15 Applications of Quantum Electrodynamics (QED) Application 1: Electron + positron annihilation into muon + anti-muon. + + e + e− µ µ− ! To order e2 in the scattering amplitude there is only one Feynman graph e p1 − µ− q1 e+ + p2 µ q2 Note the convention that the arrows on the fermion lines are always drawn in the same direction and follow the (negative) electron charge, so that the momenta assigned to the positron and µ+ are understood to be in the opposite direction from the direction of the arrows. Following the Feynman rules the matrix-element for this process (dropping the energy-momentum conserving delta-function) is: α µ β igµν γ ν δ = u (q ;λ )(ieγ ) v (q ;λ )β − v (p ;λ )(ieγ ) u (p ;λ )δ; M (µ) 1 1 α (µ) 2 2 p p 2 iε (e) 2 20 γ (e) 1 10 ( 1+ 2) + λ λ λ λ where 1; 2 are the helicities of the outgoing muons and 10 ; 20 are the helicities of the incoming electrons. Contracting indices this may be written using matrix notation as 1 ie2u q λ γµv q λ v p λ γ u p λ M = (µ)( 1; 1) (µ)( 2; 2) 2 (e)( 2 20 ) µ (e)( 1 10 ): (p1 + p2) Here the order of the terms matters - the rule is to start at the end of each fermion line and work backwards writing down each vertex or internal fermion propagator as one moves back toward the beginning of the fermion line. We need to multiply this amplitude by its complex conjugate in order to find the square-matrix- element. Using the fact that γ0 is Hermitian whereas γi are anti-Hermitian and the definition u ≡ u†γ0 (and similarly for v), we can show (using the anti-commutation relations of the γ-matrices) that uΓu0 ∗ = u0Γu; 75 where u and u0 stand for any two spinors (u or v), Γ stands for any string of γ-matrices and Γ is the string of γ-matrices in reverse order (in this case we only have one γ-matrix between the spinors.). 2 Therefore for the square-matrix-element, we have (setting (p1 + p2) = s) 4 2 † e ν µ = v (q2;λ )γ u (q1;λ )u (q1;λ )γ v (q2;λ ) jM j ≡ M M s2 (µ) 2 (µ) 1 (µ) 1 (µ) 2 λ γ λ λ γ λ u(e)(p1; 10 ) νv(e)(p2; 20 )v(e)(p2; 20 ) µv(e)(p1; 10 ) In most experiments the direction of polarisation of the final-state fermions is not measured but since these are in principle measurable we sum the square-matrix-element over final-state helic- ities. Furthermore, we usually have unpolarised incoming fermion beams so we average over incoming helicities - i.e. we sum over the helicities of all four fermions and divide by 4. Using the relations β β ∑ uα(p;λ)u (p;λ) = (p= + m)α λ= 1 and β β ∑ vα(p;λ)v (p;λ) = (=p m)α λ= 1 − we see that ∑ u(p;λ)Γv(p0;λ0)v(p0;λ0)Γu(p;λ) = Tr (=p + m)Γ =p0 m Γ λ λ − ; 0 Likewise ∑ v(p;λ)Γu(p0;λ0)u(p0;λ0)Γv(p;λ) = Tr (=p m)Γ =p0 + m Γ λ λ − ; 0 and ∑ u(p;λ)Γu(p0;λ0)u(p0;λ0)Γu(p;λ) = Tr (p= + m)Γ p=0 + m Γ λ λ ; 0 ∑ v(p;λ)Γv(p0;λ0)v(p0;λ0)Γv(p;λ) = Tr (p= m)Γ p=0 m Γ λ;λ − − 0 So that the sum over fermion helicities is reduced to the calculation of a trace of a string of γ- matrices In the case we are considering we have 4 1 e ν µ ∑ 2 q m γ q m γ p m γν p m γ M = 2 Tr =2 µ =1 + µ ) Tr (=1 + e) (=2 e) µ 4 helicities j j s − − 76 15.1 Trace Theorems Traces of long strings of γ-matrices can be performed using algebraic manipulation packages. Traces of strings of up to four γ-matrices can be done by hand, using the following properties of strings of γ-matrices, which we state without proof (they can be proved using the anti-commutation relations of γ-matrices and the cyclic property of traces). 