1 Bhabha scattering (1936)

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1 Bhabha scaering 936 Consider he scaering of an elecron and a posiron. We have already compued he annihilaion of an elecron/posiron in finding he cross-secion for producing muon pairs. I is also possible for he final paricles o be an elecron-posiron pair. In addiion, he iniial paricles may exchage a phoon wihou annihilaing. Since he paricles are diinguishable, we have only he -channel diagram for his, wihou he crossed diagram. Le he in and ou elecrons have momena p, p, respecively and he iniial and final posirons have momena k,k. Le he angle beween he iniial and final elecrons be θ. The general calculaion is lenghy; here we reric our aenion o he ulrarelaiviic case, where we can rea he elecron and posiron as massless. This means we neglec erms of order m. For GeV or TeV colliders, his erm is of E order 0 8 or 0 4, so we incur no grea error. We may drop m whenever i occurs addiively.. The scaering marix We have wo Feynman diagrams, so he marix elemen is he sum im f i im s f i + im f i where s and refer o he phoon channel. The fir, s-channel, diagram describes he annihilaion-creaion. The marix elemen is he same as we use for he muon problem, excep he paricle masses are all equal, and here we wan o call p,k he incoming momena and p,k he ougoing momena. Therefore, inead of we wrie im s f i v p ieγ α up i q η αβ ξ q αq β q ūk ieγ β v k im f s i vk ieγα up i q η αβ ξ q αq β q ū p ieγ β v k ha is, we ju need o inerchange p k in our previous resul, and we have q p + k p + k. As before, he gauge erms vanish immediaely. We replace q s, leaving The second Feynman diagram has he marix elemen im f i ū p ieγ α up i q im s f i ie s η αβ vkγ α upū p γ β v k η αβ ξ q αq β q vk ieγ β v k where p, p are he incoming and ougoing elecron, respecively; k,k are he incoming and ougoing proon or any oher fermion of charge Q, excep he elecron which would also require he u channel, respecively; q p p k k is he -channel 4-momenum; ξ allows a choice of gauge. The gauge erm does no immediaely drop in his case. We have: ū p ieγ α up i q ξ q αq β q vk ieγ β v k i ξ e ū p iq/up vkiq/v k i ξ e ū p ip/ ip/ up vk ik/ ik/ v k i ξ e ū p mup vkmv k 4im e ξ e ū p up vkv k where we use he Dirac and conjugae Dirac equaions in he middle ep. The marix elemen becomes im f i ie η αβ ū p γ α upū k γ β uk 4im e ξ e ū p up vkv k

2 where we se q. We should ge he same resul in any gauge, so we could ju se ξ, bu in any case he erm of of order m and we drop i. The resuling marix elemen is im f i im s f i + im f i ie s η αβ vkγ α upū p γ β v k + ie η µν ū p γ µ up vkγ ν v k Squaring he marix elemen leads o four erms, im f i im s f i + M s f i M f i + M f i M f s i + im f i We work hem ou one a a ime.. Marix squared: s channel We already have from he muon case. Averaging over iniial spins and summing over final spins, we found M s f i im s f i 8e4 s k k k p k p + p k m k p + k p k p + m k p + m k p + 8e4 s k k + p k + m k p + m k p + m 4 where some simplificaion occurs because of he equal masses..3 Marix squared: channel For he final erm, we have M f i ie η µν ū p γ µ up vkγ ν v k ie η ρσ v k γ ρ vkūpγ σ u p 4 η µνη ρσ ū p γ µ up vkγ ν v k v k γ ρ vkūpγ σ u p 4 η µνη ρσ ū a p γ µ ab u b p v c kγcd ν v γ ρ e f v f kū g pγgh σ u h p 4 η µνη ρσ γ µ ab u b pū g pγ σ gh u h e4 4 η µνη ρσ γ µ ab p/ + m bg γσ gh p/ + m ha γν c/ m de γρ e f k/ m f c e4 4 η µνη ρσ r γ µ p/ + mγ σ p/ + m r γ ν k/ m γ ρ k/ m p ū a p γcd ν v γ ρ e f v f k v c k where he facor 4 keeps rack of he number of imes we commue fermion fields. The races give r γ µ p/ + mγ σ p/ + m r γ µ p/γ σ p/ + mγ µ γ σ p/ + mγ µ p/γ σ + m γ µ γ σ r γ µ p/γ σ p/ + m r γ µ γ σ p λ p τr γ µ γ λ γ σ γ τ + 4m η µσ 4p λ p τ η µλ η στ η µσ η λτ + η µτ η λσ + 4m η µσ 4p λ p τ η µλ η στ η µσ η λτ + η µτ η λσ + 4m η µσ 4 p µ p σ η µσ + p µ p σ + m η µσ r γ ν k/ m γ ρ k/ m 4 k ν k ρ k k η νρ + k ν k ρ + m η νρ

