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1 Bhabha Scattering (1936) 1 Bhabha scattering (1936) Consider the scattering of an electron and a positron. We have already computed the annihilation of an electron/positron in finding the cross-section for producing muon pairs. It is also possible for the final particles to be an electron-positron pair. In addition, the initial particles may exchage a photon without annihilating. Since the particles are distinguish- able, we have only the t-channel diagram for this, without the crossed diagram. Let the in and out electrons have momenta p; p0, respectively and the initial and final positrons have momenta k;k0. Let the angle between the initial and final electrons be q. The general calculation is lengthy; here we restrict our attention to the ultrarelativistic case, where we can treat the m2 electron and positron as massless. This means we neglect terms of order E2 . For GeV or TeV colliders, this term is of order 10−8 or 10−14, so we incur no great error. We may drop m whenever it occurs additively. 1.1 The scattering matrix We have two Feynman diagrams, so the matrix element is the sum s t iM fi = iM fi + iM fi where s and t refer to the photon channel. The first, s-channel, diagram describes the annihilation-creation. The matrix element is the same as we use for the muon problem, except the particle masses are all equal, and here we want to call p;k the incoming momenta and p0;k0 the outgoing momenta. Therefore, instead of s 0 a −i qa qb b 0 iM = v¯ p (−ieg )u(p) h − (1 − x) u¯(k) −ieg v k fi q2 ab q2 we write s a −i qa qb 0 b 0 iM = v¯(k)(−ieg )u(p) h − (1 − x) u¯ p −ieg v k fi q2 ab q2 that is, we just need to interchange p0 $ k in our previous result, and we have q = p + k = p0 + k0. As before, the gauge terms vanish immediately. We replace q2 = s, leaving 2 s ie a 0 b 0 iM = h v¯(k)g u(p)u¯ p g v k fi s ab The second Feynman diagram has the matrix element t 0 a −i qa qb b 0 iM = u¯ p (−ieg )u(p) h − (1 − x) v¯(k) −ieg v k fi q2 ab q2 where p; p0 are the incoming and outgoing electron, respectively; k;k0 are the incoming and outgoing proton (or any other fermion of charge Q, except the electron which would also require the u channel), respectively; q = p− p0 = k0 −k is the t-channel 4-momentum; x allows a choice of gauge. The gauge term does not immediately drop in this case. We have: −i qa q i u¯p0(−iega )u(p) −(1 − x) b v¯(k) −iegb vk0 = (1 − x)e2u¯p0(iq=)u(p)v¯(k)(iq=)vk0 q2 q2 t2 i = (1 − x)e2u¯p0ip= − ip=0u(p)v¯(k)ik=0 − ik=vk0 t2 i = (1 − x)e2u¯p02mu(p)v¯(k)2mvk0 t2 4im2e2 = (1 − x)e2u¯p0u(p)v¯(k)vk0 t2 where we use the Dirac and conjugate Dirac equations in the middle step. The matrix element becomes 2 2 2 t ie 0 a 0 b 4im e 2 0 0 iM = h u¯ p g u(p)u¯ k g u(k) − (1 − x)e u¯ p u(p)v¯(k)v k fi t ab t2 1 where we set q2 = t. We should get the same result in any gauge, so we could just set x = 1, but in any case the term of of order m2 and we drop it. The resulting matrix element is s t iM fi = iM fi + iM fi ie2 ie2 = h v¯(k)ga u(p)u¯p0gb vk0 + h u¯p0g m u(p)v¯(k)gn vk0 s ab t mn Squaring the matrix element leads to four terms, 2 s 2 s t∗ t s∗ t 2 iM fi = iM fi + M fiM fi + M fiM fi + iM fi We work them out one at a time. 1.2 Matrix squared: s channel 2 s We already have iM fi from the muon case. Averaging over initial spins and summing over final spins, we found 4 1 s 2 8e 0 0 0 0 0 0 2 0 0 2 0 0 2 4 ∑ M fi = 2 p · p k · k − k · p (k · p) + p · k p · k − m (k · p) + (k · p) k · p + m k · p + 2m (k · p) + 2m 4 all spins s 8e4 = p · p0k · k0 + p0 · kp · k0 + m2 k0 · p0 + m2 (k · p) + 2m4 s2 where some simplification occurs because of the equal masses. 