QFT_Solution_II-2.pdf - Dylan J Temples Solution Set Two Quantum Field Theory II QFT and the Standard Model M Schwartz Contents 1 Bhabha Scattering 1.1

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Dylan J. Temples: Solution Set Two Quantum Field Theory II QFT and the Standard Model - M. Schwartz February 17, 2017 Contents 1 Bhabha Scattering. 2 1.1 s -channel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 t -channel. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3 Cross-terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.4 Differential cross-section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2 QED with Massive Photon. 9 3 QED with Fermion Coupling to Gluon. 15 3.1 Diagram 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 3.2 Diagram 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 3.3 Cross Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.4 Off-Shell Photon Lifetime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 1
Dylan J. Temples Quantum Field Theory II : Solution Set Two 1 Bhabha Scattering. Compute the differential cross section for Bhabha scattering, e + e - e + e - in the limit that the electron mass can be neglected. You should express your answer in terms of the Mandelstam in- variants s, t, u . Figure 1: The tree-level Feynman diagrams describing the process e + e - e + e - . (Left) s -channel. (Right) t -channel. The tree-level diagrams for the process e + e - e + e - are shown in figure 1, the left diagram representing the s -channel and the right diagram representing the t -channel. 1.1 s -channel. From the Feynman diagram, we can write the matrix element as i M s = ( - ie )[¯ v 2 γ μ u 1 ] ig μν q 2 ( - ie )[¯ u 3 γ ν v 4 ] = - ie 2 q 2 v 2 γ μ u 1 ][¯ u 3 γ μ v 4 ] , (1) where the propagator momentum q = p 1 + p 2 = s . We can take the modulus squared of this matrix element, noting that since M is a C -number taking the complex conjugate is equivalent to taking the hermitian conjugate: |M s | 2 = e 4 s 2 v 2 γ μ u 1 ][¯ u 3 γ μ v 4 ] ([¯ v 2 γ ν u 1 ][¯ u 3 γ ν v 4 ]) (2) = e 4 s 2 v 2 γ μ u 1 ][¯ u 3 γ μ v 4 ][¯ v 2 γ ν u 1 ] u 3 γ ν v 4 ] , (3) since each quantity in square brackets is a C -number. Performing the conjugation yields |M s | 2 = e 4 s 2 v 2 γ μ u 1 ][¯ u 3 γ μ v 4 ][¯ u 1 γ ν v 2 ][¯ v 4 γ ν u 3 ] (4) = e 4 s 2 v 2 γ μ u 1 ][¯ u 1 γ ν v 2 ][¯ v 4 γ ν u 3 ][¯ u 3 γ μ v 4 ] , (5) having noted: ν v ] = [ u γ 0 γ ν v ] = v γ ν γ 0 u = v ( γ 0 ) 2 γ ν γ 0 u = ¯ v ( γ 0 γ ν γ 0 ) u = ¯ ν u , (6) Page 2 of 20
Dylan J. Temples Quantum Field Theory II : Solution Set Two where we’ve used the facts that γ 0 is hermitian, 4 = ( γ 0 ) 2 , and γ 0 γ ν γ 0 = γ ν . Now, since a C -number is equal to its trace, we have |M s | 2 = e 4 s 2 Tr[¯ v 2 γ μ u 1 ¯ u 1 γ ν v 2 ] Tr[¯ v 4 γ ν u 3 ¯ u 3 γ μ v 4 ] (7) = e 4 s 2 Tr[ v 2 ¯ v 2 γ μ u 1 ¯ u 1 γ ν ] Tr[ v 4 ¯ v 4 γ ν u 3 ¯ u 3 γ μ ] . (8) Now we must average over the initial spins and sum over the final spins: | ¯ M s | 2 = 1 4 X spins |M| = 1 4 X r,s,r 0 ,s 0 e 4 s 2 Tr[ v r 2 ¯ v r 2 γ μ u s 1 ¯ u s 1 γ ν ] Tr[ v r 0 4 ¯ v r 0 4 γ ν u s 0 3 ¯ u s 0 3 γ μ ] , (9) let us move the sums into the traces: | ¯ M s | 2 = e 4 4 s 2 Tr " X r v r 2 ¯ v r 2 γ μ X s u s 1 ¯ u s 1 γ ν # Tr " X r 0 v r 0 4 ¯ v r 0 4 γ ν X s 0 u s 0 3 ¯ u s 0 3 γ μ # , (10) so using the completeness relations (in the m e 0 limit) we have | ¯ M s | 2 = e 4 4 s 2 Tr h / p 2 γ μ / p 1 γ ν i Tr h / p 4 γ ν / p 3 γ μ i = e 4 4 s 2 Tr h / p 2 γ μ / p 1 γ ν i Tr h / p 3 γ μ / p 4 γ ν i (11) = e 4 4 s 2 p α 2 p β 1 Tr [ γ α γ μ γ β γ ν ] p 3 α p 4 β Tr h γ α γ μ γ β γ ν i . (12) Using the identities for traces of gamma matrices, we have p α 2 p β 1 Tr [ γ α γ μ