3 Combining hese resuls, M f i 4e4 η µνη ρσ p µ p σ η µσ + p µ p σ + m η µσ k ν k ρ k k η νρ + k ν k ρ + m η νρ 4e4 p ν p ρ k ν k ρ k k η νρ + k ν k ρ + m η νρ 4e4 η νρ k ν k ρ k k η νρ + k ν k ρ + m η νρ Performing he remaining conracions, M f i + 4e4 p ν p ρ k ν k ρ k k η νρ + k ν k ρ + m η νρ + 4m η νρ k ν k ρ k k η νρ + k ν k ρ + m η νρ 4e4 p k k k + p k + m 4e4 k k 4 k k + k k + 4m + 4e4 p k k k + p k + m + 4m k k 4 k k + k k + 4m 8e4 p k k k + p k + m + m k k + m 8e4 p k + p k m m k k + m 4.4 Marix squared: fir cross erm Now consider he fir of he cross erms. We need o make sure we ve assigned he same momena o he four paricles. Bu for he s channel we ve se p as he iniial posiron momenum inead of he final. ie s η αβ vkγ α upū p γ β v k ie η µν v k γ ν vkūpγ µ u p M f s i M f i e4 4 η αβ η µν 5 4 η αβ η µν v a kγab α u b pū c p γ β cd v γe ν f v f kū g pγ µ gh u h p e4 4 η αβ η µν k/ f a γab α p/ bgγ µ gh p/ hc γβ c/ de γν e f e4 4 η αβ η µν r k/γ α p/γ µ p/ γ β k/ γ ν γab α u b pū g pγ β cd v γe ν f v f k v a kγ µ gh u h p ū c p where we have dropped all mass erms. Noice ha he fermion exchanges have inroduced an overall sign. Now evaluae he race. Fir, we eliminae he free γ-marices, using r η αβ γ α γ β r η αβ { γ α,γ β } 4. To do his, we mu bring he wo relevan gamma marices nex o each oher. Also, noice ha γ α p/ p β γ α γ β p α p/γ α p β η αβ γ β γ α p α p/γ α 3

4 and for any of p,k, p,k, p/p/ p α p β γ α γ β p α p β { γ α,γ β } p α p β η αβ p 0 The race is M f s i M f i e4 4 η αβ η µν r k/γ α p/γ µ p/ γ β k/ γ ν Using conservaion of momenum, we replace k p + k p, where M f s i M f i e4 4 η αβ η µν r k/γ α p/γ µ p/ γ β p/ + k/ p/ γ ν e4 Tp + T k T 4 p T p η αβ η µν r k/γ α p/γ µ p/ γ β p/γ ν T k η αβ η µν r k/γ α p/γ µ p/ γ β k/γ ν T p η αβ η µν r k/γ α p/γ µ p/ γ β p/ γ ν η αβ η µν r p/ γ ν k/γ α p/γ µ p/ γ β where we used he cyclic propery on T p o check ha i is relaed o T k by he cyclic subiuion, k p p k, and renaming he dummy Lorenz indices. We herefore only need o compue he fir wo. For T p, T p η αβ η µν r k/γ α p/γ µ p/ γ β p/γ ν η αβ η µν r k/γ α p µ γ µ p/ p/ p β p/γ β γ ν 4r k/p/p/ p/ η αβ r k/γ α p/ p/γ β p/ η µν r k/p/γ µ p/p/ γ ν + η αβ η µν r k/γ α γ µ p/p/ p/γ β γ ν 4r k/p/ p/p/ η αβ r k/γ α p/ p β γ β p/ p/ η µν r k/p/p µ p/γ µ p/ γ ν + η αβ η µν r k/γ α γ µ p/ p/ p/γ β γ ν Simplifying, and using p m 0, T p 8 r k/p/ 4r k/p p/ 4r k/p/p/ p/ + η αβ r k/γ α p/ γ β p 4r k/p/p/ p/ + η µν r k/p γ µ p/ γ ν + η αβ η µν r k/γ α γ µ p/γ β γ ν η αβ η µν r k/γ α γ µ p/ p/p/γ β γ ν 3 k p 4r k/p/p/ p/ 4r k/p/p/ p/ 4