1. =pp= = p pI · where I is the 4 4 unit matrix. × 2. µ γ γµ = 4I; 3. µ ν ν γ γ γµ = 2γ − or µ γ =pγµ = 2p= − 4. µ ν ρ νρ γ γ γ γµ = 4g I or µ γ =pq=γµ = 4p qI · 5. µ ν ρ σ σ ν ρ γ γ γ γ γµ = 2γ γ γ − or µ γ p=q=r=γµ = 2r=q=p= − 6. The trace of a string of an odd number of γ-matrices is zero. In particular Tr=p = 0 Tr =pq=r= = 0 f g 7. ν ν Tr γµγ = 4gµ f g or Tr p=q= = 4p q f g · 8. 2n µ1 µ2 µ2n j µ1µ j µ2 µ j 1 µ j+1 µ2n Tr γ γ γ = ∑ ( 1) g Tr γ γ − γ γ f ··· g j=2 − f ··· ··· g In particular Tr p=q=r=s= = 4(p qr s p r q s + p sq r) f g · · − · · · · 77 9. ν ν ρ Tr(γ5) = Tr γ5γµ = Tr γ5γµγ = Tr γ5γµγ γ = 0 10. ν ρ σ νρσ Tr γ5γµγ γ γ = 4iεµ or 5 µνρσ Tr γ p=q=r=s= = 4iε pµqνrρsσ Returning to the reaction e+ e µ+ µ , we have the product of two traces − ! − 1. ν µ ν ν ν µ ν µ 2 µ ν Tr q= mµ γ q= + mµ γ = Tr q= γ q= γ mµTr γ q= γ +mµTr q= γ γ m Tr γ γ 2 − 1 f 2 1 g− f 1 g f 2 g− µ f g Themiddle two terms vanish as they are traces of an odd number of γ-matrices. The remain- ing terms give (using the above formulae) ν ν ν ν 4 q qµ + qµq q q gµ m2gµ 2 1 2 1 − 1 · 2 − µ 2. ν 2 Tr (p= + me)γν (=p me)γµ = Tr =p γν =p γ + meTr γν =p γµ meTr =p γνγµ m Tr γµγν 1 2 − f 1 2 g 2 − 1 − e = 4 p ν p µ + p µ p ν p p gµν megµν 2 1 2 1 − 1 · 2 − Contracting these together and multiplying by the factor e4=(4s) we find 1 8e4 ∑ 2 p q p q p q p q m2q q m2 p p m2m2 M = 2 1 1 2 2 + 1 2 2 1 + e 1 2 + µ 1 2 2 e µ 4 helicities j j s · · · · · · − We can express this in terms of Mandelstam variables s and t using s = 2 m2 + p p = 2 m2 + q q e 1 · 2 µ 1 · 2 t = m2 +m2 2p q = m2+ m2 2p q e µ − 1 · 1 e µ − 2 · 2 u = m2 + m2 2p q = m2 + m2 2p q = 2m2 + 2m2 s t; e µ − 1 · 2 e µ − 2 · 1 e µ − − to get 1 2e4 ∑ 2 t2 st s2 m2 m2 t m2 m2 2 M = 2 2 + 2 + 4( e + µ) + 2( e + µ) 4 helicities j j s − The total cross-section is therefore 4 3 4 1 2e d q1 d q σ = 2 δ(q2 m2)θ(q0)(2π)4δ4 (p + p q q ) 2 Z π 3 π 3 2 µ 2 1 2 1 2 F s (2 ) 2Eq1 (2 ) − − − 2t2 + 2st + s2 4(m2 + m2)t + 2(m2 + m2)2 × − e µ e µ 78 The flux factor F = 2 s(s 4m2) and we can write − e p 3 d q1 1 φ = dEq1 dtd 2Eq 4 p1 1 j j and after performing the integral over q1 absorbing the energy-momentum conserving delta-function we are left with 4 e 2 2 2 2 2 2 2 σ = dφdtdEq δ s 2psEq 2t + 2st + s 4(m + m )t + 2(m + m ) π2 2 2 Z 1 1 e µ e µ 8 s s(s 4me) p1 − − − j j p 1 2 The magnitude of the incoming three-momentum p1 is s 4m . Integrating over φ and in- j j 2 − e tegrating over E to absorb the remaining delta-function we are left with the differential cross- q1 p section: dσ 2πα2 = 2t2 + 2st + s2 4(m2 + m2)t + 2(m2 + m2)2 ; dt s3(s 4m2) − e µ e µ − e (α = e2=(4π)). Of particular interest is the limit of this cross-section where s and t are both much larger than j j the masses, which may then be neglected. In that case the kinematic limit on t (obtained form the values of t for which the scattering angle in the centre-of-mass frame is 0 or π), is s < t < 0; − so that the total cross-section in this limit is πα2 0 πα2 + + 2 2 2 4 σTOT (e e− µ µ−) = dt 2t + s + 2st = ! s4 Z s 3s − Application 2: Electron-positron (Bhabha) Scattering + + e e− e e− ! In this case there are two Feynman graphs e e p1 e− e− q1 p1 − − q1 - p q p2 + + q2 2 e+ (a) e+ 2 e (b) e 79 Again the momenta for the positrons is in the opposite direction to the arrows on the fermion lines.