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ID: past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2017/4/d/solution

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Past exam of the quantum field theory course of the University of Oxford / 2017 / 4 / d / Solution by

Codex  0 2026-10-25

At fixed regulator abbreviate

A (ϵ) = \frac{a _{1}}{ϵ} + a_{2} + a_{3} ϵ + \dots, C (ϵ) = \frac{z _{1}}{ϵ} + z_{2} + z_{3} ϵ + \dots .

The coupling expansion is

g = g_{0} + A g_{0}^{2} + O (g_{0}^{3})

, whose inverse is

g_{0} = g - A g^{2} + O (g^{3})

. Using part c before removing the regulator,

β (g) γ (g) = - ϵ g_{0} (1 + 2 A g_{0}) + O (g_{0}^{3}) = - ϵ g - ϵ A g^{2} + O (g^{3}), = - ϵ g_{0} \frac{C}{1 + C g _{0}} + O (g_{0}^{2}) = - ϵ C g + O (g^{2}) .

Therefore the regulator-free coefficients at the orders determined by the supplied expansions are

β (g) = - a_{1} g^{2} + O (g^{3}), γ (g) = - z_{1} g + O (g^{2}) .

The canonical term

- ϵ g

vanishes, and the factors of

ϵ

cancel the simple poles. The finite coefficients

a_{2}, z_{2}

and higher positive powers of

ϵ

do not enter these leading regulator-free terms. Inverting the coupling is essential: substituting

g_{0} = g

too early would incorrectly give

- 2 a_{1} g^{2}

.

The calculation establishes finiteness only at the displayed perturbative orders. Higher-order pole cancellations require the higher-order coefficients, as the next part makes explicit.

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