PS8_253A_solutions.pdf - Problem Set 8 Solutions Physics 253A Fall 2018 1(a The tree level diagrams for Moller scattering e e e e are e(p1 e(p3 iM =

PS8_253A_solutions.pdf - Problem Set 8 Solutions Physics...

This preview shows page 1 - 3 out of 9 pages.

Problem Set 8 Solutions Physics 253A Fall 2018 1(a) The tree level diagrams for Moller scattering e - e - e - e - are i M = e - ( p 2 ) e - ( p 1 ) e - ( p 4 ) e - ( p 3 ) - (1) Note the relative sign, since the u -channel graph can be obtained from the t -channel graph by swapping the final state fermions. In Feynman gauge, i M = i M t - i M u (2) = ( - ie ) 2 ¯ u 3 γ μ u 1 - ig μν ( p 1 - p 3 ) 2 ¯ u 4 γ ν u 2 - ( - ie ) 2 ¯ u 3 γ μ u 2 - ig μν ( p 1 - p 4 ) 2 ¯ u 4 γ ν u 1 , (3) so that M = e 2 u 3 γ μ u 1 ) (¯ u 4 γ μ u 2 ) t - u 3 γ μ u 2 ) (¯ u 4 γ μ u 1 ) u . (4) The complex conjugate of M is M * = e 2 u 2 γ μ u 4 ) (¯ u 1 γ μ u 3 ) t - u 1 γ μ u 4 ) (¯ u 2 γ μ u 3 ) u . (5) The matrix element squared is thus |M| 2 = |M t | 2 + |M u | 2 - M t M * u - M u M * t (6) = e 4 u 3 γ μ u 1 ) (¯ u 4 γ μ u 2 ) (¯ u 2 γ ν u 4 ) (¯ u 1 γ ν u 3 ) t 2 + u 3 γ μ u 2 ) (¯ u 4 γ μ u 1 ) (¯ u 1 γ ν u 4 ) (¯ u 2 γ ν u 3 ) u 2 (7) - u 3 γ μ u 1 ) (¯ u 4 γ μ u 2 ) (¯ u 1 γ ν u 4 ) (¯ u 2 γ ν u 3 ) t u - u 3 γ μ u 2 ) (¯ u 4 γ μ u 1 ) (¯ u 2 γ ν u 4 ) (¯ u 1 γ ν u 3 ) t u . (8) Note that |M u | 2 is related to |M t | 2 by 1 2, and that M u M * t is related to M t M * u by the same transformation. So we need to compute 2 of these 4 terms and can recycle our results for the remaining 2. Next we compute the spin sums using the identity s [ u s ( p )] a u s ( p )] b = ( / p + m ) ab . Beginning with the 1 /t 2 terms, |M t | 2 1 4 X spins |M t | 2 = e 4 4 t 2 X spins u 3 ) a ( γ μ ) ab ( u 1 ) b u 4 ) c ( γ μ ) cd ( u 2 ) d u 2 ) e ( γ ν ) ef ( u 4 ) f u 1 ) g ( γ ν ) gh ( u 3 ) h (9) = e 4 4 t 2 X spins [( u 3 ) h u 3 ) a ( γ μ ) ab ( u 1 ) b u 1 ) g ( γ ν ) gh ] [( u 4 ) f u 4 ) c ( γ μ ) cd ( u 2 ) d u 2 ) e ( γ ν ) ef ] (10) = e 4 4 t 2 [( / p 3 + m ) ha ( γ μ ) ab ( / p 1 + m ) bg ( γ ν ) gh ] [( / p 4 + m ) fc ( γ μ ) cd ( / p 2 + m ) de ( γ ν ) ef ] (11) = e 4 4 t 2 Tr [( / p 3 + m ) γ μ ( / p 1 + m ) γ ν ] Tr [( / p 4 + m ) γ μ ( / p 2 + m ) γ ν ] (12) 1
Image of page 1
= e 4 4 t 2 h Tr ( / p 3 γ μ / p 1 γ ν ) + m 2 Tr ( γ μ γ ν ) i h Tr ( / p 4 γ μ / p 2 γ ν ) + m 2 Tr ( γ μ γ ν ) i , (13) where, to get to the last line, we used the fact that the trace of an odd number of gamma matrices vanishes. While it looks like it took many steps to reach the last line, with some experience one will be skip directly from line (7) to line (12). We can now evaluate these traces using the formulas Tr ( γ μ γ ν ) = 4 g μν , (14) Tr ( γ μ γ ν γ ρ γ σ ) = 4 ( g μν g ρσ - g μρ g νσ + g μσ g νρ ) . (15) This leads to |M t | 2 = 4 e 4 t 2 ( p μ 3 p 1 ν - p 1 · p 3 δ μ ν + p 3 ν p μ 1 + 4 m 2 δ μ ν ) ( p 4 μ p ν 2 - p 2 · p 4 δ ν μ + p ν 4 p 2 μ + 4 m 2 δ ν μ ) (16) = 8 e 4 t 2 ( p 1 · p 2 ) 2 + ( p 1 · p 4 ) 2 - 2 m 2 p 1 · p 3 + 2 m 4 (17) = 2 e 4 t 2 ( s 2 + u 2 + 8 m 2 t - 8 m 4 ) . (18) To reach the second line above, we used relations like p 1 · p 2 = p 3 · p 4 which follows from ( p 1 + p 2 ) 2 = ( p 3 + p 4 ) 2 . To get to the last line, we used s = ( p 1 + p 2 ) 2 = 2 m 2 + 2 p 1 · p 2 , (19) t = ( p 1 - p 3 ) 2 = 2 m 2 - 2 p 1 · p 3 , (20) u = ( p 1 - p 4 ) 2 = 2 m 2 - 2 p 1 · p 4 . (21) Having computed |M t | 2 we can write down |M u | 2 by sending 1 2, or equivalently t u : |M u | 2 = 2 e 4 u 2 ( s 2 + t 2 + 8 m 2 u - 8 m 4 ) . (22) Moving on to the next term in |M| 2 , it follows from equation (8) that M t M * u 1 4 X spins M t M * u = - e 4 4 t u Tr h ( / p 3 + m ) γ μ ( / p 1 + m ) γ ν ( / p 4 + m ) γ μ ( / p 2 + m ) γ ν i (23) = - e 4 4 t u
Image of page 2
Image of page 3

You've reached the end of your free preview.

Want to read all 9 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture