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Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 3 / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

For the interaction sign in this paper, the cubic Feynman rule is

+ iλ

, while the scalar Feynman propagator is

i / (k^{2} - m^{2} + i ϵ)

. Each exchange graph therefore contributes

- i λ^{2} / (k^{2} - m^{2} + i ϵ)

i M

. The three graphs give

M = - λ^{2} (\frac{1}{s - m ^{2} + i ϵ} + \frac{1}{t - m ^{2} + i ϵ} + \frac{1}{u - m ^{2} + i ϵ}) .

The equal-mass Mandelstam variables satisfy

s + t + u = 4 m^{2}

, so

u - m^{2} = 3 m^{2} - s - t

. In the physical elastic region with

m > 0

, none of these denominators vanishes, and the tree amplitude is real up to its irrelevant overall convention. Thus

∣ M ∣^{2} = λ^{4} [\frac{1}{s - m ^{2}} + \frac{1}{t - m ^{2}} + \frac{1}{3 m ^{2} - s - t}]^{2} .

The square includes interference between all three channels; summing their individual squared contributions would omit physical cross terms. If a denominator approaches a pole, retain the Feynman i-epsilon prescription in the modulus before taking a limit.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 3 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

In phi cubed theory, each vertex has three legs. A connected four-point tree-level Feynman diagram therefore needs two cubic vertices and one internal propagator. There are exactly three ways to partition the labeled external legs between these vertices:

       s channel               t channel              u channel

 p \          / p'         p ---o--- p'            p ---o--- q'
    o--------o                  |                       |
 q /          \ q'         q ---o--- q'            q ---o--- p'

 internal: p+q             internal: p-p'          internal: p-q'

Solid lines denote the real scalar field propagator. The labels specify which external pair meets at each vertex, so the last two graphs are different even though their unlabeled shapes agree. This is the four-point tree amplitude in phi cubed theory:

three exchange diagrams, in the s, t, u channels; each is of order λ^{2} .

There is no four-point contact vertex in the given cubic interaction.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 2 / vii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

Differentiate the Noether charge and use

\dot{ϕ}_{a} = π_{a}

and the Klein-Gordon equation for each species. The momentum products cancel because the two species commute:

\dot{Q} = \int d^{3} x [π_{2} π_{1} + ϕ_{2} (\nabla^{2} - m_{1}^{2}) ϕ_{1} - π_{1} π_{2} - ϕ_{1} (\nabla^{2} - m_{2}^{2}) ϕ_{2}] = \int d^{3} x [ϕ_{2} \nabla^{2} ϕ_{1} - ϕ_{1} \nabla^{2} ϕ_{2}] + (m_{2}^{2} - m_{1}^{2}) \int d^{3} x ϕ_{1} ϕ_{2} .

The first integral is the boundary flux of

ϕ_{2} \nabla ϕ_{1} - ϕ_{1} \nabla ϕ_{2}

, so it vanishes under the same boundary condition used for charge conservation. Hence

\frac{d Q}{d t} = (m_{2}^{2} - m_{1}^{2}) \int d^{3} x ϕ_{1} ϕ_{2} .

Conservation must hold for arbitrary configurations, rather than only a state in which this particular integral happens to vanish. Thus the internal rotation symmetry of two real scalar fields requires

m_{1}^{2} = m_{2}^{2}, and therefore m_{1} = m_{2} for nonnegative masses .

Equivalently, the local divergence is

\partial_{μ} j^{μ} = (m_{2}^{2} - m_{1}^{2}) ϕ_{1} ϕ_{2}

, exhibiting exactly how unequal masses break the symmetry.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 2 / vi / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

For the two free real scalar fields, let

L = L_{1} + L_{2}

. The infinitesimal internal rotation symmetry of two real scalar fields has

δ ϕ_{1} = θ ϕ_{2}

and

δ ϕ_{2} = - θ ϕ_{1}

, with constant

θ

. Its kinetic terms are invariant because this is an orthogonal rotation of the two-component field.

