HW10_soln

# HW10_soln - PHY 5667 Problem Set no. 10 solution Problem 1...

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Unformatted text preview: PHY 5667 Problem Set no. 10 solution Problem 1 (i) The Feynman rule obtained from L int = − λφ ¯ ψψ = − λφ ¯ ψ a δ ab ψ b is ¯ ψ a ψ b φ = − iλδ ab (ii) To get the diagram, we need to contract the field with creation and annihi- lation operators in the intial and final state. ¯ ψ a ψ b φ ← k ← k ′ a k,s ¯ a k ′ ,s ′ ← p a † φ,p i M = − iλδ ab (¯ u vector k,s ) a ( v vector k ′ ,s ′ ) b = − iλ (¯ u vector k,s ) a ( v vector k ′ ,s ′ ) a (iii) To get the total decay rate, we first have to calculate summationdisplay s,s ′ |M| 2 = λ 2 summationdisplay s,s ′ (¯ u vector k,s ) a ( v vector k ′ ,s ′ ) a (¯ v vector k ′ ,s ′ ) b ( u vector k,s ) b = λ 2 summationdisplay s ( u vector k,s ) b ¯ u vector k,s ) a summationdisplay s ′ ( v vector k ′ ,s ′ ) a (¯ v vector k ′ ,s ′ ) b = λ 2 ( / k + m ψ ) ba ( / k ′ − m ψ ) ab = λ 2 Tr[( / k + m ψ )( / k ′ − m ψ )] = 4 λ 2 (( k · k ′ ) − m 2 ψ ) To proceed, we need to know k and k ′ . Let the decaying scalar be in the rest- frame p = ( m φ , , , 0) T . Energy conservation requires E + E ′ = m φ and momen- tum conservation requires vector k = − vector k ′ . Then E = radicalBig | vector k | + m 2 ψ = radicalBig | vector k ′ | + m 2 ψ = E ′ , 1 so we find E = E ′ = m φ / 2 and | vector k | = | vector k ′ | = m φ 2 radicalBig 1 − 4 m 2 ψ /m 2 φ . Consequently summationdisplay s,s ′ |M| 2 = 4 λ 2 (( k · k ′ ) − m 2 ψ ) = 4 λ 2 ( E · E ′ + | vector k | 2 − m 2 ψ ) = 4 λ 2 parenleftBigg m 2 φ 4 (1 + 1 − 4 m 2 ψ /m 2 φ ) − m 2 ψ parenrightBigg = λ 2 ( 2 m 2 φ − 8 m 2 ψ ) = 2 λ 2 ( m 2 φ − 4 m 2 ψ ) Plugging this in the masterformula for 2-body-decay, we get for the total decay rate Γ = m φ 16 π | vector k | m φ / 2 (Bigg |M| 2 m 2 φ )Bigg cos θ f = m φ 16 π m φ / 2 radicalBig 1 − 4 m 2 ψ /m 2 φ m φ / 2 2 λ 2 ( m 2 φ − 4 m 2 ψ ) m 2 φ = λ 2 m φ 8 π parenleftBigg 1 − 4 m 2 ψ m 2 φ parenrightBigg 3 / 2 Problem 2 (i) The Feynman rules corresponding to L int = − gA μ ¯ ψ 1 γ μ ψ 2 − g ∗ A † μ ¯ ψ 2 γ μ ψ 1 are ¯ ψ 1 ,a ψ 2 ,b A μ → k 1 → k 2 ← k 3 = − igγ μ ab 2 ψ 1 ,b ¯ ψ 2 ,a A † μ → k 1 → k 2 ← k 3 = − ig ∗ γ μ ab (ii) For the decay ψ 1 → A + ψ 2 , we start the second vertex (the first vertex cannot be contracted with the creation and annihilation operators in the pro- cess and therefore does not contribute) and contract it with the creation and annihilation operators and get ψ 1 ,b ¯ ψ 2 ,a A † μ a † 1 ,p,r a 2 ,k,s a A,k ′ s ′ ← p ← k ′ ← k i M = − i g ∗ (¯ u 2 , vector k,s ) a γ μ ab ( u 1 ,vector p,r ) b ( ε ∗ vector k ′ ,s ′ ) μ where I labelled the momentum of the decaying...
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## This note was uploaded on 12/14/2011 for the course PHY 5667 taught by Professor Okui during the Fall '10 term at FSU.

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HW10_soln - PHY 5667 Problem Set no. 10 solution Problem 1...

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