2}

. In the Dirac representation of the gamma matrices, direct block multiplication gives

(γ^{0})^{2} = I_{4}, γ^{0} γ^{j} = (0 σ_{j} σ_{j} 0) = - γ^{j} γ^{0},

(1)

and

γ^{j} γ^{k} = (- σ_{j} σ_{k} 0 0 - σ_{j} σ_{k}) .

(2)

Thus the spatial anticommutators are

- 2 δ_{jk} I_{4}

, proving the Clifford relations

{γ^{μ}, γ^{ν}} = 2 g^{μν} I_{4} .

(3)

Since partial derivatives commute, their symmetric product selects the symmetric part of the gamma matrices. Multiplying the Dirac equation by

i γ^{μ} \partial_{μ} + m

gives

(i \neq \partial + m) (i \neq \partial - m) ψ = (- γ^{μ} γ^{ν} \partial_{μ} \partial_{ν} - m^{2}) ψ = - (□ + m^{2}) ψ = 0.

(4)

Hence every spinor component satisfies the Klein-Gordon equation with the same mass.

For parity, let

x_{P} = (t, - x)

and

ψ_{P} (x) = γ^{0} ψ (x_{P})

. The chain rule gives

\partial_{0} ψ (x_{P}) = (\partial_{0} ψ) (x_{P})

and

\partial_{j} ψ (x_{P}) = - (\partial_{j} ψ) (x_{P})

. Combining these signs with

γ^{j} γ^{0} = - γ^{0} γ^{j}

proves

(i \neq \partial - m) ψ_{P} (x) = γ^{0} [(i \neq \partial - m) ψ] (x_{P}) = 0.

(5)

This establishes parity symmetry in quantum field theory for the free Dirac equation, rather than merely asserting its covariance.

A positive-energy plane wave

ψ = u (p) e^{- i p \cdot x}

solves

(\neq p - m) u = 0

. Write

u = (χ, η)^{T}

. The two block equations are

(E_{p} - m) χ - (σ \cdot p) η = 0, (σ \cdot p) χ - (E_{p} + m) η = 0.

(6)

The second gives

η = (σ \cdot p) χ / (E_{p} + m)

. In the first,

(σ \cdot p)^{2} = p^{2} I_{2}

and

E_{p}^{2} - p^{2} = m^{2}

make the remaining coefficient zero. There are two independent two-component choices

χ_{s}

, so

u_{s} (p) = (χ_{s} (σ \cdot p) χ_{s} / (E_{p} + m)), ψ_{p, s} (x) = u_{s} (p) e^{- i p \cdot x} .

(7)

For a massive Dirac spinor, choose

χ_{s}

as the two rest-frame spin states along a fixed axis, for example

σ_{3} χ_{s} = \pm χ_{s}

. The label

s

distinguishes these canonical spin states after boosting; it is not automatically a helicity label.

Now

p \cdot x_{P} = p_{P} \cdot x

and

γ^{0} u_{s} (p) = u_{s} (p_{P})

for the same fixed

χ_{s}

. Therefore

ψ_{p, s} (x) ⟼ ψ_{p_{P}, s} (x) .

(8)

The parity action on canonical massive spin states leaves the canonical spin label unchanged while reversing the momentum. Spin is an axial vector under parity; its orientation is unchanged. Consequently helicity, the projection of spin along momentum, reverses. If

s

had instead been chosen to denote helicity, the transformed label would have the opposite sign.

The Dirac adjoint is

\overset{ˉ}{ψ} = ψ^{†} γ^{0}

, with

γ^{0 †} = γ^{0}

. Thus

\overset{ˉ}{ψ}_{P} (x) = [γ^{0} ψ (x_{P})]^{†} γ^{0} = ψ^{†} (x_{P}) = \overset{ˉ}{ψ} (x_{P}) γ^{0} .

(9)

The chirality matrix has

γ^{5} = i γ^{0} γ^{1} γ^{2} γ^{3} = (0 I_{2} I_{2} 0)

in this representation, so

(γ^{5})^{2} = I_{4}

and it anticommutes with every

γ^{μ}

. In particular

γ^{0} γ^{5} γ^{0} = - γ^{5}

. The parity transformation of a Dirac bilinear therefore yields

\overset{ˉ}{ψ}_{P} (x) ψ_{P} (x) = \overset{ˉ}{ψ} (x_{P}) ψ (x_{P}), \overset{ˉ}{ψ}_{P} (x) γ^{5} ψ_{P} (x) = - \overset{ˉ}{ψ} (x_{P}) γ^{5} ψ (x_{P}) .

(10)

The first bilinear is a scalar under parity, whereas the second is a pseudoscalar.

For the requested gamma matrix trace identities, use the cyclic property of the trace. Conjugation of a product of

n

gamma matrices by

γ^{5}

multiplies it by

(- 1)^{n}

, but leaves its trace unchanged. All odd products therefore have zero trace. Taking the trace of the Clifford algebra relation and using

Tr I_{4} = 4

gives

Tr (γ^{α} γ^{β}) = 4 g^{α β}

. For four factors, move the first gamma matrix successively past the next three using the anticommutators. If

A = Tr (γ^{α} γ^{β} γ^{ρ} γ^{δ})

, this gives

A = 2 g^{α β} Tr (γ^{ρ} γ^{δ}) - 2 g^{α ρ} Tr (γ^{β} γ^{δ}) + 2 g^{α δ} Tr (γ^{β} γ^{ρ}) - A .

(11)

Thus

A = 4 (g^{α β} g^{ρ δ} - g^{α ρ} g^{β δ} + g^{α δ} g^{βρ})

. Contracting with the relevant four-vectors proves all four answers:

Tr (\neq p) Tr (\neq p \neq q) Tr (\neq p \neq q \neq k) Tr (\neq p γ^{μ} \neq q γ^{ν}) = 0, = 4 p \cdot q, = 0, = 4 [p^{μ} q^{ν} + p^{ν} q^{μ} - (p \cdot q) g^{μν}] .

(12)

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Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 48 / 2 / Solution Created 2026-10-03 Updated 2026-10-14

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The derivative convention uses hypercharge Y with

Q = T_{3} + Y

. For each lepton family

l = e, μ, τ

, the left-handed fields form an SU(2) doublet

L_{l} = (ν_{l}, l)_{L}^{T}

, with

Y = - 1/2

. Their weak isospin values are

T_{3} = + 1/2, - 1/2

and their electric charges are

0, - 1

. The right-handed charged lepton

l_{R}

is an SU(2) singlet, with

T_{3} = 0

Y = - 1

and charge

- 1

. The minimal model used here has no right-handed neutrino. A sterile right-handed neutrino, if added as a singlet with

Y = 0

, would have no charged weak current interaction. Antiparticles carry the conjugate representations and opposite electric charges.

Expanding

\overset{ˉ}{L}_{l} i γ^{μ} D_{μ} L_{l}

supplies

- g \overset{ˉ}{L}_{l} γ^{μ} A_{μ}^{a} σ_{a} L_{l} /2

. The off-diagonal entries of

A^{a} σ_{a}

are

2 W^{+}

and

2 W^{-}

. With the chiral projector

P_{L} = (1 - γ^{5}) /2

, one has

\overset{ν}{ˉ}_{L} γ^{μ} l_{L} = \overset{ν}{ˉ} γ^{μ} P_{L} l

. Thus the two W boson interactions are

L_{W^{+}, l} = - \frac{g}{2 2} W_{μ}^{+} \overset{ν}{ˉ}_{l} γ^{μ} (1 - γ^{5}) l, L_{W^{-}, l} = - \frac{g}{2 2} W_{μ}^{-} \overset{ˉ}{l} γ^{μ} (1 - γ^{5}) ν_{l} .

(1)

For quarks, separate rotations diagonalize the up-type and down-type mass terms. If their left-handed rotations are

U_{uL}

and

U_{d L}

, the charged interaction contains the CKM matrix

V = U_{uL}^{†} U_{d L}

J_{had}^{μ} = \sum_{i, j} \overset{u}{ˉ}_{i} γ^{μ} (1 - γ^{5}) V_{ij} d_{j} .

(2)

There is an implicit sum over color charge. This quark mixing permits transitions between different families, rather than just the paired fields in a single doublet.

Let

J^{μ} = \sum_{l} \overset{ν}{ˉ}_{l} γ^{μ} (1 - γ^{5}) l + J_{had}^{μ}

and

κ = g / (22)

. At momenta much smaller than

m_{W}

, neglect derivatives in the heavy W boson equation. Its mass and source terms can be completed to a square:

m_{W}^{2} W_{μ}^{+} W^{- μ} - κ (W_{μ}^{+} J^{μ} + W_{μ}^{-} J^{† μ}) = m_{W}^{2} (W_{μ}^{+} - \frac{κ}{m _{W}^{2}} J_{μ}^{†}) (W^{- μ} - \frac{κ}{m _{W}^{2}} J^{μ}) - \frac{κ ^{2}}{m _{W}^{2}} J_{μ}^{†} J^{μ} .

(3)

This explicitly performs integrating out a charged vector boson. The resulting Fermi interaction is

L_{eff} = - \frac{G _{F}}{2} J_{μ}^{†} J^{μ}, \frac{G _{F}}{2} = \frac{g ^{2}}{8 m _{W}^{2}} .

