PS10-solutions - Phys 253a 1 Problem Set 10 Solutions December 2 2010 1(a The phase space integral for 1 2 is 1 2m d4 q d3 p(2)4 4(P p q 3 2E(2)3 E(2 p

# PS10-solutions - Phys 253a 1 Problem Set 10 Solutions...

• Notes
• 10
• 97% (29) 28 out of 29 people found this document helpful

This preview shows page 1 - 4 out of 10 pages.

Phys 253a 1 Problem Set 10 Solutions December 2, 2010 1. (a) The phase space integral for 1 2 is 1 2 m φ Z d 3 p (2 π ) 3 2 E p d 4 q (2 π ) 3 E q (2 π ) 4 δ 4 ( P - p - q ) This is easiest to evaluate in the center of mass frame, P = ( m φ , 0 ). If we first do the integral over q , the spatial part of the delta function enforces 0 - p - q = 0, so q = - p and E q = E p . We get 1 2(2 π ) 2 m φ Z d 3 p 4 E 2 p δ ( m φ - 2 p m 2 e + p 2 ) This is spherically symmetric in p , so we can switch to spherical coordinates 4 π 8(2 π ) 2 m φ Z p 2 dp p 2 + m 2 e δ ( m φ - 2 p m 2 e + p 2 ) = 1 8 πm φ p 2 p 2 + m 2 e p m 2 e + p 2 2 p m 2 e + p 2 = 1 2 m φ = 1 16 πm φ s 1 - 4 m 2 e m φ (b) i. The amplitude is i M = ig S u ( p ) v ( q ), which satisfies X spins |M| 2 = g 2 S Tr[( 6 p + m e )( 6 q - m e )] = 4 g 2 S ( p · q - m 2 e ) = 2 g 2 S ( m 2 φ - 4 m 2 e ) Γ = g 2 S m φ 8 π (1 - 4 x 2 ) 3 / 2 ii. The amplitude is i M = - g P u ( p ) γ 5 v ( q ), so we get X spins |M| 2 = - g 2 S Tr[ γ 5 ( 6 p + m e ) γ 5 ( 6 q - m e )] = - 4 g 2 S ( - p · q - m 2 e ) = 2 g 2 S m 2 φ Γ = g 2 S m φ 8 π p 1 - 4 x 2
Phys 253a 2 iii. The amplitude is i M = ig V μ u ( p ) γ μ v ( q ). We get 1 3 X spins |M| 2 = 1 3 g 2 V Tr[( 6 p + m e ) γ μ ( 6 q - m e ) γ ν ] X i i μ i ν = 4 g 2 V 3 ( p μ q ν + p ν q μ - ( p · q + m 2 e ) η μν ) - η μν + P μ P ν m 2 φ ! = 4 g 2 V 3 - 2 p · q + 2 ( p · P )( q · P ) m 2 φ + 3( p · q + m 2 e ) ! = 4 g 2 V 3 (2 m 2 e + m 2 φ ) Γ = g 2 V m φ 12 π (1 + 2 x 2 ) p 1 - 4 x 2 iv. The amplitude is - g A μ u ( p ) γ μ γ 5 v ( q ), so we have 1 3 X spins |M| 2 = 1 3 g 2 A Tr[( 6 p + m e ) γ μ γ 5 ( 6 q - m e ) γ ν γ 5 ] X i i μ i ν = 4 g 2 A 3 (( m 2 e - p · q ) η μν + p μ q ν + q μ p ν ) - η μν + P μ P ν m 2 φ ! = 4 g 2 A 3 ( m 2 φ - 4 m 2 e ) Γ = g 2 A m φ 12 π (1 - 4 x 2 ) 3 / 2 v. The tricky part of this one is to write down the polarization sum for an antisymmetric tensor. We’ll assume it’s transverse to pick out the spin 1 part. Then we know the polarization sum Σ μνρσ = X i i μν i * ρσ must satisfy P μ Σ μνρσ = 0 (and similarly if we dot p into any index), along with Σ μνρσ = - Σ νμρσ = Σ ρσμν . The only objects we have to build Σ with are η μν and P μ , and the only tensors we can build out of these with the correct index symmetries are η μρ η νσ - η μσ η νρ and P μ P ρ η νσ - P μ P σ η νρ + P ν P σ η μρ - P ν P ρ η μσ
Phys 253a 3 The transverseness condition implies that we should take the combination X i i μν i ρσ = η μρ η νσ - η μσ η νρ - 1 m 2 φ ( P μ P ρ η νσ - P μ P σ η νρ + P ν P σ η μρ - P ν P ρ η μσ ) (Actually, this only fixes the spin sum up to an overall scale. We must additionally check a special case, say ...) Alternatively, we can write this as X i i μν i * ρσ = C μρ C νσ - C μσ C νρ Where C μρ := η μρ - P μ P ρ m 2 φ is manifestly transverse. The amplitude is i M = - g T μν u ( p ) σ μν v ( q ). We should be a little careful about complex conjugation: ( μν v ) * = 1 2 ( u γ 0 i [ γ μ , γ ν ] v ) * = - 1 2 v i [ γ ν , γ μ ] γ 0 u = - 1 2 vi [ γ ν , γ μ ] u = μν u Thus, X spins |M| 2 = g 2 T Tr[ σ μν ( 6 q - m e ) σ ρσ ( 6 p + m e )] X i i μν i * ρσ = - g 2 T 4 Tr[[ γ μ , γ ν ]( 6 q - m e )[ γ ρ , γ σ ]( 6 p + m e )]Σ μρνσ