PS6_253A_solutions - Problem Set 6 Solutions Physics 253A Fall 2018 1(a The diagram(1 is of order e4 I find 35 diagrams contributing at this order

# PS6_253A_solutions - Problem Set 6 Solutions Physics...

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Problem Set 6 Solutions Physics 253A Fall 2018 1(a) The diagram π - φ + π + φ - (1) is of order e 4 . I find 35 diagrams contributing at this order in perturbation theory: π - φ + π + φ - π - φ + π + φ - π - φ + π + φ - π - φ + π + φ - (2) π - φ + π + φ - π - φ + π + φ - π - φ + π + φ - π - φ + π + φ - (3) π - φ + π + φ - π - φ + π + φ - π - φ + π + φ - π - φ + π + φ - (4) π - φ + π + φ - π - φ + π + φ - π - φ + π + φ - π - φ + π + φ - (5) π - φ + π + φ - φ φ π - φ + π + φ - π π π - φ + π + φ - φ π - φ + π + φ - π (6) 1
π - φ + π + φ - π - φ + π + φ - (7) π - φ + π + φ - φ π - φ + π + φ - φ π - φ + π + φ - φ π - φ + π + φ - φ (8) π - φ + π + φ - π π - φ + π + φ - π π - φ + π + φ - π π - φ + π + φ - π (9) π - φ + π + φ - φ π - φ + π + φ - φ π - φ + π + φ - π π - φ + π + φ - π (10) 1(b) A minimal set of diagrams that cancels the ξ dependence of the box diagram is π - φ + π + φ - + π - φ + π + φ - + π - φ + π + φ - (11) 1(c) To see this, first consider the following 3 diagrams (the complete set of tree level graphs for γγ φ + φ - scattering): i M μν = p 4 k p 3 μ ν + p 4 p 3 μ ν k + p 4 k p 3 μ ν (12) Let us compute these diagrams without contracting with the polarization vectors. We will not assume that the photon momenta are on shell, but we will keep the external scalars on shell: p 2 3 = p 2 4 = m 2 φ . Then i M μν = ( - ie ) 2 i ( k - 2 p 4 ) ν (2 p 3 - ) μ ( k - p 4 ) 2 - m 2 φ + ( - ie ) 2 i ( - 2 p 4 ) μ (2 p 3 - k ) ν ( - p 4 ) 2 - m 2 φ + 2 ie 2 g μν . (13) We can rewrite the denominators using the following relations: 2 - 2 · p 3 = ( - p 3 ) 2 - m 2 φ = ( k - p 4 ) 2 - m 2 φ = k 2 - 2 k · p 4 , k 2 - 2 k · p 3 = ( k - p 3 ) 2 - m 2 φ = ( - p 4 ) 2 - m 2 φ = 2 - 2 · p 4 . (14) 2
Let us check the following identity: μ M μν = - e 2 ( k - 2 p 4 ) ν (2 · p 3 - 2 ) 2 - 2 · p 3 - e 2 ( 2 - 2 · p 4 ) (2 p 3 - k ) ν 2 - 2 · p 4 + 2 e 2 ν = e 2 ( k - 2 p 4 ) ν - e 2 (2 p 3 - k ) ν + 2 e 2 ν = 2 e 2 ( k + - p 3 - p 4 ) ν = 0 , (15) by momentum conservation. Similarly, k ν M μν = - e 2 ( k 2 - 2 k · p 4 ) (2 p 3 - ) μ k 2 - 2 k · p 4 - e 2 ( - 2 p 4 ) μ (2 k · p 3 - k 2 ) k 2 - 2 k · p 3 + 2 e 2 k μ = 2 e 2 ( k + - p 3 - p 4 ) ν = 0 . (16) Now, let us return to the original diagrams: i M = p 1 p 4 p 2 k p 3 + p 1 p 4 p 2 p 3 + p 1 p 4 p 2 p 3 (17) We can write i M in terms of i M μν as follows: i M = Z d 4 k (2 π ) 4 d 4 (2 π ) 4 (2 π ) 4 δ 4 ( k + - p 3 - p 4 ) ( - ie ) 2 i (2 p 1 - ) α ( k - 2 p 2 ) β ( p 1 - ) 2 - m 2 π i Π αμ ( ) i Π βν ( k ) i M μν , (18) where we have included integrals over k and to keep things symmetric, enforcing momentum conservation with an extra delta function. The only ξ dependence is in the photon propagators: i Π μν ( p ) = - i p 2 + g μν - (1 - ξ ) p μ p ν p 2 . (19) By looking at Π μα ( ) Π νβ ( k ) M μν we see that the terms containing ξ are proportional to μ M μν = 0 or to k ν M μν = 0. This shows that the ξ dependence drops out, and the sum of these 3 diagrams is gauge invariant. 2(a) The following 3 diagrams contribute to γ π - γ π - scattering in scalar QED: i M = + p p p + q p - q ε in μ ( q ) ε out ν ( q ) + (20) = ε in μ ε out* ν ( - ie ) 2 i (2 p + q ) μ (2 p 0 + q 0 ) ν ( p 0 + q 0 ) 2 - m 2 + ( - ie ) 2 i (2 p - q 0 ) ν (2 p 0 - q ) μ ( p - q 0 ) 2 - m 2 + 2 ie 2 g μν (21) = ε in μ ε out* ν ( - ie 2 ) 2 p μ p 0 ν p 0 · q 0 - 2 p ν p 0 μ p · q 0 - 2 g μν (22) ε in μ ε out* ν M μν . (23)