(4)

The omitted terms contain derivatives and are relatively suppressed by momentum squared divided by

m_{W}^{2}

For the leptonic pion decay, Lorentz covariance restricts a vacuum-to-spin-zero current matrix element to a constant times

p^{α}

. In the rest frame, rotations eliminate the spatial components. Under parity, the temporal vector current is even, whereas the vacuum is even and the pion is odd, so its temporal matrix element equals its negative and vanishes. The temporal axial current is odd and is allowed. This proves the vacuum-to-pseudoscalar current selection rule, rather than merely assuming an axial current. Define the pion decay constant by

⟨ 0∣ \overset{u}{ˉ} γ^{α} γ^{5} d ∣ π^{-} (p)⟩ = i f_{π} p^{α} .

(5)

Then the vector-minus-axial charged weak current has matrix element

- i f_{π} V_{u d} p^{α}

, up to an irrelevant overall phase convention. Consequently the scattering amplitude for the decay can be written

M = \frac{i G _{F} f _{π} V _{u d}}{2} p_{α} \overset{u}{ˉ}_{e} (k) γ^{α} (1 - γ^{5}) v_{ν} (q) .

(6)

Use

p = k + q

, the momentum-space Dirac equation

\overset{u}{ˉ}_{e} \neq k = m_{e} \overset{u}{ˉ}_{e}

, and

\neq q v_{ν} = 0

for a massless neutrino. Since

\neq q (1 - γ^{5}) = (1 + γ^{5}) \neq q

, the neutrino contribution vanishes, leaving

M = \frac{i G _{F} f _{π} V _{u d} m _{e}}{2} \overset{u}{ˉ}_{e} (1 - γ^{5}) v_{ν} .

(7)

The fermion spin sum is, using the supplied gamma matrix trace identities,

\sum_{spins} ∣ \overset{u}{ˉ}_{e} (1 - γ^{5}) v_{ν} ∣^{2} = tr [(\neq k + m_{e}) (1 - γ^{5}) \neq q (1 + γ^{5})] = 8 k \cdot q = 4 (m_{π}^{2} - m_{e}^{2}) .

(8)

Here

(k + q)^{2} = m_{π}^{2}

k^{2} = m_{e}^{2}

, and

q^{2} = 0

. The mass term in the trace vanishes; anticommuting

γ^{5}

reduces the remaining trace to twice

tr (\neq k \neq q)

. It follows that

spins \sum ∣ M ∣^{2} = 2 G_{F}^{2} f_{π}^{2} ∣ V_{u d} ∣^{2} m_{e}^{2} (m_{π}^{2} - m_{e}^{2}) .

(9)

Using two-body decay phase space, with final momentum magnitude

(m_{π}^{2} - m_{e}^{2}) / (2 m_{π})

, gives the leptonic pseudoscalar decay width

Γ (π^{-} \to e^{-} \overset{ν}{ˉ}_{e}) = \frac{G _{F}^{2} f _{π}^{2} ∣ V _{u d} ∣ ^{2} m _{π} m _{e}^{2}}{8 π} (1 - \frac{m _{e}^{2}}{m _{π}^{2}})^{2} .

(10)

Thus the requested

m_{e}^{2} (m_{π}^{2} - m_{e}^{2})

factor is already present in the squared amplitude, and the phase space supplies one more threshold factor.

The physical explanation is helicity suppression. In the pion rest frame the two momenta are opposite. The massless antineutrino has positive helicity; zero total angular momentum would require the electron to have positive helicity too. The charged weak current instead creates a left-chiral electron, whose helicity is negative when its mass vanishes. A nonzero electron mass admits the required opposite helicity with an amplitude proportional to

m_{e}

. The decay amplitude vanishes for $m_{e} = 0$ , and the rate is suppressed by $m_{e}^{2}$ .

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Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 48 / 4 / Solution Created 2026-10-03 Updated 2026-10-14

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Put

s = q^{2} > 0

and neglect electron and final-quark masses compared with

s

. First fix a final color charge; its multiplicity will be restored for the inclusive rate. The photon-exchange scattering amplitude is, up to an overall phase,

M = \frac{e ^{2} Q _{q}}{s} [\overset{v}{ˉ} (p_{2}) γ^{μ} u (p_{1})] [\overset{u}{ˉ} (k_{1}) γ_{μ} v (k_{2})] .

(1)

Averaging over the two spins of each incident particle and summing final spins gives

\overline{∣ M ∣^{2}} = \frac{e ^{4} Q _{q}^{2}}{4 s ^{2}} tr (\neq p_{2} γ^{μ} \neq p_{1} γ^{ν}) tr (\neq k_{1} γ_{μ} \neq k_{2} γ_{ν}) .

(2)

By the gamma matrix trace identities, the first trace is

4 (p_{2}^{μ} p_{1}^{ν} + p_{2}^{ν} p_{1}^{μ} - g^{μν} p_{1} \cdot p_{2})

, and likewise for the second. Contracting the two tensors cancels the terms proportional to

(p_{1} \cdot p_{2}) (k_{1} \cdot k_{2})

, giving

\overline{∣ M ∣^{2}} = \frac{8 e ^{4} Q _{q}^{2}}{s ^{2}} [(p_{1} \cdot k_{1}) (p_{2} \cdot k_{2}) + (p_{1} \cdot k_{2}) (p_{2} \cdot k_{1})] .

(3)

In the centre-of-mass frame all four energies are

s /2

. The first pair of scalar products equals

s (1 - cos θ) /4

each, and the second pair equals

s (1 + cos θ) /4

each. Hence

\overline{∣ M ∣^{2}} = e^{4} Q_{q}^{2} (1 + cos^{2} θ)

For clarity, the printed flux hint has a factor-of-four inconsistency: when

F = 4 (p_{1} \cdot p_{2})^{2} - m_{1}^{2} m_{2}^{2} = 2 s

, the prefactor multiplying the spin-averaged amplitude is

1/ F

, not

1/ (4 F)

. Alternatively,

1/ (4 F_{0})

is correct if

F_{0} = (p_{1} \cdot p_{2})^{2} - m_{1}^{2} m_{2}^{2} = s /2

. Use this massless two-body scattering flux normalization, keeping the spin average separate.

To obtain the two-body Lorentz-invariant phase space directly, integrate the momentum-conserving spatial delta function, leaving

k_{2} = - k_{1}

. The remaining radial integral is

\int d k δ (s - 2 k) = 1/2

, so

d Φ_{2} / d Ω = 1/ (32 π^{2})

. The relativistic scattering cross-section is consequently

\frac{d σ _{one color}}{d Ω} = \frac{∣ M ∣ ^{2}}{64 π ^{2} s} = \frac{α ^{2} Q _{q}^{2}}{4 s} (1 + cos^{2} θ) .

(4)

Since

\int d Ω (1 + cos^{2} θ) = 2 π \int_{- 1}^{1} (1 + x^{2}) d x = 16 π /3

, the integrated electron-positron annihilation into a quark pair result, for this fixed color, is

σ_{one color} = \frac{4 π α ^{2} Q _{q}^{2}}{3 s} .

(5)

The first two printed formulas are thus color-stripped. Distinct final colors are orthogonal states and must be summed, producing

N_{c} = 3

. At leading order each accessible quark flavor contributes independently. Away from thresholds and narrow resonances, quark-hadron duality relates the inclusive parton rate to the observed inclusive electron-positron annihilation into hadrons, giving

σ_{had}^{(γ)} ≃ \frac{4 π α ^{2}}{3 s} 3 f ma ccess ib l e \sum Q_{f}^{2} .

(6)

This formula concerns the photon channel, assumes

s

is large compared with the relevant quark masses and the strong-interaction scale, and does not include Z boson exchange.

Beyond leading order, an exactly exclusive pair of colored quarks is not an observable asymptotic state because of confinement and hadronization. Moreover its virtual corrections have infrared divergences; real soft or collinear gluon emission cannot be excluded if a finite rate is wanted. The inclusive sum includes both the corrected two-parton channel and channels such as

q \overset{q}{ˉ} g

, with their degenerate soft and collinear limits. Real and virtual infrared divergences then cancel for an infrared-safe observable. A specified jet observable can also be meaningful if it defines an infrared-safe observable; a bare two-quark final state is not.

A convenient inclusive calculation uses the optical theorem. For the electromagnetic hadronic current

j_{μ} = \sum_{f} Q_{f} \overset{q}{ˉ}_{f} γ_{μ} q_{f}

, define

i \int d^{4} x e^{i q \cdot x} ⟨ 0∣ T j_{μ} (x) j_{ν} (0) ∣0 ⟩ = (q_{μ} q_{ν} - q^{2} g_{μν}) Π (q^{2}) .

(7)

The hadronic electromagnetic-current spectral density is obtained from its discontinuity. In this convention the hadronic R ratio is

R (s) = σ_{had} / σ_{μμ} = 12 π Im Π (s + i 0)

, where

σ_{μμ} = 4 π α^{2} / (3 s)

at this order. Cutting the forward amplitude sums all allowed hadronic intermediate states, so perturbative Quantum chromodynamics applied to this inclusive quantity implements the required sum of real and virtual contributions.

Finally, write

a = α_{s} (μ^{2})

C = 1.99 - 0.11 n_{f}

L = ln (s / μ^{2})

, and

β_{0} = 11 - 2 n_{f} /3

. The given truncated correction is

K = 1 + \frac{a}{π} + \frac{a ^{2}}{π ^{2}} (C - \frac{β _{0}}{4} L) .

