Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 301 / 1

Table of contents
- Solution 1

Solution

For a Schrödinger-picture state satisfying

i \frac{d}{d t} ∣ ψ (t) ⟩_{S} = (H_{0} + λ H_{int}) ∣ ψ (t) ⟩_{S},

define the interaction picture by

∣ ψ (t) ⟩_{I} = e^{i H_{0} t} ∣ ψ (t) ⟩_{S}, O_{I} (t) = e^{i H_{0} t} O_{S} e^{- i H_{0} t} .

Differentiation gives

i \frac{d}{d t} ∣ ψ (t) ⟩_{I} = H_{I} (t) ∣ ψ (t) ⟩_{I}, H_{I} (t) = λ e^{i H_{0} t} H_{int} e^{- i H_{0} t} .

The formal solution from

t_{0}

is the Dyson series

U_{I} (t, t_{0}) = T exp [- i \int_{t_{0}}^{t} H_{I} (s) d s] .

Through quadratic order,

U_{I} (t, t_{0}) = 1 - i \int_{t_{0}}^{t} d t_{1} H_{I} (t_{1}) - \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{1}} d t_{2} H_{I} (t_{1}) H_{I} (t_{2}) + O (λ^{3}) .

Differentiating gives

i \partial_{t} U_{I} (t, t_{0}) = H_{I} (t) [1 - i \int_{t_{0}}^{t} H_{I} (t_{2}) d t_{2}] + O (λ^{3}) = H_{I} (t) U_{I} (t, t_{0}) + O (λ^{3}),

which checks the equation to the requested order.

At the endpoints,

∣ i ⟩_{I} = e^{- i H_{0} T /2} ∣ i ⟩_{S}

and

∣ f ⟩_{I} = e^{i H_{0} T /2} ∣ f ⟩_{S}

. Therefore

T =_{I} ⟨ f ∣ U_{I} (T /2, - T /2) ∣ i ⟩_{I},

where

U_{I} (T /2, - T /2) = T exp [- iλ \int_{- T /2}^{T /2} d t \int d^{3} x ϕ_{1, I} (x) ϕ_{2, I} (x)^{3}] .

For

ϕ_{1} (p_{1}) \to ϕ_{2} (p_{1}^{'}) ϕ_{2} (p_{2}^{'}) ϕ_{2} (p_{3}^{'})

, the leading term is

A = - iλ \int d^{4} x ⟨ p_{1}^{'} p_{2}^{'} p_{3}^{'} ∣ ϕ_{1} (x) ϕ_{2} (x)^{3} ∣ p_{1} ⟩ .

There are

3!

contractions of the identical outgoing fields. With covariantly normalized states,

A = - i 3! λ (2 π)^{4} δ^{(4)} (p_{1} - p_{1}^{'} - p_{2}^{'} - p_{3}^{'}) .

Equivalently, the interaction vertex from

- λ ϕ_{1} ϕ_{2}^{3}

is

- i 3! λ

.

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