1.3 Matrix squared: t channel For the final term, we have 2 2 1 t 2 1 ie 0 m n 0 ie 0 r s 0 ∑ M fi = ∑ hmn u¯ p g u(p)v¯(k)g v k − hrs v¯ k g v(k)u¯(p)g u p 4 all spins 4 all spins t t 4 e 0 m n 0 0 r s 0 = ∑ 2 hmn hrs u¯ p g u(p)v¯(k)g v k v¯ k g v(k)u¯(p)g u p all spins 4t 4 e 0 m n 0 0 r s 0 = ∑ 2 hmn hrs u¯a p gabub (p)v¯c (k)gcdvd k v¯e k ge f v f (k)u¯g (p)gghuh p all spins 4t 4 14 e m s 0 0 n 0 0 r = (−1) ∑ 2 hmn hrs gabub (p)u¯g (p)gghuh p u¯a p gcdvd k v¯e k ge f v f (k)v¯c (k) all spins 4t e4 = h h g m (p= + m) gs p=0 + m gn k=0 − m gr (k= − m) 4t2 mn rs ab bg gh ha cd de e f f c e4 = h h tr g m (p= + m)gs p=0 + mtr gn k=0 − mgr (k= − m) 4t2 mn rs where the factor (−1)14 keeps track of the number of times we commute fermion fields. The traces give tr g m (p= + m)gs p=0 + m = tr g m p=gs p=0 + mg m gs p=0 + mg m p=gs + m2g m gs = tr g m p=gs p=0 + m2tr (g m gs ) 0 m l s t 2 ms = pl pttr g g g g + 4m h 0 ml st ms lt mt ls 2 ms = 4pl pt h h − h h + h h + 4m h 0 ml st ms lt mt ls 2 ms = 4pl pt h h − h h + h h + 4m h = 4pm p0s − p · p0h ms + p0m ps + m2h ms tr gn k=0 − mgr (k= − m) = 4k0n kr − k · k0hnr + kn k0r + m2hnr 2 Combining these results, 4 1 t 2 4e m 0s 0 ms 0m s 2 ms 0n r 0 nr n 0r 2 nr ∑ M fi = 2 hmn hrs p p − p · p h + p p + m h k k − k · k h + k k + m h 4 all spins t 4e4 = p p0 k0n kr − k · k0hnr + kn k0r + m2hnr t2 n r 4e4 − p · p0h k0n kr − k · k0hnr + kn k0r + m2hnr t2 nr 4e4 + p0 p k0n kr − k · k0hnr + kn k0r + m2hnr t2 n r 4m2e4 + h k0n kr − k · k0hnr + kn k0r + m2hnr t2 nr Performing the remaining contractions, 4 1 t 2 4e 0 0 0 0 0 0 2 0 ∑ M fi = 2 p · k p · k − k · k p · p + (p · k) p · k + m p · p 4 all spins t 4e4 − p · p0k · k0 − 4k · k0 + k · k0 + 4m2 t2 4e4 + p0 · k0(p · k) − k · k0p · p0 + p0 · kp · k0 + m2 p · p0 t2 4m2e4 + k · k0 − 4k · k0 + k · k0 + 4m2 t2 8e4 = p · k0p0 · k − k · k0p · p0 + (p · k)p0 · k0 + m2 p · p0 + m2 − p · p0−k · k0 + 2m2 t2 8e4 = p · k0p0 · k + (p · k)p0 · k0 − m2 p · p0 − m2 k · k0 + 2m4 t2 1.4 Matrix squared: first cross term Now consider the first of the cross terms. We need to make sure we’ve assigned the same momenta to the four particles. But for the s channel we’ve set p0 as the initial positron momentum instead of the final. 2 2 1 s t∗ 1 ie a 0 b 0 ie 0 n m 0 ∑ M fiM fi = ∑ hab v¯(k)g u(p)u¯ p g v k − hmn v¯ k g v(k)u¯(p)g u p 4 all spins 4 all spins s t 4 e a 0 b 0 0 n m 0 = hab hmn ∑ v¯a (k)gabub (p)u¯c p gcdvd k v¯e k ge f v f (k)u¯g (p)gghuh p 4st all spins 4 15 e a b 0 0 n m 0 0 = (−1) hab hmn ∑ gabub (p)u¯g (p)gcdvd k v¯e k ge f v f (k)v¯a (k)gghuh p u¯c p 4st all spins e4 = − h h k= ga p= g m p=0 gb k=0 gn 4st ab mn f a ab bg gh hc cd de e f e4 = − h h tr k=ga p=g m p=0gb k=0gn 4st ab mn where we have dropped all mass terms. Notice that the fermion exchanges have introduced an overall sign.
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