(8)

Differentiate with respect to

t = ln μ^{2}

, using the stated one-loop beta function

d a / d t = - β_{0} a^{2} / (4 π)

and

d L / d t = - 1

. The derivative of

a / π

cancels the explicit logarithmic derivative at order

a^{2}

. Keeping the residual derivative of the truncated expression gives

\frac{d K}{d t} = - \frac{β _{0} a ^{3}}{2 π ^{3}} (C - \frac{β _{0}}{4} L) .

(9)

Thus the principle of minimal sensitivity, applied to this expression with nonzero coupling and

β_{0} \neq = 0

, selects the one-loop stationary scale for the hadronic annihilation correction

μ_{*}^{2} = s exp [- \frac{4 ( 1.99 - 0.11 n _{f} )}{11 - 2 n _{f} /3}] .

(10)

This is a prescription based on the residual order-

a^{3}

scale dependence of an order-

a^{2}

approximation. At strictly order

a^{2}

, the derivative already vanishes for every renormalization scale; the prescription does not assert exact scale dependence of the observable. It is useful only while the selected running coupling remains perturbative. If

a = 0

β_{0} = 0

, the stated derivative imposes no unique scale.

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Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 48 / 3 / Solution Created 2026-10-03 Updated 2026-10-14

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Let the masses of

Φ

and

ϕ

M

and

μ

, respectively. Write their external four-momenta as

P = k_{1} + k_{2}

, with

P^{2} = M^{2}

and

k_{1}^{2} = k_{2}^{2} = μ^{2}

. The condition

M > 2 μ

opens the two-body decay phase space. Use mostly-minus Minkowski spacetime, the Dirac propagator

i (\neq k + m) / (k^{2} - m^{2} + i 0)

, and the convention that an amputated connected diagram contributes

i M

There is no tree-level three-scalar Feynman vertex in the displayed interaction. The leading decay amplitude has one

Φ

Yukawa interaction and two

ϕ

Yukawa interactions, so is of order

G g^{2}

. The Yukawa fermion triangle is:

Figure 1.
Fermion triangle for scalar decay with its loop-momentum routing and Yukawa vertex factors
. **Details**.

The two inequivalent cyclic orderings around the fermion loop are related by reversal and interchange of

k_{1}, k_{2}

. They give equal integrals: reversing all internal four-momenta leaves the masses and denominators unchanged, and the scalar gamma matrix trace identities below give the same numerator. Both must be included. The closed fermion loop sign supplies

- 1

, and each ordering has the product

(- 1) (- i G) (- i g)^{2} i^{3} = - G g^{2} .

(1)

There is no extra

1/2!

in the decay amplitude. Such a factor belongs to integration over the identical-particle final phase space when calculating a decay rate.

Route

a = ℓ

b = ℓ + k_{1}

c = ℓ - k_{2}

and define

D_{a} = a^{2} - m^{2} + i 0

, and similarly for

D_{b}, D_{c}

. A regulated leading decay amplitude is

i M_{loop} = - 2 G g^{2} μ_{R}^{2 ϵ} \int \frac{d ^{4 - 2 ϵ} ℓ}{( 2 π ) ^{4 - 2 ϵ}} \frac{tr [( \neq a + m ) ( \neq b + m ) ( \neq c + m )]}{D _{a} D _{b} D _{c}} .

(2)

Here dimensional regularization preserves translation of the loop momentum, and

μ_{R}

is its reference mass scale. Factors

μ_{R}^{2 ϵ}

account for the loop's dimension; dimensionally continued couplings can equivalently carry these powers. No

γ^{5}

occurs in this scalar loop, so there is no

γ^{5}

regularization ambiguity.

Using gamma matrix trace identities, traces with an odd number of gamma matrices vanish, while

tr 1 = 4

and

tr (γ^{μ} γ^{ν}) = 4 g^{μν}

. Thus the numerator simplifies completely to

N = 4 m [m^{2} + a \cdot b + a \cdot c + b \cdot c] = 4 m [m^{2} + 3 ℓ^{2} + 2 ℓ \cdot (k_{1} - k_{2}) - k_{1} \cdot k_{2}], k_{1} \cdot k_{2} = \frac{M ^{2} - 2 μ ^{2}}{2} .

(3)

For a reduction to standard scalar integrals, disregard the vanishing infinitesimals only in the numerator. Since

D_{a} + D_{b} + D_{c} = 3 ℓ^{2} + 2 ℓ \cdot (k_{1} - k_{2}) + 2 μ^{2} - 3 m^{2}

up to those infinitesimals,

N = 4 m [D_{a} + D_{b} + D_{c} + 4 m^{2} - μ^{2} - \frac{M ^{2}}{2}] .

(4)

Define the scalar bubble and scalar triangle Feynman integral by

B (s) = μ_{R}^{2 ϵ} \int \frac{d ^{4 - 2 ϵ} ℓ}{( 2 π ) ^{4 - 2 ϵ}} \frac{1}{( ℓ ^{2} - m ^{2} + i 0 ) (( ℓ + q ) ^{2} - m ^{2} + i 0 )}, q^{2} = s, C = μ_{R}^{2 ϵ} \int \frac{d ^{4 - 2 ϵ} ℓ}{( 2 π ) ^{4 - 2 ϵ}} \frac{1}{D _{a} D _{b} D _{c}} .

(5)

Cancelling each denominator in turn gives two bubbles with external invariant

μ^{2}

and one with invariant

M^{2}

. Therefore

i M_{loop} = - 8 m G g^{2} [2 B (μ^{2}) + B (M^{2}) + (4 m^{2} - μ^{2} - \frac{M ^{2}}{2}) C] .

(6)

This explicitly displays the complete gamma matrix trace simplification, the two loop orientations and the overall phase convention.

There is a renormalization qualification to this answer. For

m \neq = 0

the loop is logarithmically divergent; retaining just the unregulated four-dimensional integral would not specify a finite physical decay amplitude. To see the divergence, a Feynman parameter and a shifted loop momentum give

B (s) = \frac{i}{16 π ^{2}} [\frac{1}{ϵ ˉ} - \int_{0}^{1} d x lo g \frac{m ^{2} - x ( 1 - x ) s - i 0}{μ _{R}^{2}}] + O (ϵ), \frac{1}{ϵ ˉ} = \frac{1}{ϵ} - γ_{E} + lo g (4 π) .

(7)

The scalar triangle Feynman integral is ultraviolet finite. Hence

M_{loop, div} = - \frac{24 m G g ^{2}}{16 π ^{2} ϵ ˉ} .

(8)

A cubic counterterm from a Yukawa fermion triangle,

L_{ct} = - δ h Φ ϕ^{2} /2

, contributes

- δ h

M

and cancels the pole with

δ h_{div} = - 24 m G g^{2} / (16 π^{2} \overset{ϵ}{ˉ})

. This is an instance of counterterm closure of a massive Yukawa theory. After subtraction the physical answer is

M = - h_{R} + M_{loop, ren}

, where a renormalization condition fixes the finite cubic coupling

h_{R}

. The displayed interaction alone does not supply that condition. For example, setting

h_{R} = 0

at a specified subtraction scale defines one prescription for the loop-induced decay. The regulated expressions above remain the lowest-order loop answer without such extra data. For

m = 0

the trace, and therefore this scalar triangle contribution, vanishes identically.

Now take the Hermitian pseudoscalar Yukawa interaction with the sign printed in the PDF. Varying with respect to the Dirac adjoint gives the Dirac equation

(i γ^{μ} \partial_{μ} - m - g ϕ + i G Φ γ^{5}) ψ = 0.

(9)

Varying with respect to

ψ

and integrating by parts gives the adjoint Dirac equation

i (\partial_{μ} \overset{ˉ}{ψ}) γ^{μ} + (m + g ϕ) \overset{ˉ}{ψ} - i G Φ \overset{ˉ}{ψ} γ^{5} = 0.

(10)

The factor

i

in the coupling matters:

\overset{ˉ}{ψ} γ^{5} ψ

is anti-Hermitian, so this factor makes the interaction Hermitian for real

G

For the axial-current divergence for scalar and pseudoscalar backgrounds, set

S = m + g ϕ

and

H = G Φ

. The two equations are equivalently

γ^{μ} \partial_{μ} ψ = - i S ψ - H γ^{5} ψ, (\partial_{μ} \overset{ˉ}{ψ}) γ^{μ} = i S \overset{ˉ}{ψ} + H \overset{ˉ}{ψ} γ^{5} .

(11)

In the divergence of the axial current, use

{γ^{5}, γ^{μ}} = 0

and

(γ^{5})^{2} = 1

\partial_{μ} (\overset{ˉ}{ψ} γ^{μ} γ^{5} ψ) = (\partial_{μ} \overset{ˉ}{ψ}) γ^{μ} γ^{5} ψ - \overset{ˉ}{ψ} γ^{5} γ^{μ} \partial_{μ} ψ = (i S \overset{ˉ}{ψ} + H \overset{ˉ}{ψ} γ^{5}) γ^{5} ψ - \overset{ˉ}{ψ} γ^{5} (- i S ψ - H γ^{5} ψ) .

(12)

Thus

\partial_{μ} j_{5}^{μ} = 2 i (m + g ϕ) \overset{ˉ}{ψ} γ^{5} ψ + 2 G Φ \overset{ˉ}{ψ} ψ .

