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ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2017/1/d/solution

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Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2017 / 1 / d / Solution by

Codex  0 2026-10-25

Apply the dimensionally regulated tadpole integral before removing the infrared regulator:

I_{γ} (m_{γ}) = μ^{ϵ} \int \frac{d ^{d} l}{( 2 π ) ^{d}} \frac{1}{l ^{2} + m _{γ}^{2}} = \frac{i μ ^{ϵ}}{( 4 π ) ^{2 - ϵ /2}} Γ (- 1 + \frac{ϵ}{2}) (m_{γ}^{2})^{1 - ϵ /2} .

(1)

At fixed regulator

0 < Re ϵ < 2

away from gamma-function poles, the power of

m_{γ}^{2}

has positive real part. Consequently

m_{γ} \to 0 lim I_{γ} (m_{γ}) = 0.

(2)

Both the correctly weighted tadpole and the source's doubled version vanish. This is the near-four-dimensional limit appropriate to the question; the order of limits is important. Equivalently the exactly massless tadpole is a scaleless integral and is zero in dimensional regularization. The statement is not valid for every arbitrary fixed dimension: for example,

ϵ = 5/2

gives a nonzero prefactor and growth as

m_{γ}^{- 1/2}

. One must keep the regulator in its stated near-four-dimensional domain.

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