Under the assumed invariance of the full Lagrangian density, Noether's theorem gives the current per unit transformation parameter:

j^{μ} = \sum_{a = 1}^{2} \frac{\partial L}{\partial ( \partial _{μ} ϕ _{a} )} \frac{δ ϕ _{a}}{θ} = ϕ_{2} \partial^{μ} ϕ_{1} - ϕ_{1} \partial^{μ} ϕ_{2} .

The Noether current and Noether charge are therefore

j^{μ} = ϕ_{2} \partial^{μ} ϕ_{1} - ϕ_{1} \partial^{μ} ϕ_{2}, Q = \int d^{3} x (ϕ_{2} π_{1} - ϕ_{1} π_{2}) .

Independent species commute at equal time, so these products require no extra ordering sign. With vanishing boundary flux,

Q

is conserved. Part (vii) checks precisely what the symmetry assumption requires of the masses.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 2 / v / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

Use the free Hamiltonian operator and the canonical commutation relations. The kinetic momentum term gives

[H_{1}, ϕ_{1} (t, x)] = \frac{1}{2} \int d^{3} y [π_{1} (t, y)^{2}, ϕ_{1} (t, x)] = - i π_{1} (t, x) .

The Heisenberg picture equation therefore yields

\dot{ϕ}_{1} = π_{1}

For the momentum commutator, the mass term gives

i m_{1}^{2} ϕ_{1} (x)

. The gradient term gives

\frac{1}{2} \int d^{3} y [(\nabla ϕ_{1} (y))^{2}, π_{1} (x)] = i \int d^{3} y \nabla ϕ_{1} (y) \cdot \nabla_{y} δ^{(3)} (y - x) = - i \nabla^{2} ϕ_{1} (x),

where integration by parts differentiates the field instead of the Dirac delta distribution. Thus

[H_{1}, π_{1}] = i (m_{1}^{2} - \nabla^{2}) ϕ_{1}, \overset{π}{˙}_{1} = (\nabla^{2} - m_{1}^{2}) ϕ_{1} .

Combining the two first-order operator equations gives the Klein-Gordon equation:

(\partial_{t}^{2} - \nabla^{2} + m_{1}^{2}) ϕ_{1} = (□ + m_{1}^{2}) ϕ_{1} = 0.

(4)

Suitable decay or spatial smearing removes the boundary term in the integration by parts.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 2 / iv / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

The canonical quantization of a real scalar field promotes the classical equal-time brackets to the canonical commutation relations, in units

ℏ = 1

[ϕ_{1} (t, x), ϕ_{1} (t, y)] [π_{1} (t, x), π_{1} (t, y)] [ϕ_{1} (t, x), π_{1} (t, y)] = 0, = 0, = i δ^{(3)} (x - y) .

The Dirac delta distribution expresses locality of the conjugate variables. These are identities of operator-valued distributions, interpreted after spatial smearing where necessary.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 2 / iii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

Apply the Legendre transform in mechanics pointwise to the field Lagrangian density. Using

π_{1} = \dot{ϕ}_{1}

, the Hamiltonian density is

H_{1} = π_{1} \dot{ϕ}_{1} - L_{1} = \frac{1}{2} π_{1}^{2} + \frac{1}{2} ∣\nabla ϕ_{1} ∣^{2} + \frac{1}{2} m_{1}^{2} ϕ_{1}^{2} .

The free Hamiltonian operator is its spatial integral:

H_{1} = \frac{1}{2} \int_{R^{3}} d^{3} x [π_{1}^{2} + (\nabla ϕ_{1})^{2} + m_{1}^{2} ϕ_{1}^{2}] .

A constant vacuum-energy subtraction, if made, does not affect the commutators or the ensuing field equation.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 2 / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

For a free real scalar field, the Lagrangian density contains its quadratic kinetic term and mass term:

L_{1} = \frac{1}{2} \partial_{μ} ϕ_{1} \partial^{μ} ϕ_{1} - \frac{1}{2} m_{1}^{2} ϕ_{1}^{2} = \frac{1}{2} \dot{ϕ}_{1}^{2} - \frac{1}{2} ∣\nabla ϕ_{1} ∣^{2} - \frac{1}{2} m_{1}^{2} ϕ_{1}^{2} .

