Quantum Electrodynamics (QED), Winter 2015/16 Problem Set 3

1. (i) Show that the Mandelstam variables for A + B → 1 + 2 scattering satisfy

2 2 2 2 s + t + u = mA + mB + m1 + m2.

(ii) The general form for the differential cross-section to particle 1 is

2 −1 dσ |p1| ∂(E1 + E2) = 2 X dΩ1 16π E1E2F ∂|p1|

p 2 2 2 where F = 4 (pA · pB) − mAmB. Show that, in the centre-of-mass frame where 0 pA = (EA, p), p1 = (E1, p ) etc, one has dσ |p0| = 2 2 X. dΩ1 64π |p|(E1 + E2)

How does this simplify if m1 = m2 = mA = mB?

(iii) Now consider Compton scattering in the Lab frame where pA = (ω, k) (photon), 0 0 0 0 pB = (m, 0) (electron), p1 = (ω , k ) (photon), p2 = (E , p ) (electron). Show that 1 1 1 − = (1 − cos θ) ω0 ω m (where k0 · k = |k0||k| cos θ) and √ E0 = ω2 + ω02 − 2ωω0 cos θ + m2 and F = 4mω.

Hence show that ∂(E1 + E2) mω = 0 0 ∂|p1| E ω and ﬁnally that dσ 1 ω0 2 = 2 2 X dΩ1 64π m ω 2. Given the unpolarised matrix element for e−µ− → e−µ− scattering in the high-energy

E mµ limit is given by 2e2 X = s2 + u2 t2 show that in the centre-of-mass frame

2 2 s = (pA + pB) = ECM 2 2 2 1 t = (p1 − pA) = −ECM sin 2 θ 2 2 2 1 u = (p1 − pB) = −ECM cos 2 θ

and hence the differential cross-section is

2 dσ α 4 1 = 2 4 1 1 + cos 2 θ dΩ 2ECM sin 2 θ

where ECM is the total centre-of-mass-frame energy.

111 Quantum Electrodynamics (QED), Winter 2015/16 Problem Set 3

3. Consider the unpolarised matrix element X for Bhabha scattering

+ − + 0 0 − 0 0 e (p1, r) + e (p2, s) → e (p1, r ) + e (p2, s )

in the high-energy limit, where energies are much larger than the mass of the electron.

∗ (i) Show that the cross-term Xab + Xab, that is the interference term between the two Feynman diagrams, is given by

e4 Xab = 0 Y 16(p1 · p2)(p1 · p2)

0 0 2 where Y = −32(p1 · p2)(p1 · p2) = −8u . (ii) Given 2e2 X = t2 + u2 aa s2

use crossing symmetry to evaluate Xbb and hence show that

∗ X = Xaa + Xbb + (Xab + Xab) t2 + u2 s2 + u2 2u2 = 2e4 + + s2 t2 st u2 u2 = 4e4 + + 1 s2 t2

(iii) Using the expression in the second line above, show that, in the centre-of-mass frame, 2 4 1 4 1 dσ α 1 2 1 + cos 2 θ 2 cos 2 θ = 2 2 (1 + cos θ) + 4 1 − 2 1 dΩ 2ECM sin 2 θ sin 2 θ 4. Consider Compton scattering

γ(k, r) + e−(p, s) → γ(k0, r0) + e−(p0, s0)

We had two contributions at order e2 namely

2 0 1 iMa = −ie u¯s0 (p )/ 0 / us(p) r (p/ + k/) − m r

2 0 1 iMb = −ie u¯s0 (p )/ 0 / 0 us(p). r (p/ − k/ ) − m r

µ (i) Writing iMa + iMb = r Jµ, given p//q + /qp/ = 2p · q and (p/ − m)u(p) = 0, show that µ k Jµ = 0.

(Note that the two diagrams are not separately gauge invariant.)

222 Quantum Electrodynamics (QED), Winter 2015/16 Problem Set 3

µ (ii) Using k Jµ = 0 implies that we can write 1 X X = |iM + iM |2 2 · 2 a a rsr0s0 1 X = (µ∗ν + kµaν + bµkν) J J ∗ 4 r r µ ν rsr0s0 µ µ µ µ for arbitrary a and b . Consider a frame where k = (ω, 0, 0, ω) and hence 1 = µ µ ν ˜µ ˜µ (0, 1, 0, 0) and 2 = (0, 0, 1, 0). Show that, choosing a = b = −k where k = 1 −1 −1 2 (ω , 0, 0, −ω ), one has X µ ∗ν µ˜ν ˜µ ν µν r r − k k − k k = −η . r (iii) Writing e2 Y Y Y + Y ∗ X = aa + bb − ab ab 16 (p · k)2 (p · k0)2 (p · k)(p · k0) show that

µ ν 0 Yaa = Tr γ (p/ + k/ + m)γ (p/ + m)γν(p/ + k/ + m)γµ(p/ + m) .

How is that related to Ybb? Show also that

µ ν 0 0 Yab = Tr γ (p/ + k/ + m)γ (p/ + m)γµ(p/ − k/ + m)γν(p/ + m) .

(iv) Using the relations

Tr p//q = 4p · q Tr p//q/r/s = 4(p · q)(p · q) − 4(p · r)(q · s) + 4(p · s)(q · r) and

µ µ α α γ γµ = 4 γ γ γµ = −2γ µ α β αβ µ α β γ γ β α γ γ γ γµ = 4η γ γ γ γ γµ = −2γ γ γ and using momentum conservation to write everything in terms of the invariants p · k and p · k0 derive the leading (m → 0) terms in the expressions

0 2 4 Yaa = 32 (p · k)(p · k ) + m (p · k) + m 2 2 0 Yab = Yba = 16m 2m + p · k − p · k Using the full expressions show that " # p · k0 p · k 1 1 1 1 2 X = 2e4 + + 2m2 − + m4 − . p · k p · k0 p · k p · k0 p · k p · k0

(v) Hence, putting everything together and using the results of question 1b, show that, in the lab frame 2 0 2 0 dσ α ω ω ω 2 = 2 + 0 − sin θ . dΩ1 2m ω ω ω

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