PS9_253A_solutions.pdf - Problem Set 9 Solutions Physics 253A Fall 2018 1(a The Lagrangian for this system is 1 2 me e iD m F L = e iD 4(1 The 18

# PS9_253A_solutions.pdf - Problem Set 9 Solutions Physics...

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Problem Set 9 Solutions Physics 253A Fall 2018 1(a) The Lagrangian for this system is L = ¯ ψ e ( i /D - m e ) ψ e + ¯ ψ μ ( i /D - m μ ) ψ μ - 1 4 F 2 μν . (1) The 18 diagrams that contribute to e + e - μ + μ - at this order in perturbation theory are e - μ + e + μ - , e - μ + e + μ - (2) e - μ + e + μ - e - μ + e + μ - , e - μ + e + μ - , e - μ + e + μ - (3) e - μ + e + μ - e e , e - μ + e + μ - μ μ , e - μ + e + μ - , e - μ + e + μ - (4) e - μ + e + μ - e , e - μ + e + μ - e , e - μ + e + μ - e , e - μ + e + μ - e (5) e - μ + e + μ - μ , e - μ + e + μ - μ , e - μ + e + μ - μ , e - μ + e + μ - μ (6) 1
1(b) The minimal set of diagrams that cancels the gauge dependence of the box diagram is e - μ + e + μ - + e - μ + e + μ - (7) 1(c) To see that the sum of the above diagrams is gauge independent, consider i M μν = p 4 l k p 3 μ ν + p 4 p 3 μ l ν k (8) Let’s calculate this matrix element without contracting with polarization vectors. We will not assume that the photons are on-shell, but the external scalars are on-shell, so p 2 3 = p 2 4 = m 2 φ . Then i M μν = i ( - ie ) 2 ¯ u ( p 3 ) γ μ / k - / p 4 + m μ γ ν v ( p 4 ) ( k - p 4 ) 2 - m 2 μ + i ( - ie ) 2 ¯ u ( p 3 ) γ ν / l - / p 4 + m μ γ μ v ( p 4 ) ( l - p 4 ) 2 - m 2 μ . (9) We can use momentum conservation, ( k - p 4 ) 2 - m 2 μ = k 2 - 2 k · p 4 = l 2 - 2 l · p 3 ( l - p 4 ) 2 - m 2 μ = l 2 - 2 l · p 4 = k 2 - 2 k · p 3 , (10) to write the matrix element contracted with l μ as l μ i M μν = i ( - ie ) 2 ¯ u ( p 3 ) / l / p 3 - / l + m μ γ ν v ( p 4 ) ( p 3 - l ) 2 - m 2 μ + i ( - ie ) 2 ¯ u ( p 3 ) γ ν / l - / p 4 + m μ / lv ( p 4 ) ( l - p 4 ) 2 - m 2 μ = i ( - ie ) 2 ¯ u ( p 3 ) / l / p 3 - l 2 + / lm μ γ ν v ( p 4 ) ( p 3 - l ) 2 - m 2 μ + i ( - ie ) 2 ¯ u ( p 3 ) γ ν l 2 - / p 4 / l + / lm μ v ( p 4 ) ( l - p 4 ) 2 - m 2 μ (11) This can be simplified using / p/ k = p μ k ν γ μ γ ν = p μ k ν (2 η μν - γ ν γ μ ) = 2 p · k - / k / p (12) and ¯ u ( p )( / p - m ) = 0 , ( / p + m ) v ( p ) = 0 (13) to give l μ i M μν = - ie 2 ¯ u ( p 3 ) ( 2 p 3 · l - m μ / l - l 2 + / lm μ ) γ ν v ( p 4 ) ( p 3 - l ) 2 - m 2 μ - ie 2 ¯ u ( p 3 ) γ ν ( l 2 + / l ( - m μ ) - 2 p 4 · l + / lm μ ) v ( p 4 ) ( l - p 4 ) 2 - m 2 μ = - ie 2 ¯ u ( p 3 ) ( 2 p 3 · l - l 2 ) γ ν v ( p 4 ) - 2 p 3 · l + l 2 - ie 2 ¯ u ( p 3 ) γ ν ( l 2 - 2 p 4 · l ) v ( p 4 ) l 2 - 2 l · p 4 = ie 2 ¯ u ( p 3 ) γ ν v ( p 4 ) - ie 2 ¯ u ( p 3 ) γ ν v ( p 4 ) = 0 . (14) Note that we did not use that the photon was on-shell at any point in this calculation. Similarly we can show that k ν i M μν = i ( - ie ) 2 ¯ u ( p 3 ) γ μ / k - / p 4 + m μ / kv ( p 4 ) ( k - p 4 ) 2 - m 2 μ + i ( - ie ) 2 ¯ u ( p 3 ) / k / p 3 - / k + m μ γ μ v ( p 4 ) ( p 3 - k ) 2 - m 2 μ = - ie 2 ¯ u ( p 3 ) γ μ ( k 2 - 2 k · p 4 ) v ( p 4 ) ( k - p 4 ) 2 - m 2 μ - ie 2 ¯ u ( p 3 ) ( 2 k · p 3 - k 2 ) γ μ v ( p 4 ) ( p 3 - k ) 2 - m 2 μ = - ie 2 ¯ u ( p 3 ) γ μ v ( p 4 ) + ie 2 ¯ u ( p 3 ) γ μ v ( p 4 ) = 0 . (15) 2
Now, let us go back to the original diagrams i M = p 1 p 4 l p 2 k p 3 + p 1 p 4 p 2 p 3 (16) We can write this in terms of M μν i M = Z d 4 k (2 π ) 4 d 4 l (2 π ) 4 (2 π ) 4 δ 4 ( k + l - p 3 - p 4 ) ( - ie ) 2 i ¯ v ( p 2 ) γ μ / p 1 - / l + m e γ ν u ( p 1 ) ( p 1 - l ) 2 - m 2 π × i Π μα ( l ) i Π νβ ( k ) i M μν (17) where we have included the both the integrals over k and l to keep it symmetric, and then enforcing momentum conservation with an extra delta function. The only gauge-dependence is in the photon propagators i Π μν ( p ) = - i p 2 + g μν - (1 - ξ ) p μ p ν p 2 (18) By looking at Π μα ( l νβ ( k ) M μν we see that the terms containing ξ are (1 - ξ ) l μ l α and (1 - ξ ) k ν k β , and they vanish since l μ M