5 + η αβ η µν r k/γ α γ µ p/γ β γ ν 3 k p 8r k/ p/ p/ p/ + η αβ η µν r k/γ α γ µ p/γ β γ ν Look a η αβ η µν r k/γ α γ µ p/γ β γ ν η αβ η µν r k α γ α k/γ µ p/ η βν γ ν γ β 4r k/p/ η µν r γ µ p/γ ν k/ η αµ r γ α k/γ µ p/ + 4η µν r k/γ µ p/γ ν 4r k/p/ + 4η µν r γ µ p/γ ν k/ 6k p Therefore, T p 3 k p 6 r k/p/ + 8r k/p/ p/p/ + 3 k p 3 k p k p 0 Now compue T k, which is easier because he repeaes are closer: T k η αβ η µν r k/γ α p/γ µ p/ γ β k/γ ν η αβ η µν r k/γ α p/γ µ p/ γ β k ν γ ν k/ η αβ r k/γ α p/k/p/ γ β η αβ η µν r k/γ α p/γ µ p/ γ β γ ν k/ η αβ r k α γ α k/ p/k/p/ γ β 4r p/k/p/ k/ η αβ r γ α k/p/k/p/ γ β 4r p/k/p/ k/ 8r k/p/k/p/ 4r p/k/ p k k/p/ 8 p k r p/k/ + 4r p/k/k/p/ 3 p k and cycling k p p k we have T p 3 k p The full form of his cross erm conribuion o he squared marix elemen i herefore, M f s i M f i e4 Tp + T k T 4 p e4 0 3 p k + 3 k p 4 8e4.5 Marix squared: second cross erm The second cross erm, averaged and summed over spins is k p p k 5

6 M f i M f s i ie η µν ū p γ µ up vkγ ν v k ie s η αβ v k γ β u p ūpγ α vk We need e4 4 η αβ η µν 5 4 η αβ η µν ū a p γ µ ab u b p v c kγcd ν v γ β e f u f p ū g pγgh α v h k γ µ ab u b pū g pγcd ν v γ β e f u f p ū a p γgh α v h k v c k e4 4 η αβ η µν γ µ ab u b pū g pγgh α v h k v c kγcd ν v γ β e f u f p ū a p e4 4 η αβ η µν r γ µ p/γ α k/γ ν k/ γ β p/ Compare his o he fir cross erm, where we calculaed r γ µ p/γ α k/γ ν k/ γ β p/ r p/γ α k/γ ν k/ γ β p/ γ µ 4 η αβ η µν r k/γ α p/γ µ p/ γ β k/ γ ν 8e4 k p p k These differ only by he exchange p k, p k. Therefore,.6 Final marix squared M f i M f s i 8e4 k k k p k p The final marix squared, averaged/summed over spins, is im f i im s f i + M s f i M f i + M f i M f s i + im f i Dropping he mass erms and combining, im f i 8e4 s k k + p k + m k p + m k p + m 4 8e4 k p p k 8e4 k k k p k p 8e4 s 8e4.7 Relaiviic kinemaics + 8e4 p k + p k m m k k + m 4 k k + p k + 8e4 p k + p k k p p k k p k p + k k Consider he collision of an elecron on a posiron in he CM frame. Then he momena are p E, p k E, p p E,p k E, p 6

7 Then, dropping masses, and seing p E m E, all quaniies may be expressed in erms of E and θ: Subiuing ino he squared marix elemen, im f i E + p E m E p k E + E + p cosθ E + cosθ k k E E cosθ E cosθ E + cosθ p k E + p p E p p m E cosθ s p + k 4E 8e4 s k k + p k + 8e4 p k + p k 8e4 k p p k k p k p + k k 8 E 4E 4E cosθe cosθ + E + cosθe + cosθ 8 + E E cosθe + co cosθ 8e4 E 4E cosθe + cosθ E + cosθe E + cosθe + E cosθe + cosθ Then he differenial cross secion is dσ dω 64π s M ji 64π 4E + cos θ cosθ + cosθ + e4 cos 64π 4E θ 4 + cosθ cosθ 64π 4E + cos θ cosθ + cosθ + cos θ cosθ + cos 64π E θ + + cos4 θ sin 4 θ + cos θ cos θ sin θ 64π E + cos θ + + cos4 θ sin 4 θ This is he ulrarelaiviic limi of h Bhabha cross secion 936. cos4 θ sin θ 4 + cosθ cosθ 7