(13)

This is the classical field-equation identity requested here. It includes explicit breaking by both backgrounds; it is not a claim that a renormalized quantum axial current has no chiral anomaly when gauge interactions are present.

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Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 54 / 3 / Solution Created 2026-10-03 Updated 2026-10-14

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The middle panel shows the Feynman diagram for the leptonic W boson decay. With the charged-current Lagrangian density sign in the question, its vertex Feynman rule is

- \frac{i g}{2 2} γ^{μ} (1 - γ_{5}) .

(1)

An overall field-phase convention can change the overall amplitude sign but cannot change the decay width. With the standard Dirac field current ordering, the amplitude is

M = - \frac{g}{2 2} ϵ_{μ} (p) \overline{u} (q_{1}) γ^{μ} (1 - γ_{5}) v (q_{2}) .

(2)

The unusual order of barred fields in the printed description should be interpreted as this physical weak charged current matrix element.

Figure 1.
Higgsstrahlung, leptonic W decay and photon-mediated deep inelastic scattering
. **Details**.

For massless final fermions, the fermion spin sums are

\sum_{s} u_{s} \overline{u}_{s} = \neq q_{1}

and

\sum_{s} v_{s} \overline{v}_{s} = \neq q_{2}

. The massive-vector spin average divides by three. Thus

\overline{∣ M ∣^{2}} = \frac{g ^{2}}{24} K_{μν} T^{μν}, K_{μν} = - g_{μν} + p_{μ} p_{ν} / p^{2},

(3)

with the chiral trace

T^{μν} = Tr [\neq q_{1} γ^{μ} (1 - γ_{5}) \neq q_{2} γ^{ν} (1 - γ_{5})] .

(4)

Anticommuting

γ_{5}

through the two following gamma matrices and using

(1 - γ_{5})^{2} = 2 (1 - γ_{5})

gives its symmetric part

T_{S}^{μν} = 8 (q_{1}^{μ} q_{2}^{ν} + q_{1}^{ν} q_{2}^{μ} - g^{μν} q_{1} \cdot q_{2}) .

(5)

Its remaining part is an antisymmetric epsilon tensor from the gamma matrix trace identities; it vanishes against the symmetric polarization tensor

K

. Direct contraction gives

- g_{μν} T_{S}^{μν} = 16 q_{1} \cdot q_{2}

and

\frac{p _{μ} p _{ν}}{p ^{2}} T_{S}^{μν} = 8 [\frac{2 ( p \cdot q _{1} ) ( p \cdot q _{2} )}{p ^{2}} - q_{1} \cdot q_{2}] .

(6)

Hence

\overline{∣ M ∣^{2}} = \frac{g ^{2}}{3} [q_{1} \cdot q_{2} + \frac{2 ( p \cdot q _{1} ) ( p \cdot q _{2} )}{p ^{2}}] .

(7)

No additional averaging over hypothetical right-handed neutrino states is appropriate: the chiral projector already enforces the physical helicity coupling.

On shell,

p = q_{1} + q_{2}

p^{2} = M_{W}^{2}

and

q_{i}^{2} = 0

, so

q_{1} \cdot q_{2} = M_{W}^{2} /2

and

p \cdot q_{i} = M_{W}^{2} /2

. The squared matrix element reduces to

g^{2} M_{W}^{2} /3

. In the parent rest frame, integrate the spatial momentum delta first in two-body Lorentz-invariant phase space. The remaining radial integral is

\int d Φ_{2} = \frac{1}{( 2 π ) ^{2}} \int \frac{k ^{2} d k d Ω}{4 k ^{2}} δ (M_{W} - 2 k) = \frac{1}{8 π} .

(8)

Consequently the massless leptonic W decay width is

Γ_{e} = \frac{1}{2 M _{W}} \frac{g ^{2} M _{W}^{2}}{3} \frac{1}{8 π} = \frac{g ^{2} M _{W}}{48 π} \equiv Γ_{ℓ} .

(9)

Lepton universality gives

Γ (W^{-} \to μ^{-} \overline{ν}_{μ}) = Γ (W^{-} \to τ^{-} \overline{ν}_{τ}) = Γ_{ℓ}

in the same massless approximation. The muon and tau lepton masses produce small corrections if retained.

With an identity CKM matrix, the allowed quark channels are

d \overline{u}

and

s \overline{c}

. The corresponding third-generation channel

b \overline{t}

is closed by the top threshold. Each open channel has three orthogonal matching-color final states, so the color multiplicity in a decay width gives $Γ (d \overline{u}) = Γ (s \overline{c}) = 3 Γ_{ℓ}$ . There is no initial color average because the W boson is colorless. The tree-level W branching fractions are therefore:

Final channel	Partial width	Branching fraction, rounded
$e^{-} \overline{ν}_{e}$	$Γ_{ℓ}$	$1/9 ≃ 11%$
$μ^{-} \overline{ν}_{μ}$	$Γ_{ℓ}$	$1/9 ≃ 11%$
$τ^{-} \overline{ν}_{τ}$	$Γ_{ℓ}$	$1/9 ≃ 11%$
$d \overline{u}$	$3 Γ_{ℓ}$	$1/3 ≃ 33%$
$s \overline{c}$	$3 Γ_{ℓ}$	$1/3 ≃ 33%$

The rounded percentages sum to 99% because of rounding. The exact fractions sum to one, with total leptonic fraction

1/3

and hadronic fraction

2/3

. The total decay width is

Γ_{W} = 9 Γ_{ℓ} = \frac{3 g ^{2} M _{W}}{16 π} = \frac{3 G _{F} M _{W}^{3}}{2 π 2} .

(10)

As an illustrative numerical estimate, taking

g = 0.65

and

M_{W} = 80 GeV

gives

Γ_{ℓ} ≃ 0.224 GeV

, each open quark width

≃ 0.672 GeV

and

Γ_{W} ≃ 2.02 GeV

. These are tree-level model estimates with the stated illustrative inputs, not a fit to a measured width. Strong radiative corrections increase the inclusive quark widths.

The produced quark and antiquark radiate gluons, generating a parton shower. As their virtualities fall towards the strong-interaction scale, perturbation theory gives way to hadronization: confinement converts the colored partons into color-neutral hadrons, typically seen as two particle jets, with additional radiation sometimes producing resolved extra jets. Unstable hadrons subsequently decay; the primary quarks are not observed as free particles.

Including the CKM matrix, the charged current couples an accessible up-type flavour

u_{i}

to every kinematically accessible down-type flavour

d_{j}

. Therefore

Γ (W^{-} \to d_{j} \overline{u}_{i}) = 3 Γ_{ℓ} ∣ V_{ij} ∣^{2}, u_{i} = u, c, d_{j} = d, s, b,

(11)

The sums over

j

remain

3 Γ_{ℓ}

by row unitarity, so the massless tree-level total remains

9 Γ_{ℓ}

. Quark mixing redistributes the hadronic branching fractions; it does not create a new total color factor. If a family is above threshold, only its accessible channels enter, and the unitarity sum must not be applied to an incomplete set of final flavours.

For completeness, CKM parameter counting starts with

N^{2}

real degrees of freedom of an

N \times N

unitary matrix. This follows by counting

2 N^{2}

real entries and

N^{2}

independent Hermitian constraints in

V^{†} V = I

. The real-orthogonal subclass has

N (N - 1) /2

mixing angles; the remaining

N (N + 1) /2

parameters can be viewed as phases. Rephasing each up-type and down-type mass eigenfield changes

V_{ij}

e^{- i α_{i}} e^{i β_{j}}

. Of the

2 N

phases, the common phase does nothing, so

2 N - 1

are removable. Thus

N_{physical} = (N - 1)^{2}, N_{angles} = \frac{N ( N - 1 )}{2}, N_{CP phases} = \frac{( N - 1 ) ( N - 2 )}{2} .

(12)

For three generic nondegenerate families this is three angles and one CP-violating phase. Degenerate masses can allow additional changes of basis and reduce the count.

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Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 45 / 3 / Solution Created 2026-10-03 Updated 2026-10-14

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The product of leptonic weak charged currents in the Fermi interaction contains four-fermion cross terms such as

- \frac{G _{F}}{2} [\overset{e}{ˉ} γ^{α} (1 - γ_{5}) ν_{e}] [\overset{ν}{ˉ}_{μ} γ_{α} (1 - γ_{5}) μ] + h.c.

(1)

This produces

μ^{-} \to e^{-} + \overset{ν}{ˉ}_{e} + ν_{μ}

, and the same interaction permits neutrino-electron scattering and inverse muon decay. Pure electromagnetic interactions have no neutrino coupling and preserve the separate charged-lepton species, so cannot produce these processes by themselves. The factor

1 - γ_{5} = 2 P_{L}

fixes the chiral structure of the weak charged current.

The pion is a pseudoscalar with negative parity, while the strong-interaction vacuum is parity even. Lorentz covariance restricts a vacuum-to-pion vector-current matrix element to

a p^{α}

. In the pion rest frame only its temporal component could be nonzero, but a vector current's temporal component is parity even, whereas the pion is parity odd. The matrix element therefore equals its negative and vanishes. The temporal component of an axial current is parity odd, so its matrix element is allowed and Lorentz covariance fixes it to a constant times

p^{α}

. Defining the pion decay constant in the printed normalization gives

⟨ 0∣ J_{hadrons}^{α} ∣ π^{-} (p)⟩ = - i 2 F_{π} p^{α} .

