PS11-solutions

PS11-solutions - Phys 253a 1 Problem Set 11 Solutions...

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Unformatted text preview: Phys 253a 1 Problem Set 11 Solutions December 1, 2010 1. (a) On the previous problem set, we calculated the rate for a scalar decay to two particles ψ i ψ i and found Γ = λ 2 i m h 8 π (1- 4 x 2 i ) 3 / 2 = λ 2 m h 8 π x 2 i (1- 4 x 2 i ) 3 / 2 where λ := m h v . (b) The important point here is that the Yukawa coupling λ i proportional to the mass m i implies that the Higgs preferentially decays to heavy particles. In particular, the relative rate Γ( φ → b b ) Γ( φ → μ + μ- ) = x 2 b (1- 4 x 2 b ) 3 / 2 x 2 μ (1- 4 x 2 μ ) 3 / 2 As long as m h is somewhat bigger than 2 m b , so (1- 4 x 2 b ) 3 / 2 is of order 1, then the relative rate is of order m 2 b m 2 μ ∼ 10 3 . (c) Let the incoming momenta be p 1 ,p 2 and the outgoing momenta be p 2 ,p 4 . The amplitude is i M = ( iλ e )( iλ b )( v 2 u 1 )( u 3 v 4 ) i s- m 2 h 1 4 X spins |M| 2 = 1 4 λ 2 e λ 2 b 1 ( s- m 2 h ) 2 Tr[( 6 p 2- m e )( 6 p 1 + m e )]Tr[( 6 p 4 + m b )( 6 p 3- m b )] = 4 λ 2 e λ 2 b ( s- m 2 h ) 2 ( p 1 · p 2- m 2 e )( p 3 · p 4- m 2 b ) = λ 2 e λ 2 b ( s- m 2 h ) 2 ( s- 4 m 2 e )( s- 4 m 2 b ) The total cross section is thus σ = 4 π × 1 64 π 2 s | p f | | p i | λ 2 e λ 2 b ( s- m 2 h ) 2 ( s- 4 m 2 e )( s- 4 m 2 b ) = 1 16 πs λ 2 e λ 2 b ( s- m 2 h ) 2 ( s- 4 m 2 e ) 1 / 2 ( s- 4 m 2 b ) 3 / 2 Phys 253a 2 If the electrons are colliding with enough energy to produce b ’s, then we should probably ignore the electron mass m e relative to s . Our expression simplifies to σ = m 2 e m 2 b 64 π 2 v 4 √ s ( s- 4 m 2 b ) 3 / 2 ( s- m 2 h ) 2 = ( . 5 × 10- 3 ) 2 × 4 . 2 2 16 π × 250 4 s GeV 2 1 / 2 s GeV 2- 4 × 4 . 2 2 3 / 2 s GeV 2- 4 ( m h GeV ) 2 2 × . 389 × 10 12 fb = 10- 5 s GeV 2 s GeV 2- 4 ( m h GeV ) 2 2 fb when √ s 8GeV and we’ve used GeV- 2 = . 389mb. If we plug any of the given numbers into our formula, we get values well below 1, below 10- 8 , in fact. We only get an appreciable number if √ s- m h m h ≈ 10- 6 , and it’s pretty unlikely we’d be that lucky. This is one of the disadvantages of a lepton collider. A hadron collider, on the other hand, can bypass this difficulty. We’ll learn how next semester. (d) The dominant diagram is so called Higgs-strahlung: Z * Z h p 1 p 2 p 4 p 3 The hZZ vertex factor is ivg 2 2 cos 2 θ w g αβ , and the Ze + e- vertex is- igγ μ 2 cos θ w ( g V- g A γ 5 ) . You were allowed to assume that the Z couples like a photon, so it’s fine if you ignored the axial part of the coupling to leptons. We’ll also need to recall the massive polarization sum, ∑ μ * ν =- g μν- q μ q ν m 2 Z , and propagator Π( k 2 ) =- i k 2- m 2 Z g μν- k μ k ν m 2 Z . Phys 253a 3 The amplitude is i M = v s ( p 2 )- igγ μ 2cos θ w ( g V- g A γ 5 ) u s ( p 1 )- i ( p 1 + p 2 ) 2- m 2 Z × g μν- ( p 1 + p 2 ) μ ( p 1 + p 2 ) ν m 2 Z ivg 2 2cos 2 θ w * ν ( p 4 ) The squared amplitude is then |M| 2 = g 6 v 2 16cos 6 θ w...
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PS11-solutions - Phys 253a 1 Problem Set 11 Solutions...

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