Quantum Electrodynamics (QED), Winter 2015/16 Calculating cross sections Calculating cross sections for different processes Lepton production The simplest possible process we can consider is lepton pair production, in which an electron and position annihiliate to produce a different lepton-antilepton pair (µ or τ) labelled by l± + − + 0 0 − 0 0 e (p1; r) + e (p2; s) ! l (p1; r ) + l (p2; s ) Only one Feynman diagram contributes at tree level, and gives + 0 0 + l (p1; r¡) e (p1; r) − 0 0 − l (p2; s ) e (p2; s) 2 0 µ 0 1 0 0 M = ie u¯s (p2)γ vr (p1) 2 v¯r(p1)γµus(p2) (p1 + p2) Averaging over incoming spins and summing over outgoing spins we have 1 X X = jMj2 2 · 2 rsr0s0 4 e X 0 µ 0 0 ν 0 0 0 0 0 = 4 u¯s (p2)γ vr (p1)¯vr (p1)γ us (p2) [¯vr(p1)γµus(p2)¯us(p2)γνvr(p1)] 4(p1 + p2) rsr0s0 4 e µν ≡ 4 A Bµν 4(p1 + p2) where, given the completeness relations for u and v, we have µν µ 0 ν 0 A = Tr γ (p=1 − ml)γ (p=2 + ml) µν µ ν B = Tr [γ (p=2 + me)γ (p=1 − me)] 0 0 These two terms are related by exchanging (p1; p2) $ (−p2; −p1). Using the expression for Tr(γµ1 ··· γµn ) it is easy to show that µν 0µ 0ν 0µ 0ν µν 0 0 2 A = 4 p1 p2 + p2 p1 − g p1 · p2 + ml µν µ ν µ ν µν 2 B = 4 p1 p2 + p2 p1 − g p1 · p2 + me and hence 4 8e h 0 0 0 0 X = 4 (p1 · p1)(p2 · p2) + (p1 · p2)(p2 · p1) (p1 + p2) 2 0 0 2 2 2i + me(p1 · p2) + ml (p1 · p2) + 2meml 111 Quantum Electrodynamics (QED), Winter 2015/16 Calculating cross sections We now need to evaluate the cross-section. Let us work in the center of mass frame. We have 0 0 0 p1 = (E1; p) p1 = (E1; p ) 0 0 0 p2 = (E2; −p) p2 = (E2; −p ) Since the incoming particles have the same mass, as do the outgoing particles, in the COM frame, 0 0 conservation of energy implies E1 = E1 = E1 = E2 ≡ E. Let θ be the angle between the incoming and outgoing momenta, that is p · p0 = jpjjp0j cos θ. Finally, we note that the process can only occur if E > ml. Since ml me we have E me and we may effectively neglect terms proportional to me so that jpj ≈ E. We then find 2 0 dσ α jp j 2 0 2 2 2 = 4 E + jp j cos θ + ml dΩ COM 16E E α2 ≈ 1 + cos2 θ 16E2 2 where α = e =4π. The second line is the ultra-relativistic E ml limit where we can also neglect 0 terms proportional to ml and jp j ≈ E. Bhabha scattering The process + − + 0 0 − 0 0 e (p1; r) + e (p2; s) ! e (p1; r ) + e (p2; s ) is very similar to lepton pair production except that now two diagrams contribute to the amplitude. Let us use the same labelling of momenta and spins. We have + 0 0 + e (p1; r¢) e (p1; r) − 0 0 − e (p2; s ) e (p2; s) 2 0 µ 0 1 0 0 Ma = ie u¯s (p2)γ vr (p1) 2 v¯r(p1)γµus(p2) (p1 + p2) + 0 0 + e (p1; r£) e (p1; r) − 0 0 − e (p2; s ) e (p2; s) 222 Quantum Electrodynamics (QED), Winter 2015/16 Calculating cross sections 2 0 µ 1 0 0 0 Mb = −ie u¯s (p2)γ us(p2) 0 2 v¯r(p1)γµvr (p1) (p1 − p1) where the first amplitude is the one that enters lepton pair production (with ml = me). For simplicity, in what follows, we consider the ultra-relativistic limit where the scattering energy is much larger than me and we can effectively treat the incoming and outgoing particles as massless. Calculating the spin sums we get four terms 1 X 1 X 1 X 1 X 1 X X = jMj2 = jM j2 + jM j2 + M M∗ + M∗M 2 · 2 4 a 4 b 4 a b 4 a b rsr0s0 ∗ ≡ Xaa + Xbb + Xab + Xab We already calculated the first term in the lepton production process. In the high-energy limit, we 2 have (p1 + p2) = 2p1 · p2 and 4 2e 0 0 0 0 Xaa = 2 (p1 · p1)(p2 · p2) + (p1 · p2)(p2 · p1) (p1 · p2) 0 Since the two Feynman diagrams are related by exchanging p2 $ −p1, we immediately have 4 2e 0 0 0 0 Xbb = 0 2 (p1 · p2)(p1 · p2) + (p1 · p2)(p2 · p1) (p1 · p1) Finally