(2)

This proves both the vector-current exclusion and the axial-current form;

F_{π}

itself is a strong-interaction parameter, not fixed by symmetry.

The cross term

J_{leptons}^{α †} J_{α}^{hadrons}

gives, up to an irrelevant overall phase, the decay amplitude

M = G_{F} F_{π} p_{α} \overset{u}{ˉ}_{e} (k) γ^{α} (1 - γ_{5}) v_{ν} (q) .

(3)

Take the antineutrino mass to be zero. Since

p = k + q

\overset{u}{ˉ}_{e} \neq k = m_{e} \overset{u}{ˉ}_{e}

, and

\neq q v_{ν} = 0

, anticommutation with

γ_{5}

gives

M = G_{F} F_{π} m_{e} \overset{u}{ˉ}_{e} (k) (1 - γ_{5}) v_{ν} (q) .

(4)

The amplitude has helicity suppression by one power of the charged-lepton mass.

Using the external spin sums, with no initial spin average for a spin-zero pion,

spins \sum ∣ M ∣^{2} = G_{F}^{2} F_{π}^{2} m_{e}^{2} Tr [(\neq k + m_{e}) (1 - γ_{5}) \neq q (1 + γ_{5})] = 8 G_{F}^{2} F_{π}^{2} m_{e}^{2} k \cdot q = 4 G_{F}^{2} F_{π}^{2} m_{e}^{2} (M_{π}^{2} - m_{e}^{2}) .

(5)

The middle equality follows from

(1 - γ_{5}) \neq q = \neq q (1 + γ_{5})

(1 + γ_{5})^{2} = 2 (1 + γ_{5})

, and the gamma matrix trace identities. Summing a full massless neutrino spin basis introduces no factor of two: the current projector annihilates the inactive helicity.

For completeness, evaluate the two-body Lorentz-invariant phase space in the pion rest frame. Its three-momentum delta function sets

q = - k

, and the energy delta function is

δ (M_{π} - ∣ k ∣^{2} + m_{e}^{2} - ∣ k ∣)

. The radial root is

∣ k ∣ = \frac{M _{π}^{2} - m _{e}^{2}}{2 M _{π}} .

(6)

Combining its derivative with

1/ (4 E_{k} ∣ k ∣)

and integrating solid angle gives

\int d Φ_{2} = \frac{∣ k ∣}{4 π M _{π}} = \frac{1}{8 π} (1 - \frac{m _{e}^{2}}{M _{π}^{2}}) .

(7)

The invariant squared amplitude is independent of direction, so the supplied decay formula yields the leptonic pseudoscalar decay width

Γ_{π^{-} \to e^{-} \overset{ν}{ˉ}_{e}} = \frac{G _{F}^{2} F _{π}^{2}}{4 π} M_{π} m_{e}^{2} (1 - \frac{m _{e}^{2}}{M _{π}^{2}})^{2} .

(8)

The denominator is

4 π

because the matrix element contains

2 F_{π}

. With the alternative definition

f_{π} = 2 F_{π}

, the same width is

G_{F}^{2} f_{π}^{2} M_{π} m_{e}^{2} (1 - m_{e}^{2} / M_{π}^{2})^{2} / (8 π)

. In either convention it vanishes as

m_{e} \to 0

Replacing the electron by a muon gives the same hadronic factor and weak coupling. Hence the pion electron-to-muon decay ratio is

\frac{Γ _{π^{-} \to e^{-} \overset{ν}{ˉ}_{e}}}{Γ _{π^{-} \to μ^{-} \overset{ν}{ˉ}_{μ}}} = \frac{m _{e}^{2}}{m _{μ}^{2}} (\frac{1 - m _{e}^{2} / M _{π}^{2}}{1 - m _{μ}^{2} / M _{π}^{2}})^{2} .

(9)

This is a tree-level result with massless neutrinos. The cancellation of

F_{π}

and

G_{F}

makes the ratio a clean test of the predicted helicity suppression and electron–muon universality in the chiral weak charged current. For example, scalar or pseudoscalar four-fermion interactions could remove the charged-lepton mass factor and substantially enhance the electron mode. Radiative corrections are required for a precision experimental comparison.

With two quark generations, the charged hadronic current contains

\overset{u}{ˉ} γ_{α} (1 - γ_{5}) (cos θ_{C} d + sin θ_{C} s) .

(10)

K^{-}

contains

s \overset{u}{ˉ}

, so its annihilation into the lepton pair selects the term proportional to

sin θ_{C}

. Strong strangeness conservation excludes the

d

term from its vacuum matrix element. Its amplitude is thus multiplied by

sin θ_{C}

and its width by

sin^{2} θ_{C}

Γ_{K^{-} \to ℓ^{-} \overset{ν}{ˉ}_{ℓ}} = \frac{G _{F}^{2} F _{K}^{2}}{4 π} sin^{2} θ_{C} M_{K} m_{ℓ}^{2} (1 - \frac{m _{ℓ}^{2}}{M _{K}^{2}})^{2} .

(11)

Here

F_{K}

is the pure strong-interaction decay constant in the same square-root-of-two convention. If the pion current's quark mixing coefficient is also shown explicitly, its width carries

cos^{2} θ_{C}

; it is already absorbed if the stated pion matrix element uses the full weak current. Thus the isolated weak-factor ratio is

tan^{2} θ_{C}

, while the kaon suppression relative to unit weak coupling is the requested Cabibbo suppression factor

sin^{2} θ_{C}

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Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 44 / 2 / Solution Created 2026-10-03 Updated 2026-10-14

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Normalize the vacuum path integral by

Z [0]

. The scalar two-point correlation function is

Z [0]^{- 1} \int D ϕ ϕ (x) ϕ (0) e^{i S_{F}}

, and the spinor two-point correlation function is

Z [0]^{- 1} \int D ψ D \overset{ˉ}{ψ} ψ (x) \overset{ˉ}{ψ} (0) e^{i S_{F}}

. Vacuum boundary conditions make these time-ordered Feynman propagators. After integration by parts, the scalar quadratic action is

- \frac{1}{2} \int ϕ (- \partial^{2} + m^{2}) ϕ

The Schwinger-Dyson equation follows by integrating a functional derivative of

ϕ (0) e^{i S_{F}}

: the derivative of the insertion supplies

δ^{d} (x)

, and the action derivative supplies the kinetic operator. The analogous left Grassmann derivative calculation, or differentiation of the Grassmann Gaussian integral, gives

(- \partial^{2} + m^{2}) G_{ϕ} (x) = - i δ^{d} (x), (γ^{μ} \partial_{μ} + M) G_{ψ} (x) = - i δ^{d} (x) .

(1)

Using

\partial_{μ} \mapsto i p_{μ}

and the Clifford algebra,

(iγ \cdot p + M) (- iγ \cdot p + M) = p^{2} + M^{2}

. Consequently

G_{ϕ} (p) = \frac{- i}{p ^{2} + m ^{2} - i 0}, G_{ψ} (p) = \frac{- i ( - iγ \cdot p + M )}{p ^{2} + M ^{2} - i 0} .

(2)

The free quantum effective action is quadratic, so all scalar one-particle-irreducible vertices with

n > 2

vanish. Its two-point vertex is the stated inverse kinetic form

- p^{2} - m^{2}

For the Yukawa interaction, two vertices contribute

(- i y)^{2}

, and a closed fermion loop contributes an extra minus sign. Tracing the two spinor numerators gives

4 [M^{2} - k \cdot (k - p)]

, since the one-gamma traces vanish. Removing the overall

i

from the amplitude gives the displayed loop integral. Define

I (M) = \frac{1}{( 2 π ) ^{d} i} \int \frac{d ^{d} k}{k ^{2} + M ^{2} - i 0}, J (p; M) = \frac{1}{( 2 π ) ^{d} i} \int \frac{d ^{d} k}{( k ^{2} + M ^{2} - i 0 ) (( k - p ) ^{2} + M ^{2} - i 0 )} .

(3)

The numerator decomposition and translation invariance of dimensional regularization reduce it to

τ_{2}^{(1)} = - 4 y^{2} [(2 M^{2} + \frac{p ^{2}}{2}) J (p; M) - I (M)] .

(4)

The supplied tadpole pole is

I (M) \sim - 2 M^{2} / (ε 16 π^{2})

. A Feynman parameter combines the two bubble denominators. Shifting its loop momentum gives mass squared

M^{2} + x (1 - x) p^{2}

; differentiating the tadpole integral with respect to this squared mass gives the double-denominator pole

2/ (ε 16 π^{2})

, independent of

x

. Thus

J (p; M) \sim 2/ (ε 16 π^{2})

and

τ_{2}^{(1)} \sim - \frac{y ^{2}}{ε 16 π ^{2}} (4 p^{2} + 24 M^{2}), a = 4, b = 24.

(5)

The counterterms contribute

- A p^{2} - B

, so their minimal pole parts are

A = - \frac{4 y ^{2}}{ε 16 π ^{2}}, B = - \frac{24 y ^{2} M ^{2}}{ε 16 π ^{2}} .

