Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 312 / 1

Table of contents
- i 1
  - Solution i
- ii 1
  - Solution ii
- iii 1
  - Solution iii

i

Solution

Varying the scalar action and Fourier transforming the spatial coordinates gives the Klein-Gordon equation

ϕ_{k}^{''} + 2 \frac{a ^{'}}{a} ϕ_{k}^{'} + (k^{2} + a^{2} m^{2}) ϕ_{k} = 0.

In de Sitter spacetime,

a = - 1/ (Hη)

and

m^{2} = 2 H^{2}

, so

ϕ_{k}^{''} - \frac{2}{η} ϕ_{k}^{'} + (k^{2} + \frac{2}{η ^{2}}) ϕ_{k} = 0 .

Direct substitution shows that

η e^{- ik η}

and

η e^{ik η}

are independent solutions. Hence

ϕ (η, k) = η [C_{+} (k) e^{- ik η} + C_{-} (k) e^{ik η}] .

The mass value is the one that gives a conformally coupled scalar field in four-dimensional de Sitter spacetime.

ii

Solution

For

f (k, η) = \frac{Hη}{2 k} e^{- ik η},

the specified cosmological bulk-to-boundary propagator is

G_{R} (η, p) = f (p, η_{0}) f^{*} (p, η) = \frac{H ^{2} η _{0} η}{2 p} e^{- i p (η_{0} - η)} .

The right-right cosmological bulk-to-bulk propagator is time ordered:

G_{RR} (η_{1}, η_{2}, p) = \frac{H ^{2} η _{1} η _{2}}{2 p} e^{- i p ∣ η_{1} - η_{2} ∣} .

The mixed Wightman function in the order stated in the question is

G_{L R} (η_{1}, η_{2}, p) = f (p, η_{1}) f^{*} (p, η_{2}) = \frac{H ^{2} η _{1} η _{2}}{2 p} e^{- i p (η_{1} - η_{2})} .

The suppressed factor in each expression is

(2 π)^{3} δ^{(3)} (p + p^{'})

.

iii

Solution

The exchange graph has one quartic vertex attached to

k_{1}, k_{2}, k_{3}

, a second attached to

k_{4}, k_{5}, k_{6}

, and an internal line of momentum

k_{I} = k_{1} + k_{2} + k_{3} = - (k_{4} + k_{5} + k_{6}) .

Schematically,

(k_{1}, k_{2}, k_{3}) ⟶ ∙ -_{k_{I}} ∙ ⟵ (k_{4}, k_{5}, k_{6}) .

The Schwinger-Keldysh conjugation relation gives

B_{6}^{(LL)} = (B_{6}^{(RR)})^{*}, B_{6}^{(R L)} = (B_{6}^{(L R)})^{*} .

η_{0} = 0

only in integration limits and phases, retaining its leading explicit power from the six external propagators. The scale factors, six external propagators, and one internal propagator leave the common coefficient

P = \frac{λ ^{2} H ^{6}}{2 ^{7} k _{1} k _{2} k _{3} k _{4} k _{5} k _{6} k _{I}} .

For two right-branch vertices, split the time-ordered integral into

η_{1} > η_{2}

and

η_{2} > η_{1}

:

\int_{- \infty}^{0} d η_{1} d η_{2}, e^{i k_{L} η_{1} + i k_{R} η_{2} - i k_{I} ∣ η_{1} - η_{2} ∣} = - \frac{1}{k _{T}} (\frac{1}{E _{R}} + \frac{1}{E _{L}}) .

The two right-vertex factors contribute the compensating minus sign, so

B_{6}^{(RR)} = P η_{0}^{6} \frac{1}{k _{T}} (\frac{1}{E _{R}} + \frac{1}{E _{L}}) .

For one vertex on each branch, the two integrals factorize:

B_{6}^{(L R)} = P η_{0}^{6} \frac{1}{E _{L} E _{R}} .

Thus the quantities requested in the question are

A = B = P .

At late time these expressions are real. Summing all four in-in formalism assignments gives

B_{6} = 2 (B_{6}^{(RR)} + B_{6}^{(L R)}) = \frac{λ ^{2} H ^{6} η _{0}^{6}}{2 ^{5} k _{1} k _{2} k _{3} k _{4} k _{5} k _{6} k _{I}} \frac{k _{T} + k _{I}}{k _{T} E _{L} E _{R}} .

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