Master's Thesis in Theoretical Physics - Universiteit Utrecht

Σ ren (x + ; x − ′ ), which corresponds to the distance function ∆x2 +− defined in Eq. (6.41). Therewill be only two changes: first of all, there will be an overall minus sign because the verticescontain a sign. For the ++ case, the sign will be the same, but for the +− and −+ case thesign will be different. Secondly, remember that in the renormalization procedure we usedthe identity (6.43) for ∆x++ 2 to extract the singularity for D = 4 to a local term (see Eq.(6.69)). For the ∆x+− 2 term the local term does not appear in the identity (6.43), whichmeans that the singularity disappears completely and that the counterterm δZ 2 = 0. Thisgives as a result for the self-energies Σ ++ and Σ +−Σ ++ (x, x ′ ) = −|f | 2 (aa′ ) 3 22 1 0π 4 ✁ ∂∂ 4 [ ln 2 (µ 2 ∆x 2 ++ ) − 2ln(µ2 ∆x 2 ++ )]−|f | 2 (aa′ ) 3 22 5 π 2 ln(aa′ )i✁∂δ(x − x ′ )−|f | 2 (1 − ɛ)2 HH ′ (aa ′ ) 5 22 9 π 4 (ν 2 − 1 4)× { ✁∂∂ 2 ln 2 (K HH ′ (1 − ɛ) 2 ∆x 2 ++ ) + 2ln(aa′ )✁∂∂ 2 ln(∆x 2 ++ )} . (6.83)Σ +− (x, x ′ ) = +|f | 2 (aa′ ) 3 22 1 0π 4 ∂∂ ✁ 4 [ ln 2 (µ 2 ∆x+− 2 ) − 2ln(µ2 ∆x+− 2 )]+|f | 2 (1 − ɛ)2 HH ′ (aa ′ ) 5 2(ν 22 9 π 4 − 1 )4× { ✁∂∂ 2 ln 2 (K HH ′ (1 − ɛ) 2 ∆x 2 +− ) + 2ln(aa′ )✁∂∂ 2 ln(∆x 2 +− )} . (6.84)The retarded self-energy is simply the sum of these. Since the ++ and +− self-energies havea difference in sign, the only contributions to the retarded self-energy come from branchcuts and singularities in ∆x++ 2 and ∆x2 +− and the local term in the ++ self-energy. We usethe following relations to extract these branch cuts and singularitiesln(α∆x++ 2 ) − ln(α∆x2 +− ) = ln( y ++4 ) − ln( y +−4 ) = 2πiΘ(∆η2 − r 2 )Θ(∆η)ln 2 (α∆x++ 2 ) − ln2 (α∆x+− 2 ) = 4πi ln|α(∆η2 − r 2 )|Θ(∆η 2 − r 2 )Θ(∆η), (6.85)where ∆η = η − η ′ , r ≡ |x − x ′ | 2 and Θ is the Heaviside step function. Furthermore, Θ(∆η 2 −r 2 ) = Θ(∆η− r), since we are considering timelike distances. The retarded self-energy in Eq.(6.82) then becomesΣ ret (x, x ′ ) = −|f | 2 (aa′ ) 3 22 8 π 3 i ✁∂∂ 4 { Θ(∆η − r)Θ(∆η) [ ln|µ 2 (∆η 2 − r 2 )| − 1 ]}−|f | 2 (aa′ ) 3 22 5 π 2 ln(aa′ )i✁∂δ(x − x ′ )−|f | 2 (1 − ɛ)2 HH ′ (aa ′ ) 5 2(ν 22 7 π 3 − 1 )4{× i✁∂∂ 2 [ Θ(∆η − r)Θ(∆η)ln|α(∆η 2 − r 2 )| ] + ln(aa ′ )i✁∂∂ 2 [ Θ(∆η − r)Θ(∆η) (6.86)]} ,where I have defined α = K HH ′ (1−ɛ) 2 . Now we want to use this self-energy in the modifiedDirac equation (6.81). We first define the conformally rescaled functionχ(x) ≡ a 3 2 ψ(x). (6.87)Because the background is homogeneous and isotropic, we seek plane wave solutions of theformχ(x) = e i⃗ k·⃗x χ(η), (6.88)