(6)

Combining the kinetic terms gives wavefunction renormalization

Z_{ϕ} = 1 + A

, and combining the mass terms gives

Z_{ϕ} m_{0}^{2} = m^{2} + B

. Therefore

Z_{ϕ} = 1 - \frac{4 y ^{2}}{ε 16 π ^{2}}, m_{0}^{2} = \frac{m ^{2} + B}{Z _{ϕ}} .

(7)

These are the Yukawa scalar self-energy pole coefficients; finite parts depend on the chosen renormalization condition.

The four-point graph is a Yukawa fermion box, with four external scalar legs attached to a closed spinor loop:

Figure 1.
Fermion box with four external scalar legs in a Yukawa theory
. **Details**.

Each high-momentum dirac propagator is

O (k^{- 1})

. The product of four is

O (k^{- 4})

, and the leading gamma matrix trace identities is nonzero. The four-dimensional radial integral therefore contains

\int^{\infty} d k / k

: a logarithmic ultraviolet divergence. This local four-scalar divergence cannot be absorbed by scalar mass or field normalization. Add

- λ ϕ^{4} /4!

, and, using the stipulated four-point pole normalization, take

δ λ = - 8 y^{4} / (ε 16 π^{2})

so that

- δ λ

cancels it.

For literal cancellation of every one-loop divergence with

M \neq = 0

, counterterm closure of a massive Yukawa theory also requires the allowed scalar linear and cubic terms. A constant scalar background shifts the fermion mass to

M + y ϕ

; the divergent local fermion contribution contains a polynomial proportional to

(M + y ϕ)^{4}

. Its linear and cubic terms are not forbidden by a symmetry when the fermion mass is nonzero. A closed renormalizable family therefore has

L = - \frac{1}{2} (\partial ϕ)^{2} - \overset{ˉ}{ψ} (γ \cdot \partial + M) ψ - y \overset{ˉ}{ψ} ψ ϕ - V (ϕ), V (ϕ) = Λ + h ϕ + \frac{m ^{2}}{2} ϕ^{2} + \frac{κ}{3 !} ϕ^{3} + \frac{λ}{4 !} ϕ^{4},

(8)

with field, mass and coupling redefinitions for both scalar and spinor fields. A tadpole condition can set the renormalized

h

to zero, but its counterterm still exists. The vacuum constant is needed if vacuum energy is retained. If an exact discrete chiral symmetry is imposed with

M = 0

, the scalar potential can be even and the odd terms are forbidden; the essential new interaction is then the quartic one.

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Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 45 / 3 / Solution Created 2026-10-03 Updated 2026-10-06

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The strong-interaction matrix element between two spin-zero pseudoscalar mesons has only

p^{μ}

and

k^{μ}

available. The product of the two intrinsic parities is positive. An axial current would require a pseudovector constructed from these momenta, but an expression involving the Levi-Civita symbol needs three independent four-vectors and therefore vanishes. This is a consequence of parity conservation in the hadronic matrix element, not of parity conservation in the weak interaction. The vector current can have the two independent structures

p + k

and

p - k

. Its coefficients are Lorentz scalars; with

p^{2} = m_{K}^{2}

and

k^{2} = m_{π}^{2}

fixed, their only varying invariant is

s = q^{2}

. Hence

H^{μ} = ⟨ π^{+} (k) ∣ \overset{u}{ˉ} γ^{μ} s ∣ \overset{ˉ}{K}^{0} (p)⟩ = (p + k)^{μ} f_{+} (s) + q^{μ} f_{-} (s) .

(1)

These are the pseudoscalar-to-pseudoscalar form factors. With relativistically normalized states they are dimensionless. The vanishing axial matrix element and this decomposition explain the two equalities separately.

Write

q_{1}

and

q_{2}

for the outgoing electron and antineutrino momenta. From the Fermi interaction, an invariant scattering amplitude, up to an irrelevant overall sign or phase, is

M = \frac{G _{F} V _{u s}}{2} \overset{u}{ˉ} (q_{1}) γ_{μ} (1 - γ^{5}) v (q_{2}) [(p + k)^{μ} f_{+} (s) + q^{μ} f_{-} (s)] .

(2)

The CKM matrix element

V_{u s}

multiplies the quark current in the convention of this interaction. Let

ℓ_{μ} = \overset{u}{ˉ} (q_{1}) γ_{μ} (1 - γ^{5}) v (q_{2})

. For massless leptons, the massless Dirac equation and chirality matrix anticommutation give

q^{μ} ℓ_{μ} = \overset{u}{ˉ} (q_{1}) (\neq q_{1} + \neq q_{2}) (1 - γ^{5}) v (q_{2}) = 0.

(3)

In the second term move

\neq q_{2}

through the chiral projector before applying

\neq q_{2} v = 0

. Since

p + k = 2 p - q

, this transverse massless leptonic current gives

M = 2 G_{F} V_{u s} f_{+} (s) p^{μ} ℓ_{μ} .

(4)

The disappearance of

f_{-}

uses the massless approximation; for a massive charged lepton its contraction is proportional to the lepton mass.

Use the fermion spin sum and the supplied gamma matrix trace identities. The symmetric part of the leptonic tensor is

L_{μν} = 8 (q_{1 μ} q_{2 ν} + q_{1 ν} q_{2 μ} - g_{μν} q_{1} \cdot q_{2}) + L_{μν}^{antisym} .

(5)

The Levi-Civita symbol term is antisymmetric and drops out when contracted with

p^{μ} p^{ν}

. Thus

\sum_{spins} ∣ M ∣^{2} = 16 G_{F}^{2} ∣ V_{u s} ∣^{2} ∣ f_{+} (s) ∣^{2} [2 (p \cdot q_{1}) (p \cdot q_{2}) - m_{K}^{2} (q_{1} \cdot q_{2})] .

(6)

There is no initial-spin average because the kaon is spinless. For the integrated massless leptonic tensor, keep every factor of

2 π

explicit and define the unnormalized two-lepton Lorentz-invariant phase space

I_{μν} = \int \frac{d ^{3} q _{1}}{q _{1}^{0}} \frac{d ^{3} q _{2}}{q _{2}^{0}} δ^{(4)} (q - q_{1} - q_{2}) q_{1 μ} q_{2 ν} = \frac{π}{3} q_{μ} q_{ν} + \frac{π}{6} g_{μν} s .

(7)

The leptons are massless, so

q_{i}^{0} = ∣ q_{i} ∣

. Its trace is

g^{μν} I_{μν} = π s

. Therefore

2 p^{μ} p^{ν} I_{μν} - m_{K}^{2} g^{μν} I_{μν} = \frac{2 π}{3} [(p \cdot q)^{2} - m_{K}^{2} s] .

(8)

The three Lorentz-invariant phase-space measures and their momentum delta function contribute

1/ [8 (2 π)^{5}]

, in addition to

1/ (2 m_{K})

in the decay rate. Combining them with the spin sum gives

Γ = \frac{G _{F}^{2} ∣ V _{u s} ∣ ^{2}}{48 π ^{4} m _{K}} \int \frac{d ^{3} k}{k ^{0}} [(p \cdot q)^{2} - m_{K}^{2} q^{2}] ∣ f_{+} (q^{2}) ∣^{2}, A = \frac{G _{F}^{2} ∣ V _{u s} ∣ ^{2}}{48 π ^{4} m _{K}} .

(9)

This massless semileptonic pseudoscalar decay rate uses a two-lepton integral over future-timelike

q

and the pion integral is restricted to the physically allowed region. The null endpoint follows by continuity. The coefficient has mass dimension

- 5

, so the complete expression has mass dimension one, as a decay rate must in natural units.

In the kaon centre-of-momentum frame, put

E = k^{0}

and

κ = ∣ k ∣

. Then

s = m_{K}^{2} + m_{π}^{2} - 2 m_{K} E, (p \cdot q)^{2} - m_{K}^{2} s = m_{K}^{2} (E^{2} - m_{π}^{2}) = m_{K}^{2} κ^{2} .

(10)

The dimensionally consistent Källén function is

λ (s, m_{K}^{2}, m_{π}^{2}) = s^{2} + m_{K}^{4} + m_{π}^{4} - 2 s m_{K}^{2} - 2 s m_{π}^{2} - 2 m_{K}^{2} m_{π}^{2}, κ = \frac{λ}{2 m _{K}} .

(11)

The pion-only mass term must have fourth power: the second power printed in the PDF is dimensionally inconsistent. This repair also follows directly from squaring

E = (m_{K}^{2} + m_{π}^{2} - s) / (2 m_{K})

. Angular integration and the change of variable give

\frac{d ^{3} k}{k ^{0}} = 4 π \frac{κ ^{2}}{E} d κ = - \frac{2 πκ}{m _{K}} d s = - \frac{π λ}{m _{K}^{2}} d s,

(12)

where the negative sign reverses the endpoints. Combining this with the bracket

λ /4

yields

Γ = \frac{G _{F}^{2} ∣ V _{u s} ∣ ^{2}}{192 π ^{3} m _{K}^{3}} \int_{0}^{(m_{K} - m_{π})^{2}} d s λ (s, m_{K}^{2}, m_{π}^{2})^{3/2} ∣ f_{+} (s) ∣^{2} .

