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

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 jp1j @(E1 + E2) = 2 X dΩ1 16π E1E2F @jp1j 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 jp0j = 2 2 X: dΩ1 64π jpj(E1 + E2) How does this simplify if m1 = m2 = mA = mB? (iii) Now consider Compton scattering in the Lab frame where pA = (!; k) (photon), 0 0 0 0 pB = (m; 0) (electron), p1 = (! ; k ) (photon), p2 = (E ; p ) (electron). Show that 1 1 1 − = (1 − cos θ) !0 ! m (where k0 · k = jk0jjkj cos θ) and p E0 = !2 + !02 − 2!!0 cos θ + m2 and F = 4m!: Hence show that @(E1 + E2) m! = 0 0 @jp1j E ! and ﬁnally that dσ 1 !0 2 = 2 2 X dΩ1 64π m ! 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 = jiM + iM j2 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 = (!; 0; 0;!) 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 (! ; 0; 0; −! ), 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 ! ! ! 2 = 2 + 0 − sin θ : dΩ1 2m ! ! ! 333.

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