we have e4 Xab = 0 Y 16(p1 · p2)(p1 · p1) where X 0 µ 0 0 ν 0 Y = u¯s0 (p2)γ vr0 (p1)¯vr0 (p1)γ vr(p1)¯vr(p1)γµus(p2)¯us(p2)γνus0 (p2) rsr0s0 µ 0 ν 0 = Tr γ p=1γ p=1γµp=2γνp=2 µ Using the identities of the form γ ··· γµ, we find ν 0 0 0 0 0 0 Y = −2 Tr p=1γ p=1p=2γνp=2 = −8(p1 · p2) Tr p=1p=2 = −32(p1 · p2)(p1 · p2) To end, we substitute these results into the COM differential cross section formula. Since the final and initial masses are equal (and effectively zero in our high energy limit), we have jpj = jp0j = E. Evaluating the four momentum products we find 2 " 4 1 4 1 # dσ α 1 2 1 + cos 2 θ cos 2 θ = 2 1 + cos θ + 4 1 − 2 2 1 dΩ COM 8E 2 sin 2 θ sin 2 θ ∗ The three contributions are from Xaa, Xbb and Xab + Xab respectively. 333 Quantum Electrodynamics (QED), Winter 2015/16 Calculating cross sections Compton scattering: unpolarized Next we consider Compton scattering γ(k; ) + e−(p; s) ! γ(k0; 0) + e−(p0; s0) with unpolarized photons and electrons. Again there are two Feynman diagrams which contribute γ(k0; ¤0) γ(k; ) e−(p0; s0) e−(p; s) 2 0 0 1 M = −ie u¯ 0 (p )= =u (p) a s (p= + k=) − m s γ(k0; ¥0) γ(k; ) e−(p0; s0) e−(p; s) 2 0 1 0 M = −ie u¯ 0 (p )= = u (p) b s (p= − k=0) − m s which are related by k $ −k0. By momentum conservation we have p+k = p0 +k0 and the mass-shell conditions p2 = p02 = m2 and k2 = k02 = 0. Thus we have (p + k)2 − m2 = 2p · k and (p − k0)2 − m2 = −2p · k. The two diagrams give four contributions to the polarization sum 1 X 1 X 1 X 1 X 1 X X = jMj2 = jM j2 + jM j2 + M M∗ + M∗M 2 · 2 4 a 4 b 4 a b 4 a b e2 Y Y Y + Y ≡ aa + bb − ab ba 16 (p · k)2 (p · k0)2 (p · k)(p · k0) where X 0 0 0 0 Yaa = u¯s0 (p )= (p= + k= + m)=us(p)¯us(p)=(p= + k= + m)= us0 (p ) µ ν 0 = Tr γ (p= + k= + m)γ (p= + m)γν(p= + k= + m)γµ(p= + m) and X 0 0 0 0 0 Yab = u¯s0 (p )= (p= + k= + m)=us(p)¯us(p)= (p= − k= + m)=us0 (p ) µ ν 0 0 = Tr γ (p= + k= + m)γ (p= + m)γµ(p= − k= + m)γν(p= + m) 0 where Ybb and Yba are equal to Yaa and Yab with k $ −k . µ To calculate the traces we again repeated use identities of the form γ ··· γµ. We have, for in- stance, given the cyclic property of traces, ν 0 µ Yaa = Tr (p= + k= + m)γ (p= + m)γν(p= + k= + m)γµ(p= + m)γ = Tr (p= + k= + m)(4m − 2p=)(p= + k= + m)(4m − 2p=0) 444 Quantum Electrodynamics (QED), Winter 2015/16 Calculating cross sections Then using the identities for Tr(γµ1 ··· γµn ), after some considerable algebra, we find 0 2 4 Yaa = 32 (p · k)(p · k ) + m (p · k) + m Here we have used the fact that momentum conservation means the four-momentum products can be written in terms of three invariants: p · k, p · k0 and m2. In particular p2 = p02 = m2 k2 = k02 = 0 p0 · k = p · k0 p0 · k0 = p · k p · p0 = m2 − p · k − p · k0 k · k0 = p · k − p · k0 0 Recall Ybb is given by the same expression with k $ −k and a similar calculation gives 2 2 0 Yab = Yba = 16m 2m + p · k − p · k Finally we would like to give an expression for the cross-section. Typically Compton scattering is measured in the LAB frame. We have 0 0 0 p1 = (!; k) p1 = (! ; k ) 0 0 0 p2 = (m; 0) p2 = (E ; p ) with jkj = ! and jk0j = !0. We define θ as the angle between in the incoming and outgoing photons, 0 0 that is k · k = !! cos θ. Since photons are massless we have v12 = 1. Subsituting into the general (colinear frame) formula for the cross section we then have dσ 1 !0 @(E0 + !0)−1 = 2 0 X dΩ LAB 64π m!E @! To calculate the partial derivative we first note recall that p · k = k · k0 + p · k0. This implies the standard relation for the shift in Compton photon frequency 1 1 1 − = (cos θ − 1) ! !0 m p We also have E0 = pjk − k0j2 + m2 = !2 + !02 − 2!!0 cos θ + m2.