 Solution

ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2020/1/vii/solution

Only the term proportional to

α

k^{2}

cancels the massless denominator, leaving the standard Feynman integral

J_{D} = \int \frac{d ^{D} k}{( 2 π ) ^{D}} \frac{1}{( k ^{2} - m ^{2} + i 0 ) ^{2}} = \frac{i}{( 4 π ) ^{D /2}} Γ (2 - D /2) (m^{2})^{D /2 - 2} .

Γ (ϵ /2) = 2/ ϵ + O (1)

and

α = 1 + O (ϵ)

. Therefore

i V_{loop}^{μ}_{UV} = - i e_{D} γ^{μ} \frac{e ^{2}}{8 π ^{2} ϵ} .

The counterterm contribution is

- i e_{D} δ_{1} γ^{μ}

. Cancellation of the ultraviolet pole fixes

Z_{1} = 1 + δ_{1} = 1 - \frac{e ^{2}}{8 π ^{2} ϵ} + O (e^{4}) (D = 4 - ϵ),

where the displayed expression uses the minimal subtraction scheme; a different renormalization scheme may add finite

O (e^{2})

terms. In the frequently used convention

D = 4 - 2 ε

, the same result is

Z_{1} = 1 - e^{2} / (16 π^{2} ε) + O (e^{4})

. The sign agrees with

Z_{1} = Z_{2}

 New to topics? Read the docs here!