(13)

Thus

B = G_{F}^{2} ∣ V_{u s} ∣^{2} / (192 π^{3} m_{K}^{3}), a = 0, b = (m_{K} - m_{π})^{2}

. The lower limit is the minimum invariant mass of two massless leptons; at the upper limit the pion is at rest. The coefficient has mass dimension

- 7

, while

d s λ^{3/2}

has dimension eight. The Källén function also shows why the differential decay rate vanishes at zero pion momentum.

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Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 47 / 3 / b / Solution Created 2026-10-03 Updated 2026-10-20

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Flavour issue in the printed effective interaction. The displayed charged-current factor uses an electron field even for the muon-neutrino term. Taken literally, this introduces an extra

ν_{μ} e

charged-current interaction, contradicting the Standard Model diagrams in part (a). In the Standard Model that factor uses the charged lepton of the same flavour; for electron scattering the charged-current contribution is therefore present only for

ν_{e}

. The weak flavour selection in neutrino-electron scattering fixes the physically consistent assignment used below. The literal printed consequence is given afterwards, so the discrepancy is explicit.

At leading order in the electron mass and in the four-fermion regime, define

Γ_{L}^{α} = γ^{α} (1 - γ^{5})

. The weak neutral current yields the scattering amplitude

M_{μ} = - \frac{G _{F}}{2} [\overset{u}{ˉ}_{ν} (k^{'}) Γ_{L}^{α} u_{ν} (k)] [\overset{u}{ˉ}_{e} (p^{'}) γ_{α} (c_{V} - c_{A} γ^{5}) u_{e} (p)] .

(1)

An overall amplitude phase has no observable effect. The Fermi interaction is valid when the exchanged invariants are small compared with the squared weak-boson masses.

Average over the two initial electron spin states, and sum over final spins. There is no averaging over a sterile right-handed incoming neutrino. Using the fermion spin sum, the averaged square is

\overline{∣ M_{μ} ∣^{2}} = \frac{G _{F}^{2}}{4} Tr [\neq k^{'} Γ_{L}^{α} \neq k Γ_{L}^{β}] Tr [\neq p^{'} γ_{α} (c_{V} - c_{A} γ^{5}) \neq p γ_{β} (c_{V} - c_{A} γ^{5})] .

(2)

To display the trace contraction, define

S^{α β} (a, b) = a^{α} b^{β} + a^{β} b^{α} - g^{α β} (a \cdot b), E^{α β} (a, b) = ϵ^{α βρ σ} a_{ρ} b_{σ} .

(3)

The supplied gamma matrix trace identities give the two tensors

8 (S (k^{'}, k) + i E (k^{'}, k))

and

4 [(c_{V}^{2} + c_{A}^{2}) S (p^{'}, p) + 2 i c_{V} c_{A} E (p^{'}, p)]

. Symmetric-antisymmetric mixed contractions vanish. Set

X = (k^{'} \cdot p^{'}) (k \cdot p)

and

Y = (k^{'} \cdot p) (k \cdot p^{'})

. The remaining contractions are

S (k^{'}, k) S (p^{'}, p) = 2 (X + Y)

and

E (k^{'}, k) E (p^{'}, p) = - 2 (X - Y)

, using the stated negative epsilon-contraction sign. Their sum gives

\overline{∣ M_{μ} ∣^{2}} = 16 G_{F}^{2} [(c_{V} + c_{A})^{2} (k \cdot p) (k^{'} \cdot p^{'}) + (c_{V} - c_{A})^{2} (k \cdot p^{'}) (k^{'} \cdot p)] .

(4)

The massless Mandelstam variables satisfy

s + t + u = 0

, and the two products are

s^{2} /4

and

u^{2} /4

, respectively. Therefore

\overline{∣ M_{μ} ∣^{2}} = 4 G_{F}^{2} [(c_{V} + c_{A})^{2} s^{2} + (c_{V} - c_{A})^{2} u^{2}] .

(5)

In the centre-of-momentum frame, momentum conservation gives

E = s /2

. If

z = cos θ

is the angle cosine between

p

and

p^{'}

, then

u = - s (1 + z) /2

. The massless two-body Lorentz-invariant phase space is

d ρ_{2} = d Ω/ (32 π^{2})

, while the flux is

2 s

. Thus

\frac{d σ}{d z} = \frac{∣ M ∣ ^{2}}{32 π s} = \frac{G _{F}^{2} s}{8 π} [(c_{V} + c_{A})^{2} + \frac{1}{4} (c_{V} - c_{A})^{2} (1 + z)^{2}] .

(6)

Integrate

\int_{- 1}^{1} d z = 2

and

\int_{- 1}^{1} (1 + z)^{2} d z = 8/3

σ_{μ} = \frac{G _{F}^{2} s}{4 π} [(c_{V} + c_{A})^{2} + \frac{1}{3} (c_{V} - c_{A})^{2}] = \frac{G _{F}^{2} s}{3 π} (c_{V}^{2} + c_{V} c_{A} + c_{A}^{2}) .

(7)

Consequently the intended coefficients are

F (s) = \frac{s}{3 π}, A = 1, B = 1, C = 0.

(8)

This is the massless neutrino-electron contact cross-section.

For completeness, the literal printed muon-neutrino charged-current term would instead shift both couplings by one, exactly as in part (c):

σ_{μ, literal} = \frac{G _{F}^{2} s}{3 π} [c_{V}^{2} + c_{V} c_{A} + c_{A}^{2} + 3 (c_{V} + c_{A}) + 3] .

(9)

For arbitrary real

c_{V}, c_{A}

, its linear terms cannot be represented by the requested quadratic form with coupling-independent constants. The Standard Model result and the literal printed interaction are not the same problem.

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Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 305 / 3 / b / Solution Created 2026-10-03 Updated 2026-10-14

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The tree-level Feynman diagram contains one weak charged current vertex. Fermion arrows point along fermion-number flow, so the outgoing antiquark arrow points toward the vertex.

Figure 1.
Tree-level W-plus decay into an outgoing quark and antiquark with fermion-flow arrows
. **Details**.

For one fixed matching color, the scattering amplitude is

M = \frac{g}{2 2} V_{q q^{'}} ϵ_{μ} (p) \overset{u}{ˉ} (k) γ^{μ} (1 - γ^{5}) v (k^{'}) .

(1)

An overall phase from the vertex has no effect on the width. The fermion spin sums give

\sum u \overset{u}{ˉ} = \neq k

\sum v \overset{v}{ˉ} = \neq k^{'}

in the massless approximation. For the three initial polarizations, use a spin average of

1/3

. The gamma matrix trace identities give

T^{μν} = Tr [\neq k γ^{μ} (1 - γ^{5}) \neq k^{'} γ^{ν} (1 - γ^{5})] = 8 (k^{μ} k^{' ν} + k^{ν} k^{' μ} - g^{μν} k \cdot k^{'} + i ϵ^{μν ρ σ} k_{ρ} k_{σ}^{'}) .

(2)

The sign of the last term follows the PDF's

γ^{5}

convention and does not affect this decay: the polarization sum is symmetric. Since

p = k + k^{'}

and the daughter masses vanish,

p_{μ} T^{μν} = 0

and

k \cdot k^{'} = M_{W}^{2} /2

. Therefore

\overline{∣ M ∣^{2}}_{one color} = \frac{g ^{2} ∣ V _{q q^{'}} ∣ ^{2}}{24} (- g_{μν} + \frac{p _{μ} p _{ν}}{M _{W}^{2}}) T^{μν} = \frac{g ^{2} M _{W}^{2}}{3} ∣ V_{q q^{'}} ∣^{2} .

(3)

In the rest frame, the two-body Lorentz-invariant phase space integrates to

1/ (8 π)

: after the spatial delta function sets

k^{'} = - k

, its radial delta function fixes

∣ k ∣ = M_{W} /2

. Explicitly,

\int d Φ_{2} = \frac{1}{16 π ^{2}} \int d Ω \int_{0}^{\infty} d k δ (M_{W} - 2 k) = \frac{1}{8 π} .

(4)

The decay formula consequently gives

Γ_{one color} = \frac{∣ M ∣ ^{2}}{16 π M _{W}} = \frac{g ^{2} M _{W}}{48 π} ∣ V_{q q^{'}} ∣^{2} = \frac{G _{F} M _{W}^{3}}{6 π 2} ∣ V_{q q^{'}} ∣^{2} .

(5)

This is the printed expression, interpreted for one color. For a physical quark-flavor channel there are three orthogonal final color states. Their probabilities add; no initial color average is present for a colorless W boson. Thus

Γ_{W^{+} \to q \overset{q}{ˉ}^{'}} = N_{c} \frac{G _{F} M _{W}^{3}}{6 π 2} ∣ V_{q q^{'}} ∣^{2}, N_{c} = 3.

(6)

The PDF's formula omits this color multiplicity if read as the ordinary inclusive flavor width. It is also the familiar normalization for a colorless lepton channel.

The six physically accessible quark combinations are $u \overset{ˉ}{d}, u \overset{s}{ˉ}, u \overset{ˉ}{b}, c \overset{ˉ}{d}, c \overset{s}{ˉ}, c \overset{ˉ}{b}$ . A real on-shell W boson cannot produce a top quark; the massless approximation is applied to the accessible daughters and does not open a physically forbidden top channel. CKM matrix unitarity gives

\sum_{q^{'} = d, s, b} ∣ V_{u q^{'}} ∣^{2} = \sum_{q^{'} = d, s, b} ∣ V_{c q^{'}} ∣^{2} = 1

. Therefore the physical color-summed hadronic width at this order is

Γ_{had} = \frac{G _{F} M _{W}^{3}}{π 2} .