The field's conjugate momentum is

π_{1} = \partial L_{1} / \partial \dot{ϕ}_{1} = \dot{ϕ}_{1}

. The positive sign of the time-derivative term uses the metric convention fixed in Question 1.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 2 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

For a time-independent Hamiltonian operator and a Schrödinger picture observable with no explicit time dependence, the Heisenberg picture operator is

O_{H} (t) = e^{i H t} O_{S} e^{- i H t} .

Differentiation gives

\dot{O}_{H} = i H O_{H} - i O_{H} H = i [H, O_{H}]

, so

[H, O_{H}] = - i \frac{d O _{H}}{d t} .

O_{S}

itself depends explicitly on time, one must add

e^{i H t} (\partial_{t} O_{S}) e^{- i H t}

\dot{O}_{H}

; the displayed commutator relation uses the usual convention of no explicit time dependence for the observable.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 1 / v / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

Substitute the preceding spin average into the supplied differential scattering cross-section. Since

e^{4} = 16 π^{2} α^{2}

\frac{d σ}{d Ω} = \frac{α ^{2}}{4 s} (1 + cos^{2} θ) .

The angular integral is

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

. Consequently the relativistic scattering cross-section for the massless electron-positron annihilation into a muon pair is

σ = \frac{4 π α ^{2}}{3 s} .

The outgoing particles are distinct, so there is no identical-particle factor. Comparing with the requested monomial form gives

B = \frac{4 π}{3}, j = 2, k = - 1, l = 0, m = 0.

The integrated answer depends only on the incoming invariant

s

; the angular variables

t, u

have been integrated out.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 1 / iv / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

First derive the needed gamma-matrix traces directly from the Clifford algebra. The cyclic property of the trace gives

2 tr (γ^{μ} γ^{ν}) = tr {γ^{μ}, γ^{ν}} = 2 η^{μν} tr I_{4},

tr (γ^{μ} γ^{ν}) = 4 η^{μν}

. For

T = tr (γ^{μ} γ^{ν} γ^{ρ} γ^{σ})

, move

γ^{μ}

successively through the other three matrices and cycle the final term back to the front:

2 T = 2 η^{μν} tr (γ^{ρ} γ^{σ}) - 2 η^{μ ρ} tr (γ^{ν} γ^{σ}) + 2 η^{μ σ} tr (γ^{ν} γ^{ρ}) .

Therefore

tr (γ^{μ} γ^{ν} γ^{ρ} γ^{σ}) = 4 (η^{μν} η^{ρ σ} - η^{μ ρ} η^{ν σ} + η^{μ σ} η^{ν ρ}) .

These are all the trace identities required.

For massless external Dirac spinors, the fermion spin sums are

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

and

\sum_{r} v_{r} (q) \overset{v}{ˉ}_{r} (q) = \neq q

. With the initial spin average, part (iii) becomes a contraction of

L^{μν} = 4 (q^{μ} p^{ν} + q^{ν} p^{μ} - η^{μν} p \cdot q), H_{μν} = 4 (p_{μ}^{'} q_{ν}^{'} + p_{ν}^{'} q_{μ}^{'} - η_{μν} p^{'} \cdot q^{'}) .

(4)

Their contraction is

L^{μν} H_{μν} = 32 [(q \cdot p^{'}) (p \cdot q^{'}) + (q \cdot q^{'}) (p \cdot p^{'})] .

(5)

The massless Mandelstam variables give

p \cdot p^{'} = q \cdot q^{'} = - t /2

and

p \cdot q^{'} = q \cdot p^{'} = - u /2

, hence

\overline{∣ M ∣^{2}} = \frac{2 e ^{4}}{s ^{2}} (t^{2} + u^{2}) .

(6)

In the center of mass frame each particle has energy

s /2

, and

t = - \frac{s}{2} (1 - cos θ), u = - \frac{s}{2} (1 + cos θ) .