8 .8 Traces of gamma marices Now compue he races using he fundamenal relaion { γ α,γ β } η αβ, and he cyclic propery of he race, r A...BC r CA...B. Fir, we can show ha he race of he produc of any odd number of γ-marices vanishes by using γ 5 and {γ 5,γ a } 0, r γ α...γ β r γ α...γ β }{{} n+ r γ 5 γ 5 γ α...γ β r γ 5 γ α γ 5...γ β n+ r γ 5 γ α...γ β γ 5 n+ r γ 5 γ 5 γ α...γ β r γ α...γ β 0 For even producs, we will need races of producs of,4,6 and 8 gamma marices. r γ α γ β r γ β γ α + η αβ r γ β γ α + η αβ r r γ α γ β + 8η αβ r γ α γ β 4η αβ and r γ α γ β γ µ γ ν r γ β γ α + η αβ γ µ γ ν r γ β γ α γ µ γ ν + η αβ r γ µ γ ν r γ β γ µ γ α + η µα γ ν + η αβ r γ µ γ ν r γ β γ µ γ α γ ν η µα r γ β γ ν + η αβ r γ µ γ ν r r r γ β γ µ γ ν γ α + η να η µα r γ β γ ν + η αβ r γ µ γ ν r γ β γ µ γ ν γ α + η να r γ β γ µ η µα r γ β γ µ η µα r γ β γ ν + η αβ r γ µ γ ν γ α γ β γ µ γ ν η να r γ α γ β γ µ γ ν 4η να η β µ 4η µα η βν + 4η αβ η µν γ β γ ν + η αβ r γ µ γ ν For six, we use he simple paern o more quickly find r γ α γ β γ µ γ ν γ ρ γ σ r γ β γ α + η αβ γ µ γ ν γ ρ γ σ r γ β η αµ γ µ γ α γ ν γ ρ γ σ + η αβ γ µ γ ν γ ρ γ σ r γ β γ µ γ ν γ ρ γ σ γ α + η ασ γ β γ µ γ ν γ ρ η αρ γ β γ µ γ ν γ σ + η αν γ β γ µ γ ρ γ σ η αµ γ β γ ν γ ρ γ σ + η αβ γ µ γ ν 8

9 and from here we can use he resul for he race of four, r γ α γ β γ µ γ ν γ ρ γ σ r η ασ γ β γ µ γ ν γ ρ η αρ γ β γ µ γ ν γ σ + η αν γ β γ µ γ ρ γ σ η αµ γ β γ ν γ ρ γ σ + η αβ γ µ γ ν γ ρ γ σ 4η ασ η β µ η νρ η βν η µρ + η ρβ η µν 4η αρ η β µ η νσ η βν η µσ + η σβ η µν +4η αν η β µ η ρσ η βρ η µσ + η βσ η µρ 4η αµ η βν η ρσ η βρ η νσ + η βσ η νρ +4η αβ η µν η ρσ η µρ η νσ + η µσ η νρ or, perhaps more mnemonically, r γ α γ β γ µ γ ν γ ρ γ σ 4η αβ η µν η ρσ η µρ η νσ + η µσ η νρ 4η αµ η βν η ρσ η βρ η νσ + η βσ η νρ +4η αν η β µ η ρσ η βρ η µσ + η βσ η µρ 4η αρ η β µ η νσ η βν η µσ + η σβ η µν +4η ασ η β µ η νρ η βν η µρ + η ρβ η µν From his, if we don run ou of Greek leers, we can immediaely wrie he resul for eigh gammas: r γ α γ β γ µ γ ν γ ρ γ σ γ λ γ τ 4η αβ η µν η ρσ η µρ η νσ + η µσ η νρ 4η αµ η βν η ρσ η βρ η νσ + η βσ η νρ +4η αν η β µ η ρσ η βρ η µσ + η βσ η µρ 4η αρ η β µ η νσ η βν η µσ + η σβ η µν +4η ασ η β µ η νρ η βν η µρ + η ρβ η µν 9