(7)

If the printed one-color convention is retained for every channel, its sum is instead

G_{F} M_{W}^{3} / (3 π 2)

. In the artificial theory where all three up-type flavors, including the top, are kinematically massless, there would be nine channels and the physical sum would be

9 G_{F} M_{W}^{3} / (6 π 2)

; that is not the on-shell Standard Model channel list.

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Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 305 / 3 / b / Solution Created 2026-10-03 Updated 2026-10-05

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Use metric

(+ - - -)

ϵ^{0123} = + 1

and relativistically normalized external spinors. The Fermi interaction gives

M = \frac{G _{F}}{2} [\overset{v}{ˉ} (p_{2}) γ^{α} (P + Q γ^{5}) u (p_{1})] [\overset{u}{ˉ} (k_{1}) γ_{α} (P + Q γ^{5}) v (k_{2})],

(1)

up to an irrelevant phase. The fermion spin sums replace

u \overset{u}{ˉ}

and

v \overset{v}{ˉ}

by the slashed massless momenta. Write

C = P^{2} + Q^{2}

. The gamma matrix trace identities yield

T^{α β} (a, b) = Tr [\neq a γ^{α} (P + Q γ^{5}) \neq b γ^{β} (P + Q γ^{5})] = 4 C (a^{α} b^{β} + a^{β} b^{α} - g^{α β} a \cdot b) + 8 i PQ ϵ^{α βρ σ} a_{ρ} b_{σ} .

(2)

The symmetric-antisymmetric mixed contractions vanish. The symmetric contraction is

2 [(a \cdot c) (b \cdot d) + (a \cdot d) (b \cdot c)]

, while the two epsilon tensors contract to

- 2 [(a \cdot c) (b \cdot d) - (a \cdot d) (b \cdot c)]

. Including

G_{F}^{2} /2

and the

1/4

initial-spin average therefore gives

\overline{∣ M ∣^{2}} = 4 G_{F}^{2} [(C^{2} + 4 P^{2} Q^{2}) (p_{2} \cdot k_{1}) (p_{1} \cdot k_{2}) + (C^{2} - 4 P^{2} Q^{2}) (p_{2} \cdot k_{2}) (p_{1} \cdot k_{1})] .

(3)

The singlet color charge contractions

δ_{ij} δ_{k l}

have net factor one after summing final colours and averaging initial color charge states. Thus this is also the colour-averaged partonic result; no extra factor three is required.

In the centre-of-momentum frame, let

c_{θ} = cos θ

. The Mandelstam variables satisfy

t = - s (1 - c_{θ}) /2

and

u = - s (1 + c_{θ}) /2

, giving

(p_{2} \cdot k_{1}) (p_{1} \cdot k_{2}) = s^{2} (1 + c_{θ})^{2} /16

and

(p_{2} \cdot k_{2}) (p_{1} \cdot k_{1}) = s^{2} (1 - c_{θ})^{2} /16

. Hence

\overline{∣ M ∣^{2}} = \frac{G _{F}^{2} s ^{2}}{2} [C^{2} (1 + c_{θ}^{2}) + 8 P^{2} Q^{2} c_{θ}] .

(4)

The massless invariant flux factor is

2 s

and

d Φ_{2} = d Ω/ (32 π^{2})

, so the weak charged-current quark scattering cross-section is

F (s) = \frac{G _{F}^{2} s}{128 π ^{2}}, H_{1} = (P^{2} + Q^{2})^{2}, H_{2} = 8 P^{2} Q^{2} .

(5)

For a purely left-handed weak charged current,

P = 1

Q = - 1

, this reduces to

d σ / d Ω = G_{F}^{2} s (1 + cos θ)^{2} / (32 π^{2})

, fixing the sign of the forward term. Purely right-handed weak charged currents at both vertices have the same unpolarized angular law. The factorization convention for

F, H_{1}, H_{2}

could be rescaled by a common numerical factor; the displayed product fixes the normalization.

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Past exam of the quantum field theory course of the University of Oxford / 2022 / 2 / c / Solution 2026-10-25

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Write

u_{i} = u^{s_{i}} (p_{i})

and

v_{i} = v^{s_{i}} (p_{i})

. With

S_{f i} = δ_{f i} + i (2 π)^{4} δ^{(4)} (p_{1} + p_{2} - p_{3} - p_{4}) M_{f i}

, choose

M_{p_{1}, p_{2}, p_{3}, p_{4}}^{s_{1}, s_{2}, s_{3}, s_{4}} = g^{2} (T - A), T = (\overset{u}{ˉ}_{3} u_{1}) (\overset{v}{ˉ}_{2} v_{4}), A = (\overset{v}{ˉ}_{2} u_{1}) (\overset{u}{ˉ}_{3} v_{4}) .

(1)

The relative minus sign is fixed by the canonical anticommutation relations, not by the Mandelstam variables. It follows by crossing symmetry from the identical-fermion scattering amplitude in i, or directly by the two inequivalent external contractions of

g^{2} (\overset{ˉ}{ψ} ψ)^{2} /2

. For an explicit sign check, fix incoming state order

b_{1}^{†} d_{2}^{†} ∣0 ⟩

and outgoing order

b_{3}^{†} d_{4}^{†} ∣0 ⟩

. The direct contraction contains

(b_{3}^{†} b_{1}) (d_{2} d_{4}^{†})

, whose coefficient in the ordered product

b_{3}^{†} d_{4}^{†} d_{2} b_{1}

- 1

; the annihilation contraction

(b_{3}^{†} d_{4}^{†}) (d_{2} b_{1})

has coefficient

+ 1

. The common sign can change with the external-state convention, while their relative minus sign cannot.

Let

r = m^{2}

. The external particles are on shell, so

s + t + u = 4 r

and

p_{1} \cdot p_{2} = p_{3} \cdot p_{4} = \frac{s}{2} - r, p_{1} \cdot p_{3} = p_{2} \cdot p_{4} = r - \frac{t}{2}, p_{1} \cdot p_{4} = p_{2} \cdot p_{3} = r - \frac{u}{2} .

(2)

Use the supplied spin sums and gamma matrix trace identities. The two diagonal terms in the initial spin average are

\frac{1}{4} \sum ∣ T ∣^{2} = \frac{1}{4} Tr [(\neq p_{3} + m) (\neq p_{1} + m)] Tr [(\neq p_{2} - m) (\neq p_{4} - m)] = (t - 4 r)^{2},

(3)

\frac{1}{4} \sum ∣ A ∣^{2} = \frac{1}{4} Tr [(\neq p_{2} - m) (\neq p_{1} + m)] Tr [(\neq p_{3} + m) (\neq p_{4} - m)] = (s - 4 r)^{2} .

(4)

For the interference term, the spin sum is a single Dirac trace:

R := \sum T A^{*} = Tr [(\neq p_{3} + m) (\neq p_{1} + m) (\neq p_{2} - m) (\neq p_{4} - m)] .

(5)

Terms with an odd number of Dirac gamma matrices vanish. Expanding the remaining terms gives

\frac{R}{4} = (p_{3} \cdot p_{1}) (p_{2} \cdot p_{4}) - (p_{3} \cdot p_{2}) (p_{1} \cdot p_{4}) + (p_{3} \cdot p_{4}) (p_{1} \cdot p_{2}) + r (p_{3} \cdot p_{1} - p_{3} \cdot p_{2} - p_{3} \cdot p_{4} - p_{1} \cdot p_{2} - p_{1} \cdot p_{4} + p_{2} \cdot p_{4}) + r^{2} .

(6)

Substituting the Mandelstam variables reduces this to

R = s^{2} + t^{2} - u^{2} + 16 r u - 16 r^{2} = (s - 4 r)^{2} + (t - 4 r)^{2} - (u - 4 r)^{2} .

(7)

It is real. The initial spin average therefore yields

\overline{∣ M ∣^{2}} = g^{4} [(t - 4 r)^{2} + (s - 4 r)^{2} - \frac{R}{2}],

(8)

\overline{∣ M ∣^{2}} = \frac{g ^{4}}{2} [(s - 4 m^{2})^{2} + (t - 4 m^{2})^{2} + (u - 4 m^{2})^{2}] .

(9)

The displayed answer in the source is false. At threshold,

p_{i} = (m, 0)

, hence

s = 4 m^{2}

and

t = u = 0

. Then

\overset{v}{ˉ} u = \overset{u}{ˉ} v = 0

, while the spin sums give

\overline{∣ M ∣^{2}} = 16 g^{4} m^{4}

. The printed expression instead gives

20 g^{4} m^{4}

. This is also a counterexample in the continuous threshold limit from physical nonzero incoming momenta. At

m = 0

there is a further discrepancy: the correct result is

g^{4} (s^{2} + t^{2} + u^{2}) /2

, rather than the printed

g^{4} (3 s^{2} + 3 t^{2} - u^{2}) /2

. Changing a common phase of the scattering amplitude cannot change either comparison.

The requested

s \leftrightarrow t

symmetry nevertheless holds for the correct answer. Crossing symmetry exchanges the two ways of pairing the four external fermion lines; the momentum-independent auxiliary-field propagator supplies no channel-dependent denominator. The relative fermionic sign can reverse the crossed scattering amplitude, but disappears in its squared modulus. With