Calculating Cross Sections for Different Processes

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 = |M|2 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 = |p||p0| 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 |p| ≈ E. We then ﬁnd

2 0 dσ α |p | 2 0 2 2 2 = 4 E + |p | 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 |p | ≈ 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 = |M|2 = |M |2 + |M |2 + 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 |p| = |p0| = 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 = |M|2 = |M |2 + |M |2 + 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 = (ω, k) p1 = (ω , k ) 0 0 0 p2 = (m, 0) p2 = (E , p )

with |k| = ω and |k0| = ω0. We deﬁne θ as the angle between in the incoming and outgoing photons, 0 0 that is k · k = ωω 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 ω0 ∂(E0 + ω0)−1 = 2 0 X dΩ LAB 64π mωE ∂ω 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) ω ω0 m √ We also have E0 = p|k − k0|2 + m2 = ω2 + ω02 − 2ωω0 cos θ + m2. Thus

∂(E0 + ω0) ω0 − ω cos θ ω(1 − cos θ) + m mω = + 1 = = ∂ω0 E0 E0 E0ω0 where we have used energy conservation m + ω = E0 + ω0 and also the relation between ω and ω0 just given. Gathering all these results together we ﬁnally have

2 0 2 0 dσ α ω ω ω 2 = 2 0 + − sin θ dΩ LAB 2m ω ω ω

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