Note that by momentum conservation 2 2 p 1 r 2 2 p2 r and k 2 2p 4 k q 2 2p 3 q

Note that by momentum conservation 2 2 p 1 r 2 2 p2 r

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simplify the propagators. Note that by momentum conservation,2-2p1·=r2-2p2·randk2-2p4·k=q2-2p3·q.Now, we could go ahead and write down all the diagrams, expand everything out,and try to simplify.That’s actually pretty hard (congrats to those who man-aged it!). So let’s write things more symbolically at first and do some judiciousrearranging. LetΠμν() =-igμν+ (1-ξ)μν22be the photon propagator. We always have two photon momenta in our Feynmangraphs, which we’ll take to beandr, both going from left to right.There’sonly one loop momentum, soandrare related by momentum conservation:+r=p1+p2=p3+p4However, to keep expressions simpler, it’s useful not to solve forrin terms of(or vice versa), but instead write the loop integral as a double integral, with adelta function to enforce momentum conservation:Zd4‘ d4r(2π)4δ(p1+p2--r)For convenience, we’ll write this as simplyR. Note thatandrappear symmet-rically in the integral, so for any functionf(‘, r), we haveRf(‘, r) =Rf(r, ‘).
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Phys 253a 5
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Phys 253a 6 our sum of diagrams should be X ? = 1 2 Z V μν L ( ‘, r μρ ( νσ ( r ) V ρ,σ R ( ‘, r ) = 1 2 Z ( U μν L ( ‘, r ) + U νμ L ( r, ‘ ) + 2 ie 2 g μν ) Π μρ ( νσ ( r ) × ( U ρσ R ( ‘, r ) + U σρ R ( r, ‘ ) + 2 ie 2 g ρσ ) = 1 2 Z U μν L ( ‘, r μρ ( νσ ( r ) U ρσ R ( ‘, r ) + U νμ L ( r, ‘ μρ ( νσ ( r ) U σρ R ( r, ‘ ) + 1 2 Z U μν L ( ‘, r μρ ( νσ ( r ) U σρ R ( r, ‘ ) + U νμ L ( r, ‘ μρ ( νσ ( r ) U ρσ R ( ‘, r ) + 1 2 Z U μν L ( ‘, r ) + U νμ L ( r, ‘ ) Π μρ ( νσ ( r )(2 ie 2 g ρσ ) + 1 2 Z (2 ie 2 g μν μρ ( νσ ( r ) U ρσ R ( ‘, r ) + U σρ R ( r, ‘ ) + 1 2 Z (2 ie 2 g μν μρ ( νσ ( r )(2 ie 2 g ρσ ) = 1 2 [I + I] + 1 2 [II + II] + 1 2 [III + III] + 1 2 [IV + IV] + V = I + II + III + IV + V where in getting to the third equality, we expanded the second line and grouped terms in square brackets that are the same by index redefinitions and switching r under the integral sign. This is exactly the sum of diagrams we wanted, but we’ve managed to write it in a form that will make gauge invariance obvious. Indeed, note that V μν L ( ‘, r ) = - ie 2 (2 p 1 - ) μ ( r - 2 p 2 ) ν 2 - 2 p 1 · + (2 p 1 - r ) ν ( - 2 p 2 ) μ r 2 - 2 p 1 · r - 2 g μν r ν V μν ( l, r ) = - ie 2 ((2 p 1 - ) μ - ( - 2 p 2 ) μ - 2 r μ ) = - 2 ie 2 ( p 1 + p 2 - - r ) μ = 0 since + r = p 1 + p 2 by conservation of momentum. Similarly we find that
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