 Solution

ID: past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2024/1/c/solution

ϵ

T_{t} (ϵ) = 1 - i ϵH + O (ϵ^{2})

. The corresponding Heisenberg picture field transformation is conjugation by its inverse:

T_{t} (ϵ)^{†} ϕ (t, x) T_{t} (ϵ) = ϕ + i ϵ [H, ϕ] + O (ϵ^{2}) = ϕ + ϵ \partial_{t} ϕ + O (ϵ^{2}) = ϕ (t + ϵ, x) + O (ϵ^{2}) .

Here a gives

[H, ϕ] = - i \partial_{t} ϕ

Likewise let

T_{s} (a) = 1 + i a \cdot P + O (∣ a ∣^{2})

, with the paper's sign for its spatial translation operator. From b,

T_{s} (a)^{†} ϕ (t, x) T_{s} (a) = ϕ + i a \cdot [ϕ, P] + O (∣ a ∣^{2}) = ϕ + a \cdot \nabla ϕ + O (∣ a ∣^{2}) = ϕ (t, x + a) + O (∣ a ∣^{2}) .

These are precisely the first-order Taylor expansions. Since

H

and

P

commute in this translation-invariant theory, the finite result is

ϕ (x) = e^{i P \cdot x} ϕ (0) e^{- i P \cdot x}, P \cdot x = H t - P \cdot x .

 New to topics? Read the docs here!