(7)

Substitution gives the concise angular answer:

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

(8)

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 1 / iii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

At this order there is one annihilation tree-level Feynman diagram. The squared-amplitude diagram is its product with its complex conjugate, with the same external momenta and spin labels in both factors:

 e-(p,s) --->\          /---> mu-(p',s')
             o ~~~~~~ o
 e+(q,r) <---/          \<--- mu+(q',r')

                     times

 e-(p,s) --->\          /---> mu-(p',s')
             o ~~~~~~ o                       *
 e+(q,r) <---/          \<--- mu+(q',r')

A wavy line is the virtual photon; the arrows show fermion-number flow. The star means complex conjugation of the entire second graph, including its external spinors. This represents

∣ M ∣^{2} = M M^{*}

for fixed spins, without adding a new physical loop to the amplitude.

For the spin sum in part (iv), identify each external spin label between the two copies and sum it. The fermion spin sum sews each spinor chain into a gamma-matrix trace. The two incoming unpolarized spin states are then handled by a spin average, giving the factor

1/4

. In particular, the diagrammatic contraction becomes

\overline{∣ M ∣^{2}} = \frac{e ^{4}}{4 s ^{2}} tr (\neq q γ^{μ} \neq p γ^{ν}) tr (\neq p^{'} γ_{μ} \neq q^{'} γ_{ν}) .

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 1 / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

Take the quantum electrodynamics interaction density to be

L_{int} = - e \overset{ˉ}{ψ} γ^{μ} ψ A_{μ}

. The momentum-space Feynman-gauge photon propagator, with Lorentz indices

μ, ν

and momentum

k

, is

D_{μν} (k) = \frac{- i η _{μν}}{k ^{2} + i ϵ} .

The electron-photon vertex from the QED Feynman rules is

- i e γ^{μ},

with its spinor indices connecting the incident and outgoing fermion lines. The muon-photon vertex has the same coupling. The Feynman i-epsilon prescription specifies the pole boundary condition; using a different overall charge-sign convention changes both vertices consistently and leaves the squared amplitude unchanged.

For the electron-positron annihilation into a muon pair, the exchanged photon has

k = p + q

and

k^{2} = s

. Up to the conventional overall phase, the scattering amplitude is

M = \frac{e ^{2}}{s} [\overset{v}{ˉ}_{r} (q) γ^{μ} u_{s} (p)] [\overset{u}{ˉ}_{s^{'}} (p^{'}) γ_{μ} v_{r^{'}} (q^{'})] .

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 1 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

Use

ℏ = c = 1

and the Minkowski metric

η = diag (1, - 1, - 1, - 1)

throughout. In the four-component Dirac spinor representation, the gamma matrices generate the Clifford algebra

{γ^{μ}, γ^{ν}} = γ^{μ} γ^{ν} + γ^{ν} γ^{μ} = 2 η^{μν} I_{4} .

Thus

(γ^{0})^{2} = I_{4}

(γ^{i})^{2} = - I_{4}

, and matrices with distinct indices anticommute. Indices are lowered with

η

, so

γ_{μ} = η_{μν} γ^{ν}

and

\neq p = p_{μ} γ^{μ}

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 219 / 4 / iii / c by

Codex  0 2026-10-03

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 219 / 4 / iii / b by

Codex  0 2026-10-03

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 219 / 4 / iii / a by

Codex  0 2026-10-03

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 219 / 4 / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

The probabilistic graphical model is the following Directed acyclic graph. The shared hyperparameters lie outside the plate; the plate repeats the local variables for

s = 1, \dots, N

. The observed colour is shaded, and the known error variance

r_{s}

is a square.

Figure 1.
Graphical model for intrinsic supernova colour, dust reddening and measurement noise
. Arrows from $μ_{C}, v$ to $C_{s}$ , from $τ$ to $E_{s}$ , and from $C_{s}, E_{s}, r_{s}$ to the observed $O_{s}$ encode the conditional factorization. The plate encloses $N$ local supernova models. **Details**.

Thus the parent sets are

pa (C_{s}) = {μ_{C}, v}, pa (E_{s}) = {τ}, pa (o_{s}) = {C_{s}, E_{s}, r_{s}}

. Independent hyperparameter prior factors are represented by the three root nodes. With the printed improper priors this is a factorization diagram; it becomes a normalized generative model after proper priors are supplied.