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 Incoming links: Past exam of the quantum field theory course of the University of Oxford / 2016 / 3 / b / Solution

Past exam of the quantum field theory course of the University of Oxford / 2016 / 3 / c / Solution 2026-10-25

Use the notation

Δ_{12} = Δ (y_{1} - y_{2})

Δ_{x z} = Δ (x - z)

and

\int_{x z} = \int d^{D} x d^{D} z

. The order-

λ

term vanishes because it has five fields. At order

λ^{2}

, enumerate the eight-field Wick contractions, whose expansion denominator is

2! (3!)^{2} = 72

If the two external fields contract with each other, the remaining vacuum diagrams are a pair of cubic vertices joined by three lines, with six Wick contractions, and a pair joined by one line with a tadpole diagram at each end, with nine Wick contractions. Define their combined integrated weight by

V_{2} = \int_{x z} [\frac{1}{12} Δ_{x z}^{3} + \frac{1}{8} t^{2} Δ_{x z}] .

(1)

Thus

Z / Z_{0} = 1 + (- iλ)^{2} V_{2} + O (λ^{4})

. The numerator is

\frac{N _{2}}{Z _{0}} = Δ_{12} [1 + (- iλ)^{2} V_{2}] + (- iλ)^{2} \int_{x z} [\frac{1}{2} K_{1} (x) K_{2} (z) Δ_{x z}^{2} + \frac{1}{2} K_{1} (x) K_{2} (x) Δ_{x z} t + \frac{1}{4} K_{1} (x) K_{2} (z) t^{2}] + O (λ^{4}) .

(2)

The three latter terms have respectively 36, 36 and 18 Wick contractions. Together with the six and nine Wick contractions, they exhaust all

7!! = 105

pairings.

Dividing by

Z / Z_{0}

cancels the two vacuum diagrams, giving

⟨ Tϕ (y_{1}) ϕ (y_{2})⟩ = Δ_{12} + (- iλ)^{2} \int_{x z} [\frac{1}{2} K_{1} (x) K_{2} (z) Δ_{x z}^{2} + \frac{1}{2} K_{1} (x) K_{2} (x) Δ_{x z} t + \frac{1}{4} K_{1} (x) K_{2} (z) t^{2}] + O (λ^{4}) .

(3)

The first correction is the two-line bubble diagram, with symmetry factor

S = 2

. The second is the connected tadpole contribution to a cubic-theory two-point function: both external lines meet one vertex, joined by a single line to a second vertex bearing a tadpole diagram; its symmetry factor is also

S = 2

. Interchanging the two integrated vertices is already included in the 36 contractions and gives no extra factor.

The final term is the disconnected Feynman diagram consisting of two separate one-point tadpole diagrams, with coefficient

1/4

. It survives cancellation of vacuum bubbles, since each component touches an external point. Indeed it is precisely the leading

⟨ ϕ (y_{1})⟩ ⟨ ϕ (y_{2})⟩

. It would disappear from the connected two-point correlation function, but the question defines the full correlation function. No normal ordering or tadpole subtraction of the interaction was specified. All terms are depicted in solution, and all spacetime integrals remain unevaluated.

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Past exam of the quantum field theory course of the University of Oxford / 2016 / 3 / d / Solution 2026-10-25

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For a connected cubic Feynman diagram with

V

internal vertices,

I

internal lines and three external lines,

3 V = 2 I + 3

. Hence

V

is odd and the loop order is

L = I - V + 1 = (V - 1) /2

. Through order

λ^{3}

, there is one tree-level Feynman diagram at

V = 1

and seven labelled one-loop Feynman diagrams at

V = 3

One triangle, with one external line at each vertex. Fixing the three external labels leaves no internal automorphism, so its symmetry factor is $S = 1$ .
Three bubble diagrams on external branches of the tree, one for each choice of the distinguished external label $i$ . The other two external lines meet the remaining tree vertex. Exchanging the two parallel lines gives symmetry factor $S = 2$ for each labelled graph.
Three tadpole diagrams at the end of a chain. The middle vertex carries the distinguished external label $i$ , the other two external lines meet one end vertex, and the other end bears the loop. Reversing the two half-edges of the loop gives symmetry factor $S = 2$ for each labelled graph.

Figure 1.
All connected cubic three-point diagrams through third order: one tree and seven labelled one-loop graphs; a diagram of symmetry factor S has coefficient 1/S
. **Details**.

For an explicit check, use the propagator notation of solution and define, with

{i, j, k} = {1, 2, 3}

T B_{i} A_{i} = \int d^{D} x d^{D} z d^{D} w K_{1} (x) K_{2} (z) K_{3} (w) Δ_{x z} Δ_{z w} Δ_{w x}, = \int d^{D} x d^{D} z d^{D} w K_{i} (z) K_{j} (w) K_{k} (w) Δ_{x z}^{2} Δ_{x w}, = \int d^{D} x d^{D} z d^{D} w K_{i} (z) K_{j} (w) K_{k} (w) Δ_{x z} Δ_{z w} t .

(1)

There is no ordering of

j, k

in this notation. The complete connected correlation function through this order is

G_{3, c} = (- iλ) \int d^{D} x K_{1} (x) K_{2} (x) K_{3} (x) + (- iλ)^{3} [T + \frac{1}{2} i = 1 \sum 3 (B_{i} + A_{i})] + O (λ^{5}) .

(2)

For completeness of the classification, a one-loop connected internal graph either has a three-edge cycle, a pair of parallel edges and a bridge, or a self-loop and two bridges. The cubic degree condition forces exactly the external attachments just listed. Direct Wick contraction counting gives 1296 pairings for the triangle and 648 for each of the six other labelled graphs, divided by

3! (3!)^{3} = 1296

Connected does not mean one-particle irreducible. After amputation of external propagators, only the triangle is one-particle irreducible at this order; the six other loop graphs belong to the requested connected answer. The tadpole diagrams cannot be discarded without an additional subtraction prescription.

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