For bosonic fields, time ordering places operators with later time arguments to the left:

T {ϕ_{1} \dots ϕ_{n}} = ϕ_{π (1)} \dots ϕ_{π (n)}, t_{π (1)} \geq \dots \geq t_{π (n)} .

(1)

Normal ordering places all creation operators to the left of all annihilation operators and is denoted by colons. A Wick contraction is

contr (ϕ_{i}, ϕ_{j}) = ⟨ 0∣ T {ϕ_{i} ϕ_{j}} ∣0 ⟩ = D_{F} (x_{i} - x_{j}) .

(2)

Wick theorem states

T {ϕ_{1} \dots ϕ_{n}} =: ϕ_{1} \dots ϕ_{n} : + single contractions \sum : \dots : + double contractions \sum : \dots : + \dots,

(3)

where each sum runs over inequivalent disjoint pairings and contracted fields are replaced by their Feynman propagator.

ii

 0  0

Solution

 0  0

Write each free field as creation and annihilation parts,

ϕ_{i} = ϕ_{i}^{-} + ϕ_{i}^{+}

. Moving every annihilation part to the right produces commutators

[ϕ_{i}^{+}, ϕ_{j}^{-}]

, which are precisely the contractions. Applying the

n = 2

identity once and then normal ordering the remaining field gives

T {ϕ_{1} ϕ_{2} ϕ_{3}} =: ϕ_{1} ϕ_{2} ϕ_{3} : + D_{F} (x_{1} - x_{2}) : ϕ_{3} : + D_{F} (x_{1} - x_{3}) : ϕ_{2} : + D_{F} (x_{2} - x_{3}) : ϕ_{1} : .

(1)

No double contraction is possible for three fields. This is exactly Wick theorem at

n = 3

iii

 0  0

Solution

 0  0

For Phi-six theory, let

D_{0} = D_{F} (0)

and

D_{x y} = D_{F} (x - y)

. The Dyson series gives

Z [0] = ⟨ 0∣ T {S} ∣0 ⟩ = 1 - \frac{iλ}{6 !} \int d^{4} x ⟨ ϕ (x)^{6} ⟩_{0} + \frac{( - iλ ) ^{2}}{2 ( 6 ! ) ^{2}} \int d^{4} x d^{4} y ⟨ ϕ (x)^{6} ϕ (y)^{6} ⟩_{0} + O (λ^{3}) .

(1)

At one vertex, Wick theorem supplies

5!! = 15

pairings. At two vertices let

r

be the number of propagators joining them. It must be

0, 2, 4,

6

, and the number of contractions is

N_{r} = (r 6)^{2} r! ((5 - r)!!)^{2}, (N_{0}, N_{2}, N_{4}, N_{6}) = (225, 4050, 5400, 720) .

(2)

Therefore

Z [0] = 1 - \frac{iλ}{48} \int d^{4} x D_{0}^{3} + (- iλ)^{2} \int d^{4} x d^{4} y [\frac{D _{0}^{6}}{4608} + \frac{D _{0}^{4} D _{x y}^{2}}{256} + \frac{D _{0}^{2} D _{x y}^{4}}{192} + \frac{D _{x y}^{6}}{1440}] + O (λ^{3}) .

(3)

The four Vacuum Feynman diagram types are shown below. The

r = 0

term is two disconnected copies of the order-

λ

three-tadpole graph; the other three are connected.

Figure 1.
Vacuum diagrams in phi-six theory through second order
. At first order one six-valent vertex is paired into three tadpoles. At second order the two vertices can have two, four, or six connecting propagators, with the remaining legs paired into tadpoles.

Define

B_{1} = - \frac{iλ}{48} \int d^{4} x D_{0}^{3}

(4)

and let

B_{2}

be the sum of the

r = 2, 4, 6

terms in the second line. The disconnected

r = 0

contribution is exactly

B_{1}^{2} /2

. Hence

Z [0] = exp {B_{1} + B_{2} + O (λ^{3})},

(5)

which is the linked-cluster theorem: the logarithm of the vacuum amplitude is the sum of connected vacuum bubbles.

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 301 / 4

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