Physics 444: Quantum Field Theory Homework. 1. Compute the differential cross section, dσ/d cos θ, for unpolarized Bhabha scattering e + e e + e. Express your results in s, t and u variables. Compute the differential cross section, dσ/d cos θ, for unpolarized Møller scattering e e e e. Express your results in s, t and u variables. Relate it to the cross section of Bhabha scattering. p p' p p' k k' k k' Figure 1: Two Feynman diagrams for the s and t channels in Bhabha scattering. There are two contributing diagrams in the s and t channels see Fig. 1. The amplitudes are related by a negative sign, which we can see by considering the contractions in p, k ψ iea ν ψ ψ1 iea µ ψ 1 p, k. 1 The two channels are different by interchanging ψ 1 ψ. In Feynman gauge, the amplitude is suppressing spin indices im b = ig 1 s vkgµ up ūp γ µ vk 1 t ūp γ µ up vkγ µ vk. We want the averaged, spin-independent cross-section. Neglecting the fermion mass, we obtain M b = 1 M b = g 4 t + u 4 s s }{{} ss + u st }{{} st+ts + s + u }{{ t. 3 } tt Use P&S 4.85 for the differential cross-section in the CM frame, integrate d π, and substitute α = g /4π, s = ECM for the final expression dσ dcos θ = πα s s t + t s + u 1 t + 1. 4 s By crossing symmetry, the s diagram above becomes the u diagram in Møller scattering. To obtain the Møller differential cross-section, we interchange s u in M b, and include a phase space factor of 1/ for identical final state particles. 1
. Suppose we add to QED with electron and photon a scalar denoted by, which has Yukawa coupling to the electron denoted by ψ. The interaction terms are L int = λ ψψ e ψγ µ ψa µ 5 a Compute the 1-loop contribution with virtual scalar to the renormalization constant Z 1 vertex and Z electron self-energy. Is Z 1 = Z? Why? Write the vertex function as Γ µ = γ µ + γ µ δf 1 q. ūp γ µ δf 1 qup = iλ d d k ūp /k + mγ µ /k + mup π d k p m k m k m where k = k + q. Introducing Feynman parameters, we have ūp γ µ δf 1 qup = iλ dxdydzδx + y + z 1 d d l N π d l 3, where N = ūp /l + 1 y/q + z/p + mγ µ /l y/q + z/p + mup ūp γ µ upl d d + 1 + z m = xyq + 1 z m + zm. Performing the loop integral, we have = ūp γ µ δf 1 qup λ 4π ūp γ µ δf qup dz1 z4π/ d/ Γ d/ d Γ3 d/ 1 + z m Renormalization condition F 1 0 = 1 means δ 1 = δf 1 0. Therefore, δ 1 = Z 1 1 = λ 4π dz1 z ɛ + 1 + γ 4π 1 + z m E log 1 z m + zm 1 z m + zm δ = Z 1 arise from the renormalization of electron self-energy from the scalar loop, Σp, where the full propagator is Sp = i /p m iσ/p.
On-shell renormalization implies δ = dσ/p d/p /p=m At one loop, we compute Σ/p = λ 4π dxx/p + m ɛ γ E + log 4π 1 x m + xm x1 xp Therefore, subsituting x = 1 z, δ = Z 1 = λ 4π dz1 z ɛ γ E + log 4π z zm z m + 1 zm + z m + 1 zm. After carrying out the z integral, we verify δ 1 = δ. This is of course expected, as Z 1 = Z is a consequence of the Ward Identity or in other word gauge invariance which still holds with the introduction of this additional scalar. b Consider the renormalization of the vertex of Yukawa interaction, we must introduce another renormalization constant and write Z λ λ ψψ. Is Z λ = Z? It suffices to just compare the divergent part. There is no Ward identity for the Yukawa coupling since it is not associated with the conserved current of the U1 symmetry. Indeed, we can verify this by comparing the divergent piece of the gauge boson and scalar loop corrections to the Yukawa vertex and electron self energy functions. The correction to the self-energy function to the order of e is the same as in the original QED, we have δ e = e 4π dx x ɛ + finite, where the divergent piece of the self energy diagram from scalar loop or δ λ has been computed in part a. Now we compute the divergent pieces of the Yukawa vertex function due to scalar and photon loops. We write the vertex function as i Γ λ = i 1 + δγ λ. To the order λ, we have at one loop from scalar loop only ūp δγ λ up = iλ dz1 z d d l π d ūp l + 1 + z m up l 3, where = 1 z m + zm Renormalization condition Γ λ 0 = 1 implies δ 1 λ = λ 4π 3 4 dzz ɛ + finite
Comparing this with the δ λ. from part a, we have for the divergent pieces δ λ = λ 1 4π ɛ, δ 1λ = λ 4π 1 ɛ, δ 1λ δ λ. Next, we check the renormalization from photon loop. We compute ūp δγ λ up = ie dz1 z d d l ūp dl + 1 + z m up π d l 3, where = 1 z m. Applying renormalization condition Γ λ 0 = 1, we have δ 1 e = 4 e 4π 4 dz1 z ɛ + finite Hence, δ e = e 4π 1 ɛ, δ 1e = 8e 4π 1 ɛ, δ 1e δ e. 3. Consider the photon 4-point amplitude an amplitude with only 4 external photons at one loop. There are 6 diagrams. Show that the amplitude is finite. That is, the potential logarithmic divergences of these diagrams cancel. The Feynman diagrams are Focusing on the most divergent piece of the first diagram, p 1 p l +permutation of, 3, 4 p 3 p 4 im d 4 l Tr/ɛ 1/l/ɛ /l/ɛ 3 /l/ɛ 4 /l l + m 4 m can depends on Feynman parameters and external momenta, and it could be different for different diagrams. However, we are focusing on the case where l. Therefore, the actual form of m does not matter as it can be ignored on the denominator. Under the integration, we can make replacement l µ l ν l ρ l σ 1 4 l4 g µν g ρσ + g µρ g νσ + g µσ g ρν 4
Hence, Tr/ɛ 1 /l/ɛ /l/ɛ 3 /l/ɛ 4 /l = 1 4 l4 Tr/ɛ 1 γ µ /ɛ γ µ /ɛ 3 γ ν /ɛ 4 γ ν + /ɛ 1 γ µ /ɛ γ ν /ɛ 3 γ µ /ɛ 4 γ ν + +/ɛ 1 γ µ /ɛ γ ν /ɛ 3 γ ν /ɛ 4 γ µ = 4 3 l4 [ɛ 1 ɛ ɛ 3 ɛ 4 + ɛ 1 ɛ 4 ɛ 3 ɛ ɛ 1 ɛ 3 ɛ ɛ 4 ]. Adding up all 6 permutations permuting among, 3, and 4, we verify that the sum vanishes. We note that this result is fully expected. If this the logarithmic divergent, we must add a counter term of the form A µ A µ A ν A ν log Λ. However, this term is not gauge invariant and therefore must be absent. 5