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PS9-solutions

PS9-solutions - Phys 253a 1 Problem Set 9 Solutions...

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Unformatted text preview: Phys 253a 1 Problem Set 9 Solutions December 2, 2010 1. (a) There are lots of diagrams. Most of them are just vertex and propagator cor- rections. The other diagram we’ll deal with is the one that’s the same, but with the photon lines crossed in the middle. (b) As in problem set 6, we’ll write the loop integral as R d 4 kd 4 q (2 π ) 4 δ ( p 1 + p 2- k- q ), or R for short. Ignoring factors of i and coupling constants, the diagrams are I = Z [ v 2 γ μ 1 6 p 1-6 q- m γ ν u 1 ][ u 3 γ ρ 1 6 p 3-6 k- m γ σ v 4 ]Π μρ ( k )Π νσ ( q ) II = Z [ v 2 γ μ 1 6 p 1-6 q- m γ ν u 1 ][ u 3 γ ρ 1 6 p 3-6 q- m γ σ v 4 ]Π νρ ( q )Π μσ ( k ) Where Π μν ( k ) =- i k 2 ( η μ ν +(1- ξ ) k μ k ν k 2 ) is the photon propagator. As before, let’s define V μν L = v 2 γ μ 1 6 p 1-6 q- m γ ν u 1 + v 2 γ ν 1 6 p 1-6 k- m γ μ u 1 V ρσ R = u 3 γ ρ 1 6 p 3-6 k- m γ σ v 4 + u 3 γ σ 1 6 p 3-6 q- m γ ρ v 4 Because of the symmetry q ↔ k under R , it’s easy to check that Z V μν L V ρσ R Π μρ ( k )Π νσ ( q ) = 2(I + II) But k μ V μν L = v 2 6 k 1 6 k-6 p 2- m γ ν u 1 + v 2 γ ν 1 6 p 1-6 k- m 6 ku 1 = v 2 ( 6 k-6 p 2- m ) 1 6 k-6 p 2- m γ ν u 1 + v 2 γ ν 1 6 p 1-6 k- m ( 6 k + m-6 p 1 ) u 1 = v 2 γ ν u 1- v 2 γ ν u 1 = 0 where in getting to the second line, we’ve used the Dirac equations v 2 ( 6 p 2 + m ) = 0 and ( m-6 p 1 ) u 1 = 0. Similarly, we can show q ν V μν L = 0 k ρ V ρσ R = 0 q σ V ρσ R = 0 Thus, the gauge-dependent terms in V μν L V ρσ R Π μρ ( k )Π νσ ( q ) manifestly vanish. Phys 253a 2 2. (a) We proved in problem set 6 that the propagator for a vector boson is- i η μν- k μ k ν m 2 Z k 2- m 2 Z So the amplitude for ν e ν e → ν μ ν μ is i M = [ v 2 γ μ i ( g V + g A γ 5 ) u 1 ][ u 3 γ ν i ( g V + g A γ 5 ) v 4 ]- i ( η μν- k μ k ν m 2 Z ) k 2- m 2 Z With k = p 1 + p 2 . Now the Dirac equations 6 p 1 u 1 = 0 and v 2 6 p 2 = 0 imply k μ ( v 2 γ μ i ( g V + g A γ 5 ) u 1 ) = v 2 ( 6 p 1 + 6 p 2 ) i ( g V + g A γ 5 ) u 1 = v 2 6 p 2 i ( g V + g A γ 5 ) u 1 + v 2 i ( g V- g A γ 5 ) 6 p 1 u 1 = 0 This is just a statement of current conservation: The current that Z couples to, J μ = ψγ μ ( g V + g A γ 5 ) ψ , is conserved (by Noether’s theorem), so k μ ( v 2 γ μ i ( g V + g A γ 5 ) u 1 ) = k μ h | J μ (- k ) | e- ,p 1 ; e + ,p 2 i = 0 Similarly, the other current u 3 γ ν i ( g V + g A γ 5 ) v 4 = h e- ,p 3 ; e + ,p 4 | J ν ( k ) | i is conserved, so the k μ k ν m 2 Z part of the Z propagator doesn’t contribute. Thus, i M = [ v 2 γ μ i ( g V + g A γ 5 ) u 1 ][ u 3 γ ν i ( g V + g A γ 5 ) v 4 ]- iη μν k 2- m 2 Z Since the neutrinos are massless (so there are no other mass scales besides m Z and k ), dimensional analysis tells us that M ∼ k 2 k 2- m 2 Z Then the cross section is σ ∼ 1 s |M| 2 = s ( s- m 2 Z ) 2 which is increasing with energy when s is below m 2 Z . (Don’t worry, by the way, about infinities in...
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PS9-solutions - Phys 253a 1 Problem Set 9 Solutions...

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